# The Unapologetic Mathematician

## The Hodge Star

Sorry for the delay from last Friday to today, but I was chasing down a good lead.

Anyway, last week I said that I’d talk about a linear map that extends the notion of the correspondence between parallelograms in space and perpendicular vectors.

First of all, we should see why there may be such a correspondence. We’ve identified $k$-dimensional parallelepipeds in an $n$-dimensional vector space $V$ with antisymmetric tensors of degree $k$: $A^k(V)$. Of course, not every such tensor will correspond to a parallelepiped (some will be linear combinations that can’t be written as a single wedge of $k$ vectors), but we’ll just keep going and let our methods apply to such more general tensors. Anyhow, we also know how to count the dimension of the space of such tensors:

$\displaystyle\dim\left(A^k(V)\right)=\binom{n}{k}=\frac{n!}{k!(n-k)!}$

This formula tells us that $A^k(V)$ and $A^{n-k}(V)$ will have the exact same dimension, and so it makes sense that there might be an isomorphism between them. And we’re going to look for one which defines the “perpendicular” $n-k$-dimensional parallelepiped with the same size.

So what do we mean by “perpendicular”? It’s not just in terms of the “angle” defined by the inner product. Indeed, in that sense the parallelograms $e_1\wedge e_2$ and $e_1\wedge e_3$ are perpendicular. No, we want any vector in the subspace defined by our parallelepiped to be perpendicular to any vector in the subspace defined by the new one. That is, we want the new parallelepiped to span the orthogonal complement to the subspace we start with.

Our definition will also need to take into account the orientation on $V$. Indeed, considering the parallelogram $e_1\wedge e_2$ in three-dimensional space, the perpendicular must be $ce_3$ for some nonzero constant $c$, or otherwise it won’t be perpendicular to the whole $x$$y$ plane. And $\vert c\vert$ has to be ${1}$ in order to get the right size. But will it be $+e_3$ or $-e_3$? The difference is entirely in the orientation.

Okay, so let’s pick an orientation on $V$, which gives us a particular top-degree tensor $\omega$ so that $\mathrm{vol}(\omega)=1$. Now, given some $\eta\in A^k(V)$, we define the Hodge dual $*\eta\in A^{n-k}(V)$ to be the unique antisymmetric tensor of degree $n-k$ satisfying

$\displaystyle\zeta\wedge*\eta=\langle\zeta,\eta\rangle\omega$

for all $\zeta\in A^k(V)$. Notice here that if $\eta$ and $\zeta$ describe parallelepipeds, and any side of $\zeta$ is perpendicular to all the sides of $\eta$, then the projection of $\zeta$ onto the subspace spanned by $\eta$ will have zero volume, and thus $\langle\zeta,\eta\rangle=0$. This is what we expect, for then this side of $\zeta$ must lie within the perpendicular subspace spanned by $*\eta$, and so the wedge $\zeta\wedge*\eta$ should also be zero.

As a particular example, say we have an orthonormal basis $\{e_i\}_{i=1}^n$ of $V$ so that $\omega=e_1\wedge\dots\wedge e_n$. Then given a multi-index $I=(i_1,\dots,i_k)$ the basic wedge $e_I$ gives us the subspace spanned by the vectors $\{e_{i_1},\dots,e_{i_k}\}$. The orthogonal complement is clearly spanned by the remaining basis vectors $\{e_{j_1},\dots,e_{j_{n-k}}\}$, and so $*e_I=\pm e_J$, with the sign depending on whether the list $(i_1,\dots,i_k,j_1,\dots,j_{n-k})$ is an even or an odd permutation of $(1,\dots,n)$.

To be even more explicit, let’s work these out for the cases of dimensions three and four. First off, we have a basis $\{e_1,e_2,e_3\}$. We work out all the duals of basic wedges as follows:

\displaystyle\begin{aligned}*1&=e_1\wedge e_2\wedge e_3\\ *e_1&=e_2\wedge e_3\\ *e_2&=-e_1\wedge e_3=e_3\wedge e_1\\ *e_3&=e_1\wedge e_2\\ *(e_1\wedge e_2)&=e_3\\ *(e_1\wedge e_3)&=-e_2\\ *(e_2\wedge e_3)&=e_1\\ *(e_1\wedge e_2\wedge e_3)&=1\end{aligned}

This reconstructs the correspondence we had last week between basic parallelograms and perpendicular basis vectors. In the four-dimensional case, the basis $\{e_1,e_2,e_3,e_4\}$ leads to the duals

\displaystyle\begin{aligned}*1&=e_1\wedge e_2\wedge e_3\wedge e_4\\ *e_1&=e_2\wedge e_3\wedge e_4\\ *e_2&=-e_1\wedge e_3\wedge e_4\\ *e_3&=e_1\wedge e_2\wedge e_4\\\ *e_4&=-e_1\wedge e_2\wedge e_3\\ *(e_1\wedge e_2)&=e_3\wedge e_4\\ *(e_1\wedge e_3)&=-e_2\wedge e_4\\ *(e_1\wedge e_4)&=e_2\wedge e_3\\ *(e_2\wedge e_3)&=e_1\wedge e_4\\ *(e_2\wedge e_4)&=-e_1\wedge e_3\\ *(e_3\wedge e_4)&=e_1\wedge e_2\\ *(e_1\wedge e_2\wedge e_3)&=e_4\\ *(e_1\wedge e_2\wedge e_4)&=-e_3\\ *(e_1\wedge e_3\wedge e_4)&=e_2\\ *(e_2\wedge e_3\wedge e_4)&=-e_1\\ *(e_1\wedge e_2\wedge e_3\wedge e_4)&=1\end{aligned}

It’s not a difficult exercise to work out the relation $**\eta=(-1)^{k(n-k)}\eta$ for a degree $k$ tensor in an $n$-dimensional space.

November 9, 2009 -

## 6 Comments »

1. Nice!

Maybe it is worth mentioning that you can generalize the Hodge star to the case where the “inner product” is given by an indefinite bilinear form. The result is essentially the same with some minus signs here and there.

Comment by timur | November 9, 2009 | Reply

2. Cool post. I’ll have to go back and scour some of your other posts on this topic.

“Sorry for the delay from last Friday to today, but I was chasing down a good lead.”

Ha. That’s nothing to apologize for. Good luck to you.

Comment by John Moeller | November 9, 2009 | Reply

3. timur, you’re right, but I’m holding that as a future subject for more general geometries. I’m getting basic vector-space geometry down before getting any fancier.

Comment by John Armstrong | November 9, 2009 | Reply

4. […] a basic operation: the cross product of three-dimensional vectors. This only works out because the Hodge star defines an isomorphism from to when . We […]

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5. […] wedge product of differential forms. We have almost everything we need to define an analogue of the Hodge star on differential forms; we just need a particular top — or “volume” — form […]

Pingback by The Hodge Star on Differential Forms « The Unapologetic Mathematician | October 6, 2011 | Reply

6. […] now we can write down the Hodge star in its entirety. And in fact we’ve basically done this way back when we were talking about the Hodge star on a single vector […]

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