Exterior Algebras

April 20, 2012

This post will be very similar to last post, where I talked about symmetric algebras. Indeed, the exterior algebra is another universal construction obtained as the quotient of the tensor algebra. Among other applications, this construction is of fundamental importance in differential geometry where it is used to define differential forms, which are in turn used to generalize the machinery of calculus on $R^n$ to calculus on smooth manifolds.

Definition: The exterior algebra of an $R$ -module $M$ is the quotient of the tensor algebra $T(M)$ by the ideal $A(M)$ generated by the elements of the form $m \otimes m$ for all $m \in M.$ The exterior algebra $T(M)/A(M)$ is then denoted $\Lambda(M)$ and the image of $m_1 \otimes m_2 \otimes \cdots \otimes m_k$ in $\Lambda(M)$ is denoted by $m_1 \wedge m_2 \wedge \cdots \wedge w_k.$

The multiplication in the exterior algebra,

$(m_1 \wedge \cdots \wedge m_i) \wedge (m_1' \wedge \cdots \wedge m_j') = m_1 \wedge \cdots \wedge m_i \wedge m_1' \wedge \cdots \wedge m_j',$

is called the wedge product or the exterior product. Denote by $A^k(M)$ the set of k-tensors in $A(M)$ so $A^k(M)$ is the set of k-tensors $m_1 \otimes \cdots \otimes m_k \in T^k(M)$ for which $m_i = m_{i+1}$ for some $1 \leq i < k.$ The k’th exterior power $\Lambda^k(M)$ is the quotient $T^k(M)/A^k(M).$

By this definition, we have $m_1 \wedge \cdots \wedge m_k = 0$ in $\Lambda^k(M)$ whenever $m_i = m_{i+1}$ for some $1 \leq i < k.$ Then we have

$0 = (m+m') \wedge (m+m')$

$= (m \wedge m') + (m \wedge m') + (m' \wedge m) + (m' \wedge m')$

$= (m \wedge m') + (m' \wedge m)$

so the multiplication is anticommutative on the image of simple tensors in $\Lambda^k(M).$ i.e.:

$m \wedge m' = -m' \wedge m \quad (\text{for all } m, m' \in M).$

From these considerations, we can describe the elements of the k’th exterior power as the set of k-tensors where all elements having two equal entries have been identified with $0.$ More explicitly, $\Lambda^k(M)$ is equal to $T^k(M)$ modulo the submodule generated by all elements of the form

$m_1 \otimes m_2 \otimes \cdots \otimes m_k \quad \text{where } m_i = m_j \text{ for some } i \neq j.$

Indeed, the elements of $latex $A^k(M)$ are finite sums of k-tensors having two equal adjacent entries, so every element of $A^k(M)$ is a sum of elements of that form. Conversely, since the wedge product is anticommutative on simple tensors, we can rearrange (modulo a minus sign) any element with equal entries $m_i$ and $m_j$ for these entries to be adjacent.

The exterior algebra is also universal with respect to alternating multilinear maps. Recall that an alternating map $\varphi$ is a map such that $\varphi(x_1, \ldots, x_n) = 0$ whenever $x_i = x_j$ for some $i \neq j.$ Note that this is equivalent to

$\varphi(x_{\sigma(1)}, x_{\sigma(2)}, \ldots, x_{\sigma(n)}) = \text{sgn}(\sigma)(x_1, x_2, \ldots, x_n).$

This shows that in some sense the exterior algebra is complementary to the symmetric algebra.

Theorem (Universal property for Alternating Multilinear Maps): Let $M, N$ be two $R$ -modules for $R$ a commutative ring and $\varphi : M^k \to N$ an alternating multilinear map. Then there is a unique $R$ -module homomorphism $\phi : \Lambda^k(M) \to N$ such that $\varphi (m_1, \ldots, m_k) = \phi(m_1 \wedge \cdots \wedge m_k).$

Proof: The proof is completely analogous to the case of the exterior algebra: you take the induced homomorphism

$\tilde{\phi} (m_1 \otimes \cdots \otimes m_k) = \varphi(m_1, \ldots, m_k)$

which is then seen to be alternating. Thus, it’s kernel is in $A^k(M)$ and the map

$\phi(m_1 \wedge \cdots \wedge m_k) = \tilde{\phi}(m_1 \otimes \cdots \otimes m_k)$

is well defined.

QED

Reference: D. S. Dummit, R. M. Foote, Abstract Algebra, 3rd ed.

From → Algebra, Tensor, symmetric, exterior algebra

Exterior Algebras

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