# On diagonal part of tensor product of $C^*$-algebras

Suppose we have a $$C^*$$-algebra $$\mathcal{U}$$, Consider the $$C^*$$-subalgebra generated by elements of the form $$a\otimes a$$, what is it isomorphic to? Is it isomorphic to $$\mathcal{U}$$ itself?

• I don't think you necessarily get anything nice in general, probably since the map $a \mapsto a\otimes a$ isn't linear. However, you will in certain special cases get something nice if you look at particular "diagonal" type $C^\ast$-algebras. For instance, if $G$ is a discrete group, let $D$ be the $C^\ast$-subalgebra of $C^\ast_r(G) \otimes_{\max{}} C^\ast_r(G)$ generated by the elements $u_g \otimes u_g$ for $g\in G$. An easy consequence of Fell's absorption is that $D\cong C^\ast(G)$. – Jamie Gabe Mar 11 at 12:08
• What topology do you put on your tensor product $C^*$ algebra ? – InfiniteLooper Mar 11 at 12:29
• (Nota also that the name diagonal would have been better suited for maps of shape $a \otimes b \mapsto ab$) – InfiniteLooper Mar 11 at 12:38
• @Bleuderk minimal tensor product say – user136400 Mar 11 at 13:15

## 1 Answer

You get the symmetric part of the tensor product.

The map $$\phi: a\otimes b \mapsto b\otimes a$$ extends to an order 2 automorphism of $$\mathcal{U}\otimes\mathcal{U}$$. The set of fixed points for this $$\mathbb{Z}/2$$ action is a C*-subalgebra $$(\mathcal{U}\otimes\mathcal{U})_s$$ of $$\mathcal{U}\otimes\mathcal{U}$$. Every $$x \in \mathcal{U}\otimes\mathcal{U}$$ can be written as $$x = \frac{1}{2}(x + \phi(x)) + \frac{1}{2}(x - \phi(x))$$, expressing it as $$y + z$$ where $$\phi(y) = y$$ and $$\phi(z) = -z$$, i.e., every element is the sum of a symmetric part and an antisymmetric part.

The C*-subalgebra generated by $$\{a\otimes a: a \in \mathcal{U}\}$$ is clearly contained in $$(\mathcal{U}\otimes\mathcal{U})_s$$, and conversely, it contains, for any $$a,b \in \mathcal{U}$$, the element $$a\otimes b + b \otimes a = (a+b)\otimes (a+b) - a\otimes a - b\otimes b$$. From here one can show inductively that this C*-algebra contains everything in the algebraic tensor product that is fixed by $$\phi$$. Then anything in the tensor product is the limit of a sequence of elements of the algebraic tensor product, and taking symmetric and antisymmetric parts as above shows that any symmetric element is approximated by symmetric elements of the algebraic tensor product. Thus you get all of $$(\mathcal{U}\otimes\mathcal{U})_s$$.

For instance, if $$\mathcal{U} = C(X)$$ then you get the symmetric continuous functions in $$C(X\times X)$$.

• Does $(\mathcal{U}\otimes \mathcal{U})_{s}$ has faithful state? – user136400 Mar 12 at 6:02
• Well, every separable C*-algebra has a faithful state. If $\mathcal{U}$ doesn't have a faithful state then $(\mathcal{U}\otimes\mathcal{U})_s$ won't either. – Nik Weaver Mar 12 at 10:39
• I mean product state restricted on $({\mathcal{U}\otimes\mathcal{U}})_{s}$ – user136400 Mar 12 at 13:15
• The question I am interested to ask you that if I start with a faithful state on $\mathcal{U}$, does the product state is also faithful $(\mathcal{U}\otimes \mathcal{U})_{s}$? If you know any article regarding these please refer!! It is my urgent request. – user136400 Mar 14 at 8:18
• If $\phi$ is faithful on $\mathcal{U}$ then $\phi\otimes\phi$ is faithful on the full spatial tensor product $\mathcal{U}\otimes\mathcal{U}$, and hence also on its symmetric part. That's very easy if you just work in the GNS representation. – Nik Weaver Mar 14 at 10:31