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\ar@{–>}[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\ar@{–>}[d]^{\bar f}\ar[r]^q&\Lambda^I V\ar@{–>}[dl]^{\tilde f}\\
& W
}
\end{xy}
Since the image of $\tilde f$ is deterministic on each generator, $\tilde f$ is deterministic.

Exterior Power

Given two vector spaces $V$ and $W$, an *alternating multilinear operator* from $V^k$ to $W$ is a multilinear map
$ f: V^k \to X $
such that whenever $v_1,\cdots,v_k$ are linearly dependent vectors in $V$, then
$f(v_1,\ldots, v_k)=0$.

The *kth exterior power* of a linear space $V$, is a linear space $\Lambda^k(V)$ with alternating multilinear operator $\Lambda:V^k\to \Lambda^k(V)$ such that for any given linear space and any alternating multilinear operator $f:V^k\to W$, there is a unique linear map $\tilde f$from $\Lambda^k(V)$ to $W$ such that $\tilde f\circ \Lambda=f$ as indicated by the following commutative diagram:
\begin{xy}
\xymatrix{
V^k\ar[r]^{\Lambda} \ar[dr]_f & \Lambda^k(V)\ar[d]^{\tilde f}\\
& W
}
\end{xy}

**Construction**. Let $I$ be the linear subspace generated by $\{x_1\otimes\cdots\otimes x_k|x_1,\cdots,x_n \mbox{ are linear dependent}\}$ in $\otimes^k V$. Then $\otimes^k V/I$ is a kth exterior power.

*Proof*. Let $\bar f:\otimes^k(V)\to W$ be the linear map such that $\bar f\circ \
\otimes= f$, $q:\otimes^k(V) \to \otimes^k V/I$ be the quotient map, and $\tilde f([z])=\bar f(z)$.
\begin{xy}
\xymatrix{
V^k\ar[r]^{\otimes} \ar[dr]_f & \otimes^k(V)\ar[d]^{\bar f}\ar[r]^q & \otimes^k(V)/I\ar[dl]^{\tilde f}\\
& W
}
\end{xy}
If $z\in I$ then $z$ is a linear combination of element in $\{x_1\otimes\cdots\otimes x_k|x_1,\cdots,x_n \mbox{ are linear dependent}\}$, so $\tilde f([z])=\bar f(z)= \sum_{i=1}^n k_if(x_1^i,\cdots,x_n^i) = 0$, so $\tilde f$ is well-defined.

>Then what about the infinite exterior power?