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# nLab etale morphism of schemes

## Theorems

This entry is about étale morphisms between schemes. The term étale map is preferred in the context of topology and differential geometry, see étalé space for the topological version.

# Contents

## Definition

###### Definition

A morphism of schemes is an étale morphism if the following equivalent conditions hold

1. it is

1. smooth

2. of relative dimension $0$.

2. it is

3. it is

(A number of other equivalent definitions are listed at wikipedia.)

###### Definition

Jointly surjective collection of étale morphisms $\left\{{U}_{i}\to X\right\}$ is called an étale cover.

There is a weaker notion of a formally étale morphism.

###### Definition

A morphism is formally étale morphism if it is

###### Definition

These are sheaf-like properties, which can be formalized in the language of Q-categories (monopresheaf and epipresheaf properties on the $Q$-category of nilpotent thickenings).

## Properties

### Closure properties

• A composite of étale morphism is étale.

• The property of being étale is preserved under pullbacks along any morphism of schemes.

• A smooth map of schemes is étale iff there is an étale cover of the base scheme by open subschemes such that the pullback of the projection to each of them is an open local isomorphism of locally ringed spaces (and in particular, the pullback of the projection morphism is an étale map of the corresponding underlying topological spaces).

This disjointness picture of étale covers make them suitable for having nontrivial cohomology in situations where Zariski covers give vanishing cohomology.

### Classes of examples

###### Proposition

Let $k$ be a field. A morphism of schemes $Y\to \mathrm{Spec}k$ is étale precisely if $Y$ is a coproduct $Y\simeq {\coprod }_{i}\mathrm{Spec}{k}_{i}$ for each ${k}_{i}$ a finite and separable field extension of $k$.

This appears for instance as de Jong, prop. 3.1 i).

###### Remark

Such étale morphisms are classified by the classical Galois theory of field extensions.

###### Proposition

A ring homomorphism $A\to B$ is étale precisely if $B\simeq R\left[{x}_{1},\cdots ,{x}_{n}\right]/\left({f}_{1},\cdots ,{f}_{n}\right)$ where

• all the ${f}_{i}$ are polynomials;

• the Jacobian $\mathrm{det}\left(\frac{\partial {f}_{i}}{\partial {x}_{j}}\right)$ is a unit in $S$.

This appears for instance as de Jong, prop. 3.1 ii).

This proposition seems to be wrong for 2 reasons; first, A,B,R,S are 2 many symbols. Second, the statement sounds like “etale iff standard smooth” but only the direction “etale implies standard smooth” is true (and can be found in the stacks project). Since I couldn’t find the cited source, I couldn’t look into what the statement is supposed to be, originally. – Konrad

### As locally constant sheaves

###### Proposition

A sheaf $F$ on a scheme $X$ corresponds to an étale morphism $Y\to X$ precisely if there is an étale cover $\left\{{U}_{i}\to X\right\}$ such that each restriction

$F{\mid }_{{U}_{i}}\simeq \mathrm{LConst}{K}_{i}$F|_{U_i} \simeq LConst K_i

is isomorphic to a constant sheaf on a set ${K}_{i}$.

A proof is in (Deligne).

## Examples

• A finite separable field extension $K↪L$ corresponds dually to an étale morphism $\mathrm{Spec}L\to \mathrm{Spec}K$. These are the morphisms classified by classical Galois theory.

## References

The classical references are

• Pierre Deligne et al., Cohomologie étale , Lecture Notes in Mathematics, no. 569, Springer-Verlag, 1977.
• James Milne, Etale cohomology, Princeton Mathematical Series 33, 1980. xiii+323 pp.

Lecture notes are

Revised on December 4, 2012 18:23:25 by Konrad Voelkel? (132.230.30.143)


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# Intersection of two rhombuses

In which way do two rhombuses have to intersect to form an octagon which has four angles equal to 150◦? All sides of the octagon are equal.

-
Since you know that 4 of the angles have measure $150°$ and the sum of the measures of the interior angles of an octagon is $1080°$, consider that the other four angles might have measure $\frac{1}{4}(1080°-4\cdot150°)=120°$. That suggests using two congruent rhombuses with angles that measure $120°$ and $60°$—with congruent rhombuses, centered at the same point, and rotated 90° from one another, we will get all 8 sides of the octagon to have the same length. The octagon is the purple region of overlap in the diagram below.

format: plain


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# yet an unofficial BioEdit for OSX

To my knowledge, BioEdit is the most comprehensive biological sequence alignment editor. Most of my labmates run this software using Parallels Desktop. For some of them, BioEdit is the only reason to install Parallels Desktop.

I need to edit my alignment recently, and install it in my iMac using Wine, which is a compatibility layer for running Windows applications on POSIX-compliant OS. Although it is famous in Linux community for many years, many OSX users never heard of it.

I came out with an idea that maybe I can pack all the components into an OSX application bundle. Then I can share it with my labmates and others. I don’t need to teach them how to install and config Wine and they can just drag the App to /Applications and double click to run it.

As usual, you can install it by dragging the App to Applications folder and open it by double click. It works just like native App.
BTW: If you don’t have XQuartz installed, you need one for running BioEdit.


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Quantumness of the quantum channels

# Quantumness of the quantum channels

## Abstract

Quantum coherence is a fundamental aspect of quantum physics and plays a central role in quantum information science. This essential property of the quantum states could be fragile under the influence of the quantum operations. The extent to which quantum coherence is diminished depends both on the channel and the incoherent basis. Motivated by this, we propose a measure of nonclassicality of a quantum channel as the average quantum coherence of the state space after the channel acts on, minimized over all orthonormal basis sets of the state space. Utilizing the squared -norm of coherence for the qubit channels, the minimization can be treated analytically and the proposed measure takes a closed form of expression. If we allow the channels to act locally on a maximally entangled state, the quantum correlation is diminished making the states more classical. We show that the extent to which quantum correlation is preserved under local action of the channel cannot exceed the quantumness of the underlying channel. We further apply our measure to the quantum teleportation protocol and show that a nonzero quantumness for the underlying channel provides a necessary condition to overcome the best classical protocols.

## I Introduction

Quantum coherence arising from superposition is the most fundamental aspect of quantum mechanics and is responsible for nonclassical properties of quantum systems. The role of quantum coherence in the wide range of areas, e.g. thermodynamics, quantum algorithms, quantum correlations, biology and etc. streltsovrev2017 (); streltsovprl2015 (); streltsov distil (); napoli2016 (); anand (); hillery2016 (); shi2017 (); piani2016 (); bu2016 (); karpat2014 (); marvian2016 (); marvian2016 2 () shows the importance of its investigation. Although it is not a new concept and is as long-standing as quantum mechanics itself, the study of quantum coherence as a quantum resource theory (QRT) pleniocoherence2014 () has attracted a lot of interest recently.

On the other hand it is understood that quantum channels, the completely positive and trace preserving maps, could reduce quantum coherence of the states. Quantum channels provide a framework to investigate coherence behavior under different dynamics and noises. Cohering power and decohering power for quantum operations manikarimipour () and also classification of quantum channels based on coherence-breaking property coherencebreaking () in comparison with entanglement-breaking channels entanglementbreaking () arose from these investigations. Another important class of channels is the so-called semiclassical channels, those that map all input quantum states into ones which are diagonal in the same basis; streltsovprl2011 (). Clearly, at the result of applying a semiclassical channel the space of quantum states changes into classical one meaning that all states are distinguishable by a classical observer measuring states with respect to the channel’s diagonal basis meznaric 2013 ().

The QRT provides preliminaries to recognize the interrelations between coherence and different physical theories streltsovrev2017 (); streltsovprl2015 (); streltsov distil (); chitambar coh vs ent (); zhu 2017 (); streltsovprx2017 (); streltsov merging (); coh vs cor (); frozencoh (). Particularly, it has been shown that coherence is the resource in the process of generating entanglement in bipartite systems streltsovrev2017 (); streltsovprl2015 (). Furthermore, one of the most significant results is that quantum discord type correlations are nothing but the minimum amount of quantum coherence of the bipartite system with respect to the local product basis frozencoh (). In view of this, inability of semiclassical channels in local creation of quantum correlations is predictable as they remove all coherence of the state they applied on. Actually it has been shown that for creation of quantum correlation via local channels in finite-dimensional systems, a necessary and sufficient condition for channels is that they should not be commutativity preserving hu 2012 (). A quantum channel is said to be a commutativity preserving channel if it preserves commutativity of any two compatible states, i.e. , then hu 2012 (). Clearly semiclassical channels are commutativity preserving, so nonsemiclassicality is a necessary condition to create quantum correlations locally. Particularly in qubit systems, necessary and sufficient conditions for channels are nonunitality and nonsemiclassicality streltsovprl2011 (). These results lead us to conclude that there should be connections between the amount of nonsemiclassicality of one-party channels and quantum correlations of the bipartite states.

Furthermore, one can claim that nonunitary quantum channels generally induce decay in quantum properties of the states with respect to some orthonormal basis that the coherence of most outputs is reduced. This is at the heart of channels contraction effect on the space of states, the case of semiclassical channels is the special one making the states completely classical. In the other words, a semiclassical channel washes out quantum coherence of the input states making the states distinguishable by a classical observer. We adapt this as our notion of classicality of channels and pursue the question that how much a channel is far from this very notion of classicality. Our notion of classicality is different from the one given in meznaric 2013 () based on the concept of einselection, characterizing a classical operation as a one that commute with the completely dephasing process. Here we introduce a measure to quantify the amount of quantumness preserved by the channel, along with a natural coherence-based criterion to decide whether a given channel is semiclassical or not. We further investigate the local action of the channels and show that the quantumness of the channel provides an upper bound for the quantum correlations of the associated Choi states. Moreover, by applying the measure to quantum teleportation protocol, we show that the quantumness of the underlying channel provides a necessary, but not sufficient, condition to overcome the best classical protocols.

The paper is organized as follows. In Sec. II, we briefly review quantum coherence measures and introduce the squared -norm of coherence. In Sec. III, we introduce our method to quantify the quantumness of the channels. A closed form for the quantumness of the qubit channels and its properties are also presented in Sec. III. Section IV is devoted to study local action of the qubit channels. In particular, we identify a relation between the channel’s quantumness and the quantum correlation of its associated Choi state. Application of the presented measure to quantum teleportation protocol is presented in Sec. V. The paper is concluded in Sec. VI.

## Ii Preliminary: MEASURE FOR QUANTUM COHERENCE

Resource theory of quantum coherence is based on three ingredients; incoherent states denoted by , incoherent operations, and coherence monotones. This framework for quantum coherence provides a way to quantify coherence napoli2016 (); streltsovrev2017 (); streltsovprl2015 (); winter2016 (); zhu 2017 (); yu 2016 (); shao 2015 (); pleniocoherence2014 (); Aberg2006 (). A coherence measure is a map from the set of states to the set of non-negative real numbers, with the following conditions imposed by QRT pleniocoherence2014 (); Aberg2006 (); streltsovrev2017 (). (a) Non-negativity: and if and only if . (b) Monotonicity: for any non-selective incoherent operation, meaning that one cannot create coherence via incoherent operations. (c) Strong monotonicity: where and is an incoherent Kraus operator. This property assures that coherence cannot be increased on average using incoherent operations in case individual outputs are accessible. (d) Convexity: , means that mixing cannot increase coherence. Also it has been proved that convexity and strong monotonicity together imply monotonicity pleniocoherence2014 ().

The -norm of coherence defined by pleniocoherence2014 ()

 Cl1(ρ)=∑i,ji≠j|ρij|, (1)

is a measure satisfying the desired properties. In contrast, it is shown that the sum of the squared absolute values of all off-diagonal elements violates monotonicity pleniocoherence2014 (). However, for further use let us mention here that for qubit systems it coincides with the squared -norm

 C2l1(ρ)=[∑i,ji≠j|ρij|]2, (2)

and satisfies the properties of a bona fide measure of coherence (see Appendix A for the proof).

## Iii Quantumness of the quantum channels

Quantum coherence is an evidence of existence of the quantum properties in a given state. In order to investigate the amount of quantumness of the state space, a good candidate is to average on coherence of all states. It turns out that minimization over all orthonormal basis sets results in finding the one that most of the states behave more classically with respect to it. Obviously, without any evolution in states, the state space is so symmetric that the minimization is completely pointless. The importance of such optimization appears in case of nonunitary evolution of states and the result addresses the question that how much quantumness does a quantum channel preserve? Clearly, this is a restatement of asking “how much does a quantum channel deviate from a semiclassical one?”, as this class of operations erases quantum coherence of all states completely. Motivated by this, we define the quantumness (or equivalently the nonsemiclassicality) of a quantum channel as

 QC(E)=NCmin{|i⟩}∫C(E(ρ))dμ(ρ), (3)

where subindex denotes that the defined quantity depends on the chosen measure of coherence, and the factor is introduced in order to normalize the desired quantity at the end. Above, the integration is to average on coherence of all states. It could be straightforwardly proved, because of non-negativity condition of any bona fide measure of coherence, that and it is zero if and only if the quantum channel is semiclassical. Furthermore, at the result of integration over all input states, the equality holds and minimization over all incoherent bases leads to , where is an arbitrary unitary map, for unitary operator . So as an immediate corollary, the nonsemiclassicality of any unitary map is the same as the one of identity channel with no action on its input states, .

In the following subsection we focus on qubit channels and show that the minimization specified in Eq. (3) can be calculated analytically for the squared -norm of coherence.

### iii.1 Qubit channels

The effect of the qubit channel is an affine transformation on the Bloch sphere; , where is the Bloch vector of the input state. Here the matrix and the vector are features of the channel . Having this in mind, to be more computable, we choose the squared -norm as the measure of coherence and derive an explicit expression for the quantumness of qubit channels. To apply Eq. (3) in this case, one has to measure quantum coherence of any state with respect to an arbitrary orthonormal basis , where and . A straightforward calculation shows that

 C2l1(E(ρ)) = 4|⟨n−|E(ρ)|n+⟩|2=|r′×^n|2 (4) = |(Λr+t)×^n|2,

which can be used to obtain the quantumness of the qubit channels

 QC2l1(E) = 5234πmin^n∫|(Λr+t)×^n|2d3r (5) = 5234πmin^n∫(|Λr×^n|2+|t×^n|2)d3r.

Above we set and the minimum is taken over all unit vectors . Also the last equality follows from , arisen from the fact that for any in the integration there is a corresponding . To continue, we use for an arbitrary vector and any unit vector . Moreover, by writing and and using , we get

 QC2l1(E) = min^n(Tr[M]−^ntM^n), (6)

where . The minimum is achieved when is an eigenvector of corresponding to the largest eigenvalue, so that

 QC2l1(E) = M2+M3, (7)

where are eigenvalues of in nonincreasing order. Equation (7) provides an explicit expression for the quantumness of qubit channels.

The quantumness defined by Eq. (7) is bounded above by . To see this note from Eq. (6) that

 QC2l1(E) ≤ (Tr[M]−^ntM^n), (8)

holds for any unit vector . Recalling that the trace of a matrix is independent of the basis chosen, one can define as a set of orthonormal basis, so . This holds for any orthonormal basis, hance for any basis with we find

 QC2l1(E) ≤ 12(^nt⊥ΛΛt^n⊥+^n′t⊥ΛΛt^n′⊥)≤1. (9)

Here, the last inequality follows from the fact that singular values of cannot be greater than 1, so expectation value of on any normalized vector is bounded above by .

Interestingly, the unitary channels are the only channels that reach the maximum quantumness . Recalling that quantumness of the unitary channels is equal to the one of identity channel, we have to proof the above assertion for the identity channel. Obviously, for identity channel for which and , the bound is saturated. On the other hand, if then , implies that the channel should be unital, i.e. , as the singular values of cannot reach its maximum value unless . Putting , then there will be no condition on and , as such and the channel is identity.

In the rest of this paper, we drop the subscript from the quantumness and denote the quantumness, Eq. (6), by for the sake of brevity. In what follows we go into more details about qubit channels and their local action on two-qubit system.

## Iv Local action of the quantum channels

Equation (7) provides an analytical expression for the quantumness of qubit channels and should facilitate the investigation of questions concerning the quantum channels. In particular, it may be helpful to investigate questions concerning local actions of quantum channels such as; wether a channel with high nonclassicality will produce more quantum effects when the channel acts locally on composite systems? Is there any relation between the channel quantumness and the quantum correlation preserved under local action of the channel? To address these questions, we need to recall some important classes of quantum channels with special focus on their local action. A quantum channel is called entanglement-breaking channel, channels that wash out entanglement when applied locally entanglementbreaking (), if and only if it can be written in the so-called Holevo form holevo ()

 E(ρ)=∑nRnTr(ρFn), (10)

where is a density matrix and forms a POVM. An important subclass of the entanglement-breaking channels is the so-called quantum-classical (QC) channels holevo (), defined by Eq. (10) when where is an orthonormal basis. As it is clear from definition, any QC channel is semiclassical. On the other hand, any semiclassical channel is at least coherence-breaking, with respect to some basis, but the latter is a subset of QC channels coherencebreaking (). Accordingly, a quantum channel is quantum-classical if and only if it is semiclassical.

Considering the effect of semiclassical channels in removing quantum aspects of input states, it has been mentioned that local semiclassical channels cannot create discord type correlations. It follows from the equivalency between quantum-classical and semiclassical channels that there is a much more stronger fact; semiclassical channels not only cannot create quantum correlations locally but also destroy quantum correlations of the party they applied on correlationbreaking (). Even more, only these channels have such effect. In summary, the following equivalent statements will clarify semiclassical channels entanglementbreaking (); correlationbreaking (); (i) is a semiclassical or QC channel. (ii) is a discord-breaking channel. (iii) is a zero-discord state on part where is a maximally entangled state.

The above discussion suggests that for preserving quantum correlations in a composite system, the assumed quantum channel should retain some quantumness in the whole space of the one-party system. In the following, we will show that there is actually an interesting relation between the channel quantumness and the quantum correlation preserved under local action of the channel. Not surprisingly, such relation is based on the Choi-Jamiołkowski isomorphism, stating that there is a one-to-one correspondence between quantum channel and quantum state mathematical ().

###### Observation 1

The following relation exists between quantumness of , measured by Eq. (7), and quantum correlation of , measured by the (normalized) geometric discord

 DG(ρE)≤Q(E)≤1. (11)

To proof this we first need to assert the following lemma.

###### Lemma 1

Let be a qubit channel described by the affine parameters and . Then acts locally on the Bob’s side of a two-qubit state as

 (I⊗E)ρAB=14( 12⊗12+12⊗(Λy+t).σ+ (12) x.σ⊗12+∑ijTbijσi⊗σj),

where . Above , are the Bloch vectors of Alice and Bob, respectively, and is the correlation matrix of the initial state.

See appendix B for a proof of the above lemma.

Now, using the above lemma, the observation 1 can be straightforwardly proved by noting that: (i) For , with , we have and , so Eq. (12) reduces to

 ρE = (I⊗E)|β⟩⟨β| = 14(12⊗12+12⊗t.σ+∑ijΛ′tijσi⊗σj),

where equals to except for the sign of its second column which is opposite to the one of . (ii) The normalized geometric discord vedral () of is equal to the sum of two lower eigenvalues of . Recalling we get . Evidently HornBook1985 () which completes the proof of observation.

Now, several properties of the inequality (11) are in order. (i) Looking at two matrices and shows that if and only if (unital channels), or be an eigenvector of corresponding to its largest eigenvalue. (ii) if and only if , follows from the fact that both quantities vanish if and only if , that is to say, if and only if there exists a unit vector such that and . This was an expected result since QC channels are the only class of channels removing quantum discord and transforming any state into a quantum-classical state. (iii) Both quantities attain their maximum value if and only if is an orthogonal matrix (consequently ), i.e. for the unitary channels which transform a maximally entangled state into a maximally entangled state. (iv) Both quantities are invariant under (local) unitary maps.

Paying attention to and its common features with geometric discord we argue that, for the states defined by Eq. (IV), this quantity has the properties necessary for a good measure of quantum correlation. The duality between and , described by the Choi-Jamiołkowski isomorphism, reveals that the quantumness of is a necessary resource for quantum correlation of . In the other words, the preserved quantum correlation, measured by geometric discord, cannot exceed the quantumness of the channel. Moreover, the fact if and only if , states that a channel with nonzero quantumness will preserve some quantum correlation. This sheds light on the claim that more nonsemiclassical a quantum channel is more quantum correlation is preserved when the channel acts locally on a composite system. A semiclassical channel, on the other hand, destroys all one-party quantumness, and as such it cannot create or preserve any discord-like quantum correlation.

To demonstrate this, let us consider the amplitude damping channel characterized by the following Kraus operators nielsen ()

 K1=(100√1−γ),K2=(0√γ00). (14)

The affine parameters associated with this channel are and . Straightforward calculations show that for this channel the quantumness and geometric discord are given by

 Q(Ea)={12(6γ2−3γ+2)γ≤16,1−γγ>16, (15)

and

 DG(ρEa)={12(2γ2−3γ+2)γ≤12,1−γγ>12, (16)

respectively. Figure 1 shows the quantumness and geometric discord for this channel in terms of the channel’s parameter . As can be seen from this figure both quantumness and geometric discord are monotonically decreasing from to as channel’s parameter goes from to . Indeed, when the damping rate passes from , quantum correlation reaches the channel’s quantumness.

For entanglement-breaking channels, despite they may preserve a fraction of one-party quantumness, significant amount of quantumness should be removed. More precisely, for such channels the quantumness is bounded above as

 Q(E) ≤ 12(1−^ntΛΛt^n−∑i≠j|λiλj|)≤12, (17)

holds for any unit vector which lies in the direction of . This follows easily from Eq. (9) and noting that for entanglement-breaking channels we have entanglementbreakingqubit () , leads to . Clearly, both of inequalities in Eq. (17) cannot be saturated simultaneously, so does not definitely reach for an entanglement-breaking channel. On the other hand, for the unital channels whose necessary and sufficient condition to be entanglement-breaking is entanglementbreakingqubit (), the quantumness is bounded above by , saturated for example by a channel described by the Kraus operators , and .

Before we conclude this section, let us mention here that the quantity defined above measures the amount of quantumness preserved in the state during the action of the channel. Clearly, a nonzero value for the channel’s quantumness does not offer an ability to create quantum correlation. For example, although unital qubit channels mostly preserve quantumness of the input states, they cannot create quantum correlation locally streltsovprl2011 (). On the other hand, a nonunital channel described by the Kraus operators and creates the maximum amount of quantum discord locally correlating power (), although its quantumness is only . In summary, for creation of quantum correlations by a local channel, other additional properties rather than preserving quantumness are required.

## V Application to quantum teleportation

Consider the generalized depolarizing channel described by the Kraus operators for where , and , and are the Pauli matrices. For this channel, the affine parameters are given by and , where , and . If then and we get for quantumness of this channel .

The generalized depolarizing channel is of particular importance because of its usefulness in studying quantum teleportation. The teleportation protocol BennettPRL1993 () consists of a two-qubit states shared between two separated parties, say Alice and Bob, and an unknown qubit state where with and . The protocol is described by the generalized depolarizing channel

 EGD(ρin)=3∑i=0piσiρinσi, (18)

where and . Here where are the four maximally entangled Bell states associated with the Pauli matrices , i.e. for . Furthermore, for optimal utilization of a given entangled state as resource, one has to choose the local basis in such a way that .

To characterize the quality of the teleported state , it is useful to look at the fidelity between and . We find where is the Bloch vector of the input qubit. Moreover, the average fidelity is defined by averaging the fidelity over all possible input states , turns out that . It is clear that Alice and Bob could gain a fidelity better than (the best possible fidelity when they communicate only through classical channel), if and only if . A simple investigation shows that if then . Equivalently if then and we do not benefit the quantum advantages. However, the inverse is not correct meaning that it is possible for the channel to possesses nonzero quantumness but the average fidelity of the teleportation be less than . This implies that in order to have a teleportation protocol with fidelity better than the classical one, a nonzero quantumness for the associated channel is necessary, although it is not sufficient. For example, for the Werner state , we have and . This state is separable (disentangled) if and only if and, not surprisingly, only for such values of parameter the fidelity of the protocol is less than . However, as it is clear from Fig. 2, even for , the channel possesses a nonzero quantumness.

## Vi Conclusion

Regarding that quantum coherence is the most fundamental aspect of quantum physics, it can be used to characterize and quantify other nonclassical features identified in the emerging quantum information theory. Using this very signature of quantumness, we propose the average coherence of the channel’s outputs, minimized over all incoherent bases, as the degree of nonclassicality of the channels, . This, intuitively, quantifies the amount of quantumness preserved by a quantum channel and measures the deviation of the assumed channel from the so-called semiclassical channels. Applying the squared of as a qualified measure of coherence for qubit systems, an analytical expression is gained for any qubit channel. It is shown that attains its maximum value for any unitary map, however, in other cases it cannot reach its maximum value due to the contraction of the Bloch sphere to its associated channel’s ellipsoid, hence decreasing happens in quantum properties. By decrease in quantum properties we mean that the error of description of states using a family of classical probability distributions is getting more and more close to be vanished when the measurement is in the minimized basis. Obviously, doing maximization rather than minimization in our definition of quantumness introduces the basis that most states are noncompatible with, meaning that most states show quantumness if we perform measurement in such a basis. In the case of maximization, it could be straightforwardly proved that the contraction into totally mixed state has the lowest amount and contraction into a pure state, which is a semiclassical map, has the largest amount.

Moreover, utilizing the Choi-Jamiołkowski isomorphism between the quantum channel and Choi state , we show that the amount of quantum correlation preserved in the Choi state is bounded above by the amount of quantumness of the quantum operation. The bound is saturated both for the zero-discord states and the maximally entangled ones. Even more, the saturation occurs faithfully in a sense that one quantity attains its respective minimum (maximum) if and only if the other one attains its minimum (maximum). This, in turn, proposes to consider the quantumness of as a necessary resource for the quantum correlation of and suggests that such notion of quantumness can be considered as a quantum correlation measure for the class of Choi states. For an entanglement-breaking channel whose local action washes out all quantum entanglement of the input states, the preserved discord type quantum correlation of the Choi states cannot be larger than the washed out one, meaning that the channel should clear more than half the discord type correlations in order to remove all quantum entanglement. Due to the widespread application of the quantum channels in quantum information theory, presenting a measure to quantify nonclassicality of the quantum channels could be helpful in identifying the resources of the quantum advantages. We further introduce such application by applying our measure to quantum teleportation protocol and showing that the quantumness of the underlying channel is a necessary resource to gain fidelity better than the one offered by the best classical protocol.

## acknowledgment

This work was supported by Ferdowsi University of Mashhad under Grant No. 3/45469 (1396/10/09).

## Appendix A Proof of the monotonicity of C2l1(ρ)

Here we are going to prove properties of the squared -norm of coherence qualifying it as a coherence measure in qubit systems. To do this note that this quantity satisfies positivity and convexity as the squared of any convex function is convex too. It remains only to prove that the aforementioned quantity is strong monotone under incoherent kraus operators. In qubit case, it has been shown that a general incoherent operation admits a decomposition with at most Kraus operators streltsov2017prl ()

 K1 = (0b1a10),K2=(a200b2), (19) K3 = (a3b300),K4=(00a4b4),K5=(a5000),

where , , , and . For the input state , the squared -norm coherence equals . On the other hand

 5∑i=1piC2l1(KiρK†ipi) = 2∑i=1piC2l1(KiρK†ipi) = w(qi,q′i,ρ)C2l1(ρ),

where

 w(qi,q′i,ρ)=2∑i=12qiq′iqi(1+r3)+q′i(1−r3) (21)

and we have defined and (for ) with conditions and . In Eq. (A), the first line follows from the fact that the three last Kraus operators remove coherence of the input states. To complete the proof of monotonicity of the squared -norm, we need to show that cannot exceed 1. A straightforward calculation can be applied to see that for the parameters , the factor reaches its maximum value . This completes the proof.

## Appendix B Proof of the Lemma 1

To prove Eq. (12), we start with the Hilbert-Schmidt representation of a general two-qubit state

 ρAB=14(12⊗12 + x⋅σ⊗12 + 12⊗y⋅σ+∑ijTijσi⊗σj).

Using this and linear property of the quantum channels we get

 (I⊗E)ρAB= 14(12⊗E(12)+x.σ⊗E(12)+ (23) 12⊗E(y.σ)+∑ijTijσi⊗E(σj)).

In affine representation, one can easily shows that and , where can be used in Eq. (23) to get

 (I⊗E)ρAB=14(12⊗12+12⊗(Λy+t).σ+ x.σ⊗12+x.σ⊗t.σ+∑il(∑jΛljTij)σi⊗σl). (24)

After rearranging the terms, we arrive at Eq. (12).

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264019


text:
- Current Issue - Ahead of Print Articles - All Issues - Search - Open Access - Information for Authors - Downloads - Guideline - Regulations ㆍPaper Submission ㆍPaper Reviewing ㆍPublication and Distribution - Code of Ethics - For Authors ㆍOnline Submission ㆍMy Manuscript - For Reviewers - For Editors
 Hyper $K$-subalgebras based on fuzzy points Commun. Korean Math. Soc. 2011 Vol. 26, No. 3, 385-403 https://doi.org/10.4134/CKMS.2011.26.3.385Published online September 1, 2011 Min Su Kang Hanyang University Abstract : Generalizations of the notion of fuzzy hyper $K$-subalgebras are considered. The concept of fuzzy hyper $K$-subalgebras of type $(\alpha,\beta)$ where $\alpha,$ $\beta \in \{\in,$ ${\rm q},$ $\in\! \vee \, {\rm q},$ $\in\! \wedge \, {\rm q}\}$ and $\alpha \ne \, \, \in\! \wedge \, {\rm q}.$ Relations between each types are investigated, and many related properties are discussed. In particular, the notion of $(\in,$ $\in\! \vee \, {\rm q})$-fuzzy hyper $K$-subalgebras is dealt with, and characterizations of $(\in,$ $\in\! \vee \, {\rm q})$-fuzzy hyper $K$-subalgebras are established. Conditions for an $(\in,$ $\in\! \vee \, {\rm q})$-fuzzy hyper $K$-subalgebra to be an $(\in,$ $\in)$-fuzzy hyper $K$-subalgebra are provided. An $(\in,$ $\in\! \vee \, {\rm q})$-fuzzy hyper $K$-subalgebra by using a collection of hyper $K$-subalgebras is established. Finally the implication-based fuzzy hyper $K$-subalgebras are discussed. Keywords : hyper $K$-algebra, fuzzy hyper $K$-subalgebra (of type $(\alpha,\beta)$), support, $(\in,$ $\in\! \vee \, {\rm q})$-fuzzy hyper $K$-subalgebra, fuzzifying hyper $K$-subalgebra, implication-based fuzzy hyper $K$-subalgebra MSC numbers : 06F35, 03G25, 08A72 Downloads: Full-text PDF


text:
# Tag Info

## Hot answers tagged anomalies

16

These are all good questions. Perhaps I can answer a few of them at once. The equation describing the violation of current conservation is $$\partial^\mu j_\mu=f(g)\epsilon^{\mu\nu\rho\sigma}F_{\mu\nu}F_{\rho\sigma}$$ where f(g) is some function of the coupling constant. It is not possible to write any other candidate answer by dimensional analysis and ...

15

Nowadays there exists a more fundamental geometrical interpretation of anomalies which I think can resolve some of your questions. The basic source of anomalies is that classically and quantum-mechanically we are working with realizations and representations of the symmetry group, i.e., given a group of symmetries through a standard realization on some space ...

11

The two kinds of trace anomalies are related but distinct. The first one that you refer to is the anomaly in Weyl transformations that occurs when you put a CFT on a curved background. The CFT is still exactly conformally invariant in flat space, but this symmetry is broken by the background gravitational field. It's useful to think about CFTs in two ...

9

There are things called sigma model anomalies, see papers listed in a sample inspire database query here. Here, the anomaly is associated to the general coordinate invariance in the target space of the non-linear sigma model: the fields take values in a nontrivial manifold (and its associated vector bundles), rather than vector spaces. Classically, the ...

9

The anomalies in four dimensions are calculated from a triangular Feynman diagram with a chiral (left-right-asymmetric, when it comes to the couplings with the gauge bosons or gravitons) fermion running in the loop and three gauge bosons (and/or graviton[s]) attached at the vertices. For the Standard Model, all the gauge anomalies cancel (both leptons and ...

8

In quantum field theories it is believed that anomalies in gauge symmetries (in contrast to rigid symmetries) cannot be coped with and must be canceled at the level of the elementary fields. May be the earliest work on the subject is: C. Bouchiat, J. Iliopoulos and P. Meyer, “An Anomaly free Version of Weinberg’s Model” Phys. Lett. B38, 519 (1972). But ...

7

It is very hard to visualize these homotopy classes, since they correspond to maps $S^4\rightarrow SU(2)\approx S^3$. The homotopy groups of spheres (and any other space) are typically very difficult to calculate in generality and physicists typically ask mathematicians. But there exist simple results in the so-called "stable range" where there is a regular ...

6

This question has been posted also at http://mathoverflow.net/questions/115866/homotopy-pi-4su2z-2 with both geometric and algebraic (that was mine!) type of answers. The geometric answers tell of Pontjyagin's method of constructing explicit representations of maps to spheres. The algebraic methods tells of the answer from a general theorem which gives some ...

6

Since $SU(2)$ is topologically a three-sphere $S^3$, you can begin by investigating the homotopy groups of spheres. Unfortunately, although there are some regular results, such as $\Pi_n(S^n)=\mathbb{Z}$, and $\Pi_m(S^n)=0$ for $m<n$, I don't think there is a single method to calculate $\Pi_m(S^n)$ for $m>n$. Individual results for $m>n$ are ...

6

A more general definition of anomaly: A QFT that has no UV completion in the same dimension is anomalous. In other words, a QFT that has no well defined short distance regularization in the same dimension is anomalous. Example: A 1+1D QFT with only one right moving fermion mode is anomalous.

6

The answer in the book is almost correct albeit oversimplified. If you want to jot them down, then there are several reasons for requiring that $d=9+1$ in superstring theory. None of them however are in any way "simple" (compared to the kind of explanations that the book seems to give). Let me jot down some of the reasons that come to mind. (There may be ...

6

Anomalies (not anamolies) are a whole subject whose basics are covered by one or several chapters of almost any good enough quantum field theory textbook so it's counterproductive to retype this whole chapter here. But generally, in quantum field theory, anomalies are quantum mechanical effects breaking symmetries that exist in the classical theory – ...

5

I am not sure if this is the way you want to think about it, but I think it is worth pointing out that not having the central charge leads to a trivial quantum theory. The precise statement would be that a positive/unitary theory with c=0 has only one state, the vacuum. The details are demonstrated in J.F. Gomes. The triviality of representations of the ...

5

Well, I hope I am not oversimplifying your question because I guest that all I am going to say is well-known for you. There are symmetries which correspond to physical symmetries such as spatial rotational or translational symmetry. These symmetries are not necessary for the consistency of the theory and thus the quantum theory has not to respect the ...

5

There is no chiral anomaly/gauge anomaly if the spacetime dimension $2\ell+1$ is odd, partly because $SO(2\ell+1)$ has real or pseudo-real representations, but no complex representations. There may instead be parity anomalies in odd spacetime dimensions. In fact, there is a dimensional ladder of related anomalies $$\text{Abelian chiral anomaly in}~ ... 5 A good analogy for the difference between the two can be given in terms of two other examples of anomalies, that are possibly more familiar. Consider a field theory with a global symmetry, take U(1) for simplicity. At the classical level, the equations of motion lead to the existence of a conserved current (Noether's theorem). At the quantum level, the ... 5 The worldsheet Weyl anomaly in bosonic string theory is an example. More generally in any dimension you can have trace and Weyl anomalies that break scale or conformal invariance, even in systems with only bosons. 4 If gauge symmetries are not fake, but are real symmetries, then， when they are anomalous, it simply means that the theory just does not have the symmetries. The theory is still well defined, at least. If gauge symmetries are fake and representing redundancy in our description, then， when they are anomalous, it means that the theory is inconstant. So ... 4 1) The axial vector current j^{\mu 5} is a pseudovector$$j^{\mu 5}~:=~\overline{\psi}\gamma^{\mu}\gamma^5\psi~=~j^{\mu}_R-j^{\mu}_L,\qquad j^{\mu}_{R,L}~:=~ \overline{\psi}_{R,L}\gamma^{\mu}\psi_{R,L}, \psi_{R,L}~:=~P_{R,L}\psi,\qquad P_{R,L} ~:=~\frac{1\pm\gamma^5}{2} . $$The 4-divergence d_{\mu}j^{\mu 5} is a pseudoscalar. That the axial ... 4 I'm pretty sure the answer is "no." There's a conformal anomaly in any even number of dimensions. For instance, in 4 dimensions it's the statement that T^\mu_\mu is nonzero for a CFT on a general curved background, equal to coefficients a and c times the Euler density and Weyl^2 terms. But there's no central extension of the conformal algebra the way ... 4 Your question is very interesting. I would like to mention something along the line of your question, but perhaps from another viewpoint. Recently there are some better understanding along the thinking between (1)"whether a theory is free from anomaly (the anomaly matching condition satisfied)," (2)"whether the symmetry of a theory is on-site symmetry," ... 3 The existence of anomalies is almost always accompanied by an extension of the gauge group commutation relations. The case of non-Abelian axial anomaly is may be the most known case. The abstract gauge group algebra:  [G_a(x), G_b(y)] = if_{ab}^{c} \delta^N(x-y) (N is the number of dimensions), is not realized at the quantum field level When N=1, ... 3 Related Why do some anomalies (only) lead to inconsistent quantum field theories What if we work without the gauge redundancy, with just the physical polarizations, what goes wrong then if we don't make sure the would-be gauge symmetry is not anomalous? This is a very natural question I wondered some time ago. In electrodynamics, if you want to ... 3 In general the chiral non-Abelian anomaly does not vanish. It is proportional to the three dimensional symmetric tensor$$ d_{ABC} = \mathrm{tr}(T_A\{ T_B T_C\}) This tensor vanishes in the particular case of $SU(2)$, but it is nonvanishing for $SU(N)$, for $N>2$. It should be also mentioned, however, that in the special case of $SU(2)$ there is a ...

2

In the case of an ungauged symmetry, the product of the exponents of the action functional corresponding of all sectors of the theory needs not to be a scalar, i.e., there can be still a net anomaly as far as its symmetry is not gauged. In this case, the exponent of the action functional will be a section of a line bundle, and the wave function(als) ...

2

There is a very simple and enlightening explanation due to N.V.Gribov given in his following conference article and also beautifully explained by Dmitri Kharzeev in the following arxiv article(section 1). Gribov's argument doesn't involve the heavy machinery of quantum field theory. He actually proves that in the case of colinear electric and magnetic ...

2

One can posit mathematical string theories in any dimensions of any kind. However, I do not understand why there are ten dimensions and not just any other number? The specific dimensions arise from the requirements of the known physics encapsulated in the Standard Model and other data coming from particle physics, plus the requirement of General ...

2

(1) String Theory is a very mathematical theory based on some natural assumptions, and this ends up relating Quantum Mechanics and General Relativity, as we want. Some of the equations in String Theory, however, have a proportionality constant $c$ in it, called the central charge. And when we manipulate these equations and set them equal to each other, we ...

2

I think 't Hooft and Kugo are solving different problems. 't Hooft addresses the issue that the anomaly involves a topological term. As a result, in perturbation theory there is no theta dependence and the anomaly equation by itself does not solve the $U(1)$ problem. He shows that topological objects (semi-classically, instantons) generate theta ...

Only top voted, non community-wiki answers of a minimum length are eligible


text:
# Inline math messes up vertical spacing

Often when I use math inside a paragraph, the spacing becomes ajar. This particularly happens when I use exponents or "choose" notation (for combinations).

Minimal working example below. The issue can be much worse with other setups with different fonts and font sizes.

\documentclass{article}
\usepackage{lipsum}
\begin{document}
\lipsum[1][1-3] There are $|A|={N \choose k}^2$ different options. \lipsum[1][1-3]
\end{document}

How it looks:

• TeX increases the spacing when the lines would otherwise overlap. You can change some parameters to control this, but how do you actually want to handle this situation? Jan 29 at 14:19
• Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.
– Community Bot
Jan 29 at 14:22
• unrelated but use \binom{N}{k} not {N \choose k} in latex Jan 29 at 14:44
• One solution woulod be to increase the spacing on every line. See tex.stackexchange.com/questions/819/double-line-spacing Jan 29 at 23:00
• @JohnKormylo with a highly underrated answer. Would get an upvote if it were an official answer :)
– ABC
Jan 30 at 0:31

I suggest you familiarize yourself with the smallmatrix environment that's provided by the amsmath package.

It is straightforward to create a dedicated macro -- called \smallchoose in the following example -- that builds on the smallmatrix environment to typeset an expression that's noticeably smaller, along the vertical axis, than either {N \choose k} or \binom{N}{k}. Indeed, the result of $\smallchoose{N}{k}^2$ is so small as to no longer affect the line spacing.

\documentclass{article}
\usepackage{lipsum}    % filler text
\usepackage{amsmath} % for 'smallmatrix' env.
\newcommand\smallchoose[2]{%
\bigl(\!\begin{smallmatrix}#1\\#2\end{smallmatrix}\!\bigr){}}

\begin{document}

${N\choose k}^2$ $\binom{N}{k}^2$ $\smallchoose{N}{k}^2$

\bigskip
\lipsum[1][1-3]
There are $|A|={N \choose k}^2$ different options.
\lipsum[1][1-3]

\medskip
\lipsum[1][1-3]
There are $|A|=\smallchoose{N}{k}^2$ different options.
\lipsum[1][1-3]
\end{document}

Just for historical context, I note that my copy of The Printing of Mathematics, (O.U.P, 1954) has this paragraph about binomial coefficients:

If I render that into LaTeX, you can see that the layout is rather better than OUP managed in 1954.

\documentclass{article}
\usepackage{amsmath}
\usepackage[textwidth=324pt]{geometry}
\begin{document}
\textit{Binomial coefficients} are now usually written with the notation $\binom{n}{r}$,
though this, as here, breaks the line awkwardly when it appears in the
text, as it may well do in enumerating the number of members in a set.
If this symbol is printed with $5\frac12$-point letters as $\scriptstyle\binom{n}{r}$, it is less quickly read,
especially with a more elaborate form such as $\scriptstyle\binom{mn+p-1}{n}$. The older forms
${}^nC_r$, $C^n_r$, which still survive, also have this disability and wastefully add
the intrusive $C$.  It is sugested that the symbol $(n\mathbin{!}r)$, which has been
used, might be tried more frequently.
\end{document}

I quite like (n!r) but I am not sure it will ever catch on.

• Agree. Another alternative is to write it as a 2-argument function nCr, which is perhaps deemed inelegant but at any rate clear. Jan 30 at 15:49
• +1 for the LaTeX typesetting of that classic. Jan 31 at 0:47

You have got many nice LaTeX answers, and hopefully resolved your problem. Let me give you one possible (partly new, tuning in progress) way to handle this in ConTeXt. (It will not help you in this case, but another TeX.SE user asked me to add it here. Maybe we will see something similar in LaTeX in the near future? And maybe it will help somebody else?)

As others have pointed out, extra space is usually added between lines when needed, in the sense that the depth of the top line is too big and/or the height of the bottom line is too big. In your example the main problem is the superscript 2 that sits high (as you will see it is placed lower in ConTeXt, probably due to some other font parameter setup).

Now, the standard model, described above, takes the whole lines in consideration when deciding on adding some extra skip or not between lines. With so-called profiling we act locally, to see if there really is some clash going on. In your example, just above the superscript 2 there is nothing that has a depth, so in principle the lines could sit close together. Let us see it in action with one example:

The top paragraph is set without profiling. The lines are shown with help of \showmakeup[line]. In the second example, profiling is switched on with \setupalign[profile]. We also use a helper (enabled with \enabletrackers[profiling.lines.show]) to show where it was activated. As you see, all extra space above the line with the "problematic" superscript 2 is gone. In the line below it is not, since that would mean that the left parenthesis in the binomial would be too close to the capital I on the line below. Let us also see how it looks without the helpers:

This mechanism is of course adding some overhead to the compilation, but it has turned out to be negligible (One reason is that it only has to check lines where the extra skip is added, and that is usually not too many.) It is possible to configure the profiling (amount and granularity) with \setuplineprofile.

It does not really fit:

If the p on the line above had been slightly to the left it would overlap the 2. Tex makes a decision based on the whole line. You can allow overlaps in individual cases with \smash (2) or for all lines with \lineskiplimit (3) or adjust the layout so the problem does not occur (4) and (5)

\documentclass{article}
\usepackage{lipsum,amsmath}
\showoutput
\showboxdepth=4
\begin{document}

1. \lipsum[1][1-3] There are $|A|=\binom{N}{k}^2$ different options. \lipsum[1][1-3]

2. \lipsum[1][1-3] There are \smash{$|A|=\binom{N}{k}^2$} different options. \lipsum[1][1-3]

{\setlength{\lineskiplimit}{-5pt}
3. \lipsum[1][1-3] There are $|A|=\binom{N}{k}^2$ different options. \lipsum[1][1-3]

}

4. \lipsum[1][1-3] There are $|A|=\binom{N}{k}^2$ different options. \lipsum[1][1-3]

5. \lipsum[1][1-3] There are $|A|=\binom{N}{k}{}^2$ different options. \lipsum[1][1-3]

\end{document}

• Another possibility is to increase the \baselineskip, although this obviously affects far more of the document. Nonetheless, if this formulation occurs frequently throughout, it may be an appropriate adaptation. Jan 29 at 15:37

If you have many such objects in your document, I'd consider increasing the interline space.

For the particular case, however, it's the exponent that gets in the way, so I can suggest lowering it.

\documentclass{article}
\usepackage{amsmath}
\usepackage{lipsum}

\makeatletter
\NewDocumentCommand{\Tbinom}{mme{^}}{%
\IfNoValueTF{#3}{\tbinom{#1}{#2}}{%
\begingroup
\sbox\z@{$\m@th\tbinom{#1}{#2}$}%
\usebox{\z@}\raisebox{\dimexpr\ht\z@-\height}{$\m@th^{\mkern-1mu #3}$}%
\endgroup
}%
}
\makeatother

\begin{document}

\lipsum[1][1-3] There are $|A|=\Tbinom{N}{k}^{2}$ different options. \lipsum[1][1-3]

\end{document}

The same, but with \linespread{1.05} before \begin{document}

• +1. Would you say that the advantage of \Tbinom{N}{k}^{2} over \binom{N}{k}{}^{2} lies in better placement of the exponent?
– Mico
Jan 29 at 17:44
• @Mico Yes, that was my first attempt, but the exponent comes up too low. Jan 29 at 20:26
• I appreciate this suggestion!
– ABC
Jan 30 at 0:33


text:
## Installing Cordova

To install Cordova:

sudo apt-get install nodejs

NodeJs is an Open Source server environment.

sudo apt-get install npm

Npm is a registry package.

sudo apt-get install git

Git is is a decentralized versioning system.
sudo npm install -g cordova

This line will install Cordova.

Once done, if you execute:

cordova

You should already getting an answer from the CLI that Cordova is installed. Then create a folder, name it like you want and go within it, once there execute:

cordova create hello com.example.hello HelloWorld

hello is the name of your project, HelloWorld the name of your app.

Then you modify everything that you have to modify within the www folder as you would do for any website.

Then you need to indicate on which platform you wish to build it for:

cordova platform add android

In this case it will add all the necessary files to Cordova in order to create apps for Android.

Then you need to build the file to have it on your smartphone, in the case of Android it will be the .apk file:

cordova build --android

It may be possible that you are getting a message such as this one:

"Failed to find 'ANDROID_HOME' environment variable. Try setting it manually.
Failed to find 'android' command in your 'PATH'. Try update your 'PATH' to include path to valid SDK directory."

This message means that the Android SDK is not installed yet on your computer. So you need to go on the Android official website and download and install it on your computer: https://developer.android.com/studio. There are packages for each operating system.

Once the file downloaded, unzip it where you want on your computer.

In my case this file is now accessible as: /home/myusername/sdk-tools-linux-4333796

Then you need to configure ANDROID_HOME, in my case it will be:

export ANDROID_HOME=$HOME/sdk-tools-linux-4333796 export PATH=$PATH:$ANDROID_HOME/tools export PATH=$PATH:\$ANDROID_HOME/platform-tools

Once done, you will see that they will ask you to install Android Studio as well.

it will generate normally the .apk file. And then you are good to go.

##### How to launch the emulator with Cordova?

Last modified: Wednesday, 27 November 2019, 9:27 PM


text:
How to make a section header span the whole page in a twocolumn document

I am writing a book in twocolumn format. Section headers appear in their respective column, and for most of the book, that is fine for me.

But for a specific chapter of the book, I would prefer the section header to span both columns. Only the section header, not the subsection headers, and only in that specific chapter of the book.

Is there a way to achieve this?

-
multicol package may help you. –  xport Jul 22 '11 at 7:58
@xport: I've read about the multicol package, and that it is more flexible in this way, but I also read that the multicol environment isn't really meant to render a whole document in columns. The book is somewhat extensive / complex... are you serious about that suggestion? (Best make it an answer so I could vote for it...) –  DevSolar Jul 22 '11 at 8:00
How to force column break in multicol answer may help you too. Don't forget to up vote if you are satisfied with it. :-) –  xport Jul 22 '11 at 8:01
I cannot repost my answer because a link is NOT regarded as an answer by SE/SO policy, so I put in comments. :-) –  xport Jul 22 '11 at 8:03
You can use multicols environment as many as you want in the document. –  xport Jul 22 '11 at 8:10

if it starts on a new page use the optional argument of \twocolumn

\documentclass[twocolumn]{article}
\usepackage{lipsum}
\begin{document}
\twocolumn[\section{I can find a tool to convert PDF to
EPS in my neither bathroom nor kitchen}]

\lipsum[1-2]

\section{bar}
\lipsum[1-5]

\end{document}

-
Excellent! This is what I've been looking for (but obviously not hard enough). –  DevSolar Jul 22 '11 at 8:38
@Dev: Are you sure that each section in your document starts on a new page? :-) –  xport Jul 22 '11 at 9:00
@xport: No, but all of them that I need to stretch across columns. ;-) Besides, I've been doing some testing of your multicol suggestion, and it somewhat messes up my column layout. Herbert's suggestion worked perfectly right out-of-the-box. Sorry. ;-) –  DevSolar Jul 22 '11 at 9:04

Herbert's answer with \twocolumn[...] makes each section starts on a new page. If you don't like this behavior, then use multicol package that is very flexible.

\documentclass{article}
\usepackage[a3paper,margin=2cm]{geometry}
\usepackage{lipsum,multicol}
\title{Introduction to \LaTeX}
\author{xport}
\begin{document}
\maketitle
\tableofcontents
\section{I can find a tool to convert PDF to EPS in my neither bathroom nor kitchen. Normal mode.}
\begin{multicols}{2}
\lipsum[1]
\lipsum[2-3]
\end{multicols}

\section{Jesus is the absolute expression of perfection. With column break.}
\begin{multicols}{2}
\lipsum[1]
\columnbreak
\lipsum[2-3]
\end{multicols}

\section{Jesus is the absolute expression of perfection. Filling from left to right column.}
\begin{multicols*}{2}
\lipsum[1]
\lipsum[2-4]
\end{multicols*}

\lipsum[1]
\end{document}

Notes:

1. Use \raggedcolumns globally in the preamble if you don't want the bottom baselines of all columns to be aligned. The default is \flushcolumns, so TEX will normally try to make both the top and bottom baselines of all columns align.
-
I just scared my coworkers with a guffaw when I realized there's actually a package for lorem ipsum... :-D –  DevSolar Jul 22 '11 at 8:30
Try load blindtext package and use \blindtext or Blinddocument in your body. –  xport Jul 22 '11 at 8:35


text:
# Search by Topic

#### Resources tagged with Working systematically similar to Tree Tops:

Filter by: Content type:
Age range:
Challenge level:

### There are 126 results

Broad Topics > Using, Applying and Reasoning about Mathematics > Working systematically

### Problem Solving, Using and Applying and Functional Mathematics

##### Age 5 to 18 Challenge Level:

Problem solving is at the heart of the NRICH site. All the problems give learners opportunities to learn, develop or use mathematical concepts and skills. Read here for more information.

### Gr8 Coach

##### Age 11 to 14 Challenge Level:

Can you coach your rowing eight to win?

### Pole Star Sudoku 2

##### Age 11 to 16 Challenge Level:

This Sudoku, based on differences. Using the one clue number can you find the solution?

### The Naked Pair in Sudoku

##### Age 7 to 16

A particular technique for solving Sudoku puzzles, known as "naked pair", is explained in this easy-to-read article.

##### Age 11 to 16 Challenge Level:

Four small numbers give the clue to the contents of the four surrounding cells.

### One Out One Under

##### Age 14 to 16 Challenge Level:

Imagine a stack of numbered cards with one on top. Discard the top, put the next card to the bottom and repeat continuously. Can you predict the last card?

### 9 Weights

##### Age 11 to 14 Challenge Level:

You have been given nine weights, one of which is slightly heavier than the rest. Can you work out which weight is heavier in just two weighings of the balance?

### LOGO Challenge - Triangles-squares-stars

##### Age 11 to 16 Challenge Level:

Can you recreate these designs? What are the basic units? What movement is required between each unit? Some elegant use of procedures will help - variables not essential.

### Difference Sudoku

##### Age 14 to 16 Challenge Level:

Use the differences to find the solution to this Sudoku.

### Olympic Logic

##### Age 11 to 16 Challenge Level:

Can you use your powers of logic and deduction to work out the missing information in these sporty situations?

### More on Mazes

##### Age 7 to 14

There is a long tradition of creating mazes throughout history and across the world. This article gives details of mazes you can visit and those that you can tackle on paper.

### Instant Insanity

##### Age 11 to 18 Challenge Level:

Given the nets of 4 cubes with the faces coloured in 4 colours, build a tower so that on each vertical wall no colour is repeated, that is all 4 colours appear.

### Twin Line-swapping Sudoku

##### Age 14 to 16 Challenge Level:

A pair of Sudoku puzzles that together lead to a complete solution.

### Magic Caterpillars

##### Age 14 to 18 Challenge Level:

Label the joints and legs of these graph theory caterpillars so that the vertex sums are all equal.

### Ratio Sudoku 2

##### Age 11 to 16 Challenge Level:

A Sudoku with clues as ratios.

### Intersection Sudoku 1

##### Age 11 to 16 Challenge Level:

A Sudoku with a twist.

### Twin Corresponding Sudoku

##### Age 11 to 18 Challenge Level:

This sudoku requires you to have "double vision" - two Sudoku's for the price of one

### Crossing the Bridge

##### Age 11 to 14 Challenge Level:

Four friends must cross a bridge. How can they all cross it in just 17 minutes?

### Integrated Sums Sudoku

##### Age 11 to 16 Challenge Level:

The puzzle can be solved with the help of small clue-numbers which are either placed on the border lines between selected pairs of neighbouring squares of the grid or placed after slash marks on. . . .

### Twin Corresponding Sudokus II

##### Age 11 to 16 Challenge Level:

Two sudokus in one. Challenge yourself to make the necessary connections.

### Spot the Card

##### Age 14 to 16 Challenge Level:

It is possible to identify a particular card out of a pack of 15 with the use of some mathematical reasoning. What is this reasoning and can it be applied to other numbers of cards?

### Twin Chute-swapping Sudoku

##### Age 14 to 18 Challenge Level:

A pair of Sudokus with lots in common. In fact they are the same problem but rearranged. Can you find how they relate to solve them both?

### Magnetic Personality

##### Age 7 to 16 Challenge Level:

60 pieces and a challenge. What can you make and how many of the pieces can you use creating skeleton polyhedra?

### LOGO Challenge - the Logic of LOGO

##### Age 11 to 16 Challenge Level:

Just four procedures were used to produce a design. How was it done? Can you be systematic and elegant so that someone can follow your logic?

##### Age 11 to 14 Challenge Level:

You need to find the values of the stars before you can apply normal Sudoku rules.

### Tea Cups

##### Age 7 to 14 Challenge Level:

Place the 16 different combinations of cup/saucer in this 4 by 4 arrangement so that no row or column contains more than one cup or saucer of the same colour.

### Integrated Product Sudoku

##### Age 11 to 16 Challenge Level:

This Sudoku puzzle can be solved with the help of small clue-numbers on the border lines between pairs of neighbouring squares of the grid.

### Wallpaper Sudoku

##### Age 11 to 16 Challenge Level:

A Sudoku that uses transformations as supporting clues.

### Warmsnug Double Glazing

##### Age 11 to 14 Challenge Level:

How have "Warmsnug" arrived at the prices shown on their windows? Which window has been given an incorrect price?

### Latin Squares

##### Age 11 to 18

A Latin square of order n is an array of n symbols in which each symbol occurs exactly once in each row and exactly once in each column.

### Seasonal Twin Sudokus

##### Age 11 to 16 Challenge Level:

This pair of linked Sudokus matches letters with numbers and hides a seasonal greeting. Can you find it?

### Cinema Problem

##### Age 11 to 14 Challenge Level:

A cinema has 100 seats. Show how it is possible to sell exactly 100 tickets and take exactly £100 if the prices are £10 for adults, 50p for pensioners and 10p for children.

### LOGO Challenge - Sequences and Pentagrams

##### Age 11 to 18 Challenge Level:

Explore this how this program produces the sequences it does. What are you controlling when you change the values of the variables?

### Tetrahedra Tester

##### Age 11 to 14 Challenge Level:

An irregular tetrahedron is composed of four different triangles. Can such a tetrahedron be constructed where the side lengths are 4, 5, 6, 7, 8 and 9 units of length?

### When Will You Pay Me? Say the Bells of Old Bailey

##### Age 11 to 14 Challenge Level:

Use the interactivity to play two of the bells in a pattern. How do you know when it is your turn to ring, and how do you know which bell to ring?

### Difference Dynamics

##### Age 14 to 18 Challenge Level:

Take three whole numbers. The differences between them give you three new numbers. Find the differences between the new numbers and keep repeating this. What happens?

### Sticky Numbers

##### Age 11 to 14 Challenge Level:

Can you arrange the numbers 1 to 17 in a row so that each adjacent pair adds up to a square number?

### Rainstorm Sudoku

##### Age 14 to 16 Challenge Level:

Use the clues about the shaded areas to help solve this sudoku

### Rectangle Outline Sudoku

##### Age 11 to 16 Challenge Level:

Each of the main diagonals of this sudoku must contain the numbers 1 to 9 and each rectangle width the numbers 1 to 4.

### Bent Out of Shape

##### Age 14 to 18 Challenge Level:

An introduction to bond angle geometry.

### Inky Cube

##### Age 7 to 14 Challenge Level:

This cube has ink on each face which leaves marks on paper as it is rolled. Can you work out what is on each face and the route it has taken?

### First Connect Three for Two

##### Age 7 to 14 Challenge Level:

First Connect Three game for an adult and child. Use the dice numbers and either addition or subtraction to get three numbers in a straight line.

### Building with Longer Rods

##### Age 7 to 14 Challenge Level:

A challenging activity focusing on finding all possible ways of stacking rods.

### I've Submitted a Solution - What Next?

##### Age 5 to 18

In this article, the NRICH team describe the process of selecting solutions for publication on the site.

### Twin Corresponding Sudoku III

##### Age 11 to 16 Challenge Level:

Two sudokus in one. Challenge yourself to make the necessary connections.

### Multiples Sudoku

##### Age 11 to 14 Challenge Level:

Each clue in this Sudoku is the product of the two numbers in adjacent cells.

### Star Product Sudoku

##### Age 11 to 16 Challenge Level:

The puzzle can be solved by finding the values of the unknown digits (all indicated by asterisks) in the squares of the $9\times9$ grid.

### Colour Islands Sudoku

##### Age 11 to 14 Challenge Level:

An extra constraint means this Sudoku requires you to think in diagonals as well as horizontal and vertical lines and boxes of nine.


text:
# Hereditary ring

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

In mathematics, especially in the area of abstract algebra known as module theory, a ring R is called hereditary if all submodules of projective modules over R are again projective. If this is required only for finitely generated submodules, it is called semihereditary.

For a noncommutative ring R, the terms left hereditary and left semihereditary and their right hand versions are used to distinguish the property on a single side of the ring. To be left (semi-)hereditary, all (finitely generated) submodules of projective left R-modules must be projective, and to be right (semi-)hereditary all (finitely generated) submodules of projective right submodules must be projective. It is possible for a ring to be left (semi-)hereditary but not right (semi-)hereditary, and vice versa.

## Examples

• Semisimple rings are easily seen to be left and right hereditary via the equivalent definitions: all left and right ideals are summands of R, and hence are projective. By a similar token, in a von Neumann regular ring every finitely generated left and right ideal is a direct summand of R, and so von Neumann regular rings are left and right semihereditary.
• An important example of a (left) hereditary ring is the path algebra of a quiver. This is a consequence of the existence of the standard resolution (which is of length 1) for modules over a path algebra.

## Properties

• For a left hereditary ring R, every submodule of a free left R-module is isomorphic to a direct sum of left ideals of R and hence is projective.[2]

## References

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• {{#invoke:citation/CS1|citation

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• {{#invoke:citation/CS1|citation

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• {{#invoke:citation/CS1|citation

|CitationClass=citation }}


text:
library(tidyverse)
library(forcats)
library(ggrepel)
library(smodels)
library(cleanNLP)
library(glmnet)

theme_set(theme_minimal())
options(dplyr.summarise.inform = FALSE)
options(width = 77L)

## Applications of all this stuff

Now that we are through the first project and about halfway through the semester, I thought it would be helpful to take a quick step back and put into perspective what we have learned so far.

Hopefully it is fairly clear that if you want to work in a field that performs data analysis, it is generally useful to understand the basic terminology of predictive models and how to present this information in a short but informative way. I also hope that you feel that the structure of the course has helped teach you some of these basics and give space to practice and hone your skills actually applying and presenting the results.

What might be less obvious is why some of the specific techniques we are looking at may be useful in your future endeavours outside of research in natural language processing, linguistics, or a related field.

To help motivate the more general applicability of these techniques to various fields, now that you know the basics, here are four concrete reasons that the approaches for text analysis we are learning may be more directly useful than expected:

1. Lots of variables The distinguishing feature of our core set of models (penalized regression) is the need to do regression of multi-class classification when there are a very large number of variables. This is a very common problem in many (most?) domains. For example: genetics, engineering, marketing, accounting, finance, operations research, and supply chain management. Text is a good example to work with in class because the features and results should be familiar to anyone familiar with the English language. Most of the other examples require understanding a lot of domain knowledge that will not be common to everyone in the class.
2. Feature Construction In order to do text analysis we need to construct features; they are not given to us directly. Some can be created directly from the data, but most often useful features come from selectively summarizing the data from the tokens table. Again, this is very common in industry and research applications, where the most interesting data points are observered at a different frequency or unit of measurement than the response variable. For example, think of computing a credit score based on summary statistics selected from a person’s financial history or determining a patient’s prognosis based on a history of medical records and lab reports. As with the lots of variables point, text is just a useful example of this kind of analysis that does not require specialized domain knowledge.
3. Text is actually quite common A surprising amount of data being produced and analyzed in industry and research applications contains at least some free text fields. Think, for example, of trying to detect indicators of fraud when investigating a collection emails, or automatically detecting themes or signaling problems by doing text analysis on call center logs. In the age of large datasets, the amount of free text used in all kinds of applications will likely only grow. Being able to know the specific techniques to work with is just another benefit of focusing on textual data.
4. False Binary: Supervised/Unsupervised We have not yet talked about unsupervised learning, a kind of machine learning that is not focused on predictive models. This will be the focus on Project 3 and Project 4. Text datasets lend themselves to both modes of analysis simultaneously; this is another common feature of real-world applications that is often lost in other textbook examples of machine learning applications.

I hope that these examples help show further links between and possible applications of what we are learning in class. It should also be mentioned that I find working with textual data to also be just a bit more fun (at least, for most of us) than orienting the class around other tabular datasets.

## Feedback from Project 1 and looking towards Project 2

I was overall very impressed with the presentations for the first project. There were virtually no major mistakes and it was clear that a lot of time and energy went into the project. I really appreciate that and hope it was a productive and not-to-painful (or even enjoyable?) experience. I have called out specific things to keep in mind for the second project, mostly minor things, in your feedback. I am happy to discuss further in your breakout groups today.

The second project has the same general instructions and format, but the data and classification task has a larger scope. You’ll need to be more careful to focus your final analysis on just a small set of interesting things. I am not evaluating the scope of what you did, but rather how well you were able to tell a story about something interesting in the data. I recommend the following rough schedule:

• Today, have everyone download the data and start building a few models. Discuss as a group what you are seeing and perhaps break out different tasks for each of you to look at for next week.
• In class next Tuesday, after my notes on a new ML technique, talk about the different results and formulate a plan for what big things you want to focus on for the project write-up and presentation.
• In class next Thursday, after my notes on another new ML technique, make a plan to put the different pieces together. Take note and ask about other visualizations or other things you’d like to do with the data that you may not be obvious from my notes.

My guess is that you will find Project 2 much harder. We will built out more techniques for working with this kind of data as we head into Project 3. Just do your best with the tools we have at hand!


text:
## Illustration regarding payment of surcharge in challan form VAT-2B.

In view of calculation of amount for Challan-2, 2A and 2B is as follows :-

Suppose sale is Rs. 1 lakh

Tax @ 5% = 5000

additional tax @ 10% of tax = Rs. 500

Total Tax is Rs. 5500

Suppose a client comes to you and tell you that his total Tax + Surcharge is = Rs. 5500, then you can calculate VAT and Surcharge as follows :-

5500 x 10/11 = 5000 – VAT

5500/11 = 500 – Surcharge

Now you have to pay Rs. 5500 in which is 5000 is VAT and 500 is Surcharge. This can be paid as follows :-
On Challan-2 = Rs. 4900 (you can deposit in cash or through cheque/draft in favour of Assistant Excise and Taxation Commissioner)

On Challan 2A = Rs. 500 (you can deposit in cash or through cheque/draft in favour of Punjab Municipal Fund)

On Challan 2B = Rs. 100 (you can deposit in cash or through cheque/draft in favour of Punjab Municipal Infrastructure Development Fund)

These figures represent percentage wise as follows :-

On Challan 2 = 98% of VAT (excluding surcharge) (i.e. Rs. 4900)
On Challan 2A = 10% of VAT (excluding surcharge) (i.e. Rs. 500)
On Challan 2B = 2% of VAT (excluding surcharge) (i.e. Rs. 100)

[Note : Practically the above solution is the only solution. Although if we go by the exact wording of the law, - tax, Municipal Fund and Municipal Infrastructure Development Fund cannot be calculated properly.]

See here relevant notification on form VAT 2B


text:
Lunch tray and torque

1. gamesandmore

32
A lunch tray is being held in one hand, as the drawing illustrates. The mass of the tray itself is 0.180 kg, and its center of gravity is located at its geometrical center. On the tray is a 1-kg plate of food and a 0.350-kg cup of coffee. Obtain the force T exerted by the thumb and the force F exerted by the four fingers. Both forces act perpendicular to the tray, which is being held parallel to the ground.

I don't even have a clue where to start??
Here is what I tried though, but I'm not sure:
Sum of Counter-clockwise Torques = Sum of Clockwise Torques
Tf = Tplate + Tcup + Ttray
then:
Ff(.1m) = (1kg*9.80m/s^2)(.24m)+(.35kg*9.80m/s^2*.38m)+(.18kg*9.80m/s^2*.2m)
and ended up with F being equal to: 40.1... but I'm not sure?

2. gamesandmore

32
Anyone??? I need this as soon as I can.

3. PhanthomJay

6,151
You have somewhat of an idea of the correct approach, but when you sum torques about the left end, you have to include the torque provided by the thumb as well as the fingers. It is better to sum torques about the thumb, because then that force will not enter into the torque equation because it has no torque. Then apply Newton 1 in the y direction to determine the thumb force.

4. gamesandmore

32
im confused... sum up at the left end?

Last edited: Nov 28, 2006
5. PhanthomJay

6,151
Newton's first law requires that the torques must sum to 0 (Clockwise torques = counterclockwise torques); AND, the forces in the vertical direction must sum to 0 (Forces up = forces down). Now try summing torques about the thumb and see what you get for Ff. You'll have to do a little math to get the proper distances from the load points to the thumb, but so what?

Last edited: Nov 28, 2006
6. gamesandmore

32
I understand the part about newton's first law...
But I cannot figure out what you mean...
This is due in 40 minutes and I'm stressing out about it.

7. PhanthomJay

6,151
You had the right approach, stop stressing. Look at the clockwise torques about the thumb. I'll start you off withthe cup of coffee. Its torque about the thumb is

0.35(9.8)(.330) = 1.13

Now do the tray and plate torques about the thumb, and add 'em up with the coffee torque detrmined above, then set the result equal to the fingers counterclockwise torque of Ff(0.50), and solve for Ff. Remember, the distance to use in finding the torques is the distance form the load to the thumb.

8. gamesandmore

32
alright, I got F: 79.0 N
Thank you!!!!

T I am confused on right now... would it be the same value as F?

F = T + Fplate + Fcup + Ftray?
edit: I tried that and got 64 as an answer, sound right?

Last edited: Nov 28, 2006
9. PhanthomJay

6,151
I didn't check your math, but once you get F, then apply Newton 1: weight of tray plus weight of cup plus weight of plate plus the thumb force T = F. Clock is ticking, I usually don't do all the work.....

10. PhanthomJay

6,151
Gee, now you got me stressed. How we doing for time? Check your math. .35(9.8)(.33) + 1(9.80)(.19) + (.18)(9.8)(.150) = F(.050). Solve for F. , then solve for T.

11. gamesandmore

32
Yes, I'm sorry I didn't reply again.
Thank You very much, I got full credit for the problem. :)

12. Haibane

5
I apologize for bringing up an old thread but I have more or less the same problem with this equation, even after looking at it and trying to solve it.

It is the same except that the mass of the tray 0.200 kg, and the coffee is 0.250 kg.

Now Ive tried working it out, factoring in the fact that the cup of thumb is at a distance of .0600m which means using this as the axis Id have to remove .0600 for every lever arm which gives me.

(F*.04) = ((9.8*1kg)*.04)+((9.8*.250)*.32)+((19.8*.180)*.34) which leads to

(F*.04) = 1.764 + .784 + .59976 or just (.60) which = 3.148 and then is divided by .04 to = 78.7.... when the correct answer for F supposed to be 70.6.

What am I missing or doing wrong? This problem is driving me up the wall

-Haibane

Last edited: Dec 6, 2006

6,151


text:
## Encyclopedia > Meridian, Mississippi

Article Content

# Meridian, Mississippi

Meridian is a city located in Lauderdale County, Mississippi of which it is the county seat. As of the 2000 census, the city had a total population of 39,968.

History Meridian was founded in 1854 at crossing point of the Mobile & Ohio railroad line by the Vicksburg & Montgomery railroad line; originally known as Sowashee Station, and officially incorporated as Meridian in 1860.

The town was burned in 1864 by General William Tecumseh Sherman.

By 1900 the population of Meridian was approximately 25,000 people.

Geography Meridian is located at 32°22'29" North, 88°42'15" West (32.374841, -88.704160)1.

According to the United States Census Bureau, the city has a total area of 118.8 km² (45.9 mi²). 116.9 km² (45.1 mi²) of it is land and 1.9 km² (0.8 mi²) of it is water. The total area is 1.64% water.

Demographics As of the census of 2000, there are 39,968 people, 15,966 households, and 10,026 families residing in the city. The population density is 342.0/km² (885.9/mi²). There are 17,890 housing units at an average density of 153.1/km² (396.5/mi²). The racial makeup of the city is 43.99% White, 54.37% African American, 0.17% Native American, 0.60% Asian, 0.03% Pacific Islander, 0.28% from other races, and 0.57% from two or more races. 1.08% of the population are Hispanic or Latino of any race.

There are 15,966 households out of which 31.1% have children under the age of 18 living with them, 36.2% are married couples living together, 23.3% have a female householder with no husband present, and 37.2% are non-families. 33.2% of all households are made up of individuals and 14.0% have someone living alone who is 65 years of age or older. The average household size is 2.39 and the average family size is 3.06.

In the city the population is spread out with 27.2% under the age of 18, 9.9% from 18 to 24, 26.6% from 25 to 44, 19.7% from 45 to 64, and 16.5% who are 65 years of age or older. The median age is 35 years. For every 100 females there are 84.0 males. For every 100 females age 18 and over, there are 77.0 males.

The median income for a household in the city is $25,085, and the median income for a family is$31,062. Males have a median income of $29,404 versus$19,702 for females. The per capita income for the city is \$15,255. 28.6% of the population and 24.6% of families are below the poverty line. Out of the total people living in poverty, 40.8% are under the age of 18 and 22.0% are 65 or older.

Sites of Interest The city's Grand Opera House is remarkable for being preserved little changed from when it opened in 1890.

Meridian was the birthplace of early country music star Jimmie Rodgers. Highland Park houses a Jimmie Rodgers Museum, as well as a 19th century carousel.

External Links City of Meridian website (http://www.meridianms.org/)

All Wikipedia text is available under the terms of the GNU Free Documentation License

Search Encyclopedia
 Search over one million articles, find something about almost anything!

Featured Article
 Quantum entanglement ... as $\rho = \sum_i w_i |\alpha_i\rangle \langle\alpha_i|$. All the above results for density matrices and the quantum entropy remain valid ...


text:
# generalization of the discreteness of Hecke groups to general reductive groups

Consider the subgroup $$G_{\lambda}$$ of $$SL_2(\mathbb R)$$ generated by $$N_{\lambda} = \begin{bmatrix} 1 & \lambda \\ 0 & 1 \end{bmatrix}$$ and $$S = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$$ where $$\lambda>0$$. Then it's known by Hecke that $$G_{\lambda}$$ is discrete if and only if $$\lambda \geq 2$$ or $$\lambda=2 \cos(\frac{\pi}{n})$$ where $$n \geq 3$$ is an integer. They are named for Hecke, and are used by Hecke to study modular forms.

Is there is a generalization of this fact to more general groups? For example, consider a subgroup of $$SL_n(\mathbb R)$$ generated by several elementary matrices and some involutions, when is it discrete?

What about $$p$$-adic analogues?

• Even when you consider two unipotent matrices in $SL(2,C)$ this problem does not have a good answer (it is undecidable in certain sense). mathoverflow.net/questions/109967/… – Misha Feb 8 at 20:35
• Even the group generated by $\begin{pmatrix} 1 & \lambda \\ 0 & 1 \end{pmatrix}$ inside $\operatorname{SL}_2(\mathbb Q_p)$ is not discrete. (In $\operatorname{SL}_2(\mathbb F_p((t)))$ it's finite, although I don't know if your group $G_\lambda$ still is.) – LSpice Feb 9 at 0:16
• @Misha Thank you for that answer.. How about other groups? For example $SL_3(\mathbb R)$? – zzy Feb 11 at 20:58
• @zzy: I do not know for sure about $SL(3,R)$. For "positive relatively Anosov" subgroups (a very special class of discrete subgroups) a description should be possible. But for general discrete subgroups I am very skeptical. However, not enough is known about discrete subgroups of $SL(3,R)$ at this point to prove anything definitive. – Misha Feb 11 at 21:24


text:
## Random Points on a Sphere (Generating Random Sequences III)

October 12, 2010

Last week, we discussed how we could generate uniform random points in a triangle using a (tiny) bit of linear algebra, mostly the parallelogram rule, and a random variable uniform on $[0,1]$. It required a tiny bit of math and was computationally very simple.

This time, let us see how we can generate uniformly distributed random points on a sphere.


text:
# Itemize Environment: Change only size of the bullet not the text

I would like to change only the size of the bullet not the text of item as mentioned in the question. I tried following approach but it changes the bullet size as well as text size. Thanks in advance for your help.

MWE:

\documentclass{article}
\usepackage{enumitem}

\begin{document}
{\large
\begin{itemize}
\item Item 1
\item Item 2
\item Item 3
\item Item 4
\end{itemize}
}
\end{document}

• Bigger or smaller? – egreg Mar 26 '16 at 18:29
• @egreg: I would like to see for \large size. – Santosh Mar 26 '16 at 18:32
• The reason why {\large...} doesn't work (well, it works, but it changes anything) is that the scope of \large is restricted to the itemize environment, but since there is no other font change (usually) this has an effect on all typesetting. – user31729 Mar 26 '16 at 18:48

The optional argument of enumitem can be used with label={\large\textbullet}, but this will not produce the best result always, perhaps, tiny bullets may have to be raised a little bit.

As can be seen, the font size of the normal text isn't changed.

There's the font=... option as well, meant for the label font (which can be 'any' command, from font size change to complete font shape change)

\documentclass{article}
\usepackage{enumitem}

\begin{document}

{ Tiny bullets: \tiny \textbullet}
\begin{itemize}[label={\tiny\raisebox{1ex}{\textbullet}}]
\item Item 1
\item Item 2
\item Item 3
\item Item 4
\end{itemize}

{Normally sized bullets: \textbullet}
\begin{itemize}
\item Item 1
\item Item 2
\item Item 3
\item Item 4
\end{itemize}

{Large bullets: \Large \textbullet}
\begin{itemize}[label={\large\textbullet}]
\item Item 1
\item Item 2
\item Item 3
\item Item 4
\end{itemize}
\end{document}


text:
# pi definition

• noun:
• The sixteenth letter for the Greek alphabet. See Table at alphabet.
• Mathematics A transcendental quantity, around 3.14159, represented because of the icon π, that conveys the ratio for the circumference to your diameter of a circle and appears as a continuing in a lot of mathematical expressions.
• An amount of kind that has been jumbled or thrown together randomly.
• The name regarding the sixteenth letter associated with the Classical and contemporary Greek alphabets and the seventeenth in Old Greek.
• An irrational and transcendental constant representing the proportion of circumference of a Euclidean circle to its diameter; around 3.1415926535897932384626433832795; usually written π.
• Metal kind that is spilled, combined together, or disordered. Also referred to as pie.
• The inorganic orthophoshate ion; -- emblematic used in biochemistry.
• quite a few kind confusedly mixed or unsorted.
• A Greek page (Π, π) corresponding on Roman letter p.
• The page π, Π, as regularly denote the number or quotient approximately articulating the proportion associated with circumference of a circle to its diameter; additionally, the quotient or the ratio itself. The worthiness for the quotient pi, to twenty decimal places, is 3.14159265358979323846 (see note). The quantity pi is an irrational number, for example. it may not be expressed while the quotient of two integers. Additionally, it is a transcendental number, for example. it cannot be expressed as a root of an algebraic equation with a finite range terms; and using this fact follows the impossibility for the quadrature regarding the group by purely algebraic procedures, or by the aid of a ruler and compass.
• Printing-types blended collectively indiscriminately; type in a confused or jumbled condition or mass.
• The name of Greek letter Π, π, corresponding on Roman P, p.
• The name of a symbol (π) found in geometry the ratio for the circumference of a circle to its diameter. or 3.1415927: first so used by Euler.
• the ratio for the circumference to the diameter of a circle; approximately corresponding to 3.14159265358979323846...
• the sixteenth letter of this Greek alphabet
• somebody who can be used as a detective to collect information
• an antiviral medicine used against HIV; interrupts HIV replication by binding and preventing HIV protease; usually used in combo with other medicines
• the scientist in charge of an experiment or research study
• verb-transitive:
• To jumble or mix up (type).
• To put into a mixed and disordered problem, as kind; to combine and disarrange the kind of.
• verb-intransitive:
• To become jumbled.
• verb:
• To spill or combine printing type. Additionally, "to pie".
• maybe not area of the typical font character set; specially, non-Roman kind or symbols as opposed to level alphanumeric Roman type.
• abbreviation:
• pica (conventionally, 12 points = 1 pica, 6 picas = 1 inches)
• piaster
• others:
• To reduce (printing-types) to a situation of pi.

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### Anomalous Magnetic Moment of Electron in Chern-Simons QEDAnomalous Magnetic Moment of Electron in Chern-Simons QED

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 Author Das, Ashok ♦ Perez, Silvana Source arXiv.org Content type Text File Format PDF Date of Submission 2003-11-15 Language English
 Subject Domain (in DDC) Computer science, information & general works ♦ Natural sciences & mathematics ♦ Physics Subject Keyword High Energy Physics - Theory ♦ High Energy Physics - Phenomenology ♦ physics:hep-ph ♦ physics:hep-th Abstract We calculate the anomalous magnetic moment of the electron in the Chern-Simons theory in 2+1 dimensions with and without a Maxwell term, both at zero temperature as well as at finite temperature. In the case of the Maxwell-Chern-Simons (MCS) theory, we find that there is an infrared divergence, both at zero as well as at finite temperature, when the tree level Chern-Simons term vanishes, which suggests that a Chern-Simons term is essential in such theories. At high temperature, the thermal correction in the MCS theory behaves as $\frac{1}{\beta} \ln \beta M$, where $\beta$ denotes the inverse temperature and $M$, the Chern-Simons coefficient. On the other hand, we find no thermal correction to the anomalous magnetic moment in the pure Chern-Simons (CS) theory. Description Reference: Phys.Lett.B581:182-192,2004 Educational Use Research Learning Resource Type Article Page Count 13


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## Archive for October 23rd, 2011

### Definitions

October 23, 2011

By now you will have seen several definitions in lectures. Many of them will be written in the form

Definition. A blah is …

That is, the definition is displayed and the word being defined is in italics (or underlined if somebody is writing by hand). Sometimes, one doesn’t bother with the display, and simply says, during a discussion, “We define a blah to be …”

What is likely to have been emphasized less is that there are several different kinds of definition. In this post I’d like to enumerate some of them and give examples. It’s very much worth being aware, each time you meet a definition, what kind it is.
(more…)


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# Planar embeddings of planar graph

Let $G=(V,E)$ be a planar graph, only specified by its set of vertices and edges. Suppose $|V|=n$. According to Fary's theorem, there exists a planar embedding of $G$ with straight line segments. Then this embedding can be represented by a set of points in the plane (each $v\in V$ is represented by a unique point $p_v$ and there is a line segment between $p_u$ and $p_v$ iff $(u,v)\in E$)

Questions :

1. Is it possible to find this set of points in polynomial time (i.e. $n^{O(1)}$) ?
2. What if we force the points to have integer coordinates ?
3. Can we force the maximum (euclidian) distance between any two points to be $n^{O(1)}$ ?

Such statements could be very useful for proving lower bounds for some geometrical problems

• All three of your questions are answered in the page you link. – Peter Taylor Jun 20 '18 at 14:50
• I'm not sure, in this article they never mention the complexity of computing this planar embedding in polynomial time. My questions are not about existence but time complexity – Mathieu Mari Jun 25 '18 at 12:48
• In 'Related Results' they do mention being able to solve in linear time for a particular case. – koverman47 Jul 25 '18 at 15:08