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Hi, my apologies for a rather non-specific question. I wonder if there is a general set of conditions under which operators are commutative in functional analysis. Most that I've found is that "operators are, in general, not commutative". Is there any reference someone could point me to for some kind of review or special cases in which commutativity is established (or forbidden)? Thanks!

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By "operators are commutative", do you mean that two operators $F,G$ commute with each other, or do you mean an operator $F$ that commutes with all other operators? –  Harry Gindi Apr 10 '10 at 23:46
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I can't imagine something at the level of generality you're asking, but if you're just trying to get a little intuition about operators commuting or not commuting, I recommend starting with the n-by-n matrix case. E.g., two matrices commute if they share a spanning set of eigenvectors (i.e., they can be simultaneously diagonalized). Conversely, a matrix with n distinct eigenvalues will commute only with matrices that share its eigenvectors. –  Jonas Meyer Apr 11 '10 at 0:14
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I think, as Jonas says, that you really have to try and understand the matrix case a bit better, in order to work out what you yourself mean by "conditions under which operators commute. I mean, they commute if they commute; your question, although it's a natural impulse, is IMHO just not sufficiently well-defined to admit a good answer here. Examples first; "big picture" afterwards. –  Yemon Choi Apr 11 '10 at 0:26
    
Thanks to all for comments - I was thinking if there is a way to predict a priori whether F,G will commute. But I agree that the question is too general. Thanks for the suggestion - I will try start by examining the case where the operators are represented as nxn matrices... –  crippledlambda Apr 11 '10 at 0:33
    
This is an exercise in Serge Lang's Algebra on page 593, if you prefer doing exercises from a book. This is the key fact for proving the uniqueness of the Jordan-Chevalley decomposition (which is the next exercise). –  Harry Gindi Apr 11 '10 at 1:37
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3 Answers 3

up vote 6 down vote accepted

One obvious but important observation is that, for operators on a $n$-dimensional vector space over a field, if $1 < n < \infty$, we have $AB \neq BA$ generically. In other words, consider the commutativity locus $\mathcal{C}_n$ of all pairs of $n \times n$ matrices $A,B$ such that $AB = BA$ as a subset of $\mathbb{A}^{n^2}$. This is clearly a Zariski closed set -- i.e., defined by the vanshing of polynomial equations. It is also proper: take e.g. $A = \left[ \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right] \oplus 0_{n-2}$ and $B = \left[ \begin{array}{cc} 0 & 1 \\ 0 & 1 \end{array} \right] \oplus 0_{n-2}$. Since $\mathbb{A}^{n^2}$ is an irreducible variety, $\mathcal{C}_N$ therefore has dimension less than $N^2$. This implies that over a field like $\mathbb{R}$ or $\mathbb{C}$ where such things make sense, $\mathcal{C}_N$ has measure zero, thus giving a precise meaning to the idea that two matrices, taken at random, will not commute.

One could ask for more information about the subvariety $\mathcal{C}_N$: what is its dimension? is it irreducible? and so forth. (Surely someone here knows the answers.)

I would guess it is also true that for a Banach space $E$ (over any locally compact, nondiscrete field $k$, say) of dimension $> 1$, the locus $\mathcal{C}_E$ of all commuting pairs of bounded linear operators is meager (in the sense of Baire category) in the space $B(E,E) \times B(E,E)$ of all pairs of bounded linear operators on $E$.

Kevin Buzzard has enunciated a principle that without further constraints, the optimal answer to a question "What is a necessary and sufficient condition for $X$ to hold?" is simply "X". This seems quite applicable here: I don't think you'll find a necessary and sufficient condition for two linear operators to commute which is nearly as simple and transparent as the beautiful identity $AB = BA$.

Still, you could ask for useful sufficient conditions. Diagonalizable operators with the same eigenspaces, as mentioned by Jonas Meyer above, is one. Another is that if $A$ and $B$ are both polynomials in the same operator $C$: this shows up for instance in the Jordan decomposition.

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Let $P,Q$ be operators on complex Hilbert space. If there is an operator $T$ and polynomials $p,q$ so that $P = p(T)$ and $Q = q(T)$, then $P,Q$ commute.

More generally, in a setting where the functional calculus works, if there are any two functions $p,q$ so that $P = p(T)$ and $Q = q(T)$, then $P,Q$ commute. For example, if $T$ is Hermitian (or more generally, normal), and $p,q$ are just measurable functions on the complex plane, then $p(T)$ and $q(T)$ are defined and commute.

As an aside, you have to decide whether you want BOUNDED operators or not, and proceed accordingly.

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Hi, I suppose this book will be useful for you:

Banach Algebra Techniques in Operator Theory (R. Douglas)

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-1. This is a very general reference, and it's not clear why the author would find it helps with his question (although I don't find the original author's question a very meaningful one). –  Yemon Choi Apr 11 '10 at 10:16
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