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The following fairly specific question comes up in a bordism computation I'm trying to do:

Are there compact $\mathbb Z$-oriented $4k+2$ dimensional manifolds with boundary $M$ such that $im(H_{2k+1}(M; \mathbb Z/2)\to H_{2k+1}(M, \partial M; \mathbb Z/2))$ has odd dimension as a $\mathbb Z/2$ vector space?

Clearly the answer is no if $k=0$. Also, I can show that this can't happen if $\partial M=\emptyset$ using a combination of Poincare duality and the universal coefficient theorem. But I haven't been able to rule out the possibility if the boundary is non-empty, or to construct examples.

Thanks.

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  • $\begingroup$ If you're working mod 2 why do you care that they are oriented? $\endgroup$
    – Dylan Wilson
    Commented Jan 27, 2011 at 2:52
  • $\begingroup$ Dylan, $RP^6$ (or $RP^2$) is a counterexample, with empty boundary, even. $\endgroup$
    – Ben Wieland
    Commented Jan 27, 2011 at 4:02
  • $\begingroup$ Fair question. I'm interested in bordism groups of Z/2-Witt spaces. If a Z/2 Witt space is defined to only be Z/2 oriented, then I can compute those bordism groups, which are isomorphic to Z/2. But it's not clear that a Z/2 Witt space shouldn't be defined to be Z-oriented but satisfy the Z/2 Witt condition. Thus I would like to compute such a bordism group. I can show that it's 0 or Z/2. The possible invariant is the middle dimensional Z/2 intersection homology pairing in W(Z/2)=Z/2. This question would determine whether there is a Witt space with only point singularities representing 1. $\endgroup$
    – Greg Friedman
    Commented Jan 27, 2011 at 5:01

2 Answers 2

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I claim it is not possible. The image is the rank of $H_{2k+1}(M;\mathbb Z_2)/rad$, where $rad$ is the radical of the intersection form on $H_{2k+1}(M;\mathbb Z_2)$.

The intersection form on $H_{2k+1}(M;\mathbb Z_2)/rad$ is hyperbolic, i.e. has a "symplectic" basis, therefore this vector space has even dimension.

Let me try to prove that it is hyperbolic: the tricky point is to show that all classes square to zero, i.e. $\langle x^2,[M,\partial M]\rangle =0$ for all $x\in H^{2k+1}(M,\partial M;\mathbb Z_2) $.

Now $\langle x^2,[M,\partial M]\rangle =\langle \beta Sq^{2k}x,[M,\partial M]\rangle=\langle Sq^{2k}x,\beta [M,\partial M]\rangle=0$ where $\beta$ denotes the cohomology respectively homology Bockstein.

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  • $\begingroup$ The fact that it's not possible leads to an interesting corollary: given an oriented vector bundle $\xi$ of rank $2k+1$ over a connected closed manifold of dimension $2k+1$, then the top Stiefel-Whitney class $w_{2k+1}(\xi)$ is trivial. The proof: the zero section map $M \to M^{\xi}$ (the target is the Thom space) represents the Euler class. It also coincides with Jeff's map $H_{2k+1}(D(\xi)) \to H_{2k+1}(D(\xi),S(\xi))$. QED. Is this a known result? I couldn't find it in the literature. $\endgroup$
    – John Klein
    Commented Jan 28, 2011 at 16:10
  • $\begingroup$ @Martin: can you explain why the kernel is the radical? $\endgroup$
    – John Klein
    Commented Jan 28, 2011 at 22:24
  • $\begingroup$ @John's first comment: If the bundle is oriented, M needs to be oriented as well. The Euler class of an odd-dimensional bundle is known to be 2-torsion, so even the Euler class is zero here. (Hatcher has an exercise in the vector bundle book that for every oriented $(2k+1)$-dimensional bundle $\xi$ one has $e(\xi)=\tilde{\beta}w_{2k}(\xi)$.) $\endgroup$
    – Martin O
    Commented Jan 29, 2011 at 0:08
  • $\begingroup$ @John's second comment: the adjoint of the intersection form $H_{2k+1}(M;\mathbb Z_2) \times H_{2k+1}(M;\mathbb Z_2)\to\mathbb Z_2$ is given by $H_{2k+1}(M;\mathbb Z_2) \to H_{2k+1}(M,\partial M;\mathbb Z_2)\cong H^{2k+1}(M;\mathbb Z_2)$ where we used Poincaré duality. The kernel of the map is the radical of the intersection form, i.e. all classes which have zero intersection with everything. $\endgroup$
    – Martin O
    Commented Jan 29, 2011 at 0:14
  • $\begingroup$ Very nice argument. Thank you. Hmmm. I wonder if it can be adopted to intersection homology using Goresky's intersection cohomology operations. I'm not sure if the Bockstein's would work out, though, because of the intricacies of coefficients in intersection homology (in particular, $IC(X;G)$ is not necessarily the same as $IC(X)\otimes G$) $\endgroup$
    – Greg Friedman
    Commented Jan 30, 2011 at 6:35
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Martin O's answer is very nice. So in an oriented $2n$-manifold with $n$ odd the mod $2$ self-intersection of any $n$-dimensional mod $2$ homology class is $0$.

Looking for a more geometric explanation of that, or anyway an explanation with no Steenrod operations in sight, I came up with the following (which is also related to John Klein's comment):

Let's assume that the given class is represented by an immersed $n$-manifold $M$. The mod $2$ self-intersection number is then the evaluation on the mod $2$ fundamental class of $M$ of the mod $2$ Euler class of the normal bundle of the immersion. So it comes down to the following:

Claim: Let $n$ be odd and suppose that $M$ is a closed $n$-dimensional manifold and $E$ is a rank $n$ vector bundle such that the total space of $E$, considered as a $2n$-manifold, is orientable. Then the mod $2$ Euler class of $E$ is $0\in H^n(M;\mathbb Z/2)$.

Proof: A rank $n$ vector bundle has a twisted integral Euler class, which belongs to $H^n(M;\Gamma)$, where $\Gamma$ is the coefficient system (locally isomorphic to $\mathbb Z$) associated with $w_1(E)$, the obstruction to orientability of $E$. The mod $2$ Euler class is the mod $2$ reduction of this, so it suffices if this twisted integral class is $0$. The (twisted) integral Euler class of a vector bundle of odd rank is always killed by $2$ (this is a standard fact in the oriented case, and it seems clear in the twisted case, too), so it suffices if the group $H^n(M;\Gamma)$ is torsion-free. But by Poincare duality it is isomorphic to $H_0(M;\mathbb Z)$, since $E$ and the tangent bundle of $M$ have the same orientability obstruction by hypothesis.

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  • $\begingroup$ Tom is there a way to generalize this to cycles not represented by manifolds? $\endgroup$
    – Greg Friedman
    Commented Jan 30, 2011 at 6:39
  • $\begingroup$ Well, there's a nontrivial theorem of Thom, from his big 1954 paper on cobordism, to the effect that the canonical map from unoriented smooth bordism to mod 2 singular homology is surjective. (The analogue for oriented bordism and integral homology is false.) $\endgroup$
    – Tom Goodwillie
    Commented Jan 30, 2011 at 15:58

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