A theorem of Grothendieck states that any smooth reductive algebraic group over a field $k$ admits a maximal torus over $k$. My question concerns what happens for schemes.

Let $S$ be a scheme and let $G$ be a smooth reductive group scheme over $S$. Does $G$ admit a maximal torus over $S$?

Given that this is Grothendieck we are talking about, I imagine if he knew how to prove the result over schemes he would have done so, in particular I'm willing to believe that the answer is "no". However I don't know any explicit counter-examples.

I have a specific application in mind where my scheme $S$ is not too badly behaved, for example I can assume that $S$ is Noetherian, affine, regular with $\mathrm{Pic}(S)=0.$ If anyone knows any positive results in such special cases, I would also be very interested.

  • $\begingroup$ What sources have you consulted? The most standard treatment of reductive schemes seems to be Demazure's Expose XIX in SGA3 (see section 2): math.jussieu.fr/~polo/SGA3. Here the usual assumption is that $S$ is an arbitrary scheme. $\endgroup$ Commented May 2, 2014 at 15:50
  • $\begingroup$ I have been mostly reading Brian Conrad's notes on reductive group schemes. Here he has a result which says that a maximal torus exists étale locally (this is also in SGA3 I believe). However I don't know of an example which illustrates that one really does need to work étale locally. $\endgroup$ Commented May 2, 2014 at 16:10
  • $\begingroup$ I hope this is ok, but since there is some contention about the answer provided, I decided to unaccept it and offer a bounty. If anybody is able to conclusively answer my question, I would be most obligued. $\endgroup$ Commented May 6, 2014 at 7:24
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    $\begingroup$ Working etale locally for a maximal torus is indeed not necessary: Zariski locally suffices, see [SGA3, XIV, 3.20] (and the (*) footnote there in case it is of interest). $\endgroup$ Commented May 6, 2014 at 13:00
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    $\begingroup$ I don't know this stuff and was just trying to reconcile contradictory claims. The abelianization of a Borel doesn't depend on the choice of Borel and thus descends everywhere. In the quasi-split case, the torus of the Borel is better than other tori, though not necessarily a subgroup over general bases. In the general case, I think this construction yields the torus of the quasi-split inner form, not remotely like a subgroup of the original group, even in the field case (contrary to my previous claim). I don't know the purpose. $\endgroup$ Commented May 10, 2014 at 15:40

3 Answers 3


Counterexamples exist even over $S = \mathrm{Spec} \mathbb{Z}$. See, for instance, Lemma 1.1 and Example 6.2 of http://math.stanford.edu/~conrad/papers/redgpZ.pdf or other places of that paper. If $\mathscr{G}$ in Example 6.2 there had a maximal torus over $\mathbb{Z}$, then that maximal torus would be split, contradicting the anisotropy of $G_{\mathbb{R}}$.


I might well be missing something here, but:

Consider $S = \mathbb{P}^2$ and $E$ the tangent bundle to $\mathbb{P}^2$. Set $G = GL(E)$. If $T$ is a maximal torus of $G$ then,for every point $x \in \mathbb{P}^2$, we have a maximal torus $T_x$. The eigenspaces of $T_x$ form two points in $\mathbb{P}(E)$; let $\Lambda \subset \mathbb{P}(E)$ be the set of eigenspaces of the maximal torii. Then $\Lambda \to \mathbb{P}^2$ is a double cover. Since $\mathbb{P}^2$ is simply connected, $\Lambda$ has two connected components. Let $L_1$ and $L_2$ be the sub-line bundles of $E$ spanned by these components. Then $E = L_1 \oplus L_2$. But a standard computation with Chern classes shows that $E$ is not the direct sum of two line bundles.

Of course, this example is neither affine nor has vanishing Pic, as you last paragraph requests.

I now have an regular affine example with vanishing Pic. Take $X = \{ (a,b,c,x,y,z) : ax+by+cz=1 \} \subset \mathbb{C}^6$. I first note that $X$ is simply connected: Projection onto the $(x,y,z)$ plane reveals $X$ to be a rank $2$ affine bundle over $\mathbb{C}^3 \setminus \{ 0 \}$, so $X$ is homotopy equivalent to $\mathbb{C}^3 \setminus 0$, or to $S^5$.

In this answer, Steve Lansberg shows that $\mathcal{O}(X)$ is a UFD, so all line bundles on $X$ are trivial, but that $X$ possesses a nontrivial vector bundle $E$ of rank $2$. As in the previous answer, if $GL(E)$ had a maximal torus, the eigenspaces of that torus would give a double cover of $X$. Since $X$ is simply connected, that double cover is trivial and the torus is split. Then $E = L_1 \oplus L_2$. Since $L_1$ and $L_2$ are line bundles on $X$, they are trivial, but then $E$ is trivial, a contradiction.

  • $\begingroup$ What do you mean when you say that the eigenspaces of $T_x$ form two points in $\mathbb{P}(E)$? I am having trouble seeing a canonical decomposition of $E_x$ into eigenspaces. $\endgroup$
    – S. Carnahan
    Commented May 6, 2014 at 16:22
  • $\begingroup$ Given a scheme $X$, a rank $r$ vector bundle $E$ on $X$, and a torus $T \subset GL(E)$, we can consider the closed subscheme of $\mathbb{P}(E)$ given by $\{ (x,v) \in \mathbb{P}(E) : t \cdot v = v \ \forall_{t \in T_x} \}$. (Equality as points in projective space, so proportionality as elements of the vector bundle.) If I am not confused, in each fiber $\mathbb{P}(E_x)$ this should be a reduced zero dimensional scheme of length $r$. (So, if we work over the complex numbers, $r$ points over each closed point.) $\endgroup$ Commented May 6, 2014 at 16:29
  • $\begingroup$ Or, less formally, by assumption we have a torus $T_x$ in each fiber $GL(E_x)$. The elements in those torii all share a common eigenspace decomposition of $E_x$. $\endgroup$ Commented May 6, 2014 at 16:34
  • $\begingroup$ Ok, sorry. I was confusing monomial transformations (from the char. lattice) with linear transformations. $\endgroup$
    – S. Carnahan
    Commented May 6, 2014 at 16:43

I doubt that this is true even in the simplest case $G=GL(E)$, where $E$ over $S$ is a nondecomposable vector bundle.

  • $\begingroup$ Are you able to prove that this is the case? $\endgroup$ Commented May 2, 2014 at 16:11
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    $\begingroup$ A maximal torus T in GL(E) over S (with E a vector bundle) gives a decomposition of the vector bundle into line bundles. You can see this, even working with S a smooth variety over the complex numbers, since the eigenspaces for T give a local decomposition of E into 1-dim spaces. $\endgroup$
    – Marty
    Commented May 2, 2014 at 23:44
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    $\begingroup$ @Marty: I don't get it . This only works if $T$ is split. $\endgroup$ Commented May 5, 2014 at 9:50
  • $\begingroup$ @Laurent: Yes I see your point. Are you able to fix it? $\endgroup$ Commented May 6, 2014 at 7:21
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    $\begingroup$ One can deduce that $T$ is split in this setting for, e.g., $S = \mathbb{A}^3_{\mathbb{C}} - \{(0, 0, 0)\}$ and $E$ a nontrivial vector bundle on $S$ (for the existence of which, see Sasha's answer in mathoverflow.net/questions/35788/…). I am not sure though how Zariski local decomposition into line bundles is of help for the desired counterexample. $\endgroup$ Commented May 6, 2014 at 13:28

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