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?

  • $\begingroup$ Without the stably trivial assumption on the tangent bundle it is easy to find examples of smooth affine varieties which are not parallelizable but which are holomorphically paralellizable. For example, consider a general curve of genus $g>1$ and remove a point. $\endgroup$
    – naf
    Jul 1 '11 at 15:35
  • $\begingroup$ Dear ulrich, you are absolutely right. I find it amazing that every holomorphic vector bundle on a non compact Riemann surface (and in particular on a complex affine algebraic curve) is trivial, whereas on an affine curve of genus $g\geq 1$ already the Picard group is huge, essentially as big as that of the completed curve. I gave an explicit example just to be specific and because I found it funny to mention a down-to-earth conic in algebraic geometry, a branch of mathematics that has the unfortunate reputation of being very sophisticated and abstract... $\endgroup$ Jul 1 '11 at 17:17

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.

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    $\begingroup$ Great question and great answer! $\endgroup$ Jul 1 '11 at 10:46
  • $\begingroup$ Dear ulrich, thank you very much for your great answer. It seems that your reference is in a Russian journal to which I have no access. Is there a non-gated online reference to the article (I don't mind that it is in Russian) or some translation or some secondary reference? $\endgroup$ Jul 1 '11 at 13:44
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    $\begingroup$ On a lighter note, it is quite amusing that you mention the tangent bundle of the real 2-sphere. I gave a talk centered on the Serre-Swan theorem three weeks ago and this example was at the center of my survey. More generally, I keep mentioning that example to whoever wants to listen to me (and to some others, too) and now I can prove that I'm not the only one to mention it, in case my colleagues start to think that this is some pathological obsession of mine:-) $\endgroup$ Jul 1 '11 at 13:49
  • $\begingroup$ And on a darker note, this confirms my old rant that it is an unforgivable omission that Suslin never got a Fields Medal. Shame on the relevant Fields Committee(s)! $\endgroup$ Jul 1 '11 at 14:34
  • $\begingroup$ Dear Georges, sorry I don't any other reference. There is an English translation of Suslin's paper but it is, I think, behind a paywall. $\endgroup$
    – naf
    Jul 1 '11 at 15:17

There is a huge amount of work on these kinds of questions (I am aware of them since one of my colleagues, Satya Mandal, works on related topics). For example, this recent paper:

shows that stably free modules on smooth affine threefolds over alg. closed fields of char. not $2,3$ are free.

Another interesting relevant issue for your new question is when you can split a rank $1$ free off a projective module of rank $d=\dim R$ (Serre showed you can always do if the rank $>d$). Nori outlined a program to find the obstruction for this, something now called the Euler class group. You can find some relevant information and references at the second paper here:

(Local Coefficients and Euler Class Groups)



  • $\begingroup$ Dear Hailong, I particularly appreciate your two links, since I had no idea about what was being done in that subject right now, nor about who was involved. So I am very lucky to have someone as qualified as you to guide me: thanks a lot ! $\endgroup$ Jul 1 '11 at 17:42
  • $\begingroup$ Dear Georges, you are too kind, I just work at the same place with someone who knows this stuff! I am sure there are people who know a lot more about your second question. May be some of them will join MO soon. $\endgroup$ Jul 1 '11 at 21:44

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