I'm reading through Exterior Differential Systems by Bryant, Chern, Gardner, Goldschmidt, and Griffiths.

They say,

Let $e_i$ be a base of $V$ and $\omega^k$ its dual base, so that $$\langle e_i,\omega^k\rangle = \delta^k_i, \quad 1\le i,k \le n.$$ Then an element $\xi\in\Lambda^p(V)$ can be written $$ \xi=1/p!\sum a^{i_1\cdots i_p}e_{i_1}\wedge\cdots \wedge e_{i_p} $$ and an element $\alpha \in \Lambda^p(V^\ast)$ as $$ \alpha=1/p!\sum b_{i_1\cdots i_p}\omega^{i_1}\wedge \cdots \wedge \omega^{i_p}. $$ [where] the coefficients $a^{i_1\cdots i_p}$ and $b_{i_1\cdots i_p}$ are supposed to be anti-symmetric in any two of their indices, so that they are well defined. It follows that any multivector is a linear combination of decomposable multivectors.

Where does the $1/p!$ term come from?

I'm also unsure about the notation for the coefficients $a^{i_1\cdots i_p}$ and $b_{i_1\cdots i_p}$. How are these defined?


The wedge products in the sum are a basis of the exterior algebra, so this is just saying that they span i.e. everything is a sum of them. But, for example, $e_1\wedge e_2=-e_2\wedge e_1$, so the sums above, which are apparently over all $i_1,\cdots, i_p$, have a lot of replication. Because there are $p!$ permutations of $p$ objects, if the coefficients $a^{i_1\cdots i_p}$ are antisymmetric, then there will be $p!$ replications of them. So for convenience the sum is divided by $p!$ and then the coefficients are the same as the basis obtained from summing over ordered tuples $(i_1,\cdots,i_p)$, which is actually the usual basis. If you didn't do this then the coefficients above would deviate from the standard basis by a factor of $p!$, which would not be the end of the world but it was not the choice made here.


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