rules for operator commutativity? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T23:54:48Z http://mathoverflow.net/feeds/question/20968 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/20968/rules-for-operator-commutativity rules for operator commutativity? crippledlambda 2010-04-10T23:44:18Z 2010-04-12T17:27:51Z <p>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!</p> http://mathoverflow.net/questions/20968/rules-for-operator-commutativity/20994#20994 Answer by Soheil Malekzadeh for rules for operator commutativity? Soheil Malekzadeh 2010-04-11T09:39:59Z 2010-04-11T09:39:59Z <p>Hi, I suppose this book will be useful for you:</p> <p>Banach Algebra Techniques in Operator Theory (R. Douglas)</p> http://mathoverflow.net/questions/20968/rules-for-operator-commutativity/21007#21007 Answer by Pete L. Clark for rules for operator commutativity? Pete L. Clark 2010-04-11T13:38:17Z 2010-04-12T17:09:22Z <p>One obvious but important observation is that, for operators on a $n$-dimensional vector space over a field, if $1 &lt; n &lt; \infty$, we have $AB \neq BA$ <em>generically</em>. 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. <code>$A = \left[ \begin{array}{cc} 1 &amp; 1 \\ 0 &amp; 1 \end{array} \right] \oplus 0_{n-2}$</code> and <code>$B = \left[ \begin{array}{cc} 0 &amp; 1 \\ 0 &amp; 1 \end{array} \right] \oplus 0_{n-2}$</code>. 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.</p> <p>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.)</p> <p>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 <strong>meager</strong> (in the sense of Baire category) in the space $B(E,E) \times B(E,E)$ of all pairs of bounded linear operators on $E$. </p> <p>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 $AB = BA$.</p> <p>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. </p> http://mathoverflow.net/questions/20968/rules-for-operator-commutativity/21129#21129 Answer by Gerald Edgar for rules for operator commutativity? Gerald Edgar 2010-04-12T17:27:51Z 2010-04-12T17:27:51Z <p>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. </p> <p>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. </p> <p>As an aside, you have to decide whether you want BOUNDED operators or not, and proceed accordingly.</p>