Let $M$ and $N$ be two von Neumann algebras. A natural way to define $M\overline{\otimes}N$ is the von Neumann algebra generated by the algebraic tensor product $M\otimes N$ in $B(H\otimes K)$ where $M\subseteq B(H)$ and $N\subseteq B(K).$ Is this tensor product same as the minimal $C^*$-algebraic tensor product of $M$ and $N$? If so how to prove this? What is the definition of minimal infinite $C^*$-algebraic tensor product of von Neumann algebras $(M_n,\tau_n)_{n\geq 1},$ where $\tau_n$'s are normal faithful tracial state?
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$\begingroup$ So, your question is whether $M \bar\otimes N$ is equal to $M \otimes_\min N$? That is not true in general. $\endgroup$– Adrián González PérezMar 21, 2018 at 16:02
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1$\begingroup$ For a counterexample you can use $M = N = \ell^\infty(\mathbb{N})$. You have that $M \bar\otimes N = \ell^\infty(\mathbb{N} \times \mathbb{N}) = C(\beta (\mathbb{N} \times \mathbb{N}))$ and $M \otimes_\min N = C(\beta \mathbb{N} \times \beta \mathbb{N})$, where $\beta X$ is the Stone-Cech compactification of $X$, but $\beta (\mathbb{N} \times \mathbb{N}) \neq \beta \mathbb{N} \times \beta \mathbb{N})$. $\endgroup$– Adrián González PérezMar 21, 2018 at 16:12
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$\begingroup$ Mateusz has already explained the details, but the following "big picture" comment might be helpful. The simplest way to understand why the two tensor products might be different: von Neumann algebras are SOT closures, hence they are at least as big as the norm closures and should usually be strictly bigger. So the vN tensor product, being by definition a SOT closure, should be bigger than the norm closure which is the Cstar minimal tensor product. $\endgroup$– Yemon ChoiMar 21, 2018 at 19:02
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As Adrián remarked in the comments, the answer is negative. Let me give a somewhat inexplicit explanation of this. It actually fails even in the abelian case. Indeed, let $M=N = L^{\infty}[0,1]$. Note that we have a multiplication map $m: L^{\infty}[0,1]\otimes L^{\infty}[0,1] \to L^{\infty}[0,1]$ (on the algebraic tensor product) given by $f\otimes g \mapsto fg$. We can view each element of $L^{\infty}[0,1]\otimes L^{\infty}[0,1]$ as an element of the von Neumann algebraic tensor product $L^{\infty}[0,1]\overline{\otimes} L^{\infty}[0,1]\simeq L^{\infty}[0,1]^2$, i.e. a function of two variables. The multiplication map in this setting looks like this: we have a function $\sum_{i=1}^{k} f_i(x)g_{i}(y)$ and map it to the function $\sum_{i=1}^{k} f_i(x)g_i(x)$. So, if $F\in L^{\infty}[0,1]\otimes L^{\infty}[0,1]$, then $m(F)(x) = F(x,x)$. The multiplication map extends to the minimal tensor product -- this can be checked by hand.
On the other hand, if the minimal and von Neumann algebraic tensor products agreed in this case, we would have a multiplication map $m: L^{\infty}[0,1]^2\to L^{\infty}[0,1]$, which takes acts as restriction to the diagonal. Since the diagonal is of measure zero, there is no way this map can be well defined.
As for the definition of the infinite tensor product, you do the following. Use the trace $\tau_n$ to obtain a GNS representation of $M_n$ on $L^{2}(M_n)$ equipped with a cyclic vector $\Omega_n$. Now form the the infinite tensor product of Hilbert spaces $L^{2}(M_n)$ with respect to the stabilising sequence $(\Omega_n)$. Now the infinite (algebraic) tensor product acts on the resulting Hilbert space, so we may close it to obtain a von Neumann algebra.
answered Mar 21, 2018 at 16:18
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$\begingroup$ .I am really a beginner. So let me tell you what I understand by von Neumann algebra tensor product of two von Neumann algebras. So there can be two definitions it seems. 1.If $M_i\subseteq B(H_i)$ for $i=1,2.$ Then consider the von Neumann algebra generate by the algebraic tensor products $M_1\otimes M_2$ in $B(H_1\otimes H_2).$ Call it $M_1\overline{\otimes}M_2.$ Define a linear functional on $M_1\otimes M_2$ as $\tau(x_1\otimes x_2):=\tau(x_1)\tau(x_2).$ This will become a trace on $M_1\overline{\otimes }M_2$? (I am not sure) $\endgroup$– MathbuffMar 21, 2018 at 17:16
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$\begingroup$ @ Wasilewski.The second definition which is minimal $C^*$-algebraic tensor product is following. By GNS representation we have $\pi_1:M_1\to B(L^2(M_1))$ as $\pi_1(x)\Lambda(y)=\Lambda_1(xy).$ $\Lambda_1:M_1\to L^2(M_1)$ is the natural inclusion map and $L_2(M_1)$ is the completion of $M_1$ w.r.t. inner product $<x,y>-\tau_1(xy^*).$ So define $\pi:M_1\otimes M_2\to B(L^2(M_1)\otimes L^2(M_2))$ as $m\otimes n\to \pi_1(m)\otimes \pi_2(n)$ and $M_1\otimes _{min}M_2$ is the von Neumann algebra generated by $\pi(M_1\otimes M_2)$ in $B(L^2(M_1)\otimes L^2(M_2))$. IS it okay? $\endgroup$– MathbuffMar 21, 2018 at 17:25
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1$\begingroup$ What people usually mean by minimal tensor product is the $C^{\ast}$-algebra generated by the algebraic tensor product. Note that both the von Neumann algebraic and the minimal tensor products do not depend on the representations $M_i\subset B(H_i)$; any two would yield the same tensor product. $\endgroup$ Mar 21, 2018 at 17:37
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1$\begingroup$ Yes, that's what I'm saying. I don't understand your last sentence though. Why should the $C^{\ast}$-algebraic tensor product contain the von Neumann algebra generated by $M_1$ and $M_2$? It's the other way round. $\endgroup$ Mar 21, 2018 at 18:13
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1$\begingroup$ The $C^{\ast}$-algebra generated by $M_1$ and $M_2$ is usually not a von Neumann algebra. The next sentence is correct, and this is exactly what I said in the previous comment; the von Neumann algebra is bigger. $\endgroup$ Mar 21, 2018 at 19:56