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
![\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}](/images/math/2/4/0/2400d5d29b4b1a201991db701c9a42e3.png)
by the Hirzebruch signature theorem and severeal property of the -genus. In particular the surgery obstruction depends only on the bundle over
.
Now
is the sum of
and
in
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](/images/math/6/2/5/6255fb61ac0ebbaec05bc4ef8de61bd9.png)
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
![\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,](/images/math/4/8/1/4812d8321d698df60375d7e1c6188e1c.png)
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.