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 $mathcal{C}$.


  • 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

3 Answers

3 Answers

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 $tau$.

Correct answer by Vadim Alekseev on January 16, 2021

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

Add your own answers!

Related Questions

Lorentzian vs Gaussian Fitting Functions

5  Asked on January 1, 2022 by jimmidyjoo


Motives under de-singularization

0  Asked on January 1, 2022 by user161697


Ask a Question

Get help from others!

© 2022 All rights reserved.