📂Representation Theory

Tensor Product of Vector Spaces Defined by the Universal Property

Definition¹

Let $U$ and $V$ be finite-dimensional vector spaces. A tensor product of $U$ and $V$ is a vector space $W$ together with a bilinear map $\phi : U \times V \to W$, that is, an ordered pair $(W, \phi)$ satisfying the following property.

• For any vector space $X$ and any bilinear map $\psi : U \times V \to X$, there exists a unique linear map $\tilde{\psi} : W \to X$ satisfying the following.

  $$ \tilde{\psi} \circ \phi = \psi $$

  $$ \begin{matrix} U \times V & \overset{\phi}{\longrightarrow} & W \\ \psi \searrow & & \swarrow \tilde{\psi} \\ & X & \end{matrix} $$

Put simply, $(W, \phi)$ is a tensor product of $U$ and $V$ if, for the fixed $\phi$, whatever $(X, \psi)$ is given, there exists exactly one linear map $\tilde{\psi}$ such that $\tilde{\psi} \circ \phi = \psi$. Usually, the term tensor product refers to the vector space $W$ satisfying the above definition, and (since it is in fact unique) it is denoted as follows.

$$ W = U \otimes V $$

Explanation

The above definition does not refer to one particular space; rather, it defines the rule by which a given pair $(W, \phi)$ becomes a tensor product. A condition of this kind, which describes the object to be defined not through its concrete structure but solely through its relations with maps, is called a universal property.

Suppose we want to "multiply" vectors of two vector spaces $U$ and $V$ to produce a new vector. For it to deserve the name of a product, at the very least the distributive law must hold and scalars must be able to come out of the product. This means that, with one variable held fixed, it must be linear in the other variable, and such a map is precisely a bilinear map. One can construct as many such maps $\psi$ as one likes, but if every $\psi$ can be decomposed in the form $\psi = \tilde{\psi} \circ \phi$ with respect to a fixed $\phi$, then $\phi$ could be called something fundamental among the bilinear maps defined on $U \times V$. The theorem below guarantees the existence and uniqueness of such a tensor product $\phi$.

Constructive Definition

The tensor product can be defined as an object satisfying the universal property as above, or it can be defined by presenting a concrete structure. That structure is the function space constructed from the Cartesian product of finite sets $\Gamma_{1}$ and $\Gamma_{2}$, and it is also called a tensor product.

$$ \mathbb{C}^{\Gamma_{1}} \otimes \mathbb{C}^{\Gamma_{2}} := \mathbb{C}^{\Gamma_{1} \times \Gamma_{2}} $$

Theorem

Let $U$ and $V$ be finite-dimensional vector spaces.

(a) Existence: A tensor product $(W, \phi)$ of $U$ and $V$ exists.

(b) Uniqueness: The tensor product of $U$ and $V$ is unique. That is, if $(W_{1}, \phi_{1})$ and $(W_{2}, \phi_{2})$ are both tensor products of $U$ and $V$, then there exists a unique isomorphism $\Phi : W_{1} \to W_{2}$ satisfying the following.

$$ \Phi \circ \phi_{1} = \phi_{2} $$

(c) Basis and Dimension: If $(W, \phi)$ is a tensor product of $U$ and $V$, $\left\{ e_{1}, \dots, e_{n} \right\}$ is a basis of $U$, and $\left\{ f_{1}, \dots, f_{m} \right\}$ is a basis of $V$, then the following set is a basis of $W$.

$$ \left\{ \phi(e_{j}, f_{k}) : 1 \le j \le n, 1 \le k \le m \right\} $$

Therefore, the dimension of the tensor product is as follows.

$$ \dim (U \otimes V) = \dim(U) \times \dim(V) $$

$\phi (e_{j}, f_{k})$ is commonly denoted by $e_{j} \otimes f_{k}$.