# Talk:Surgery obstruction map I (Ex)


$\displaystyle \begin{array} {rl} \mathrm{sign}(M)-\mathrm{sign}(X) & = \langle L(TM), [M] \rangle - \langle L(TX),[X]\rangle \\ & = \langle L(\nu_M)^{-1}, [M] \rangle - \langle L(TX),[X]\rangle \\ & = \langle L(\overline{f}^*(\eta))^{-1}, f_*([X]) \rangle - \langle L(TX),[X]\rangle \\ & = \langle f^*L(\eta)^{-1}, f_*([X]) \rangle - \langle L(TX),[X]\rangle \\ & = \langle L(\eta)^{-1},[X]\rangle - \langle L(TX),[X]\rangle \end{array}$

by the Hirzebruch signature theorem and several properties of the $L$$L$-genus. In particular the surgery obstruction depends only on the bundle over $X$$X$. Now $(\xi\oplus\xi,\phi * \phi)$$(\xi\oplus\xi,\phi * \phi)$ is the sum of $(\xi,\phi)$$(\xi,\phi)$ and $(\xi,\phi)$$(\xi,\phi)$ in $\mathcal N (X)=[X,G/O]$$\mathcal N (X)=[X,G/O]$ with respect to the Whitney sum. Moreover

$\displaystyle \theta(-(\xi,\phi))+\theta(-(\xi,\phi)) - \theta(-(\xi\oplus\xi,\phi * \phi)) = 2 \langle L(TX\oplus\xi), [X] \rangle - \langle L(TX\oplus\xi\oplus\xi), [X] \rangle -\langle L(TX),[X]\rangle$

If this is non-zero, then the surgery obstruction is not a group homomorphism with respect to the Whitney sum.

As an example take $X=\mathbb H P^2$$X=\mathbb H P^2$:

There are fiber homotopically trivial bundles on $X$$X$ corresponding to classes in $[X,G/TOP]$$[X,G/TOP]$ which restrict to any given class in $[S^4,G/Top]$$[S^4,G/Top]$, as follows from the Puppe sequence with $\pi_7(G/TOP)=0$$\pi_7(G/TOP)=0$. From another exercise we know that on $S^4$$S^4$ we have such vector bundles with first Pontryagin class $48k$$48k$ times the generator of $H^4(S^4)$$H^4(S^4)$. This means that on $X$$X$ we have a vector bundle $\xi$$\xi$ with $p_1(\xi)=48$$p_1(\xi)=48$ whose sphere bundle is fiber homotopically trivial, by a fiber homotopy equivalence $\phi$$\phi$. We compute

$\displaystyle 2 \langle L(TX\oplus\xi), [X] \rangle - \langle L(TX\oplus\xi\oplus\xi), [X] \rangle -\langle L(TX),[X]\rangle = c \langle p_1(\xi)^2 ,[X] \rangle \ne 0,$

where the constant $c$$c$ can be computed from the L-genus to be $-1/9$$-1/9$.

So the surgery obstruction is not a group homomorphism with respect to the Whitney sum.