User oliver nash - MathOverflow

What is the ring structure of the complex topological K-theory of a non-singular complex quadric?

2013-05-18T16:20:33Z

I would like to know the ring structure of $K(Q_n)$ explicitly where $Q_n \subset \mathbb{P}^{n+1}$ is the non-singular $n$-dimensional complex quadric and $K(Q_n) = K^0(Q_n)$ is the complex topological $K$-theory of $Q_n$ (with analytic topology). What is it? Can anyone provide a reference?

Ideally, I would like an expression for $K(Q_n)$ with generators for which I know how to compute the Chern character. Also, as it happens, I am most interested in the case where $n$ is odd.

To my great surprise, I have not been able to find this in the literature (with one qualification, see below). I am certain that such a basic calculation must be well-known to experts but I have been unable to find it. Hence this question.

While I have been aware of the basics of $K$-theory for years, this is the first time I have really had to work with it so I am very inexpert. In the last week I grokked relevant-seeming chunks of the books of Karoubi, Atiyah, Hatcher but I'm still quite green.

Motivation
My interest in this ring arose in the course of a problem I have been thinking about but I am hoping that the fact that $K(Q_n)$ is such a basic object will be sufficient motivation to justify this question.

What I do know

Since the quadric has a cell decomposition with only even-dimensional cells, $K^1$ vanishes and $K = K^0$ is free Abelian with rank equal to the number of cells (and generators corresponding to the cells). For $n$ even this rank is $n+2$ (because there is an extra cell in middle dimension) for $n$ odd, it is $n+1$.
The ordinary cohomology $H^* = H^{\rm even}$ is of course free Abelian of the same rank (Lefshetz tells us the restriction map from $H^*(\mathbb{P}^{n+1}, \mathbb{Z})$ is an isomorphism in all dimensions below $2n+2$ except middle dimension for $n$ even). The Chern character thus embeds $K$ as a maximal-rank lattice inside $H^*(Q_n, \mathbb{Q})$ . However it is not the same as the lattice $H^*(Q_n, \mathbb{Z})$ .
Since we know $H^*(Q_n, \mathbb{Q})$ as a ring, it might be satisfactory to know the images of the Chern character on a set of generators of $K(Q_n)$.
The cases $n=1, 2, 4$ are easy since $Q_n$ is respectively $S^2$, $S^2\times S^2$, $G(2, 4)$ (the complex Grassmannian).
$Q_n$ is diffeomorphic to the real oriented Grassmannian $\tilde G(2, n+2)$ and so is a homogeneous space $SO(n+2)/SO(2)\times SO(n)$. There are tools for calculating $K$-theory for homogeneous spaces pioneered by Atiyah and Hirzebruch (I believe). Subsequently Hodgkin introduced a spectral sequence which seems to allow relatively straightforward (if lengthy) calculation in many cases, including $Q_n$.
I managed to find a paper where the above technique is apparently used to calculate $K(\tilde G(k, n))$ for general $k, n$: Sankaran, Zvengrowski "K-theory of Oriented Grassmann Manifolds", Math. Slovaca 47(3). It looks right though I would probably be tempted to work from first principles myself than to specialize their results to my $k=2$ case.

Bottom line
Surely I am missing the obvious here? I find it astonishing that I should need to use the methods of Atiyah-Hirzebruch-Hodgkin for such a simple space. Perhaps if I thought more carefully about $\mathbb{P}^{n+1}/Q_n$ or $Q_{n+1}/Q_n$ (bearing in mind natural cell decompositions) then I could use the exact sequences either for the pairs $Q_n \subset \mathbb{P}^{n+1}$ or $Q_{n} \subset Q_{n+1}$ to work this out?

I am tempted to believe the reason I cannot find this in the literature is that it is so trivial. What am I missing?

Answer by Oliver Nash for Rotation in Hyperkähler manifolds

2012-11-07T21:46:49Z

This is really just a comment but it doesn't quite fit so I'll make it an answer.

It seems worth commenting on the difference between the infinitesimal and local/global versions of this question.

Let's say we're in real dimension $4n$. Infinitesimally, i.e., on the tangent space at any point, orthogonal complex structures compatible with the orientation are parameterised by $SO(4n)/U(2n)$ (where the compatibility with orientation is to require the Pfaffian be positive). Since $\dim(SO(4n)/U(2n)) = 2n(2n-1) > 2$ for $n > 1$ it is clear that the property in the question fails infinitesimally for $n > 1$. The claim could thus only hold locally (or indeed globally) if the condition of being integrable somehow miraculously restricted to the hyperkahler $S^2 \subset SO(4n)/U(n)$ at each point, which is false. The easiest counter examples are any complex structure on $\mathbb{H}^n$ not in the standard hyperkahler family as in Paul Reynolds's very helpful comment.

In dimension $4$ things are slightly more interesting since $SO(4)/U(2) \simeq S^2$ so infinitesmally in 4 dimensions all the relevant complex structures are a linear combination of any hyperkahler triple. However the coefficients $a_i$ are not constant in general. To settle the matter we must therefore exhibit an integrable complex structure for which $a_i$ are non-constant. I confess I cannot think of a trivial example off the top of my head but at first glance this paper appears to discuss the matter, at least locally. (Certainly they exist.)

Answer by Oliver Nash for 2D Problems Which are Easier to Solve in 3D

2012-09-10T15:39:59Z

A not-so-serious answer; hopefully what it lacks in depth it makes up for by being elementary.

Suppose we forget Pythagoras's theorem and define a binary operation on positive reals by sending $(a, b)$ to the length of the hypotenuse of the right-angled triangle with side lengths $a, b$ forming the right angle.

The associativity of this operation is trivial in three dimensions but not so in two.

I came across this here: D. Bell, "Associative Binary Operations and the Pythagorean Theorem", The Mathematical Intelligencer, Vol. 33, No. 1 (2011), 92-95, DOI: 10.1007/s00283-010-9171-6 http://www.springerlink.com/content/r8t12847357j1ln7/

Apparently it is also mentioned here: L. Berrone, "The Associativity of the Pythagorean Law", The American Mathematical Monthly, Vol. 116, No. 10, Dec., 2009 http://www.jstor.org/discover/10.2307/40391255?uid=3738232&uid=2129&uid=2&uid=70&uid=4&sid=21101035566283

Answer by Oliver Nash for Do there exist modern expositions of Klein's Icosahedron?

2012-02-05T19:55:51Z

I got interested in this subject last year and just got round to writing up some notes which I hope may be of use.

I also have a python script which implements the Klein's icosahedral solution of the quintic linked from this page.

Answer by Oliver Nash for Magic trick based on deep mathematics

2010-06-14T15:25:22Z

I hope this is contribution is appropriate; I think that a nice puzzle based on Hamming codes discussed a little here: http://ocfnash.wordpress.com/2009/10/31/yet-another-prisoner-puzzle/

is the following:

A room contains a normal 8×8 chess board together with 64 identical coins, each with one “heads” side and one “tails” side. Two prisoners are at the mercy of a typically eccentric jailer who has decided to play a game with them for their freedom. The rules are the game are as follows.

The jailer will take one of the prisoners (let us call him the “first” prisoner) with him into the aforementioned room, leaving the second prisoner outside. Inside the room the jailer will place exactly one coin on each square of the chess board, choosing to show heads or tails as he sees fit (e.g. randomly). Having done this he will then choose one square of the chess board and declare to the first prisoner that this is the “magic” square. The first prisoner must then turn over exactly one of the coins and exit the room. After the first prisoner has left the room, the second prisoner is admitted. The jailer will ask him to identify the magic square. If he is able to do this, both prisoners will be granted their freedom

Comment by Oliver Nash

2013-06-09T22:17:56Z

It's a long time since I read it but if I remember correctly, I think you might find the following paper useful (unless you've already read it): Hilton, P., Pedersen, J., "Catalan numbers, their generalization, and their uses" Math. Intelligencer 13(2) (1991). (Incidentally Google seems to be able to find a version outside a paywall.)

Comment by Oliver Nash

2013-05-19T11:40:42Z

Thanks Dylan, all three of these are very helpful remarks. If I don't receive an answer by tomorrow I think I'll just bash it out myself, most likely using Atiyah-Hirzebruch as you suggest in (iii).

Comment by Oliver Nash

2013-04-10T18:16:50Z

It is maybe worth mentioning that for projective varieties, the full tensor product corresponds to the JOIN of X with another copy of itself after appropriate embeddings. Eisenbud's "Commutative Algebra: With a View Toward Algebraic Geometry" has a helpful discussion on page 304: books.google.ie/…

Comment by Oliver Nash

2012-11-27T10:02:07Z

@Igor, I don't think so. For $k > 1$, $H^2(S^{2k}) = 0$.

Comment by Oliver Nash

2012-11-11T10:18:00Z

This isn't exactly physical or geometric but I've always liked the interpretation of $1/\zeta(2)$ as the (limiting) probability that two independent, uniform random natural numbers are coprime. And indeed likewise for $k > 2$ different natural numbers.

Comment by Oliver Nash

2012-07-02T10:46:49Z

The notes I peddled in this answer: mathoverflow.net/questions/9474/… may also be of interest and hopefully are less a painful read.

Comment by Oliver Nash

2012-04-17T10:39:46Z

Indeed a PDF of the paper is available here: maths.tcd.ie/~onash/pity_the_prisoners_files/…

Comment by Oliver Nash

2012-02-15T13:18:56Z

I realize you are merely sketching things in this very helpful answer but one minor point seems worth adding. The equations of motion of the classical Yang-Mills action you mention are in fact $d_\omega \star F_\omega = 0$. A connection solving the (anti-)self-dual equations $F_\omega = \pm \star F_\omega$ solves $d\star F_\omega = 0$ by the Bianchi identity. However such connections are just the (anti-)self-dual solutions and correspond to global minima of the action. The action can have other critical points which are not global minima.