Talk:Surgery obstruction map I (Ex)

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(Created page with "<wikitex>; We take $X=\mathbb H P^2$ and consider various bundle reductions of the normal bundle: The normal map $id_X$ gives the base point of $\mathcal N (X)$. The surgery o...")
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<wikitex>;
<wikitex>;
We take $X=\mathbb H P^2$ and consider various bundle reductions of the normal bundle:
We take $X=\mathbb H P^2$ and consider various bundle reductions of the normal bundle:
The normal map $id_X$ gives the base point of $\mathcal N (X)$.
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The normal map $id_X$ gives the base point of $\mathcal N (X)\cong [X,G/TOP]$.
The surgery obstruction of a normal map $M\to X$ covered by $\nu_M\to \xi$ equals
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An element of $[X,G/TOP]$ is given by a bundle $\xi$ together with a fiber homotopy trivialization $\phi$.
$$ sign(M)-sign(X)=\langle L(-\xi),[X]\rangle -1,$$
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Under the isomorphism $\mathcal N (X) \cong [X,G/O]$, the pair $(\xi,\phi)$ corresponds to a normal map $M\to X$ covered by $\nu_M\to \nu_X\oplus \xi$.
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The surgery obstruction of a normal map $M\to X$ covered by $\nu_M\to \eta$ equals
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$$ sign(M)-sign(X)=\langle L(-\eta),[X]\rangle -1,$$
so it depends only on the bundle over $X$.
so it depends only on the bundle over $X$.
There are fiber homotopically trivial bundles on $X$ corresponding to classes in $[X,G/TOP]$
There are fiber homotopically trivial bundles on $X$ corresponding to classes in $[X,G/TOP]$
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collapses.
collapses.
From [[Fibre_homotopy_trivial_bundles_(Ex)|another exercise]] we know that on $S^4$ we have such vector bundles with first Pontryagin class $48k$ times the generator of $H^4(S^4)$.
From [[Fibre_homotopy_trivial_bundles_(Ex)|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 vector bundles $\xi_1,\xi_2$ whose sphere bundles are fiber homotopically trivial, by fiber homotopy equivalences $\phi_i$. Then $(\xi_1\oplus\xi_2,\phi_i * \phi_2)$ is the sum of $(\xi_1,\phi_1)$ and $(\xi_2,\phi_2)$ in
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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 $\phi$.
$\mathcal N (X)=[X,G/O]$ with respect to the Whitney sum.
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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 we compute
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We compute
$$\sigma(\xi_1,\phi_1)+\sigma(\xi_2,\phi_2) - \sigma(\xi_1\oplus\xi_2,\phi_i * \phi_2)
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$$\sigma(-(\xi,-\phi))+\sigma(-(\xi,\phi)) - \sigma(-(\xi\oplus\xi,\phi * \phi))
= \langle L(TX\oplus\xi_1), [X] \rangle + \langle L(TX\oplus\xi_2), [X] \rangle - \langle L(TX\oplus\xi_1\oplus\xi_2), [X] \rangle -\langle L(TX), [X] \rangle. $$
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= 2 \langle L(TX\oplus\xi), [X] \rangle - \langle L(TX\oplus\xi\oplus\xi), [X] \rangle -1
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= c \langle p_1(\xi)^2 ,[X] \rangle \ne 0$, $$
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and so the surgery obstruction is not a group homomorphism with respect to the Whitney sum.
</wikitex>
</wikitex>

Revision as of 22:44, 29 May 2012

We take X=\mathbb H P^2 and consider various bundle reductions of the normal bundle: The normal map id_X gives the base point of \mathcal N (X)\cong [X,G/TOP]. 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/O], 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 -1,

so it depends only on the bundle over X. 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], since the corresponding Atiyah-Hirzebruch spectral sequence collapses. 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. 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. We compute

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

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