jump to navigation

Passage from compact Lie groups to complex reductive groups November 25, 2010

Posted by Joel Kamnitzer in Algebraic Geometry, representation theory, things I don't understand.
trackback

Once again, I’m preparing to teach a class and needing some advice concerning an important point. I’m teaching a course of representation theory as a followup to an excellent course on compact Lie groups, taught this semester by Eckhard Meinrenken. In my class, I would like to explain transition from compact Lie groups to complex reductive groups, as a first step towards the Borel-Weil theorem.

A priori, compact connected Lie groups and complex reductive groups, seem to have little in common and live in different worlds. However, there is a 1-1 correspondence between these objects — for example U(n) and GL_n(\mathbb{C}) are related by this correspondence. Surprisingly, it is not that easy to realize this correspondence.

Let us imagine that we start with a compact connected Lie group K and want to find the corresponding complex algebraic group G. I will call this process complexification.

One approach to complexification is to first show that K is in fact the real points of a real reductive algebraic group. For any particular K this is obvious — for example S^1 = U(1) is described by the equation x^2 + y^2 = 1. But one might wonder how to prove this without invoking the classification of compact Lie groups. I believe that one way to do this is to consider the category of smooth finite-dimensional representation of the group and then applying a Tannakian reconstruction to produce an algebraic group. This is a pretty argument, but perhaps not the best one to explain in a first course. A slightly more explicit version would be to simply define G to be Spec (\oplus_{V} V \otimes V^*) where V ranges over the irreducible complex representations of K (the Hopf algebra structure here is slightly subtle).

In fact, not only is every compact Lie group real algebraic, but every smooth map of compact Lie groups is actually algebraic. So the
the category of compact Lie groups embeds into the category of real algebraic groups. For a precise statement along these lines, see this very well written
MO answer by BCnrd.

A different approach to complexification is pursued in
Allen Knutson’s notes and in Sepanski’s book. Here the complexification of K is defined to be any G such that there is an embedding K \subset G(\mathbb{C}) , such that on Lie algebras \mathfrak{g} = \mathfrak{k} \otimes_{\mathbb{R}} \mathbb{C} . (Actually, this is Knutson’s definition, in Sepanski’s definition we first embed K into U(n) .) This definition is more hands-on, but it is not very obvious why such G is unique, without some structural theorems describing the different groups G with Lie algebra \mathfrak{g} .

At the moment, I don’t have any definite opinion on which approach is more mathematically/pedagogically sound. I just wanted to point out something which I have accepted all my mathematical life, but which is still somewhat mysterious to me. Can anyone suggest any more a priori reasons for complexification?

About these ads

Comments

1. David Speyer - November 27, 2010

It seems to me that, from any of these perspectives, the hard part is to show that a compact Lie group has any nontrivial finite dimensional representations. In the Tannakian persective, how do you know the tensor category isn’t trivial; in Allen/Sepanski’s description, you define it in by assuming that K has a unitary rep.

What is the easiest way to get to this fact?

Ben Webster - November 27, 2010

David- Isn’t any unitary irreducible of a compact group finite dimensional for simple topological reasons? Assuming that it’s easy; just decompose $L^2(G)$.

2. Peter J McNamara - November 28, 2010

David – Any Lie group has an adjoint representation. If you want a faithful representation of a compact Lie group, take a direct sum of adjoint tensor central character for appropriate central characters.

Ben – I’m not sure about this simple topological reasons part, I thought you needed to use something like spectral theory of compact operators. eg, if f \in L^2(G) is invariant under the adjoint action of G, then it is central, and acts by a compact operator on L^2(G). You find the irreps inside the finite-dimensional eigenspaces for non-zero eigenvectors.

3. David Speyer - November 28, 2010

@Ben Maybe? I don’t know the argument which you are sketching.

@Peter Isn’t there a difficult step being left out there, where you have to relate characters of the Lie algebra and characters of the group? It seems to me that, a priori, I could have \mathfrak{g}/[\mathfrak{g}, \mathfrak{g}] = \mathbb{R}^n for n>0, and yet \mathrm{Hom}(G, S^1) = (0).

4. Peter J McNamara - November 28, 2010

The part of my comment that involves central characters is complete nonsense and I don’t know how to fix it.

comment that belongs on meta.sbseminar: Ben’s comment has no number!

5. Allen K. - November 29, 2010

a) There’s a subtlety you didn’t mention — there’s a unique _affine_ complexification. S^1 can be complexified to an elliptic curve. (I suppose R has two different complexifications too!)

b) Eckhard.

c) Ben’s numberless comment bugs me too!

d) Fixing the central characters argument would be much like proving Lie’s theorem, no?

6. Emmanuel Kowalski - November 30, 2010

About using L^2(G): part of the not-so-obvious part is to define L^2(G) and check that it _is_ a unitary representation… Constructing from scratch the Haar measure of a general compact group is not so easy. And even the continuity of the regular representation on L^2(G) is not purely formal.

7. Joel Kamnitzer - December 2, 2010

Allen – Thanks for catching the spelling mistake.

As far as the affine property goes, I am using the (reasonably standard) terminology that algebraic group means affine algebraic group.

8. Charles Rezk - December 2, 2010

I’ve got a couple of naive questions about the relationship between these things.

a) What is the difference betweeen (1) the 2-category of connected compact Lie groups, and (2) the 2-category of connected complex reductive groups? (A 2-morphism between homomorphisms is given by conjugation by an element.) My guess is that they are nearly equivalent; but I’ve never seen anything like a precise statement. (They aren’t exactly equivalent; the center of $GL_n$ is bigger than the center of $U(n)$.)

b) What if we drop the condition that the groups be connected?

Ben Webster - December 2, 2010

I hope you mean that its numberlessness bothers you, not its content (admittedly, its content is half-baked). When I’m in the WP dashboard, there’s a “reply” option to a comment which adds a numberless comment. I’m not sure what the point of not putting a number is. The point for me is to not to have to go to the page in order to add the comment.

9. Ben Webster - December 2, 2010

Charles- If you consider the 2-categories where 1-morphisms are isomorphisms between semi-simple compact groups, and isomorphisms between semi-simple complex groups, then you get an equivalence. This is essentially just the fact that compact forms are self-normalizing. If you allow tori or non-isomorphisms, then this won’t work, since you have have non-compact normalizers to images of compact stuff in these situations.

10. Bart - December 2, 2010

Page 71 of J.P. Serre’s book “Complex Semisimple Lie Algebras” states: “the construction of the affine algebra $A_G$ of $G$ is easy: its elements are the continuous functions $f$ on $K$ whose translates span a finite dimensional vector space. Note that this algebra has a canonical real structure; it therefore defines an algebraic group $L$ over $\mathbb{R}$. The set of real points of $L$ is $K$; its set of complex points $G$.”

In Serre’s book, $K$ is assumed to have a semisimple Lie algebra, but does that matter?

11. Joel Kamnitzer - December 3, 2010

Bart – Thanks for your comment. Serre’s construction is a nicer version of the construction G = Spec \otimes V \otimes V^* , I mentioned in my post. This is because a function having a finite-dimensional translate is equivalent to it being the matrix coefficient of a finite-dimensional representation.

I don’t believe that K needs to be compact in this construction.

I guess I am agreeing now with David’s first comment that once you know that K has lots of finite-dimensional representations, then it is not so hard to show that it is algebraic and has a complexification.

This suggests that there might be some way to directly prove the following theorem:
If K is a closed connected Lie subgroup of U(n) , then it is an algebraic subgroup, i.e. it is a (real) subvariety.

12. Evan Jenkins - December 4, 2010

I really like the argument in the article on Tannaka duality by Joyal and Street because it really lets you separate out the part of the proof that is “purely formal” (in that it works not just for compact Lie groups but for arbitrary topological monoids), the part that relies crucially on analytic facts about compact groups, and the (extremely small!) part that actually requires the word “Lie.”

Here is a sketch of their proof. Given any topological monoid M, one can form the following two gadgets. The Tannaka monoid \mathcal{T}(M) consists of those endomorphisms of the forgetful functor U: \mathrm{Rep}(M) \to \mathrm{Vect} that are monoidal and self-conjugate. (It is straightforward to show that \mathcal{T}(K) is a compact group for a compact group K.) The algebra of representative functions R(M) is the subalgebra of C(M) spanned by the matrix coefficients of representations. This is a bialgebra under the coproduct \Delta f(x, y) = f(xy).

There’s a natural map j: M \to \mathrm{Spec}_{\mathbb{R}}(R(M)), which corresponds to the evaluation. There’s also a natural map \pi: M \to \mathcal{T}(M) coming from the action of M on representations. The map \pi can be extended to an operator-valued Fourier transform \mathcal{F} : \mathrm{Spec}_{\mathbb{R}}(R(M)) \to \mathcal{T}(M), which is an isomorphism of topological monoids.

We’d like to show that \pi is an isomorphism (this is Tannaka-Krein duality). This is not true in general, but when M = K is a compact group, we can bring all of our favorite analytic results to bear. Peter-Weyl immediately implies that \pi is injective. Surjectivity is trickier, but it ultimately boils down to leveraging the orthogonality relations to show first that \pi induces an equivalence between the representation theories of K and \mathcal{T}(K) and then that this forces the image of \pi to be all of \mathcal{T}(K).

The only bit that’s missing is showing that R(K) is finitely generated. This will be true if K admits a faithful representation. For a compact Lie group, the set of kernels of representations has a minimal element (since they are closed submanifolds), so Peter-Weyl tells us that there’s a representation with zero kernel. (I suppose this also requires the fact that continuous maps between Lie groups are smooth.)

While this may not be the most accessible proof for a first course, it might be essentially the only way to do it that does not involve already knowing about the structure theory of reductive groups (for instance, to get the uniqueness statement in the Knutson/Sepanski approach you mention). I think this is the price for trying to remove the analytic parts of the proof; the idea of Tannaka duality should probably in principle underlie any proof of this result, and you have the option of doing it in the (hopefully) familiar compact setting or the unfamiliar algebraic setting.

[\operatorname seemed to be causing trouble; I changed them all to \mathrm . DES]

13. BCnrd - December 4, 2010

Dear Joel: If you look in Chapter 3 of Bourbaki LIE, they have a section on “complexification”. If $G$ is a connected real Lie group (in the smooth or real-analytic sense, say), one can consider morphisms from $G$ to the underlying real Lie group of a connected complex Lie group. They define a *complexification* of $G$ to be an initial such morphism, and prove that this always exists and has the expected Lie algebra (if I remember correctly). What they do not address is whether this is interesting in specific cases (e.g., no serious examples discussed), and they don’t address when the canonical map to the complexification has trivial kernel (need not). But one can give direct arguments (using facts about compact groups) to show that for compact $G$ it has trivial kernel. Although one doesn’t have a “complexification” operation on real-analytic manifolds akin to what one has on the algebraic side (where extension of scalars always makes sense), the Bourbaki viewpoint is a nice one (e.g., it is not ad hoc) provided that it is supplemented with finer results that address how it works out in various interesting cases (such as compact Lie groups).

14. Jim Humphreys - December 8, 2010

@Brian: Your comment is apt, but keep in mind that the anonymous Bourbaki group hardly ever contaminated what they did by serious examples. At least their so-called exercises contain some examples, but their pure and often welcome rigorous theoretical viewpoint is uncompromised by the inconvenient question students usually ask: Why?

15. Allen Knutson - January 24, 2011

Constructing from scratch the Haar measure of a general compact group is not so easy.

Constructing it for compact Lie groups is, though. Pick a volume form at the origin, left-translate to get a left-invariant volume form, use it to define an orientation, integrate, get some finite positive number automagically. Rescale by that to get the left-invariant volume form of total volume 1.

16. David Speyer - February 14, 2011

Joel asks whether there is a direct proof of the following:

If K is a closed connected Lie subgroup of U(n) , then it is an algebraic subgroup, i.e. it is a (real) subvariety.

I decided to give this a try this morning. Here is my best effort. I am missing the following lemma:

Lemma 1: A closed subgroup of U(n) is a smooth submanifold.

Assuming that, here is the proof. Let g be an element of U(n) not in K. Our goal is to construct a polynomial p such that p|_K is 1 and p(g) \neq 1. This will present K as an intersection of infinitely (even uncountable many) polynomials. One could then use a further argument to show that K is a finite intersection of polynomials, but I won’t do that.

Let L be the closure, in U(n), of the subgroup generated by K and g. So L is a closed subgroup of U(n). Let V be the (2n)-dimensional representation g \oplus \overline{g} of U(n): the direct sum of the standard rep and its dual.

Lemma 2: Let K \subsetneq L \subseteq U(n) be a chain of closed Lie subgroups of U(n). For m sufficiently large, there is a vector in V^{\otimes m} which is fixed by K and not by L.

Proof: Let \chi be the trace function on U(n). The character of V is then \chi + \overline{\chi}; call this function T. Note that T(\mathrm{Id}) = 2n, that |T(g)| <2n everywhere else on K, and that T drops off quadratically near the identity.

The space of K fixed vectors of V^{\otimes m} has dimension: \int_K T^m \mu_K where \mu_K is Haar measure normalized to have volume 1. For m large, the dominant contribution to the integral is from a small neighborhood of the identity.

Let k = \dim K. Near the identity, this integral is well approximated by
\displaystyle{a_K \int_{\mathbb{R}^k} (2n e^{-(t_1^2 + \cdots t_k^2)})^m dt_1 \cdots d t_k \approx a_K (2n)^m (\sqrt{2 \pi} m^{-1/2})^k}.
Here a_K is a constant that’s hard to describe correctly.

If \dim L = \ell is greater than \dim K=k, then the space of fixed vectors for L has dimension growing like (2n)^m / m^{\ell/2}, which drops off faster than (2n)^m / m^{k/2}, so there is a vector fixed by K and not by L.

If K and L have the same dimension, let e be the index of K in L. Then a_L = a_K/e so, again, for m large, there are more vectors fixed by K than by L. QED

OK, that proves the lemma. Let v \in V^{\otimes m} be fixed by K and not by \ell. So v is not fixed by g. Normalize v to have length 1. Let p(g) = \langle gv, v \rangle (the Hermitian inner product). Then p is 1 on K and not at g.

But, written out explicitly, p(g) is a polynomial in the entries of g and their complex conjugates. And, since g is unitary, the entries of \overline{g} are polynomials in the entries of g, and in (\det g)^{-1}. Clearing out determinants from the denominator, K is distinguished from g by a polynomial.

17. Joel Kamnitzer - February 14, 2011

Wow, David. Nice argument. Just in time too (for my course).

The Lemma 1 you mention is a standard fact: see for example Brocker-tom Dieck Theorem 3.11.

What is a_k and how do you show the part about the integral being well-approximated?

18. David Speyer - February 14, 2011

Here are the rough details. Since you are teaching the course, you get to find all the two’s and pi’s :).

Let K be a subgroup of U(n), with lie algebra \mathfrak{k} \subseteq \mathfrak{u}(n). There are two reasonable volume forms on K. One is the Haar measure of K, normalized to have integral 1. The other is to take the standard Riemann metric on U(n) (translate the Killing form) and restrict it to K. A Riemann metric gives a natural volume form. The constant a_K is the ratio of these two, or something like it. In particular, if K has finite index in L, then the forms gotten by restriction are the same. But the Haar measures obey \mu_L = (1/e) \mu_K. So that’s why a_L = a_K/e.

Now, approximating the integral. Choose small balls B in K, and b in \mathfrak{k}, where exp is an isomorphism. Find \delta such that |T|<2n-\delta on K \setminus B and such that e^{-\langle h,h \rangle } < 1-\delta/2n on \mathfrak{k} \setminus b, where \langle , \rangle is the invariant bilinear form gotten by restriction from \mathfrak{u}(n).

Then, if I haven't screwed up, \int_{K \setminus B} T^{m} = O((2n - \delta)^m) and \int_{\mathfrak{k} \setminus b} (2n e^{-\langle h,h \rangle })^m  = O((2n - \delta)^m). That deals with the nondominant terms.

(continued)

19. David Speyer - February 14, 2011

For the dominant terms, you need to bound \int_{h \in b} \left( T(e^h)^m - (2n e^{-\langle h,h \rangle })^m \right). We should be able to write T(e^h) = (2n) e^{-\langle h,h \rangle + O(|h|^4)}, where I may well have the constants wrong. So we need to deal with (2n)^m \int_b e^{- m \langle h,h \rangle + O(m |h|^4)} = (2n)^m \int_b e^{- m \langle h,h \rangle} (1 + O(m |h|^4)).
Make the change of coordinates h'=\sqrt{m} h and see that your error term is O(1/m) times your main term.

Approximating integrals of the form \int f^m, as m \to \infty, and where f has an isolated simple critical point, is a common task. The approach is always, near that critical point, to write f = f_0 e^{-Q + O(C)}, where Q is a positive definite quadratic and C has cubic order. One then makes a linear change of variable to get \int f_0^m e^{-Q} (1+\cdots), where the ellipses is a series that decays with m. I thought that Etingof told me this was called the “stationary phase approximation”, but wikipedia seems to think that deals with oscillatory integrals.

20. David Speyer - February 14, 2011

One bug in the preceeding: |T|=2n at -\mathrm{Id}. The easiest fix is probably to use that 2n+1 dimensional representation: g \mapsto g \oplus 1 \oplus \overline{g}.

21. Scott Carnahan - March 16, 2011

The phrase “stationary phase approximation” is sometimes used when “method of steepest descent” is the more appropriate choice.

22. Li Zhong - March 21, 2011

The easiest argument I have seen about finite-dimensionality for irreducible representations of compact groups is provided by Serge Lang in his book on $SL(2,\mathbf{R})$. It assumes existence of Harr measures and finiteness of volume of $K$, and nothing else.

23. David Speyer - March 21, 2011

That is nicely written, thanks! Section II.2, for those who want to look it up.


Sorry comments are closed for this entry

Follow

Get every new post delivered to your Inbox.

Join 691 other followers

%d bloggers like this: