Hi Math StackExchange,

Let $A$ be a commutative, infinite dimensional, unital, *-algebra represented by bounded operators on a Hilbert space $H_A$. Next let $B$ be a finite non-commutative *-algebra represented on a Hilbert space $H_B$.

Is it true in general that the product $A\otimes B$ is represented by $\textbf{bounded}$ operators on $H_A\otimes H_B$? How could I show this?


Remark: this has nothing to do with particular $\ast$-subalgebras. Given $S$ bounded on $H_A$ and $T$ bounded on $H_B$, there is a canonical way to define $S\otimes T$ bounded on $H_A\otimes H_B$ with, moreover, $\|S\otimes T\|= \|S\|\,\|T\|$. So if your representations are denoted by $\pi_A:A\longrightarrow B(H_A)$ and $\pi_B:B\longrightarrow B(H_B)$, we obtain in a canonical way a representation $\pi_A\otimes \pi_B:A\odot B\longrightarrow B(H_A\otimes H_B)$ of the algebraic tensor product $A\odot B$ such that $\|(\pi_A\otimes \pi_B)(a\otimes b)=\|\pi_A(a)\otimes \pi_B(b)\|=\|\pi_A(a)\|\| \pi_B(b)\|$.

Let us denote $H_A\odot H_B$ the algebraic tensor product, whose completion with respect to the inner product defined by $(x\otimes y,x'\otimes y'):=(x,x')(y,y')$ is $H_A\otimes H_B$. In particular, note that $\|x\otimes y\|=\|x\|\,\|y\|$.

If $S\in B(H_A)$ and $T\in B(H_B)$, then the bilinear map $(x,y)\longrightarrow Sx\otimes Ty$ factors to a linear map from $H_A\odot H_B$ to $H_A\otimes H_B$ by the universal property of the algebraic tensor product. It is denoted by $S\otimes T$ and it satisfies $$ (S\otimes T)(x\otimes y)=Sx \otimes Ty\qquad\forall x\in H_A, \forall y\in H_B. $$

Claim: the operator $S\otimes T: H_A\odot H_B\longrightarrow H_A\otimes H_B$ has norm $\|S\otimes T\|=\|S \|\,\|T\|$. Hence it extends uniquely to a bounded operator $S\otimes T\in B(H_A\otimes H_B)$ with the same norm.

The second part is immediate and is the linear case of the unique extension of uniformly continuous functions from dense sets. So we need only check the first part. Since the unit ball $\{z\in H_A\odot H_B\,;\,\|z\|\leq 1\}$ contains $\{x\otimes y\,;\, \|x\|=\|y\|=1\}$, we have $$ \|S\otimes T\|=\sup_{\|z\|\leq 1}\|(S\otimes T)z\|\geq \sup_{\|x\|=\|y\|=1}\|(S\otimes T)(x\otimes y)\|=\sup_{\|x\|=1}\|Sx\|\sup_{\|y\|=1}\|Ty\|=\|S\|\,\|T\|. $$ The other direction is more delicate. The trick is to observe that $$ S\otimes T=(S\otimes I)(I\otimes T). $$ So it only remains to show that $\|S\otimes I\|\leq \|S\|$. By symmetry, this will also show that $\|I\otimes T\|\leq \|T \|$ whence $ \|S\otimes T\|\leq \|S\otimes I\|\,\|I\otimes T\|\leq \|S\|\,\|T\|$ as desired.

Take an arbitrary vector $z=\sum_{j=1}^n x_i\otimes y_j$ in $H_A\odot H_B$. By bilinearity, we can replace the $y_j$'s by an orthonormal basis of their span. So we can assume $(y_j)$ is orthonormal, whence $(x_j\otimes y_j)$ and $(Sx_j\otimes y_j)$ are orthonormal as well. Then $$ \|(S\otimes I)z\|^2=\sum_{j=1}^n \|Sx_j\otimes y_j\|^2=\sum_{j=1}^n \|Sx_j\|^2\|y_j\|^2\leq \|S\|^2\sum_{j=1}^n \|x_j\|^2\|y_j\|^2 $$ $$ =\|S\|^2\sum_{j=1}^n \|x_j\otimes y_j\|^2=\|S\|^2\|z\|^2. $$ This concludes the proof of the claim.

| cite | improve this answer | |

Given two C$^*$-algebras $A$ and $B$ represented on Hilbert spaces $H_A$ and $H_B$, we can always represent their (algebraic) tensor product $A \otimes B$ by bounded operators on $H_A \otimes H_B$. We can do this whether they are commutative or not, infinite-dimensional or not.

Let's start by showing that $a \otimes id_{H_B}$ is a bounded operator on $H_A \otimes H_B$ for $a \in A$. The same proof will show that $id_{H_A} \otimes b $ is a bounded operator on $H_A \otimes H_B$ for $b \in B$. We will then have that $$ \Vert a \otimes b \Vert \leq \Vert a \otimes id_{H_B} \Vert \cdot \Vert id_{H_B} \otimes b \Vert = \Vert a \Vert \cdot \Vert b \Vert, $$ for all $a \in A$ and $b \in B$. By the triangle inequality, it will then follow that every operator in the algebraic tensor product can be represented by bounded operators on $H_A \otimes H_B$.

Suppose that $\xi \in H_A \otimes H_B$; we can write it as $\sum_i x_i \otimes y_i$, where the vectors $y_i \in H_B$ are mutually orthogonal and of norm 1, i.e., $(y_i, y_j)=\delta_{ij}$ and $\Vert y_i \Vert_2 =1$.

Let's apply $a \otimes id_{H_A}$ to $\xi$. Then $$ \Vert (a \otimes id_{H_A})(\sum_i x_i \otimes y_i) \Vert_2= \Vert \sum_i a x_i \otimes y_i \Vert_2, $$ and, by the definition of the norm on $H_A \otimes H_B$, this is equal to $$ \Vert \sum_i a x_i \Vert_2. $$ By the definition of the operator norm, we have $$ \Vert \sum_i a x_i \Vert_2 \leq \Vert a \Vert \cdot \Vert \sum_i x_i \Vert_2. $$ We have shown that $$ \Vert a \otimes id_{H_B} \Vert \leq \Vert a \Vert. $$ (In fact, it is easy to show that they are equal.) So by the remarks above, we are done.

We have defined a norm on $A \otimes B$, the norm that they have as operators on $H_A \otimes H_B$. This norm is called the spatial tensor product norm or the minimal (C$^*$-algebraic) tensor product. If you search for these terms, you will be able to find a great deal of information about this norm.

By the way, at first glance, it may look like I am assuming that $id_{H_A} \in A$ or $id_{H_B} \in B$; I am not assuming that. I am only using that $a \otimes id_{H_B}$ and $id_{H_A} \otimes B$ make sense as bounded linear operators on $H_A \otimes H_B$.

The above discusses the norm obtained by letting $A \subseteq B(H_A)$ and $B \subseteq B(H_B)$ act in the "natural" way on $H_A \otimes H_B$. (Here $B(H)$ means the bounded linear operators on the Hilbert space $H$.) Given two C$^*$-algebras of bounded operators on a Hilbert space, there are, in general, many different C$^*$-norms that can be placed on $A \otimes B$ and many different ways of representing them as bounded operators on some Hilbert space $K$. However, if $A$ is a commutative C$^*$-algebra or if $B$ is a finite dimensional C$^*$-algebra, then there is only C$^*$-norm that can be placed on $A\otimes B$, the spatial tensor product norm described above. For more on this, search for information about nuclear C$^*$-algebras.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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