# Tensor powers at the service of humanity

There is a standard convention in mathematics to denote the set of functions from set $X$ to set $Y$ by $Y^X$. This looks a lot like $\mathbb{R}^n$, with the only difference that $n$ is a natural number whereas $X$ is a set. The motivation of this post is to treat $n$ as a set to uncover some parallels between functions and vector spaces.

## $\mathbb{R}^n$ as a set of functions

Let $[n]$ denote a set of $n$ elements. One can think of it as the set $\{ 1, 2, \dots , n \}$. Be careful though, because $[n]$ is not a subset of $\mathbb{N}$—its elements are not natural numbers but some arbitrary objects, which we simply enumerated for convenience. For example, $ \cap  = \emptyset$ and $ \cup  = $.

The key observation allowing us to unify notation is that we can view the vector space $\mathbb{R}^n$ as the set of functions from $[n]$ to $\mathbb{R}$,

Indeed, if $x \in \mathbb{R}^n$, you can query $x$ for its $k$-th coordinate, $x(k)$. That means $x$ can be viewed as a function $x : [n] \rightarrow \mathbb{R}$. The set of all such functions $\mathbb{R}^{[n]}$ is a vector space isomorphic to $\mathbb{R}^n$.

Let’s see this idea in action on several examples.

## Direct product

Consider the direct product of vector spaces $\mathbb{R}^{} \times \mathbb{R}^{}$. It is easy to see that

The exponential notation is very suggestive, as you may have noticed. It allows one to manipulate vector spaces using intuition from natural numbers.

## Tensor product

Consider the tensor product of vector spaces $\mathbb{R}^{} \otimes \mathbb{R}^{}$. It is easy to see that

In case of the tensor product, dimensionalities multiply. Interesting! We know that exponents multiply when one takes a power of a power (i.e., $(a^b)^c = a^{bc}$). Can we understand the tensor product in this way? It turns out we can. Check it out yourself.

$$\begin{equation*} \mathbb{R}^{} \otimes \mathbb{R}^{} \cong \left( \mathbb{R}^{} \right)^{} = \left\{  \rightarrow \left\{  \rightarrow \mathbb{R} \right\} \right\}. \end{equation*}$$

Note that parentheses are important, because $(2^3)^4 = 2^{12}$ whereas $2^{(3^4)} = 2^{81}$.