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I was recently informed by a source of the following fact:

Theorem 1: The linking form on an orientable smooth 5-manifold $M$ is alternating if and only if $M$ is spin$^{\mathbb{C}}$.

Question 1: Does anybody know a reference/attribution for this fact? I've poked around online and in the library, and asked my usual go-to experts, but found nothing. (This page gives a reference to a criterion of Wall in the simply-connected case, but not the general one.)

I'm not in a position to be able to ask my source for a reference, but I'm pretty confident that Theorem 1 is correct because I think I have a proof. In fact the proof gives a criterion for the linking form on any orientable, odd-dimensional topological manifold to be alternating, which specializes to the above fact for smooth 5-manifolds.

Question 2: Is such a criterion already known? Written down? Is it remotely interesting?

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    $\begingroup$ I think Wall's criterion ($w_2 = 0$) applies whenever $H_1$ is torsion-free, not just when the manifold simply connected. $\endgroup$ Feb 6, 2018 at 16:59
  • $\begingroup$ Is there something wrong with the following example? Consider $M = \mathbb{CP}^2 \times S^1$. Then $H^3(M; \mathbb{Z}) = \mathbb{Z}$, so the domain of the linking form is actually $0$, which I think qualifies it to be alternating. On the other hand, $w_2(M)$ is the reduction of $c_1(\mathbb{CP}^2)$, which is certainly not $0$. $\endgroup$
    – user84144
    Feb 6, 2018 at 23:42
  • $\begingroup$ You're absolutely right, I was being sloppy. What I should have said is that $w_2(x) = b(x, x)$ where $b$ is the linking form, $x \in H_2(M; \mathbb{Z})_{\text{tors}}$ and $w_2$ is viewed as a map $H_2(M; \mathbb{Z}) \to \mathbb{Z}_2$ via the isomorphism $H^2(M; \mathbb{Z}_2) \cong \operatorname{Hom}(H_2(M; \mathbb{Z}), \mathbb{Z}_2)$ by the Universal Coefficient Theorem (as $H_1(M; \mathbb{Z})$ is torsion-free, there is not $\operatorname{Ext}$ term). So the correct statement is that $b$ is alternating iff $w_2 : H_2(M; \mathbb{Z})_{\text{tors}} \to \mathbb{Z}_2$ is zero. $\endgroup$ Feb 7, 2018 at 14:11
  • $\begingroup$ In particular, spin implies the linking form is alternating, but as your example demonstrates, the converse is not true. $\endgroup$ Feb 7, 2018 at 14:12

2 Answers 2

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This is not an answer, but too long for a comment, and hopefully of some use.

I believe the theorem you mention should be true for any 5-dimensional Poincaré duality space $X$. You can define the linking form on the torsion group $H^3(X,\mathbb{Z})_{\rm tor}$ (which is isomorphic to $H_2(X, \mathbb{Z})_{\rm tor}$ via the Poincaré duality structure) by setting $(x,y) = x' \cup y$, where $x' \in H^2(X,\mathbb{Q}/\mathbb{Z})$ maps to $x$ under the boundary map $\partial: H^2(X,\mathbb{Q}/\mathbb{Z}) \to H^3(X,\mathbb{Z})$ associated to the short exact sequence $0 \to \mathbb{Z} \to \mathbb{Q} \to \mathbb{Q}/\mathbb{Z} \to 0$.

It should then be true that $(x,x) = (x,\partial \overline{w}_2) = x'\cup \partial \overline{w}_2$, where $\overline{w}_2$ is the image in $H^2(X,\mathbb{Q}/\mathbb{Z})$ of the second Stiefel-Whitney class $w_2 \in H^2(X,\mathbb{Z}/2)$ of the Spivak normal bundle. This would imply that the linking form is alternating if and only if $\partial \overline{w}_2$ vanishes in $H^3(X,\mathbb{Z})_{\rm tor} \subseteq H^3(X,\mathbb{Z})$, which, in turn, is equivalent to $w_2$ lifting to $H^2(X,\mathbb{Z})$. In the case of smooth manifolds this is the same as admitting a ${\rm spin}^{\mathbb{C}}$-structure.

For simply-connected 5-dimensional Poincaré duality spaces this is proved in Stöcker's paper from 1982 entitled "On the structure of 5-dimensional Poincaré duality spaces", but I was not able to find a proof for the general case.

It's possible that there is a small gap in the literature here, so it sounds to me that if you have a proof (in either the smooth, topological or Poincaré setting) it would be great if you write it up and publish it. I would certainly be interested to read it.

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    $\begingroup$ Thanks for pointing out Stocker's paper, which I was not aware of. At the risk of blowing my cover -- the writeup already exists (sort of) in arxiv.org/pdf/1706.00151.pdf, cf. section 7.3; that paper really focuses on a different situation. $\endgroup$
    – user84144
    Feb 9, 2018 at 13:47
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The final sentence of Theorem 1 of Browder's Remark on the Poincaré Duality Theorem states that a five-dimensional space $X$ with Poincaré duality has $H_2(X) = F + T + T + \mathbb{Z}_2$, where $F$ is free and $T$ is torsion, if and only if $w_3 \neq 0$. If $b$ denotes the linking form and $x$ is the generator of $\mathbb{Z}_2$, then $b(x, x) = \frac{1}{2}$ so $b$ is not alternating. Therefore, if $b$ is alternating, then $w_3 = 0$ which is equivalent to $w_2$ having an integral lift (which is equivalent to spin${}^c$ in the smooth case).

I don't know of a reference for the converse. However, if $H_2(X)$ has no two-torsion, then it is spin${}^c$ and the fact that the linking form is alternating follows from the classification of non-singular anti-symmetric linking forms; see here.

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  • $\begingroup$ There is a small subtlety here to take into account: the $w_3$ appearing in Browder's theorem lives in $H^3(X,\mathbb{Z}/2)$, and is equal to $Sq^1w_2$. The vanishing of $w_3$ is hence not equivalence to $w_2$ having an integral lift, but to $w_2$ having a lift to $H^2(X,\mathbb{Z}/4)$. The obstruction to $w_2$ having an integral lift is a certain class in $H^3(X,\mathbb{Z})$ which lifts $w_3$. $\endgroup$ Jun 12, 2018 at 9:47
  • $\begingroup$ @YonatanHarpaz: You're right. I realised this last night, but I couldn't think of an example where $w_3 = 0$ but $W_3 \neq 0$. Do you know of such an example? $\endgroup$ Jun 12, 2018 at 12:44
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    $\begingroup$ Such an example would provide an answer to this question. $\endgroup$ Jun 12, 2018 at 15:55

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