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Let $\phi$ define a $*$-automorphism from the matrix algebras $M_n(\mathbb{C})$ to $M_n(\mathbb{C})$ such that $\phi(I) = I$. Is it true that any such map $\phi$ can be represented as $\phi(x) = U x U^{\dagger}$ (where $U$ is a suitable unitary matrix)? If not, what is the most general expression?

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It is redundant to require $\phi(I)=I$. –  Jonas Meyer Aug 30 '10 at 19:25
    
how can this be shown? –  kett Aug 30 '10 at 20:41
    
Let $A$ be an element of $M_n$, and let $B=\phi^{-1}(A)$. Then $\phi(I)A=\phi(I)\phi(B)=\phi(IB)=\phi(B)=A$. Similarly, $A\phi(I)=A$, so $\phi(I)$ is an identity for $M_n$. More generally, if $f:R\to S$ is a surjective ring homomorphism and $R$ is unital, then $f(1_R)$ is an identity for $S$, and the only difference in the proof is that you take $B\in f^{-1}(A)$ in case $f$ is not injective. –  Jonas Meyer Aug 30 '10 at 20:56
    
Another way to see it in the $M_n(\mathbb{C})$ case is to take a family $\{e_i\}$ of $n$ mutually orthogonal projections (which necessarily add to I, because the sum is a projection with rank $n$). Then the images also form a family $\{\phi(e_i\}$ of orthogonal projections, and the sums is a projection with trace $n$, that is $I$. –  Martin Argerami Sep 1 '10 at 14:43

5 Answers 5

up vote 3 down vote accepted

Here is one generalization:

Every $*$-automorphism of the algebra of compact operators on a Hilbert space is conjugation by a unitary operator on that space.

Using the fact that the algebra of compact operators is irreducible, this can be seen as a special case of:

Every irreducible $*$-representation of the algebra of compact operators on a Hilbert space is unitarily equivalent to the identity representation.

A proof can be found for instance in Section 1.4 of Arveson's An invitation to C* algebras. Another proof of the first assertion that gives more information can be found in Proposition 1.6 of Raeburn and Williams's Morita equivalence and continuous trace C*-algebras.

The first part is still true if you take all bounded operators instead of only the compact ones. (And these are the same thing in the finite dimensional case.)

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If $\phi$ is a $*$-automorphism then $\psi:A\mapsto\phi(\overline A)$ is a $\mathbb{C}$-automorphism. By the Skolem-Noether theorem every $\mathbb{C}$-automorphism of $M_n(\mathbb{C})$ is inner, that is of the form $\psi(A)=UAU^{-1}$. This must commute with the $*$-operation: $A\mapsto\overline{A}^t$. This leads to $UAU^{-1} =\overline{U^t}^{-1}A\overline{U^t}$ for all $A$. Thus implies that $U$ and $\overline{U^t}^{-1}$ are the same up to a constant multiple. By multiplying $U$ by a constant we may make $U$ unitary.

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As an alternative to Robin Chapman's solution, I would like to state Exercise 7.8 from Rørdam's, Larsen's and Laustsen's "Introduction to the K-theory of C*-algebras":

For every unital AF-algebra $A$ there is a short exact sequence $$ 1\to\overline{\mathrm{Inn}}(A)\to\mathrm{Aut}(A)\to\mathrm{Aut}(K_0(A))\to 1, $$ where $\overline{\mathrm{Inn}}(A)$ denotes approximately inner automorphisms and $\mathrm{Aut}(K_0(A))$ denotes group automorphisms preserving the unit class and the positive cone in $K_0(A)$.

If $A$ is the matrix ring, then $\mathrm{Aut}(K_0(A))$ is trivial and hence every automorphism of $A$ is approximately inner. Since $A$ is separable, every approximately inner automorphism is the pointwise limit of a sequence of inner automorphisms. And I think the finite-dimensionality of $A$ implies that the pointwise limit of a sequence of inner automorphisms is again inner.

Using the statement above, one immediately sees that, for instance, $\mathbb C\oplus\mathbb C $ possesses an automorphism which is not approximately inner.

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Does anyone know how I can actually prove that the pointwise limit of a sequence of inner automorphisms of a finite-dimensional C*-algebra is again inner? –  Rasmus Bentmann Aug 30 '10 at 19:31
    
At a guess: compactness? (take a cluster point/limit of your sequence of implementing elements) –  Yemon Choi Aug 30 '10 at 19:49
    
Yemon's argument works, because your sequence of unitaries is bounded in a finite dimensional space, and so it has a convergent subsequence. On the other hand, already between the answers and the comments there are several proofs that any automorphism of a finite-dimensional C$^*$-algebra is inner. So all you need to convince yourself about is that the pointwise limit of inner automorphisms is an automorphism, and that's easy. –  Martin Argerami Sep 1 '10 at 15:04
    
@M. Argerami: Well, I don't want to use that any automorphism of a finite-dimensional C∗-algebra is inner because that's what I want to prove in the end. –  Rasmus Bentmann Sep 1 '10 at 20:02
    
I see. My bad, then. Regarding the last statement in your answer, if your $A$ is abelian, then the only inner automorphism is the identity, and then of course the only approximately inner automorphism is the identity; and a simple example of a non-inner automorphism of $\mathbb{C}\oplus\mathbb{C}$ is the ``flip'', $(a,b)\mapsto(b,a)$. –  Martin Argerami Sep 2 '10 at 0:20

Another proof can be obtained using that $M_n(\mathbb{C})$ is singly generated (and finite-dimensional). So $M_n(\mathbb{C})=C^*(s)$ for some $s$ (the shift, for example). Now, of course, $\phi(s)$ is a generator for the image. And by Spetch's theorem, $\phi(s)$ and $s$ are unitarity equivalent (because $\phi$ is multiplicative and it preserves the trace). Then there exists a unitary $U\in M_n(\mathbb{C})$ with $\phi(s)=UsU^{-1}$. If you now take any $a\in M_n(\mathbb{C})$, we have $a=\sum_{j=0}^{n-1} \alpha_js^j+\sum_{j=1}^{n-1}\beta_j(s^*)^j$, for coefficients $\alpha_j,\beta_j$, and so $$ \phi(a)=\sum_{j=0}^{n-1} \alpha_j\phi(s)^j+\sum_{j=1}^{n-1}\beta_j\phi(s^*)^j $$ $$ =\sum_{j=0}^{n-1} \alpha_j((UsU)^{-1})^j+\sum_{j=1}^{n-1}\beta_j(Us^*U^{-1})^j=UaU^{-1} $$

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Yet another proof would be to consider a system $(e_{kj})$ of matrix units in $M_n(\mathbb{C})$, coming from some orthonormal basis (the canonical one, say). It is then easy to check that $(\phi(e_{kj}))$ is another system of matrix units, and so it corresponds to another orthonormal basis. The unitary implementing the change of basis is the one implementing $\phi$.

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