# Various problems concerning Exterior calculus

1. Apr 8, 2006

### 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.

Last edited: Apr 9, 2006
2. Apr 9, 2006

### Oxymoron

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

3. Apr 9, 2006

### 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.

4. Apr 9, 2006

### Cexy

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 ;)

5. Apr 9, 2006

### Oxymoron

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. :)

Last edited: Apr 9, 2006
6. Apr 9, 2006

### George Jones

Staff Emeritus
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

Last edited: Apr 9, 2006
7. Apr 10, 2006

### 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?

8. Apr 10, 2006

### 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.

9. Apr 11, 2006

### Doodle Bob

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.

10. Apr 11, 2006

### 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 $\omega$and 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, $g_{(p)}$:

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

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

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

Last edited: Apr 11, 2006
11. Apr 11, 2006

### 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?

Last edited: Apr 11, 2006
12. Apr 11, 2006

### Doodle Bob

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 $\iota_X \omega$

Last edited: Apr 11, 2006
13. Apr 11, 2006

### Cexy

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).

14. Apr 21, 2006

### 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

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

15. Apr 23, 2006

### Doodle Bob

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}.

16. Apr 23, 2006

### Oxymoron

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.

Last edited: Apr 23, 2006
17. Apr 24, 2006

### 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.

18. Apr 26, 2006

### Oxymoron

$$\omega = e_1\wedge \dots \wedge e_p$$

$$X^{\flat} = e_n$$

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