# How to find the class in H^*(M)?

1. Oct 14, 2011

### kakarotyjn

This is an exercise 6.43 from Bott's Topology book.I'm not asking it's solution,but ask something relative to it.

Let $$\pi :E \to M\$$ be an oriented rank 2 bundle.$$\Phi = d(\rho (r)\frac{{d{\theta _\alpha }}}{{2\pi }}) + \frac{1}{{2\pi i}}d(\rho (r){\pi ^*}\sum\limits_\gamma {{\rho _\gamma }d\log {g_{\gamma \alpha }})} \$$ is the explicit formula for the Thom class.As we know wedging with the Thom class is an isomorphism $$() \wedge \Phi :{H^*}(M) \to H_{cv}^{* + 2}(E)\$$ .The problem is to find the class u on M such that
$${\Phi ^2} = \Phi \wedge {\pi ^*}u{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} in{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} H_{cv}^*(E){\kern 1pt}$$

My confusion is: What does $${\Phi ^2}\$$means?Does it mean $${\phi ^2}=\phi \wedge \phi$$? If it is yes,then this is my caculation $$\Phi = d\rho (r)\frac{{d{\theta _\alpha }}}{{2\pi }} + \frac{1}{{2\pi i}}d\rho (r){\pi ^*}\sum\limits_\gamma {{\rho _\gamma }d\log {g_{\gamma \alpha }}} + \frac{1}{{2\pi i}}\rho (r){\pi ^*}\sum\limits_\gamma {d{\rho _\gamma }d\log {g_{\gamma \alpha }}} \$$ and $${\Phi ^2} = 2d\rho (r)\frac{{d{\theta _\alpha }}}{{2\pi }}\frac{1}{{2\pi i}}\rho (r)({\pi ^*}\sum\limits_\gamma {d{\rho _\gamma }d\log {g_{\gamma \alpha }})} + 2\frac{1}{{2\pi i}}d\rho (r)({\pi ^*}\sum\limits_\gamma {{\rho _\gamma }d\log {g_{\gamma \alpha }}} )\frac{1}{{2\pi i}}\rho (r)({\pi ^*}\sum\limits_\gamma {d{\rho _\gamma }d\log {g_{\gamma \alpha }})} + {(\frac{1}{{2\pi i}}\rho (r){\pi ^*}\sum\limits_\gamma {d{\rho _\gamma }d\log {g_{\gamma \alpha }}} )^2}\$$,did I do it right?How can I find the class u?Could somebody give me hints?

Thank you!

2. Oct 14, 2011

### mathwonk

Loring Tu wrote that book with Bott, and I bet he would be happy to answer this if you email him at Tufts.

3. Oct 15, 2011

### kakarotyjn

Oh,Thank you!That was a great hint!