Why does the definition of a unitary t-design use tensor products?

Question

A t-design $X$ is defined by the following equation

$$\frac{1}{|X|} \sum_{U \in X} U^{\otimes t} \otimes\left(U^*\right)^{\otimes t}=\int_{U(d)} U^{\otimes t} \otimes\left(U^*\right)^{\otimes t} d U$$

where the right hand side uses the Haar measure for $dU$. Why does this definition make sense compared to simply replacing $U^{\otimes t}$ with $U^t$. I am trying to understand why we invoke the action of $U$ on $t$ copies of the space instead of the action of $U^t$ on one copy of the space. What is the motivation behind that?

Answer 1

It's one of the equivalent definitions. A more common way to define it, that matches the spherical and complex projective design definitions, is by introducing the equation $$ \frac{1}{|X|}\sum_{U\in X} p(U) = \int_{\mathbb{U}(d)} p(U) {\rm d}\mu(U), $$ for any homogeneous polynomial $p$ of degree $(t,t)$, which means that any monomial in $p$ has exactly $t$ ordinary and $t$ conjugate variables (whose values are elements of $U$).

From this you can derive $$ \frac{1}{|X|}\sum_{U\in X} (U^{\otimes t})M(U^{\otimes t})^\dagger = \int_{\mathbb{U}(d)} (U^{\otimes t})M(U^{\otimes t})^\dagger {\rm d}\mu(U) $$ for any matrix $M$ acting on the tensor product space $H^{\otimes t}$, and then the definition in question.