## Tensor Algebra

**Definition**. Suppose $V$ is a vector space, the tensor algebra of $V$ is defined as an algebra $TV$ with a linear map $T:V\to TV$ such that for any algebra $A$ and any linear map $f:V\to A$ there exists a unique algebra homomorphism $\tilde f:TV\to A$ such that $f=\tilde f\circ T$ as indicated by the commutative diagram below.
\begin{xy}
\xymatrix{
V\ar[r]^T\ar[rd]_f & TV\ar@{–>}[d]^{\tilde f}\\
& A }
\end{xy}

**Existence**. $TV$ is defined as $\oplus_{n\in \mathbb N}\otimes^n V$ and $T$ is defined as a map send a vecctor $v$ to an assignment of $v$ to 1 and 0 to others.

*Proof*. Since $\tilde f\circ T(v)=f(v)$, $\tilde f(\otimes_{j}v_j)=\Pi_j f(v_j)$ where $\otimes_{j}v_j$ is an element of $\otimes^n V_j$, $f(\oplus_{i} w_i)=f(\sum_i w_i\chi_i)=\sum_i f(w_i)$. Hence $\tilde f$ is unique if it exists.

Let $\tilde f_i$ be the uqnique linear map as indicated by the following commutative diagram:

\begin{xy}
\xymatrix{
\Pi^i V\ar[r]^\otimes\ar[rd]_{\Pi^i f} & \otimes^i V\ar@{–>}[d]^{\tilde f_i}\\
&W}
\end{xy}
Define $\tilde f=\sum_i \tilde f_i .$ We only have to prove $\tilde f$ preserves product. $\tilde f((\oplus_n u_n)(\oplus_n v_n))=\tilde f(\oplus_n \sum_{m+l=n} u_m\otimes v_l)=\sum_n \sum_{m+l=n}\tilde f_n(u_m\otimes v_l)=\sum_n\sum_{m+l=n} \tilde f_m(u_m) \tilde f_l(u_l)=\sum_m \tilde f_m(u_m)\sum_l \tilde f_l(v_l)=\tilde f(\oplus_n u_n)\tilde f(\oplus_n v_n).$

## Infinite Tensor Product

**Definition**. Suppose $(V_i)_{i\in I}$ is a family of linear spaces. 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}
What is $\oplus_{i\in I}V_i$? Just a subspace of the Cartesian product of $(V_i)_{i\in I}$.

**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
$f(\sum_{k=1}^nl_i\oplus_{i\in I}v_i^k)=f(\sum_{k=1}^nl’_i\oplus_{i\in I}{v’}_i^k)$.

**Tensor Algebra**. A character of an infinite set $S$ is that $\newcommand{\card}{\operatorname{card}}$ its cardinal number $\card(S)$ satisfies
$$\card(S)^2:=\card(S\times S)=\card(S).$$
So we can define a product in $\oplus_{C\leq \card(2^S)}\otimes_{i\in C} V.$
For $u_D\in \otimes_{i\in D} V, w_E\in \otimes_{i\in E} V$, define $$u_D\cdot w_E:=u_D\otimes w_E\in \otimes_{\card(D\sqcup E)} V,$$
and for $u=\oplus_{D\leq \card(2^S)} u_D$ and $w=\oplus_{E\leq \card(2^S)} w_E$, define $$u\cdot v:=\oplus_{C\leq \card(2^S)}\sum_{\card(D\sqcup E)=C} u_D\otimes w_E.$$
In order to let the sum in the above formula make sense, we can define $\oplus_i V_i=\{(v_i)|\{i|v_i\neq 0\} \mbox{ is a finite set. }\}$, or import norms.