Antisymmetrization and forms.

A reference for this material is Chapter 14 of John M. Lee. Introduction to smooth manifolds. Second edition. Grad. Texts in Math., 218. Springer, New York, 2013. xvi+708pp. ISBN: 978-1-4419-9981-8.

Let \(V\) be a finite-dimensional vector space. Then

\[ (V \otimes \cdots \otimes V ) ^{\ast } \cong V^{\ast } \otimes \cdots \otimes V^{\ast } . \]

The left hand side is in correspondence with the set of \(k\)-multilinear maps

\[ V \times \cdots \times V \to \mathbb {R} . \]

The correspondence can be made a little more explicit. For \(\alpha _1 , \ldots , \alpha _k \in V^{\ast }\), one has the multilinear map \(V \times \cdots \times V \to \mathbb {R} \) given by

\[ (v_1, \ldots , v_k) \mapsto \alpha _1 (v_1) \cdots \alpha _k (v_k) . \]

This depends multilinearly on the \(\alpha _i\)’s and induces the isomorphism

\[ V^{\ast } \otimes \cdots \otimes V^{\ast } \cong Mult_k (V , \ldots , V ; \mathbb {R} ). \]

Let \(S_k\) denote the \(k\)-th symmetric group (the group of bijections of the set \(\{ 1, \ldots , k \}\)).

Note that for \(\eta \in (V^{\ast } )^{\otimes k}\), the effect of \(A_k\) at the level of \(k\)-multilinear maps is

\begin{equation}\label {eq:antisymmetrization} A_k (\eta ) (v_1, \ldots , v_k) = \frac {1}{k!} \sum _{\sigma \in S_k } \text {sign} (\sigma ) \, \eta ( v_{\sigma (1)} , \ldots , v_{\sigma (k) } ). \end{equation}

Solution:

Fix \(v _1, \ldots , v_k \in V\) and \(\tau \in S_k\). Then

\begin{align*} A_k ( \eta )(v_{\tau (1)}, \ldots , v_{\tau (k) }) & = \frac {1}{k!} \sum _{\sigma \in S_k} \text {sign}(\sigma ) \, \eta ( v_{\sigma \tau (1)} , \ldots , v_{\sigma \tau (k)}) \\ & = \frac {1}{k!} \sum _{\theta \in S_k } \text {sign} (\tau ) \, \text {sign} (\theta ) \, \eta (v_{\theta (1) } , \ldots , v_ {\theta (k )}) \\ & = \text {sign} (\tau )\, A_k ( \eta ) (v_1, \ldots , v_k) . \end{align*}

This shows \(A_k (\eta ) \) is antisymmetric. The second claim follows directly from (eq:antisymmetrization).

We denote by \(\pi : (V^{\ast } )^{\otimes k} \to \Lambda ^k (V ^{\ast } ) \) the projection. We write

\[ \pi (\alpha _1 \otimes \cdots \otimes \alpha _k ) : = \frac {1}{k! } \alpha _1 \wedge \cdots \wedge \alpha _k . \]
Solution:
\begin{align*} \pi A_k ( \alpha _1 \otimes \cdots \otimes \alpha _k ) & = \frac {1}{k!} \frac {1}{k!} \sum _{\sigma \in S_k} \text {sign}(\sigma ) \, v_{\sigma (1) } \wedge \cdots \wedge v_{\sigma (k) } \\ & = \frac {1}{k!} \frac {1}{k!} \sum _{\sigma \in S_k} v_{1 } \wedge \cdots \wedge v_{k } \\ & = \frac {1}{k!} v_{1 } \wedge \cdots \wedge v_{k } \\ & = \pi ( v_1 \otimes \cdots \otimes v_k ). \end{align*}
Solution:

The space of antisymmetric \(k\)-multilinear maps is precisely \(A_k ( (V^{\ast })^{\otimes k} )\). Then

\[ \pi (A_k (V^{\ast })^{\otimes k}) = \pi ( ( V^{\ast })^{\otimes k} ) = \Lambda ^k (V^{\ast } ) . \]
This shows that \(\pi \) is surjective. By the HW, the spaces have the same dimension, and \(\Lambda ^k (V^{\ast } )\) has basis
\[ \{ \eta _{i_1} \wedge \cdots \wedge \eta _{i_k} \,\vert \, 1 \leq i_1 < \cdots < i_k \leq n \} , \]
where \(\{ \eta _1, \ldots , \eta _n \} \subset V^{\ast } \) is the dual basis of a basis \(\{ e_1, \ldots , e_n \} \subset V\).

It will be convenient to have the following in the future.

Solution:

For \(v_1 , \ldots , v_{k + \ell } \in V\), one has

\begin{align*} &A_{k + \ell } ( A_k (\omega ) \otimes A_{\ell } (\eta ) ) (v_1, \ldots , v_{k + \ell }) \\ & = \frac {1}{(k+ \ell )! } \sum _{\sigma \in S_{k + \ell } } \text {sign}(\sigma ) \, A_k (\omega )( v_{ \sigma (1) } , \ldots , v_{\sigma (k)}) A_{\ell } (\eta ) ( v_{\sigma (k + 1 )}, \ldots , v_{\sigma ( k + \ell )} ) \\ & = \frac {1}{(k+ \ell )! \, k ! \, \ell !} \sum _{\substack {\sigma \in S_{k + \ell } , \\ \tau \in S_k , \\ \theta \in S_{\ell }}} \text {sign}(\sigma ) \, \text {sign}(\tau ) \, \text {sign}(\theta ) \, \omega ( v_{ \sigma \tau (1) } , \ldots , v_{\sigma \tau (k)}) \eta ( v_{\sigma (k + \theta (1) )}, \ldots , v_{\sigma ( k + \theta (\ell ))} ) \\ & = \frac {1}{(k + \ell )! } \sum _{\sigma \in S_{k + \ell } } \text {sign}(\sigma ) \omega (v_{\sigma (1)}, \ldots , v_{\sigma (k)}) \eta (v_{\sigma (k + 1)}, \ldots , v_{\sigma (k + \ell )})\\ & = A_{k + \ell } (\omega \otimes \eta ) ( v_1, \ldots , v_{k + \ell }) . \end{align*}

For \(\omega \in \Omega ^k (M)\), in local coordinates we can write

\[ \omega = \sum _{1 \leq i_1 < \ldots < i_k \leq n } \omega _{i _1, \ldots , i_k} dx^{i_1} \wedge \cdots \wedge dx ^{i_k} . \]

Note that for \(j_1, \ldots , j_k \in \{ 1, \ldots , n \}\), we have

\begin{align*} dx^{i_1} \wedge & \cdots \wedge dx^{i_k} (\tfrac {\partial }{\partial x ^{j_1}}, \ldots , \tfrac {\partial }{\partial x ^{j_k}}) \\ & = k! A_k (dx^{i_1} \otimes \cdots \otimes dx^{i_k} ) (\tfrac {\partial }{\partial x ^{j_1}}, \ldots , \tfrac {\partial }{\partial x ^{j_k}}) \\ & = \sum _{\sigma \in S_k} \text {sign}(\sigma ) \, dx^{i_1} \left ( \tfrac {\partial }{\partial x ^{j _{\sigma (1)}}} \right ) \cdots dx^{i_k} \left ( \tfrac {\partial }{\partial x ^{j _{\sigma (k)}}} \right ) . \end{align*}

If the sets \(\{ i_1 , \ldots , i_k \}\) and \(\{ j_1, \ldots , j_k \}\) are not the same, then each summand is zero. If the sets are the same, since the \(i_{\ell }\)’s are distinct, there is a unique element \(\sigma \) such that \(j_{\sigma (1)} < \ldots < j_{\sigma (k)}\). In that case, that is the only summand that survives, and the result is \(\text {sign}(\sigma ) \in \{ -1, 1 \} \).