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the situation i want to talk about is the following:

$(H_1,\varphi_1),(H_2,\varphi_2)$ irreducible representation of a $C^*$-algebra $A$. A bounded operator $T:H_1\rightarrow H_2$ such that $T\varphi_1(a)=\varphi_2(a)T$ for all $a\in A$

I am asked to prove the following implication: If $\varphi_1$ and $\varphi_2$ are NOT unitarily equivalent, then $T=0$


My ideas for so far: Proof by contradiction: Suppose $T\neq 0$, then $T^*$ exists and is bounded. I want to prove that $T$ is unitary because then we have the situation that both representations are equivalent (unitarily). But therefore i have to show that $T$ is surjective and that $(Tx,Ty)=(x,y)$ for all $x,y$, but this seems to be very difficult or is there something which i don't see and which makes the proof easier? I only know more that non-zero vectors are cyclic vectors since the representations are irreducible but i have no idea how to use it. Or: is the direct implication the better choise?

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You have $$ T^*T\varphi_1(a)=T^*\varphi_2(A)T=\varphi_1(a)T^*T. $$ (for the second equality, note that $\varphi_1(a)T^*=T^*\varphi_2(a)$ by taking adjoints on your original equality).

So $T^*T$ commutes with $\varphi_1(a)$ for all $a\in A$. As $\varphi_1$ is irreducible, $T^*T$ commutes with every operator in $B(H_1)$, so it is a scalar: $T^*T=\lambda I$ for some $\lambda\in\mathbb C$. If $T\ne0$, then $\lambda\ne0$.

A similar argument shows that $TT^*$ is scalar, and since the nonzero elements in the spectra of $T^*T$ and $TT^*$ agree, we get that $TT^*=\lambda I$.

From the positivity of $T^*T$ we get that $\lambda>0$. Then $V=T/\sqrt\lambda$ is a unitary that conjugates $\varphi_1$ and $\varphi_2$.

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  • $\begingroup$ I do not see how $T^*T$ commutes with every operator in $\mathcal{B}(H_1)$ unless the representations are faithful and surjective. Is this assumption part of the definition of irreducibility? $\endgroup$ – Nanashi No Gombe May 7 '17 at 10:48
  • $\begingroup$ Irreducible is enough. It means that the image is strongly dense. $\endgroup$ – Martin Argerami May 7 '17 at 11:05
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This is Schur's lemma: the kernel of $T$ is an invariant subspace of $H_1$, hence $0$ or $H_1$ by irreducibility.

  • If $0$ then T is injective. (However in infinite dim, this does not imply surjective) But the intertwining relation shows that Im(T) is an non zero invariant subspace of $H_2$, i.e. the whole $H_2$.

    T is then bijective, and to get a unitary operator, take the polar decomposition of T, the partial isometry part will turn out to be unitary. (not $T$ itself)

  • else is what we want!

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