Talk:Surgery obstruction map I (Ex)
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If $X$ is a manifold, then the normal map $id_X$ gives the base point of $\mathcal N (X)$. | If $X$ is a manifold, then the normal map $id_X$ gives the base point of $\mathcal N (X)$. | ||
An element of $[X,G/TOP]$ is given by a bundle $\xi$ together with a fiber homotopy trivialization $\phi$. | An element of $[X,G/TOP]$ is given by a bundle $\xi$ together with a fiber homotopy trivialization $\phi$. | ||
− | Under the isomorphism $\mathcal N (X) \cong [X,G/TOP]$, the pair $(\xi,\phi)$ corresponds to a normal map $M\to X$ covered by $\nu_M\to \nu_X\oplus \xi$. | + | Under the isomorphism $\mathcal N (X) \cong [X,G/TOP]$, the pair $(\xi,\phi)$ corresponds to a normal map $f:M\to X$ covered by $\overline{f}:\nu_M\to \nu_X\oplus \xi$. |
Assume that the dimension is $4k$ and that $X$ is simply connected. Then the surgery obstruction of a normal map $M\to X$ covered by $\nu_M\to \eta$ equals | Assume that the dimension is $4k$ and that $X$ is simply connected. Then the surgery obstruction of a normal map $M\to X$ covered by $\nu_M\to \eta$ equals | ||
− | + | \begin{align*} | |
− | + | \mathrm{sign}(M)-\mathrm{sign}(X) &= \langle L(TM), [M] \rangle - \langle L(TX),[X]\rangle \\ | |
+ | &= \langle L(\nu_M)^{-1}, [M] \rangle - \langle L(TX),[X]\rangle \\ | ||
+ | &= \langle \overline{f}^*(L(\eta))^{-1}, f_*([X]) \rangle \langle L(TX),[X]\rangle \\ | ||
+ | &= \langle L(\eta)^{-1},[X]\rangle - \langle L(TX),[X]\rangle, | ||
+ | \end{align} | ||
+ | by the Hirzebruch signature theorem and severeal property of the $L$-genus. In particular the durgery obstruction depends only on the bundle over $X$. | ||
Now $(\xi\oplus\xi,\phi * \phi)$ is the sum of $(\xi,\phi)$ and $(\xi,\phi)$ in $\mathcal N (X)=[X,G/O]$ with respect to the Whitney sum. | Now $(\xi\oplus\xi,\phi * \phi)$ is the sum of $(\xi,\phi)$ and $(\xi,\phi)$ in $\mathcal N (X)=[X,G/O]$ with respect to the Whitney sum. | ||
Moreover | Moreover |
Revision as of 18:11, 31 May 2012
If is a manifold, then the normal map gives the base point of . An element of is given by a bundle together with a fiber homotopy trivialization . Under the isomorphism , the pair corresponds to a normal map covered by . Assume that the dimension is and that is simply connected. Then the surgery obstruction of a normal map covered by equals \begin{align*}
\mathrm{sign}(M)-\mathrm{sign}(X) &= \langle L(TM), [M] \rangle - \langle L(TX),[X]\rangle \\ &= \langle L(\nu_M)^{-1}, [M] \rangle - \langle L(TX),[X]\rangle \\ &= \langle \overline{f}^*(L(\eta))^{-1}, f_*([X]) \rangle \langle L(TX),[X]\rangle \\ &= \langle L(\eta)^{-1},[X]\rangle - \langle L(TX),[X]\rangle,
\end{align} by the Hirzebruch signature theorem and severeal property of the -genus. In particular the durgery obstruction depends only on the bundle over . Now is the sum of and in with respect to the Whitney sum. Moreover
If this is non-zero, then the surgery obstruction is not a group homomorphism with respect to the Whitney sum.
As an example take :
There are fiber homotopically trivial bundles on corresponding to classes in which restrict to any given class in , as follows from the Puppe sequence with . From another exercise we know that on we have such vector bundles with first Pontryagin class times the generator of . This means that on we have a vector bundle with whose sphere bundle is fiber homotopically trivial, by a fiber homotopy equivalence . We compute
where the constant can be computed from the L-genus to be .
So the surgery obstruction is not a group homomorphism with respect to the Whitney sum.