# Covariant and exterior derivatives

1. Aug 23, 2010

### Rasalhague

Reading Roger Penrose's The Road to Reality, I wondered what is the relationship/difference between these? Can one be expressed simply in terms of the other? The exterior derivative seems to be only defined for form fields. He says the covariant derivative of a scalar field (0-form field) is the same as its exterior derivative. Is this the case for higher dimensional forms, or can the covariant derivative of a form field be different from its exterior derivative? The rules Penrose gives for them look different on some points, but are there differences that aren't accounted for by the antisymmetric property of forms? In the following, alpha and beta are tensors, phi a scalar.

*

Covariant derivative:

$$(1)\enspace \nabla (\alpha + \beta)=\nabla \alpha + \nabla \beta$$

$$(2)\enspace \nabla (\alpha \cdot \beta)= (\nabla \alpha) \cdot \beta + \alpha \cdot (\nabla \beta)$$

$$(3)\enspace \nabla \phi = \mathrm{d}\phi=\frac{\partial \phi}{\partial x^i} \mathrm{d}x^i$$

(The dot in (2) stands for a contraction of some number of pairs of indices, including possibly a contraction of no indices, i.e. a tensor product.)

*

Exterior derivative:

$$(1)\enspace \mathrm{d}(\alpha + \beta)=\mathrm{d} \alpha + \mathrm{d} \beta$$

$$(2)\enspace \mathrm{d}(\alpha \wedge \beta)=\mathrm{d} \alpha \wedge \beta + (-1)^p \, \alpha \wedge \mathrm{d} \beta$$

$$(3)\enspace \mathrm{d}(\mathrm{d} \alpha)=0$$

*

I guess one difference would be that in general $\nabla (\nabla \phi) \neq 0 = \mathrm{d}(\mathrm{d}\phi)$.

Last edited: Aug 24, 2010
2. Aug 24, 2010

### haushofer

The exterior derivative sends p-forms to (p+1)-forms, the covariant derivative sends (p,q)-tensors to (p+1,q)-tensors (the p here gives the "number of covariant indices"). For symmetric connections, so no torsion, one can replace the partial derivative in the exterior derivative with a covariant one.

Besides this I'm not aware of any relationship between them.

3. Aug 24, 2010

### Rasalhague

To take the example of a 1-form, is this right?

$$\mathrm{d}(f_k \mathrm{d}x^k) = \frac{\partial f_k}{\partial x^j} \mathrm{d}x^j \mathrm{d}x^k=\nabla(f_k)\mathrm{d}x^k$$

Then the covariant derivative replaces the partial derivatives and corresponding basis 1-forms (rather than just the partial derivatives), or, to put it another way, the exterior derivative would have the same effect as the covariant derivative if the latter was restricted to operate only on the coordinates of the 1-form but not the basis 1-forms. Whereas the true covariant derivative of the 1-form would be

$$\nabla(f_k\mathrm{d}x^k)=\nabla (f_k) \mathrm{d}x^k+f_k \nabla(\mathrm{d}x^k)$$

$$=\frac{\partial f_k}{\partial x^j}\mathrm{d}x^j\mathrm{d}x^k+f_k \frac{\partial (\mathrm{d}x^k)}{\partial x^j} \mathrm{d}x^j\mathrm{d}x^k$$

with an extra term dependent on how the basis field is changing.

4. Aug 26, 2010

### Rasalhague

In his diagram notation, Penrose denotes the covariant derivative with a circle around a tensor and with "connector string" dangling down from the circle. Does anyone know what his notation for the exterior derivative is in this system?

5. Aug 26, 2010

Covariant derivative is a very general concept. In general it operates on "sections of fibre bundles" endowed with connections (infinitesimal parallel transport law). Here, for Penrose, it can operate on tensor fields. Differential forms are particular cases of tensor fields. One way to see that acting on forms it reduces to exterior derivative is to take a general expression for a covariant derivative of a tensor field with a bunch of covariant and contravariant indices and amuse yourself finding that for forms (antisymmetric covariant tensors) all extra terms cancel. But this happens only with so called Levi-Civita connection in Riemannian geometry. If there is torsion present, than it is not true. Penrose did not mention it.

Last edited: Aug 26, 2010
6. Aug 26, 2010

### lavinia

I think that covariant derivatives are along curves. exterior derivatives are not.

Also I think that exterior derivative is only defined in the tangent bundle whereas connections can be defined on any vector bundle.

Last edited: Aug 26, 2010
7. Aug 26, 2010

You are right. However if you have covariant derivatives, you can use them in such a way that that they map p-forms into (p+1)-forms. For instance, for one form $$\omega_{\mu}$$

$$\nabla_{\mu}\omega_{\nu}-\nabla_{\nu}\omega_{\mu}=(d\omega)_{\mu\nu}$$

And yes, covariant derivatives are for general vector bundles. You also have exterior covariant derivatives for vector-valued forms. For scalar-valued forms they coincide with simple exterior derivatives.

P.S. I do not know how to use tex on this forum. What is the format? Where it is explained? I need to learn it! ... OK. I got it.

Last edited: Aug 26, 2010
8. Aug 26, 2010

### Rasalhague

In this book, Penrose treats the word connection as a synonym of covariant derivative. Is that how you understand it?

I tried it with a 1-form. Let

$$\pmb{\omega}=\omega \, \mathrm{d}x^\mu$$

Then

$$\mathrm{d}\pmb{\omega} = \left ( \frac{\partial \omega_\mu}{\partial x^\nu}\mathrm{d}x^\nu \right ) \wedge \mathrm{d}x^\mu = \frac{\partial \omega_\mu}{\partial x^\nu}\mathrm{d}x^\nu \wedge \mathrm{d}x^\mu$$

Is that right? For the covariant derivative, I get

$$\nabla \pmb{\omega} = \nabla (\omega_\mu) \, \mathrm{d}x^\mu + \omega_\mu \nabla (\mathrm{d}x^\mu)$$

$$= \frac{\partial \omega_\mu}{\partial x^\nu}\mathrm{d}x^\nu \wedge \mathrm{d}x^\mu + \omega_\mu \nabla (\mathrm{d}x^\mu)$$

where the final term would vanish iff the coordinate basis 1-form field, and hence the coordinate basis tangent vector field, was constant. Now, I'm unsure how to interpret this. On the one hand, there is a difference, and the we can't expect the basis to be constant. On the other hand, the difference seems to be coordinate dependent, but I thought these derivatives were supposed to be independent of coordinates, which makes me think I've made a mistake, or else the final term in the last equation really does reduce to zero in general. If so, how can we tell?

$$\omega_\mu \nabla (\mathrm{d}x^\mu) = \omega_\mu \frac{\partial (\mathrm{d}x^\mu)}{\partial x^\kappa} \mathrm{d}x^\kappa = ... ?$$

Well, the partial should be 0 wrt every coordinate not equal to mu, and if it was a tangent vector would it be of unit magnitude tangent to the coordinate curve $x^\mu$. So maybe it's something analogous to that. And scaled by $\omega_\mu$. I don't know.

9. Aug 26, 2010

### lavinia

But this works only on the tangent bundle since exterior derivative is not defined on an arbitrary vector bundle?

Also is this formula is correct for an arbitrary connection, no restrictions?

10. Aug 26, 2010

### Rasalhague

My (extremely shaky) understanding was that the covariant derivative turns any (p,q) tensor field into a (p,q+1) tensor field. If you then specify a curve, you can contract $\nabla (T)$ with a tangent vector tangent to the curve at any point. And the exterior derivative acts on a q-form field (field of antisymmetric tensors of valence (0,q) to give a (q+1)-form field. So the exterior derivative is defined on the cotangent bundle, and various other bundles that I don't have the terminology for yet, but--if I've got this right--not on the tangent bundle. I was wondering whether the covariant derivative of a q-form field would ever or always be the same as its exterior derivative.

11. Aug 26, 2010

First of all, a connection (sometimes called 'affine connection') on the tangent bundle (for vector fields) determines naturally, by demanding the Leibniz rule to hold, a connection, and thus a covariant derivative, for all tensors. Differential forms can be considered as a special case of tensors (totally antisymmetric covariant). Therefore you can take a covariant derivative of a differential form. This will not be a differential form, it will be a covariant tensor, alright, but not totally antisymmetric. But then you can antisymmetrize it (sum over all permutations of indices with proper factors). You will get a differential form this way. If the connection has no torsion, then the result coincides with exterior derivative.

That is the idea.

12. Aug 26, 2010

### lavinia

So this only makes sense for special connections - and only on the tangent bundle.

13. Aug 26, 2010

### Rasalhague

(Assuming no torsion...)

Ah, great. If I've understood correctly, this answers my other question too, about Penrose's diagram notation. I guess another way of writing it would be:

$$2\nabla_{[\mu} \omega_{\nu]} = (\mathrm{d}\omega)_{\mu\nu}$$

from which we can generalise to any number of indices:

$$k! \, \nabla_{[\mu_1} \omega_{\mu_2...\mu_k]} = (\mathrm{d}\omega)_{\mu_1...\mu_k}$$

Last edited: Aug 26, 2010
14. Aug 26, 2010

Right. But be very careful. In mathematical literature you will find two different conventions for the exterior derivative. For instance in your first equation you have factor 2. In some textbooks you will find it lacking!

For instance here is an extract from Kobayashi-Nomizu, Foundations of Differential Geometry, Vol. I p. 149

Last edited: Aug 26, 2010
15. Aug 26, 2010

### Rasalhague

Thanks, arkajad. What, only two different conventions? That's spectacularly consistent for mathematical literature ;-)

16. Aug 28, 2010

### Rasalhague

Ivan Avramidi defines the exterior derivative $\mathrm{d}\alpha$ of a p-form $\alpha$ by its components thus:

$$(\mathrm{d}\alpha)_{\mu_1...\mu_{p+1}} = (p + 1) \partial_{[\mu_1}\alpha_{\mu_2...\mu_{p+1}]}$$

Why is there no term due to variation of the basis forms? (Do these notes implicitly assume a constant basis field, and could that be the reason? I'm thinking of his statement: "The Euclidean metric is given just by the Kronecker delta symbol," which is only true in constant orthonormal coordinates.)

Why is the coefficient before the partial sign neither $(p + 1)!$, in agreement with the one convention, nor $1$, in agreement with the other? (Is this a typo, a third convention, or only apparently different from the definitions in posts #13 and #14?)

Last edited: Aug 28, 2010
17. Aug 28, 2010

The formula expresses exterior derivatives in terms of components. Try to derive it yourself and you will know the answer for your both questions. If you get stuck at some point - ask for help. But first try.

18. Aug 28, 2010

### Rasalhague

Derive Avramidi's definition from one of these?

$$(\mathrm{d}\alpha)_{\mu_1...\mu_{p+1}}=\frac{1}{p+1!} \epsilon^{\mu_1...\mu_{p+1}} \nabla_{\mu_1} \alpha_{\mu_2...\mu_{p+1}}$$

or

$$(\mathrm{d}\alpha)_{\mu_1...\mu_{p+1}}=\epsilon^{\mu_1...\mu_{p+1}} \nabla_{\mu_1} \alpha_{\mu_2...\mu_{p+1}}$$

where the latter, for example, is

$$\epsilon^{\mu_1...\mu_{p+1}} (\partial_{\mu_1} \alpha_{\mu_2...\mu_{p+1}} +(\alpha_{\mu_2...\mu_{p+1}} \nabla(\mathrm{d}x^{\mu_{2}}\otimes...\otimes \mathrm{d}x^{\mu_{p+1}}))_{\mu_1...\mu_{p+1}})$$

To get further, I probably need to read up more about the covariant derivative and how to calculate Cristoffel symbols. My one previous attempt at that I got about half of the coefficients right (a draw!), so that's something I need to sort out anyway. My first priority though has been to get some of these definitions straight, and then move on to computations and what can be derived from the definitions. It certainly helps to know what various conventions there are, so that I'm not reading a bit of one author's definition, half understanding it, and trying to supplement my half-understanding with another author's conflicting definition.

Question 2. Looking at Avramidi's,

$$(\mathrm{d}\alpha)_{\mu_1...\mu_{p+1}}=\frac{p+1}{(p+1)!} \epsilon^{\mu_1...\mu_{p+1}} \partial_{\mu_1} \alpha_{\mu_2...\mu_{p+1}}$$

$$=(\mathrm{d}\alpha)_{\mu_1...\mu_{p+1}}=\frac{1}{p!} \epsilon^{\mu_1...\mu_{p+1}} \partial_{\mu_1} \alpha_{\mu_2...\mu_{p+1}}$$

it seems that either $\epsilon^{\mu_1...\mu_{p+1}} (\alpha_{\mu_2...\mu_{p+1}} \nabla(\mathrm{d}x^{\mu_{2}}\otimes...\otimes \mathrm{d}x^{\mu_{p+1}}))_{\mu_1...\mu_{p+1}})=0$ (either because a constant frame has been implicitly assumed, or in general, perhaps by some symmetry involving the permutation symbol or some symmetry between the components of alpha and the covariant derivative of the basis p-forms, or for some other reason) and he's using a third convention where, instead of 1 or (p+1)!, he inserts a coefficient of (p+1), or $\epsilon^{\mu_1...\mu_{p+1}} (\alpha_{\mu_2...\mu_{p+1}} \nabla(\mathrm{d}x^{\mu_{2}}\otimes...\otimes \mathrm{d}x^{\mu_{p+1}}))_{\mu_1...\mu_{p+1}})$ always has a value such that the whole expression on the right hand side of one or other of the first two equations in this post is equal to the right hand side of his eq. (57) defining the exterior derivative.

19. Aug 29, 2010

I see your problems. Look at that (taken from Penrose):

Now, when Penrose writes "Some people", it may well mean "most people". So, please, tell me whether you want to learn using the convention of Roger Penrose, or the convention of "some people". Or, perhaps, you want to learn both. Mixing two different conventions will necessarily lead to confusion. If not today, then tomorrow or after-tomorrow.

20. Aug 29, 2010

### Rasalhague

I'd like to learn both. If it helps to pick one convention to concentrate on begin with, I'd pick what most people use. When I see an expression in which Penrose has used an "is proportional to" sign, I'd like to know where most people would put the 1/p! to make it an equality.

Avramidi has the coefficient 1/p! in his definition of antisymmetrisation, eq. (20), where p is the number of arguments (indices) of the tensor being antisymmetrised. In the first part of eq. (57), which gives a component expression for the exterior derivative, the square brackets indicate antisymmetrisation, which for him here must include the coefficient 1/((p+1)!), since the number of arguments of the tensor being antisymmetrised is p+1. But unless I'm mistaken, he also, as well as antisymmetrising, multiplies the whole expression by p+1. Does this account for the missing term (due to variation of the basis with respect to position), making connection equal to coordinate connection:

$$\nabla_{[\mu_1} \alpha_{\mu_2...\mu_{p+1}]}=(p+1)\partial_{[\mu_1} \alpha_{\mu_2...\mu_{p+1}]} \enspace ??$$

If so, how, and is that a further difference between Avramidi's convention and Penrose's, besides the difference you highlighted?

21. Aug 29, 2010

You may like to verify the second equality in his formula (57) by expanding both sides for p=1 and p=2. This will give you some intuition of what is going on.

22. Aug 29, 2010

### Rasalhague

No intuition yet, but please let me know if this reveals what it is that I'm getting wrong, or indeed whether I'm right about anything.

In the first version, I get, for p = 1,

$$(\mathrm{d}\alpha)_{\mu_1\mu_2} = (\partial_{\mu_1} \alpha_{\mu_2}-\partial_{\mu_2} \alpha_{\mu_1})$$

But for p = 2,

$$(\mathrm{d}\alpha)_{\mu_1\mu_2\mu_3} = \frac{1}{2}(\partial_{\mu_1} \alpha_{\mu_2\mu_3}+\partial_{\mu_3} \alpha_{\mu_1\mu_2}+\partial_{\mu_2} \alpha_{\mu_3\mu_1}-\partial_{\mu_3} \alpha_{\mu_2\mu_1}-\partial_{\mu_1} \alpha_{\mu_3\mu_2}-\partial_{\mu_2} \alpha_{\mu_1\mu_3})$$

If the coordinate connection is, in this instance, (a) equal to the covariant derivative, or (b) not equal and not proportional to it, then it seems to me that Avramidi's definition agrees with neither of the two conventions you mentioned: (i) antisymmetrise (as Avramidi defines antisymmetrisation), (ii) antisymmetrise and multiply by (p+1)! If the coordinate connection is not equal to the covariant derivative, but differs from it by a factor of (p+1), then Avramidi's definition is equivalent to (i). If the coordinate connection is not equal to the coordinate derivative, but differs from it by a factor of 1/p!, where p is the number of arguments of $\alpha$, rather than the number of arguments of $\mathrm{d} \alpha$, then Avramidi's definition is equivalent to (ii), unless there are arthithmetic mistakes on my part or conceptual errors relating to the definition of one or more of: antisymmetrisation, the exterior derivative, the covariant derivative, the relation of one or both of these to partial differentiation of coordinates, etc.

I don't understand the notation used in the second version. For p = 1, it seems to be saying

$$(\mathrm{d}\alpha)_{\mu_1 \mu_2} = \partial_{\mu_1} \alpha_{\mu_1\mu_0\mu_2\mu_2}-\partial_{\mu_2} \alpha_{\mu_1\mu_1\mu_3\mu_2}$$

which is meaningless, since alpha only has one argument, and the number of indices on each side don't match. And even if they did, since this is suppose to be antisymmetric, the repetion of indices would make it equal to zero.

Last edited: Aug 29, 2010
23. Aug 29, 2010

So far so good. Now, take into account the fact that $$\alpha$$ is antisymmetric and collect together the same terms. Simplify.

As for your second question, notice that $$\mu_q$$ is absent in $$\alpha$$ on the right in the last expression in (57). You will have two terms, one for q=1,one for q=2. Understand his formula as: skip $$\mu_q$$.

24. Aug 29, 2010

### Rasalhague

For p = 2,

$$(p+1) \, \partial_{[\mu_1} \alpha_{\mu_2...\mu_{p+1}]}$$

$$=\frac{1}{2}(\partial_{\mu_1} \alpha_{\mu_2\mu_3}+\partial_{\mu_3} \alpha_{\mu_1\mu_2}+\partial_{\mu_2} \alpha_{\mu_3\mu_1}-\partial_{\mu_3} \alpha_{\mu_2\mu_1}-\partial_{\mu_1} \alpha_{\mu_3\mu_2}-\partial_{\mu_2} \alpha_{\mu_1\mu_3})$$

$$=\partial_{\mu_1} \alpha_{\mu_2\mu_3}+\partial_{\mu_3} \alpha_{\mu_1\mu_2}+\partial_{\mu_2} \alpha_{\mu_3\mu_1}$$

which is equal to Avramidi's other expression in (57),

$$\sum_{q=1}^{3}(-1)^{q-1}\partial_{\mu_q}\alpha_{\mu_r\mu_s}$$

where r and s are numbers from {1,2,3}, not equal to q or each other, r < s,

$$=\partial_{\mu_1}\alpha_{\mu_2\mu_3}-\partial_{\mu_2}\alpha_{\mu_1\mu_3}+\partial_{\mu_3}\alpha_{\mu_1\mu_2}$$

$$=\partial_{\mu_1}\alpha_{\mu_2\mu_3}+\partial_{\mu_2}\alpha_{\mu_3\mu_1}+\partial_{\mu_3}\alpha_{\mu_1\mu_2}$$

Assuming this generalises to any possible value of p, Avramidi's definition differs from Penrose's,

$$(\mathrm{d}\alpha)_{\mu_1...\mu_{p+1}}=\partial_{[\mu_1} \alpha_{\mu_2...\mu_{p+1}]}$$

by a factor of p+1, and from the definition in #13 by a factor of (p+1)/((p+1)!). Is that right?

I just had a thought about the other question. The Cristoffel symbols, i.e. the part of the component expression of the covariant derivative that depends on variation of the basis field, can be expressed in terms of derivatives of the components of the metric tensor field. Could it be that the symmetry of the metric tensor causes them to be symmetric too? Wikipedia says the Christoffel symbols are symmetric in their lower indices in the absence of torsion, which suggests they're not entirely symmetric, but could it be something to do with whatever symmetry they do have that causes them to vanish? (This is assuming that Penrose's definition agrees with Kobayashi-Nomizu; does it?)

25. Aug 29, 2010

A generic covariant derivative, even compatible with metric, will have $$\Gamma^{\mu}_{\nu\sigma}$$ nonsymmetric in the lower indices. There will be many such covariant derivatives. But if you demand it being symmetric (zero torsion), then it is unique, and is given by Christoffel symbols expressed by the derivatives of the metric.