On the bicommutant of an operator

Question

What would be the idea behind the so called bicommutant in connection to multiplicity theory and Neumann algebras (Conway's exposition in "A course in functional...")?

To me it just looks like a rather obscure construction to obtain something nice. But I find it hard to believe that there wasn't any intuition involved when inventing thing concept.

Answer 1

Let $\mathcal{H}$ be a Hilbert space and let $M$ be an operator on $\mathcal{H}$. Write $\{M\}'$ for the commutant of $M$ and $\{M\}''$ for the bicommutant of $M$. As observed by MarianoSuárez-Álvarez in the comments, an operator $B$ is in the bicommutant of $M$ if and only if every reducing subspace of $M$ is also a reducing subspace of $B$. Intuitively, $B$ has a 'simpler' structure than $M$, since it has more reducing subspaces. In fact, if $M$ is normal, then the von Neumann algebra generated by $M$ (and therefore also the bicommutant of $M$) is given by $\{f(M) \, \vert \, f\in L^\infty(\sigma(M))\}$, i.e. the bicommutant consists of all functions of $M$. In probability theory, the analogous result is that a random variable $Y$ is measurable with respect to the $\sigma$-algebra generated by another random variable $X$ if and only if there is a measurable function $f$ such that $Y=f(X)$.

I think the situation is most clear when one considers simply a normal matrix $M$ in $\mathcal{H}=\mathbb{C}^n$. In that case, we have the

Spectral Theorem: Let $M$ be a normal operator in $\mathcal{H}=\mathbb{C}^n$. Then $\mathcal H$ has an orthogonal decomposition $\mathcal{H}=\mathcal{H}_1\oplus\ldots\oplus\mathcal{H}_N$ such that $M=\mathrm{diag}(\lambda_1 I_1,\ldots,\lambda_N I_N)$, where, for $1\leqslant j \leqslant N$, $I_j$ denotes the identity in $\mathcal{H}_j$ and $\lambda_1,\ldots,\lambda_N$ denote the distinct eigenvalues of $M$.

It is therefore straightforward to compute both the commutant and the bicommutant.

Proposition: We have $C\in \{M\}'$ if and only if $C=\mathrm{diag}(C_1,\ldots,C_N)$, where, for $1\leqslant j \leqslant N$, $C_j$ is a linear operator in $\mathcal{H}_j$.

Proof: If $CM=MC$, then we have, for all $j,k\in\{1,\ldots,N\}$, $$ \lambda_k C_{jk} = \lambda_j C_{jk}. $$ It follows that $C_{jk}=0$ unless $j=k$.

Proposition: We have $B\in \{M\}''$ if and only if $B=\mathrm{diag}(\mu_1 I_1,\ldots, \mu_N I_N)$, where, for $1\leqslant j \leqslant N$, $\mu_j\in\mathbb{C}$.

Proof: Since $M\in \{M\}'$, we have $B\in\{M\}'$ and therefore $B=\mathrm{diag}(B_1,\ldots,B_N)$. Furthermore, if $BC=CB$ for $C\in\{M\}'$, then we have, for all $1\leqslant j \leqslant N$, $$ B_{j}C_{j} = C_j B_{j}. $$ But if $B_j$ commutes with every linear operator $C_j$ in $\mathcal{H}_j$, then $B_j=\mu_j I_j$ for some $\mu_j\in\mathbb{C}$, see the question A linear operator commuting with all such operators is a scalar multiple of the identity..

Corollary: $B\in\{M\}''$ if and only if there is a polynomial $p$ such that $B=p(M)$.

Proof: Let $p$ be the Lagrange polynomial such that $p(\lambda_j)=\mu_j$.