Is the volume form on an oriented Riemannian manifold parallel?

This question actually has two parts:

1. Am I correct that the Riemannian connection induces a connection on the k-th exterior power of the cotangent bundle, for any k? What I had in mind was: view the Riemannian connection as a connection on the frame bundle and then think of the k-th exterior power of the cotangent bundle as the associated bundle to the frame bundle under the natural action of GL(n) on the k-th exterior algebra of (R^n)*.

2. If you view the volume form of an oriented Riemannian manifold as a section of the n-th exterior power of the cotangent bundle, is it always parallel with respect to this connection?

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1.) Yes; the extension to pure k-covectors is given by the Liebniz identity $$\nabla \left(X_1 \wedge \cdots \wedge X_k\right) = \sum_k X_1 \wedge \cdots \wedge \nabla X_k \wedge \cdots \wedge X_k$$ Extend to all of $\bigwedge^k T^*M$ by linearity.

2.) Yes again. This is easiest to see by thinking about parallel transport. Given a curve $\gamma$ from $p$ to $q$ and an orthogonal frame $\Psi_p, \Psi_q$ at each point, parallel transport along $\gamma$ determines a unique orthogonal transformation $O$ such that $$P^\nabla_\gamma \Psi_p = O \cdot \Psi_q$$ This means that any invariants of the group $O(n)$ are parallel, such as the volume form.

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The volume form is not an $O(n)$ invariant, though. It's an $SO(n)$ invariant. However they have the same Lie algebra and this is all that the coavariant derivative is probing. In other words, just parallel transport along a null-homotopic loop and you're set. – José Figueroa-O'Farrill Dec 26 '09 at 23:00
Good point! I guess I should have said "volume density"... – Matt Noonan Dec 26 '09 at 23:37
Are the signs in your formula for (1) correct? I think they only appear if one insists in applying $\nabla$ only in the first factor. – Mariano Suárez-Alvarez Dec 27 '09 at 0:00
@Mariano: You're right, of course. I've edited the answer to reflect the correct (lack of) signs. – Matt Noonan Dec 31 '09 at 19:09

If you think about the exterior bundle as an associated bundle of the positive oriented frame bundle, then the highest power $\Lambda^n T^\star M$ correponds to the trivial representation $\rho\colon So(n)\to R^*=Gl(R).$ The volumne form is just the "constant" section $[f,1]\in \Gamma(SO(M)\times_\rho R).$ Hence it is parallel by the definition of the covariant derivative on associated bundles of frame bundle.

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These are important questions. When one starts to define connections, one usually imposes the Leibnitz condition on it. But I am proceeding too fast. First, let's recall what a connection is. Here we come to one very confusing point, at least, it was for me, when I started learning differential geometry: There are various ways of defining connections. Just to name some of them:

1. Define the Levi-Civita connection on a (Riemannian) manifold. The Levi-Civita connection is an affine connection that preserves the metric structure, i.e., $\nabla g = 0$ and that is torsion-free, i.e., $\nabla_{X}Y-\nabla{Y}X = [Y, X]$, where $[. , .]$ denotes the Lie-bracket of two elements of $\Gamma(TM)$. Then one usually starts by showing that, by this definition, the Levi-Civita connection (hereafter called LC connection) is uniquely determined.

2. However, one can even generalize this approach by at first considering linear connections, then metric ones and then affine ones. After all that one can turn to the derivation of the Levi-Civita connection. Personally, I like this approach pretty much because it shows very well why the LC connection is so important. Namely, it eases computations for the Lie-derivative which normally does not require the notion of the LC connection at all. Also there are many fascinating mathematical objects, such as Killing fields which can be calculated (or at least defined) effectively, if one introduces the notion of the LC connection.

3. One can even start more generally, by considering fiber bundles. This approach was taken by the French mathematician Charles Ehresmann (1905-1979). This approach has one fundamental advantage: Generality, however, it lacks the easy-to-grasp-effect (I like to assign a definition this value if it has a very intuitive meaning). Interestingly, this approach does yield the LC connection as a special case when one reduces fiber bundles to vector bundles and considers the bundles $\pi : TM \longrightarrow M$ as a special case and requires the base manifold $M$ to be equipped with a symmetric, positive definite 2-times covariant tensor field, i.e., the metric tensor $(g_{ij})_{i,j}$.

Why am I writing all this? Simply to sum up what we already know and how beautifully, at least, in my opinion, all these definitions harmonize.

To come to your question (I apologize for relying so much on your patience):

If one has finally obtained a manifold that is equipped with a Riemannian metric, then one goes further and asks what the term orientation means for general Riemannian manifolds. Although for submanifolds of $\mathbb{R}^{n}$ one has a very intuitive picture of what orientability means. We could expand this picture by making use of the Nash Embedding theorem, but it turns that this is not necessary.

Example: Consider for instance the so-called Möbius strip. It is a typical example of a non-orientable two-dimensional submanifold of the Euclidean 3-space.

Then, one can ask further what it means for a manifold to be orientable. Consider now the tangent and the cotangent bundle. It is now possible to associate to each vector space a certain orientation if one introduces one starting point for that. Consider a basis that is called "unit basis". Define the orientation of this basis to be positive. Then, one imposes further conditions on the unit basis, namely orthonormality w.r.t the Riemannian metric. This way, one can define an orientation on the cotangent bundle $TM^{*}$ as well (Duality principle).

Now we simply ask what the Riemannian volume form is: It is certainly $SO(n)$-invariant, because we have chosen an orientation and require the 1-forms that constitute a basis of $TM^{*}$ to be positively oriented (c.f. what I said above).

OK, now to question one: Via the Leibnitz rule and the duality principle it is possible to define an induced connection on the cotangent bundle and, by that way, further expanding this definition to the k-th exterior power of the cotangent bundle.

Namely, from the tensor product spaces can be obtained the k-th exterior cotangent space. I would like to refer you to the book Introduction to Kähler Manifolds which contains and excellnt treatment of exactly this. Via the equivalence relation one then breaks down the Leibnitz rule to the k-h exterior power cotangent bundle. The rules is exactly what Matt has written above.

Let's come to question two: We know that the volume form is $SO(n)$ invariant, where $n=dim(M)$ Since we define the volume form $\omega := dx^{1} \wedge \dots \wedge dx^{n}$, with the $dx^{i}$ for $i=1, \dots, n$ to be the product of the basis of the cotangent bundle, we can make the following thought.

We know how to "differentiate" the basis w.r.t to the LC connection. When we calculate all that, we finally obtain that the volume form is parallel. This was how I learned it. Of course one can use representation theory here which is, in my eyes, very interesting and also more elegant.

If there are questions, don't hesitate to contact me.

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For anyone using this answer as a starting point for learning about connections, I just wanted to mention that (unlike gggg gggg) I find the idea of an Ehresmann connection to be much easier to grasp than the other notions of connection mentioned here. In other words, the "easiest place to start" may not be the same for everybody! – Vectornaut Sep 13 '12 at 16:08