2
$\begingroup$

The statement of the Novikov conjecture is a bit esoteric. Does the following simplified conjecture have any known counterexamples?

C: For a smooth closed 4n-fold $M$, the Pontryagin numbers are homotopy invariant. That is given $f: M \to N$ a homotopy equivalence with $N$ likewise closed and smooth 4n-fold, all Pontryagin numbers of $M,N$ coincide. As far as I can see this is not implied by the Novikov conjecture.

Improve this question
$\endgroup$
5
  • 3
    $\begingroup$ That is very false. Roughly speaking, surgery theory says if $M$ is a simply connected manifold that you can make a new manifold $N$ homotopy equivalent to $M$ with new Pontrjagin classes, whatever you want them to be (as long as it satisfies the Hirzebruch signature formula). (eg, an infinite family equivalent to $S^2\times S^4$ parameterized by $p_1$ and another infinite family for $\mathbb C\mathbb P^4$, parameterized by either $p_1$ or $p_2$, which are related by Hirzebruch.) The Novikov conjecture measures the extent to which this fails in the presence of fundamental group. $\endgroup$
    – Ben Wieland
    Commented Jun 15, 2019 at 2:29
  • 2
    $\begingroup$ @BenWieland Besides calculating the signature correctly, the Pontryagin numbers formed from the chosen Pontryagin classes have to satisfy a set of congruence relations as well, as given by the Hattori-Stong theorem. Let me know if I've misunderstood your claim. $\endgroup$
    – Aleksandar Milivojević
    Commented Jun 15, 2019 at 3:34
  • $\begingroup$ @BenWieland, Do you have a reference please? Also do you mean $S^4 \times S^4$? The signature formula is in dim 4n. $\endgroup$
    – Yasha
    Commented Jun 16, 2019 at 19:29
  • 1
    $\begingroup$ The situation simplifies in dim not a multiple of 4 by not having the signature obstruction, which is why I gave $S^2\times S^4$. In the case of $S^4\times S^4$, the mflds are, roughly, parameterized by $p_1\in H^4=Z^2$, with $p_2$ determined by the signature formula. Exercise: consider 4d vector bundles over $S^4$, ie, $O(4)$ bundles. parameterize them. Take the space of unit vectors to get $S^3$ bundles over $S^4$. Infinitely many have the homology of $S^3\times S^4$. Compute $p_1$ and see that anything is possible. Harder: show that infinitely many are homotopy equivalent to $S^3\times S^4$ $\endgroup$
    – Ben Wieland
    Commented Jun 17, 2019 at 21:13
  • $\begingroup$ Sounds, good. Leave it as an answer, and add some references please. I looked at Novikov's paper and he says something similar but also without references ("it is well known kinda thing"). I guess all this is indeed well known to experts but they are vanishing, so making this explicit would be of great help to some people. $\endgroup$
    – Yasha
    Commented Jun 19, 2019 at 20:56

0

Reset to default

You must log in to answer this question.