4
$\begingroup$

Let $M\in \mathcal{B}(F)^+$ and $S_1,S_2\in \mathcal{B}(F)$.

I claim that if $S_1$ and $S_2$ are $M$-self adjoint (i.e. $MS_1=S_1^*M$ and $MS_2=S_2^*M$) such that $S_1S_2=S_2S_1$, then $\exists\,(X,\mu)$, $\varphi_1,\varphi_2\in L^\infty(\mu)$ and a unitary operator $U:F\longrightarrow L^2(\mu)$, such that $$U(MS_k)U^*h=\varphi_kh,\;\forall h\in F,\,k=1,2.$$ Do you think that this claim is true?

Thanks.

Improve this question
$\endgroup$
6
  • 2
    $\begingroup$ Your assumptions imply that $(MS)^* = S^*M = MS$, so $MS$ is self-adjoint, which shows that your claim is true, right? (I'm not quite convinced, however, that $(X,\mu)$ can be chosen as a ($\sigma$-)finite space, in particular if $F$ is not assumed to be separable.) $\endgroup$
    – Jochen Glueck
    Commented Dec 1, 2017 at 10:11
  • $\begingroup$ Thank you. Do you mean that $\mu(X)$ can be equal to $\infty$? $\endgroup$
    – Schüler
    Commented Dec 1, 2017 at 10:16
  • 1
    $\begingroup$ Well, that depends. If $X$ can be chosen to be $\sigma$-finite, then $X$ can also be chosen to be finite by a simple change-of-density argument. However, I doubt that $X$ can always be chosen to be $\sigma$-finite in the spectral theorem (in particular, if $F$ is not separable). To be sure, I suggest to look it up in a textbook or a monograph where the spectral theorem is treated. $\endgroup$
    – Jochen Glueck
    Commented Dec 1, 2017 at 10:26
  • $\begingroup$ I have edited my question. $\endgroup$
    – Schüler
    Commented Dec 1, 2017 at 12:55
  • $\begingroup$ The answer will be negative if (and only if) you can find $S_1,S_2$ commuting, with $MS_i$ self-adjoint for both $i$, but with $MS_1$ not commuting with $MS_2$. $\endgroup$
    – Matthew Daws
    Commented Dec 1, 2017 at 13:28

1 Answer 1

Reset to default
9
$\begingroup$

There is a 2x2 matrix counter-example.

$$ M = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}, \quad S_1 = \begin{pmatrix} 1 & 2 \\ 1 & 4 \end{pmatrix}, S_2 = \begin{pmatrix} a & 2c \\ c & d \end{pmatrix}. $$ Then $$ MS_1 = \begin{pmatrix} 1 & 2 \\ 2 & 8 \end{pmatrix}, \quad MS_2 = \begin{pmatrix} a & 2c \\ 2c & 2d \end{pmatrix} $$ so $MS_1$ is self-adjoint, and $MS_2$ is when $a,c, d\in\mathbb R$. Then $$ S_1S_2 = \begin{pmatrix} a + 2c & 2c+2d \\ a+4c & 2c+4d \end{pmatrix}, \quad S_2S_1 = \begin{pmatrix} a+2c & 2a+8c \\ c+d & 2c+4d \end{pmatrix}. $$ So these commute when $d = a + 3c$. But then $$ MS_2 = \begin{pmatrix} a & 2c \\ 2c & 2a+6c \end{pmatrix} $$ Tediously multiplying out shows these commute only when $a=c$.

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ What is $b$? I get that the commutator of $MS_1$ and $MS_2$ vanishes when $a=c$. $\endgroup$
    – Igor Khavkine
    Commented Dec 1, 2017 at 15:51
  • $\begingroup$ Yes, well spotted. $b=2c$ (from some paper calculation I did). Answer corrected. $\endgroup$
    – Matthew Daws
    Commented Dec 1, 2017 at 16:43

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .