It is mentioned on Wikipedia that every unital *-homomorphism $\Phi:M_i\to M_j$ is necessarily of the form $\Phi(a)=U^*(a\otimes I_r)U$ for some unitary $U$ and some $r$. (Here $M_i$ are the $i\times i$ complex matrices and $I_r$ is the $r\times r$ identity.)

No proof or reference is given.

How is this proven? (A reference to a textbook is also OK for me.)

I will need to formalize the proof in a computer-aided theorem prover, so an elementary proof would be preferred. (By elementary I mean not more than introductory graduate-level textbook level, maybe.)


  • This answer seems to touch the question but I don't understand it. (It uses "Morita equivalence" which as far as I can tell is quite a power-tool in this context.)
  • Information about whether something similar holds also for the bounded operators on a larger Hilbert space would be appreciated.

Theorem: For Hilbert spaces $H,K$, every normal unital *-homomorphism $\Phi:\mathcal B(H)\to\mathcal B(K)$ is of the form $\Phi(a)=U(a\otimes 1_{K_0})U^∗$ for some Hilbert space $K_0$ and some unitary $U$.

Here is a super down-to-earth proof, from a functional analysis / operator-algebras perspective. I'll start by working in the infinite-dimensional setting. Let $H,K$ be Hilbert spaces, write $\newcommand{\mc}{\mathcal}\mc B(H)$ for the algebra of all bounded operators on $H$, and $\mc K(H)$ for the compact operators. Let $\Phi:\mc B(H)\rightarrow\mc B(K)$ be a unital $*$-homomorphism which is normal (aka weak$^\ast$-continuous). As $\mc K(H)$ is weak$^\ast$-dense in $\mc B(H)$, $\Phi$ is completely determined by its restricton to $\mc K(H)$, That $\Phi$ is unital corresponds to $\Phi:\mc K(H)\rightarrow\mc B(K)$ being non-degenerate meaning that $\{ \Phi(\theta)(\xi) : \theta\in\mc K(H), \xi\in K \}$ has a dense linear span in $K$.

If $H$ is finite-dimensional, of course $\mc B(H) = \mc K(H) \cong M_i$ where $i$ is the dimension of $H$.

So, consider a non-degenerate $*$-homomorphism $\Phi:\mc K(H)\rightarrow\mc B(K)$. For $\xi,\eta\in H$ write $\theta_{\xi,\eta}$ for the rank-one operator $\alpha\mapsto (\alpha|\eta) \xi$. Then $\theta_{\xi,\eta}^* = \theta_{\eta,\xi}$ and $\theta_{\xi,\eta} \theta_{\xi_1,\eta_1} = (\xi_1|\eta) \theta_{\xi,\eta_1}$. Here I write $(\cdot|\cdot)$ for the inner-product on $H$. Fix a unit vector $\xi_0\in H$, and consider $\theta_{\xi_0, \xi_0}$ which is a projection (self-adjoint idempotent). So $p = \Phi(\theta_{\xi_0, \xi_0})$ is also a projection. Let $K_0\subseteq K$ be the (closed) subspace forming the image of $p$.

Define $U:H\odot K_0 \rightarrow K$ by $$ U(\xi\otimes\alpha) = \Phi(\theta_{\xi,\xi_0})(\alpha), $$ and extend by linearity. I write $\odot$ for the algebraic tensor product, and will write $\otimes$ for the (completed) Hilbert space tensor product. Then \begin{align*} ( U(\xi\otimes\alpha) | U(\eta\otimes\beta) ) &= (\Phi(\theta_{\xi,\xi_0})(\alpha) | \Phi(\theta_{\eta,\xi_0})(\beta)) \\ &= (\Phi(\theta_{\xi_0, \eta} \theta_{\xi,\xi_0})(\alpha) | \beta ) \\ &= (\xi|\eta) (\Phi(\theta_{\xi_0,\xi_0})(\alpha) | \beta ) = (\xi|\eta) (p(\alpha) | \beta ) \\ &= (\xi|\eta) (\alpha | \beta ). \end{align*} Thus $U$ is an isometry, and so extends to $H\otimes K_0$. Now compute \begin{align*} U^* \Phi(\theta_{\xi,\eta}) U(\xi_1\otimes\alpha) &= U^* \Phi(\theta_{\xi,\eta}) \Phi(\theta_{\xi_1,\xi_0})(\alpha) \\ &= (\xi_1|\eta) U^*\Phi(\theta_{\xi,\xi_0})(\alpha) \\ &= (\xi_1|\eta) U^*U(\xi\otimes\alpha) \\ &= \theta_{\xi,\eta}(\xi_1) \otimes \alpha. \end{align*} So $U^*\Phi(\theta_{\xi,\eta})U = \theta_{\xi,\eta}\otimes 1$ and so by linearity and continuity, $U^*\Phi(\theta)U = \theta\otimes 1$ for each $\theta\in\mc K(H)$.

If we can show that $U$ has dense range, it must be onto (as it's an isometry), and so will be a unitary, and so $UU^*=1$ and so $\Phi(\theta) = U(\theta\otimes 1)U^*$ as required.

If $\xi_1$ is another vector, we see that $$ \Phi(\theta_{\xi,\xi_1})(\alpha) = \Phi(\theta_{\xi,\xi_0})\Phi(\theta_{\xi_0,\xi_1})(\alpha) = U(\xi \otimes \beta), $$ say, where $\beta = \Phi(\theta_{\xi_0,\xi_1})(\alpha)$. Letting $\xi, \xi_1,\alpha$ vary, taking linear span, and using non-degeneracy, we see that $U$ does indeed have dense range.

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    $\begingroup$ This is the way to do it. In one sentence: look at what happens to the rank 1 matrices. $\endgroup$ – Nik Weaver Apr 14 at 14:21
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    $\begingroup$ I guess this is the difference between being an algebraist and an analyst :). This proof I can't understand but the other two (well one was mine), I can. I guess I had better finally learn this stuff :(. +1 $\endgroup$ – Benjamin Steinberg Apr 14 at 15:28
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    $\begingroup$ As someone with occasional pretensions to both guilds: isn't the common theme between @BenjaminSteinberg 's answer and Matt's "use the matrix units from the algebra you're mapping out of to chop up the Hilbert space on the target side"? $\endgroup$ – Yemon Choi Apr 14 at 15:34
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    $\begingroup$ One question about notation: when you say "which is normal (aka weak*-continuous)", do you mean that "normal homomorphism" is defined as "weak*-continuous homomorphism", or is weak*-continuity a consequence of being normal (in which case, which of the many meanings of the word normal is used here)? Also just to make sure: The theorem you show is: "For Hilbert spaces $H,K$, every normal unital *-homomorphism $\Phi:\mathcal B(H)\to\mathcal B(K)$ is of the form $\Phi(a)=U(a\otimes 1)U^*$ for some unitary $U$ and some Hilbert space $K_0$ (on which $1$ operates)", right? $\endgroup$ – Dominique Unruh Apr 14 at 16:53
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    $\begingroup$ Found the answer to the normality question (Conway, A course in operator theory, Def 46.1). Normality in this sense is a notional that applies to positive linear maps between von Neumann algebras. And by Coro 46.5, a positive linear map is normal iff it is weak* continuous. — So I think I got it all, and this is exactly the kind of elementary I had hoped for! And it covers the infinite case. Perfect. :) $\endgroup$ – Dominique Unruh Apr 14 at 17:31

Here is a proof using the Skolem-Noether theorem. One formulation of is that any two unital embeddings of a simple algebra like $M_i$ into $M_j$ are conjugate by an invertible matrix. This follows because a homomorphism $\psi\colon M_i\to M_j$ is a representation of dimension $j$. All representations of $M_i$ are isomorphic to $(\mathbb C^i)^k$ for some $k\geq 0$ and so any two $j$-dimensional representations are equivalent, that is, simultaneously conjugate (also $i\mid j$).

Now we have one star-embedding $\gamma\colon M_i\to M_j$ given by $a\mapsto \begin{bmatrix} a & 0\\ 0 & I\end{bmatrix}$. Let $\psi\colon M_i\to M_j$ be another. Then by Skolem-Noether, we have an invertible matrix $P$ with $P\gamma(a)P^{-1}=\psi(a)$ for all $a\in M_i$. Now I will show that we can replace $P$ by a unitary. First note that $P\gamma(a^*)P^{-1}=\psi(a^*)$ and so using that $\gamma$ and $\psi$ are $*$-homomorphisms, we get $\psi(a)=(P^{-1})^*\gamma(a)P^*$. It follows that $P^*P\gamma(a)=\gamma(a)P^*P$. We can write $P^*P=\begin{bmatrix}A & B \\ C & D\end{bmatrix}$ with $A,B,C,D$ block matrices and we deduce $Aa=aA$, $aB=B$, $C=Ca$ for all $a\in M_i$. We deduce that $A=\lambda I$ for some scalar $\lambda$ and $B=0=C$ (taking $a=0$). So $P^*P=\begin{bmatrix} \lambda I & 0\\ 0 & D\end{bmatrix}$ and $D^*=D$, $\lambda$ is a positive real as $P^*P$ is positive definite (since $P$ is invertible). Also, $D$ is a positive definite Hermitian matrix and so has a positive definite Hermitian square root which I will denote $\sqrt{D}$ abusively with $\sqrt{D}^*\sqrt{D}=D$.

Let $$U=P\begin{bmatrix} \frac{1}{\sqrt{\lambda}}I & 0\\ 0 & \sqrt{D}^{-1}\end{bmatrix}.$$ I claim that $U$ is unitary and conjugates $\gamma$ to $\psi$. First note that $$U^*U = \begin{bmatrix} \frac{1}{\sqrt{\lambda}}I & 0\\ 0 & (\sqrt{D}^*)^{-1}\end{bmatrix}P^*P\begin{bmatrix} \frac{1}{\sqrt{\lambda}}I & 0\\ 0 & \sqrt{D}^{-1}\end{bmatrix} =I$$ from the construction and hence is unitary.

On the other hand, $U^*=U^{-1} = \begin{bmatrix} \sqrt{\lambda}I & 0\\ 0 & \sqrt{D}\end{bmatrix}P^{-1}.$ Therefore, $$U\begin{bmatrix} a & 0 \\ 0 & I\end{bmatrix}U^*=P\begin{bmatrix} a & 0 \\ 0 & I\end{bmatrix}P^{-1}=\psi(a)$$ as required.


$\DeclareMathOperator{\End}{End}\DeclareMathOperator{\Hom}{Hom}\newcommand{\C}{\mathbb{C}}$The Morita theory argument says the following: For $V$ a finite-dimensional vector space, there is an equivalence of categories (in particular, a bijection of isomorphism classes) between vector spaces and (left) $\End(V)$-modules. In one direction, we send a vector space $W$ to $V\otimes W$; in the other direction, we send a $\End(V)$-module $M$ to $\Hom_{\End(V)}(V,M)$. The isomorphism $V\otimes\Hom_{\End(V)}(V,M)\to M$ is given by the evaluation map $v\otimes f\mapsto f(v)$.

In your situation, the algebra homorphism $\Phi$ turns $\C^j$ into a $\End(\C^i)\cong M_i$-module. The $\End(\C^i)$-module $\C^i$ is cyclic, i.e. generated by a single nonzero vector, which we can take to be the first basis vector; then $U\cong\Hom_{M_i}(C^i,\C^j)$ is the subspace of $\C^j$ which is annihilated by all elements $\Phi(M)$, where $M$ is a matrix whose first column vanishes. One can use elementary linear algebra to find a basis of this space, which must have cardinality $\frac{j}{i}$ (in particular, this number must be an integer), giving rise to an isomorphism $\phi:\C^{j/i}\cong U$. The resulting map $\C^i\otimes\C^{j/i}\cong \C^i\otimes U\cong \C^j$ sends $v\otimes w$ to $\Phi(M_v)(\phi(w))$, where $M_v$ is the matrix which has $v$ in the first column and $0$ everywhere else (by definition of $U$, we can actually put anything into the other columns). Expressing this linear isomorphism in the natural bases of both sides gives rise to an element of $GL(j)$ which intertwines $\Phi$ with $\operatorname{id}\otimes I_r$.

All of this should go through exactly the same way in the presence of hermitian metrics on your vector spaces, provided that you restrict to $*$-homomorphisms. If $V$ is infinite-dimensional, one has to take the correct completion of the tensor product; I think the explicit map I defined above should still work.


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