# The Unapologetic Mathematician

## Inner Products on Differential Forms

Now that we’ve defined inner products on $1$-forms we can define them on $k$ forms for all other $k$. In fact, our construction will not depend on the fact that they come from a metric at all.

In fact, we’ve basically seen this already when we were just dealing with vector spaces and we introduced inner products on tensor spaces. Pretty much everything goes just as it did then, so going back and reviewing those constructions will pay dividends now.

Anyway, the upshot: we know that we can write any $k$-form as a sum of $k$-fold wedges, so the bilinearity of the inner product means we just need to figure out how to calculate the inner product of such $k$-fold wedges. And this works out like

$\displaystyle\begin{aligned}\langle \alpha_1\wedge\dots\wedge \alpha_k,\beta_1\wedge\dots\wedge \beta_k\rangle&=\frac{1}{k!}\frac{1}{k!}\sum\limits_{\pi\in S_k}\sum\limits_{\hat{\pi}\in S_k}\mathrm{sgn}(\pi\hat{\pi})\langle \alpha_{\pi(1)}\otimes\dots\otimes \alpha_{\pi(k)},\beta_{\hat{\pi}(1)}\otimes\dots\otimes \beta_{\hat{\pi}(k)}\rangle\\&=\frac{1}{k!}\frac{1}{k!}\sum\limits_{\pi\in S_k}\sum\limits_{\hat{\pi}\in S_k}\mathrm{sgn}(\pi\hat{\pi})\langle \alpha_{\pi(1)},\beta_{\hat{\pi}(1)}\rangle\dots\langle \alpha_{\pi(k)},\beta_{\hat{\pi}(k)}\rangle\\&=\frac{1}{k!}\frac{1}{k!}\sum\limits_{\pi\in S_k}\sum\limits_{\hat{\pi}\in S_k}\mathrm{sgn}(\pi^{-1}\hat{\pi})\langle \alpha_1,\beta_{\pi^{-1}(\hat{\pi}(1))}\rangle\dots\langle \alpha_{k},\beta_{\pi^{-1}(\hat{\pi}(k))}\rangle\\&=\frac{1}{k!}\frac{1}{k!}\sum\limits_{\pi\in S_k}\sum\limits_{\sigma\in S_k}\mathrm{sgn}(\sigma)\langle \alpha_1,\beta_{\sigma(1)}\rangle\dots\langle \alpha_k,\beta_{\sigma(k)}\rangle\\&=\frac{1}{k!}\sum\limits_{\sigma\in S_k}\mathrm{sgn}(\sigma)\langle \alpha_1,\beta_{\sigma(1)}\rangle\dots\langle \alpha_k,\beta_{\sigma(k)}\rangle\\&=\frac{1}{k!}\det\left(\langle\alpha_i,\beta_j\rangle\right)\end{aligned}$

Now let’s say we have an orthonormal basis of $1$-forms $\{\eta^i\}$ — a collection of $1$-forms such that $\langle\eta^i,\eta^j\rangle$ is the constant function with value $1$ if $i=j$ and $0$ otherwise. Taking them in order gives us an $n$-fold wedge $\eta^1\wedge\dots\wedge\eta^n$. We can calculate its inner product with itself as follows:

$\displaystyle\begin{aligned}\langle\eta^1\wedge\dots\wedge\eta^n,\eta^1\wedge\dots\wedge\eta^n\rangle&=\frac{1}{n!}\det\left(\langle\eta^i,\eta^j\rangle\right)\\&=\frac{1}{n!}\det\left(\delta^{ij}\right)=\frac{1}{n!}\end{aligned}$

We’ve seen this before when talking about the volume of a parallelepiped, but it still feels like this should have volume $1$. For this reason, many authors will rescale the inner products on $k$-forms to compensate. That is, they’ll define the inner product on $\Omega^k(U)$ to be the determinant above, rather than $\frac{1}{k!}$ times the determinant like we wrote. We’ll stick with this version, but remember that not everyone does it this way.

### Like this:

Posted by John Armstrong | Differential Geometry, Geometry

## 2 Comments	»

1. [...] say that is an orientable Riemannian manifold. We know that this lets us define a (non-degenerate) inner product on differential forms, and of course we have a wedge product of differential forms. We have almost everything we need to [...]

Pingback by | October 6, 2011 | Reply

2. [...] these inner products are those induced on differential forms from the metric. This doesn’t quite hold, but we can show that it does hold “up to [...]

Pingback by | October 21, 2011 | Reply

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