Killing Chern classes - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T02:53:44Z http://mathoverflow.net/feeds/question/10474 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/10474/killing-chern-classes Killing Chern classes algori 2010-01-02T06:57:34Z 2010-01-03T00:25:34Z <p>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 $E'\to BG$.</p> <p>The class $a$ gives an element $a'\in E^{0,2k-1}_2$ for some $k$. Assume that $d_i(a')=0,i&lt; 2k$. 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 $b$?</p> <p>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 $H^{{\bullet}}(E',\mathbf{Q})\cong H^{{\bullet}}(B,\mathbf{Q})$.</p> <p>This is probably standard, but for some reason I don't see how to prove it nor can construct a counter-example off hand.</p> <p>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</p> <p>$$(E\times EG)/H\to (E\times EG)/G$$</p> <p>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 $H^{\bullet}(BG,\mathbf{Q})\to H^{\bullet}(BH,\mathbf{Q})$.</p> http://mathoverflow.net/questions/10474/killing-chern-classes/10497#10497 Answer by Tyler Lawson for Killing Chern classes Tyler Lawson 2010-01-02T17:43:36Z 2010-01-02T17:43:36Z <p>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 <code>$E_{2k}$</code>-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 <code>$d_{2k}$</code>-differential on this class in $H^{2k-1}(G)$ is the "corresponding generator" in $H^{2k}(BG)$.</p>