# Commutant of the conjugations by unitary matrices

Let $$\mathcal{L}(\mathbb{C}^{n \times n})$$ denote the algebra of all linear mappings from $$\mathbb{C}^{n \times n}$$ to $$\mathbb{C}^{n \times n}$$ and let $$\mathcal{C} \subseteq \mathcal{L}(\mathbb{C}^{n \times n})$$ denote the subalgebra of all $$\phi \in \mathcal{L}(\mathbb{C}^{n \times n})$$ which satisfy $$\phi(U^*AU) = U^*\phi(A)U$$ for all $$A \in \mathbb{C}^{n \times n}$$ and all unitary $$U \in \mathbb{C}^{n \times n}$$ (in other words, $$\mathcal{C}$$ is the commutant of the set of all conjugations by unitary matrices).

Question. Is there an explicit description of $$\mathcal{C}$$?

Of course, there is some freedom for interpretation of the word "explicit"; I would be most happy with a set of mappings in $$\mathcal{L}(\mathbb{C}^{n \times n})$$ which spans $$\mathcal{C}$$.

Remarks:

• Clearly, the identity $$\operatorname{id}_{\mathbb{C}^{n \times n}}$$ is an element of $$\mathcal{C}$$.

• The operator $$\tau: \mathbb{C}^{n \times n}\to \mathbb{C}^{n \times n}$$ given by $$\tau(A) = \operatorname{tr}(A) \cdot \operatorname{id}_{\mathbb{C}^{n \times n}}$$ is an element of $$\mathcal{C}$$ (where $$\operatorname{tr}(A)$$ denotes the trace of the matrix $$A$$).

• The span of $$\operatorname{id}_{\mathbb{C}^{n \times n}}$$ and $$\tau$$ is a subalgebra of $$\mathcal{C}$$ (since $$\tau^2 = n\tau$$), but I don't know whether $$\mathcal{C}$$ is larger than this span.

• The conjugation representation of $\mathrm U(n)$ on $\mathbb M_n(\mathbb C)$ decomposes into irreducible representations, and the algebra of $\mathrm U(n)$-equivariant maps is completely determined by this decomposition: it is a sum of matrix algebras, one for each irreducible component, of the size equal to the multiplicity of that irreducible component. Now, there is, for instance, $\mathfrak{su}(n)$ as an irreducible subspace (and $\mathbb C\cdot 1$, as well). I don't know the decomposition of $\mathfrak{su}(n)^\perp$ off the top of my head, but it should be in the literature. Nov 30, 2019 at 10:38
• I just realised that $\mathfrak{su}(n)$ is a real, not complex, subrep'n. Its complex counterpart is $\mathfrak{sl}(n)$, and it's irreducible (see answer below). Nov 30, 2019 at 11:22

Building up on my comment, I can now give the complete answer. The space of matrices can be decomposed as follows: $$\mathbb M_n(\mathbb C) = \mathbb C\cdot\mathrm{id}\oplus \mathfrak{sl}(n),$$ where $$\mathfrak{sl}(n) = \{X\in\mathbb M_n(\mathbb C)\mid \mathrm{Tr}(X) = 0\}.$$

Thus, the conjugation representation of $$\mathrm{U}(n)$$ decomposes as the sum of a trivial representation and the conjugation repesentation on $$\mathfrak{sl}(n)$$. The latter is irreducible as a complex representation of $$\mathrm{U}(n)$$ because:

• the complexification of $$\mathfrak{u}(n)$$ is $$\mathfrak{gl}(n)$$, and
• the Lie algebra $$\mathfrak{sl}(n)$$ is a simple complex Lie algebra.

Therefore the algebra of linear $$\mathrm U(n)$$-equivariant maps is isomorphic to $$\mathbb C\oplus \mathbb C$$. The elements $$(1,0)$$ and $$(0,1)$$ are just orthogonal projections to $$\mathbb C\cdot \mathrm{id}$$ and its orthogonal complement $$\mathfrak{sl}(n)$$.

So, the space $$\mathcal C$$ from the question is indeed spanned by $$\mathrm{id}_{\mathbb M_n}$$ and $$\tau$$.

• Thanks a lot for your answer! Could you specify what you mean by $\mathfrak{u}(n)$ (as opposed to $U(n)$)? Nov 30, 2019 at 21:20
• The Lie algebra of $\mathrm{U}(n)$: $\mathfrak u(n) = \{X\in\mathbb M_n(\mathbb C)\mid X^* = -X\}$. Dec 1, 2019 at 9:18

$$\mathcal{C}$$ is simply the span of the two maps that you noted (the identity and the trace) -- there is nothing else in the commutant.

One (admittedly somewhat roundabout) way of seeing this is to notice that if you unpack $$\phi$$ into an $$n^2 \times n^2$$ matrix $$\Phi$$ in the "usual" way (i.e., instead of thinking of it as a linear transformation acting on matrices, think of it as a matrix acting on their vectorizations), then your commutation relation is equivalent to $$(U \otimes \overline{U})\Phi(U \otimes \overline{U})^* = \Phi$$ for all unitary $$U \in \mathbb{C}^{n\times n}$$ (here $$\overline{U}$$ is the entrywise complex conjugate of $$U$$).

This is the defining property of something called an isotropic state from quantum information theory, and it is well-known (see this paper, for example) that all matrices with this property are linear combinations of the identity matrix and the "maximally entangled state" $$\rho = \sum_{i,j=1}^n \mathbf{e}_i\mathbf{e}_j^* \otimes \mathbf{e}_i\mathbf{e}_j^*$$ (where $$\{\mathbf{e}_i\}$$ is the standard basis of $$\mathbb{C}^n$$). These two matrices correspond to the trace linear map and the identity linear map, respectively, once you "un-vectorize" everything.

• +1 Thank you very much! After I while of thinking I finally chose to accept Vadim Alekseev' answer (although your post answers my question as well) since he was a bit earlier and since his answer "forced" me to learn a few facts about Lie algebras ;-). Anyway, I'm very grateful to you for pointing out the connection to quantum information theory! Dec 3, 2019 at 16:45

I posted (nearly) this on MSE, so if it doen't belong here plese remove.

Let $$\Phi\in\mathcal{L}(\mathbb{C}^{n \times n})$$ and suppose that for any unitary conjugation operator $$\mathcal{U}\in\mathcal{L}(\mathbb{C}^{n \times n})$$, $$\mathcal{U}\Phi=\Phi\mathcal{U}$$. Let $$\{e_1,e_2,\dots,e_n\}$$ denote the standard orthonormal basis for $$\mathbb{C}^n$$, let $$e_{ij}$$ denote the $$n\times n$$ matrix with $$1$$ in the $$i$$th row $$j$$th column and zeros elswhere, and let $$E_{ij}=\Phi(e_{ij})$$. Claim, there exists $$r,s\in\mathbb{C}$$ such that, $$$$E_{ij}=\begin{cases} re_{ij} & \text{ if } i\neq j, \text{ and }\\ % re_{ij}+sI & \text{ if } i= j, \end{cases}\tag{1}$$$$ so that for $$B\in\mathbb{C}^{n\times n}$$, $$\Phi(B)=rB+\mathrm{tr}(B)sI$$. Accordingly, fix $$1\leq i,j\leq n$$ with $$i\neq j$$, let $$K_{ij}$$ denote the span of $$\{e_i,e_j\}$$, and let $$P_{ij}$$ denote the orthogonal projection onto $$K_{ij}$$. Notate the compressions to $$K_{ij}$$ by $$\begin{gather*}\epsilon_{ii}=P_{ij}E_{ii}\Bigm|_{K_{ij}}= \begin{bmatrix} a_{ii} & b_{ii} \\ c_{ii} & d_{ii} \end{bmatrix}\qquad % \epsilon_{ij}=P_{ij}E_{ij}\Bigm|_{K_{ij}}= \begin{bmatrix} a_{ij} & b_{ij} \\ c_{ij} & d_{ij} \end{bmatrix}\\ % \epsilon_{ji}=P_{ij}E_{ji}\Bigm|_{K_{ij}}= \begin{bmatrix} a_{ji} & b_{ji} \\ c_{ji} & d_{ji} \end{bmatrix}\qquad % \epsilon_{jj}=P_{ij}E_{jj}\Bigm|_{K_{ij}}= \begin{bmatrix} a_{jj} & b_{jj} \\ c_{jj} & d_{jj} \end{bmatrix} % \end{gather*}$$

Let $$U_{1}$$, $$U_{2}$$, and $$U_{3}$$ be the unitary matrices which fix the orthogonal complement of $$K_{ij}$$ with action on $$K_{ij}$$ given by $$u_{1}=\begin{bmatrix} i & 0 \\ 0 & 1\end{bmatrix}$$, $$u_{2}=\begin{bmatrix} 0 & 1 \\ 1 & 0\end{bmatrix}$$, and $$u_{3}=\frac{1}{\sqrt 2}\begin{bmatrix} 1 & 1 \\ -1 & 1\end{bmatrix}$$ respectively. Note that for $$A=\begin{bmatrix} a & b \\ c & d\end{bmatrix}\in\mathbb{C}\times\mathbb{C},$$

$$u_1Au_1^\dagger=\begin{bmatrix} a & -ib \\ ic & d\end{bmatrix}\quad u_2Au_2^\dagger=\begin{bmatrix} d & c \\ b & a\end{bmatrix}\quad u_3Au_3^\dagger= \frac12\begin{bmatrix} a+b+c+d & -a+b-c+d \\ -a-b+c+d & a-b-c+d \end{bmatrix}$$

For $$k=1,2,3$$, $$U_kP_{ij}=P_{ij}U_k$$ so that $$\mathcal{U}_{k}\Phi= \Phi\mathcal{U}_{k}$$ implies for $$\ell,m\in\{i,j\}$$ , $$$$\label{eq:compression} u_k\epsilon_{\ell m}u_k^\dagger= P_{ij}\Phi(U_ke_{\ell m}U_k^\dagger)\Bigm|_{K_{ij}}\tag{2}$$$$ With $$k=1$$, equation (2) shows that the off-diagonal entries of $$\epsilon_{ii}$$ and $$\epsilon_{jj}$$ equal zero and that all entries of $$\epsilon_{ij}$$ and $$\epsilon_{ji}$$ except for $$b_{ij}$$ and $$c_{ji}$$ must equal zero. Since $$\mathcal{U}_2(e_{ii})=e_{jj}$$, $$a_{ii}=d_{jj}$$ and $$d_{ii}=a_{jj}$$, and since $$\mathcal{U}_2(e_{ij})=e_{ji}$$, $$b_{ij}=c_{ji}$$. With these identities, it follows that $$2u_3\epsilon_{ij}u_3^\dagger= \begin{bmatrix}b_{ij} & b_{ij} \\ -b_{ij} & -b_{ij} \end{bmatrix}$$. Further, since $$2U_3e_{ij}U_3^\dagger=e_{ii}+e_{ij}-e_{ii}-e_{ii}$$, $$\begin{bmatrix}b_{ij} & b_{ij} \\ -b_{ij} & -b_{ij} \end{bmatrix}= \begin{bmatrix} a_{ii}-d_{ii} & b_{ij} \\ -b_{ij} & d_{ii}-a_{ii}\end{bmatrix},$$ so that $$a_{ii}-d_{ii} = b_{ij}$$. Letting $$r=b_{ij}$$ and $$s=d_{ii}$$ one has, $$\epsilon_{ii}=\begin{bmatrix} s+r & 0 \\ 0 & s \end{bmatrix}\quad % \epsilon_{ij}=\begin{bmatrix} 0 & r \\ 0 & 0 \end{bmatrix}\qquad % \epsilon_{ji}=\begin{bmatrix} 0 & 0 \\ r & 0 \end{bmatrix}\qquad % \epsilon_{jj}=\begin{bmatrix} s & 0\\ 0 & s+r \end{bmatrix}\qquad %$$ Letting $$i,j$$ run through all unequal pairs yields equation (1).

Remark. Using similar techniques one can show the following. Let $$\Phi\in\mathcal{L}(\mathbb{C}^{n\times n})$$ and suppose that for any orthogonal conjugation operator $$\mathcal{O}\in\mathcal{L}(\mathbb{C}^{n\times n})$$, $$\mathcal{O}\Phi=\Phi\mathcal{O}$$. There exists $$r,s,t\in\mathbb{C}$$ such that, for $$B\in\mathbb{C}^{n\times n}$$, $$\Phi(B)=rB+sB^\top+\mathrm{tr}(B)tI$$.