Talk:Surgery obstruction map I (Ex)
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.