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Let $A$ and $B$ two $C^*$-algebras, $H_1$ and $H_2$ complex Hilbert spaces and $\pi_1:A\to B(H_1)$, $\pi_2:B\to B(H_2)$ two $*$-representations. Then there is a $*$-representation $\pi_1\otimes \pi_2:A\odot B\to B(H_1\otimes H_2)$, defined as follows:

1.We have $\pi_1\odot \pi_2:A\odot B\to B(H_1)\odot B(H_2)$, $a\otimes b\mapsto \pi_1(a)\otimes \pi_2(b).$ ($\odot$ denotes the tensor product as $*$-algebras)

2.We have the canonical embedding $$\iota: B(H_1) \odot B(H_2)\to B(H_1 \otimes H_2),$$ where $T\in B(H_1)$ and $S\in B(H_2)$ will be mapped to $T\otimes S\in B(H_1\otimes H_2)$ with $T\otimes S(v\otimes w)=T(s)\otimes S(w)$ ($H_1 \otimes H_2$ is the tensor product as a Hilbert space).

Finally, we set $\pi_1\otimes \pi_2:=\iota \circ \pi_1\odot \pi_2$.

Conversely, I'm searching for an example which demonstrates that if you have a $*$-representation $\pi:A\odot B\to B(H_1\otimes H_2)$, then $\pi$ is does not have to be in the form of $\pi_1\otimes \pi_2$, where $\pi_1:A\to B(H_1)$, $\pi_2:B\to B(H_2)$ are two $*$-representations.

I.e. the claim is: there are $*$-representations $\pi:A\odot B\to B(H_1\otimes H_2)$ on the $*$-algebraic tensor product of $A$ and $B$, which are not induced by $*$-representations of $A$ and $B$.

One suggestion is to choose the flip: $A=B=B(\ell^2)$, $H_1=H_2=\ell^2$, and $$\pi:B(\ell^2)\odot B(\ell^2)\to B(\ell^2\otimes \ell^2),$$ $$T\otimes S\mapsto S\otimes T.$$ But I'm stuck to give a formal proof that it is such an example (although I think it can't be too difficult...). Do you have an idea how to do it?

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3 Answers 3

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Suppose for simplicity that $A$ and $B$ are unital. If now $\pi = \pi_1 \otimes \pi_2$ then the operators $\pi(a \otimes 1)$ commute with all the operators of the form $id \otimes B$ with $B \in B(H_2)$. Fix a unit vector $f \in H_2$ with ONprojection $Q$ onto it. Then for $\psi \in H_1$ we have $\pi(A \otimes 1) \psi \otimes f = \pi_1(A)\psi \otimes f$. This allows to reconstruct $\pi_1(A)$ from $\pi$. Exchanging the roles of $1$ and $2$, we can reconstruct $\pi_2$.

Taking this as idea, and given an arbitrary rep $\pi$ with the property that $\pi(a \otimes 1)$ commutes with $id \otimes B(H_2)$ and $\pi(1 \otimes b)$ commutes with $B(H_1) \otimes id$, we take two rank one projection $P \in B(H_1)$ and $Q \in B(H_2)$ and consider $\pi_1$ and $\pi_2$ defined by $\pi(a \otimes 1) Q = \pi_1(a) \otimes Q$ and $\pi(1 \otimes b) P = 1 \otimes \pi_2(b) P$. (this is well-def!) This should give you the two reps (here you need the fact that things commute) you are looking for, proving that the above condition is sufficient and necessary.

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    $\begingroup$ Dear Dr Waldmann. Could you kindly provide some clarification? Do you mean $$ \pi(a \otimes 1_{B}) \circ ({\operatorname{Id}_{\mathcal{H}_{1}}} \otimes Q) = {\pi_{1}}(a) \otimes Q $$ and $$ \pi(1_{A} \otimes b) \circ (P \otimes \operatorname{Id}_{\mathcal{H}_{2}}) = P \otimes {\pi_{2}}(b) $$ in the second paragraph? I don’t see how one can compose $ \pi(a \otimes 1_{B}) $ with $ Q $ alone, or $ \pi(1_{A} \otimes b) $ with $ P $ alone. Finally, are you saying that these formulas define $ \pi_{1} $ and $ \pi_{2} $ in a consistent manner? Thank you! $\endgroup$
    – Transcendental
    Commented Sep 5, 2016 at 3:27
  • $\begingroup$ @Transcendental I was a bit sloppy, the composition is meant as you say. If the projections $P$ and $Q$ are rank one, then the $\pi_1$ and $\pi_2$ are well-def, ie. consistently defined. $\endgroup$
    – Stefan Waldmann
    Commented Sep 5, 2016 at 14:16
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Let $A=B=c_{\circ} \subseteq B(\ell^2)$. Let $\sigma:A \otimes B \to B(\ell^2 \otimes \ell^2)$ such that $\sigma(f \otimes g)=fg \otimes 1$. It is not hard to show that $\sigma$ is a representation.

On contrary we assume that, there are representations $\pi, \rho: c_{\circ} \to B(\ell^2)$ such that $\sigma=\pi \otimes \rho$. Thus for each $g \in c_{\circ}$ we have
$$g \otimes 1=\sigma(1 \otimes g)=\pi(1) \otimes \rho(g)=1 \otimes \rho(g)$$ which is contradiction.

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  • $\begingroup$ The same argument should work for $l^{\infty}$ instead of $c_0$, I guess (since $c_0$ is non-unital). $\endgroup$
    – user62639
    Commented Sep 20, 2016 at 7:38
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The following is a bit more elementary (and purely algebraic) in nature than the previous examples. Take $A=B=\mathbb{C}^2$. Identify $A\otimes B\cong\mathbb{C}^4$. Observe that for trivial linear algebra reasons, any tensor product representation of $A\otimes B$ generates a C*-algebra in its range that has dimension 0, 1, 2, or 4. Take $H_1=\mathbb{C}^2=H_2$ and consider any representation that generates a $3$-dimensional C*-algebra, such as $$ \pi: A\otimes B\cong\mathbb{C}^4\to B(H_1\otimes H_2)\cong M_2\otimes M_2 \cong M_4, \pi(\lambda_1,\lambda_2,\lambda_3,\lambda_4) = \operatorname{diag}(\lambda_1,\lambda_1,\lambda_3,\lambda_4)\otimes 1. $$ Alternatively, $H_1=\mathbb{C}^2, H_2=\mathbb{C}$ also does the trick via $$ \pi: \mathbb{C}^2\otimes\mathbb{C}^2\to M_2, \pi(e_i\otimes e_j) = \operatorname{diag}(\delta_{1,i},\delta_{2,j}). $$ Although the image here has dimension 2, a small calculation shows that this cannot be of product form, either.

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  • $\begingroup$ Hi Gabor. There is a typo. It is $B(H_1\otimes H_2)\cong M_4$, not $M_4\otimes M_4$ $\endgroup$
    – user62639
    Commented Oct 15, 2016 at 15:36

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