If we consider the AdS-Schwarzschild manifold, defined by $M^n=[s\_0,\infty)\times\mathbb{S}^{n-1}$ equipped with the Riemannian metric 
$$\overline{g}=\frac{1}{1-ms^{2-n}+s^2}ds\otimes ds+s^2g\_{\mathbb{S}^{n-1}},$$
where $m>0$ is a fixed positive number, $s\_0$ is the unique positive solution of the equation $1+s\_0^2-ms\_0^{2-n}=0$ and $g\_{\mathbb{S}^{n-1}}$ is the standard round metric on the unit sphere $\mathbb{S}^{n-1}$. The scalar curvature of $M$ is equals $-n(n-1)$.


My question is:



> 
> There exists some spin structure on $M$ ?
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> 
> 


Since the Hyperbolic space is a particular case when $m\to0$ and the hyperbolic space admits a spin structure, is natural asking for a spin structure on the AdS- Schwarschild space?



Spin Structure on AdS- Schwarzschild manifold
The existence or nonexistence of a spin structure on a smooth manifold $M$ is a topological question, in that it does not depend on the choice of a Riemannian metric on $M$. Spin structures are obstructed by the second [Stiefel-Whitney class](https://en.wikipedia.org/wiki/Stiefel%E2%80%93Whitney_class) $w\_2\in H^2(M;\mathbb Z/2)$, in that $M$ admits a spin structure iff $w\_2(TM) = 0$.


Since $M^2 = [s\_0,\infty)\times S^{n-1}$, its tangent bundle decomposes as


$$TM^3 \cong \pi\_1^\*T([s\_0,\infty)) \oplus \pi\_2^\*TS^{n-1},$$


where $\pi\_1,\pi\_2$ are the projections from $M^n$ onto its two components. Since $[s\_0,\infty)$ is contractible, its tangent bundle is trivial. The tangent bundle to $S^{n-1}$ is stably trivial, since embedding $S^{n-1}\hookrightarrow\mathbb R^n$ realizes $\underline{\mathbb R}^n \cong T\mathbb R^n|\_{S^{n-1}} = TS^n\oplus\nu$, where $\nu$ is the normal bundle to the embedding, and $\nu$ is a trivial bundle.


Since $w\_2$ is a stable characteristic class, it vanishes on stably trivial vector bundles, so $w\_2(M^n) = 0$, and therefore $M^n$ admits a spin structure.


