- #1
kakarotyjn
- 95
- 0
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 [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!
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!
