text:
# $M$ is free $R$-module $\iff$ $M$ has $R$-basis

We will define the free $$R$$-modules.

Definition. Let $$R$$ be a ring with $$1_R$$ and $$F$$ an left $$R$$-module. We call $$F$$ free $$R$$-module, if $$F=\bigoplus_{i\in I} R_i$$ where $$R_i:=\langle b_i \rangle \cong _RR,\ \forall i \in I$$ and $$I$$ is a set of indexes (finite or infinite).

We will try to prove the following theorem.

Theorem. Let $$R$$ be a ring with $$1_R$$ and $$M$$ an left $$R$$-module. The following are equivalent.

1. $$M$$ is a left, free $$R$$-module.
2. $$M$$ has basis.

Proof. $$1.\implies 2.$$ According to our definition, $$M=\bigoplus_{i\in I} R_i$$, where $$R_i:=\langle b_i \rangle \cong _RR,\ \forall i \in I$$. and $$I$$ is a set of indexes (finite or infinite).

We define the set $$S:=\{e_i:=(\delta_{i,\lambda} )_{\lambda\in I}: i\in I\}\subseteq \bigoplus_{i\in I} R,$$ where $$\delta_{i,\lambda} =1_R$$, if $$\lambda=i$$ and $$\delta_{i,\lambda} =0_R$$ otherwise. Then, it's easy to observe that the set $$S$$ is $$R$$-basis for $$\bigoplus_{i\in I} R$$ has an $$R$$-basis and $$\bigoplus_{i\in I} R_i \cong \bigoplus_{i\in I} R = M$$, thus $$M$$ has an $$R$$-basis.

$$2.\implies 1.$$ We suppose that $$S:=\{e_i \in M :i \in I \} \subseteq M$$ is an $$R$$-basis for $$M$$. Obviously, the key is to define an $$R$$ module homomorphism $$\phi:M\longrightarrow \bigoplus_{i\in I} R.$$

And now my questions.

Questions. (1) Is the first part okey?

(2) The index set $$I$$ may be infinite, so does $$S$$. Then, does every element $$m\in M=\langle S \rangle$$ have a unique expression, as finite linear $$R$$-combination, in the form $$m=\sum_{k=1}^{n} r_k e_k$$, where $$n\in \Bbb N , r_k \in R, e_k \in S$$?

(3) Which the $$R$$-module isomorphism?

Thank you.

• I'm guessing that the definition is using the internal direct sum (which is a relation between submodules, despite the functional and equality notation) and not the external direct sum which has formal elements $\sum_{i \in I} \lambda_i e_i$. – Daniel Schepler Nov 29 '18 at 21:52
• @DanielSchepler Yes, indeed. But the two direct sums are in some way equivalent. So, it'doesn't really matter. Do you agree? Note that this definitions is taken form Rotman's book, Advanced Modern Algebra, AMS, 2012. – Chris Nov 29 '18 at 21:59
• Well, even when you have a canonical isomorphism $\alpha$, it's usual to make applications of that isomorphism explicit - so instead of $S := \{ e_i \ldots \}$ you would say something like $S := \{ \alpha(e_i) \ldots \}$. So, under the canonical isomorphism of the external direct sum $\bigoplus_{i\in I} R$ with $R$ under the hypothesis of $R$ being the internal direct sum of submodules $\langle b_i \rangle \simeq R$, what would $e_i$ be sent to? – Daniel Schepler Nov 29 '18 at 22:50
• Okay, let's take the external. It's easy then to do the transfromation. But my question is, shall we sent $m:=\sum_{i=1}^{n}r_i e_i \longmapsto \phi(m)= \phi (\sum_{i=1}^{n}r_i e_i) :=(r_1,...,r_n)$? And should these sums be finite? – Chris Nov 29 '18 at 22:58
• I'd say more specifically, my suggestion is to prove $M = \bigoplus_{i\in I} \langle b_i \rangle$ and $\langle b_i \rangle \simeq_R R$ for all $i \in I$ $\Leftrightarrow$ $\{ b_i \mid i \in i \}$ is a basis for $M$. (The part requiring $\langle b_i \rangle \simeq_R R$ is essential to rule out modules like $R / I$ where $I$ is a proper nontrivial ideal.) – Daniel Schepler Nov 30 '18 at 0:37


text:
## Material Buying Guide for Cement

The construction of the house involves more than 100 different materials. You will have to spend lot of time towards buying all those materials. You have to exercise care when you are buying all the materials.

Among them, cement is the most important raw material used in the construction of the house. It is available in grey colour in bag. One bag of cement contains 50kg weight of cement. Buying cement also involves lot of stages like selection of right type of cement, searching for the right supplier of the cement.

This process of buying any material is time consuming and demanding. It is not too easy as one may think. All these have been discussed in detail which you can read here.

Courtesy - Shutterstock

#### A) Personnel:

##### 01. Is Cement Need or Desire?

Of course, the cement is one’s need to construct the house, but you have to check whether the purpose for which you are buying cement is your need or desire. i.e. whether construction of a swimming pool in the house is a need or desire?

##### 02. Why You Need the Cement for Constructing a House?

It is most the useful binding material. Hence it is used for different purposes i.e. RCC work, plastering, flooring, filling cracks, making of compound wall, septic tank, etc. The purpose will define the type of cement you need. You will need different types for different purposes.

##### 03. Quality and Specification of Cement:
Courtesy - Shutterstock

i) There are many types of cement available in the market to suit your need. Although there are major three types of cement used in construction like:

• Ordinary Portland Cement (OPC)
• Portland Pozzolana Cement (PPC)
• White Cement

ii) You have also to decide whether you really want some specific cement like sulphate resistant cement, low heat cement, etc. To know more about special cement click here. You will need to decide whether to buy OPC or PPC.

##### 04. Study of Product Literature:
Courtesy - Gujarat Ambuja Cement Ltd.

When you buy cement, you should read all the properties which are generally described on the bags of cement and on the manufacturer’s website. To know more about product literature click here.

#### B) Finance:

##### 01. Budget:

Cement is essential and must for the construction of the house, so you can’t economise on the cost of the cement. But first, you should fix the overall budget of the house. And if you have limited budget try to save on other material but not on cement.

e.g. Your budget for fixing vitrified tiles is Rs. 10,000 and cost of cement is Rs. 2000 and tiles is Rs.8000. You can’t save on cement. You will have to save Rs.1000 in tiles only.

##### 02. Quantity you want to Buy:

You have to estimate the right quantity. i.e. how many cement bags do you need? It depends on which purpose you want to use the cement, i.e. in flooring, in plastering, etc.

##### 03. From where you can buy the Cement?

Once you estimate the quantity of cement, then you should find right supplier or manufacturers of the cement i.e. whether it is directly available from the manufacturers or manufacturer’s sales depot who provides testing like cube test, etc.  According to that, the price of the cement may be different.

##### 04. Cost of Transport including Taxes:
Courtesy - Shutterstock

You can transport the cement bag by truck if you are buying in bulk otherwise you can use other small vehicle. The cost of transporting cement bag depends upon the location of site from the cement plant or manufacturer’s depot. You should also include all the taxes like service tax, loading, unloading, etc.

#### C) Quality & Brand:

##### 01. Popular Brand of Cement:
Courtesy - Gujarat Ambuja Cement Ltd.
Courtesy - Ultratech Cement

You can use the cement of some popular brands like Ultratech, Ambuja, AAC, Birla, JK Laxmi, Jaypee, Binani, etc. Some mini cement plant is also available in the market.

##### 02. Testing:

Once you are buying the cement, you have to check the quality of cement. For that, you can do some field test and laboratory test if the work is big and you are buying in bulk. To know more about the field test of cement click here.

You may also like:

##### 03. Certification of Cement:

In India, for any manufacturers getting ISI is compulsory, including other cement plants. So you just check whether on your cement bag, written ISI mark or not.

D) Life of Material:

Courtesy - Shutterstock
• Cement does not last long. It is not recommended to use cement more than 3 months old.
• For how long period you can use the cement? For that you can go through the following link:
##### Can I Use One Year Old Cement Bag?

So you should buy only that quality that you can use immediately, particularly in monsoon or when monsoon is near, and of course you have enough space to store the cement. Cement loses strength with time.

#### E) Conclusion:

It is thus keeping in view all above factors; one has to buy the right cement bag as per their need. From above factor, we can conclude that

• Select the right type and grade of cement and its manufacturer or brand.
• If you are buying a huge number of cement bags, try to place a written order and this written order should include the specification, quantity of cement, time of delivery, terms of payment including advance, etc.
• When you receive the cement on site, please check the quantity that you ordered.
• Check whether cement bags do not have any damages like wear and tear of bag or lumps of cement.
• Once you are satisfied with the quality as well as the quantity of cement bags as you ordered, make the payment as per your contract and obtain the receipt of payment.
Courtesy - Shutterstock
• Preserve all the bills and cement bags literature given by the manufacturer. Keep them safe; you may need them in future in case of any emergency.

## Material Exhibition

Explore the world of materials.

format: plain


text:
# Ohm's Law - coaxial conducting cylinder, find resistance and surface charge density

## Homework Statement

Consider two coaxial conducting cylinders with radii a and 3a and length L. The region a<r<2a is filled with a material of conductivity σ1, and the region 2a<r<3a has conductivity σ2. (Assume ε12o.) The inner cylinder is held at potential V0 and the outer cylinder at V=0, so there is a radial current I.
(a) Determine the resistance.
(b) Determine the surface charge density on the boundary at r=2a.

## Homework Equations

V=IR, R=pL/A, R=V/I=L/σA

## The Attempt at a Solution

How do I do this? This probably is a simple problem, but I am having a hard time visualizing the path to the answer. Please help. :-)

Related Introductory Physics Homework Help News on Phys.org
gneill
Mentor

If you're given the conductivity of a material then you also have the resistivity. Imagine adding thin cylindrical shells one at a time and adding up the total resistance (i.e. there's an integration involved).

What's the resistance of a thin shell? You can calculate the surface area, you have its thickness (dr -- I did say thin!). Add 'em up!


text:
Brouwer's Fixed Point Theorem

# Brouwer's Fixed Point Theorem

Recall from one of the results on The Induced Mapping from the Fundamental Groups of Two Topological Spaces page that if $S^1$ is not a retract of $D^2$. We will use this result to prove the famous Brouwer's fixed point theorem.

 Theorem 1 (Brouwer's Fixed Point Theorem): Every continuous function from the closed unit disk onto itself has a fixed point. That is, if $f : D^2 \to D^2$ is a continuous then there exists a point $(x, y) \in D^2$ such that $f(x, y) = (x, y)$.
• Proof: Let $f : D^2 \to D^2$ be a continuous function and suppose that no fixed point exists. That is, for every $(x, y) \in D^2$ we have that $f(x, y) \neq (x, y)$.
• Since $f(x, y) \neq (x, y)$ there must exist a unique line starting at $f(x, y)$, passing through $(x, y)$, and then hitting the boundary of $D^2$ (the unit circle $S^1$. Define a function $g : D^2 \to S^1$ for each $(x, y) \in D^2$ by $g(x, y)$ will be the unique point on the boundary of $S^1$.
• Then $g$ is a clearly a continuous function, and $g \circ \mathrm{in} = \mathrm{id}_{S^1}$. So $g$ is a retraction of the unit circle $S^1$ with the closed unit disk $D^2$.
• But this is impossible from the result stated at the top of the page. So there exists and $(x, y) \in D^2$ such that $f(x, y) = (x, y)$. $\blacksquare$
 Theorem 2 (Brouwer's Fixed Point Theorem for $D^n$): Every continuous function $f : D^n \to D^n$ has a fixed point.

Here, $D^1$ is the closed unit interval, $D^2$ is of course the closed unit disk, $D^3$ is the closed unit ball, etc…

The proof of Theorem 2 is analogous to that of Theorem 1, so we will omit it.

 Theorem 3 (General Brouwer's Fixed Point Theorem): If $A \subset \mathbb{R}^n$ is homeomorphic to $D^n$ then every continuous function $f : A \to A$ has a fixed point.


text:
# Testing for normality - low p-value

I've a dataset with about 12k values given. It looks like this:

When I try to test for normality I get following results, where p-value is extremely slow:

NormaltestResult(statistic=93.328975616353148, pvalue=5.4183922830109284e-21)
ShapirotestResult(0.9910582304000854, 9.806942512599108e-26)

(values for distribution curve fitting)

Searching best parameters for distribution gennorm (error 1.63000230512e-06)
Searching best parameters for distribution norm (error 2.16497464839e-06)
sumsquare_error
gennorm         0.000002
norm            0.000002

Am I wrong about thinking it is normally distributed because I get such low p-values?

• If I remember correctly, I think this has been dealt with numerous times on site; I'll try to dig up one of the threads but here's a precis. 1. Your distribution will not be exactly normal. (Ever.) 2. you have a huge sample, so even trivial differences will lead to very low p-values. 3. Testing goodness of fit doesn't tell you about the suitability of using a normal model -- with large samples, very low p-values don't necessarily indicate a problem. – Glen_b Jan 23 '16 at 5:23
• Thank you! Please add your comment as answer so I can accept it. Currently I do non-linear regression with other features - non-normality is no problem. – x4k3p Jan 23 '16 at 5:26
• If I don't find any of the duplicates I believe are already here, I will make it an answer. – Glen_b Jan 23 '16 at 5:31
• This isn't actually a duplicate, but one part of the answer there makes similar points. Still looking for a near-duplicate of this question. – Glen_b Jan 23 '16 at 5:51


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# How to graph distribution of Order statistics?

Is there a software that can graph the pdfs and Cds of an arbitrary number of order statistics or is there some code such software?

How to do it?

I'm trying to understand the distribution of order statistics easily.

• The order statistics of the uniform distribution have beta-function pdfs, so those are the easiest to plot and to understand. Then the $k$ of $n$ order statistic for an arbitrary continuous distribution with cdf $F$ is distributed as $F^{-1}$ of the corresponding order statistic for the uniform distribution. – Matt F. Dec 17 '19 at 13:25

You should be able to do that in most statistical or graphing software. Some examples:

The minimum of $$n$$ iid observations from some distribution is an order statistic. Its density function (assuming a continuous distribution) can be calculated by (in R)

dmin  <-  function(x, N=1, basename="norm", ..., log=FALSE) {
dfun  <-  get(paste("d", basename, sep=""))
pfun  <-  get(paste("p", basename, sep=""))
logdens  <-  log(N) + (N-1)*pfun(x, ..., lower.tail=FALSE, log.p=TRUE) + dfun(x, ..., log=TRUE)
retval  <-  if (log) logdens else exp(logdens)
retval
}

And we can use it to graph the density of the minimum of 10 iid standard normals:

 plot( function(x) dmin(x, N=10, "norm"),  from=-5, to=5, col="blue", main="density of min of 10 iid N(0, 1)")

For more general order statistics we can do

dorder <-  function(x, N=1, r=1, basename="norm", ..., log=FALSE)  {
pfun  <-  get(paste("p", basename, sep=""))
dfun  <-  get(paste("d", basename, sep=""))
stopifnot(r <= N)

logdens  <-  -lbeta(r, N-r+1)+(r-1)*pfun(x, ..., log.p=TRUE) +(N-r)*pfun(x, ..., lower.tail=FALSE, log.p=TRUE) + dfun(x, ..., log=TRUE)
retval  <-  if (log) logdens else exp(logdens)

retval
}

As an example the density of 3rd order stat for a sample of 10 standard normals:

 plot( function(x) dorder(x, N=10, 3,  "norm"),  from=-5, to=5, col="blue", main="density of 3rd order stat of 10 iid N(0, 1)")

EDIT

You probably needs to study the theory of order statistics and their distributions some more. There are many posts here, you could look through this list. Wikipedia is also use full and has good references.

As for your questions in comment, 1) Since for any $$r < s$$ we have $$X_{(r)}\le X_{(s)}$$ the same must hold for their expectations (assuming they exist.) 2) is definitely wrong.

• Is the distribution of the first order statistics have a mean always less than that of the last order sttaitsics? Are higher order statistics' distribution simply the right shifting of the lower order statistics? – user262965 Dec 17 '19 at 14:34
• @Numbers, yes to the first question, no to the second, and you can verify these with the order statistics of the uniform distribution. – Matt F. Dec 18 '19 at 5:02


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# $L^\infty$ bounds of Steklov eigenfunctions and spectrum stability under domain variation

created by bucur on 01 Apr 2019

[BibTeX]

Submitted Paper

Inserted: 1 apr 2019
Last Updated: 1 apr 2019

Year: 2019

Abstract:

We give a practical tool to control the $L^\infty$-norm of the Steklov eigenfunctions in a Lipschitz domain in terms of the norm of the $BV$-trace operator. The norm of this operator has the advantage to be characterized by purely geometric quantities. As a consequence, we give a spectral stability result for the Steklov eigenproblem under geometric domain perturbations and several examples where stability occurs. In particular we deal with geometric domains which are not equi-Lipschitz, like vanishing holes, merging sets, approximations of inner peaks.


text:
# [Python] How to write this better

## Recommended Posts

glBender    156
Alright, so I've figured out that there is almost always a better way to do things with Python than I'm already doing. Just for practice, I decided to write a function to calculate the correct change(in US coins) to give to a person given a cash amount. This was the simplest way I could figure out to do it, any ideas for improvements?

def calc(amount):
# pairs the amounts with the names
types = [(25,'quarters'),(10,'dimes'),(5,'nickels'),(1,'pennies')]
for t in types:                    #for each kind of coin(greatest to least)
count = (amount)/t[0]          #how many coins will go into that amount
if count > 0:                  #if at least one
print str(count) +' '+t[1] #print out the number
amount -= count*t[0]       #subtract that amount from the total

##### Share on other sites
I really don't like using tuples like that (ie. using hardcoded indices), so one suggestion I have is:

def calc(amount):    # pairs the amounts with the names    types = [(25,'quarters'),(10,'dimes'),(5,'nickels'),(1,'pennies')]    for coinValue, coinName in types:       #for each kind of coin(greatest to least)        count = (amount)/coinValue          #how many coins will go into that amount        if count > 0:                  	    #if at least one            print count, coinName           #print out the number            amount -= count*coinValue       #subtract that amount from the total

Makes things a lot more explicit :).

EDIT: Don't need str() when printing count, and separating by comma adds in a ' '.

[Edited by - cow_in_the_well on May 7, 2007 2:12:49 AM]

##### Share on other sites
glBender    156
Oh okay, I didn't realize I could set up a loop like that... But I guess that makes sense. Thanks for the tips


text:
coexpennan

2021-02-22

Evaluate the following integrals. Include absolute values only when needed.
$\int {3}^{-2x}dx$

Bentley Leach

$I=\int {3}^{-2x}dx$

$-2dx=dt$
$dx=\frac{dt}{-2}$
$I=\frac{-1}{2}\int {3}^{t}dt$
$I=\frac{-1}{2}\left(\frac{{3}^{t}}{\mathrm{ln}|3|}\right)+C$
$I=\frac{-1}{2}\left(\frac{{3}^{-2x}}{\mathrm{ln}3}\right)+C$
$I=\frac{-1}{2\left(\mathrm{ln}{3.3}^{2x}\right)}+C$

Do you have a similar question?


text:
# Devoted to high culture

This puzzle is part 4 of Gladys' journey across the globe. Each part can be solved independently. Nevertheless, if you are new to the series, feel free to start at the beginning: Introducing Gladys.

Dear Puzzling,

Today I visited a rather unusual religious site. The people here seem highly spiritual, yet at the same time extremely tolerant. Unfortunately I was not allowed to witness their ceremonies myself, but I did get to take a tour of the church during the day.

Wish you were here!

Across
1. From the creators of Ice Age | Unpaid trainee
2. Irish singer | Joker actor
3. Small image | Gemstone
4. Politician who would be sober tomorrow
5. Ballad's Chevy | English county
6. Couch | Piece of artillery
7. Frantic | Stitch up
8. Mediterranean island | Funeral container

Down
1. German empire | Fantasy monster
2. 2.54 centimetres | Ethiopian language
3. Happy | Japanese politician
4. Author Fleming | Repeat aloud
5. Open with a key | City with its own gulf
6. Name spelt with PH instead of V | Signified by Σ
7. Basque separatists | Central, interior
8. Small bread | Urban area

the INTERNATIONAL CHURCH OF CANNABIS (yes, that's a real thing) in Denver, CO

The gray numbers

are meant to be interpreted as a gridless Slitherlink where blanks are considered 0. The solution to this Slitherlink encloses some of the letters of the crossword, which when read from top to bottom and left to right give us Gladys's location.

Completed grid:

the title mentioning "high" culture and the letter in the flavortext saying that the people there are spiritual but tolerant.

I think I got the crossword part:

  12345678
j us
1 riointern
2 enyaleto
3 iconopal
4 churchill
5chasekent
6sofacannon
7 rabidsew
8 creteurn
enm

But I'm puzzled as for what to do next...

It looks like some of the numbers in slitherlink can't be right, like the 1s in the top row - they already have more than one line next to them...


text:
# Please help me to understand how tags end up visible when you start typing a different but related term in the tagging window of a question post

I've been active in Stack Exchange for several years and posted many questions, and while I'm not stupid per se I have never figured out how the site's tagging system works. It's a black box to me.

I know that there is ample information in metaFAQ but what I need is an explanation that a five year old could understand.

The background here is that

1. I have recently asked some questions with the tag and I plan to ask some more.
2. In parallel, I have answered that I like the idea of a more specific pyrotechnics tag.
3. "explosive bolt" ends up being used as a generic term for a variety of pyrotechnic devices. recent example and Wikipedia's Frangible nut begins "Not to be confused with Explosive bolt."

Question: If I wanted eventually for a pyrotechnics tag to exist, and then to appear as a suggestion when someone started typing explosive-bol... in the tag-adding box (because it's a common catch-all term even though not correct), should I start adding an explosive-bolts now as a strategy toward that end? Are there other ways to try to make that happen? Is it something that a benevolent moderator can do easily without explosive-bolt ever being a tag of its own, or even then does the tag need to exist first? Is there something about users having enough reputation on one or both tags that comes into play here?

I"m so confused!

Anything that can't be set up as a differential equation and solved with Wolfram Alpha or scipy.integrate is a mystery to me.

• I don't know for sure, but my experience is that tag completion (both on the [tags page[(space.stackexchange.com/tags)) and when composing a question) try to match what you have typed in, with the name of the tag, its tag synonyms, and the usage text. So it really helps to have good usage text. Mar 21 at 4:07
• You will probably get better answers asking on the main meta site. Mar 21 at 4:08
• @DrSheldon thanks but I recently got beat up there (i.e. severely down voted +3/-10) for asking something that I voluntarily closed as duplicate which I assume means "don't ask duplicates!" I'm looking for kinder, gentler folks here who may have read through and understood the extensive but sometimes fragmented knowledge base there and re-explain it in a way I can understand, as carefully explained in the first two sentences of this question.
– uhoh
Mar 21 at 4:15
• mostly unrelated, but a really great read in Buddhism SE meta: Does downvoting an answer qualify as divisive and harsh speech?
– uhoh
Mar 21 at 4:16

Synonyms can be made without a tag by that name existing. Please do not create tags just for the purpose of them becoming synonyms.

• Okay that's good news, thanks!
– uhoh
Mar 22 at 14:53
• @uhoh There's actually a process for users to create synonyms without mod intervention too, but it's very intensive so it's usually faster to ask on meta: space.stackexchange.com/tags/mars/synonyms
– called2voyage Mod
Mar 22 at 16:34
• For whatever reason, there's no way for a mod to approve a synonym suggestion, even though mods can unilaterally create synonyms using the merge tag tool (even if no tag exists for the synonym).
– called2voyage Mod
Mar 22 at 16:35
• Yes, that addresses "Is there something about users having enough reputation on one or both tags that comes into play here?" I'm confident that I'll never understand all the ins and outs of how tags work, so I'll stick to meta discussions when something comes up.
– uhoh
Mar 23 at 0:14


text:
## March 29, 2007

### This Week’s Finds in Mathematical Physics (Week 248)

#### Posted by John Baez

In week248 of This Week’s Finds, see movies of coronal mass ejections, auroras, and tornados on the Sun!

Then, continue reading the Tale of Groupoidification — in which we see how spans of groupoids arise naturally in geometry.

The picture above is taken by the recently launched satellite Hinode. It shows filaments of plasma moving along the magnetic field lines around a sunspot.

Here’s some more eye candy: pictures of coronal loops taken by TRACE – the Transition Region and Coronal Explorer:

Posted at March 29, 2007 2:44 AM UTC

TrackBack URL for this Entry:   http://golem.ph.utexas.edu/cgi-bin/MT-3.0/dxy-tb.fcgi/1219

### Re: This Week’s Finds in Mathematical Physics (Week 248)

I’m really liking this Tale you’re spinning. Looking forward to more.

Posted by: John Armstrong on March 29, 2007 4:05 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 248)

I’m glad to see incidence geometry as spans has gotten off the ground. ;)

Posted by: David Roberts on March 29, 2007 6:32 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 248)

Very interesting. I’m looking forward to seeing where this is going.

Hope you don’t mind my pointing out a couple of typos: in the definition of “span of groupoids”, you have “it should preserve preserves all the structure…”

Then a few lines later in the example you have “… we can rotate the whole picture and get a new point and a new line containing a new plane”. Not sure if this one is a typo, or just my misunderstanding: shouldn’t it be “contained in a new plane” (i.e. the points of the line are a subset of the points of the plane, so the line is contained within the plane)? Or am I just misreading this sentence?

Posted by: logopetria on March 29, 2007 8:38 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 248)

All the typos you caught are really typos! I’ll fix them. Thanks.

I actually love it when people point out typos and other mistakes in This Week’s Finds — I like to get these things fixed.

Posted by: John Baez on March 29, 2007 5:50 PM | Permalink | Reply to this

### Magmas?

I am confused about the term groupoid. The name suggests something which is almost a group, and that was apparently the original definition: a group but the binary product is not everywhere defined. Wikipedia mentions both this definition and your categorical one. Are they equivalent? There was some discussion about groupoids versus magmas at WP.

### Re: Magmas?

Another thing that’s not clear: I remember discussion about “groupoids” being the thing to use for studying “semi-symmetries”. IIRC, a semi-symmetry was something like the finite pattern “||||||||” being almost symmetric - in the sense of mapping almost onto itself - under translation by one bar-separation, with the issue being that for the finite pattern the end bar doesn’t have anything to map onto. This was related to the fact that you could have a tranlsation morphism (?) T which was defined for all the bars except the final one, and likewise for the inverse T.

Is this a separate concept that also goes under the name groupoid, a different application of the same mathematical gadget “groupoid” or an aspect to the story about spans and groupoids? I’ve noodled about with the framework in the TWF and as far as I can see your morphisms in the definition of a groupoid could fit with the partial symmetry, but then I can’t see how with your F and G functors you don’t end up with “full” symmetries. (Maybe it’s that in your concrete example of incidence geometry all symmetry-type things are full symmetries?)

Whilst trying to see if I understood the concepts properly, I tried to come up with another concrete groupoidfication example. I think if you work with numbers you can define “not being coprime” as the relationship and your witness being one of the non-unity common factors. Then, say, multiplying everything by a value X again gives you a valid relation, ie, it’s a symmetry. But maybe dividing by a value X is a semi-symmetry: when your witness is divisible by X your statement remains valid, but in some cases you witness and all the other possible witnesses won’t be divisible by X, so it’s not a “full symmetry”. Is this on the right lines?

Posted by: dave tweed on March 29, 2007 10:16 AM | Permalink | Reply to this

### Re: Magmas?

Replying to myself about the obvious flaw: whilst multiplying by a value X keeps all existing relationships valid (ie, 4 and 6 with witness 2 becomes 8 and 12 with witness 4 when multiplied by 2), it also makes everything that wasn’t “not coprime” into “not coprime”, so it can’t be a global symmetry.

Posted by: dave tweed on March 29, 2007 10:24 AM | Permalink | Reply to this

### Re: Magmas?

Alan Weinstein’s Groupoids: Unifying Internal and External Symmetry gives a good little introduction to groupoids. You can see that the two Wikipedia definitions Thomas mentions are equivalent. (The other sense of ‘groupoid’ - merely a set with a binary operation - is clearly different.)

A situation similar to dave tweed’s finite number of bars is also treated. This is a finite patch of tiles, whose symmetries are considered in terms of a transformation groupoid (page 746).

Posted by: David Corfield on March 29, 2007 12:28 PM | Permalink | Reply to this

### Re: Magmas?

I am confused about the term groupoid.

The term was originally used by Oystein Ore back in the 1930s to refer to a set with a binary operation. That usage is still quite common; its MSC classification number is 20N02. What is now called just “groupoid” used to be called a “Brandt groupoid”. You can find the noncategorical definition in, for instance, R.H. Bruck’s classic Survey of Binary Systems.

I’m not sure when exactly it happened, but eventually these just came to be called “groupoids”. That might be due to the influence of Bourbaki. The replacement term “magma” for sets with a binary operation is certainly a Bourbakism (Bourbaki-ism?).

Folks who work a lot in quasigroups, loops, and the like (MSC 20N05) still tend to use “groupoid” in its classical sense. I work in that area, but have coauthored a paper with one of the main proponents (Weinstein) of the modern sense of the term, so I switched to “magma”. I still get annoyed referee reports insisting I switch back to “groupoid”.

Posted by: Michael Kinyon on March 29, 2007 5:57 PM | Permalink | Reply to this

### Re: Magmas?

It sounds like you’re talking about two ways to define a groupoid: as a set with partially defined multiplication and totally defined inverses satisfying some axioms, and as a category where morphisms have inverses. As the Wikipedia explains, these definitions are equivalent if one gets the axioms right in the first approach.

There’s also an unrelated, archaic definition of groupoid: ‘a set equipped with an arbitrary binary operation’. I urge everyone to avoid this confusingly different usage. The preferred term for such an entity is ‘magma’, which nicely captures the feel of this raw, primeval concept.

Posted by: John Baez on March 29, 2007 6:02 PM | Permalink | Reply to this

### Re: Magma…?

John Baez wrote:
The preferred term for such an entity is ‘magma’

Yes, but do you prefer “magmas” or “magmata” for the plural? :-)

Posted by: Michael Kinyon on March 29, 2007 6:35 PM | Permalink | Reply to this

### Re: Magma…?

Oh the jokes to be made…

I’ll just say that since I’m a proponent of “topoi”, I’d have to say “magmata”.

Incidentally, while I like the “primal” explanation, I think the original sense of the term actually captures it better. Magma is what’s left over after a semi-liquid substance (say, treacle) has had the water pressed or evaporated away. A magma is what’s left of a semigroup after we boil off the associativity, like a rack is the “wrack and ruin” after we rip composition from a group to leave only conjugation.

Posted by: John Armstrong on March 29, 2007 7:17 PM | Permalink | Reply to this

### Re: Magma…?

So what is left of a category after we boil off the associativity? A reflexive digraph? Maybe it needs a more “primal” name.

Posted by: Michael Kinyon on March 29, 2007 7:33 PM | Permalink | Reply to this

### Re: Magma…?

Michael wrote:

So what is left of a category after we boil off the associativity?

Then you’ve got the multi-object analogue of a magma. Since the multi-object analogue of a group is a groupoid, and similarly for ringoids and algebroids, what you’ve got is a… magmoid!

All part of the general philosophy of oidization.

Posted by: John Baez on March 29, 2007 7:44 PM | Permalink | Reply to this

### Re: Magma…?

JB wrote:
All part of the general philosophy of oidization.

Alan Weinstein and I used to call it “oidification” in our email exchanges, but I won’t quibble.

Posted by: Michael Kinyon on March 29, 2007 8:03 PM | Permalink | Reply to this

### Re: Magma…?

Hmm.. to keep the chemistry theme maybe we should go back and start talking about groupoxides as examples of oxidation?

Posted by: John Armstrong on March 29, 2007 10:13 PM | Permalink | Reply to this

### Re: Magmas?

The campaign to rename groupoids-in-the-other-sense to magmas really bothers me. Almost no one who studies the other kind of groupoids calls them magmas. Since they named them groupoids first, who are we to tell them they can’t call them that anymore? It’s one of the purest exercises of power politics in mathematics that I’ve seen.

Posted by: Walt on April 5, 2007 5:56 PM | Permalink | Reply to this

### Re: Magmas?

Brandt used the term gruppoid in 1926 in our sense of category with inverses, clearly as a variant on gruppe. I wonder when the first translation as groupoid occurs.

Would you be as bothered if those interested in the other sense of groupoid insisted that German speakers use gruppoid their way?

Posted by: David Corfield on April 5, 2007 8:50 PM | Permalink | Reply to this

### Re: Magmas?

I’m happy with both fields using the term groupoid. I’m happy if the people who work with non-Brandt groupoids spontaneously say “Hey, magma is a great term. Let’s use that.” I’m not happy with practitioners of one field dictating the terminology of the other field.

Posted by: Walt on April 6, 2007 4:19 AM | Permalink | Reply to this

### Re: Magmas?

Walt wrote:

Since they named them groupoids first, who are we to tell them they can’t call them that anymore? It’s one of the purest exercises of power politics in mathematics that I’ve seen.

Ah yes — those big bad category theorists, bossing everyone around! Yet another example of how category theorists run the show in American academia. Not.

Seriously, it’s very common in mathematics for people to urge their favorite terminology on other people. However, as long as this takes the form of public rhetoric, it’s the exact opposite of ‘power politics’, because there’s no ‘power’ backing up the argument except the power of persuasion. Anyone can tell anyone else what to do – but in the end, everyone can do whatever they want! I find nothing wrong with this.

We only have a case of ‘power politics’ if, for example, a referee refuses to accept a paper unless the author changes his terminology. This I would find more obnoxious (though sometimes justifiable, e.g. if somebody calls prime numbers ‘composite’.)

As for my little joke about ‘category theorists running the show in American academia’, it’s worth noting that category theory is so downtrodden that the NSF almost completely refuses to fund work on this subject. Which is why you can count the official ‘category theorists’ in the US on one hand. In Europe the situation is different. So: funding is where power politics really comes into play…

Posted by: John Baez on April 5, 2007 9:06 PM | Permalink | Reply to this

### Re: Magmas?

I don’t know of a single other example of where people outside of a field, people who have no interest in that field, have renamed the main concept of the field. Is there really another example? I find that surprising, but I’m prepared to be corrected.

As unpopular it may be with the NSF, category theory is a much more fashionable branch of mathematics than binary operations. (And in my opinion, rightfully so. Groupoids-as-binary-operations is not a field near and dear to my heart.)

In certain hands, rhetoric is power. Bourbaki is so influential, that people will accept the term magma as the rightful term out of ignorance. Look at the discussion on Wikipedia about whether to call them magmas or groupoids – it mainly consists of people who know nothing about the existing literature on the subject, but who’ve heard that Bourbaki or someone influenced by Bourbaki called them magmas, so we should call them magmas. (Someone actually used the argument that if it was good enough for Serre, then it’s good enough for them. How is that not power?)

In your own comment, you treat the name change as a fait accompli. If you knew nothing about the subject other than your comment, you would come away with the impression that the actual name that everyone uses now is magma, and that calling them a groupoid is “archaic”. You are an influential figure. You don’t think you have the power to shape perception of mathematical reality?

Posted by: Walt on April 6, 2007 4:15 AM | Permalink | Reply to this

### Re: Magmas?

Walt wrote:

In certain hands, rhetoric is power.

[…]

You don’t think you have the power to shape perception of mathematical reality?

I hope I do, and I’m certainly not shy of trying to do it. It isn’t the kind of power that I associate with your phrase ‘pure power politics’. I’m not rejecting papers or withholding funding from people who use the term ‘groupoid’ in some way I don’t like. I’m just advocating my views! But you’re right, I shouldn’t just tell people to use ‘groupoid’ the way I like. I should explain my reasons.

I’ve done it before — but I’ll do it again:

I think the term ‘groupoid’ is a poor descriptor of ‘a set with a binary operation’. A set with a binary operation is so much less than a group, less than a semigroup or even a monoid, less than a loop or even a quasigroup, that calling it a ‘groupoid’ — something like a group — is misleading. A set with a binary operation is really not very much like a group at all! It’s something very basic, though. ‘Magma’ seems as good a term as any.

A category with inverses, on the other hand, really is very much like a group, so the name ‘groupoid’ is very fitting. Many basic results about groups extend nicely to groupoids! Furthermore, the term fits into a nice set of analogies:

group groupoid

ring ringoid

algebra algebroid

which has the potential to become a major pattern in mathematics: replacing structures that are secretly one-object categories by their many-object versions.

Posted by: John Baez on April 6, 2007 6:29 PM | Permalink | Reply to this

### Re: Magmas?

I don’t think there’s anything wrong with calling Brandt groupoids just groupoids, just I don’t have any problem with the fact that ring theorists mean one thing by “algebra”, and universal algebraists mean another. I object to the idea that the terminology of a field can be imposed from the outside. For example, algebraists have normal subgroups while differential geometers have normal bundles. It would bother me if group theorists announced “we’ve renamed normal bundles to volcanic bundles”. It would bother me more the more influential algebraists were relative to differential geometers.

Sets-with-binary-operations is unfashionable a field as I can imagine. I just picture someone who is unfortunate enough to fall in love with such a field, who ends up in a non-tenure track job at Podunk U, and then is told, “We don’t think your field is very interesting. Plus, we changed all of your terminology.”

Posted by: Walt on April 7, 2007 4:33 AM | Permalink | Reply to this

### Re: Magmas?

Walt wrote:

Sets-with-binary-operations is unfashionable a field as I can imagine. I just picture someone who is unfortunate enough to fall in love with such a field, who ends up in a non-tenure track job at Podunk U, and then is told, “We don’t think your field is very interesting. Plus, we changed all of your terminology.”

For what it’s worth, I’m far more interested in sets-with-binary-operations than most mathematicians. So, my desire to call sets-with-binary-operations ‘magmas’ is simply due to a desire for appropriate terminology. And, I don’t think the first name something is given is likely to be the best name it’ll ever get! We need to throw out old names occasionally — if we didn’t, we’d still be using Roman numerals. How do we decide when to do this? The right approach is to let people argue about what’s right and see what happens.

Posted by: John Baez on April 7, 2007 7:38 AM | Permalink | Reply to this

### category theoretic funding

I have to say, judging from Urs’ reports, category theoretic mathematical physics seems to be blooming in Europe. And I remember John Power telling us in Minneapolis that things were better for category theoretic computer science in Europe.

What evidence is there on the pure front? And is there an explanation for this continental difference? The dominance of your pragmatic philosophy?

Posted by: David Corfield on April 6, 2007 11:00 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 248)

Presumably at some point one will need to generalise from the two-ness of spans. E.g., if you’re interested in a type of flag composed of figures in several dimensions. Is there a term for, say, a three-legged span? Then would a degroupoidification form a kind of three-dimensional matrix?

This span business doesn’t seem totally unrelated to what statisticians are doing when they study the correlation of two variables, e.g., a sample of people and a pair of maps to weights and heights. Again one can look at higher-order correlations.

Posted by: David Corfield on March 29, 2007 12:49 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 248)

Then would a degroupoidification form a kind of three-dimensional matrix?

Perhaps ‘tensor’ would have been better.

Posted by: David Corfield on March 29, 2007 1:12 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 248)

David wrote:

Perhaps ‘tensor’ would have been better.

You got it. Jim, Todd and I use the awkward term ‘multispan’ to mean an object $S$ equipped with $n$ different morphisms

$f: S \to X_i$

By the process of degroupoidification, multispans of groupoids give rank-$n$ tensors. These turn out to be very useful.

The curious thing is that normally, rank-$n$ tensors are covariant in some slots and contravariant in others. When we groupoidify, the distinction between covariant and contravariant disappears.

I’ll let you guess why.

But you’re making me give away some of the fun to come…

Posted by: John Baez on March 29, 2007 6:12 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 248)

This might be more “giving away some of the fun”, but there’s an asymmetry in the definition of groupoidification given so far that I don’t quite understand. Specifically, I don’t see why we’re only replacing relations between X and Y with spans S:X->Y, and not cospans C:X->Y (i.e. in the case of sets, a set C with functions from X and Y).

In one of the examples last week, you said “every Russian has exactly one favorite Frenchman and one favorite Englishwoman” making that Russian a witness to a relation between that Frenchman and that Englishwoman. But likewise, we may suppose that every Frenchman and every Englishwoman has a favourite Russian; if Gérard and Emma are both fans of Dostoevsky, doesn’t he witness a relation between them? [Moreover, if we change “favourite Russian” into “favourite weepy movie” it might end up being a very important relation between them!]

So does everything work out the same if we swap cospans for spans? I strongly suspect it doesn’t, but I haven’t worked out why not yet. If not, and if ‘multispans’ (and presumably ‘multicospans’) are worth studying, can we also get anything interesting out of things that have arrows to some of the sets and from the others?

Of course, if this is all part of the fun that you want to save for later, I’d be content with an infuriatingly tantalising hint instead of more detail!

Posted by: logopetria on March 29, 2007 8:37 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 248)

Well cospans are just spans in the opposite category, so when we decategorify we should get the same set, no? In some cases (knot theory) cospans seem to be more natural, but I don’t think it actually changes much. Of course, I could be wrong…

Posted by: John Armstrong on March 29, 2007 10:11 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 248)

Logopetria wrote:

So does everything work out the same if we swap cospans for spans? I strongly suspect it doesn’t, but I haven’t worked out why not yet.

Ah, interesting! Cospans play a conceptually very different role in this theory than spans, but they’re also important. We’ll see them fairly soon, I hope.

Of course, if this is all part of the fun that you want to save for later, I’d be content with an infuriatingly tantalising hint instead of more detail!

Okay, that’s what I’ll do:

We’ll often want to work with groupoids ‘over a base’: that is, groupoids equipped with a morphism to a fixed groupoid $B$. A span in the category of groupoids over a base will involve both a span of groupoids and a cospan of groupoids! Imagine a commutative diamond with downwards-pointing arrows, with $B$ at the base.

Why work with groupoids ‘over a base’? I claim that when we study groupoids on which a fixed group (or groupoid) acts, we’re studying groupoids over a base.

The details of how this works may serve as a fun puzzle, if you’re not already familiar with it.

By the way: one reason I’m dropping so many ‘infuriatingly tantalizing hints’ in response to questions is that if I blow all my expository energy answering questions, I’ll never finish telling the Tale!

Posted by: John Baez on March 31, 2007 3:15 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 248)

David wrote:

Think of all the fun to be had groupoidifying equations such as $A V_k = S_{i j} T_{i j k}$, plugging indices together by equivalence of groupoid, is it?

Yes, it’s lots of fun! Next, imagine that all the coolest sorts of linear algebra — like the theory of simple Lie algebras, quantum groups, and their representations, turns out to groupoidify to give a fascinating new view on incidence geometry! Even more fun!

More technically:

Indices are plugged together by the groupoidified version of the ‘inner product’ — for a quick explanation see page 7 of Jeff Morton’s talk at CT ‘07.

This, by the way is the answer to my puzzle about why we don’t need to worry about the difference between covariant and contravariant tensors: the vector spaces obtained by degroupoidification are all Hilbert spaces, so we can raise and lower indices at will — so it’s no sin to leave them all lowered, as you’ve done.

But we’ll have to be patient.

You can be impatient if you like: knowing someone’s impatient makes it a little easier, psychologically at least, to find more time to write This Week’s Finds.

And perhaps when done, Klein 2-geometry will appear quite straightforward.

The groupoidification project, which was largely pushed through by Jim and Todd, has a curious similarity to our Klein 2-geometry project. You’ll keep seeing this more and more as I go on. But, the groupoidification project has been a lot more sucessful so far in making contact with other branches of math. Maybe that’s because it had a more solid conceptual motivation. In our Klein 2-geometry work, we were just messing around, trying to categorify everything in sight — sort of a scattershot approach.

Posted by: John Baez on April 1, 2007 10:45 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 248)

In our Klein 2-geometry work, we were just messing around, trying to categorify everything in sight — sort of a scattershot approach.

But it was scattershot at a barn door from twenty paces, wasn’t it? Or at leasting finding sand in the Sahara.

Posted by: David Corfield on April 6, 2007 11:08 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 248)

I just caught one of those dizzying glimpses you sometimes offer us. Think of all the fun to be had groupoidifying equations such as $A.V_k = S_{ij}T_{ijk}$, plugging indices together by equivalence of groupoid, is it?

But we’ll have to be patient. And perhaps when done, Klein 2-geometry will appear quite straightforward.

Posted by: David Corfield on March 30, 2007 12:12 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 248)

Regarding my last paragraph, what should we make of efforts to introduce probabilistic ideas in topics such as algebraic topology. E.g., Betti numbers of random manifolds looks at the expectations of Betti numbers. It begins

In various fields of applications, such as topological robotics, configuration spaces of mechanical systems depend on a large number of parameters, which typically are only partially known and often can be considered as random variables. Since these parameters determine the topology of the configuration space, the latter can be viewed in such a case as a random topological space or a random manifold. To control such a system one has to understand geometry, topology and control theory of random manifolds.

One of the most natural notion to investigate is the mathematical expectation of the Betti numbers of random manifolds. Clearly, these average Betti numbers encode valuable information for engineering applications; for instance they provide an average lower bound for the number of critical points of a Morse function (i.e. observable) on such manifolds.

If measures can be placed on Grassmanians, as in Klain and Rota’s wonderful lectures, might there be scope in the groupoidified version?

Posted by: David Corfield on April 2, 2007 9:29 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 248)

Talking of symmetry, and weather on astronomical bodies, have you seen the giant hexagon on Saturn?

Posted by: Dan Piponi on March 29, 2007 10:16 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 248)

I saw this linked from Bad Astronomy in one of the comments. Am I horribly off base to assume it’s likely from a spherical harmonic with roughly hexagonal symmetry? It really does look a lot like some of their graphs that I’ve seen.

Posted by: John Armstrong on March 29, 2007 10:50 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 248)

Rossby waves on Earth are vaguely similar, and apparently have 4-6 ‘lobes’. But it’s hard to imagine something like this maintaining a regular 6 sided figure over 25000km for (at least) tens of years.

Posted by: Dan P on March 30, 2007 6:03 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 248)

I’d always assumed those spectacular pictures of coronal loops as something that required billions of dollars worth of equipment to see, and indeed the pictures you show did. So I was amazed to find that you can buy a hydrogen-alpha telescope for about \$500 and see them for yourself. Obviously it’s not quite the same as a view from a satellite, but through a H-α filter you can clearly see solar prominences and as often as not you can also see that they form faint wispy loops. It’s quite amazing to see these things with your own eyes!

Posted by: Dan Piponi on March 29, 2007 10:31 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 248)

I’ve been on a solar binge since reading Week 248, and Dan P’s telescope comment. If any cafe goers are interested the New York amateur astronomy soc will have some H-alpha telescopes set up in Central Park next Saturday (weather permitting).

The Sun is indeed quite having its day in itself. There was an interesting BBC article today too. (And some nice sound samples to download)

Posted by: Allan E on April 21, 2007 4:47 AM | Permalink | Reply to this

### Wilson lines

I think I got the groupoid idea. The set of open Wilson lines is a groupoid, because two Wilson lines can only be glued if they have a common endpoint. In contrast, the set of gauge transformations is a group. So one can have a group acting on a groupoid.

### Re: This Week’s Finds in Mathematical Physics (Week 248)

It would have been nice if material on E8 had been covered in week 248 rather than week 247: it would have been a rather spooky coincidence, but as it was it wasn’t to be.

Posted by: Ian Stopher on March 30, 2007 10:25 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 248)

Maybe you didn’t read all the way to the end of week248?

Posted by: John Baez on March 31, 2007 3:28 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 248)

Groupoids, functors and natural transformations form a 2-category. Presumably spans of groupoids form some sort of a weak 2-category?

If the groupoids are topological groupoids or Lie groupoids, then choosing the arrows to be functors is much too restrictive. The problem is that fully faithful and essentially surjective functors may have no continous inverses. So one has to localize at such functors (variously called essential equivalences or weak equivalences). The most common solutions seems to be: replace functors by bibundles. The price is that one loses associativity of 1-arrows. That is, groupoids, bibundles and equivariant maps are a weak 2-category… I presume this is coming up in the new installments?

Posted by: Eugene Lerman on March 31, 2007 2:54 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 248)

If we are going to demand composition of arrows, can anyone explain this puzzle in Sloane’s? It is asbout semigroups, not groupoids, so there is more structure, not less, but still puzzling.

A071536 Extensions to a semigroup of the (categorical) composition of arrows in the complete directed graph on n labeled nodes.

1, 1, 3, 3, 147

OFFSET

0,3

COMMENT

Terms obtained, especially the 147, lend some support to the peculiar conjecture that the composition in any category can be extended to a total, associative operation, that is to a semigroup (not, of course, a monoid).

EXAMPLE

For n=2, arrows 0:0->0, 1:1->1, f:0->1, g:1->0, the self-dual solution (commutes with reversing arrows) has multiplication table (rows and columns indexed by 0, 1, f, g in order) with rows: 0 f f 0; g 1 1 g; 0 f f 0; g 1 1 g.

KEYWORD

hard,nonn,nice

AUTHOR

F. Lockwood Morris (lockwood(AT)ecs.syr.edu), May 29 2002

Posted by: Jonathan Vos Post on March 31, 2007 5:31 AM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 248)

Incidence is very interesting indeed. I am looking forward to the relation between incidence and Hecke operators/algebra. Especially since I heard that these operators are closely related to more common objects like combinatorial laplacians (discrete laplacian operators on a graph, depending on the indidence matrix!). So while laplacians (on discrete or continuus domains) are very relevant (and understood) for graphs, physics and spectral geometry (in noncommutative geometry length is even defined using the spectrum of the Dirac operator), Hecke operators seem (to me…) to be analogous to this in arithemetic and algebraic geometry?? At least, this seems to be suggested in James Arthur: The Principle of Functoriality”, Bulletin of the AMS v.40 no. 1 October, 2002. So, if this is true I hope we might get some basic intuitive understanding of arithmetic/algebraic theories (like Langlands program, Mordell-Faltings etc.) using concepts better known to ordinary applied mathematicians (graphs, laplacians and geometry). Just posing a very naive question: in geometry/graph theory the idea of hyperbolicity (genus of the surface>1 ) is very important and it is related to the (spectrum of the) laplacian, I believe that in arithmetic/algebraic geometry (e.g. Mordell-Faltings theorem) the same phenomenon pops up? Genus>1 means a significant change in the behavior (numbers of solutions) of the algebraic solutions. Is this also related to the incidence structure and a relevant Hecke algebra? Just curious….

Posted by: lauret on March 31, 2007 4:25 PM | Permalink | Reply to this

### Re: This Week’s Finds in Mathematical Physics (Week 248)

This relation to Hecke operators should be pretty interesting. I once grew fond of the fact that lots of “duality” transformations in the literature are all examples of 2-linear maps i.e. (morphisms of 2-vector space) obtained by having a vector bundle on a span (also, now, known as a bi-brane) and performing pull-push through this correspondence: I once listed some examples in Fourier-Mukai, T-Duality and other linear 2-Maps. The Hecke transformation in geometric Langlands is among these, at least in its “classical limit”.

And in fact that makes good sense and fits very nicely into the very big picture, parts of which is the picture drawn by Kapustin/Witten, which becomes, quite clear once you match it to the general theory of 2-vectorial duality in 2-dimensional QFT, as hinted at in Duality and defects in rational conformal field theory.

Posted by: urs on March 31, 2007 10:17 PM | Permalink | Reply to this
Read the post The n-Café Quantum Conjecture
Weblog: The n-Category Café
Excerpt: Why it seems that quantum mechanics ought to be the de-refinement of a refined theory which lives in one categorical degree higher than usual.
Tracked: June 8, 2007 6:26 PM

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text:
# Do points inside the PPF curve always mean productive inefficiency?

If a firm produces combinations of goods along the PPF curve, it has achieved its productive efficiency. And when a firm reaches productive efficiency, it means that all factors of production have been fully utilized. And another way of measuring productive efficiency is to look at its average cost. If a firm has reached productive efficiency, it means that the firm is producing at the lowest average cost.

But what about the points that fall inside the PPF?

Let’s say A (15,10) is on the PPF curve, and B (15,5) lies beneath it. Assume that a firm is producing at point B, and the reason why it isn’t producing at a higher level is that there’s still some cash deposit in its business account that hasn’t been withdrawn yet.

Since cash deposit is NOT a factor of production, even the firm is producing at points beneath the PPF, it doesn’t necessarily imply that the firm has not utilized all factors of production. Where did I go run?

## 1 Answer

Well the problem here is in a set up. It is inconsistent to say that firm is producing below PPF not due to underutilization of factors of production but due to cash, since cash will not enter production function, or if you assume that cash is here used to buy factors of production it is inconsistent to say that firm fully utilizes all avaiable factors while having excess cash.

For example, take very simple production function $$F=L^{1/2}$$ and firm has access to 100 units of labor, with all the standard background assumptions. I used this simplistic case because here PPF is just one point instead of set of points, and in this case the PPF would just be point F=10. If firm would be using 100 units of labor it would be operating at PPF and if not it would be operating below PPF. However, you cant just arbitrary announce that this firm is producing F=8, due to not using all cash if cash does not enter as factor of production. That is simply inconsistent with having production function $$F=L^{1/2}$$

Furthermore, let us assume that firm does not just have 100 units of labor but that it has 100 dollars and has to purchase labor on labor market for 1 dollar per labor. Well in this case if firm has still some excess cash, for example if it purchased only 70 units of labor, then it still does not uses all factors of production in most efficient manner because given its 100 dollars it still has 30 units of labor avaiable for hire that it could use. Consequently, the problem in the setting above is that it just arbitrary announces that firm is utilizing all factors of production as much as possible while not being at PPF. That is like asking, from math we know that unit circle is given by all points that satisfy equation $$1=x^2+y^2$$, but I assume point (100,10) lies on unit circle and it clearly does not satisfy the equation. It is just inconsistent reasoning.

• @ 1muflon1♦ thank you for your answer. To make no mistake about it, you are suggesting that if I don't consider cash in the production function, then I can't just announce that the firm hasn't achieved its productive efficiency because there's still excess cash. Thus, in this case, the points lying beneath the PPF simply indicates that the firm is not producing in the most efficient way, or not using the best technology. Either way, it's not due to the excess cash. Did I get you wrong? – Underwood Jan 20 at 12:59
• @Underwood yes and in addition even if you assume firm uses cash to purchase factors, it would not be appropriate to say that firm uses all factors of production that are available to it until it is not able to purchase more. – 1muflon1 Jan 20 at 13:12
• @Underwood the key to understand this is to understand the definition of productive efficiency, productive efficiency is a situation in which the economy or an economic system could not produce any more of one good without sacrificing production of another good and without improving the production technology. In a perfectly competitive market you can prove that any such allocation above will also be such where the average total cost for each good is minimized, but that result wont hold necessarily in other market structures. In addition, min. of cost does not imply firm has free cash. – 1muflon1 Jan 20 at 16:59
• in a model of perfect competition with no opportunity cost all profits (including accounting ones) are zero and firm is not left with anything extra after minimizing all cost. Efficiency, especially productive efficiency is not the same as maximizing profit. In fact in non-competitive market structures often the way how firm achieve profit is to not produce as much as possible given productive efficiency, but to lets say underproduce to increase market price, like a monopoly would (i.e. monopoly will not produce at point where MC=P, but at point where MR=P). – 1muflon1 Jan 20 at 17:02
• @Underwood you can prove it by setting up a mathematical model with perfectly competitive firms, no market imperfections and host of other standard assumptions and examining allocation of factors in equilibrium - at least that is the gist of it. That summary sounds simple but the model itself is of course more complex and requires a good command of math and definitely too long to just walk you through it in the comments, but you can see it in most graduate level micro textbooks. – 1muflon1 Jan 20 at 18:01


text:
# Interpretation of Mann-Kendall statistics

I am doing a trend analysis for vegetation dynamics using Mann-Kendall test for trend detection and Sen's slope for computing the magnitude of the trend. There are 6 sites inside a conservation area and 6 outside. Below is is an example for one variable. I need some help with the interpretation.

1. I understand that the negative sign of the score implies a negative trend. But what about the score itself, what does that mean? How high the number should be for me to decide it's a highly positive or a negative trend?

2. What about the p value? Does it decide whether the trend is statistically significant?

3. What does the tau coefficient mean here?

4. I have such trends for 60 sites. What more can I do to compare trends inside and outside conservation areas?

Please suggest relevant literature if possible.

P value < 0.05 tells that there is (monotonic)trend and if $\tau$ is +ve, increasing trend, and if $\tau$ is -ve, decreasing trend p value >0.05, no monotonic trend, away from monotonic trend.


text:
Christopher Hammond

Print publication date: 2015

Print ISBN-13: 9780198738671

Published to Oxford Scholarship Online: August 2015

DOI: 10.1093/acprof:oso/9780198738671.001.0001

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(p.469) Appendix 7 Group theory in crystallography

Source:
The Basics of Crystallography and Diffraction
Publisher:
Oxford University Press

From our reading of Chapter 4 in this book, we understand the term point group of a crystal simply to denote the ‘collection and arrangement’ of symmetry elements that together describe its symmetry and distinguish it from all the other 31 point groups. Similarly, we understand the term space group of a crystal to denote the distribution and arrangements of symmetry elements in space that describe its three-dimensional pattern and distinguish it from all the other 229 space groups.

But the word group has a precise and much deeper meaning. We now need to make clear the distinction between the terms symmetry element and symmetry operation. A symmetry operation is a movement of body (a molecule or crystal) in which, after the operation has been carried out, is in an equivalent configuration: that is to say that it is indistinguishable from the original configuration, though not necessarily identical to it. For example, rotation of a pencil by $60∘$ about the pencil-lead direction or axis results in an equivalent configuration—and of course, repeating the process a total of six times returns us to the original configuration. The symmetry element (the hexad axis) is the geometrical line about which these symmetry operations are carried out. The same applies to all the other symmetry elements that we have learned about: their existence is demonstrated (as with the pencil) by showing that the appropriate symmetry operations exist.

Group theory in crystallography is concerned with the number and distribution of symmetry operations in a crystal and the relationships between them. By its means, the 32 point groups and the 230 space groups can be formally identified and enumerated. In the words of Vainshtein (1981) ‘Group theory … enables one to reveal the deeper meaning and regularities of the symmetry concept itself’. It also has important applications in the analysis of the symmetry properties of atoms and molecules, as well as crystals. However, largely because of the mathematical formalism and abstract terms in which group theory is inevitably expressed, the subject is generally by-passed in most books on crystallography—either because of the ‘difficulty’ of the subject or because it is only needed in such particular applications as those mentioned earlier. Another practical hindrance for the crystallographer is that the Schönflies’ notation for point groups and space groups is generally used rather than the Hermann–Mauguin notation (see Note and Table A7.4).

Here, we will describe group theory as it applies to crystallographic point group symmetry in the very simplest terms, with only a brief mention of space group symmetry. Readers with a sufficient mathematical background and who wish to pursue the subject further will find the books listed in Further Reading useful.

(p.470) A7.1 Group theory: axioms and multiplication tables

We begin by reconsidering the symmetry operations between the pair of right and left hands shown in Fig. 4.5. We see that a reflection (symbol $m)$ in Fig. 4.5(b) plus a rotation of 180° (a diad axis, symbol 2, perpendicular to the mirror plane) is equivalent to a centre of symmetry or inversion monad axis, symbol $1ˉ$ as shown in Fig. 4.5(a). Writing this algebraically we may put m.2 = $1ˉ$ where the dot means the product or combination of one symmetry operation (m) followed by another (2) giving rise to a centre of symmetry ($1ˉ)$. Conversely, we may write $1ˉ$.2 = m. Furthermore, the order in which the symmetry operations are carried out gives the same result, e.g. m.2 = 2.$m=1ˉ$, etc. (This is only the case for the so-called Abelian groups, named after the Norwegian mathematician, Niels Henrik Abel 1802–1829.) Finally, if we find the product or combination of a mirror symmetry operation with itself, we ‘get back where we started’; e.g. the operation of a reflection m (right to left hand) followed by another reflection m in the same plane returns us back to a right hand. This combination of symmetry operations is equivalent to the operation of a monad axis of symmetry, symbol 1 (see Section 2.3, p. 59), i.e. a rotation of $360∘$ (about any axis) that brings the crystal into coincidence with itself. Writing this algebraically: m.m = 1 and similarly 2.2 = 1 and $1ˉ.1ˉ$ = 1.

In group theory the symmetry operations m, 2, $1ˉ$, and 1 are called a set of elements of the group and they fulfil the first requirement or first axiom, of group theory, namely that the combination or product of any two elements (including an element with itself) is also an element of the set. As we see, these four symmetry operations do indeed accord with this axiom. They are also in accord with the second axiom of group theory, namely the combination of elements is associative. This means that if we combine three elements, the sequence of combination gives the same result, e.g. for m, 2 and $1ˉ$, (m.2).$1ˉ=1ˉ.1ˉ=1$, and m.(2.$1ˉ)=m.m=1$.

The third axiom introduces a new idea. It may be stated as follows: the set of elements contains a single element known as the identity, which has the property that, when it is combined with any other element, the result is that same element. Here, the identity is the monad axis, symbol 11. Clearly, in the previous example, 1.$m=m$ etc. and thus the third axiom is fulfilled. Finally, the fourth axiom may be stated: for each element in the set there exists another unique element called its inverse, such that the combination of the element and its inverse is the identity. In this example, each of the symmetry elements in the set is its own inverse: we may consider the combination m.m = 1 to mean, as indicated earlier ‘reflection one way followed by reflection back again’, or 2.2 = 1 to mean ‘a rotation of 180° clockwise followed by a rotation of 180° anti-clockwise’ (which is equivalent of course to two clockwise or anti-clockwise rotations).

Fig. A7.1. (a) The multiplication table for monoclinic point group 2/m and (b) the corresponding stereographic projection. Note that, as with space group symbols, the symbols ○ and are related by a reflection or a centre of symmetry.

The combinations, or products, of a set of elements in a group can be set out in the form of a matrix array, known as a multiplication table or ‘Cayley’s Square’ (after the English mathematician Arthur Cayley 1821–1895) as shown in Fig. A7.1(a).

(p.471) Notice that all the combinations of symmetry elements in the table contain the original four symmetry elements, i.e. the first axiom is fulfilled. This multiplication table (said to be of order 4 since there are four elements in the set), represents monoclinic point group symmetry 2/m that is also represented by a stereogram (see Section 12.5.1) in Fig. A7.1(b).

Fig. A7.2. Multiplication tables for (a) point group 2 and (b) point group m.

We can now write down the multiplication tables of monoclinic point group 2 (elements 1 and 2) and monoclinic point group m (elements 1 and m), both are of order 2.

These two groups are called subgroups of group 2/m; conversely group 2/m is the supergroup of these subgroups. A considerable part of the application of group theory to crystallography consists of first devising simple groups and then finding supergroups based upon them by adding the appropriate symmetry operations. In this way, complicated groups are constructed from simple ones. We can proceed in this way to write down the multiplication tables of all the 32 point groups—which are of higher order of course—the greater the number of the symmetry operations present. However, we shall only do this for four further examples, orthorhombic point groups 222 and mm2, trigonal point group 32, and tetragonal point group 4.

Fig. A7.3. (a) The multiplication table for orthorhombic point group 222 and (b) the corresponding stereographic projection.

For point group 222 the symmetry operations are 1, 2x, 2y, 2z; the subscripts referring to the crystallographic axis along which the corresponding diad axis lies.

(p.472) Notice that a rotation of 180° about x followed by a rotation of 180° about y equals a rotation of 180° about z, i.e. 2x.2y = 2z, and so on.

Fig. A7.4. (a) The multiplication table for orthorhombic point group mm2 and (b) the corresponding stereographic projection.

For point group mm2 the symmetry operations are 1, mx, my, and 2z.

Notice that a reflection in a plane perpendicular to x followed by a reflection in a plane perpendicular to y equals a rotation of 180° about z, i.e. mx.my = 2z. Indeed, we encountered this equivalence in our derivation of the ten plane point groups (Section 2.3) (p.473) where we found (Fig. 2.3(3)) that the operation of two perpendicular mirror lines ‘automatically’ generates a diad axis of symmetry along their intersection.

Notice also the similarity between the multiplication tables for point groups 222 and mm2, i.e. 1$≡$1, $mx≡$2x, $my≡$2y, $2z≡$2z. These two groups are said to be isomorphic (meaning ‘equal form’) since there is a unique one-to-one correspondence between the elements of the two groups. The groups in Fig. A7.2 are also isomorphic.

Fig. A7.5. (a) The multiplication table for trigonal point group 32 and (b) the corresponding stereographic projection.

For point group 32 (quartz) the symmetry operations are 1, 3, 32, 2x, 2y, 2u. These are indicated on the stereographic projection in Fig. A7.5(b). Here, we have to distinguish a symmetry operation and its inverse: 3 represents an anti-clockwise rotation of 120° and 32 a clockwise rotation of 120° (equivalent to an anti-clockwise rotation of 240), i.e. 3.32 = 1 (fourth axiom) and 3.3 = 32. (The symbol 32 should not be confused with the symbol 32 that denotes a screw triad axis.) Also, unlike our previous examples, the order of combination is important (non-Abelian). For example, 3.2x = 2y and 2x.3 = 2u; i.e. depending on the order of combination, the operation of an anti-clockwise rotation of 120° about the z-axis combined with a rotation of 180° about the x-axis gives a rotation of 180° either about the y-axis or the u-axis. The multiplication table is given in Fig. A7.5(a), the order of combination being ‘column times row’.

Fig. A7.6. (a) The multiplication table for tetragonal point group 4 and (b) the corresponding stereographic projection.

For point group 4, the symmetry operations (about the z-axis) are 1, 4 (90° anti-clockwise), 42 (180° anti-or-clockwise = 2), and 43 (270° anti-clockwise or 90° clockwise). These operations are shown in the multiplication table and stereographic projection in Fig. A7.6(a) and (b).

Notice that for each symmetry operation there is an inverse, i.e. 4$−1$ = 43, (4$2)−1$ = 42 and (4$3)−1$ = 4. We may check that the second axiom (the associative law) holds, e.g. 4.(42.43) = 4.4 = 42 and (4.42).43 = 43.43 = 42

(p.474) A note on nomenclature for symmetry operations

The operations of rotation about a symmetry axis, both rotation axes and screw axes, are sometimes called proper operations, or operations of the first sort because they do not result in a change of hand of the asymmetrical unit. Conversely, the operations of inversion, reflection, and glide-reflection, are called improper operations, or operations of the second sort since they do result in a change of hand. This distinction is important in the classification of point groups (see Table 3.1) and space groups.

A7.2 Matrix representation of symmetry operations: character tables

Matrices were used (see Section 5.8) to describe the transformations of plane indices and zone axis symbols from one axis system to another. Here, we use matrices to describe symmetry operations and their combinations: in effect we substitute the algebraic process of matrix multiplication for the geometrical process of successively applying symmetry operations. This in turn leads to a new (matrix) way of representing point groups. Note that in what follows we refer the matrices to axes x, y, and z.

The identity (monad symmetry operation), symbol 1 or E

Suppose we have a point with coordinates x, y, z. The identity transforms the point ‘into itself’, i.e. $x→x$, $y→y$, $z→z$. In matrix notation this is:

$Display mathematics$

(p.475) Reflection mx (i.e. in the y–z plane, perpendicular to the x-axis)

Here, a point (x, y, z) becomes ($xˉ$, y, z). (p.476) In matrix notation this is:

$Display mathematics$

Inversion, symbol $1ˉ$

Here, a point (x, y, z) becomes ($xˉ$, $yˉ$, $zˉ)$. In matrix notation this is:

$Display mathematics$

Rotation operations

Fig. A7.7. Coordinate transformation of a point following an anticlockwise rotation, angle [Greek] phi.

First, consider a general rotation of $ϕ∘$ anti-clockwise about the z-axis (Fig. A7.7). A point (x1, y1, $z1)$ becomes a point (x2, y2, z2) (or vector $r1$ becomes vector $r2)$. These are related by the equations:

(I leave it to the reader to work out these equations. Hint: consider the x1 and y1 components of vector $r1$ separately, rotate both anti-clockwise and find x2 and y2 from the sums.)

In matrix notation this is:

$Display mathematics$

and:

$Display mathematics$

Notice (a) that these matrices are the transposes of each other (i.e. rows and columns interchanged) and (b) that the inverse of each matrix (see Section 5.10) is simply equal to the transpose. They are said to be conjugate.

For $ϕ$ = 180° the anti-clockwise and clockwise matrices are identical.

$Display mathematics$

For $ϕ$ = 90° the matrices are:

$Display mathematics$

$Display mathematics$

Combinations of symmetry operations; here we give two examples:

$Display mathematics$

Matrix representation of point groups

Consider for example point group $2/m$ (Fig. A7.1). We now represent the symmetry operations 1, 2z, mz, and $1ˉ$ by the matrices:

(p.477)

$Display mathematics$

and for point group mm2 (Fig. A7.4), we now represent the symmetry operations 1, mx, my, and 2z by the matrices:

$Display mathematics$

Each matrix corresponds to a single operation in the group, which may be summarized by considering their operations on a point $xyz$. For example, for point group mm2 above:

$Display mathematics$

mm2

1 (or $E)$

mx

my

2z

x

1

$−1$

1

$−1$

y

1

1

$−1$

$−1$

z

1

1

1

1

xy

1

$−1$

$−1$

1

Table A7.1 Character table for point group mm22.

mm2

1($E)$

mx

my

2z

A1

1

1

1

1

z

$x2$, $y2$, $z2$

A2

1

$−1$

$−1$

1

Rz

xy

B1

1

$−1$

1

$−1$

x, Ry

xz

B2

1

1

$−1$

$−1$

y, Rx

yz

The results may be set out in the form of a table.

Also included in the table is the effect of the symmetry operations on the product function xy (simply multiply x and y in each of the columns here). As can be seen, xy transforms differently to x, y, and z. If we do the same for the product functions xz, yz, $x2$, $y2$, and $z2$, we find that we simply repeat the combinations already in the table. For example, yz transforms like y and xz transforms like x. Hence, in summary, (p.478) there are four ways in which the functions transform in point group mm2. They are tabulated in the form of a character table (not to be confused with a multiplication table—Section A7.1).

The four rows A1, A2, B1, and B2 are called representations of the point group mm2 and Rz, Ry, and Rx are axial vectors. However, at this point our introduction to character tables must stop, leaving a reader to pursue these topics in the books listed in Further Reading.

A7.3 Symmorphic and enantiomorphous space groups

The 230 space groups are derived by combining the 32 point group symmetries with the appropriate Bravais lattices and then taking into account the additional space groups arising from the existence of the translational symmetry elements—glide planes and screw axes. It is, in essence, the same procedure by which we derived the 17 plane groups (Sections 2.4 and 2.5) but is obviously more complicated. In group theory, the procedure is to combine the multiplication table of each point group with the appropriate lattice translations. Consider the general symbol for a direction (Section 5.7):

$Display mathematics$

Table A7.2 The 73 symmorphic space groups (based on Burns and Glazer, 2013).

Space groups—based on point groups:

System

Bravais lattice

Non-centrosymmetric enantiomorphous (total 24)

Non-centrosymmetric non-enantiomorphous (total 25)

Centrosymmetric non-enantiomorphous (total 24)

Cubic

P

P23, P432

$P4ˉ$3m

Pm $3ˉ$, Pm$3ˉ$m

I

I23, I432

$I4ˉ$3m

Im $3ˉ$, Im $3ˉ$m

F

F23, F432

$F4ˉ$3m

Fm $3ˉ$, Fm $3ˉ$m

Tetragonal

P

P4, P422

$P4ˉ$, P4mm, $P4ˉ$2m, $P4ˉm$2

P4/m, P4/mmm

I

I4, I422

$I4ˉ$, I4mm, $I4ˉ$2m, $I4ˉm$2

I4/m, I4/mmm

Orthorhombic

P

P222

Pmm2

Pmmm

I

I222

Imm2

Immm

$C(A)$

C222

Cmm2, Amm2

Cmmm

F

F222

Fmm2

Fmmm

Trigonal

R

R3, R32

R3m

$R3ˉ$, $R3ˉm$

P

P3, P312, P321

P3m1, P31m

$P3ˉ$, $P3ˉ$1m, $P3ˉ$ m1

Hexagonal

P

P6, P622

$P6ˉ$, P6mm, $P6ˉm$2, $P6ˉ$2m

P6/m, P6/mmm

Monoclinic

P

P2

Pm

P2/m

C

C2

Cm

C2/m

Triclinic

P

P1

$P1ˉ$

For an ‘infinite’ crystal the integers u, v, w take any value from -$∞$ to +$∞$ and specify the coordinates of the origins of every unit cell. Each set of such integers u, v, w constitute a group, i.e. they are in accord with the four axioms stated here except (1) the rule of combination is addition and (2) the identity is zero. From this, multiplication tables may be constructed but the mathematical procedures involved (which begin so simply!) are too sophisticated to be covered in detail in this book. Instead we derive the 73 symmorphic space groups by combining the appropriate point groups and Bravais lattices. For example, the five cubic point groups (23, 432, $4ˉ$3m, $m3ˉ$, and $m3ˉm)$ combined with the three (P, I, and $F)$ Bravais lattices give rise to 15 symmorphic space groups. In other systems there is more than one way in which the point group and Bravais lattice can be combined. For example, the trigonal point group 3m combined with the hexagonal P lattice gives rise to two space groups, P3m1 and P31m (recall the two possible plane groups, p31m and p3m1 in Table 2.6). In the orthorhombic system point group mm2 may be combined with the C-centred lattice with the diad axis either along the C-axis (space group Cmm2) or along the a- (equivalently b-) axis (space group Amm2, equivalently Bmm2). The 73 symmorphic space groups are listed in Table A7.2. As can be seen, they are almost equally divided between the three groups of point groups (see Table 3.1).

The 24 symmorphic space groups based on the 11 centrosymmetric point groups correspond to the 24 Patterson space groups (Section 13.3.2) and include the eleven Laue groups (Table 9.1) with the appropriate/permissible cell centrings.

The space groups may also be divided into those based on:

1. 1. The eleven non-centrosymmetric enantiomorphous point groups. These space groups, total 65, which were first worked out by Sohncke in 1867 (Section 4.6), (p.479) are sometimes called the axial space groups and are important in that proteins and nucleic acids (which do not possess mirror planes or inversion axes of symmetry) crystallize in these space groups. They include the eleven enantiomorphous pairs (Table A7.3).

2. 2. The ten non-centrosymmetric non-enantiomorphous point groups, total 73 space groups; the same number (coincidence?) as the symmorphic space groups.

3. 3. The eleven centrosymmetric (non-enantiomorphous) point groups, total 92 space groups.

Table A7.3 The eleven pairs of enantiomorphous space groups.

 P31 and P32 P41 and P43 P61 and P65 P62 and P64 P3112 and P3212 P3121 and P3221 P4122 and P4322 P41212 and P43212 P6122 and P6522 P6222 and P6422 P4132 and P4332

(p.480) A note on Schönflies’ notation for symmetry elements, point groups, and space groups

Schönflies treated the crystallographic symmetries as mathematical ‘groups’ of operations. He called a group containing one rotation operation and its powers a cyclic group, symbol C. For example, the operations of a four-fold axis, symbol C4 are the rotations $1,π2,π,3π2$ which are all ‘powers’ of the fundamental rotation $π2$. A mirror (reflection) operation, symbol σ, normal to the rotation axis he symbolized with a superscript h. A mirror operation parallel to the rotation axis he symbolized with the superscript v.

As noted in Section 4.3 (see p. 104), Schönflies used rotation-reflection or alternating axes (rather than rotation-inversion (or simply inversion/axes)) symbolized by S (from Sphenoidisch). Next, we have the groups with normal diad axe(s) symbolized by D (from Diëdergruppe). Those ‘D’ groups that also include a plane of symmetry normal to the rotation axis are symbolized, as before, by the superscript h and those with a diagonal plane of symmetry are symbolized by the superscript d. In the cubic system the symbols T and O are used to represent tetrahedral and octahedral symmetry, respectively.

The Schönflies symbol may also include the letter i denoting a centre of symmetry, so Ci is a synonym for S2 (=$1ˉ)$ and $C3i$ is a synonym for S6 (= $3ˉ)$. We may also note that the $5ˉ$ inversion axis that describes icosahedral (non-crystallographic) symmetry (Section 4.9) corresponds to Schönflies’ $S10$ rotation-reflection axis.

Table A7.4 Hermann–Mauguin and Schönflies point group symbols.

Non-centrosymmetric point groups

Centrosymmetric point groups (non-enantiomorphous)

System

Enantiomorphous

Non-enantiomorphous

Cubic

T(23), O (432)

$Td(4ˉ$3$m)$

$Th(m$3), $Oh(m3ˉm)$

Tetragonal

C4(4), D4(422)

$S4(4ˉ)$, $C4v$(4mm), $D2d(4ˉ$2$m)$

$C4h4m$, $D4h4mmm$

Orthorhombic

D2(222)

$C2v$(mm2)

$D2h$(mmm)

Trigonal

C3(3), D3(32)

$C3v$ (3m)

$S6(3ˉ)$, $D3d(3ˉ$m)

Hexagonal

C6(6), D6(622)

$C3h(6ˉ)$, $C6v$(6mm), $D3h(6ˉm$ 2)

$C6h(6m)$, $D6h(6mmm)$

Monoclinic

C2(2)

Cs, $C1h(m)$

$C2h(2m)$

Triclinic

C1 (1)

$Ci(1ˉ)$

The corresponding Schönflies and Hermann–Mauguin point group symbols are set out in Table A7.4 (arranged in the same form as Table 3.1).

Clearly, the Hermann–Mauguin notation is to be far preferred since it contains the essential point group symmetry information. This is even more the case for Schönflies’ space group notation in which the space groups, isomorphous with each point group, are simply numbered sequentially (presumably in the order in which he derived them). For example, the five space groups isomorphous with point group 23 (Nos. 195–199 in the International Tables) are denoted T1(P23), T2(F23), T3(I23), T4(P213), and T5(I213).

Notes:

(1) The identity is also given the symbol E (German: Einheit)

(2) This is the character table for the water molecule. Its use in describing the atomic vibrations of the water molecule may be found (e.g.) in Hargittai & Hargittai (1995) and Ladd (2014).


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# How to effectively find a starting point solving a non-linear equation?

I have the following equation (the Kurz-Giovanola-Trivedi model [1]) $$v^2 \frac{\pi^2 \Gamma}{P^2 D^2} + v \frac{mC_0(1-k)\xi}{D[1-(1-k)Iv(P)]} + G = 0,$$ where $$Iv(P)=P \cdot \exp(P) \cdot E(P)$$, $$P=\frac{Rv}{2D}$$, $$E(P)=\int_{P}^{\infty} \frac{\exp(-x)}{x} dx$$ is the exponential integral function, $$\xi=1-\frac{2k}{[1+(\frac{2\pi}{P})^2]^{\frac{1}{2}}-1+2k}$$, $$v$$ and $$R$$ are unknowns, and everything else is constant. To solve this equation, I use MATLAB (the core function is fzero) and suppose that $$v$$ is an array within a range $$(10^{-2};10^2)$$. The fzero function attempts to find a zero of one equation with one variable and might be called with either a one-element starting point or a two-element vector (starting interval).

My question is: how to effectively find a starting point?

I have started by solving this equation relative to $$P$$, and it was relatively easy to determine starting points $$P_0$$ for different intervals of the $$v$$ array by trial-and-error method. After finding $$P$$, it was easy to find $$R$$.

However, now I need to solve the equation relative to $$R$$ directly, without the $$P$$-step. The trial-and-error method is not the best option here, probably due to the type of the $$R(v)$$-dependence (please see below).

I recommend using a global optimization solver.

You should be able to solve this, or at least try, using the BMIBNB global optimization solver https://yalmip.github.io/solver/bmibnb/ under YALMIP https://yalmip.github.io/tutorial/basics/ under MATLAB.

In particular, note that YALMIP supports the exponential integral, expint, https://yalmip.github.io/command/expint/

For this problem:

Declare the variables v and R as sdpvar.

Enter the equation, and lower and upper bounds on the variables, all inside [] , as the constraints in the 1st argument of the optimize command

Set the objective to [] as the 2nd argument in the optimize command.

Set 'bmibnb' as the solver using sdpsettings as the 3rd argument in the optimize command.

You do not need a starting value. You will need a local nonlinear solver installed under MATLAB which will be called by BMIBNB, as well as a MILP (Mixed Integer Linear Programming) solver to be called by BMIBNB.

• I upvoted with the note that, this might be very expensive for their problem. I would suggest, in addition to Mark's suggestion, to somehow linearize system, do one step and use it as an initial condition. For example, consider the following; $$v_0^2 \frac{\pi^2 \Gamma}{P^2 D^2} + v_0 \frac{mC_0(1-k)\xi}{D[1-(1-k)Iv(P)]} + G = 0$$ to obtain a guess for $R$ and $$v^2 \frac{\pi^2 \Gamma}{P_0^2 D^2} + v \frac{mC_0(1-k)\xi}{D[1-(1-k)Iv(P_0)]} + G = 0$$ to obtain a guess for $v$. $v_0$ and $P_0$ (hence $R_0$) can be chosen arbitrarily, $1$ probably would work. Or you can do multiple iterations... Aug 5, 2020 at 21:56


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# 섬유강화 복합재의 $G_ {IC}$ 결정을 위한 일인자방법

Lee, Gyeong-Yeop;Go, Seung-Gi
이경엽;고승기

• Published : 1996.11.01
• 12 23

#### Abstract

The work factor approach was applied to determine $G_ {IC}$ of fiber reinforced composites (AS4/3501) from a single unidirectional (0-deg) DCB specimen. Elastic work factors of DCB specimen for three different symmetrical staking sequences were derived from a simple bending theory and a finite element method. The results showed that elastic work factors calculated from both methods were comparable each other. In particular, the elastic work factor of DCB specimen with symmetrical stacking sequence is independent of stacking sequence. The $G_ {IC}$ determined from the work factor approach was compared with that determined by the compliance method. The results showed that the work factor approach and the compliance method produce comparable results of $G_ {IC}$. Thus, $G_ {IC}$ can be determined from a single DCB specimen using the work factor approach.