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I have an idea for a possible counterexample to the noncommutative Stone-Weierstrass problem. A good answer to the following question would really help.

Let $\mathcal{A}$ be the C*-algebra of $2\times 2$ complex matrices, let $\mathcal{B}$ be the C*-subalgebra of $2\times 2$ diagonal matrices, and let $v$ and $w$ be the unit vectors $\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$ and $\left(\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right)$, respectively. Then $f: A \mapsto \langle Av,v\rangle$ and $g: A \mapsto \langle Aw,w\rangle$ are pure states on $\mathcal{A}$. They are distinct, but $f(A) = g(A)$ for all $A \in \mathcal{B}$: we say that $\mathcal{B}$ fails to separate $f$ and $g$.

I would like to find a C*-algebra $\mathcal{A}'$ which (unitally) contains $\mathcal{A}$ together with a C*-subalgebra $\mathcal{B}'$ which contains $\mathcal{B}$, such that $\mathcal{B}'$ separates any extensions $f'$ and $g'$ of $f$ and $g$ to states on $\mathcal{A}'$. This could be trivially accomplished by setting $\mathcal{A}' = \mathcal{B}' = \mathcal{A}$, so let me add one more condition. Let $P = \left[\matrix{1&1\cr1&1}\right]$ and observe that the distance from $P$ to $\mathcal{B}$ is $1$. (Evaluation in the $(1,2)$ matrix entry is a norm one linear functional on $\mathcal{A}$ which takes the value $1$ on $P$ and is zero on $\mathcal{B}$. Conversely, the distance from $P$ to the identity matrix is $1$.) To prevent trivial solutions to my problem, I impose the additional requirement that the distance from $P$ to $\mathcal{B}'$ must still be $1$.

Can we find a C*-algebra $\mathcal{A}'$ which unitally contains $\mathcal{A}$ together with a C*-subalgebra $\mathcal{B}'$ which contains $\mathcal{B}$, such that (1) the distance from $P$ to $\mathcal{B}'$ is $1$ and (2) $\mathcal{B}'$ separates any two states on $\mathcal{A}'$ which extend $f$ and $g$, respectively?

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  • $\begingroup$ Would you be interested in a partial solution where $\mathcal B'$ only separated pure extensions of $f$ and $g$? $\endgroup$ – Chris Ramsey Aug 2 '15 at 17:50
  • $\begingroup$ @ChrisRamsey: Sure, although I ultimately will need the stronger result. $\endgroup$ – Nik Weaver Aug 2 '15 at 20:57
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The answer to my question is no. Let $\mathcal{A} = M_2$ be the algebra of $2\times 2$ complex matrices, let $\mathcal{B}$ be the subalgebra of diagonal matrices, let $\mathcal{A}'$ be a C*-algebra which unitally contains $\mathcal{A}$, and let $\mathcal{B}'$ be a C*-subalgebra of $\mathcal{A}'$ which contains $\mathcal{B}$.

$\mathcal{A}$ contains the matrix units $e_{11}$, $e_{12}$, $e_{21}$, $e_{22}$, and hence so does $\mathcal{A}'$. So we have $\mathcal{A}' \cong M_2\otimes \mathcal{C}$ for some other unital C*-algebra $\mathcal{C}$. Also, since $\mathcal{B}$ contains $e_{11}$ and $e_{22}$, we have $$\mathcal{B}' = e_{11}\mathcal{B}'e_{11} + e_{11}\mathcal{B}'e_{22} + e_{22}\mathcal{B}'e_{11} + e_{22}\mathcal{B}'e_{22} = \mathcal{B}'_{11} + \mathcal{B}'_{12} + \mathcal{B}'_{21} + \mathcal{B}'_{22}.$$ We have $P = e_{11} + e_{12} + e_{21} + e_{22}$. There are two cases.

Case 1. $\mathcal{B}'_{12}$ contains an element $x$ satisfying $\|x - e_{12}\| < 1$. Then $\mathcal{B}'$ contains the element $y = e_{11} + x + x^* + e_{22}$, and we have $$\|y - P\| = \|(x - e_{12}) + (x^* - e_{21})\| = \|x - e_{12}\| < 1.$$ So in this case the distance from $\mathcal{B}'$ to $P$ is strictly less than 1.

Case 2. We have $\|x - e_{12}\| \geq 1$ for all $x \in \mathcal{B}'_{12}$. By the Hahn-Banach theorem, we can find a norm one linear functional on $e_{11}\mathcal{A}'e_{22}$ which takes $e_{12}$ to $1$ and vanishes on $\mathcal{B}'_{12}$. By the isomorphism $e_{11}\mathcal{A}'e_{22} \cong \mathcal{C}$, this yields a state $h$ on $\mathcal{C}$. Then $f\otimes h$ and $g\otimes h$ are two states on $\mathcal{A}'$ which respectively extend $f$ and $g$, but $\mathcal{B}'$ does not separate them, because they agree on $\mathcal{B}'_{11} \subseteq e_{11}\mathcal{A}'e_{11}$ and $\mathcal{B}'_{22} \subseteq e_{22}\mathcal{A}'e_{22}$, and they both vanish on $\mathcal{B}'_{12}$ and $\mathcal{B}'_{21}$. So in this case $\mathcal{B}'$ fails to separate all extensions of $f$ and $g$.

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This is long. I originally thought this would work but it does not separate all extensions of $f$ and $g$ (see Nik's comment).

Consider the unital embedding $\mathcal A = M_2 \subset M_4$ given by $$A = \left[\begin{array}{cc}a&b\\c&d\end{array}\right] \mapsto A \otimes I_2 = \left[\begin{array}{cccc} a &&b\\&a&&b \\ c&&d\\&c&&d\end{array}\right]$$ and for $\gamma_n = -\sqrt \frac{n-1}{n} + \frac{i}{\sqrt{n}}\in\mathbb T$ let $$\mathcal B_n \equiv \left\{ \left[\begin{array}{cccc}a &&b \\ &a&&\gamma_nb\\c&&d\\&\bar{\gamma}_nc&&d\end{array}\right] : a,b,c,d\in \mathbb C\right\}$$ Then $\mathcal B$ is embedded in $\mathcal B_n$ which is a C$^*$-subalgebra of $M_4$. Finally, define $\mathcal B' \equiv \oplus_{n=1}^\infty \mathcal B_n \subset \oplus_{n=1}^\infty M_4 \equiv \mathcal A'$ with $A\in \mathcal A$ embedded as $\oplus_{n=1}^\infty (A\otimes I_2) \subset \mathcal A'$.

Claim 1: The distance from $P$ to $\mathcal B'$ is 1.

Let $\gamma_\infty = -1 = \lim_{n\rightarrow \infty} \gamma_n$ and define $\mathcal B_\infty$ in the same manner as above. What is the distance between $\mathcal B_\infty$ and $P\otimes I_2$? This is the same as calculating the minimum norm of $\left[\begin{array}{cc} 1+b&\\&1-b\end{array}\right], b\in\mathbb C$ which is 1 since $2 = |1 + b + 1 - b| \leq |1 + b| + |1 -b|$. Because of the definition of these matrix algebras it is straightforward to see that $\lim_{n\rightarrow \infty} d(\mathcal B_n, P\otimes I_2) = d(\mathcal B_\infty, P\otimes I_2) = 1$. Therefore, $$d(\mathcal B', \oplus_{n=1}^\infty (P\otimes I_2)) = 1$$

Claim 2: $\mathcal B'$ separates any two states on $\mathcal A'$ extending $f$ and $g$.

Let $v' = (v_1,v_2,v_3,v_4)\in \mathbb C^4$ such that $\langle (A\otimes I_2) v',v'\rangle = \langle Av,v\rangle$ for all $A\in M_2$. An easy calculation gives that $\|(v_1,v_2)\| = \frac{1}{\sqrt{2}}$ and $(v_1,v_2) = (v_3,v_4)$. The $w'$ case follows similarly except with $(w_1,w_2) = -(w_3,w_4)$.

Now for the general case, $\mathcal A'$ is naturally represented on the Hilbert space $\mathcal H = \oplus_{n=1}^\infty \mathbb C^4$. Let $v', w'\in \mathcal H$ be unit vectors such that $f'(C) = \langle Cv',v'\rangle$ and $g'(C) = \langle Cw',w'\rangle$ are states extending $f$ and $g$ respectively. From the previous paragraph they must have a particular form, namely $v' = (r_1,s_1,r_1,s_1, r_2,s_2,r_2,s_2,\cdots)$ and $w' = (t_1,u_1,-t_1,-u_1, t_2,u_2, -t_2,-u_2,\cdots)$.

Define $B_n = \left[\begin{array}{cccc}0&&1\\&0&&\gamma_n \\ \gamma_n&&0\\ &1&&0\end{array}\right] \in \mathcal B_n$ and let $B' = \oplus_{n=1}^\infty B_n \in \mathcal B'$. Now, $$Im\left\langle B_n\left(\begin{array}{c} r_n\\s_n\\r_n\\s_n\end{array}\right),\left(\begin{array}{c} r_n\\s_n\\r_n\\s_n\end{array}\right)\right\rangle = Im(1+\gamma_n)(\|r_n\|^2 + \|s_n\|^2) = \frac{1}{\sqrt{n}}(\|r_n\|^2 + \|s_n\|^2)$$ whereas $$Im\left\langle B_n\left(\begin{array}{c} t_n\\u_n\\-t_n\\-u_n\end{array}\right),\left(\begin{array}{c} t_n\\u_n\\-t_n\\-u_n\end{array}\right)\right\rangle = Im(1+\gamma_n)(-\|t_n\|^2 - \|u_n\|^2) = -\frac{1}{\sqrt{n}}(\|t_n\|^2 + \|u_n\|^2).$$ Hence, $$Im\langle B'v',v'\rangle = \sum_{n=1}^\infty \frac{1}{\sqrt{n}}(\|r_n\|^2+\|s_n\|^2) > 0$$ and $$Im\langle B'w',w'\rangle = -\sum_{n=1}^\infty \frac{1}{\sqrt{n}} (\|t_n\|^2 + \|u_n\|^2) < 0.$$ Therefore, $f'(B') \neq g'(B')$.

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    $\begingroup$ This is really clever, but it doesn't quite work --- your characterization of states on $\mathcal{A}'$ is faulty. There are states which aren't vector states. Let $\phi_n$ and $\psi_n$ be the states corresponding to the sequences $(r_1, s_1, \ldots)$ and $(t_1, u_1, \ldots)$ where $r_n = u_n = 1$ and all other terms are zero. Then you see that $\mathcal{B}'$ just barely separates them, so that if $\phi$ and $\psi$ are respectively cluster points of the sequences $(\phi_n)$ and $(\psi_n)$ then $\mathcal{B}'$ doesn't separate $\phi$ and $\psi$. $\endgroup$ – Nik Weaver Aug 2 '15 at 2:57
  • $\begingroup$ Oh dear, you are absolutely correct. $\endgroup$ – Chris Ramsey Aug 2 '15 at 10:52
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I showed in my other answer that there are no $\mathcal{A}'$ and $\mathcal{B}'$ such that ${\rm dist}(\mathcal{B}',P) = 1$ and $\mathcal{B}'$ separates any states on $\mathcal{A}'$ which extend $f$ and $g$. In a comment Chris Ramsey asked about a weaker result where $\mathcal{B}'$ separates all pure extensions of $f$ and $g$. This can be done. The construction is similar to the one given by Chris in his answer.

Let $\mathcal{A}' = M_2 \oplus M_2 \oplus M_2$, with the embedded copy of $\mathcal{A}$ given by $\{A \oplus A \oplus A: A \in M_2\}$. Thus $P$ embeds as the element $\left[\begin{matrix}1&1\cr 1&1\end{matrix}\right] \oplus \left[\begin{matrix}1&1\cr 1&1\end{matrix}\right] \oplus \left[\begin{matrix}1&1\cr 1&1\end{matrix}\right]$. Let $\mathcal{B}' = \left\{\left[\begin{matrix}a&b\cr c&d\end{matrix}\right] \oplus \left[\begin{matrix}a&b\omega\cr c\omega^2&d\end{matrix}\right] \oplus \left[\begin{matrix}a&b\omega^2\cr c\omega&d\end{matrix}\right]: a,b,c,d \in \mathbb{C}\right\}$ where $\omega$ is a primitive cube root of unity. This is just the embedded copy of $\mathcal{A}$ conjugated by a unitary, so it is a C*-subalgebra of $\mathcal{A}'$. Setting $b = c = 0$ shows that it contains the embedded copy of $\mathcal{B}$.

(1) ${\rm dist}(\mathcal{B}',P) = 1$. This is because any element of $\mathcal{B}'$ has upper right entries $b$, $b\omega$, $b\omega^2$, at least one of which must lie in the left half-plane of $\mathbb{C}$. Thus any element of $\mathcal{B}'$ minus the embedded $P$ has a matrix entry which is at least $1$ in absolute value.

(2) $\mathcal{B}'$ separates the pure extensions of $f$ from the pure extensions of $g$. Any pure state on $M_2 \oplus M_2 \oplus M_2$ must be a pure state on one of the summands and zero on the other two summands. So the pure extensions of $f$ are $f \oplus 0 \oplus 0$, $0 \oplus f \oplus 0$, and $0 \oplus 0\oplus f$, and similarly for $g$. Apply these extensions to the element $\left[\begin{matrix}0&1\cr 0&0\end{matrix}\right] \oplus \left[\begin{matrix}0&\omega\cr 0&0\end{matrix}\right] \oplus \left[\begin{matrix}0&\omega^2\cr 0&0\end{matrix}\right] \in \mathcal{B}'$. The pure extensions of $f$ take the values $\frac{1}{2}$, $\frac{\omega}{2}$, $\frac{\omega^2}{2}$ and the pure extensions of $g$ take the values $-\frac{1}{2}$, $-\frac{\omega}{2}$, $-\frac{\omega^2}{2}$, and the two sets do not overlap. So any pure extension of $f$ and any pure extension of $g$ applied to this element yield different results.

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