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Why the ICM is not as good an idea as it might sound September 3, 2009

Posted by Ben Webster in conferences.

Jim Humphreys’s comment reminded me of one of my rants that has yet to be bloggified. The non-hyperbolic title is above; those who are amused rather than annoyed by my hyperbolic titles can imagine it was called “why the ICM is a scam.”

Now, as usual, you should take this with big fat grain of salt; I’ve only been to one ICM, during one of the unhappiest periods of my recent life (if you don’t believe me, just wait for the post on why Denmark is a scam), so I’m doing a pretty unscientific extrapolation. That said, I’ve been to one ICM, and it didn’t make me all that eager to go back.

As far as I can tell, average conference quality as a function of size is pretty close to strictly deceasing; of course, this is somewhat debatable for small values of N, though I could easily believe the “conferences” of most consistent quality are between 2 people who have something to interest to say to each other. Another way to say this is that adding most people to most conferences will make them worse (especially when one considers quality of experience for a fixed person, rather than some more overall measure), since that person will just get in the way of the people who have something to say to each other talking (this may sound weird, but you can only meet so many people at a conference, so you want to make sure you meet the right ones). This effect becomes much worse when you consider than conferences have finite resources, not just in terms of funding, but also in terms of things like personal space in the venue and places to get lunch conveniently. In particular, some barrier is crossed when size of the conference raises beyond the number of people you can easily have a conversation with during the duration of the conference (somewhere around 30 or 40 for a week-long conference).

I think this effect is particularly strong for relatively young people. Once you’ve been around long enough, almost any conference you’re willing to go to will have plenty of people you know, and so it’s easier to navigate around the people who are in the way; at this point, I would even believe that optimal conference would increase a bit, since it raises the probability that specific people you know will be there.

By this logic, one would expect the ICM to be the worst of all conferences, and this is certainly what my experience bears out. The ICM, by virtue of it’s size, must be put in a venue ill suited to a math conference (the ICM and Joint Meetings are the only conferences I’ve ever seen with no blackboards), must charge a large conference fee, make it almost impossible to get lunch near the venue and ironically, is a rather difficult place to meet new people. After all, at a conference with 30 people, you can probably have an interesting mathematical conversation with almost anyone. The ICM, on the other hand, tends to surround you with people with whom you have very little mathematical commonality.

Not that I’m calling for the disbanding of the ICM, just pointing out that it’s more something that I (and I suspect many other people) appreciate the existence of more than I actually enjoy participating in. I suspect I’ll go again some time in the future, but I’m in no hurry.


1. Joel Kamnitzer - September 3, 2009

Dude, relax. Your advisor is one of the plenary speakers!

2. Andy P. - September 3, 2009

Conferences like the ICM and the Joint Meetings are lousy places for establishing new collaborations or things like that. However, I must say that frequently the plenary talks are very nice, and I really enjoy hearing good talks about subjects that I don’t get to think about very often. Plus I usually manage to run into people I don’t see very often.

The big meetings are almost the opposite of the very focused conferences, where I usually don’t enjoy the talks much but I get to chat with other people in my area and learn what they’ve been working on.

3. Scott Carnahan - September 4, 2009

I definitely agree that the ICM is awful for networking (I basically spent the week hanging out with people I already knew), but I can imagine that if you were at an institution that didn’t attract top-flight seminar speakers, there could be enormous mathematical benefits just from going to the talks. This is perhaps slightly diminished by the existence of resources like the arXiv.

The lack of interesting lunch near the Madrid conference center was a little disappointing, but I have fond memories of raiding the local supermercados.

4. Theo - September 4, 2009

What’s the matter with Denmark?

5. Scott Morrison - September 4, 2009

Somehow I trust the ingenuity of the locals wanting to feed me more next time than last time. That is to say, hopefully lunch will be better.

6. Mikael Vejdemo Johansson - September 4, 2009

@Theo – It’s full with Danes for one thing! ;-)

7. Ben Webster - September 4, 2009

Joel- True, and congratulations to him.

Scotts- It’s true that the Madrid situation was probably uniquely bad, and that going to grocery store and having a picnic was not a bad solution. However, I think it’s a representative example of the logistical problems.

Theo- Mostly the weather, the food and the exchange rate. But it’s not really that I have a vendetta against Denmark; my point was actually that fall 2006 was not a happy time for me, which any one who’s talked to me about my time in Denmark found out the hard way.

Mikael- Word.

8. Greg Kuperberg - September 4, 2009

What’s the matter with Denmark?

I’ve heard that there is something rotten in it. Although when I was there, it was really very nice.

If I may interject with a math question: I think that if K/F is a finitely generated field extension, then the algebraic subfield of K is also finitely generated and therefore finite-dimensional. I don’t know that it is so hard to prove, but it is difficult for me to make my own reasoning reliable. Does anyone know a reference?

I use this fact (if indeed it is a fact) to show that if Gamma is a finitely generated, indiscrete subgroup of PSU(a,b), then it has an elliptic element of infinite order. This is the converse to what I use in this paper that I am writing, so I don’t really need it either, but again, a reference would be cool.

9. Greg Kuperberg - September 4, 2009

Sorry, I should have said a dense subgroup, not an indiscrete subgroup.

10. Matthew Emerton - September 4, 2009

Re: Greg’s question from 8.

Lang’s book on algebraic geometry (quite out of fashion now) has a useful discussion of finitely generated field extensions at the beginning, if I’m remembering correctly. (There are probably many other references, but this is the one that — hopefully correctly — I remember.)

11. David Ben-Zvi - September 5, 2009

Ben – my experience at an ICM was very different. I went to the Berlin ICM (1998) while a grad student, and found it incredibly inspiring. The plenary speakers (almost) without exception delivered spectacular talks, far better than the norm of math talks. The section talks were more on par with normal conference talks, but the plenary speakers (for example Voevodsky, Shor, Sarnak, Diaconis, Miwa) really put in serious effort to communicate to a wider audience. The Fields medalists also gave great broader-audience talks, except for my favorite talk of the conference, by Kontsevich, which was very specialized but to me supremely exciting. I was also lucky to hover around the right experts at the right time so as to be able to join a “secret” seminar Kontsevich gave, which was very exciting.
In any case the range of top rate math was wonderful – I would strongly recommend going to someone midway in grad school or a young postdoc, when you are likely to have enough background to get something out of a lot of talks in many areas but are not yet transformed into a grumpy narrow old person who only cares about their field..

As for mingling, I met and befriended a whole lot of young mathematicians.. maybe the trick is to find someone very social who seems to know everyone (in my case, Niranjan Ramachandran), and in any case to consciously avoid just sticking around with the same three people you already know going in.

12. David Speyer - September 5, 2009


I think I can prove your result.

Lemma 1: Let k be a field and R a sub-k-algebra of k^{alg}. Then R is a field.

Proof: For any nonzero r \in R, we must show that r^{-1} is in R. The minimal polynomial of r is of the form \sum_{i=1}^n a_i r^i=0, with a_0 \neq 0. So r^{-1} = -a_0^{-1} \sum a_i r^{i-1}.

Lemma 2: Let L/K be algebraic, and be finitely generated as an extension of fields. Then L is finite dimensional over K.

Proof: Let t_1, t_2, …, t_r be generators for L as a field over K. Let R be the K-algebra generated by the t_i. By Lemma 1, R is a field, so R = L. If d_i is the degree of t_i over K, then a standard argument shows that R is spanned as a K-vector space by the monomials \prod t_i^{e_i}, with 0 \leq e_i \leq d_i.

Now, the main proof: let L/k be the finitely generated field extension in question, and let \ell be the algebraic closure of k in L. Since L is finitely generated over \ell, we can write L/\ell as a pure transcendental extension, \ell(x_1, \ldots, x_r), followed by a finite extension. Note that the x_i are also algebraically independent over k, so we have k(x_i) \subset \ell(x_i) \subset L.

Now, L/k(x_i) is algebraic, because each step in the tower is clearly algebraic. Also, L/k(x_i) is finitely generated as a field extension, because L/k is. So, by lemma 2, L has finite dimension as a k(x_i) vector space. But, clearly, \mathrm{dim}_k \ell = \mathrm{dim}_{k(x_i)} \ell(x_i) \leq \mathrm{dim}_{k(x_i)} L. So \mathrm{dim}_k \ell is finite, as desired.

13. Jim Humphreys - September 5, 2009

Certainly some of the older books like Lang’s “Algebraic
Geometry” or Weil’s “Foundations…” (which I don’t have at
hand) include enough field theory to serve as references
for Greg’s question. A more recent text “Field Theory”
by S. Roman (Springer GTM 158) has a detailed proof in
Theorem 3.3.5 of the basic result, using ideas like those
just written down by David: “a subfield in a finitely generated field extension is also f.g. over the base field”
(hence is finite dimensional if algebraic). Alternatively,
this finite generation result is just stated as Exercise 4
on page 374 of Lang’s “Algebra” (third edition) for those
who crave brevity.

As to my original comment on ICM 2010, I remain fairly
neutral, having never attended any of these congresses.
There have always been a lot of other competing uses
for time and travel money.

14. Greg Kuperberg - September 5, 2009

David’s proof is very nice, of course. It would be tempting for me to say that it was roughly the same as my proof. But actually I’m left wondering what my proof was and whether I really had one.

The citation in Roman’s book is ideal. Thanks very much to Jim Humphreys. The citation to Lang is more cumbersome because it is not clear what you might use from the section preceding the exercise. (Maybe it would be clear if I had the book next to me, but I am just browing in Amazon.)

While I have the attention of people such as David and Jim, I have one more frustrating missing “basic fact” in my paper to be. I have read in several places that the only real forms of the Lie algebra sl(n,\C) are sl(n,\R) and su(a,b) with a+b = n (and maybe sl(n,\C) itself, depending on your definition). I need an algebraic groups version of this fact. Suppose that G is a subgroup of SL(n,\C) which is real Zariski closed and complex Zariski dense, but which is not all of SL(n,\C). Then I think that the Lie algebra of G is one of the forms listed. But my intuition isn’t very good for the technicalities of the algebraic geometry here, so again, is there a good reference?

15. Greg Kuperberg - September 5, 2009

Okay, here is at least most of an argument: If G in SL(n,\C) is real Zariski closed, then it is the group of real points of an algebraic variety over \R, so it has a complexification G(\C). The inclusion of G in SL(n,\C) extends to a homomorphism from G(\C) to SL(n,\C), and it is is a corollary of Chevallay’s theorem on constructible images that the image of G(\C) is Zariski closed. Since G is Zariski dense by hypothesis, G(\C) surjects onto SL(n,\C). In particular, the Lie algebra g(\C) surjects onto sl(n,\C).

Again, even when I’m right, I’m just not practiced enough to know when to fill in more details. But I believe this reduces matters to the Lie algebra level, where the classification of precomplexifications is known.

16. Scott Carnahan - September 6, 2009

Greg, I’m not an expert in algebraic groups, but I think you need to specify G with a little more precision. Does G already have an algebraic group structure, or is it just a collection of complex points in SL_n? If G is a real algebraic group such that complexified Lie(G) is isomorphic to sl_n, then Lie(G) has one of the forms you specified, and the question is which real algebraic groups have those Lie algebras.

17. Greg Kuperberg - September 6, 2009

Scott: That was the part that I was struggling with. Suppose that G is a subset of SL(n,\C) which is a group, which is closed in the real Zariski topology on SL(n,\C), and whose closure in the complex Zariski topology is all of SL(n,\C). Does it follow that G is a real algebraic group such that either Lie(G) = sl(n,\C) or Lie(G) \tensor \C = sl(n,\C)?

18. Scott Carnahan - September 6, 2009

Okay, I had some vocabulary difficulties, but I think I’ve sorted them out. The set-theoretic group SL_n(\C) has a topology generated by complex points of Zariski opens in the scheme SL_{n,\mathbb{C}}, which you call the complex Zariski topology, and it has another topology generated by real points of Zariski opens in the Weil restriction \text{Res}_\mathbb{R}^\mathbb{C} SL_{n,\mathbb{C}}, which you call the real Zariski topology. Since G is a closed subgroup of the real points of the Weil restriction, it is the group of real points of a real algebraic subgroup of the Weil restriction. This works in general for perfect fields, and my reference is the first page of Algebraic Groups and Discontinuous Subgroups.

For the Lie algebra part, you can note that the Lie algebra of the Weil restriction is a real vector space with a complex structure, and you can ask that it be spanned as a complex vector space by the real Lie subalgebra Lie(G).

19. Jim Humphreys - September 6, 2009

I also have problems dechiphering language Greg uses to
describe his question, but probably Scott’s version is
reasonable. I have little hands-on experience with the
hybrid of real semisimple Lie and algebraic groups
(and their Lie algebras). But I have generally found the
route worked out by Borel and Tits for reductive
groups over arbitrary fields to be a fairly satisfactory
compromise between group schemes and concrete real
or complex matrix groups. The older language of Weil,
or Chevalley (especially when he wrote his volumes 2,3
on Lie groups but then abandoned the project), or the
version of Hochschild when working over infinite fields
doesn’t really adapt well enough for finite, local, and
other fields. One useful reference: Borel’s later lecture
notes (Math Reviews 2007h:22003), along with his books
and the paper by Tits in the 1965 Boulder conference
volume referenced by Scott. I’m one of the dwindling
band of survivors of that five week conference organized
mainly by Borel, which I couldn’t appreciate well at the
time while still in graduate school. Not a leisurely

Currently I’m sheltering at home near Smith College from
the area onslaught of 30,000 students. But I did sneak
over by foot to the rail trail trestle over the river, to
watch traffic jams on the parallel Coolidge Bridge. The
riverscape along the other side is almost pristine.
This year the weather gods have been very kind to us.

20. Jim Humphreys - September 6, 2009

I guess the word I wanted is “deciphering”.

21. Greg Kuperberg - September 6, 2009

The set-theoretic group SL_n(\C) has a topology generated by complex points of Zariski opens in the scheme…and it has another topology generated by real points of Zariski opens in the Weil restriction…

Right, this is what I am talking about. I think that part of the reason that my question may have been confusing is that I was using the old, pre-Grothendieck interpretation of the Zariski topology. But of course I don’t object to post-Grothendieck descriptions, when I can understand them.

For the Lie algebra part, you can note that the Lie algebra of the Weil restriction is a real vector space with a complex structure, and you can ask that it be spanned as a complex vector space by the real Lie subalgebra Lie(G).

Yes, this is what I ask for, but on this point we may have been talking past each other. I’m asking for this property of Lie(G) as a corollary of the hypotheses on G, and not as a hypothesis itself. In particular, I would think that it follows from the fact that G is both real Zariski closed and complex Zariski dense. Complex Zariski dense means that G intersects every Zariski open in the scheme SL(n,\C) over \C.

The argument that I think I have for this question requires Chevallay’s theorem that the image of a quasiprojective variety under a regular map is constructible. According to the literature, it is a corollary of Chevallay’s theorem that if the regular map is a morphism between affine algebraic groups, then its image is a closed subvariety.

22. In which I celebrate Labor Day by ripping off others « Portrait of the Mathematician - September 7, 2009

[…] We have less than a year to go until the next ICM, and speculation about the prizewinners is starting to kick into gear over at Not Even Wrong. Ngo Bau Chao looks like the best bet for a Fields next year, but the really interesting thing is the very good possibility that the Nevanlinna prize might finally be awarded to a woman, with nearly half the qualified TCS sectional speakers, plus the token TCS plenary speaker (Irit Dinur), being women. But Ben Webster suggests that the ICM is maybe not such a good idea. […]

23. Scott Carnahan - September 7, 2009

If the corollary of Chevalley’s theorem is the one given in Brian Conrad’s paper just after the theorem, then I do not believe it is applicable to your situation. You have to be careful to avoid mixing the category of varieties over R with the category of varieties over C. If you do want to mix them, you have to go through the appropriate functors.

24. Greg Kuperberg - September 7, 2009


I have in mind this Chevalley’s theorem, and corollary 1.4(a) in Borel’s GTM 126: “Let G’ be a k-group and alpha:G -> G’ a morphism. Then alpha(G) is a closed subgroup of G; and it is defined over k if alpha is defined over k.”

The morphism in this case is the one from G(\C) to SL(n,\C), induced by the inclusion of G into the Weil restriction of SL(n,\C). I believe that this corollary 1.4(a) implies the spanning condition for Lie(G).

25. Jim Humphreys - September 8, 2009

Greg – The language used to formulate your original
question still confuses me a little, so I might be
oversimplifying it. In any case the question looks fairly
concrete and should not require sophisticated machinery.
(Or ideas from Brian Conrad’s paper, which I sort-of
reviewed for Math Reviews and which deals with a much
broader setting for “algebraic groups”.) I’m tempted to
say that your given real Lie group is just an
\mathbb{R}-form of SL(n,\mathbb{C})
so that its Lie algebra is just one of the well-known
\mathbb{R}-forms of the simple complex Lie
algebra of type A, ranging from compact type to split
type. The work of Borel and Tits on classification and
structure of k-forms for arbitrary fields of definition is
a more sophisticated outgrowth of the older work on
real and complex groups and uses more explicitly the
framework of Galois cohomology and restriction of
scalars, etc. The later sections of Borel’s lecture notes
which I mentioned convey the flavor of the classical work
for real and complex groups in fairly readable fashion.
Transitions between Lie groups and algebraic groups
are not easy to quote from the literature, but Borel at
least tried to bridge the language gap.

26. Greg Kuperberg - September 8, 2009

I’m tempted to say that your given real Lie group is just an \R-form of SL(n,\C) so that its Lie algebra is just one of the well-known \R-forms of the simple complex Lie algebra of type A, ranging from compact type to split type.

Yes, that is what I was tempted to say all along, although it didn’t help that I didn’t list the \R-forms properly in my first message on this. Your Borel reference also looks very useful! It may be a better reference for several things that I need. However, I don’t think it has the specific technical thing that I was looking for: If G is complex Zariski dense, then it contains an \R-form as a finite-index subgroup. But I think that I have a proof of that small lemma using Borel’s other book.

27. Scott Carnahan - September 8, 2009

I think I understand what you’re trying to do now. Let’s replace your set-theoretic group with the unique real-algebraic group that sits inside \text{Res}^\mathbb{C}_\mathbb{R} SL_{n,\mathbb{C}} whose real points are the elements of your group. I’ll call this group G. Since base change is left adjoint to Weil restriction, the inclusion canonically induces a homomorphism of complex algebraic groups \phi: G_\mathbb{C} \to SL_{n,\mathbb{C}} (as you mentioned a few comments ago). The unit of the adjunction endows the Lie algebra of G with the structure of a totally real subspace of the Lie algebra of G_\mathbb{C}, and the latter Lie algebra is the complex span of the former.

\phi induces a map on complex Lie algebras. Let’s suppose it is not surjective. Then the image of \phi is a complex algebraic subgroup of positive codimension, and the union of base-changed elements of G(\mathbb{R}) has Zariski closure inside this subgroup. This contradicts your “complex Zariski dense” condition, so we get Lie(G_\mathbb{C}) \twoheadrightarrow sl_{n, \mathbb{C}}.

Unfortunately, this doesn’t seem to restrict your group to the list you want. I think there are real subgroups of \text{Res}^\mathbb{C}_\mathbb{R} SL_{n,\mathbb{C}} of relatively small codimension.

28. Greg Kuperberg - September 8, 2009

Hi Scott. Thanks for bearing along with me. It is true that my list of real Lie subalgebras of sl(n,\C) that span it over \C was incomplete. Even so, although I don’t have a complete picture of this yet, I think that they are either all of sl(n,\C) or half of it.

29. David Speyer - September 8, 2009

Dear Greg and Scott: I lost the details of your conversation a while ago, but do I understand that you have reduced the problem to classifying real Lie subalgebras of sl(n, C) which span it over C?

If so, then I think Greg is right that they are all (1/2)-dimensional, except for sl(n, C) itself. Here is a sketch of a proof:

Let V be such a subalgebra. Consider W:=V \otimes_{\mathbb{R}} \mathbb{C}, as a subalgebra of \mathfrak{sl}(n,\mathbb{C}) \otimes_{\mathbb{R}} \mathbb{C}. Now, \mathfrak{sl}(n,\mathbb{C}) \otimes_{\mathbb{R}} \mathbb{C} is isomorphic to \mathfrak{sl}(n,\mathbb{C}) \oplus \mathfrak{sl}(n,\mathbb{C}). Let’s call these summands L_1 and L_2 (so I don’t have to type as much!)

I believe that the condition that V spans over \mathbb{C} says that W surjects onto each of L_1 and L_2 under projection. In other words, for any g \in L_1, we can lift to some (g,h) \in W. Since W is a Lie subalgebra, bracketing with (g,h) must take W into itself. But bracketing with anything takes L_1 into itself. So we deduce that W \cap L_1 is an ideal of L_1. Since L_1 is simple, this means that L_1 \cap W is either empty (in which case W is half dimensional) or is all of L_1. Similarly, if W is more than half dimensional then W \supset L_2.

In summary, if W has more than (1/2) the dimension of L_1 \oplus L_2, then it is all of L_1 \oplus L_2.

Am I missing anything?

30. Greg Kuperberg - September 8, 2009

David: This time I really did have a proof, but yours is better.

There is a wonderful theorem of Philip Hall that says that if G is a finite product of finite simple groups G_k, and if H is a subgroup that surjects onto each factor, then H is a “diagonal” subgroup in a suitably generalized sense. In other words, up to isomorphisms of the factors, H is a product of diagonals.

This theorem and its proof apply equally well to a finite direct sum of finite-dimensional simple Lie algebras. In fact you found most of the proof. Once you verify the surjection property, you immediately get the dimension condition for W.

Indeed you immediately get that there are two kinds of half-dimensional W when n is at least 3: Those that come from of an inner automorphism of sl(n,\C), and those that come from the outer automorphism. This type of analysis is a basic step in the classification of real simple Lie algebras.

31. Scott Carnahan - September 8, 2009

Thanks David. I think the part I missed was the step that V spanning over C implies W surjecting on each factor. For some reason, I thought it might be possible to surject onto just one of the factors. Is there a Galois automorphism that switches them?

32. Greg Kuperberg - September 8, 2009

Scott: Suppose that A is a complex algebra or Lie algebra and A_\R is A as a real algebra, i.e., the forgetful functor from \C-vector spaces to \R-vector spaces applied to A. Then A_\R tensor_\R \C is isomorphic as a \C-algebra to A oplus A, and the Galois automorphism of the left factor A_\R indeed switches the two summands.

This works in general for vector spaces or algebras or any such structures over a field K which is a Galois extension of another field F. It’s more fun to work out what happens when K is a non-Galois extension. The heart of the matter is the following question: Suppose that K is a non-Galois extension of F. Then what is the F-algebra K tensor_F K? (Note that it is two different questions, depending on whether K is separably non-Galois or purely inseparable.)

This is something that fans of 0-dimensional schemes, a.k.a. dots, like to think about.

33. Daniel Moskovich - September 8, 2009

I don’t know whether I would specifically criticize the ICM (never having been to one), but there certainly seems to be a critical size for conferences, beyond which nothing really gets done.
A few dozen people seems to work nicely, and if you get the right group together it can be wonderful! I don’t know about collaborations, but a conference of just the right size which develops the right sort of atmosphere can just be tremendous fun! And one feels less socially isolated caring about mathematical things, because all these other cool people care about similar things too!

34. David Speyer - September 9, 2009

I checked Fulton and Harris and, according to the table on page 433 (section 26.1), Greg only missed one case: if n is even, then sl_{n}(\mathbb{C}) is the complexification of sl_{n/2}(\mathbb{H}). Here \mathbb{H} is the quaternions, and sl means that the real part of the trace is zero.

For n=2, I note that this is isomorphic to su_2, but I assume that it is not in general or Fulton and Harris wouldn’t have listed it seperately.

35. Greg Kuperberg - September 9, 2009

Right, to summarize, the general list of real forms is sl(n,\R), su(p,q), and sl(n/2,\H). In general they are all different.

When n = 2, there are not one but two cute coincidences: su(2) is isomorphic to sl(1,\H), and sl(2,\R) is isomorphic to su(1,1).

36. JSE - September 10, 2009

“There is a wonderful theorem of Philip Hall that says that if G is a finite product of finite simple groups G_k, and if H is a subgroup that surjects onto each factor, then H is a “diagonal” subgroup in a suitably generalized sense. In other words, up to isomorphisms of the factors, H is a product of diagonals.”

Just dropping in to second this; I find I use this fact surprisingly often.

37. Florian - September 10, 2009

Do you have a reference for this theorem of Hall’s?


38. Greg Kuperberg - September 10, 2009

author = “Philip Hall”,
title = “The {Eulerian} functions of a group”,
journal = “Quart. J. Math.”,
volume = 7,
year = 1936,
pages = “134–151”}

I learned of the reference from a paper of Dunfield and Thurston.

The proof is a good graduate problem in finite group theory. (Indeed, it has been said that until Feit-Thompson, most results in finite group theory were relatively short.) Here is the version posted in my paper (which should be public in a few hours in the arXiv):

Lemma. Suppose that each of G_1, G_2, \ldots, G_\ell is a minimal simple Lie group or a non-abelian finite simple group, and suppose that H \subseteq G = G_1 \times G_2 \times \cdots \times G_\ell is a closed subgroup that surjects onto each factor G_k. Then H is a diagonal subgroup of G.

39. abm - September 10, 2009

Dear Greg: What does the phrase “minimal simple Lie group” mean in your statement of the Lemma? The only possible meanings for the word “minimal” in this context that I can think of either exclude nearly all simple lie groups or are redundant.

40. Greg Kuperberg - September 10, 2009

It means centerless; maybe I should have used that term. Technically speaking, a simple Lie group is not in general a simple group, but rather a Lie group whose Lie algebra is simple. For any fixed Lie algebra, they are all covering spaces of the minimal/centerless one. Also when you take a covering space of a Lie group, it is always a central extension.

A finite group which is analogous to a “simple” Lie group is called quasisimple.

41. Noah Snyder - September 10, 2009

Arxiv version of Greg’s paper that you’ve all be discussing is up.

42. Greg Kuperberg - September 10, 2009

Thanks, Noah!

43. Florian - September 11, 2009

Dear Greg: thank you very much for the reference! Now the statement (and proof) is clear too, that’s very nice. It’s interesting that I can’t find Hall’s paper on Mathscinet, only at Zentralblatt (too early?).

Incidentally, I think there’s a word missing in the abstract of your preprint. Also, is “connectness” a word or is it a typo?

44. Florian - September 11, 2009

P.S.: isn’t “minimal simple Lie group” the same as “adjoint simple Lie group”?

45. Greg Kuperberg - September 11, 2009

Florian: Your questions are good; I wish I knew your full name.

It may well be the same as an “adjoint” simple Lie group, but I did not know that term and even now I can only find it in a couple of papers. I think that “centerless” simple Lie group is a more common term, and maybe the one that I should have used.

46. Scott Carnahan - September 11, 2009

I believe “adjoint form” is the more commonly used phrase for semisimple groups (Borel definitely used it), but “centerless” seems to have an advantage in terms of clarity.

47. Emmanuel Kowalski - September 12, 2009

“adjoint” (form or group or type) is definitely the terminology in the algebraic group literature (Borel’s book and Springer’s use it). It’s meaning is that is (isomorphic to) the image of the adjoint representation of the group on the Lie algebra. Note that in positive characteristic, the kernel of the adjoint representation of a semisimple algebraic group is not necessarily the same as the center (this is mentioned in those two books if I remember right, with an example in one of them, which I can’t recall).

48. Greg Kuperberg - September 12, 2009

I am working in characteristic 0, so for me it doesn’t really matter. However, in the general context of Hall’s lemma, centerless is more correct than adjoint. The point is that “centerless simple” means really and truly simple, while “simple” means quasisimple.

49. Florian - September 12, 2009

Sorry about hiding my last name, here’s a link to my web page.

“Centerless” sounds good to me.

I have also seen G “adjoint”, for G a semisimple algebraic group over a field, defined by Z(G) = 1, regardless of the characteristic. (E.g., Jantzen’s book on reps. of alg. gps.) But here, Z(G) must be considered as group scheme, which may be non-reduced. This definition should be equivalent to the one Emmanuel Kowalski gave.

Example: G = SL_p in characteristic p. This is not adjoint, even though on the level of geometric points the centre is trivial. We have Z(G) = mu_p.

50. Emmanuel Kowalski - September 12, 2009

So this is another good example of why scheme-theoretic viewpoints are important even for fairly concrete questions…

A quick look at SGA3 (for which a proper LaTeX re-edition organized by P. Gille and P. Polo is approaching completion, see this page for the current drafts of the various exposés) yields indeed that a semisimple algebraic group scheme over a base S is adjoint if and only if the center, as S-group scheme, is trivial (if I inter[polate right the notation; this is Exposé XXII, Proposition 4.3.5, (i)).

(By the way, as JSE pointed out, Hall’s Theorem or variants thereof are used very often in number theory, though we tend to name it using some combination of the names Goursat, Kolchin and Ribet, with Serre sometimes thrown in; various books of Katz contain versions for Lie algebras and groups).

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