# Killing Chern classes

Let $G$ be a compact connected Lie group and let $E\to B$ be a principal $G$-bundle. Suppose $a$ is a rational cohomology class of $E$ such that its pullback $b$ under an orbit inclusion map $G\to E$ is one of the standard multiplicative generators of $H^{{\bullet}}(G,\mathbf{Q})$ . Let $E'=E\times EG/G$ be the Borel construction (corresponding to the action of $G$ on $E$) and let $(E^{pq}_r,d_r)$ be the Leray spectral sequence corresponding to the fiber bundle $E'\to BG$.

The class $a$ gives an element $a'\in E^{0,2k-1}_2$ for some $k$. Assume that $d_i(a')=0,i< 2k$. Is it true that $d_{2k}(a')$ is what has remained in $E_{2k}$ of the multiplicative generator of $H^{{\bullet}}(BG,\mathbf{Q})$ corresponding to $b$?

For simplicity one can assume $G=U(n)$, in which case what remains in $E_\infty$ of the generator of $H^{{{\bullet}}}(BG,\mathbf{Q})\cong E^{{\bullet},0}_2$ corresponding to $b$ is precisely the $k$-th Chern class of $E$, under the natural isomorphism $H^{{\bullet}}(E',\mathbf{Q})\cong H^{{\bullet}}(B,\mathbf{Q})$.

This is probably standard, but for some reason I don't see how to prove it nor can construct a counter-example off hand.

upd: here is a weaker version, which would be easier to (dis)prove: take $G=U(n)\times H$ where $H$ is another Lie group and suppose that the pullback of $a$ to $G$ is the canonical generator of $H^{\bullet}(U(n),\mathbf{Q})\subset H^{\bullet}(G,\mathbf{Q})$ in degree $2k-1$. Is it true that $d_{2k}(a')$ is mapped to zero under the mapping of the spectral sequences induced by the pullback of $E'$ to $BH$ i.e. by the map

$$(E\times EG)/H\to (E\times EG)/G$$

To prove this it would suffice, of course, to show that $d_{2k}(a')$ is represented in $E_2$ by a class in $$H^{\bullet}(BG,\mathbf{Q})\cong H^{\bullet}(BU(n),\mathbf{Q})\otimes H^{\bullet}(H,\mathbf{Q})$$ that is mapped to zero under $H^{\bullet}(BG,\mathbf{Q})\to H^{\bullet}(BH,\mathbf{Q})$.

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The inclusion $G \to E$ induces a map $EG \to E'$ of spaces over $BG$, where the map of fibers is $G \to E$, and there is a map backwards of Serre spectral sequences. Because $a$ lifts a standard generator of the cohomology of $G$, and $a'$ is a cycle up to the $E_{2k}$-page, the differential of this element maps to the differential of the standard generator in the spectral sequence $H^p(BG;H^q(G)) \Rightarrow H^{p+q}(*)$. The $d_{2k}$-differential on this class in $H^{2k-1}(G)$ is the "corresponding generator" in $H^{2k}(BG)$.
Thanks, Tyler! The map $EG\to E'$ induces an isomorphism of the $q=0$ rows of thet $E_2$ pages of the spectral sequences: both are just the $H^*(BG,\mathbf{Q})$. Why is the map of the $E_{2k}$ pages injective on the zero row? (If it isn't, then we can't deduce the image of $a'$ in $E^{2k,0}_{2k}$ from its image in the spectral sequence of $EG\to BG$.) Or did you mean something else? –  algori Jan 2 '10 at 18:22
For $G$ a product of $U(n)$'s (the case I'm mainly interested in) it is possible to fix the generators of $H^*(BG,\mathbf{Q})$ as follows: for a single $U(n)$ the generator in degree $2l$ is the $l$-th Chern class of the tautological bundle. These are well defined if the Chern classes are required to satisfy the usual axioms (the only ambiguity is the sign of the $c_1$ of the tautological bundle on $\mathbf{P}^1(\mathbf{C})$). Then, if we have a product of several $U(n)$'s, we take the tensor products of the generators constructed above. There should be something similar should for general $G$ –  algori Jan 2 '10 at 21:55
Actually, I don't have to be that rigid about the generators. I've updated the posting including a weaker version of the problem. Very roughly speaking, it says that if $G$ is a product and $a$ comes from one factor, then its differential is zero restricted to $B$ of the other factor. –  algori Jan 3 '10 at 0:31