Exterior algebras

Motivation

Suppose $V$ is an $n$ -dimensional $F$ -vector space and $B = {v_{1}, \dots, v_{n}}$ is basis for $V$ . Since $V$ is the free $F$ -module on the set ${v_{1}, \dots, v_{n}}$ , each linear transformation $T : V \to V$ corresponds to a unique choice of vectors $w_{1}, \dots, w_{n} \in V$ , namely the images of each of the basis vectors $v_{i}$ . Moreover, the determinant is a function that assigns to $T$ a single value $det (T) \in F$ . With the basis $B$ fixed, this determinant can be viewed as a function

det : V \times \dots \times V \to F

that assigns to each $n$ -tuple $(w_{1}, \dots, w_{n})$ the determinant of the corresponding linear transformation.

This determinant function is characterized by three nice properties:

It is alternating: If we swap the positions of two of the vectors in the $n$ -tuple, the determinant changes sign.
It is multilinear: If we fix all but one vector in the $n$ -tuple, the resulting function $V \to F$ is $F$ -linear.
It is 1 on the identity transformation.

One immediate consequence of the first property is that if $w_{i} = w_{j}$ for any distinct $i, j$ then the determinant of that $n$ -tuple is zero. Combined with the second property, it follows that the determinant of an $n$ -tuple $(w_{1}, \dots, w_{n})$ is zero whenever the vectors are linearly dependent.

This alternating multilinear function seems like something very close to the tensor algebra or symmetric algebra construction. Let's begin by looking for an analogue of the tensor algebra for which we have the additional property that $m \cdot m = 0$ for all $m \in M$ .

The construction

Definition of exterior algebra

Let $R$ be a commutative ring (with unity) and $M$ be an $R$ -module. The exterior algebra of $M$ is the $R$ -algebra obtained by taking the quotient of the tensor algebra $T (M)$ by the ideal $A (M)$ generated by all elements of the form $m \otimes m$ for $m \in M$ . The exterior algebra $T (M) / A (M)$ is denoted $⋀ (M)$ and the image of $m_{1} \otimes m_{2} \otimes \dots \otimes m_{k}$ in $⋀ (M)$ is denoted $m_{1} \land m_{2} \land \dots \land m_{k}$ .

As with the symmetric algebra, the ideal $A (M)$ is generated by homogenous elements and hence is a graded ideal. It follows that $⋀ (M)$ is a graded ring and the homogeneous component of degree $k$ is

\overset{k}{⋀} (M) = T^{k} (M) / A^{k} (M) .

This $R$ -module is called the $k^{th}$ exterior power of $M$ . Note that since $A (M)$ is generated (as an ideal) by degree 2 homogenous elements, we have $A^{0} (M) = A^{1} (M) = 0$ and hence have $\overset{0}{⋀} (M) = R$ and $\overset{1}{⋀} (M) = M$ .

Notational note

Anecdotally, it seems common to drop the parentheses when dealing with the exterior algebra, and instead write simply $⋀ M$ and $\overset{k}{⋀} M$ . Dummit & Foote does not do this, however, so I have stuck with the parentheses here. (It also matches our notation for the tensor and symmetric algebras.)

The alternating property

The multiplication in $⋀ (M)$ is given by

(m_{1} \land \dots \land m_{i}) \land (m_{1}^{'} \land \dots \land m_{j}^{'}) = m_{1} \land \dots m_{i} \land m_{1}^{'} \land \dots \land m_{j}^{'} .

This is called the wedge (or exterior) product.

This multiplication is alternating in the following sense. By construction, we have $m_{1} \land \dots \land m_{k} = 0$ in $\overset{k}{⋀} (M)$ whenever $m_{i} = m_{i + 1}$ for any $i$ . We claim that we then have anti-commutativity for simple wedges; i.e., for every $m, m^{'} \in M$ we have

m \land m^{'} = - (m^{'} \land m) .

To see this, observe that by construction we have

(m + m^{'}) \land (m + m^{'}) = 0.

Expanding out the left-hand side gives

(m \land m) + (m \land m^{'}) + (m^{'} \land m) + (m^{'} \land m^{'}) = 0.

The first and last wedges are zero by construction. The claim thus follows.

Warning

This anti-commutativity does not extend to arbitrary products. For example,

\begin{aligned} m \land (n_{1} \land n_{2}) & = (m \land n_{1}) \land n_{2} \\ = - (n_{1} \land m) \land n_{2} \\ = - n_{1} \land (m \land n_{2}) \\ = n_{1} \land (n_{2} \land m) \\ = (n_{1} \land n_{2}) \land m . \end{aligned}

So $m$ and $n_{1} \land n_{2}$ commute.

A universal property of the exterior algebra

As with the tensor algebra and symmetric algebra functors, there is a functor from the category of $R$ -modules to the category of those $R$ -algebras $A$ with the property $a^{2} = 0$ for all $a \in A$ .

A universal property of the symmetric algebra

Let $(R -Alg)_{0}$ be the category of $R$ -algebras $A$ with the property that $a^{2} = 0_{A}$ for every $a \in A$ , and let $U : (R -Alg)_{0} \to R -Mod$ be the forgetful functor. Then there is a functor $⋀ : R -Mod \to (R -Alg)_{0}$ and a natural bijection

τ_{M, A} : {Hom}_{(R -Alg)_{0}} (⋀ (M), A) \tilde{\to} {Hom}_{R -Mod} (M, U (A)) .

In other words, the functor $⋀$ is a left adjoint of the forgetful functor $U$ .

One last time, as with the tensor algebra and symmetric algebra functors, we can now deduce many properties of $⋀ (M)$ :

It is an $R$ -algebra (in which $a^{2} = 0$ for all $a \in ⋀ (M)$ ) that we can associate to the $R$ -module $M$ ;
The construction is functorial, so that if $f : M \to N$ is an $R$ -module morphism then there is a corresponding $R$ -algebra morphism $⋀ (f) : ⋀ (M) \to ⋀ (N)$ ;
The identity $R$ -algebra morphism $⋀ (M) \to ⋀ (M)$ corresponds to an $R$ -module morphism $M \to U (⋀ (M))$ . Classically, this is viewed as an inclusion $M ↪ ⋀ (M)$ onto the degree 1 component. (This is a component of the unit of the adjunction.)
Since $⋀$ is a left adjoint it commutes with all colimits; in particular, it commutes with coproducts (which again are called direct sums in these categories).

Examples

Suppose $V$ is an $n$ -dimensional vector space over a field $F$ with basis ${v_{1}, \dots, v_{n}}$ . When $0 \leq k \leq n$ , the set of vectors
${v_{i_{1}} \land \dots \land v_{i_{k}} ∣ 1 \leq i_{1} < \dots < i_{k} \leq n}$
is a basis for $\overset{k}{⋀} (V)$ . When $k > n$ the $R$ -module $\overset{k}{⋀} (V)$ is trivial.

This same statement is true more generally when $R$ is a commutative ring and $M$ is a free $R$ -module of rank $n$ .

Continuing the previous example, suppose $ϕ : V \to V$ is any linear endomorphism of $V$ . For every $k$ we then have an $F$ -linear transformation
$\overset{k}{⋀} (ϕ) : \overset{k}{⋀} (V) \to \overset{k}{⋀} (V) .$
When $k = n$ , the $F$ -vector space $\overset{n}{⋀} (V)$ is one-dimensional with basis vector $v_{1} \land v_{2} \land \dots \land v_{n}$ and
$\overset{n}{⋀} (ϕ) (v_{1} \land \dots \land v_{n}) = ϕ (v_{1}) \land \dots \land ϕ (v_{n}) = D (ϕ) \cdot (v_{1} \land \dots v_{n})$
for some scalar $D (ϕ) \in F$ . One can verify that this function $D$ satisfies the three axioms for a determinant function and hence $D (ϕ) = det (ϕ)$ .

Suggested next note

Symmetric and alternating tensors