7
$\begingroup$

In this paper, Friedland shows (in Lemma 3.4) that if $\phi$ is an isomorphism of coherent algebras, then there exists a unitary $U$ such that $$ \phi(M) = UMU^\dagger$$ for all $M$. I am wondering if the same is true of any trace-preserving isomorphism $\phi'$ between two self-adjoint unital matrix algebras as long as $\phi'(M^\dagger) = \phi'(M)^\dagger$.

Further comments:

A coherent algebra is a unital (contains identity) matrix algebra containing the all ones matrix which is additionally closed under conjugate transpose (i.e. is self-adjoint) and closed under Schur (entrywise) multiplication. An isomorphism of coherent algebras is an algebra isomorphism that also preserves conjugate transposition and Schur products.

In his proof, Friedland says that since a coherent algebra is semisimple and has a representation of the form given in equation (2.2) of the paper, it suffices to show that any such isomorphism $\phi$ preserves the trace map.

If I understand correctly, then the fact that a coherent algebra is semisimple and has a representation of the form given in (2.2) is a consequence of the fact that it is self-adjoint (closed under conjugate transpose).

Am I correct in my understanding? In other words, is it true that if $\phi$ is a trace-preserving isomorphism of unital self-adjoint matrix algebras such that $\phi(M^\dagger) = \phi(M)^\dagger$, then there is a unitary matrix $U$ such that $\phi(M) = UMU^\dagger$ for all $M$?

$\endgroup$
  • $\begingroup$ Could you specify what you mean by a self-adjoint unital algebra? This may sound like a silly question but presumably you want to rule out examples of the following form: take ${\mathbb C}[t]/ \langle t^{N+1}\rangle$ and use conjugation of complex numbers as your involution $\endgroup$ – Yemon Choi Mar 10 '16 at 11:34
  • $\begingroup$ @YemonChoi Sorry, I am not very familiar with this area. I was referring only to subalgebras of the algebra of $n \times n$ complex matrices. By self-adjoint I mean closed under conjugate transpose and by unital I mean containing the identity matrix. $\endgroup$ – David Roberson Mar 10 '16 at 11:42
  • $\begingroup$ OK, that makes sense, I just wanted to clarify. $\endgroup$ – Yemon Choi Mar 10 '16 at 12:01
4
$\begingroup$

It is true that every unital self-adjoint algebra $A$ is semisimple. For $A$ contains no non-zero nilpotent right ideal (given any such ideal $I$ and any $M \in I$, we have trace($MM^{\ast}) =0$ since $MM^{\ast} \in I$ is nilpotent, and this forces $M = 0$). Thus $A$ is semisimple, and is a direct sum of full matrix algebras.

If $A$ and $B$ are both unital self adjoint subalgebras of a full matrix algebra $C$, and $\phi: A \to B$ is a trace preserving isomorphism which respects $\ast$, then $A$ and $B$ have the same size full matrix algebra summands with the same multiplicities, since the sizes of the full matrix algebra summands of $A$ are the traces of the primitive idempotents of $Z(A)$ and likewise for $B$.

The primitive idempotents of $Z(A)$ are self-adjoint, for if $e$ is one such, then so is $e^{\ast}$, and $ee^{\ast}$ has positive real trace so certainly $ee^{\ast} \neq 0$. Hence the (commuting) primitive idempotents of $Z(A)$ may be simultaneously diagonalized by a single unitary matrix $V$, and likewise for $B$ ( with another unitary matrix $W$).

Hence it suffices to consider the case (possibly replacing $A$ and $B$ by unitary conjugates) that $\phi(e) = e$ for each primitive idempotent $e$ of $Z(A)$. But then $eA = \phi(eA)$ for each such $A$, and $\phi$ induces an automorphism of the full matrix algebra $eA$ which respects the Hermitian adjoint. Any such automorphism is induced by conjugation of a unitary matrix in $eA$, so the desired conclusion holds.

$\endgroup$
  • $\begingroup$ Is this a well-known result/is there something I can reference for the result? $\endgroup$ – David Roberson Jul 25 '16 at 22:15
  • 1
    $\begingroup$ It is a combination of well-known results ( structure of semisimple algebras combined with an involution on the algebra) but I am not sure where is the best place for a single reference. $\endgroup$ – Geoff Robinson Jul 26 '16 at 6:07

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.