Let $G$ be an algebraic group over an algebraically closed field. If $H$ and $K$ are closed subgroups and one of them is connected, then their commutator $[H,K]$ is also connected.

Is there an easy way to see this fact?

The proof that I see in Springer's book on Linear Algebraic Groups is very long-winded. (Of course, he obtains many results on the way). My question is, is the given statement true for any topological group? If yes, does that proof apply to the algebraic groups' case?

(Remark: The definition of a topological group includes the Hausdorff property, whereas algebraic groups are not so).


5 Answers 5


The case for $G$ a topological group doesn't look hard; it just comes down to two facts:

  • A union of connected sets which have a point in common is also connected. For example, suppose $K$ is a connected subgroup and $H$ is any set. For each fixed $h \in H$, the set of commutators $[h, K]$ is the image of $K$ under the continuous map $G \to G$ that takes $g$ to $[h, g]$; this image is connected and contains the identity (since $g = e$ belongs to $K$). But then the set of all commutators $[h, k]$ is a union of connected sets $[h, K]$ each containing the identity, so it too is connected.

  • If $S \subset G$ is a connected set containing the identity, then so is the group it generates. This is on the same principles as before, that the image of a connected set under a continuous map is connected, and the union of connected sets with an element in common is connected. For example, the set $T = S \cup S^{-1}$ is connected, and then each $T \cdot T \cdot \ldots \cdot T$ is connected, being the image of the connected set $T \times T \times \ldots \times T$ under the multiplication map $G^n \to G$, and finally the union of all these sets gives the group generated by $S$.

Now apply the second fact to the connected set of commutators coming from the first fact, to show that the subgroup $[H, K]$ is connected.

  • 4
    $\begingroup$ It is somewhat subtle, but an algebraic group is not a topological group for the Zariski topology. This is because for a topological group one requires that the multiplication map $\mu:G \times G \to G$ be continuous for the product topology, while for algebraic groups one puts on the Zariski topology on $G \times G$, which does not coincide with the product topology in general. So it is not clear to me why the argument for topological groups should imply the statement for algebraic groups. $\endgroup$
    – Guntram
    Sep 2, 2012 at 5:24
  • 2
    $\begingroup$ Right, I was only addressing one of the questions of the OP. There is no reason I can think of why the statement for topological groups should imply the statement for algebraic groups. $\endgroup$
    – Todd Trimble
    Sep 2, 2012 at 6:04
  • $\begingroup$ I missed the subtle point that Guntram mentioned. Thanks for pointing it out. $\endgroup$ Sep 2, 2012 at 15:31

There are actually three questions here. The first is: Is there an easy way to see this fact?, to which the answer is (almost certainly) no. Keep in mind that both Chevalley and Borel came to algebraic groups from a background in Lie groups and were therefore well aware of what worked for a topological group. The third question has the same answer: whether the topological group result applies to the algebraic group case. "Long-winded" answers may be a necessary evil here. Keep in mind too that the applications of the foundational results behind your questions deal with whether certain subgroups of algebraic groups are closed and/or connected. There is some essential interaction between those properties.

Todd has worked out the answer to the second question, concerning topological groups. Here the basic results are fairly old, so probably a similar proof is written down somewhere in the literature. But I'll elaborate on some of the other answers and comments about the essential difference in the algebraic group setting.

1) Algebraic groups (here assumed affine) are given the Zariski topology; in particular, irreducible sets are the more natural refinement of the topological notion of connected. In fact, Chevalley's original arguments about commutator groups and such just used the term "irreducible". It's true that for an algebraic group, which has only finitely many irreducible components (all disjoint), the notions "irreducible" and "connected" coincide. But this can cause confusion at times. In any case, the Zariski product topology isn't the usual one, so all topological arguments involving products and continuous maps have to be rethought in algebraic geometry.

2) In his 1951 second volume in French on Theorie des groupes de Lie, Chevalley tried out a framework for algebraic groups which proved later to be inadequate in prime characteristic especially (so he changed gears). But he did rethink all the foundational material related to connectedness, which led classically to connectedness of familiar linear groups. His II.7 contains the prototype, hard to read now, of the basic argument.

3) The notes by Bass of Borel's 1968 Columbia lectures Linear Algebraic Groups adopted more modern algebraic geometry but avoided most scheme language due to time constraints. Chevalley's "long-winded" argument is recast here in I.2.2 and I.2.3 (where part (a) is used to get part (b)). These ideas are crucial for several applications. Coming this early in the theory and used for example to treat solvability, the proofs necessarily rely on first principles. (Borel's expanded second edition, Springer GTM 126, leaves this material unchanged.)

4) No substantive changes are made in my 1975 book GTM 21 and in Springer's 1981 Birkhauser text. In my book, see 7.5 and 17.2, while in Springer's book see 2.2.6-2.2.8.

  • $\begingroup$ Thank you very much for your enlightening views. Point (1) was especially instructive. But in (4) - references, your book has no 7.5 and 17.2 is about universal enveloping algebras. Are you sure you wanted to point to that? $\endgroup$ Sep 3, 2012 at 18:25
  • $\begingroup$ @Abhishek: Sorry for the wrong GTM number, which I've edited. (I meant my second book in 1975, with the same title Linear Algebraic Groups as the books by Borel and Springer.) $\endgroup$ Sep 3, 2012 at 20:52

It seems to me that in a topological group $G$ if $H$ is a connected subset containing the identity and $K$ is any subset then the subgroup generated by all commutators $hkh^{-1}k^{-1}$ is connected. $G$ does not to be Hausdorff, and $H$ and $K$ do not have to be closed or to be subgroups.

The ingredients in the proof are: (1) The image of a connected set under a continuous map is connected. (2) The union of a set of connected sets is connected if they all have some point in common. (3) The subgroup generated by a connected set is connected.

  • $\begingroup$ Sorry, Tom, I didn't see your answer while I was typing. $\endgroup$
    – Todd Trimble
    Sep 2, 2012 at 3:33
  • $\begingroup$ What are the connected sets whose union you are considering in (2)? $\endgroup$ Sep 2, 2012 at 3:37
  • $\begingroup$ Actually, is what you wrote true if the connected subset does not contain the identity? If for example $H$ and $K$ were both 1-element sets, then it seems $[H, K]$ could be a discrete set, unless I'm missing something. $\endgroup$
    – Todd Trimble
    Sep 2, 2012 at 3:39
  • $\begingroup$ Todd, yes, I was unconsciously assuming $H$ contained the identity. Fixed now. $\endgroup$ Sep 2, 2012 at 12:12
  • $\begingroup$ The proof that I had in mind was exactly the one that Todd gives in detail in his answer. $\endgroup$ Sep 2, 2012 at 15:58

Using classical varieties (and classical points only), since $G^n$ does not have the product topology (in the algebraic group setting) it isn't clear what can be useful for $G$ concerning "topological" statements (as in Todd's answer) concerning the product topology on the subset $T \times T$ inside $G \times G$ when $T$ is just some random subset of $G$ (not yet known to be constructible). So although topological groups provide valuable intuition that can sometimes be transported to the case of algebraic groups (which are of course not themselves topological groups in general), in this case the central issue is not addressed by thinking about topological groups.

The purpose of the longer delicate arguments one finds in the basic textbooks on algebraic groups is that the commutator subgroup is reached in "finitely many steps" (even without connected hypotheses on $H$ or $K$, which is very important for applications) and so is constructible. It is for constructible $T$ that $T \times T$ with the "right" topology (inherited from $G \times G$) is connected when $T$ is connected, etc. The hard part therefore involves a problem which doesn't arise in the topological group setting (unless one poses finer topological question, such as closedness of $(H,K)$ under some reasonable hypotheses, which is a deeper problem than mere connectedness).

For an arbitrary (not necessarily constructible) connected subset $T$ of $G$ is the subset $T \times T$ inside $G \times G$ (the latter given the Zariski topology) connected?

  • $\begingroup$ What do you mean by `constructible'? $\endgroup$ Sep 2, 2012 at 15:30
  • 1
    $\begingroup$ Chevalley's theorem on preservation of constructibility under images pervades the beginning of the theory of algebraic groups. The constructible subsets of a noetherian topological space are the finite unions of locally closed sets. Check Wikipedia under "constructible set" (though the example there is weak, since it is even locally closed; a better example is the subset of the plane given by the union of the origin and the complement of the $x$-axis). By the way, the end of your posted question made me think that you were well aware that algebraic groups aren't topological groups. $\endgroup$
    – grp
    Sep 2, 2012 at 16:07

This fact is Corollary 16.4.1 on page 40 of the course notes from (the first semester of) Brian Conrad's course on linear algebraic groups. In the notes there is a proof of this fact is modern language, over an arbitrary field. The notes can be found here: http://math.stanford.edu/~conrad/252Page/handouts/alggroups.pdf


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.