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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.

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?

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Three approaches: (i) Adams computed the $K$ theory of real projective spaces a while ago, and there is a Serre-type spectral sequence for any cohomology theory (take ordinary cohomology of the total space with coefficients in the $K$-theory of the fiber, in this case), so you can write down some fiber sequences to try and compute the $K$ theory of unoriented Grassmannian $G(2, n)$, and then go after the oriented one. (ii) You could use the fact that we already know what the $K$-theory of $BSO(2)$ is, and try to understand the difference between this and the approximations by Grassmannians, –  Dylan Wilson May 18 '13 at 18:03
(iii) There's always the Atiyah-Hirzebruch SS. –  Dylan Wilson May 18 '13 at 18:03
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). –  Oliver Nash May 19 '13 at 11:40
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1 Answer

up vote 2 down vote accepted

I guess I might as well answer my own question as it might help somebody in the future. I ended up working this out using the methods of Hodgkin. (The Atiyah-Hirzebruch SS leaves one with a series of extension problems and so only gives the Abelian group structure.)

In fact I have written up a careful proof of this in the appendix to this paper: http://arxiv.org/abs/1308.0949 so I will just give the statement here and refer to the paper for the proof (especially as the proof, though fairly routine, is quite long).

Hodgkin's results allow one to compute the K-theory of homogeneous spaces $G/H$ when $\pi_1(G)$ is torsion-free. For this reason we represent the quadric $Q$ as: $$Q = \frac{Spin(n+2)}{Spin^c(n)}$$ where $Spin^c(n)$ is the double cover of $SO(2)\times SO(n) \subset SO(n+2)$ in $Spin(n+2)$.

Given this, representations of $Spin^c(n)$ give vector bundles on $Q$ and hence classes in $K(Q)$. There is one representation (and hence class in $K(Q)$) which I wish to highlight. To define it, we consider the double cover $Spin(2)\times Spin(n)$ of $Spin^c(n)$. Now if $RSO(2) \simeq \mathbb{Z}[t, t^{-1}]$ is the representation ring of $SO(2)$ then $RSpin(2) \simeq \mathbb{Z}[t^{1/2}, t^{-1/2}]$. Also, for $n$ odd, $Spin(n)$ has the unique spin representation $\delta$ (of dimension $2^{(n-1)/2}$). Neither $t^{-1/2}$ nor $\delta$ descends to a representation of $Spin^c(n)$ but their product does. We let $X$ be the bundle on $Q$ associated to the representation $t^{-1/2}\delta$ of $Spin^c(n)$. With this defined, we can state:

Let $Q \subset \mathbb{P}^{n+1}$ be an $n$-dimensional non-singular quadric ($n \ge 3$, odd). Let $L = \mathcal{O}(1)-1 \in K(Q)$ and let $X$ be the bundle defined above, then:

  • $1, L, L^2, \cdots, L^{n-1}, X$ are a $\mathbb{Z}$-basis for the torsion-free ring $K(Q)$
  • $L^{n+1} = 0$
  • $LX = 2^{(n+1)/2} - 2X$
  • $2^{(n+1)/2}X = 2^n - 2^{n-1}L + \cdots + 2L^{n-1} - L^n$ (this is equivalent to the previous bullet but shows why we need $X$ instead of $L^n$)

There is also a slightly-complicated formula for $X^2$ which I will suppress and a similar statement for $n$ even except that now there are two bundles $X^+$, $X^-$.

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