OK, folks. In attempting to rise to Bens's challenge (to express the Maxwell field eqns in terms of differential forms), I find I have reached a block (Alpha knows about this - I told him privately I had resolved it, but I now see I hadn't).

On a prosaic level, it boils down to this: Some sources will tell you that a 1-form, for example, is a co-vector; others will tell you it is a co-vector field. So what gives?

Let's ignore the "differential" part of the definitions of p-forms, and state my problem in the simplest, or rather, most naive, possible terms:

Suppose that M is a differentiable manifold, and let p,\; q \in M. Define the tangent spaces T_pM\;T_qM, these being the spaces of all vectors tangent to the specified points on our manifold.

Now note this vital point: Elements (vectors) in T_pM know absolutely nothing about elements in T_qM.

Let's now define the space of all co-tangent vectors at a point as T_p*M etc. We will likewise assume that the vectors (co-vectors, recall) "inhabiting" different co-tangent spaces cannot talk to each other either.

Now, we remind ourselves what a co-tangent vector is; it is a linear functional on a tangent vector that returns a scalar, which may (or may not) be an inner product.

So. The vector space T_p*M is a space of linear functionals. (Notice there is no essential difference between "functional" and a "function" - it's just a fancy way of recording what the arguments are).

These guys, by the definition above, are totally promiscuous. That is, a co-vector, which is just a functional, in T_p*M is as happy to act on a vector in T_pM as it is to act on a vector in T_qM

Likewise, the vector space T_pM is an equal slut - it is quite happy to be acted on by elements of T_p*M,\; T_q*M,\;...; in each of the above scenarios, these unseemly couplings will give birth to a scalar, or rather a tensor of one rank lower than each of the participants.

So, recall that a vector field is simply a choice of a single vector from each of the T_pM, likewise the co-vector field. Then, I think by my reasoning above (if that's what it is), there is no distinction that needs to be made between a 1-form as an element in a vector space, and a 1-form as a vector field. Whether this will apply to 2-forms, 3-forms etc. I strongly doubt

Ughh! This is an ugly way of thinking, and I cannot convince myself it is anywhere near correct, but it is the only way I can think to bring the different definitions of 1-forms into register

This is driving me crazy; can anyone please help?

It is not true that T^*_pM can act on T_qM if p\neq q. If \omega\in T^*M and u\in TM, then the scalar field \omega(u) is obtained by pointwise application of \omega on u, it has nothing to do with coupling between points.

temur: I thank for you response. With respect, I am not sure you saw my problem,

So, although I hadn't consulted Wikipedia when I made my OP, it nicely illustrates the problem, I thinkIn linear algebra, a one-form on a vector space is the same as a linear functional on the space. ......
In differential geometry, a one-form on a differentiable manifold is a smooth section of the cotangent bundle. Explicitly, a one-form on a manifold M is a smooth mapping of the total space of the tangent bundle of M to R whose restriction to each fibre is a linear functional on the tangent space. Symbolically,

\alpha : TM \rightarrow {\mathbf R},\quad \alpha_x = \alpha|_{T_xM}: T_xM\rightarrow {\mathbf R}This is the problem I distilled from my various texts.

Notice that the second mapping, i.e. the restriction of the domain of \alpha in the above necessarily leads to the notion of a co-vector space T_x*M \ni \alpha_x dual to the vector space T_xM - this is familiar enough.

But what of the first mapping in the Wiki entry? All we know is that there is an element \alpha \in T*M that acts on TM.

How can we find a rule that says that, of necessity, only elements in T_x*M can act on elements in T_xM, bearing in mind that T_xM, \quad T_yM,...\subset TM and T_x*M,\quad T_y*M,... \subset T*M? From this formulation, which is widely replicated, the above restriction is, in a sense, "optional".

I cannot see a rule that says this restriction must, and always does, apply. Certainly no text I have seen gives one.

Which was the thrust of my OP

I agree something needs to be clarified, but I am not sure the particular way you are posing the problem really catch the heart of the issue.

The cotangent space T^*_xM is defined to be the dual of T_xM, so without additional structures the elements of T^*_xM cannot act on elements of T_yM if x\neq y. In the Wiki definition it does not say the first map is always a one form, but only those first maps satisfying the property involving the second map.

The cotangent space T^*_xM is defined to be the dual of T_xM, so without additional structures the elements of T^*_xM cannot act on elements of T_yM if x\neq y. Yes, I agree, but this is merely to re-state my problem. In the Wiki definition it does not say the first map is always a one form, .

Oh, but it does, or at least strongly implies it... Explicitly, a one-form on a manifold M is a smooth mapping of the total space of the tangent bundle of M to R

\alpha : TM \rightarrow {\mathbf R}How else would one interpret this? (and don't tell me that the restriction is part of the definition, since this makes no sense; it's like saying "I define a function f:R \to R only when f|_Z: Z \to R". That is, my real-valued function only makes sense if I feed it an integer; no sane person would do this)

I stress that Wikipedia is not one one my usual references, but it does illustrate the ambiguities in the written texts I have here.

Look, I don't care that much, in the sense that, if some physics jock tells me that he always interprets a 1-form as a vector field, and some mathman tells me he always interprets a 1-form as a vector, I can work around it.

I just need to know. But from the above muddled Wiki entry, I don't see how you can have it both ways

...and don't tell me that the restriction is part of the definition, since this makes no sense; it's like saying "I define a function f:R \to R only when f|_Z: Z \to R". That is, my real-valued function only makes sense if I feed it an integer; no sane person would do this)


I am telling you this: the restriction is part of the definition. It's more like saying "I call a function f:R^2 \to R a good function only when f|_{\{x\}\times R}: R \to R is linear for every x\in R".

Doh. I am going to say this before anyone else gets in first:

My stupidity at times is truly breath-taking! Look;

A differential p-form is a smooth section of the union of the p-th exterior power of the co-tangent spaces on a manifold. This union is called the exterior co-tangent bundle and may be written as

\bigcup \nolimits_{x \in M}\Lambda^p(T^*_x)

So the first exterior power of any vector space is defined to be \Lambda^1(V) = V. And obviously, the union V^* \cup V^* = V^*. And, if I allow myself the liberty of using the term "section" for the case that V^* is not a fibre is the cotangent bundle, I find immediately that a "section" of V^* is simply an element in that space, i.e. a linear functional, a co-vector, a dual vector.

So the Wikipedia article was right, temur was right, everybody was right, except me!

*blush*

The other folly I may as well own up to, as long as I am wearing my hair shirt, is my arrogant refusal to look up the original (algebraic) definitions. So, a 1-form in Euclidean 3-space is defined as

\omega = Adx+Bdy+Cdz, where A,\;B,\;C are \mathbb{R}-valued functions of 3 variables. And looking in my high-school text, I find a vector field defined as

F = P\text{i} + Q\text{j} + R\text{k}, where again the coefficients on the unit vectors are scalar functions of 3 variables.

One sees immediately there is a 1-1 correspondence between the two, so one is to some extent at least, entitled to think of an arbitrary 1-form as a bit like a field. So whether it is a field or not is simply a matter of semantics.

Are you referring to the flat and sharp operators?

I don't think so, but I'm not sure. I had rather assumed they had something to do with raising and lowering tensor indices. Like T^{\mu \nu} = g^{\mu \alpha}g^{\nu \beta} T_{\alpha \beta} (summation implied), with the obvious inverse operation.

Which reminds me I wanted to say something about the Hodge (star) operator, a cunning little beast is ever there was one. But for now, I'd better get some work done, or there'll be trouble.......

Anyway, whatever. Let's continue chatting about differential forms, as it seems they are the happening thing in Physics.

Consider the notation. By writing \Lambda^p(V) I am defining a space of p-vectors. What does this notation mean?

Well, let's break it down. \Lambda^1(V_n) says that, given an old-fashioned vector space of dimension n (recall this simply means there are n linearly-independent basis vectors), then I choose 1 at a time of such vectors, and parcel them up to form \Lambda^1(V_n).

Then obviously, there can be no difference between the original V and this parcel; hence the definition \Lambda^1(V_n) = V_n. It is customary to write the dimension of the space of 1-forms as \dim =\begin{pmatrix}n\\ 1\end{pmatrix} =n. Since \Lambda^1(V_n)= V_n, is this just a fancy way of writing \frac{n}{1}? Noooo. We'll see why in a bit, but first........

.....let's work back. \Lambda^0(V_n) means choose no basis vectors from V. So if this is to mean anything at all (and it need not!), it means that this guy has to be a bunch of scalars. And fact, it is so defined.

Now let's go forwards again, and look at \Lambda^2(V_n). This says, given n basis vectors in V, choose 2 at a time from the entire set, and parcel theses choices up as before. So the issue now is: how many such choices are there?

Pause for thought......

That's right, there are \frac{n(n-1)}{2} such choices. Then even I can see that \dim = \begin{pmatrix}n \\ 2\end{pmatrix} is not just \frac{n}{2}.

So in general then, \dim \Lambda^p(V_n) = \frac{n(n-1)}{p} \equiv \begin{pmatrix}n \\ p\end{pmatrix}.

It should be obvious from the foregoing that there can be no creature \Lambda^p(V_n) where p > n. Keep this firmly in mind, as it tells us something rather profound about the algebra of these spaces. Which I shall come to shortly, but now I gotta run

I would have replied earlier but I've spent the last 4 days visiting a few mates and having lots of beer and BBQs.
My stupidity at times is truly breath-taking! Look;I emailed a professor in a US uni to ask for clarification about something to do with Bianchi identities in a paper of his and didn't realise what essentially came down to \epsilon_{133} = 0. Repeated indices on an antisymmetric tensor? Duuuuuhhhh!
I don't think so, but I'm not sure. I had rather assumed they had something to do with raising and lowering tensor indices. Like T^{\mu \nu} = g^{\mu \alpha}g^{\nu \beta} T_{\alpha \beta} (summation implied), with the obvious inverse operation.Yes, the metric maps a space to it's dual (and the inverse metric vice versa). I would imagine that provided the metric is everywhere defined and non-singular then the covariant and contra-variant formulations can be put on equal footings
Well, let's break it down. \Lambda^1(V_n) says that, given an old-fashioned vector space of dimension n (recall this simply means there are n linearly-independent basis vectors), then I choose 1 at a time of such vectors, and parcel them up to form \Lambda^1(V_n). Thats' creepy. I was reading exactly the relevent section of Nakahara yesterday morning!

Close to copy and paste :

An r-form \eta is a real values form acting on vectors, \eta : TM \wedge \ldots \wedge TM \to \mathbb{R}. We can generalise this operator so that we can differentiate the r-form \phi \in \Omega^{r}(P) \otimes V where dim(V) = n, \phi : TP \wedge \ldots \wedge TP \to V. \phi = \sum_{\alpha=1}^{n} \phi^{\alpha} \otimes e_{\alpha} where \{e_{\alpha}\} is a basis for V and \phi^{\alpha} \in \Omega^{r}(P).
So in general then, \dim \Lambda^p(V_n) = \frac{n(n-1)}{p} \equiv \begin{pmatrix}n \\ p\end{pmatrix}.

It should be obvious from the foregoing that there can be no creature \Lambda^p(V_n) where p > n. Keep this firmly in mind, as it tells us something rather profound about the algebra of these spaces. Which I shall come to shortly, but now I gotta runMmm..... rings of forms...[Homer slobering noise]

I would have replied earlier but I've spent the last 4 days visiting a few mates and having lots of beer and BBQs. Oh, the hard life of a student! Dear, dead days.....

Si I have a question, which probably stupid, but, having lost all sense of pride, I'll ask anyway. But to set in context for the general reader, let me offer a correction - well, not really a correction, 'cos I wasn't wrong, more like a clarification.

I stated, correctly, that the dimension of the space \Lambda^2V_n of 2-forms is given by \dim=\begin{pmatrix}n \\ 2\end{pmatrix} = \frac{n(n-1)}{2}. However this fails to reveal something rather splendid.

So let me recast the RHS as \frac{n!}{p!(n-p)!}. Then is is immediately obvious that the dimensions of the spaces \Lambda^pV_n and \Lambda^{n-p}V_n are the same. And we know that vector spaces of equal dimension are isomorphic. In the present case, this isomorphism is given by the Hodge (star) operator. Cool

Let's now turn to the exterior derivative. In a nutshell, this operator, called reasonably enough d is a mapping \Lambda^pV_n \to \Lambda^{p+1}V_n, which when you see it written down, is as easy as pie.

So the question is this. There must surely arise situations where p+1 = q and p= n-q. Is it, in this circumstance, permissible to use d and Hodge interchangeably? Or is the difference between them rather more profound? On the face of it there doesn't seem to be.

Eeek, "on the face of it" is never profound! Ah well.....

I stated, correctly, that the dimension of the space \Lambda^2V_n of 2-forms is given by \dim=\begin{pmatrix}n \\ 2\end{pmatrix} = \frac{n(n-1)}{2}. However this fails to reveal something rather splendid.
You got that the dimensionality of the p-forms is \begin{pmatrix}n \\ p\end{pmatrix}. And you got that you only have p in 0 to n. But what's \sum_{p=0}^{n}\begin{pmatrix}n \\ p\end{pmatrix}? 2^{n}

So if you take the union of all the various p-forms, is \Lambda^{\ast}(T^{\ast}M) = \bigcup \Lambda^{p}(T^{\ast}M) , then you have that \textrm{dim}\left(\Lambda^{\ast}(T^{\ast}M)\right) = 2^{n} .

And for an added bonus, d, the exterior derivative is now now an endomorphism on \Lambda^{\ast}(T^{\ast}M). And it's always nicer to have an operator which maps a space to itself.
So the question is this. There must surely arise situations where p+1 = q and p= n-q. Is it, in this circumstance, permissible to use d and Hodge interchangeably? Or is the difference between them rather more profound? On the face of it there doesn't seem to be.Do you mean "Is [d,\ast]=0?" or do you mean "Is \ast = d?"

The former can be true in special circumstances but not in general, even in particular dimensions. The former isn't. If the dimensionality of the spaces and forms are just right, you can construct a notion where d or \ast are endomorphisms. For instance, if dim = n = 2m, then \ast is an automorphism on any m-forms. Infact, since \ast^{2} \propto 1 and if \ast : A \to A, you should be able to use it to split the m-forms into two seperate spaces depending on the eigenvalues of \ast's eigenforms.

It is true that vector spaces of equal dimension are isomorphic, but there is no natural isomorphism that cries out to be "the one" unless there is an additional structure. So the Hodge star operator is something you choose (e.g. there can be many Hodge star oeprators on the same manifold), or induced by something extra such as a Riemannian metric. But the exterior derivative is defined without any additional structure.

Besides this, the Hodge star operator is an isomorphism, while the exterior derivative has a kernel in general.

Right, you two, I agree. An equally compelling argument (in your favours) would be that the Hodge operator depends upon an inner product, whereas the exterior derivative doesn't.

Ah well, I was just flying a kite.

I realized last night that one can do some interesting things rather simply with the exterior derivative. I was also told that, if you put this together with the Hodge, then things get super-dooper amazing. I don't fully understand it yet, but will persevere.

Using inner products and d makes for some awesome stuff, because you can then define d* (or d^{\dag}) such that if you have some appropriatedly defined inner product ( , ) then you can define d* by ( da , b ) = ( a , d*b ).

Doing some integration by parts and what not you find that if d : \Lambda^{p}(TM) \to \Lambda^{p+1}(TM) then, so d d^{\dag} : \Lambda^{p}(TM) \to \Lambda^{p}(TM) and d^{\dag} d : \Lambda^{p}(TM) \to \Lambda^{p}(TM). In general [d,d^{\dag}] \not =0, but that only sweetens things.

For instance, dd*+d*d is (barring a few details you can find in the opening chapters of Joyce's 'Riemannian Holonomy Groups and Calibrate Geometry' or someone's 'Notes on Seiberg-Witten Equations') the Laplacian, if we're working on a Euclidean system. If it's Lorentzian, then it's the wave operator!

And in true Home Shopping Channel style, but wait, there's more!. We know that dd = 0 and so d*d*=0. So dd^{\dag}+d^{\dag}d = (d+d^{\dag})^{2}. And so we have a square root to the wave operator. Precisely what Dirac needed to construct his Dirac equation for spinors!

So d+d* is the Dirac operator and since d is naturally covariant, so is d*. Everything is very nicely expressed in terms of differential forms!

Aren't the ranges of d and d* are different? How do you add them?

You work in the ring of forms. That's why I mentioned constructing the ring via the union of the various p-form spaces. Then you can consider d an endomorphism, not a map between two inequivalent spaces. Then you can add d and d* in a consistent way, since they both map \Lambda^{\ast} \to \Lambda^{\ast}

From this you can do great things like show any harmonic form is in closed.

An harmonic form is a zero of the Laplacian, so H\alpha = (dd^{\dag} + d^{\dag}d)\alpha= 0. So (H\alpha,\alpha) = (dd^{\dag}\alpha,\alpha) + (d^{\dag}d\alpha,\alpha) = (d^{\dag}\alpha,d^{\dag}\alpha) + (d\alpha,d\alpha) = 0

But by virtue of your inner product (d^{\dag}\alpha,d^{\dag}\alpha) \geq 0 and (d\alpha,d\alpha) \geq 0 so d^{\dag}\alpha = d\alpha = 0

So if \alpha is harmonic, \alpha is closed and co-closed.

Yes, differential geometry truely rocks.

Cool! Can you give a hint to how to connect d+d* to spinors?

Oh my word, this stuff is such fun! Don't you just love it when a whole area of mathematics opens its legs for you (in a crude manner speaking)?

So. Given a 0-form in \mathbb{R}^3,\; f(x,y,z) one easily sees that df = \nabla f, the gradient.

{Edit:For what follows, it might be instructive for the general reader (assuming there are any!) if I ran through this,

Given f(x,y,z) it is obvious we need 3 partial differential equations, and in this simple case, we know we must add them together. So one has, from intermediate calculus, that df = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy +\frac{\partial f}{\partial z}dz = \nabla f. This is the definition of the gradient 1-form, often, not quite accurately referred to as a vector}

In like fashion one finds that, for the 1-form \omega, that d\omega is the curl of a vector field, and that, given a 2-form \alpha that d\alpha is the divergence.

This is simply a matter of working the definition d: \Lambda^p(V_n) \to \Lambda^{p+1}(V_n)

The familiar identities curl (grad) = div (curl) = 0 follow instantly from the Lemma of Poincaré; d^2 =0. (Incidentally, when I was first offered this without proof, I assumed someone was off their head, and I was pretty sure it wasn't me. But of course, by setting, say d(df)=0 one sees easily that this is true since second order mixed partial derivatives commute!)

The Laplacian in \mathbb{R}^3 is, if anything, even lovelier. Let the Hodge operator be defined as \ast: \Lambda^p(V_n) \to \Lambda^{n-p}(V_n). Then in \mathbb{R}^3, if f \in \Lambda^0(V_n), then df=\frac{\partial f}{\partial x}dx+ \frac{\partial f}{\partial y}dy+ \frac{\partial f}{\partial z}dz, a 1-form.

Then \ast df =\frac{\partial f}{\partial x}dy \wedge dz+ \frac{\partial f}{\partial y}dz \wedge dy+ \frac{\partial f}{\partial z}dx \wedge dy, a 2-form (3 - 1 = 2!)

Then applying "d" again I find d \ast df = (\frac{\partial^2 f}{\partial x^2}+ \frac{\partial^2 f}{\partial y^2}+ \frac{\partial^2 f}{\partial z^2})dx\wedge dy \wedge dz = \nabl^2 f\;dx \wedge dy \wedge dz, a 3-form.

Since we are in 3-space, the final application of Hodge brings us back to a n - p = 3 - 3 =0-form space, whereby \ast d \ast df = \nabl^2 f. But now that I know all this, I don't need all those unwieldy partials!

Now how sweet is that!! Don't you just love it?

Temur et al: I found a paper with an appendix for spinor reps in Minkowski space, might be a clue?:
http://theses.ulb.ac.be/ETD-db/collection/available/ULBetd-09162004-120651/unrestricted/these.pdf

Thanks! looks like a nice thesis.

Right, I know what an algebra is, but put "Clifford" in front of it, and I'm lost.

The appendix isn't much to do with what you're chasing here though, is it? No differential forms, just matrices and algebras? Does it sort of show a derivation of the Dirac matrices for spinor reps, from general SO(n) "first principles"?