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Here is a topological question which seems quite elementary. The answer to this question may be useful e.g. in estimating the orders of the automorphism groups of some algebraic varieties and in computing the cohomology of some moduli spaces.

Let $K$ be a compact Lie group, $T$ a maximal torus of $K$ and $V$ a complex vector space of dimension $d$. (If one wishes, instead of $K$ and $T$ one can consider a complex reductive group and its Borel subgroup.) From a representation $R:T\to GL(V)$ we can construct a homogeneous vector bundle with total space

$$E=K\times_T V.$$

(I.e., we identify $(kt,v)$ with $(k,R(t)v$ for all $k\in K, t\in T, v\in V$.)

Set $E_0$ to be $E$ minus the zero section. This space is fibered over $K/T$ with fiber $V$ minus the origin.

Suppose the Euler (=top Chern) class of the vector bundle is zero. Then $H^{2d-1}(E_0,\mathbf{Z})\cong \mathbf{Z}$, because $K/T$ has only even-dimensional cells. Take a generator $a$ of $H^{2d-1}(E_0,\mathbf{Z})$. We can identify the integral cohomology of $E_0$ with $H^*(K/T,\mathbf{Z})\otimes\Lambda(a)$ where $\Lambda$ stands for exterior $\mathbf{Z}$-algebra.

We have the natural group action map $K\times E_0\to E_0$. The cohomology of $E_0$ has no torsion, so we can identify $H^*(K\times E_0,\mathbf{Z})$ with $H^*(K,\mathbf{Z})\otimes H^*(E_0,\mathbf{Z})$ using the projections. The latter can be written as $$H^*(K,\mathbf{Z})\otimes H^*(K/T,\mathbf{Z})\otimes\Lambda(a).$$

The pullback of $a$ under the action map is $1\otimes a+x\otimes 1$ for some $x\in H^{*}(K,\mathbf{Z})\otimes H^{*}(K/T,\mathbf{Z})$. Question: find $x$. The weights of the representation are assumed to be known; for simplicity one can assume that $K=U(n)$ or $SU(n)$.

For real cohomology I know how to reduce the problem to linear algebra using Lie algebra cohomology. But the result I am able to get in this way is not very illuminating. And moreover, I have no idea how to extract the integral structure out of it.

For projectivized (and not spherized) bundles, the question becomes trivial. But this does not seem to help.

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up vote 5 down vote accepted

I'll answer in the case of K = U(n) because I'm less familiar with the cohomology of the other compact Lie groups. Let BK and BT be the classifying spaces.

The representation of T on V gives rise to a vector bundle on BT with total space F, and complement of the zero section F0. There is an Euler class c in H*(BT). As H*(BT) is a polynomial algebra in these cases, there are basically two cases: either the Euler class is zero, or it is not a zero divisor. There is a map from K/T to BT which is, up to homotopy, the "quotient" by the action of K.

In the first case, you will actually find that your element 'a' in H*(E0) is pulled back from H*(F0), where the action of K is trivial.

In the second case, you have an exact sequence (the Gysin sequence) $$ 0 \to H^\*(BT) {\mathop\to^c} H^\*(BT) \to H^\*(F_0) \to 0 $$ of modules over H*(BK), which is actually a free resolution.

I find it easier to describe the "coaction" you want in terms of an action on homology. The space E0 is the pullback of F0 along the map from K/T to BT. An Eilenberg-Moore argument finds that there are isomorphisms $$ H\_\*(K) \cong Ext_{H^\*(BK)} (\mathbb Z, \mathbb Z) $$


$$ H\_*(E\_0) \cong Ext\_{H^\*(BK)} (H^\*(F\_0), \mathbb Z). $$

(I have to be a little careful about the gradings; these Ext-groups are bigraded according to a grading on the ring and a grading on the Ext-group, and the elements in Ext^s actually have their grading shifted down by s.)

The coaction on homology you describe is actually just the Yoneda pairing on Ext. For a specific choice of Euler class this is something that you can work out by writing down a resolution of Z over H*(BK) and using that to find generators for your Ext-algebra.

I don't know if an example would be more helpful.

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Thanks, Tyler! There is one thing though that is not clear to me. A spectral sequence gives the (co)homology only up to a filtration. How does one prove that the Yoneda pairing is indeed the homology of $K$ acting on the homology of $E_0$, and not just one Eilenberg-Moore spectral sequence acting on the other one? E. g. the Leray spectral sequences corresponding to the maps to $K/T$ fail to detect the non-trivial part of the (co)action. – algori Nov 19 '09 at 5:57
Absolutely this is usually only a statement up to filtration/extension problems, but in this case the spectral sequence being acted on has no indeterminacy whatsoever. When K is U(n), the Eilenberg-Moore spectral sequence for the homology of $E_0$ is concentrated in $Ext^0$ and $Ext^1$, with no possible additive extensions, and the product of elements in $Ext^r$ and $Ext^s$ is in $Ext^{r+s}$ up to indeterminacy from higher filtrations - so the Yoneda product really tells the whole story. For more general Lie groups the cohomology of T may be a more complicated module over that of K. – Tyler Lawson Nov 19 '09 at 14:02

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