Squares and Motzkins

Greg Smith gave an awesome colloquium here last week about his paper with Blekherman and Velasco on sums of squares.

Here’s how it goes.  You can ask:  if a homogeneous degree-d polynomial in n variables over R takes only non-negative values, is it necessarily a sum of squares?  Hilbert showed in 1888 that the answer is yes only when d=2 (the case of quadratic forms), n=2 (the case of binary forms) or (n,d) = (3,4) (the case of ternary quartics.)  Beyond that, there are polynomials that take non-negative values but are not sums of squares, like the Motzkin polynomial

X^4 Y^2 + X^2 Y^4 - 3X^2 Y^2 Z^2 + Z^6.

So Greg points out that you can formulate this question for an arbitrary real projective variety X/R.  We say a global section f of O(2) on X is nonnegative if it takes nonnegative values on X(R); this is well-defined because 2 is even, so dilating a vector x leaves the sign of f(x) alone.

So we can ask:  is every nonnegative f a sum of squares of global sections of O(1)?  And Blekherman, Smith, and Velasco find there’s an unexpectedly clean criterion:  the answer is yes if and only if X is a variety of minimal degree, i.e. its degree is one more than its codimension.  So e.g. X could be P^n, which is the (n+1,2) case of Hilbert.  Or it could be a rational normal scroll, which is the (2,d) case.  But there’s one other nice case:  P^2 in its Veronese embedding in P^5, where it’s degree 4 and codimension 3.  The sections of O(2) are then just the plane quartics, and you get back Hilbert’s third case.  But now it doesn’t look like a weird outlier; it’s an inevitable consequence of a theorem both simpler and more general.  Not every day you get to out-Hilbert Hilbert.

Idle question follows:

One easy way to get nonnegative homogenous forms is by adding up squares, which all arise as pullback by polynomial maps of the ur-nonnegative form x^2.

But we know, by Hilbert, that this isn’t enough to capture all nonnegative forms; for instance, it misses the Motzkin polynomial.

So what if you throw that in?  That is, we say a Motzkin is a degree-6d form

expressible as


P^4 Q^2 + P^2 Q^4 - 3P^2 Q^2 R^2 + R^6

for degree-d forms P,Q,R.  A Motzkin is obviously nonnegative.

It is possible that every nonnegative form of degree 6d is a sum of squares and Motzkins?  What if instead of just Motzkins we allow ourselves every nonnegative sextic?  Or every nonnegative homogeneous degree-d form in n variables for n and d less than 1,000,000?  Is it possible that the condition of nonnegativity is in this respect “finitely generated?”






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What correlation means

From Maria Konnikova’s New Yorker piece on Randall Munroe and what makes science interesting:

In a meta-analysis of sixty-six studies tracking interests over time (the average study followed subjects for seven years), psychologists from the University of Illinois at Urbana–Champaign found that our interests in adolescence had only a point-five correlation with our interests later in life. This means that if a subject filled out a questionnaire about her interests at the age of, say, thirteen, and again at the age of twenty-one, only half of her answers remained consistent on both.

I think it’s totally OK to not say precisely what correlation means.  It’s sort of subtle!  It would be fine to say the correlation was “moderate,” or something like that.

But I don’t think it’s OK to say “This means that…” and then say something which isn’t what it means.  If the questionnaire was a series of yes-or-no questions, and if exactly half the answers stayed the same between age 13 and 21, the correlation would be zero.  As it should be — 50% agreement is what you’d expect if the two questionnaires had nothing to do with each other.  If the questionnaire was of a different kind, say, “rate your interest in the following subjects on a scale of 1 to 5,” then agreement on 50% of the answers would be more suggestive of a positive relationship; but it wouldn’t in any sense be the same thing as 0.5 correlation.  What does the number 0.5 add to the meaning of the piece?  What does the explanation add?  I think nothing, and I think both should have been taken out.

Credit, though — the piece does include a link to the original study, a practice that is sadly not universal!  But demerit — the piece is behind a paywall, leaving most readers just as unable as before to figure out what the study actually measured.  If you’re a journal, is the cost of depaywalling one article really so great that it’s worth forgoing thousands of New Yorker readers actually looking at your science?




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That Jeff Koons feeling

Jed Perl opposes Jeff Koons:

The sculptures and paintings of this fifty-nine-year-old artist are so meticulously, mechanically polished and groomed that they rebuff any attempt to look at them, much less feel anything about them.

But four paragraphs later:

Koons knows how to capitalize on the guilty pleasure that the museumgoing public takes in all his mixed messages. He knows how to leave people feeling simultaneously ironical, erudite, silly, sophisticated, and bemused.

Does Koons make people feel things, or does he not?  Or are irony, erudition, silliness, sophistication, and bemusement feelings that don’t count as feelings?

Jed Perl writes well but I find his judgment strange.  About Jeff Koons I have no opinion.  But I remember his name because of the piece he wrote about Francis Bacon, which seems to suggest that people like Bacon not because of anything in the paintings, but because the artist sports a biography and attitude that appeals to mushy-minded would-be avant-gardists.  “The Bacon mystique,” Perl writes, “is not grounded in his paintings so much as in a glamorous list of extenuating circumstances.”

To me this makes no sense.  I went to a small museum which was showing some of Bacon’s paintings and I was knocked over by them.  Whoa, what is that?  I had no idea who he was, or whether he was glamorous, or whether it was cool to like him.

I think it’s OK to say (as Perl also does, later in that piece) that Bacon is a stupid painter and only people who are stupid about painting like his paintings.  But it’s crazy to deny that people actually do like Bacon’s paintings, as such, not just the idea of Bacon’s paintings, or the idea of being the kind of person who likes Bacon’s paintings.

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So… yeah

Lately CJ has a habit of ending every story he tells by saying

“So… yeah.”

I first noticed it this summer, so I think he picked it up from his camp counselors. What does it mean? I tend to read it as something like

“I have told my story — what conclusions can we draw from it? Who can say? It is what it is.”

Is that roughly right? Per the always useful Urban Dictionary the phrase is

“used when relating a past event and teller is unsure or too lazy to think of a good way to conclude it”

but I feel like it has more semantic content than that. Though I just asked CJ and he says it’s just his way of saying “That’s all.” Like “Over and out.”

So yeah.

Squeeze! Squeeze!

I hope the world never runs out of awesome Earl Weaver stories.

I saw Earl Weaver put on a suicide squeeze bunt, in Milwaukee. It worked. Everybody asked him, ‘Wait, we thought you told us you didn’t even have a sign for a suicide squeeze, because you hated it so much.’ Earl said, ‘I still don’t.’ I asked him, ‘How did you put it on then?’ He said, ‘I whistled at Cal Ripken, Sr., my third base coach. Then I shouted at him, ‘Squeeze! Squeeze! Then I motioned a bunt.’ I said, ‘Paul Molitor was playing third. Didn’t he hear you?’ Earl said, ‘If he did, I’m sure he thought there was no way we were putting it on, or I wouldn’t have been yelling for it.’

This is from the Fangraphs interview with the greatest announcer of our time, Jon Miller. His memoir, Confessions of a Baseball Purist, is full of great stuff like this. I didn’t know until just this second that it had been reissued by Johns Hopkins University Press.

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The 1979 Houston Astros hit only 49 home runs

49 home runs! That’s nuts. They hit more triples than home runs. Their home run leader was Jose Cruz, who hit 9. In September they went 20 straight games without hitting a home run, the longest such streak in modern baseball. And that was after they went 15 games without hitting a rome run in July!

Must have been a pretty bad team, right? But no! They won 89 games and finished second, just a game and a half behind the Reds. That 15 game homerless streak in July? They went 11-4 in those games.

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Benson Farb’s ICM talk

One of the things I’ve been spending a lot of time on mathematically is problems around representation stability and “FI-modules,” joint with Tom Church, Benson Farb, and Rohit Nagpal.  Benson just talked about this stuff at the ICM, and here it is:

In the latest stable representation theory news, Andy Putman and (new Wisconsin assistant professor!) Steven Sam have just posted an exciting new preprint about the theory of representations of GL_n(F_p) as n goes to infinity; this is kind of like the linear group version of what FI-modules does for symmetric groups.  (Or, if you like, our thing is their thing over the field with one element….!)  This is something we had hoped to understand but got very confused about, so I’m looking forward to delving into what Andy and Steven did here — expect more blogging!  In particular, they prove the Artinian conjecture of Lionel Schwartz.  Like I said, more on this later.

Boyhood: one more note

I thought of one more small thing, concerning the last scene.

Continue reading

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Breuillard’s ICM talk: uniform expansion, Lehmer’s conjecture, tauhat

Emmanuel Breuillard is in Korea talking at the ICM; here’s his paper, a very beautiful survey of uniformity results for growth in groups, by himself and others, and of the many open questions that remain.

He starts with the following lovely observation, which was apparently in a 2007 paper of his but which I was unaware of.  Suppose you make a maximalist conjecture about uniform growth of finitely generated linear groups.  That is, you postulate the existence of a constant c(d) such that, for any finite subset S of GL_d(C),  you have a lower bound for the growth rate

\lim |S^n|^{1/n} > c(d).

It turns out this implies Lehmer’s conjecture!  Which in case you forgot what that is is a kind of “gap conjecture” for heights of algebraic numbers.  There are algebraic integers of height 0, which is to say that all their conjugates lie on the unit circle; those are the roots of unity.  Lehmer’s conjecture says that if x is an algebraic integer of degree n which is {\em not} a root of unity, it’s height is bounded below by some absolute constant (in fact, most people believe this constant to be about 1.176…, realized by Lehmer’s number.)

What does this question in algebraic number theory have to do with growth in groups?  Here’s the trick; let w be an algebraic integer and consider the subgroup G of the group of affine linear transformations of C (which embeds in GL_2(C)) generated by the two transformations

x -> wx


x -> x+1.

If the group G grows very quickly, then there are a lot of different values of g*1 for g in the word ball S^n.  But g*1 is going to be a complex number z expressible as a polynomial in w of bounded degree and bounded coefficients.  If w were actually a root of unity, you can see that this number is sitting in a ball of size growing linearly in n, so the number of possibilities for z grows polynomially in n.  Once w has some larger absolute values, though, the size of the ball containing all possible z grows exponentially with n, and Breuillard shows that the height of z is an upper bound for the number of different z in S^n * 1.  Thus a Lehmer-violating sequence of algebraic numbers gives a uniformity-violating sequence of finitely generated linear groups.

These groups are all solvable, even metabelian; and as Breuillard explains, this is actually the hardest case!  He and his collaborators can prove the uniform growth results for f.g. linear groups without a finite-index solvable subgroup.  Very cool!

One more note:  I am also of course pleased to see that Emmanuel found my slightly out-there speculations about “property tau hat” interesting enough to mention in his paper!  His formulation is more general and nicer than mine, though; I was only thinking about profinite groups, and Emmanuel is surely right to set it up as a question about topologically finitely generated compact groups in general.







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Notes on Boyhood

Richard Linklater’s Boyhood is certainly the best movie I’ve seen this year, likely the best movie I’ll see this year.  But I don’t see a lot of movies.  After the spoiler bar, some notes on this one.  I meant to write this right after I saw it, but got busy, so no doubt I’ve forgotten some of what I meant to say and gotten other things wrong. Continue reading

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