## Michigan panel on math teaching in the age of internet forumsSeptember 28, 2012

Posted by David Speyer in Uncategorized.

Sorry for disrupting non-Michigan people with this, but on Monday Oct. 1, at 5:40 PM, I will be on a panel in East Hall B844 on teaching math in the age of google and math.SE. I haven’t been given a list of topics, but I’m hoping to talk about both how to do accurate evaluation of our own students, and how to ethically help other institutions’ students. All levels of participant, from undergraduate up to full professor, are welcome to attend. I assume there will be an official announcement going out, but I thought people interested in the topic might see it better here.

## Bill Thurston 1946-2012August 23, 2012

Posted by David Speyer in Uncategorized.

Bill Thurston died on the evening of August 21st. His son Dylan writes “He was surrounded by family, and went very peacefully, after a fight with melanoma since spring 2011. Please pass this on as appropriate.”

I knew Professor Thurston only through his writing, first in publication and recently on Mathoverflow. It was always a joy. I have avoided more routine obligations than I care to admit by reading and rereading his papers. He believed that mathematics was a fundamentally human task, and that his goal was not simply to provide the reader with a bulleted list of truths, but to provide a picture and an intuition that made them obvious.

I had the thought to organize a blogfest, where various math bloggers would write up expositions of some aspect of his work. But I ran into an obstacle: What subject did I think I could explain better than he did? So, instead, here are some of my favorites, for your pleasure and inspiration:
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## Quick Mathematica squibAugust 14, 2012

Posted by David Speyer in Uncategorized.

I have spent a lot of time manipulating multivariate polynomials in Mathematica, and have been very frustrated at trying to make Coefficient behave the way I think it should. I think Coefficient[2x+3xy, x] should be 2, Mathematica thinks it should be 2+3y. Or rather, I don’t think that this should be the default, but it is very often my desired behavior and I can’t find an option to tell Mathematica that’s what I want.

I just wrote a little snippet to do this for me, and I’m copying it here so I can reuse it in the future:

myCoeff[poly_, vars_, exp_] :=      If[Max[exp] == 0, poly, Coefficient[poly, Apply[Times, vars^exp]]] /.        Map[(# -> 0) &, vars]

For example, myCoeff[2 x + 3 x*y, {x, y}, {1, 0}] is 2, just as I want. Note that myCoeff[2k*x+2x*y, {x,y}, {1,0}] is 2k; so you can still easily have Mathematica treat some variables as constants.

UPDATE Code simplified a bit, because I realized that creating useless replacement rules wouldn’t do any harm.

## El Naschie loses libel suit against NatureJuly 11, 2012

Posted by David Speyer in Uncategorized.

A British court has just ruled against Mohammed El Naschie in his libel case against Nature. To quote from the decision:

My conclusions are that the article is substantially true … and that it was the product of responsible journalism.

To my mind, this basically finishes the story of El Naschie the man, but does not conclude the question of how such a journal was allowed to remain at Elsevier for so long.

Past El Naschie threads have been unpleasant and unproductive, so I’m leaving comments closed on this one. Thanks to Mike Usher at Math 2.0 for making me aware of this news.

## Why aren’t my prime pictures pretty?July 3, 2012

Posted by David Speyer in Uncategorized.

This is a followup to my post What if primes hated to start with nine? and you might want to read that post for context. This post describes my failure to create a nice illustration of the effect of the zero of $\zeta$ at $1/2 + 14.134725 i$ on the distribution of the primes. Of course, the answer might be really dumb — I might just have a coding error, or the range of primes I was using (around $10^9$) may not be large enough. But I suspect there is something interesting here, and I’d like to know what. The basic question of this post is “Why does the left hand graph look as random as the right hand one?”

UPDATE I have a prettier picture!

Posted by David Speyer in Uncategorized.

There are a number of sums over primes whose asymptotic behavior is easy to determine. For example, we know that
$\log n! = \sum_p \sum_k \lfloor \frac{n}{p^k} \rfloor \log p$
and also $\log n! = n \log n - n + O(\log n)$. A bit of rearrangement and some elementary bounds yields
$\sum_{p \leq n} \frac{n}{p} \log p = n \log n + O(n)$
so
$\sum_{p \leq n} \frac{\log p}{p} = \log n + O(1). \quad (1)$

On the other hand, the estimate
$\sum_{p \leq n} 1 = \frac{n}{\log n} + O\left( \frac{n}{(\log n)^2} \right) \quad (2)$
is the Prime Number Theorem, whose proof was a major triumph of nineteenth century mathematics.

If you spend a lot of time reading about the Prime Number Theorem, you’ll see a lot of messing around with sums over primes, and it isn’t clear which ones are elementary and which ones are deep. This post is about a heuristic I found a while ago to separate the two: Would the given asymptotic still hold if primes hated to have first digit nine?

## ABC conjecture rumorJune 12, 2012

Posted by Scott Carnahan in Algebraic Geometry, Number theory.

There is a rumor circulating here in Japan, to the effect that S. Mochizuki has proved the ABC conjecture. My understanding is that blogs are for spreading such things.

Apparently, he had predicted some years ago that he would finish a proof in 2012, so I suppose this is an “on-time delivery”. It is certainly no secret that his research program has been aiming at the conjecture for several years.

Our very own Noah Snyder did some original work on the function field version of this conjecture, when he was a high-school student.

Update (Sept 4, 2012): This rumor seems to be true! You can find the four “Inter-universal Teichmuller Theory” papers on the very bottom of his papers page.

## Bizarre persecution of Russian mathematicianMay 16, 2012

Posted by David Speyer in Uncategorized.

I don’t fully understand this story, but it is time sensitive. Misha Verbitsky is a complex geometer who is also an active Russian political blogger.
Recently, he was arrested by the Russian government while attempting to board a flight out of the country. It turns out he was convicted in absentia of violating the trademark of a man named Igor Pugach, by using an image of Pugach to illustrate a blogpost criticizing Pugach. Verbitsky was fined 300,000 rubles (approximately 9,700 dollars) and will not be able to leave Russia until he pays the sum. Verbitsky’s blog is here; due to my inability to read Russian I have not located the post in question. A newspaper article on the issue is here.

Obviously, this violates all sorts of legal norms. Under US trademark law, you simply can’t be convicted of a trademark violation if you are not selling something. You could be convicted of a copyright violation, but this case would likely be fair use. Moreover, you certainly can’t have a large judgment made against you in absentia, except after extraordinary efforts have been made to notify you of the proceedings. I don’t know Russian law, but I am quite willing to say this is immoral, and expect that it is also illegal. (There is also a profoundly dumb statement by Pugach’s spokesperson that simply wearing a beard similar to Pugach’s is a trademark/copyright violation but, since that doesn’t appear to be the basis of the court ruling, I am going to lump that under “people say remarkably dumb things sometimes”.)

There is a petition to free Verbitsky here directed to Putin. I tend to be skeptical of the value of internet petitions, but it’s a start. I have also submitted this story to Boing Boing and the Volokh Conspiracy, in the name of boosting the signal.

I’d be interested in suggestions as to what we should be doing to get this story off the internet and to the attention of people who care. My next thought is to contact the US State Dept, or ask my Senators to do so. Does anyone know whether Verbitsky is a US Citizen? I assume our government will care more if this is the case. (Wikipedia reports that he lived in the US in the 90′s, and lives in Russia while travelling frequently now, but doesn’t give citizenship.)

## Fun with y^2=x^p-xMay 3, 2012

Posted by David Speyer in Algebraic Geometry, characteristic p, Number theory.

Here’s a basic example that comes up if you work with elliptic curves: Let $p$ be a prime which is $3 \mod 4$. Let $E$ be the elliptic curve $v^2=u^3-u$ over a field of characteristic $p$. Then $E$ has an endomorphism $F(u,v) = (u^p, v^p)$. It turns out that, in the group law on $E$, we have $F^2 = [-p]$. That is to say, $F(F(u,v))$ plus $p$ copies of $(u,v)$ is trivial.

I remember when I learned this trying to check it by hand, and being astonished at how out of reach the computation was. There are nice proofs using higher theory, but shouldn’t you just be able to write down an equation which had a pole at $F(F(u,v))$ and vanished to order $p$ at $(u,v)$?

There is a nice way to check the prime $3$ by hand. I’ll use $\equiv$ for equivalence in the group law of $E$. Remember that the group law on $E$ has $-(u,v) \equiv (u,-v)$ and has $(u_1,v_1)+(u_2,v_2)+(u_3,v_3) \equiv 0$ whenever $(u_1, v_1)$, $(u_2, v_2)$ and $(u_3, v_3)$ are collinear.

We first show that

$\displaystyle{ F(u,v) \equiv (u-1, v) - (u+1, v) \quad (\dagger)}$

Proof of $(\dagger)$: We want to show that $F(u,v)$, $(u+1,-v)$ and $(u-1,v)$ add up to zero in the group law of $E$. In other words, we want to show that these points are collinear. We just check:

$\displaystyle{ \det \begin{pmatrix} 1 & u^3 & v^3 \\ 1 & u-1 & -v \\ 1 & u+1 & v \end{pmatrix} = 2 v (v^2-u^3+u) = 0}$

as desired. $\square$.

Use of $(\dagger)$: Let $(u_0, v_0)$ be a point on $E$. Applying $F$ twice, we get

$\displaystyle{ F^2(u_0,v_0) \equiv F \left( (u_0-1,v_0) - (u_0+1,v_0) \right)}$

$\displaystyle{ \equiv (u_0-2,v_0) - 2 (u_0,v_0) + (u_0+2,v_0)}$.

Now, the horizontal line $v=v_0$ crosses $E$ at three points: $(u_0, v_0)$, $(u_0-2, v_0)$ and $(u_0+2, v_0)$. (Of course, $u_0 -2 =u_0+1$, since we are in characteristic three.) So $(u_0-2, v_0) + (u_0, v_0) + (u_0+1, v_0) \equiv 0$ and we have

$\displaystyle{F^2(u_0, v_0) \equiv -3 (u_0, v_0)}$

as desired. $\square$.

I was reminded of this last year when Jared Weinstein visited Michigan and told me a stronger statement: In the Jacobian of $y^2 = x^p-x$, we have $F^2 = [(-1)^{(p-1)/2} p]$, where $F$ is once again the automorphism $F(x,y) = (x^p, y^p)$.

Let me first note why this is related to the discussion of the elliptic curve above. (Please don’t run away just because that sentence contained the word Jacobian! It’s really a very concrete thing. I’ll explain more below.) Letting $C$ be the curve $y^2 = x^p-x$, and letting $p$ be $3 \mod 4$, we have a map $C \to E$ sending $(x,y) \mapsto (y x^{(p-3)/4}, x^{(p-1)/2})$, and this map commutes with $F$. I’m going to gloss over why checking $F^2 = [(-1)^{(p-1)/2} p]$ on $C$ will also check it on $E$, because I want to get on to playing with the curve $C$, but it does.

So, after talking to Jared, I was really curious why $F^2$ acted so nicely on the Jacobian of $C$. There are some nice conceptual proofs but, again, I wanted to actually see it. Now I do.

## Mathematicians take a standApril 8, 2012

Posted by Scott Morrison in elsevier, evil journals.

Douglas Arnold and Henry Cohn have just posted to the arXiv their article Mathematicians take a stand, which will also appear in the Notices of the AMS shortly.

In it they describe the background to the Elsevier boycott, and make a case for more people joining in. It might appear at this point that the boycott is losing steam, but I’m pretty sure this is not the case. A lot has been happening, although too much of it is “behind the scenes”. Elsevier has made some concessions, although these so far seem to mostly miss the point. There’s a rumour of more to come. Two weeks ago some representatives met with the Journal of Number Theory’s editorial board, to discuss their concerns. The meeting was inconclusive, but it seems the editorial board there remains unsatisfied with Elsevier. (It appears that the minutes of that meeting are googleable…)

I hope that the prestigious mathematics journals still with Elsevier (particularly the three discussed in Arnold and Cohn’s article: Advances, the Journal of Algebra, and the Journal of Number Theory) make sure they get what they need from Elsevier. Right now, they have the support of the mathematical community, and I hope they will not be timid about making demands. I’m not sure what’s already been discussed in private, but this would be my list:

• Legal ownership of the journals to be transferred to the editorial board or their chosen representatives, with Elsevier kept on as the publisher on a contract basis.
• Open access to the historical archives (everything older than 5 years), with clear rules and ideally an open license, allowing access for indexing, preservation and analysis.
• Prices reduced to no more than twice the per page cost of comparable journals published by professional societies, and a clear mechanism for libraries to achieve proportional reductions in bundle costs by dropping unwanted journals.

I don’t realistically expect that Elsevier will agree to any of these right away. But, nevertheless, I hope we demand this of them. At the moment their interests are simply diametrically opposed to ours — their commercial interest is to restrict access to our work, while our interest is to have the widest possible dissemination. Until they clearly see that we’re willing to take our toys and go home, I think there’s little hope of Elsevier taking meaningful steps.

Finally, it would be great  to hear from editors at Elsevier journals about the impact of the boycott. Has Advances seen a decrease in quantity or quality of submissions? If so, what are they going to do about it?