Exterior algebras

Motivation

Suppose $V$ is an $n$ -dimensional $k$ -vector space and $B = {v_{1}, \dots, v_{n}}$ is basis for $V$ . Since $V$ is the free $k$ -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 k

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 $V \to k$ is $k$ -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 $m_{1} \otimes m_{2} \otimes \dots \otimes m_{i}$ in $⋀ (M)$ is denoted $m_{1} \land m_{2} \land \dots \land m_{i}$ .

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{i}{⋀} (M) = T^{i} (M) / A^{i} (M) .

This $R$ -module is called the $i^{th}$ 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 $\overset{i}{⋀} M$ . 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 $m_{1} \land \dots \land m_{i} = 0$ in $\overset{i}{⋀} (M)$ 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

Like the tensor and symmetric algebras, the exterior algebra construction is indeed (the object function of) a functor left adjoint to a certain forgetful functor. Unlike the previous two constructions, however, the correct category from which we are forgetting is not so obvious. So instead let's first note the following.

For each $R$ -module $M$ , define a functor $F_{M} : R - Alg \to S e t$ as follows:

The object function: Let $U : R - Alg \to R - Mod$ denote the forgetful functor. For each $R$ -algebra $A$ , let $F_{M} (A)$ be the set of $R$ -module morphisms $f : M \to U (A)$ that satisfy $f (m)^{2} = 0$ in $A$ for every $m \in M$ .
The arrow function: For each $R$ -algebra morphism $ϕ : A \to A^{'}$ , define $F_{M} (ϕ) : F_{M} (A) \to F_{M} (A^{'})$ by "composition with $U (ϕ)$ ". In other words, for each $R$ -module morphism $f : M \to U (A)$ that satisfies $f (m)^{2} = 0$ for all $m \in M$ , let $F_{M} (ϕ) = U (ϕ) \circ f : M \to U (A^{'})$ .

It turns out that the $R$ -algebra $⋀ (M)$ represents the functor $F_{M}$ , in the sense that there is a natural isomorphism

τ : {Hom}_{R - Alg} (⋀ (M), -) \tilde{\Rightarrow} F_{M} .

In other words, there are natural bijections

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

More concretely, for every $R$ -algebra $A$ and $R$ -module morphism $f : M \to U (A)$ such that $f (m)^{2} = 0$ for all $m \in M$ , there exists a unique $R$ -algebra morphism $h : ⋀ (M) \to A$ through which $f$ "factors", i.e., such that $f$ equals the composition of the inclusion of $M$ into $⋀ (M)$ (as the degree 1 component) followed by $h$ (really $U (h)$ ).

You usually see this property expressed informally as "Maps from $M$ to algebras $A$ with $a^{2} = 0$ extend uniquely to maps from $⋀ (M)$ ."

It's also true that this construction is functorial, in that we can define an "exterior algebra functor"

⋀ : R - Mod \to R - Alg .

However, this functor is not adjoint to any functor (that I know of). What is true is that you can define a new type of algebra, sometimes called a super-commutative $R$ -algebra^[1], and that our construction is really a functor from the category of $R$ -modules to the category of super-commutative $R$ -algebras, left adjoint to the evident forgetful functor from super-commutative $R$ -algebras back to $R$ -modules. But that's a bit too much for our current interests, so let's move on to some examples.

Examples

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

This same statement is true more generally when $R$ is a commutative ring and $M$ is a free $R$ -module of rank $n$ .^51701e
Continuing the previous example, suppose $ϕ : V \to V$ is any linear endomorphism of $V$ . For every $i$ we then have a $k$ -linear transformation
$\overset{i}{⋀} (ϕ) : \overset{i}{⋀} (V) \to \overset{i}{⋀} (V) .$
When $i = n$ , the $k$ -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 $D (ϕ) \in k$ . 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

I've also seen them called graded-commutative superalgebras and other word-salad combinations. ↩︎