# 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 $$tau^2 = ntau$$), but I don’t know whether $$mathcal{C}$$ is larger than this span.

MathOverflow Asked by Jochen Glueck on January 16, 2021

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

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 $$mathbb Coplus mathbb C$$. The elements $$(1,0)$$ and $$(0,1)$$ are just orthogonal projections to $$mathbb Ccdot mathrm{id}$$ 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 $$tau$$.

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

Let $$Phiinmathcal{L}(mathbb{C}^{n times n})$$ and suppose that for any unitary conjugation operator $$mathcal{U}inmathcal{L}(mathbb{C}^{n times n})$$, $$mathcal{U}Phi=Phimathcal{U}$$. Let $${e_1,e_2,dots,e_n}$$ denote the standard orthonormal basis for $$mathbb{C}^n$$, let $$e_{ij}$$ denote the $$ntimes n$$ matrix with $$1$$ in the $$i$$th row $$j$$th column and zeros elswhere, and let $$E_{ij}=Phi(e_{ij})$$. Claim, there exists $$r,sinmathbb{C}$$ such that, $$begin{equation} E_{ij}=begin{cases} re_{ij} & text{ if } ineq j, text{ and }\ % re_{ij}+sI & text{ if } i= j, end{cases}tag{1} end{equation}$$ so that for $$Binmathbb{C}^{ntimes n}$$, $$Phi(B)=rB+mathrm{tr}(B)sI$$. Accordingly, fix $$1leq i,jleq n$$ with $$ineq j$$, 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 $$u_{1}=begin{bmatrix} i & 0 \ 0 & 1end{bmatrix}$$, $$u_{2}=begin{bmatrix} 0 & 1 \ 1 & 0end{bmatrix}$$, and $$u_{3}=frac{1}{sqrt 2}begin{bmatrix} 1 & 1 \ -1 & 1end{bmatrix}$$ respectively. Note that for $$A=begin{bmatrix} a & b \ c & dend{bmatrix}inmathbb{C}timesmathbb{C},$$

$$u_1Au_1^dagger=begin{bmatrix} a & -ib \ ic & dend{bmatrix}quad u_2Au_2^dagger=begin{bmatrix} d & c \ b & aend{bmatrix}quad u_3Au_3^dagger= frac12begin{bmatrix} a+b+c+d & -a+b-c+d \ -a-b+c+d & a-b-c+d end{bmatrix}$$

For $$k=1,2,3$$, $$U_kP_{ij}=P_{ij}U_k$$ so that $$mathcal{U}_{k}Phi= Phimathcal{U}_{k}$$ implies for $$ell,min{i,j}$$ , $$begin{equation} label{eq:compression} u_kepsilon_{ell m}u_k^dagger= P_{ij}Phi(U_ke_{ell m}U_k^dagger)Bigm|_{K_{ij}}tag{2} end{equation}$$ 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 $$2u_3epsilon_{ij}u_3^dagger= begin{bmatrix}b_{ij} & b_{ij} \ -b_{ij} & -b_{ij} end{bmatrix}$$. 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 $$Phiinmathcal{L}(mathbb{C}^{ntimes n})$$ and suppose that for any orthogonal conjugation operator $$mathcal{O}inmathcal{L}(mathbb{C}^{ntimes n})$$, $$mathcal{O}Phi=Phimathcal{O}$$. There exists $$r,s,tinmathbb{C}$$ such that, for $$Binmathbb{C}^{ntimes n}$$, $$Phi(B)=rB+sB^top+mathrm{tr}(B)tI$$.

Answered by Edwin Franks on January 16, 2021

$$mathcal{C}$$ 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 $$U in mathbb{C}^{ntimes n}$$ (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" $$rho = sum_{i,j=1}^n mathbf{e}_imathbf{e}_j^* otimes mathbf{e}_imathbf{e}_j^*$$ (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.

Answered by Nathaniel Johnston on January 16, 2021

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