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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.

flag	asked Jan 26 2011 at 23:04 Greg Friedman 2,074●4●14

	If you're working mod 2 why do you care that they are oriented? – Dylan Wilson Jan 27 2011 at 2:52
	Dylan, $RP^6$ (or $RP^2$) is a counterexample, with empty boundary, even. – Ben Wieland Jan 27 2011 at 4:02
	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. – Greg Friedman Jan 27 2011 at 5:01

Martin O · Accepted Answer · 2011-01-28 00:57:04Z UTC

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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.

answered Jan 28 2011 at 0:57

Martin O
341●2●4

	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. – John Klein Jan 28 2011 at 16:10
	@Martin: can you explain why the kernel is the radical? – John Klein Jan 28 2011 at 22:24
	@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 $e(\xi)=\tilde{\beta}w_{2k}(\xi)$.) – Martin O Jan 29 2011 at 0:08
	@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. – Martin O Jan 29 2011 at 0:14
	Thanks Martin, for both explanations. – John Klein Jan 30 2011 at 0:21

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Tom Goodwillie · Answer 2 · 2011-01-29 04:38:30Z UTC

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.

Are there oriented $4k+2$ manifolds such that $im(H_{2k+1}(M; Z/2)\to H_{2k+1}(M, \partial M; Z/2))$ has odd dimension?

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