Various problems concerning Exterior calculus

[Oxymoron]

Im currently learning some Exterior calculus which Im hoping will help me with my honours project.

The problem I am stuck at is the following.

\mbox{Show that } i_X\ast \omega = \ast(\omega \wedge X^{\flat})

where X^{\flat} is the one-form related to the vector field X by the metric, and \omega is some p-form. Also, i_X is meant to be the interior derivative. Note: this is not a homework problem, but intended as a discussion thread on the concepts concerned with problems of this type. In effect, such a thread could help me understand more about what is going on so that I may be able to show what I have written.


But the main reason for posting here is that I would like to understand how I can use Hodge star operators, exterior derivatives, and musical isomorphisms to define an operation which is identical to all the classical vector calculus operations, and in particular curl. I believe it helps me more to discuss with other people.

Any discussion on any of the material that I have mentioned would be greatly recieved.

Cheers.

[Oxymoron]

I suppose I should start with the following question:

If I have a vector field X, then what am I doing to it by 'flatting' it, X^{\flat}?

[Oxymoron]

I just read that for a semi-Riemannian manifold M, there is a canonical \mathcal{F}(M)-linear isomorphism between tensor fields of the type (r,s) and of the type (r-1,s+1),

T^{(r,s)} \cong T^{(r-1,s+1)}..

If we take some tensor field A \in T^{(r,s)}(M) then the value of A^{\flat} \in T^{(r-1,s+1)}(M) on arbitrary one-forms and vector fields is define by

A^{\flat}(\theta^1,\dots,\theta^{r-1},X_1,\dots,X_{s+1}) = A(\theta^1,\dots,X^*_b,\dots,\theta^{r-1},X_1,\dots,X_{b-1},X_{b+1},\dots,X_{s+1})

where X^*_b is the one-form metrically equivalent to X_b.

So if I have a tensor, say

R(X,Y,\omega)

which acts on, say, 2 vector fields and 1 one-form such as in this case. I could also say

R(X,Y,Z^{\flat})

since I cannot insert a vector field into the third slot because the third slot is reserved for forms. So I can simply flat a vector field and insert it.

I think this is also called type-changing.

[Cexy]

But the main reason for posting here is that I would like to understand how I can use Hodge star operators, exterior derivatives, and musical isomorphisms to define an operation which is identical to all the classical vector calculus operations, and in particular curl. I believe it helps me more to discuss with other people.
If you have a one-form \omega then the components of the exterior derivative acting on it are

(d\omega)_{ab} = \partial_{[a}\omega_{b]}

So for example, the 12 component is \partial_1\omega_2 - \partial_2\omega_1. This looks uncannily similar to the curl. If we then use the Hodge star operator, assuming that we have a Euclidean metric, then we send the 12 component to the 3 component etc, and so we can write the curl as *d\omega.

Similarly we can write the divergence of a one-form as *d*\omega, and we can write the gradient of a scalar field \phi (a function) as d\phi.

When we're using a Euclidean metric we don't have to differentiate between forms and vectors, because the metric is g_{ab}=\delta_{ab}. So anything that works for forms also works for vectors. This is why you'll often see all indices written as 'down' indices when you're working in Euclidean space.

This is only a physicist's view, not a mathematician's, so I've probably missed some detail ;)

[Oxymoron]

Posted by Cexy

This is only a physicist's view, not a mathematician's, so I've probably missed some detail ;)

What you wrote is interesting actually.

You said that the gradient of a scalar field can be written as

\mbox{grad}\phi = d\phi

but the exterior derivative d can only act on forms can they not? So we have a useful operator on forms which corresponds to the grad operator in vector calculus. So Im guessing we can also define a similar operation on vectors by simply sharping the whole thing:

\mbox{grad}V = (d\phi)^{\sharp}

And you also wrote that the divergence of a vector field V is

\mbox{div} V = \ast d \ast \omega^{\flat}

where \omega is a one-form. Note: the one form in this definition is flatted because we want the divergence of a vector field. (Is this ok?) In three dimensions, the index of the one-form is 1. Then we star it and get 3-1 = 2 index. Then ext. diff. and we get (3-1)+1 = 3. Then star it again and get and index of 3-((3-1)+1) = 0. This tells us that the divergence of a one-form gives a zero-form, or a scalar.

So with the definition of curl, we begin with a one-form. Ext. diff. it and get 1+1 =2. And then star it and get 3 - (1+1) = 1. This tells us that the curl operator on a one-form returns a one-form. Which is all good. :)

[George Jones]

If I have a vector field X, then what am I doing to it by 'flatting' it,

Yikes, I can't believe I'm doing this. I must truly be addicted to physics forums. I probably won't be able to make any more replies for at least another 48 hours.

Let X be a vector field. Use the metric g to define the covector field (1-form) X^{\flat} associated with X: for every vector field Y

X^{\flat} \left( Y \right) := g \left( X , Y \right).

This is the abstract version of index lowering for physicists. To see this, let \left\{ e_{1}, \dots , e_{n} \right\} be a set of basis vector fields, and let \left\{ \omega^{1}, \dots , \omega^{n} \right\} be the associated dual basis of 1-forms. Write g_{ij} = g \left( e_{i} , e_{j} \right).

Write X in terms of the basis vector fields,

X = X^{i} e_{i},

and X^{\flat} in terms of the basis 1-forms,

X^{\flat} = X_{i} \omega^{i},.

Then,

X^{\flat} \left( e_{j} \right) = X_{i} \omega^{i} \left( e_{j} \right) = X_{i} \delta_{j}^{i} = X_{j}.

But, by definition,

X^{\flat} \left( e_{j} \right) = g \left( X , e_{j} \right) = g \left( X^{i} e_{i} , e_{j} \right) = X^{i} g \left( e_{i} , e_{j} \right) = X^{i} g_{ij}.

Combining these results gives

X_{j} = X^{i} g_{ij}.

This is the reason for the flat notation. Just as flatting a musical note lowers it by a semitone, flatting here lowers indices

Regards,
George

[Oxymoron]

Ok, so I must interpret the RHS of my very first equation in post #1 as the wedge product of a 1-form and a 1-form associated with X. So its a wedge product of two 1-forms?

[garrett]

Hi Oxy, I'm pretty sure you're going to have to use the definition of the Hodge dual in terms of components and the permutation symbol in order to prove your equation (1). It's a bit of a mess.

[Doodle Bob]

Hi Oxy, I'm pretty sure you're going to have to use the definition of the Hodge dual in terms of components and the permutation symbol in order to prove your equation (1). It's a bit of a mess.

I'm not quite sure about that. Due to multilinearity, we can assme that omega is an elementary p-form, i.e. \omega = r{e_1}^*\wedge\dots\wedge{e_p}^*, where {e1,...,en} is an ordered, orthnormal basis of V and \{{e_1}^*,\dots,{e_p}^*\} is the dual basis. r is some real number.

Then there are two cases either \omega(X)=0 or not. Either case allows us to assume that X is in \{{e_1},\dots,{e_p}\}
or in \{{e_{p+1}},\dots,{e_n}\}.

Note *\omega = r^{-1}{e_{p+1}}^*\wedge\dots\wedge{e_n}^*, and similarly for others...

This should cut down the messiness considerably.

[Oxymoron]

This is my understanding of the Hodge star operator, it may be flawed:

Let \Omega^p(M) denote the space of p-forms on a manifold M. We can turn this space into a graded algebra by incorporating the wedge operation (or exterior product):

\wedge\,:\,\Omega^p(M) \times \Omega^q(M) \rightarrow \Omega^{p+q}(M)

which is an operation that takes a p-form \omegaand a q-form \phi and gives a p+q form \omega \wedge \phi. We also have

\omega \wedge \phi = (-1)^{pq}\phi\wedge\omega

The interior derivative encompasses the process of contraction and anti-symmetrizing. That is, take a p-form (which is a map on a set of p vector fields to give a function), then contract on one vector field, and antisymmetrize.

If X is a vector field, then the interior derivative with respect to X is

i_X \,:\,\Omega^p \rightarrow \Omega^{p-1}

where p-forms \omega get mapped to i_X\omega.

The interior derivative of a wedge product of p and q forms is

i_X(\omega\wedge\phi) = i_X\omega \wedge\phi + (-1)^p\omega\wedge i_X \phi

and we also have the following property

i_X\circ i_X = 0


The vector space \Omega^p(M) is isomorphic to \Omega^{n-p} with no natural isomorphism. By establishing a metric on the manifold we unearth a preferred isomorphism called the Hodge dual operator.

So take a manifold M which has a metric g on it. Firstly, we have

g = g_{ab}\theta^a\otimes \theta^b

which means that a covariant metric tensor gives us a contravariant metric tensor. So we also have

g_{(1)}=g^{ab}X_a\otimes X_b

where g^{ab}g_{ab} = \delta^a_c, and g_{(1)} is a metric on 1-forms. We can extend this to a metric on p-forms, g_{(p)}:

g_{(p)}(\alpha_1\wedge\alpha_2\wedge\dots\wedge\al pha_p,\beta_1\wedge\beta_2\wedge\dots\wedge\beta_p ) = \det[g_{(1)}(\alpha_i,\beta_i)]

that is, it is the determinant of the matrix of scalar products of the 1-form factors.

The Hodge star is an operator which takes p-forms to (n-p)-forms, or we could say

\omega \mapsto \ast\omega

Notice that \ast\omega is a unique form such that if we act on it my some other p-form, \phi \in \Omega^p(M) we get

\phi\wedge\ast\omega = g_{(p)}(\phi,\omega)\Omega(M)

[Oxymoron]

So if I have a p-form \omega which can be written as

\omega = re^*_1\wedge \dots \wedge e^*_p

where r is a real number, and e* are dual to the orthonormal basis e.
Hodge starring w gives an n-p vector where n is the dimension of the manifold.

So an arbitrary element of the manifold can be written in terms of n basis elements with metric. Then *w has n-p elements:

\ast\omega = re^*_{p+1} \wedge \dots \wedge e^*_n

which is exactly what DB wrote. Now the interior derivative takes a p-form and gives a p-1 form, so

i_X\ast\omega = re^*_{p+1}\wedge\dots\wedge e^*_{n-1}


So if I can show that this equals the RHS of (1) then I'd have shown it?

[Doodle Bob]

Now the interior derivative takes a p-form and gives a p-1 form, so

i_X\ast\omega = re^*_{p+1}\wedge\dots\wedge e^*_{n-1}


So if I can show that this equals the RHS of (1) then I'd have shown it?

The above is assuming that X={e_n}, i.e. \iota_X \omega=0. There is a seperate case if \iota_X\omega\neq0. In this case, you can assume that X={e_1}.

Otherwise you seem to have the gist of it.

N.B. I originally had r^{-1} as the coefficient for *w, but that's not right. Plus I forgot that w is not necessarily a 1-form, so I've changed \omega(X) to \iota_X \omega

[Cexy]

What you wrote is interesting actually.

You said that the gradient of a scalar field can be written as

\mbox{grad}\phi = d\phi

but the exterior derivative d can only act on forms can they not?
What I was trying to get across is that when we're working with the Euclidean metric, it doesn't matter whether we're talking about forms or vectors. Because the Euclidean metric satisfies g_{ab}=1 for a=b and g_{ab}=0 otherwise, the i component of a flattened vector (or vector field) has the same value as the i component as the vector itself - in physics speak, it doesn't matter whether the index is up or down.

Something else that you might like to ponder is what the operator d*d acting on a scalar field corresponds to, in the cases where we have a Euclidean metric and a Lorentzian metric (one negative component in diagonal form, and the others positive).

[Oxymoron]

So I have on the RHS

\ast(\omega\wedge X^{\flat})

and I have to show that this equals

re^*_{p+1} \wedge \dots \wedge e^*_{n-1}

in terms of components.

So \omega = re^*_1\wedge \dots \wedge e^*_p

Then (\omega\wedge X^{\flat}) = re^*_1 \wedge \dots \wedge e^*_{p+1}

Then \ast(\omega\wedge X^{\flat}) = re^*_{p+2} \wedge \dots \wedge e^*_n

Can I simply subtract 1 from the indices and claim this equals

re^*_{p+1} \wedge \dots \wedge e^*_{n-1}?

[Doodle Bob]

So I have on the RHS

\ast(\omega\wedge X^{\flat})

and I have to show that this equals

re^*_{p+1} \wedge \dots \wedge e^*_{n-1}

in terms of components.

So \omega = re^*_1\wedge \dots \wedge e^*_p

Then (\omega\wedge X^{\flat}) = re^*_1 \wedge \dots \wedge e^*_{p+1}

Then \ast(\omega\wedge X^{\flat}) = re^*_{p+2} \wedge \dots \wedge e^*_n

Can I simply subtract 1 from the indices and claim this equals

re^*_{p+1} \wedge \dots \wedge e^*_{n-1}?

Certainly not. The problem is that you're confusing what your assumption about X is. On the RHS, you have X=e_n, but in the work above you are assuming that X=e_{p+1}.

[Oxymoron]

Posted by Doodle Bob

Certainly not.

You're right. It should be

(\omega\wedge X^{\flat}) = e_1 \wedge \dots \wedge e_p \wedge e_1 \wedge \dots \wedge e_{p-1}

Actually I dont like this either. I missing bits. Should it be...

(\omega\wedge X^{\flat}) = \omega_{i_1\dots i_p}X^{_{i_1\dots i_p}}\mbox{d}x^{i_1}\wedge\dots\wedge\mbox{d}x^{i_ p}\wedge\mbox{d}x^{i_1}\wedge\dots\wedge\mbox{d}x^ {i_{p-1}]

But how on Earth am i going to get that to be a rank (n-p-1) form? Which is the case on the LHS.

[Doodle Bob]

no that too is incorrect. X is a vector, so X-flat will be a 1-form. I think you have forgotten the whole strategy here: first assume we can find a basis e={e1 ... en} such that X is in e and w is a simple p-form with respect to e.
The general case will follow from multi-linearity of both sides of the equation.

[Oxymoron]

\omega = e_1\wedge \dots \wedge e_p

X^{\flat} = e_n

(\omega\wedge X^{\flat}) = e^1\wedge\dots\wedge e^p\wedge e^n