# Abelianization of GL(H)

This is related to Theo's question about the abelianizations of finite dimensionsal Lie groups.

I am interested in a specific (infinite-dimensional) case of the above question. Let H be an infinite-dimensional Hilbert space and GL(H) represent the bounded two-sided invertible operators on H. Is there a nice description of the commutator subgroup G (the group generated be elements of the form ABA^{-1}B^{-1}) and the abelianization GL(H)/G?

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The solution to Problem 240 in Halmos is: "On an infinite-dimensional Hilbert space, the commutator subgroup of the full linear group is the full linear group itself."

It is mentioned, although the details are not given, that every invertible operator is the product of two(!) commutators. Reference: A. Brown and C. Pearcy, Multiplicative commutators of operators, Can. J. Math. 18 (1966) 737-749.

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By polar decomposition, it suffices to show every unitary is a commutator and every positive self-adjoint operator is a commutator. This can probably be done with spectral theory, though I don't see how exactly. – Eric Wofsey Oct 27 '09 at 0:06
Polar decomposition is how Halmos proceeds. He relies on some tricks to show that every invertible normal operator is the product of two commutators. – Faisal Oct 27 '09 at 0:29

Here's a positive result towards this answer. This doesn't feel like something that could be adapted to a general proof, but it at least is an easy result backing up the result quoted in Halmos' book.

Consider a unitary, diagonalisable operator, U, on a separable Hilbert space, $H$. So there is some countable orthonormal basis on which $U$ is diagonal. Index it by the integers and let $S$ be the shift operator (which is unitary). Let $\lambda_i$ be the corresponding eigenvalues of $U$. For each $i$, let $\mu_i$ be defined by $\mu_0 = 1$ and $\mu_{i+1}/\mu_i = \lambda_i$. Let $V$ be the operator which acts on $e_i$ by $\mu_i$. As the $\lambda_i$ were all unitary, $V$ is a unitary operator. Then

$$V^{-1}S^{-1}VSe^i = V^{-1}S^{-1}Ve_{i+1} = \mu_{i+1}V_{-1}S^{-1}e_{i+1} = \mu_{i+1}V^{-1}e_i = \mu_{i+1}/\mu_i e_i = \lambda_ie_i = Ue_i$$

So $V^{-1}S^{-1}VS = U$.

In particular, one can get $\lambda I$ this way for $\lambda \in S^1$.

As I said, this feels too specific to adapt to a general proof, but it is a simple example.

Edit: In a similar way, we can get $\lambda I$ for any non-zero $\lambda$. Let's stay with separable Hilbert spaces for simplicity. To make things clear, let $H$ be the Hilbert space we're interested in, and let $H'$ be a standard reference Hilbert space. Let $H_0 \subseteq H$ be a closed subspace of infinite dimension and codimension. Split the complement as a sum $H_1 \oplus H_2 \oplus \dots$ where each $H_i$ is an infinite dimensional closed subspace. So $H \cong \bigoplus H_j$ and each $H_j$ are isomorphic to each other and to $H'$. Choose $A \in GL(H')$. Let $S_1$ be the switch operator that swaps $H_{2i}$ with $H_{2i+1}$. Let $T_1 = (A,I,A,I,\dots)$. Then $T_1 S_1 T_1^{-1} S_1^{-1}$ is $(A,A^{-1},A,A^{-1},\dots)$. Now let $S_2$ be the switch operator that swaps $H_{2i-1}$ with $H_{2i}$ and leaves $H_0$ alone. Let $T_2 = (I,A,I,A,\dots)$. Then $T_2 S_2 T_2^{-1} S_2^{-1} = (I,A,A^{-1},A,A^{-1},\dots)$. So multiplying these together, we get $(A,I,I,I,\dots)$.

Thus for any splitting $H \cong H_0 \oplus H_0^\perp$, and $A \in GL(H_0)$, $B \in GL(H_0^\perp)$ we can get

$$\begin{bmatrix} A & 0 \\\\ 0 & B \end{bmatrix}.$$

In particular, $\lambda I$ for any $\lambda \in \mathbb{C}^*$.

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I think this is more or less the argument Halmos gives to prove that every unitary is a commutator - one has to do a more fiddly transfinite version, but this seems to be the heart of the matter – Yemon Choi Oct 27 '09 at 21:45

I agree with Eric's belief, albeit for a "cheat" reason: if the abelianization were non-trivial then you'd get a notion of determinant for GL, and that seems unlikely to me.

For what it's worth, according to Google Books this is Problem 240 in Halmos' "A Hilbert space problem book", although I can't see the answer. Problem 239 says that every unitary can be realized as ABA-1B-1 for some bounded invertible operators A and B, which is also suggestive.

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Well, the question is more subtle than it might seem on the first sight: Finite von Neumann algebras do have a notion of determinant (Fuglede-Karlson determinant), but B(H) is not a finite von Neumann algebra. See my answer: mathoverflow.net/questions/2681/abelianization-of-glh/… – Dmitri Pavlov May 1 '10 at 9:36
Interesting - I know (a bit) about the F-K determinant, but not about the commutator result described in your answer. I didn't bring up the FK-determinant precisely because B(H) is only semifinite not finite, and I wasn't sure to what extent if any things could be generalized to the semifinite case. – Yemon Choi May 1 '10 at 19:18
Correction: It's called Fuglede-Kadison determinant, not Fuglede-Karlson. FK determinant can be generalized to the semifinite case: For any semifinite von Neumann algebra M we have a (non-canonical) trace defined on a certain dense subalgebra of M and an associated determinant defined on a certain dense subgroup of GL(M). (Here dense means dense in σ-weak topology.) – Dmitri Pavlov May 2 '10 at 4:22

Incidentally, the question can be generalized to arbitrary von Neumann algebras instead of B(H) and is closely related to its analog for Lie algebras: What is the set of all linear combinations of commutators [a, b] = ab-ba, where a and b belong to some von Neumann algebra M? It turns out that it consists of all elements with traceless finite part (i.e., if we decompose M as K×L, where K is finite and L is properly infinite, then an element m=k+l is a linear combination of commutators if and only if tr(k)=0). Likewise, an element of the multiplicative group is a product of commutators aba^{-1}b^{-1} if and only if the Fuglede-Kadison determinant of its finite part is 1: det(k)=1 if m=k+l.

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I'm not entirely sure it's always the case for bounded operators, but let's first assume our operators are representable as exp X and exp Y for some trace-class operators. Then [X, Y] should have non-zero trace which, I think, is impossible. So you won't be able to get anything whose det is not 1, and by the standard reasoning you can get everything whose det is 1. So I think you end up with the same R^* as with finite-dimenisonal case.

You're reproducing a version of the first algebraic K-group of C. Here's an interesting reference.

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This is quite a different question than computing K_1 -- the group K_1(C) is related to GL(C), which is the direct limit of the finite-dimensional GL_n(C). While GL(C) injects into GL(H), its closure consists of (some) compact operators. In other words, it is far from even being dense. – Andy Putman Nov 9 '09 at 21:01
Indeed, I confused two different things. – Ilya Nikokoshev Nov 9 '09 at 21:20
As we're in a Hilbert space, I think that the "(some)" is redundant: the closure in the norm topology of finite rank operators is the space of compact operators - unless I'm missing something. – Loop Space Nov 11 '09 at 10:35