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This question is similar to this previous one but I think it is harder.

Let $X$, $Y$, $Z$, and $W$ be $2\times 2$ Hermitian matrices. Can we always find $\theta,\phi \in [0,\pi/2]$ and $2\times 2$ unitaries $U$ and $V$ such that $$SUXU^*S + CVYV^*C$$ and $$SUZU^*S + CVWV^*C,$$ are both scalar multiples of the identity matrix, where $S = {\rm diag}(\sin \theta, \sin \phi)$ and $C = {\rm diag}(\cos \theta, \cos\phi)$?

I'm guessing this is true. Maybe there is a homotopy reason as in the previous question. I think an answer in either direction would likely lead to a solution to this other previous question.

Edit: another way to ask this is, given $X$, $Y$, $Z$, and $W$, can we find $2\times 2$ matrices $A$ and $B$ such that $$AA^* + BB^*$$ $$AXA^* + BYB^*$$ $$AZA^* + BWB^*$$ are all scalar multiples of the identity (and the first one is not zero).

Edit 2: another reformulation. Let $\alpha$ and $\beta$ be the column vectors of $A^*$ and $\gamma$ and $\delta$ the column vectors of $B^*$ in the first edit. Then by looking at matrix entries of the expressions given there we see that the conditions become $$\langle X\alpha, \alpha\rangle + \langle Y\gamma, \gamma\rangle = \langle X\beta, \beta\rangle + \langle Y\delta, \delta\rangle$$ and $$\langle X\alpha,\beta\rangle + \langle Y\gamma,\delta\rangle = 0,$$ plus similar conditions with $Z, W$ or $I_2,I_2$ in place of $X,Y$. The problem is to find $\alpha,\beta,\gamma,\delta \in \mathbb{C}^2$, not all zero, satisfying these equations.

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  • $\begingroup$ I made a grammatical edit for clarity - hope that's OK: I changed "(display 1) and (display 2) where S=... and C=... are both multiples of I" to "(display 1) and (display 2) are both multiples of I where S=... and C=..." $\endgroup$ – Anthony Quas Jan 23 '18 at 4:32
  • $\begingroup$ Thank you. I agree that it's more readable this way. $\endgroup$ – Nik Weaver Jan 23 '18 at 5:14
  • $\begingroup$ My understanding is that the action of $G=GL(2,\mathbb{C})$ on 2x2 Hermitian matrices ($H\mapsto AHA^*$ for $A\in G$) plays a role here. I wonder what is known about it (it should be well-known...) $\endgroup$ – Dima Pasechnik Jan 31 '18 at 10:18
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Too long for a comment, maybe someone with a respectable CAS and the skill to use it can try. MATLAB's symbolic toolbox returns "Warning: Cannot find explicit solution" from the obvious/naive attempt

%% Parameters
syms x0 x1 x2 y0 y1 y2 z0 z1 z2 w0 w1 w2;   % all assumed complex
X = [x1,x0;conj(x0),x2];    % Hermitian
Y = [y1,y0;conj(y0),y2];    % Hermitian
Z = [z1,z0;conj(z0),z2];    % Hermitian
W = [w1,w0;conj(w0),w2];    % Hermitian
%% Variables
syms a11 a12 a21 a22 b11 b12 b21 b22 L M;   % all assumed complex
A = [a11,a12;a21,a22];
B = [b11,b12;b21,b22];
%% Equations
C = A*A'+B*B'-eye(2);
D = A*X*A'+B*Y*B'-L*eye(2);
E = A*Z*A'+B*W*B'-M*eye(2);
eqns = [C(1,1) == 0, C(1,2) == 0, C(2,1) == 0, C(2,2) == 0,...
    D(1,1) == 0, D(1,2) == 0, D(2,1) == 0, D(2,2) == 0,...
    E(1,1) == 0, E(1,2) == 0, E(2,1) == 0, E(2,2) == 0];
%% Solution
S = solve(eqns,[a11 a12 a21 a22 b11 b12 b21 b22 L M])
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  • $\begingroup$ I saw your comment this morning, and I've been struggling with learning how to use a computer algebra system ...getting an answer this way wouldn't really satisfy me because I'd want to know why. Still, at the very least simply knowing which answer is correct would surely be helpful. $\endgroup$ – Nik Weaver Jan 24 '18 at 19:16
  • $\begingroup$ BTW, I think you want to declare $x1$, $x2$, $y1$, $y2$, etc. real. Also note that if $M$ is Hermitian then we can express "$M$ is a scalar matrix" by just saying $M(1,1)=M(2,2)$ and $M(1,2) = 0$, if that helps. $\endgroup$ – Nik Weaver Jan 24 '18 at 19:19
  • $\begingroup$ Argh...you’re right about the diagonals $\endgroup$ – Steve Huntsman Jan 25 '18 at 1:42
  • $\begingroup$ Anyway, I installed the "Maxima" CAS and learned enough about it to ask it this question, but even when I asked it to solve a baby version of the problem it hung indefinitely... $\endgroup$ – Nik Weaver Jan 25 '18 at 16:54
  • $\begingroup$ Maxima is not good for these kind of things; you need to use a dedicated algebra CAS such as Singular, or Macaulay2, or Magma to have a fighting chance of an answer here. $\endgroup$ – Dima Pasechnik Jan 29 '18 at 12:26
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First let $P$ be the space of nonzero quadruples $(X,Y,Z,W)$ of Hermitian matrices, and let $Q$ be the set of sextuples $(X,Y,Z,W,A,B)$ where $(X,Y,Z,W)\in P$ and $(A,B)\neq(0,0)$ and the three matrices mentioned are multiples of the identity. There is a projection map $\pi\colon Q\to P$, and we want to know whether this is surjective.

Let $P_1$ be the subspace of $P$ where $\|X\|^2+\|Y\|^2+\|Z\|^2+\|W\|^2=1$, and let $Q_1$ be the subspace of $Q$ where $(X,Y,Z,W)\in P_1$ and $AA^*+BB^*=I$. Because our equations are appropriately homogeneous, it is equivalent to ask whether the projection $\pi\colon Q_1\to P_1$ is surjective. Note that $P_1$ and $Q_1$ are compact, so $\pi(Q_1)$ is closed in $P_1$. Thus, if there are any points not in $\pi(Q_1)$, then they form a nonempty open set, and we can hope to find a counterexample by random search.

For fixed $(X,Y,Z,W)\in P$ it works out that we are trying to solve 12 equations in 16 variables, so it is reasonable to hope for a solution. I generated 1000 random examples and in each case set four of the 16 variables arbitrarily and solved numerically for the remaining 12, with success in every case. This convinces me that the answer is probably positive.

Maple code is as follows:

with(LinearAlgebra):
r := () -> rand(-100..100)()/10.;
rand_hermitian := proc()
 local a,b,c,d;
 a := r(); b := r(); c := r(); d := r();
 <<a|b+I*c>,<b-I*c|d>>;
end:

dagger := (M) -> map(conjugate,Transpose(M));

assume(a1::real,a2::real,a3::real,a4::real,
       a5::real,a6::real,a7::real,a8::real,
       b1::real,b2::real,b3::real,b4::real,
       b5::real,b6::real,b7::real,b8::real);

A := <<a1+a2*I|a3+a4*I>,<a5+a6*I|a7+a8*I>>;
B := <<b1+b2*I|b3+b4*I>,<b5+b6*I|b7+b8*I>>;

for i from 1 to 1000 do
 if modp(i,10) = 0 then print(i); fi;
 W := rand_hermitian();
 X := rand_hermitian();
 Y := rand_hermitian();
 Z := rand_hermitian();
 MM := {A.dagger(A) + B.dagger(B),
        A.X.dagger(A) + B.Y.dagger(B),
        A.Z.dagger(A) + B.W.dagger(B)}:
 MM := map(M -> op(expand([M[1,2],M[2,1],M[1,1]-M[2,2]])),MM): 
 MM := map(z -> op([Re(z),Im(z)]),MM) minus {0}:
 sol := fsolve({a1-1,a8,b2,b8,op(MM)});
 if not(type(subs(sol,[a1,a2,a3,a4,a5,a6,a7,a8,b1,b2,b3,b4,b5,b6,b7,b8])
             [numeric$16])) then
  print("problem");
  break;
 fi;
od:
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  • $\begingroup$ I really appreciate this. That's the same conclusion I came to, but in my case based only on a small number of examples worked by hand. $\endgroup$ – Nik Weaver Jan 24 '18 at 22:25

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