Consider a hypersurface $X=V(f) \subset \mathbb A^n\_{\mathbb C}$, where $f(T\_1, T\_2,\ldots,T\_n)\in \mathbb C[T\_1,Y\_2,\ldots, T\_n]$ is a polynomial .  

Assume that $X$ is smooth, i.e. that $df(x)\neq 0 \;$ for all $x\in X$ . My question is simply whether $X $ is parallelizable i.e. whether its tangent bundle $T\_X$ is algebraically trivial. 


I've asked a few friends and their answer was unanimously "no, why should it be?", but they couldn't provide a counter-example. Here are a some considerations which might show that the question is not so ridiculous as it looks. 


We have the exact sequence of vector bundles on $X$


$$0 \to T\_X\to T\_{A^n\_{\mathbb C}}|X\to N(X/A^n\_{\mathbb C})\to 0$$


Now, the normal bundle $N(X/A^n\_{\mathbb C})$ is trivial (trivialized by $df$) and the restricted bundle $T\_{A^n\_{\mathbb C}}|X$ is trivial because already $T\_{A^n\_{\mathbb C}}$ is trivial. Moreover the displayed exact sequence of vector bundles splits, like all exact sequences of vector bundles, because we are on an affine variety. So we deduce (writing $\theta$ for the trivial bundle of rank one on $X$)
$$\theta^n=T\_X\oplus \theta $$
In other words the tangent bundle is stably trivial, and this is already sufficient to deduce (by taking wedge product) that $\Lambda ^{n-1}T\_X=\theta$ (hence the canonical bundle $K\_X=\Lambda ^{n-1}T\_X^\ast$ is also trivial). This suffices to prove that indeed for $n=2$ the question has an affirmative answer: every smooth curve in $A^2\_{\mathbb C}$ *is* parallelizable.


Another argument in favor of parallelizability is that there are no analytic obstructions: O. Forster has proved a result for complex analytic manifolds which implies that analytically (and of course differentiably) our hypersurface is parallelizable: $T\_{X\_{an}}=\theta \_{an}^{n-1}$.This is why I choose $\mathbb C$ as the ground field: the question makes perfect sense over an arbitrary *algebraically closed* field but I wanted to be able to quote the related analytic result.[ As ulrich remarks, parallelizability can't be deduced over the non-algebraically closed field $\mathbb R$, as shown by a 2-sphere]


**Edit** ulrich's great reference not only answers my question but seems to yield more results in the same direction. For example consider a smooth complete intersection: $X=\{ x\in \mathbb C^n|f\_1(x)=f\_2(x)=\ldots=f\_k(x)=0 \} $ with the $f\_i$'s polynomials and the $df\_i(x)$'s linearly independent at each $x\in X$ . Then, just as above, the normal bundle is trivial and the tangent bundle is stably trivial: $\theta^n=T\_X\oplus \theta ^{k} $  

So Suslin's incredible theorem again allows us to conclude that $X$ is parallelizable. 


However not all affine smooth algebraic varieties are parallelizable: for example the complement of a smooth conic in $\mathbb P^2(\mathbb C)$ is a smooth *affine* variety (Veronese embedding !) but is not even differentiably parallelizable. I wonder if these differentiable obstructions are the only ones preventing algebraic parallelizability of smooth algebraic subvarieties of $\mathbb C^n$. Any thoughts, dear friends? 



Is every algebraic smooth hypersurface of affine space parallelizable?
Yes. Suslin has proved that every stably trivial vector bundle of rank $n$ on an affine variety of dimension $n$ over an algebraically closed field is trivial. See:


Suslin, A. A. Stably free modules. Mat. Sb. (N.S.) 102(144) (1977), no. 4, 537–550, 632. 


Note that this is not true over arbitrary fields; for example, the tangent bundle of the $2$-sphere (given by $x^2 + y^2 + z^2 - 1 = 0$) over $\mathbb{R}$ is not trivial (since it is not so even topologically). However, it is true over finite fields and also over $C\_1$ fields of characteristic $0$. See: 


Bhatwadekar, S. M. A cancellation theorem for projective modules over affine algebras over $C\_1$-fields. J. Pure Appl. Algebra 183 (2003), no. 1-3, 17–26.


