Two different spin structures of the real projective space $\Bbb RP^3$

Question

It is known that every orientable 3-manifold has a spin structure, because its tangent bundle is trivial. Also it is known that if a manifold $X$ has a spin structure, then the number of distinct spin structures is equal to the order of $H^1(X;\Bbb Z_2)$. In particular, the real projective space $\Bbb RP^3$ has exactly two different spin structures, say $s_1,s_2$. Is there a self diffeomorphism of $\Bbb RP^3$ that interchanges these two spin structures, i.e. is there a diffeomorphism $f:\Bbb RP^3\to \Bbb RP^3$ satisfying $f^* s_1=s_2$?

Answer 1

I'm editing this to reflect the discussion.

You might think that there cannot be an automorphism of ${\mathbb R}P^3$ that acts non-trivially on the degree one cohomology. But the bijection between the spin structures and degree one cohomology is not canonical, it's a torsor. So it may be possible for an automorphism to swap the two spin structures.

To fix conventions, a spin structure is a structure on a manifold with a given orientation. The space ${\mathbb R}P^3$ is orientable because $3$ is odd, and there is an orientation reversing diffeomorphism which gives a bijection between the two spin structures for one and the two spin structures for the other. So we fix an orientation.

Now in Section IV.B.(i) of Dabrowski and Trautman, "Spinor structures on spheres and projective spaces", in the second to last paragraph they give a topological way to distinguish the two spin structures, so there cannot be an orientation preserving diffeomorphism swapping them.

They then go on to make the claim that the two spin structures are related by an orientation reversing isometry. This I do not understand, and it misled me to give my original answer.

Answer 2

Here's an argument that doesn't require knowing that the mapping class group of $\mathbb{RP}^3$ is trivial.

I will exhibit two spin structures $s_\pm$ on $\mathbb{RP}^3$ that have distinct Rokhlin invariant. If $s$ is a spin structure on a 3-manifold $Y$, we pick a compact spin 4-manifold $W$ bounding $(Y,s)$: the Rokhlin invariant is defined to be $\sigma(W) \in \mathbb Z/16\mathbb Z$. (The "classical" Rokhlin invariant, defined for integer homology spheres, lives in $\mathbb Z/2\mathbb Z \cong 8\mathbb Z/16\mathbb Z$.)

$\mathbb{RP}^3$ is obtained both as $+2$- and $-2$-surgery along an unknot in $S^3$. Let $W_\pm$ be the corresponding handlebodies: these are obtained by attaching a $\pm2$-framed 4-dimensional 2-handle to the 4-ball along the unknot.

Both $W_+$ and $W_-$ are spin 4-manifolds, since they have $H_1 = 0$ and even intersection form (hence $w_1 = w_2 = 0$). Moreover, the spin structure is unique. Therefore we have two spin structures $s_\pm$ on $\mathbb{RP}^3$ obtained by restricting the unique spin structure on $W_\pm$ to $\mathbb{RP}^3$. Now, since $\sigma(W_\pm) = \pm1$, the two spin structures have distinct Rokhlin invariants, so they are not isomorphic. (Technically, we're choosing identifications $\partial W_\pm \to \mathbb{RP}^3$, but the argument below works anyway and pulling back the restrictions, but the argument doesn't really care.)

(Since $W_+ = -W_-$, maybe all of this has something to do with what Dabrowski and Trautman are talking about.)

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Maybe let me add a remark that we don't really need the full strength of the Rokhlin theorem here, but it's enough to know a bit about even unimodular forms (namely that their signature is divisible by 8) to show that the "mod 8 Rokhlin invariant" distinguishes $s_+$ and $s_-$.