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 !



$Spin^c$ structure on the mapping torus of an automorphism of the torus
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 $\overline{\beta(z)} = \pmatrix{ b\_{11} & b\_{12} & 0 \cr b\_{21} & b\_{22} & 0 \cr 0 & 0 & 1 }$


