Talk:Surgery obstruction map I (Ex)
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= 2 \langle L(TX\oplus\xi), [X] \rangle - \langle L(TX\oplus\xi\oplus\xi), [X] \rangle -1 | = 2 \langle L(TX\oplus\xi), [X] \rangle - \langle L(TX\oplus\xi\oplus\xi), [X] \rangle -1 | ||
= c \langle p_1(\xi)^2 ,[X] \rangle \ne 0, $$ | = c \langle p_1(\xi)^2 ,[X] \rangle \ne 0, $$ | ||
− | + | where the constant $c$ can be computed from the L-genus to be $-1/9$. | |
+ | |||
+ | So the surgery obstruction is not a group homomorphism with respect to the Whitney sum. | ||
+ | |||
</wikitex> | </wikitex> |
Revision as of 22:00, 29 May 2012
We take : 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 . The surgery obstruction of a normal map covered by equals
so it depends only on the bundle over . 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 . Now is the sum of and in with respect to the Whitney sum. 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.