What is it?

http://en.wikipedia.org/wiki/Hodge_dual

might help, though i haven't a clue what it means.

:-)

any two vector spaces which have the same dimension are isomorphic. To see this, just choose bases for the two vector spaces, say, {e_i} for V₁ and {f_i} for V₂. Then define your linear operator by its action on each basis vector: let T(e_i)=f_i

then it's easy to see that T, expressed in these bases is simply the identity matrix, which is invertible, so therefore T is an isomorphism.

the problem is that these choices of basis are completely arbitrary. If you wanted to use this isomorphism to do some calculation, then the result of the calculation would depend on your choice of basis, and you have no way of nailing down that freedom to get a single answer for your calculation.

one says that this isomorphism is not <b>canonical</b>

Now sometimes, when the vector spaces have extra structure, this structure allows you to to choose a single isomorphism, without the ambiguity of choosing a basis.

for example, the vector space V and its dual space V* have the same dimension, and so are isomorphic (well, if they are finite dimensional). If V has a bilinear form <-,->, then we can create an operator that takes v \in V to <-,v> in V*. This operator is linear. If the bilinear form is nondegenerate, then this operator is invertible, and therefore an isomorphism.

Notice that we could have used the more concrete and easier to understand operation of declaring our isomorphism to be that operator which takes the vector defined by components vⁱ to the covector defined by components v_i. however, this isomorphism makes use of a basis, so it's not readily apparent that this isomorphism is independent of basis (i.e. canonical). it actually is. it's the same isomorphism as the basis independent formula above. but with this formula we do have to check that it is basis independent.

This isomorphism is what is going on when you raise and lower indices in relativity, or when you take two kets in QM, turn one of them into a bra, and sandwich them together. Or when you take the Poisson bracket of a function in classical mechanics.

so anyway, with any vector space, we can form the exterior algebra (if the vector space in question is the cotangent space of an oriented (pseudo-)Riemannian manifold, then these are called differential forms). The k^th exterior power of V has dimension (n choose k) = n!/k!(n-k)! = (n choose n-k)

so the k^th exterior power and the n-k^th exterior power of V both have the same dimension and are therefore isomorphic. if we want to find a canonical isomorphism, then we will need to give V an inner product and an orientation.

so what is the canonical isomorphism? well it is given by that formula in the link that Boris2 posted: <-/\*v,vol>=<-,v>

with that definition it's a little hard to see what the hodge star does, but it is easy to see that the definition is basis independent, that it depends on the inner product and the volume form (which depends on the orientation).

there is also the easier to understand, but not basis independent definition, which is also in that link: *(e₁/\.../\e_k)=e_k+1/\.../\e_n

with this definition, it is hard to see that the Hodge dual is independent of basis, and that it depends on the inner product and the orientation, but it is easy to actually calculate the dual of some vector.

The Hodge dual is used in Maxwell's equations dF=0 and *d*F=j. you know that Maxwell's equations have to depend on the metric, because they have to be able to give you Coulomb's Law, for example, and Coulomb's Law depends on distance squared. the metric gives you the distance. The Hodge dual is also used to change Stoke's theorem in its simple differential forms language into the divergence theorem and curl theorem that you know from vector calculus. The Laplacian operator is built with the Hodge star. The Hodge dual is also crucial to the Hodge decomposition theorem, which is a pretty useful theorem. Also, in Yang-Mills theory in 4 dimensions, it's possible to have *F=&plusmn;F in euclidean spacetime or *F=&plusmn;iF in Minkowski spacetime. these (anti-)selfdual fields are local minima of the action, and are called instantons.

In the case of the electromagnetic field, or the Yang-Mills field (only in four dimensional spacetime, though) there is a very physical way to understand the Hodge dual. It exchanges the electric field for the magnetic field, and vice versa.

You had to ask, didn't you, oxymoron. ;)

I must say, most of that went over my head.

Forget university... lethe's-lectures here I come.

Okay, when and in which course did you learn all that!? I'd like to know if you don't mind.

Another question:

What does invariant mean? Putting this into context; I am trying to understand the invariant nature of Maxwell's equations (the symmetrised Maxwell's equations - actually you might want to explain this as well! I don't fully understand the difference between the ordinary Maxwell equations (the ones which say that the curl of E is the negative time rate of change of the magnetic field etc...) to the Dirac symmetrized Maxwell's equations.

Well, to be invariant under some transformation means basically that if you apply some transformation, say translation, to something that it does not change. For example, if you add to the magnetic vector potential the gradient of a scalar function, the magnetic field is said to be invariant under that gauge transformation. Also, the eigenvalues of a linear operator is invariant under a change of basis (unitary transformation), however the eigenvectors themselves are not, you would have to determine the eigenvectors in the new basis. It means exactly what it sounds like, you just have to specify how something is invariant.

Forget university... lethe's-lectures here I come.

Okay, when and in which course did you learn all that!? I'd like to know if you don't mind.

Hmm... well I would expect a good course in differential topology to cover this stuff, only I took a course in differential topology, and it didn't. I was actually a little annoyed with that course.

I wouldn't expect to learn the Hodge dual in a physics course. Well, i mean, in some QFT books, they introduce the Hodge dual of F_&mu;&nu; in Minkoswki space, which they define to be &epsilon;_{&mu;&nu;&rho;&sigma;}F^&rho;&sigma;. But that is the most you will ever see of the Hodge dual in any physics book that i know of.

so, i never learned this in a course. I learned in through self-study.


Another question:

What does invariant mean? Putting this into context; I am trying to understand the invariant nature of Maxwell's equations (the symmetrised Maxwell's equations - actually you might want to explain this as well! I don't fully understand the difference between the ordinary Maxwell equations (the ones which say that the curl of E is the negative time rate of change of the magnetic field etc...) to the Dirac symmetrized Maxwell's equations.
By Dirac symmetrized Maxwell's equations, you mean Maxwell's equations with a monopole term, right?

Well, if you look at Maxwell's equations in vacuo (i.e. with no matter present), then they are symmetric under Hodge duality, right? dF=0 and d*F=0. one nice thing about these symmetric forms is that if you do find a self-dual (or anti-self-dual) solution, then it is automatically a solution to Maxwell's equations.

anyway, i'm not sure i know what your question means. invariant means transforms under the trivial representation of some group, i.e. means doesn't change. Maxwell's equations are <i>covariant</i> under the Poincaré group, not <i>invariant</i>. this means that both sides of the equation transform in the same representation of the Poincaré group. Like, F is a 2-form, which means it transforms in an irreducible tensor product representation. *F is also a 2-form, d*F is a 3-form, and *d*F is a 1-form. therefore, as long as j is also a 1-form, then *d*F=j is a Lorentz covariant equation. But neither side of the equation is <i>invariant</i>.

Perhaps what you mean is that the equations are invariant under Hodge duality? Only i don't think that is true (unless for some reason the magnetic current were dual to the electric current...)