Monthly Archives: November 2010

Why SMB is no longer with the Hampshire College Summer Studies in Mathematics

Just came across this, from my old REU buddy Sarah-Marie Belcastro.  I’m old enough to remember when Hampshire, Ross’s program at OSU, and (to some extent) Math Olympiad training were the only places a high school kid could get formal training in really advanced math.  Hampshire has done a lot of good for a lot of kids over the years, especially countercultural kids who love math but not regimentation, contests, and grades.  (I myself was a super-compliant establishment type who loved contests and grades, so I did MOP.)   It’s a shame to hear they’re having management problems.

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Quomodocunquizing

The name of this blog, almost featured in the New York Times.  (Thanks to Terry for pointing this out.)

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In which I am impressed by Biddy Martin’s political savvy

The University of Wisconsin, like all big public institutions, faces a future of declining state support.  And, like all big public institutions, we have to figure out how to keep doing our jobs despite that.  Chancellor Martin’s proposal is a New Badger Partnership, under which UW would be allowed to set its own tuition, as Michigan does.  The sticker price of a Wisconsin education would go up, and the extra revenue would be plowed back into financial aid in order to keep college affordable for middle-class students and their parents.

I don’t know enough about higher education policy to comment on the merits of the plan.  But it’s kind of a work of political genius, isn’t it?  To the Democrats in the state government, Martin can say “UW needs to do much, much more to give working families a chance at a world-class education, even if rich Chicago parents take a hit. ” And to the Republicans, she can say, “I came here to run this university like a business, and that means charging market rates.”

Why it’s genius:  because she’s right on both counts!  The ultimate free-market dream is differential pricing:  charging each customer the maximum they’re willing to pay.  Most businesses don’t get to examine their customers’ bank balance before naming a price.  But UW does.  If the university can be more capitalist than the capitalists, redistribute wealth downward, and reduce our dependence on legislative whim, all at the same time, why shouldn’t we?

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Modeling lambda-invariants by p-adic random matrices

The paper “Modeling λ-invariants by p-adic random matrices,” with Akshay Venkatesh and Sonal Jain, just got accepted by Comm. Pure. Appl.  Math. But I forgot to blog about it when we finished it!  (I was a little busy at the time with the change in my personal circumstances.)

Anyway, here’s the idea.  As I’ve already discussed here, one heuristic for the Cohen-Lenstra conjectures about the p-rank of the class group of a random quadratic imaginary field K is to view this p-part as the cokernel of g-1, where g is a random generalized symplectic matrix over Z_p.  In the new paper, we apply the same philosophy to the variation of the Iwasawa p-adic λ-invariant.

The p-adic λ-invariant of a number field K is closely related to the p-rank of the class group of K; in fact, Iwasawa theory more or less gets started from the theorem that the p-rank of the class group of K(\zeta_{p^n}) is

\lambda n + \mu p^n + \nu

for some constants \lambda, \mu, \nu when n is large enough, with \mu expected to be 0 (and proved to be 0 when K is quadratic.)  On the p-adic L-function side, the λ-invariant is (thanks to the main conjecture) related to the order of vanishing of a p-adic L-function.  On the function field side, the whole story is told by the action of Frobenius on the p-torsion of the Jacobian of a curve, which is specified by some generalized symplectic matrix g over F_p.  The p-torsion in the class group is the dimension of the fixed space of g, while the λ-invariant is the dimension of the generalized 1-eigenspace of g, which might be larger.  It’s also in a sense more natural, depending only on the characteristic polynomial of g (which is exactly what the L-function keeps track of.)

So in the paper we do two things.  On the one hand, we study the dimension of the generalized 1-eigenspace of a random generalized symplectic matrix, and from this we derive the following conjecture: for each p > 2 and r >= 0,  the probability that a random quadratic imaginary field K has p-adic λ-invariant r is

p^{-r} \prod_{t > r} (1-p^{-t}).

Note that this decreases like p^{-r} with r, while the p-rank of the class group is supposed to be r with probability more like p^{-r^2}.  So large λ-invariants should be substantially more common than large p-ranks.

The second part of the paper tests this conjecture numerically, and finds fairly good agreement with the data. A novelty here is that we compute p-adic  λ-invariants of K for small p and large disc(K); previous computational work has held K fixed and considered large p.  It turns out that you can do these computations reasonably efficiently by interpolation; you can compute special values L(s,chi_K) transcendentally for many s; given a bunch of these values, determined to a certain p-adic precision, you can compute the initial coefficients of the p-adic L-function with some controlled p-adic precision as well, and, in particular, you can provably locate the first coefficient which is nonzero mod p.  The location of this coefficient is precisely the λ-invariant.  This method shows that, indeed, large λ-invariants do pop up!  For instance, the 3-adic λ-invariant of Q(\sqrt{-956238}) is 14, which I think is a record.

Some questions still floating around:

  • Should one expect an upper bound \lambda \ll_\epsilon D_K^\epsilon for each odd p?  Certainly such a bound is widely expected for the p-rank of the class group.
  • In the experiments we did, the convergence to the conjectural asymptotic appears to be from below.  For the 3-ranks of class groups of quadratic imaginary fields, this convergence from below was conjectured by Roberts to be explained by a secondary main term with negative coefficient.  Roberts’ conjecture was proved this year — twice!  Bhargava, Shankar, and Tsimerman gave a proof along the lines of Bhargava’s earlier work (involving thoughful decompositions of fundamental domains into manageable regions, and counting lattice points therein) and Thorne and Taniguchi have a proof along more analytic lines, using the Shintani zeta function.  Anyway, one might ask (prematurely, since I have no idea how to prove the main term correct!) whether the apparent convergence from below for the statistics of the λ-invariant is also explained by some kind of negative secondary term.
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Why was 6 afraid of 7?

Because 7 had exotic differentiable structures!

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How to ruin your Thanksgiving

Eat this week’s magnificent Ian’s special:  pizza with turkey, cranberries, green beans, and fried onions.  It’s very unlikely your family will serve you anything as good as this.

 

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The space of unknots and the space of unknotted ropes

Here’s something I didn’t know.  Suppose we consider the space K of “long knots” — embeddings of R into R^3, which send t to (t,0,0) whenever |t| > 1.  By closing up the large |t| ends of the arc, you get a knot in S^3.

The path component of K containing the embedding of R along the x-axis is the space of long unknots.  Hatcher proved, as a consequence of his proof of Smale’s conjecture, that the space of long unknots is contractible!  Hatcher also proved that the space of short unknots is homotopic to the Grassmannian of 2-planes in 4-space.  (I take it this follows from the contractibility of the space of long unknots, but didn’t think about it.)  Moreover, Hatcher proves in the unpublished “Topological Moduli Spaces of Knots” that every connected component of K is a K(\pi,1).  When the component corresponds to a torus knot, the fundamental group is Z: for a hyperbolic knot, it is ZxZ.

How do we get a circle in the space of long knots?  Hatcher makes a lovely “moving bead” argument on p.3 of the linked preprint.  Let S be a “short knot” of the given type, i.e. an embedding of S^1 into S^3.  For each point x on S, draw a small bead B around x; then the complement of B in S^3 looks like R^3, and the segment of the knot outside the bead is a long knot.  Now let the bead slide around the knot.  For each x, you get a long knot in the same isotopy class, and this gives a circle contained in the given connected component of K.

I was thinking about this because of Greg Buck‘s very interesting colloquium yesterday.  Greg is interested in the space of knots with positive thickness — what you might call the space of knotted ropes.  Let’s just think about the unknot.  For any knot K in R^3, we can define the radius of K to be the minimal distance between a point x on K and another point on K which lies on the plane through x perpendicular to the knot.  Let U_r be the space of unknots of radius r.  U_0 is just the space considered by Hatcher above, a Gr(2,4).  As we increase r, the space U_r  gets smaller — at some point, it vanishes entirely.  Buck presented a physical argument that U_r is disconnected for some intermediate values of r:  that is, he passed around a  thick closed rope which couldn’t be untangled, but whose meridian is an unknot.

Beyond that, how does the topology of U_r vary with r?  I have no idea — but what a beautiful question!  It is, of course, very reminiscent of questions about configurations of hard discs in a box discussed here earlier.

Buck’s talk centered on an energy functional \phi on the space of knots, which blows up when the knot gets very close to acquiring a self-intersection (i.e. when the radius gets small.)  You might think of \phi as measuring “distance from the boundary of moduli space.”  You might even hope that \phi would be some kind of Morse function on the components of the space of knots!  Indeed, Hatcher expresses a desire for exactly such a function, which would provide a retract of the infinite-dimensional space of knots of a given isotopy type to some finite-dimensional submanifold of minimal energy whose topology we could hope to understand.  At least in some simple cases, Buck’s energy seems to behave like such a function; he showed us magnificent movies of a crimped, tangled knot flowing along the energy gradient to a handsome, easy to grasp, maximal-radius representative of its isotopy class.  Great!

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Cabbage apples mustard, cheese strawberries pineapples

Something fast to eat on cold nights:

some shredded cabbage (in a bag is fine)

2-3 apples

mustard (we use spicy brown)

red wine vinegar

Procedure:  put some olive oil in a pan, chop the apples and fry them with a little pepper.  Throw the shredded cabbage on top of it and fry further.  Put in some mustard and stir.  Then toss a glug of red wine vinegar in the pan, cover, and let steam a few minutes.  Good with chopped up turkey kielbasa in it.

CJ asked me also to blog about his strawberry pineapple cheese pie.  (This is a layer of shredded cheese on a plate, with frozen strawberries and pineapples on top, microwaved until melted and thawed respectively.)

What does Republican state government mean for the University of Wisconsin?

Well, we have to talk about the election a little, right?

Governor-elect Scott Walker today, telling the regents of UW not to expect any increase in allocation from the state:

“It isn’t just always about more money. It’s going to be about finding ways to take the dollar we have, finding ways with flexibility, innovation and creativity, to apply those dollars in the best way possible to meet those goals.”

Jim Doyle has been in the Capitol ever since I moved to Wisconsin.  So I really have no sense of what a Republican governor, state senate, and state assembly means for UW-Madison and the UW system.  Will the university lose whole departmentsIs embryonic stem cell research at UW kaput?  Will Walker back Chancellor Martin’s plan to charge market-rate tuition?  (The Daily Cardinal says yes.)

Give predictions in comments.  Or tell me about your own state university system, and how it fares under Republican governance.

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How many people will vote on Facebook?

As I write this, about 4.7 million Americans have clicked the “I voted” button on Facebook.  As best I can tell, there are between 125 and 150 million Facebook accounts in the US, and about 90% of these belong to people of voting age.  Of course, lots of people have Facebook accounts but don’t use them.  Facebook says half of registered users log in every day.  So let’s say there’s a pool of about 60 million people who might be expected, if they voted, to click the Facebook button.  Turnout in the election is expected to be about 50% — though of course it will be lower among 18-24s, who are disproportionately represented on Facebook.  How many voters will Facebook tally by the end of polling?

Of my friends, 13% have voted.

 

 

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