1
$\begingroup$

Let $V$ be a finite dimensional vector space and let $T(V)$ denote the tensor algebra $$T(V) = \bigoplus_{k=0}^\infty T^k(V)$$ where $$T^k(V) = \underbrace{V \otimes \cdots \otimes V}_k.$$ Let $I(V)$ be the two-sided ideal of $T(V)$ generated by the set $$\{v \otimes v | v \in V\}.$$ Then the exterior algebra of $V$, denoted $\bigwedge(V)$, is defined as the quotient $$\bigwedge(V) = T(V)/I(V).$$ I understand all of this so far. That is, an element of $\bigwedge(V)$ is an equivalence class of elements of elements in $T(V)$ that differ by any element of the form $a \otimes v \otimes v \otimes b$ for $v \in V$ and $a, b \in T(V)$.

What is confusing me is how any this relates to the wedge product. Based on what I have read, the image of any element in $T(v)$ under this quotient map should be a wedge product. For example, $$v_1 \otimes \cdots \otimes v_k \mapsto v_1 \wedge \cdots \wedge v_k.$$ I do not see how the wedge product appears. What does this quotient and tensor products have to do with the wedge product? More generally, what is the motivation behind carrying out this quotient at all?

Also as a side question, some textbooks use the notation $\bigwedge^*(V)$ to denote the exterior algebra of $V$. Is this an alternate notation to $\bigwedge(V)$ or does this instead mean $\bigwedge(V^*)$?

$\endgroup$
2
  • 1
    $\begingroup$ What is your definition of the wedge product? That the sign changes if we exchanges two factors in the product? $\endgroup$ Dec 6, 2023 at 8:07
  • $\begingroup$ @VladimirLysikov Yes that is the definition I am using. $\endgroup$
    – CBBAM
    Dec 6, 2023 at 8:10

1 Answer 1

1
$\begingroup$

Note that $$a \otimes x \otimes y \otimes b + a \otimes y \otimes x \otimes b = a \otimes (x+y) \otimes (x+y) \otimes b - a \otimes x \otimes x \otimes b - a \otimes y \otimes y \otimes b $$ That is, $a \otimes x \otimes y \otimes b \equiv - a \otimes y \otimes x \otimes b \pmod {I(V)}$. This means that if we exchange two factors in a tensor product, the sign of the corresponding element of $\bigwedge(V)$ changes. So if we take the subspace of $\bigwedge(V)$ spanned by products of $k$ elements, it is isomorphic to the wedge power $\bigwedge^k V = \underbrace{V \wedge \dots \wedge V}_{\text{$k$ times}}$.

The notation $\bigwedge^\bullet V$ is used to emphasise that the exterior algebra is graded and its graded parts are wedge products $\bigwedge^k V$, we replace a parameter $k$ with bullet to say that we consider the graded object with the homogeneous parts given by different $k$.

$\endgroup$
2
  • $\begingroup$ Thank you! So the set we are quotenting by is exactly what we need to impose the anticommutative property. $\endgroup$
    – CBBAM
    Dec 6, 2023 at 8:32
  • 2
    $\begingroup$ Yes. It is defined in this non-obvious way because it works better in characteristic $2$, where $1=-1$ (and also it is shorter to write). $\endgroup$ Dec 6, 2023 at 8:36

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .