# ...Prove Their Worth...

"Problems worthy of attack
prove their worth
by hitting back." - Piet Hein

A kind of running diary and rambling pieces on my struggles with assorted books, classes, and other things, as they happen. You must be pretty bored to be reading this...

## Thursday, August 29, 2002

Ah, the Hodge star operator. What did I ever do without it? Not much! I promised to write about it a long, long time ago, when I was beating my head against a wall trying to grasp it, but haven't gotten around to it until now. Mea culpa.

So what is it? Well, it took me quite a few days to figure it out, and I'm going to try to give a brief overview of my picture of it at this point. First, we'll need the concept of differential forms. If you don't know what those are, none of this will probably make sense. Briefly (I talk about this more in previous posts), a '1-form' is just something that eats a vector field and craps out a number. A 1-form looks like this: df. If you have a function f, you can take its exterior derivative, and get an object called df, naturally enough. On R^n, this is df = [Partial derivative with respect to u] * f * dx^u, where Einstein summation convention is used, and the dx^u's are basis 1-forms on R^n.

Anyway, we've got 1-forms. If you take the exterior derivative of a 1-form, you get a 2-form, and so on. Also, there's something called a wedge product that you can do to forms (and vectors, too!). It's meant to be a generalization of the idea of a cross product of vectors. This just has to satisfy u /\ v = -v /\ u, and u /\ v has to form 'an algebra'. But never mind the technical details! The point is, if you wedge two 1-forms together, you can get a two-form, and so on.

(By the way, 2-forms look like so: dx/\dy (i.e.,two one-forms wedged together, as I said above), and say 3-forms look like dx /\ dy /\ dz.)

Now, if you take a 1-form, and take its exterior derivative, you get a 2-form, as I said above. But, if you actually try this on R^3 (meaning, good ol' space), you get something that looks a hell of a lot like a curl! Similarly, if you wedge two one forms together on R^3, you get a 2-form, you get something that closely resembles a cross product!

Except it doesn't look exactly the same. The difference is, first of all, is that when people say 'curl' and 'cross product', they are talking about vectors or vector fields, not 1-forms. But this is actually not a complete disaster, because if the manifold we're playing on is equipped with a metric, we can use that to make 1-forms into vector fields, and so everything's cool. Except it's not.

The problem is that, as I noted above, we're not dealing with 1-forms, we're dealing with 2-forms. And those aren't the same as one-forms, and they most definetly aren't the same as vectors! (Or we could be dealing with '2-vectors' - meaning two vectors wedged together, but the point is the same: the result of the wedge product of two vectors ain't a vector, it's a 2-vector.)

So, we need a way to convert two-forms to one-forms, and vice versa. This is where the Hodge star operator comes in. Visually (on R^3), a 2-form can (with caveats) be pictured as a little 'area'. That is, say you picture a 1-form as a little arrow (huge caveats here), and you put two of them end-to-end, they form part of a parallelogram. That parallelogram 'is' a 2-form. The Hodge star operator will convert that area into an arrow, which will be perpendicular to the two other arrows.

The full-blown definition goes something like this. A Hodge star operator, denoted by *, on an n-dimensional manifold M equipped with with a metric and an orientation, is a linear operator from p-forms to (n-p)-forms on M. (Example: Make M = R^3. Then n=3, and if m is say 2, then *: 2-forms -> (3-2)-forms = 1-forms.)

The Hodge * operator has to satisfy v /\ *u = vol, where v and u are p-forms on M, *u is the 'Hodge dual' of u, that is, a (n-p) form, is the inner product of v and u (you need a metric to define it), and vol is a volume form on M. Whew.

A 'metric' is just a way of taking 'inner products' (aka 'dot products'). Given two vectors (or other things), a metric will spit out a number. Metrics have many, many uses! You need a metric to define distances, and to define volumes, and areas, and angles, and other things. For instance, if you want to be able to tell if two things are 'perpendicular', you need to define 'perpendicular' first, and that calls for a metric! (If the inner-product of two unit vectors is 1, they're perpendicular - remember, 'inner-product' is just a fancy way of saying 'dot product'.)

A 'volume form' is a funny thing. On an n-dimensional manifold M, a volume form is a 'nowhere vanishing n-form'. But what it really is is a way to measure volumes! Or at least, that's one use. On R^3, an example of a volume form would be dx/\dy/\dz. You'd integrate it 'over M' to get the volume of M, hence the name. There's another issue hiding here, and that's 'orientations'. Turns out, to properly talk about a volume form, you need an 'orientation'. That's just a way of telling between left and right, really. The 'standard volume form' on R^3 is dx/\dy/\dz. dy/\dx/\dz is also a volume form, and it's just as good, but it's in the opposite orientation from the conventional standard, as it turns out. The way to tell if something's in the standard orientation is to see if it's an even or odd permutation away from the standard. (This is actually a theorem, but a very handy one.) For example, on R^3, we have dx/\dy/\dz. Let's relabel that as dx^1/\dx^2/\dx^3. The 'standard' is then (1, 2, 3). An odd permutation of that, for instance (1,2,3) -> (2,1,3) -> (2,3,1) -> (3,2,1), where we had to flip two numbers three times, an odd number, gives a 'negative' orientation, dx^3/\dx^2/\dx^1. If we were to do an even number of 'flips', we'd get a positive permutation - meaning, it gives a 'right and left' the same as the standard, instead of the opposite way.

Anyway, I won't go into actually calculating Hodge duals, because that's generally something of a pain in the ass, but I'll give a few examples: *dx = dy/\dz, *dy = dz/\dx, *(dx/\dy) = dz, *(dy/\dx) = -dz, etc.

So the Hodge star operator can turn 2-forms into 1-forms. (And 3-forms into 0-forms, which are just functions, and vice versa.) This is what we needed! Now we can take the wedge product of two 1-forms, apply the Hodge star, and get a 1-form. Or take the exterior derivative of a 1-form, and apply the Hodge star. And if we do that, then what only looked sort of like curls and cross products will be curls and cross products!

What's more, the exterior derivative of two-forms on R^3 looks like the divergence of a vector field, except it's a 3-form. Applying the Hodge star, we actually get something that is the divergence. And it turns out that the exterior derivative of a function is just a way of talking about its gradient without using a dot product. So, first of all, it turns out that on R^3, the exterior derivative is a grand generalization of gradients, curls, and divergences, and second, hodge stars let us pretend two different kinds of forms are 'the same'. (Well, ok, we need a metric in there to get from forms to vectors, but whatever - we need a metric to define the Hodge star operator anyway.)

Now, as I've briefly mentioned a few times, a wedge product is a generalization of a cross product, but to actually get a 'cross product', we need a Hodge star, and for that, we need a metric and an orientation on our manifold, in the case of the cross product, R^3. There's actually a nice visual reason why we need a metric and an orientation to define a cross product.

People often say 'a cross product of two vectors is a vector perpendicular to both of the input vectors, and its direction is given by the right hand rule'. There's a lot hiding under the bonnet here! First, you can only claim the result of a cross product of two vectors is a vector, and not some kind of '2-vector' in R^3. See, it turns out that the 'algebra' formed by the wedge products on an n-dimensional vector space V, called "/\^pV", and it has n!/[p!(n-p)!] dimensions.

The thing of it is, if V is 3-dimensional, that is, n = 3, then /\^2V (the 'algebra of two-forms', if you will) has dimension 3!/2!1! = 6/2 = 3, which is again 3-dimensional. So we can, if we want, pretend the wedge product of two vectors is a vector, instead of thinking of it as it really is, which is a 2-vector - they have the same number of dimensions, and using a star operator, we can turn one into the other and forget we ever used it. But this only works for n=3! 3D is special that way!

Anyways, here's where the visual stuff comes in. The wedge product of two vectors in 3-space 'looks' like a parallelogram formed from the two vectors. Now, if we want to change that to look like a cross product, we'll need to somehow change that parallelogram into a vector, and what's more, it'll need to be perpendicular to both of our original vectors. Not only that, but of course there's two such perpendicular vectors in 3-space (pointing in opposite directions to each other), and we need to pick one.

Well, we need a hodge star to do all that. The metric comes in getting things 'perpendicular', and a choice of orientation comes in to pick which of the two possible vectors we'll call 'the' cross product.

Whew. My fingers are tired, I'm confused, and so on. Watch out: having just proof-(ha!)-read over this, I think I made a muddle of distinguishing 1-forms and vectors. Untangling that mess is left as an exercise for the sadomasochistic reader. If you want, you can make this resemble what I 'promised' I'd deliver yesterday, by adding a supply of your own sex, violence, and bridges. Also Shakespeare.

Disclaimer: This is from memory, muddled, confused, and unlikely to be particularly educational (or entertaining, but hey, one never knows.) It's likely riddled with errors, too, both intentional (where I fudged stuff to avoid explaning tricky stuff) or unintentional (where I either just don't get it, or simply screwed up.) If you want the real thing, check out Baez and Munian's book, which is excellent, has far more details, and is very clear.