Various problems concerning Exterior calculus

[Oxymoron]

Im currently learning some Exterior calculus which I am 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 I am 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 $\omega$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

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

 

[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 staring 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 separate 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 don't 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]

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

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

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