# 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. – Vadim Alekseev Nov 30 '19 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). – Vadim Alekseev Nov 30 '19 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 $U(n)$)? – Jochen Glueck Nov 30 '19 at 21:20
• The Lie algebra of $\mathrm{U}(n)$: $\mathfrak u(n) = \{X\in\mathbb M_n(\mathbb C)\mid X^* = -X\}$. – Vadim Alekseev Dec 1 '19 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! – Jochen Glueck Dec 3 '19 at 16:45