Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Is the following true:

For every unit vectors $x_1,..., x_n$, $y_1,..., y_n$ in $\mathbb{C}^k$

there exist a Hilbert space $H$, unitary operators $U_1,...,U_n$ and $V_1,...,V_n$ in $B(H)$ and unit vectors $x,y \in H$,

such that $[U_i,V_j]=0$ for every $1\leq i,j\leq n$ and $$\langle x_i, y_j \rangle = \langle U_i V_j x,y\rangle \qquad\text{for every $1\leq i,j\leq n.$}$$

EDIT: What about the same question, but with $H$ - finite dimensional?

share|improve this question
5  
TeX note: Please use \langle and \rangle for angle brackets. Inequalities are not brackets. I fixed the markup for you here. –  Harald Hanche-Olsen Jan 26 '11 at 15:49
1  
A stupid comment, but what happens with $n=2, 3, ...$? –  Zen Harper Jan 26 '11 at 16:35
1  
@Zen: k=1 is OK, $|x_i|=|y_j|=1$, $U_i=x_i$, $V_j=y_j$. –  Kate Juschenko Jan 26 '11 at 17:53
1  
There is a strictly stronger statement that also has a chance of being true: Given $n$ unit vectors $x_1,\dots,x_n \in \mathbb{C}^n$, let $x=\sqrt{\frac1n} \left[1,\dots,1\right]^T \in\mathbb{C}^n$ be the normalized all-ones vector. Then there exist unitary $n \times n$ diagonal matrices $D_1,\ldots,D_n$ (with $D_1=I$, if desired) such that $\langle x_i,x_j \rangle = \langle D_ix,D_jx \rangle$. –  Tracy Hall Jan 27 '11 at 8:57
3  
Are you aware of the fact that the answer is positive when the initial vectors belong to $\mathbb{R}^k$ ? This follows from the existence of a $k$-subspace in $B(R^{2^k})$ in which every matrix is a multiple of an orthogonal matrix (Clifford algebras give that); this was used by Tsirelson to connect Grothendieck theorem and Bell inequalities. See tau.ac.il/~tsirel/download/qbell87.html –  Guillaume Aubrun Feb 4 '11 at 8:22

1 Answer 1

This is not an answer to Kate's question, but a remark that is too long for a comment. I answers (negatively) Tracy's comment. I am sure all this is clear to Kate.

The content of my comment is that you cannot take all the $U_i$'s and $V_j$'s commuting, and this is related to Grothendieck's inequality (see Pisier's very nice recent survey).

Indeed, then the $C^*$-algebra generated by the $U_i$'s and the $V_j$'s would be commutative and thus isomorphic to some $C(X)$ for a compact space $X$, and the linear functional $A\mapsto \langle Ax,y\rangle$ would correspond to a (signed) measure $\mu$ of total mass at most $1$ on $X$.

This would mean that you could write $\langle x_i,y_j\rangle$ as $\int f_i g_j d\mu$ where $f_i$ and $g_j \in C(X)$ have modulus one. Multiplying $g_j$ by the Radon-Nikodym derivative of $\mu$ with respect to $|\mu|$ we could as well assume that $\mu$ be positive. But Grothendieck's inequality says that such a decomposition is in general not possible unless $\sup_i \|f_i\|_2 \sup_j \|g_j\|_2\geq K_G$ where $K_G>1$ is Grothendieck's constant (complex case) and $\|\cdot\|_2$ is the $L^2$ norm with respect to $\mu$.

In fact (see remark 2.12 here), it is even known that, in general, one cannot take diagonal $D_i$'s as in Tracy Hall's comment unless $\sup_i \|D_i\|_2 \geq c n \log n$ for some $c$. Here for a diagonal matrix $D$, by $\|D\|_2$ I mean $(\sum_k |D_k|^2)^{1/2}$.

share|improve this answer
    
Thanks, Mikael, Gilles' servey was exactly my motivation! –  Kate Juschenko Jan 27 '11 at 15:28
    
The question is related to the H'-norm. Most likely the question has negative solution, but I just don't know how to find it. –  Kate Juschenko Jan 27 '11 at 15:30
    
By the way, H' and max norm coincides on $E\otimes E$, where $E$ is generated by 2-unitaries $U_1$ and $U_2$ in $C^*(\mathbb{F}_{\infty})$. –  Kate Juschenko Jan 27 '11 at 15:32
    
What is H'-norm? –  Mikael de la Salle Jan 27 '11 at 15:36
    
I just checked Gilles survey, he added this question to the remark on the page 42. Actually, it is not exactly the same question, but the equality of H' and max will follow from it. –  Kate Juschenko Jan 27 '11 at 15:39

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.