exterior power

The $n^{th}$ exterior power of a vector space $V$ is closely related to the tensor product.

We can think of them as antisymmetric tensors.

Definition

The $n^{th}$ exterior power of a vector space $V$ over a field $k$ is a $k$-vector space ${\wedge }^n(V)$ and an antisymmetric multi-linear map

$$\iota :V^n⟶{\wedge }^n(V)$$

that satisfies the following universal property:

For every $k$-vector space $W$, and an antisymmetric multi-linear map

$$\phi :V\to W$$

there exists a unique linear (NOT MULTI-LINEAR) map $\bar{\phi }:{\wedge }^n(V)\to W$ such that this diagram commutes:

Notation: we denote $\iota (v_1,\dots ,v_n)=v_1\wedge \cdots \wedge v_n$

Explicit construction

Using the tensor product (with its explicit construction) we can build the wedge product. Note $\iota :V^n⟶{\wedge }^n(V)$ is a multi-linear map, so by the universal property of the tensor product, there exists a linear map $\Psi :V^{\otimes n}\to {\wedge }^{n}V$ such that

commutes.

$\Psi$ is surjective, and

$$\ker \Psi =\text{span}\{v_1\otimes \cdots \otimes v_n-\text{sgn}(\sigma )v_{\sigma (1)}\otimes \cdots \otimes v_{\sigma (n)}\}{v}_1,\dots ,v_n\in V,\sigma \in S_n$$

Therefore, by the first isomorphism theorem

$${\wedge }^n(v)\cong V^{\otimes n}/\ker \Psi$$

Other identifications

multi-linear maps from dual

We can identify the exterior product with the vector-space of all multi-linear maps

$$(V^{\ast }{)}^k⟶\mathbb{F}$$

This looks like

$$\begin{array}{rl}\iota :V^k& \to {\wedge }^k(V)\\ (v_1,\dots ,v_k)& \mapsto \left(\begin{array}{rl}\iota (v_1,\dots ,v_k):(V^{\ast }{)}^k& ⟶\mathbb{F}\\ ({\epsilon }_1,\dots ,{\epsilon }_n)& ⟼\sum _{\sigma \in S_n}\text{sgn}(\sigma ){\epsilon }_1(v_{\sigma (1)})\cdots {\epsilon }_k(v_{\sigma (k)})\end{array}\right)\end{array}$$

Alternative constructions/ definitions

We can define this using the free vector space just like the tensor product just using a different quotienting set.

todo add more here about how to construct it.

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References