0
$\begingroup$

Let $F$ be a closed manifold. What are the Pontryagin numbers on $E=F\times S^1$? More generaly, let $E$ be a closed manifold which is a fiber bundle over $S^1$ (with fiber $F$). $E$ is also called mapping torus. What are the Pontryagin numbers on $E$?

When $E$ is 4-dimensional, the signature of such a fiber bundle over $S^1$ is zero, which implies that the corresponding Pontryagin number is zero. I wonder if all the Pontryagin numbers of closed orientable mapping tori in any dimensions are always zero.

$\endgroup$
5
$\begingroup$

You are asking about vanishing of the Pontryagin numbers, and hence the (rational) cobordism class, of the mapping torus of a diffeomorphism. According to M. Kreck, (Bordism of diffeomorphisms Bull. Amer. Math. Soc. Volume 82, Number 5 (1976), 641-789, with details in Cobordism of odd-dimensional diffeomorphisms. Topology 15 (1976). 353-361) any cobordism class in the kernel of the signature homomorphism contains a mapping torus. So the signature in dimension 4 is the only obstruction, but that's not true in higher dimensions.

The special case of $S^1 \times M$ is the boundary of $D^2 \times M$, so of course its Pontryagin numbers vanish.

$\endgroup$
5
  • $\begingroup$ Do you mean the Pontryagin numbers of some closed orientable mapping tori are not zero? $\endgroup$ – Xiao-Gang Wen Apr 28 '14 at 2:55
  • $\begingroup$ Yes, that's apparently what Kreck's theorem implies; it seems a bit counterintuitive to me. Have you looked in the referenced papers? I don't know a specific example, but perhaps they have one. Kreck also wrote a book (Lecture Notes in Mathematics, 1069) on the subject. $\endgroup$ – Danny Ruberman Apr 28 '14 at 11:44
  • $\begingroup$ Indeed, Kreck's theorem is very helpful. Thanks for refs. I asked the question to make sure that I understand Kreck's theorem (I am a physicist). Kreck has obtained the cobordism group of mapping tori for dimension greater then 4. The cobordism group of closed 4-dim mapping tori is calculated in another paper by Melvin, and is found to be 0. I wonder do you know what is the cobordism group of closed 3-dim mapping tori (or any refs)? Thanks! $\endgroup$ – Xiao-Gang Wen Apr 28 '14 at 15:49
  • $\begingroup$ Let's be careful about the terminology. What you call the cobordism group of n+1 dimensional mapping tori is usually called the cobordism group of diffeomorphisms of n-manifolds. (This is clearer, since in your terminology you might be concerned about the cobordism class of the manifold arising as a mapping torus.) The cobordism group of diffeomorphisms of surfaces was computed by F. Bonahon: Cobordism of automorphisms of surfaces, Ann. Scient. Ec. Norm. Sup. 16 (1983), 237-270. numdam.org/item?id=ASENS_1983_4_16_2_237_0. $\endgroup$ – Danny Ruberman Apr 28 '14 at 16:04
  • $\begingroup$ Thank you very much for the ref! Do you happen to know the generators of the 3-dim cobordism group? What characteristic classes can detect the cobordism group (in 3-dim and also in higher dimensions)? $\endgroup$ – Xiao-Gang Wen Apr 29 '14 at 12:29
1
$\begingroup$

The answer is 0. The proof of the first case is as follows. Consider a family of product Riemannian metrics $ g_{t}=t^{2}ds^{2}+h$, $0<t<1$, on E, where h is a Riemannian metric on F. The straight forward calculation shows that the sectional curvature (or the norm of curvature operator) of each $g_{t}$ is bounded by a constant C independent of t, i.e. $|Rm(g_{t})|<C$. Now the Chern-Weil theory says that for any Pontryagin number P we have $$ P= \int_{E} R_{t} dvol_{g_{t}} ,$$ where $R_{t}$ is a polynomial of curvature operators. When t tends to 0, we have $$ |\int_{E} R_{t} dvol_{g_{t}} | < C Vol(E, g_{t}) \rightarrow 0.$$ However the Pontryagin number P is an integer independent of t, and thus it must be 0.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.