## Exterior Power

**Definition**. Suppose $I$ is an index set, $V$ and $W$ are linear vector spaces, a multilinear map $f:\oplus_{i\in I} V\to W$ is called an *alternating operator* if $f((v_i)_{i\in I})=0$ whenever $(v_i)_{i\in I}\in \oplus_{i\in I} V$ is a family of linearly dependent vectors in $V$, here $\oplus_{i\in I} V$ merely means a subspace of $\Pi_{i\in I} V$. The I-*exterior power* of a linear space $V$ with a linear space $\Lambda^I V$ with an alternating operator $\Lambda$ such that for any linear space $W$ and an alternating operator $f$, there exists an alternating operator $\tilde f$ such that the following diagram is commutative:
\begin{xy}
\xymatrix{
\oplus_{i\in I} V\ar[r]^{\Lambda} \ar[rd]_f&\Lambda^I V\[email protected]{–>}[d]^{\tilde f}\\
& W
}
\end{xy}

**Existence**. Let $L$ be the linear subspace of $\otimes^I V$ gernerated by $\{v_1\otimes\cdots \otimes v_n |v_1,\cdots, v_n \mbox{ are linearly dependent. }\}$, then $\otimes^I V/L$ with $q\circ \otimes$ (where $q$ is the quotient map) is the I-*exterior power* of $V$.

*Proof*. For any alternating operator $f:V\to W$, we know that there exists some linear map $\bar f$ s.t. the left triangle commutative in the following diagram. Define $\tilde f$ as the map induced by $\bar f$, i.e., $$\tilde f(q(x))=\bar f(x)\forall x\in \otimes^I V.$$
\begin{xy}
\xymatrix{
\oplus_{i\in I} V\ar[r]^{\otimes} \ar[rd]_f&\otimes^I V\[email protected]{–>}[d]^{\bar f}\ar[r]^q&\Lambda^I V\[email protected]{–>}[dl]^{\tilde f}\\
& W
}
\end{xy}
Since the image of $\tilde f$ is deterministic on each generator, $\tilde f$ is deterministic.

## 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\[email protected]{–>}[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\[email protected]{–>}[d]^{\tilde f_i}\\
&W}
\end{xy}
Define $\tilde f=\sum_i \tilde f_i .$ We only have to prove $\tilde f$ preserves product. $\tilde f((\oplus_n u_n)(\oplus_n v_n))=\tilde f(\oplus_n \sum_{m+l=n} u_m\otimes v_l)=\sum_n \sum_{m+l=n}\tilde f_n(u_m\otimes v_l)=\sum_n\sum_{m+l=n} \tilde f_m(u_m) \tilde f_l(u_l)=\sum_m \tilde f_m(u_m)\sum_l \tilde f_l(v_l)=\tilde f(\oplus_n u_n)\tilde f(\oplus_n v_n).$