This is an exercise 6.43 from Bott's Topology book.I'm not asking it's solution,but ask something relative to it.(adsbygoogle = window.adsbygoogle || []).push({});

Let [tex] \pi :E \to M\ [/tex] be an oriented rank 2 bundle.[tex] \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 }})} \ [/tex] is the explicit formula for the Thom class.As we know wedging with the Thom class is an isomorphism [tex] () \wedge \Phi :{H^*}(M) \to H_{cv}^{* + 2}(E)\ [/tex] .The problem is to find the class u on M such that

[tex] {\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} [/tex]

My confusion is: What does [tex] {\Phi ^2}\ [/tex]means?Does it mean [tex] {\phi ^2}=\phi \wedge \phi [/tex]? If it is yes,then this is my caculation [tex] \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 }}} \ [/tex] and [tex] {\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}\ [/tex],did I do it right?How can I find the class u?Could somebody give me hints?

Thank you!

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

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