# Cobordism and finite sheeted covers of manifolds

Let $M$ be an oriented manifold, not necessarily compact. Let $M'$ be a (finite) $k$-sheeted cover and let $\pi:M'\longrightarrow M$ be the covering map.

Question 1 : Is it true that $M'$ is (oriented) cobordant to $k$ disjoint copies of $M$?

Question 2 : If the answer to the above is true then let $W$ be such a cobordism. Is there a map $\widetilde{\pi}:W\longrightarrow M$ such that $\widetilde{\pi}$ restricts to the natural maps on each end?

I would guess that the answers to these questions should be well-known. However, any references, proofs or counter-examples(?) would be helpful.

EDIT : From the discussion/answers below, it is clear that the answer to (1) is positive when $M$ is a closed, oriented manifold. The characteristic numbers determine the oriented cobordism class of $M$. However, when $M$ is an open manifold, these numbers do not make immediate sense. I'm particularly interested in both questions in this scenario.

• Well, for the double cover $S^2 \to \mathbb RP^2$, $S^2$ is not cobordant to 2 copies of $\mathbb RP^2$. But maybe you wanted oriented manifolds? – Sam Gunningham Sep 5 '13 at 19:58
• @ Sam - Thanks! I made the necessary changes. – Somnath Basu Sep 5 '13 at 20:12
• Since signature is a cobordism invariant, I suggested the "1-sheeted cover" $\mathbb{C}P^2\#\mathbb{C}P^2\to\mathbb{C}P^2\#\overline{\mathbb{C}P^2}$ (the cover's signature is 2 whereas the base's signature is 0), but of course there is no such diffeomorphism $\mathbb{C}P^2\to\overline{\mathbb{C}P^2}$ so this ramble wouldn't make any sense. – Chris Gerig Sep 5 '13 at 20:47
• $S^2$ IS cobordant to $2$ copies of $RP^2$. Both are null-cobordant. – András Szűcs Sep 5 '13 at 21:56

The answer is yes to the first question. This follows from the fact that characteristic numbers determine completely cobordism classes. Actually both questions have two versions: an oriented and an unoriented one.

The fact that the characteristic numbers are multiplied by $k$ under a $k$-fold covering follows from the definition of characteristic numbers through vector fields. (Each Pontrjagin class and Stiefel Whitney class can be obtained by taking some, properly chosen number of generic vector fields and consider the points, where the dimension of the space spaned by them drops by (at least) a given number compered with the possible maximum. (This given number is $1$ for $W$ and 2 for the $P$ classes.) The set of these points is a cycle dual to the corresponding characteristic class. Hence the characteristic numbers can be obtained as the number of intersections of such cycles. Now by the covering map we can pull back the vector fields, we can pull back the cycles, and clearly the number of intersections of the cycles in the covering manifold will be $k$ times that in the base manifold.

• The fact that characteristic numbers (Pontrjagin and Stiefel-Whitney numbers) determine cobordism classes completely is true for compact, oriented manifolds. What is a reference for this being true for manifolds which are not necessarily compact? – Somnath Basu Sep 5 '13 at 20:41
• @Somnath: How would you define cobordisms for noncompact manifolds? – Misha Sep 5 '13 at 20:43
• For $k=2$ the unoriented version is trivial. (It claims that $M'$ is a boundary. Take the line bundle over $M$ associated to the double covering $M' \to M.$ Then $M'$ is the sphere bundle of this line bundle. It bounds the corresponding ball-bundle.) – András Szűcs Sep 5 '13 at 20:50
• For $k=2$ the oriented version is not so trivial in an elementary way, but Dold gave a nice, simple proof. – András Szűcs Sep 5 '13 at 20:52

Here is Dold's argument showing that if $M' \to M$ is a double cover and an orientation preserving map, than $M'$ is oriented-cobordant to two copies of $M$.

Let $V \subset M$ be a codimension $1$ submanifold representing the $Z_2$-homology class dual to $W_1$ of the double covering $M' \to M.$ Let $N$ be the closed tubular neighbourhood of $V$ in $M.$ Note, that if we take two copies of $M \setminus int N$ and attach them appropriately along the boundaries, then we obtain $M'.$

Now take two copies of the cylinder $M\times [0,1]$ and attach them along $N \times 1$ to each other so that from the two copies of $M \times {1}$ we obtain $M'$. We obtain a compact, oriented manifold with boundary, the boundary consisists of two copies of $M \times {0}$ and one copy of $M'$

Its not true in complex cobordism. In Quillen's paper Elementary proofs of some results of cobordism theory using Steenrod operations", Section 4, he computes the complex cobordism class of a principal $\mathbb{Z}/k$-bundle $p : E \to B$ in terms of the associated line bundle $L \to B$ (given by $\mathbb{Z}/k \subset \mathbb{C}^\times$).

The answer is that $[p : E \to B] = [k](e(L))/e(L) \in MU^0(B)$, where $[k](x)$ is the $k$-series of the formal group law for complex cobordism. You can easily cook up examples, such as manifold approximations to skeleta of $B\mathbb{Z}/k$, for which this is not $k\cdot [\mathrm{id} : B \to B]$.

Presumably you wanted an answer in oriented cobordism. Perhaps the above counterexample survives under $MU^*(-) \to MSO^*(-)$?

• I am surprised that this fails; because complex bordism is torsionfree, and cobordism classes are detected by Chern numbers. Chern numbers are multiplicative for finite covers (the simplest proof I know uses Chern-Weil Theory). Where is the mistake? – Johannes Ebert Sep 6 '13 at 14:32
• Ah, I forgot to say: this is addressing Question 2. For Question 1 it is true, by the argument you give. – Oscar Randal-Williams Sep 6 '13 at 15:09