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I am curious to know what the commutator subgroup of the group of (bounded) invertible linear operators on a complex Hilbert space is?

Note that by "commutator subgroup" I mean the subgroup generated by the commutators $[A,B]=ABA^{-1}B^{-1}$ of bounded invertible linear operators $A,B.$

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1 Answer 1

up vote 4 down vote accepted

Update: The question you ask about is Problem 240 in Halmos, A Hilbert space problem Book, Springer GTM 16.

Solution 240. On an infinite-dimensional Hilbert space, the commutator subgroup of the full linear group is the full linear group itself.

(generalizing the result given in the initial answer, see below). The solution is given on page 348f and refers to several earlier exercises.

The proof is close to the one given in the paper by Brown and Pearcy mentioned below but it is somewhat simplified in that a slightly less precise statement is proved (but it still is a bit lengthy), so I content myself with pointing out that it is quite reminiscent of the Eilenberg swindle.


The case in which the Hilbert space is separable and infinite-dimensional was solved in

Arlen Brown, Carl Pearcy, Multiplicative commutators of operators, Canad. J. Math. 18 (1966), 737–749, MR200720.

  1. Theorem 5 on page 746 states that a normal invertible operator is a commutator if and only if it is not of the form $\lambda \cdot \mathrm{id} + C$ with $C$ a compact operator and $|\lambda| \neq 1$ a scalar.

  2. From this it follows from polar decomposition and their earlier (somewhat technical) results that every invertible operator is the product of two commutators (Corollary 4.7 on page 746).

Hence:

The group of invertible operators on a separable infinite-dimensional Hilbert space is perfect, i.e. its own commutator subgroup.

The proofs are a bit too involved to expand upon here.


Note: The paper of Brown and Pearcy is quite famous because it was crucially used by Lawrence G. Brown and Claude Schochet in their paper $K_{1}$ of the compact operators is zero, Proc. Amer. Math. Soc. 59 (1976), 119–122, MR0412863 to prove what the title of their paper says.


Added: To complete the discussion of separable complex Hilbert spaces, I'm incorporating a (now deleted) comment into my answer:

If $n \geq 1$ the commutator subgroup of $GL_{n}(\mathbb{C})$ is $SL_{n}(\mathbb{C})$. The proof is straightforward:

  1. The determinant of a commutator is $1$, so it must be in the special linear group;
  2. The special linear group is generated by elementary matrices;
  3. Every elementary matrix is a commutator.
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That's interesting. Thank you, I appreciate it. –  Ehsan M. Kermani Aug 30 '11 at 14:27
    
You're welcome. I've now found that the answer in the infinite-dimensional case doesn't depend on separability and the argument is well-explained in Halmos's book linked to in the update part of the answer. I should have thought of looking there earlier. –  t.b. Aug 30 '11 at 16:11
    
@all: if someone knows a more streamlined exposition of this results (or related ones) than what's in Halmos, I'd be glad to know about it. –  t.b. Aug 31 '11 at 5:13

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