7
$\begingroup$

In Jim Davis's paper "Manifold aspects of the Novikov conjecture" (Surveys on surgery theory, vol 1, pages 195-224) he writes down (Theorem 6.5) a sort-of converse to the Novikov conjecture. He writes:

Theorem: Let $M$ be a closed, oriented, smooth manifold of dimension $\ge5$, together with a map $f\colon M\to B\pi$ to the classifying space of a discrete group, inducing an isomorphism on fundamental groups. Given any cohomology class $${\mathcal L}={\mathcal L}_1+{\mathcal L}_2+\cdots$$ in $H^{4\ast}(M; {\mathbb Q})$ with the ${\mathcal L}_i\in H^{4i}(M; {\mathbb Q})$, if $f_\ast({\mathcal L}\cap [M])=0$ in $H_{n-4\ast}(B\pi;{\mathbb Q})$, then there is a nonzero integer $R$ so that, for any multiple $r$ of $R$, there is a homotopy equivalence $h\colon M\to N$ of closed smooth manifolds so that $$h^\ast({\mathcal L}_N)-{\mathcal L}_M=r{\mathcal L},$$ where ${\mathcal L}_M$ and ${\mathcal L}_N$ are the total Hirzebruch $L$-classes of $M$ and $N$.

There are many versions of the Novikov conjecture. One version states that, if $h\colon N\to M$ is an orientation-preserving homotopy equivalence between closed orientable manifolds, then for any discrete group $\pi$ and any map $f\colon M\to B\pi$, we have $$f_\ast\circ h_\ast({\mathcal L}_N \cap [N])=f_\ast({\mathcal L}_M\cap [M])$$ in $H_\ast(B\pi; {\mathbb Q})$.

Another version is the following: Suppose that $M$ is a closed oriented manifold of dimension $4k+i$ with fundamental group $\pi$. Let $f\colon M\to BG$ be a map and suppose that $\alpha\in H^i(B\pi; {\mathbb Q})$. Then the higher signature given by \begin{equation*} \mathrm{sig}_\alpha(M)\equiv \langle f^\ast(\alpha)\cup {\mathcal L}_M, [M]\rangle \end{equation*} is an oriented rational homotopy invariant.

Note: The map $f\colon M\to B\pi$ induces an isomorphism on the fundamental group in the theorem, but the Novikov conjecture does not require an isomorphism.

My question is: how does the theorem in his paper serve as a (partial) converse to Novikov? Maybe it's just a matter of playing with cup and cap products and sliding symbols around but I couldn't get things to match up. Thanks for reading. [Apologies for any typos.]

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.