# Unital *-homomorphisms between matrices

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.)

Notes:

• 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.

• This is the way to do it. In one sentence: look at what happens to the rank 1 matrices. – Nik Weaver Apr 14 at 14:21
• 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 – Benjamin Steinberg Apr 14 at 15:28
• 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"? – Yemon Choi Apr 14 at 15:34
• 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? – Dominique Unruh Apr 14 at 16:53
• 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. :) – 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.