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?
 A: 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 $C(X\times X)$.
