## Infinite Tensor Product

**Definition**. Suppose $(V_i)_{i\in I}$ is a family of linear spaces. The direct sum of $(V_i)_{i\in I}$ is defined as $\oplus_{i\in I}V_i=\{(v_i)_{i\in I}|\{i|v_i\neq 0\}\mbox{ is a finite set }\}$, and the tensor product of $(V_i)_{i\in I}$ is defined as a linear space $\otimes_{i\in I} V_i$ with a multilinear operator $\otimes:\oplus_{i\in I}V_i\to \otimes_{i\in I} V_i$ such that for any given linear space $W$ and a multilinear operator $f:\oplus_{i\in I}V_i\to W$, there is a unique linear operator $\tilde f:\otimes_{i\in I} V_i\to W$ such that the following diagram is commutative:
\begin{xy}
\xymatrix{
\oplus_{i\in I}V_i\ar[r]^{\otimes} \ar[dr]_f& \otimes_{i\in I} V_i\ar[d]^{\tilde f}\\
& W
}
\end{xy}
**Existence**. Let $L(\oplus_{i\in I} V_i)$ denotes the set of the multilinear operators on $\oplus_{i\in I} V_i$. Define $\otimes_{i\in I} v_i\in L(L(\oplus_{i\in I} v_i))$ by $$\otimes_{i\in I} v_i(f)=f(\oplus_{i\in I} v_i),$$ and define $\otimes_{i\in I} V_i$ to be the subspace generated by $\{\otimes_{i\in I} v_i|v_i\in V_i\forall i\in I\}$ in $L(L(\oplus_{i\in I} V_i))$.
For any given multilinear operator $f:\oplus_{i\in I}V_i\to W$, in order to make the diagram commutative $\tilde f$ must be defined by $$\tilde f(\otimes_{i\in I}v_i)=f(\oplus_{i\in I}v_i).$$
And $\tilde f$ is also well-defined in this way:
Suppose $\sum_{k=1}^nl_i\otimes_{i\in I}v_i^k= \sum_{k=1}^nl'_i\otimes_{i\in I}{v'}_i^k$.
For any $\tau\in L(W)$, $\tau \circ f\in L(\oplus_{i\in I}V_i)$, so
$\sum_{k=1}^nl_i\otimes_{i\in I}v_i^k(\tau\circ f)=\sum_{k=1}^nl'_i\otimes_{i\in I}{v'}_i^k(\tau\circ f)$, i.e.,
$\tau(f(\sum_{k=1}^nl_i\oplus_{i\in I}v_i^k))=\tau(f(\sum_{k=1}^nl'_i\oplus_{i\in I}{v'}_i^k))$, hence
$f(\sum_{k=1}^nl_i\oplus_{i\in I}v_i^k)=f(\sum_{k=1}^nl'_i\oplus_{i\in I}{v'}_i^k)$.
**Remark**. The proof seems the same as the finite case~