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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.

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    $\begingroup$ 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. $\endgroup$ – Vadim Alekseev Nov 30 '19 at 10:38
  • $\begingroup$ 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). $\endgroup$ – Vadim Alekseev Nov 30 '19 at 11:22
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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 $\tau$.

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  • $\begingroup$ Thanks a lot for your answer! Could you specify what you mean by $\mathfrak{u}(n)$ (as opposed to $U(n)$)? $\endgroup$ – Jochen Glueck Nov 30 '19 at 21:20
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    $\begingroup$ The Lie algebra of $\mathrm{U}(n)$: $\mathfrak u(n) = \{X\in\mathbb M_n(\mathbb C)\mid X^* = -X\}$. $\endgroup$ – Vadim Alekseev Dec 1 '19 at 9:18
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$\mathcal{C}$ 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.

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  • $\begingroup$ +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! $\endgroup$ – Jochen Glueck Dec 3 '19 at 16:45

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