6
$\begingroup$

Suppose I'm given a finite set of possibly unbounded commuting self-adjoint operators $T_i : \mathfrak H \supset \mathscr D(T_i)\to \mathfrak H, i = 1 , \dots , N$ on a Hilbert space (in the sense of commuting resolvents $\forall i, j: \exists z_i \in \rho (T_i) , z_j \in \rho (T_j):[R_{T_i} (z_i ) , R_{T_j} (z_j)] = [R_{T_i} (\overline{z_i} ) , R_{T_j} (z_j)] = 0$). In linear algebra this implies that these operators can be simultaneously diagonalized.

Is there any general rigorous notion of simultaneous diagonalization in this functional analysis setting? Intuitively I would expect something like a projection-valued measure

$$E: \mathscr B(\mathbb R^N ) \to \mathbb C^N \otimes \mathfrak L(\mathfrak H)$$

supported on $\sigma (T_1) \times \dots \times \sigma (T_N)$, such that $E^{(T_i)} (A) = E(\mathbb R \times \dots \times \mathbb R \times \underbrace{A}_{i} \times \mathbb R \times \dots \times \mathbb R )$.

Or maybe a guarantee that there is a common measure space $(\Omega , \mu )$ and a common unitary transformation $U : \mathfrak H \to L^2(\Omega, \mathrm d\mu )$ such that

$$\forall i : \exists f_i \in \text{Meas}(\Omega, \mathrm d\mu): T_i = U^* M_{f_i} U$$

Is there anything like that?

$\endgroup$
5
$\begingroup$

Apply the SNAG (Stone-Naimark-Godement-Ambrose) theorem to the unitary group generated by these operators.

$\endgroup$
  • $\begingroup$ Wow, what an elegant theorem. Thank you very much, this is totally what I'm looking for! $\endgroup$ – jacques Mar 15 '17 at 18:46
3
$\begingroup$

Another way to do it is to consider the bounded operators $(T_i + iI)^{-1}$, check that these are normal and commute so they can be simultaneously represented as multiplication operators on an $L^2$ space, and then untransform to represent the original $T_i$ as unbounded multiplication operators.

$\endgroup$
  • $\begingroup$ Hm, okay, but how do I know this holds for bounded operators? $\endgroup$ – jacques Mar 15 '17 at 19:01
  • $\begingroup$ This is the spectral theorem. E.g. see Theorem 5.4.5 of my book. $\endgroup$ – Nik Weaver Mar 15 '17 at 19:41
3
$\begingroup$

There is a very nice treatment of this question in the textbook by Konrad Schmüdgen "Unbounded Self-adjoint Operators on Hilbert Space", see in particular Theorem 5.23 p. 103.

While there is no problem constructing a unique product measure $\mu\otimes\nu$ for $\sigma$-finite measure spaces $(X,\mathcal{M},\mu)$ and $(Y,\mathcal{N},\nu)$ the analogous question for projection-valued measures encounters a snag unless $X$ and $Y$ are sufficiently nice like $\mathbb{R}^n$. See the remark after Theorem 4.10 in the same book and the reference to the article by Birman, Vershik and Solomjak.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.