Talk:Surgery obstruction map I (Ex)

Revision as of 23:07, 29 May 2012

If $<wikitex>; 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 $\phi$. Under the isomorphism $\mathcal N (X) \cong [X,G/TOP]$, the pair $(\xi,\phi)$ corresponds to a normal map $M\to X$ covered by $\nu_M\to \nu_X\oplus \xi$. The surgery obstruction of a normal map $M\to X$ covered by $\nu_M\to \eta$ equals $$ sign(M)-sign(X)=\langle L(-\eta),[X]\rangle - \langle L(TX),[X]\rangle,$$ so it 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. Moreover $$\theta(-(\xi,-\phi))+\theta(-(\xi,\phi)) - \theta(-(\xi\oplus\xi,\phi * \phi)) = 2 \langle L(TX\oplus\xi), [X] \rangle - \langle L(TX\oplus\xi\oplus\xi), [X] \rangle -\langle L(TX),[X]\rangle $$ If this is non-zero, then the surgery obstruction is not a group homomorphism with respect to the Whitney sum. As an example take $X=\mathbb H P^2$: There are fiber homotopically trivial bundles on $X$ corresponding to classes in $[X,G/TOP]$ which restrict to any given class in $[S^4,G/Top]$, as follows from the Puppe sequence with $\pi_7(G/TOP)=0$. From [[Fibre_homotopy_trivial_bundles_(Ex)|another exercise]] we know that on $S^4$ we have such vector bundles with first Pontryagin class k$ times the generator of $H^4(S^4)$. This means that on $X$ we have a vector bundle $\xi$ with $p_1(\xi)=48$ whose sphere bundle is fiber homotopically trivial, by a fiber homotopy equivalence $\phi$. We compute $$ 2 \langle L(TX\oplus\xi), [X] \rangle - \langle L(TX\oplus\xi\oplus\xi), [X] \rangle -\langle L(TX),[X]\rangle = c \langle p_1(\xi)^2 ,[X] \rangle \ne 0, $$ where the constant $c$ 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>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 $\phi$. Under the isomorphism $\mathcal N (X) \cong [X,G/TOP]$, the pair $(\xi,\phi)$ corresponds to a normal map $M\to X$ covered by $\nu_M\to \nu_X\oplus \xi$. The surgery obstruction of a normal map $M\to X$ covered by $\nu_M\to \eta$ equals

$$\displaystyle sign(M)-sign(X)=\langle L(-\eta),[X]\rangle - \langle L(TX),[X]\rangle,$$

so it 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. 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 $X=\mathbb H P^2$:

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