5
$\begingroup$

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 $H^{\bullet}(BG,\mathbf{Q})\to H^{\bullet}(BH,\mathbf{Q})$.

$\endgroup$

1 Answer 1

3
$\begingroup$

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 $H^{2k}(BG)$.

$\endgroup$
4
  • $\begingroup$ 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? $\endgroup$
    – algori
    Jan 2, 2010 at 18:22
  • 1
    $\begingroup$ Right, but "the" corresponding element, as I usually understand it, in the cohomology of BG is not well-defined - it's only unique up to the indeterminacy from the images of differentials in the spectral sequence associated to G -> EG -> BG. Did you have some specific corresponding element in mind? $\endgroup$ Jan 2, 2010 at 19:57
  • $\begingroup$ 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 $G$ $\endgroup$
    – algori
    Jan 2, 2010 at 21:55
  • $\begingroup$ 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. $\endgroup$
    – algori
    Jan 3, 2010 at 0:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.