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Let $\alpha$ be an orientation-preserving automorphism of the torus $T^2 = S^1 \times S^1.$ Since the mapping torus $M_{\alpha} = T^2 \times [0, 1] / (x, 0) \sim (\alpha(x), 1)$ is an orientable compact 3-dimensional manifold, it is $spin^c.$

But is there a $spin^c$ structure on $M_{\alpha}$ that restricts to the trivial $spin^c$ structure on $T^2 \times [1/4, 3/4]$ ?

Thank you !

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  • $\begingroup$ Actually, every $spin^c$-structure on $T^2\times (0,1)$ is the restriction of one on $M_\alpha$. The $spin^c$-structures form a torsor for $H^2$, and the restriction map $H^2(M_\alpha)\to H^2(T^2\times (0,1))$ is surjective. $\endgroup$
    – Tim Perutz
    Commented Aug 16, 2011 at 22:48

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Yes. Moreover, there's a trivialization of the tangent bundle of $M_\alpha$ that restricts to the standard trivialization of the tangent bundle of $T^2 \times [1/4,3/4]$. By "the standard" trivialization I mean one that's invariant under the action of $T^2$. You construct it by hand -- take the invariant one and apply your gluing map $(x,0) \sim (\alpha(x),1)$. The gluing map does not preserve the trivialization on the boundary, so on one end of the boundary you need to rotate it a little, to ensure that it is preserved.

So what I'm saying is you can do it for $spin$ structures, and every $spin$ structure induces a $spin^c$ structure, giving you what you want.

edit in response to Alain's comment: I will describe a trivialization of the tangent bundle of $T^2 \times [0,1]$ which descends to a trivialization of the tangent bundle of $M_\alpha$, and this works for any $\alpha \in SL_2 \mathbb Z$.

Think of $T^2$ as $\mathbb R^2 / \mathbb Z^2$. So I will define the trivialization up on the space $\mathbb R^2 \times [0,1]$. Coordinates in $\mathbb R^2 \times [0,1]$ will be given by $(x,y,z)$. A trivialization of the tangent space to $(x,y,z)$ is a matrix in $GL_3(\mathbb R)$. So what we need to do is describe a matrix $A_{(x,y,z)} \in GL_3(\mathbb R)$ which respects all the identifications on the boundary (and tangent spaces) and which depends continuously on $(x,y,z)$.

$SL_2 \mathbb Z \subset GL_2(\mathbb R)$ is in the path-component of the identity matrix, so given $\alpha \in SL_2 \mathbb Z$, let $\beta : [0,1] \to GL_2(\mathbb R)$ be the path such that $\beta(0) = I$ and $\beta(1) = \alpha$. Then $A_{(x,y,z)} = \overline{\beta(z)}$ is our trivialization of the tangent bundle of $\mathbb R^2 \times [0,1]$, and this descends (by design) to the tangent bundle of $M_\alpha$. In the above, if $\beta(z) = \pmatrix{ b_{11} & b_{12} \cr b_{21} & b_{22} }$, let $\overline{\beta(z)} = \pmatrix{ b_{11} & b_{12} & 0 \cr b_{21} & b_{22} & 0 \cr 0 & 0 & 1 }$

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  • $\begingroup$ @ Ryan: Can you be more specific on "rotating a little"? Since $\alpha$ is not isotopic to the identity (assuming $\alpha\neq Id$), I don't see how you do it. $\endgroup$
    – Alain Valette
    Commented Aug 16, 2011 at 21:52
  • $\begingroup$ You've got much more flexibility when it comes to trivializations of the tangent bundle -- the $SL_2 \mathbb Z$ restriction is only for the gluing map. Anyhow, I hope the edit helps. $\endgroup$
    – Ryan Budney
    Commented Aug 16, 2011 at 22:53
  • $\begingroup$ @ Ryan, after editing: OK, thank you for the explanation. Meanwhile I also learned of a theorem by Steenrod, that every orientable 3-manifold is parallelizable! $\endgroup$
    – Alain Valette
    Commented Aug 17, 2011 at 12:28

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