Parallel volume forms

by Willie Wong

In the previous post on Newton-Cartan theory and Galilean geometry, I showed that the Galilean manifolds admit a preferred volume form. After some discussion with my old officemate Pin Yu, and a bit of digging on the internet, I found that this notion of a parallel volume form is a rather well-developed one in classical differential and affine geometry.

Equiaffine manifolds
(Reference: Affine Differential Geometry by Katsumi Nomizu and Takeshi Sasaki, around page 14) We start with two definitions.

Definition 1
Given an affine smooth manifold (M,\nabla) where \nabla is a torsion-free connection, we say that it is locally equi-affine if at every point p\in M there exists a neighborhood on which one can define a non-vanishing top form (in other words a local volume form) that is parallel under the connection.
Definition 1′
The triple (M,\nabla,\omega) where M is a smooth n-dimensional manifold, \nabla is a torsion free affine connection, and \omega is a smooth volume form is said to be an equi-affine manifold, or that (\nabla,\omega) defines an equi-affine structure on M, if \nabla\omega \equiv 0.

Observe that if (M,\nabla) is locally equi-affine and simply connected, then the standard partition of unity argument can be used to patch together a parallel volume form \omega making (M,\nabla,\omega) an equi-affine manifold.

Observe that in the proof of Proposition 8 in yesterday’s post, the computation in local frame only establishes that a Galilean manifold is locally equi-affine. To upgrade to a global parallel volume form without assuming simply-connected requires using the fact that the local volume form factors into a time-like part \epsilon_t which is parallel and a trace-part \mu_\sigma which can be normalized globally using the SO(n) structure \sigma. This will be discussed more in the next section when we talk about general vector bundles.

The importance of local equi-affine structures lies in the following fact

Fact 2
For an arbitrary affine manifold (M,\nabla) with a torsion-free connection, the associated Ricci curvature Ric(X,Y) = c(R(\cdot,X)Y) is not necessarily a symmetric tensor.

In the pseudo-Riemannian case, the Ricci tensor is symmetric due to the fact that the metric induces a canonical isomorphism of TM \leftrightarrow T^*M which carries elements of so(p,q) (acting on TM) to a two-form. This combined with first Bianchi identity gives the pairwise exchange symmetry of the Riemann tensor and hence the symmetry of the Ricci tensor. We can, however, divorce the symmetry of Ricci from the metric

Lemma 3
The following two conditions are equivalent for an affine manifold with a torsion-free connection:

  • The Ricci tensor is symmetric.
  • The manifold is locally equi-affine.

Proof. Fix \tilde\omega some local volume form. For any local vector field X, \nabla_X\tilde\omega is another fully anti-symmetric tensor, and hence can be written locally as a multiple of \tilde\omega by some function. Since this map from vector field to the associated function is clearly tensorial, we have that \nabla\tilde\omega = \lambda\otimes\tilde\omega. So using the Leibniz rule, we have that \nabla^2\tilde\omega = (\nabla \lambda)\otimes \tilde\omega + \lambda\otimes\lambda\otimes\tilde\omega. Now take the antisymmetric part in the first two indices

Equation 4
(\nabla\wedge\nabla)\tilde\omega = (\nabla\wedge \lambda)\otimes\tilde\omega + (\lambda\wedge\lambda)\otimes\tilde\omega = d\lambda \otimes\tilde\omega

where we used that \nabla is torsion free and that \lambda is a real one-form. On the other hand, the antisymmetric part of the Hessian operator \nabla^2 is related to the Riemann tensor. A direction computation (for covariant derivatives of (0,n)-tensors) shows

(\nabla\wedge\nabla)_{XY}\tilde\omega = \tilde\omega \cdot R(Y,X)

where the dot denotes sum of contractions of the contravariant component of R(Y,X) against every covariant component of \tilde\omega. Or, to be more precise

[\tilde\omega\cdot R(Y,X)](Z_1,\ldots Z_n) = \sum_{i = 1}^n (-1)^{i-1}\tilde\omega(R(Y,X)Z_i,Z_1,\ldots\hat{Z}_i \ldots,Z_n)

A simple computation (using a frame in TM let’s say) using the antisymmetry of \omega we obtain that \tilde\omega\cdot R(Y,X) = c(R(Y,X)) \tilde\omega. So this tells us that

Equation 5
d\lambda(X,Y) = c(R(Y,X))

Now, observe that the first Bianchi identity (again using the torsion free condition) gives

R(\cdot,X)Y - R(\cdot,Y)X + R(X,Y) = 0

Taking the contraction, we have

c(R(Y,X)) = Ric(X,Y) - Ric(Y,X)

And so Equation 5 gives

Equation 6
\displaystyle d\lambda = \bigwedge(Ric)

where the right hand side denotes the antisymmetric part of Ricci tensor. With Equation 6 prepared, we can prove the Lemma. Suppose \tilde\omega = \omega is a parallel volume form that defines the local equi-affine structure. Then \lambda = 0 \implies d\lambda = 0. By Equation 6 this implies that Ricci must be symmetric. To go the reverse way, suppose Ricci is symmetric, then let \tilde\omega be an arbitrary volume form. Equation 6 implies that \lambda is closed. On a possibly smaller, simply-connected neighborhood, \lambda is exect, so there exists some function u such that du = \lambda. A direct computation then shows that for \omega = e^{-u}\tilde\omega,

\nabla\omega = -\nabla u \otimes e^{-u}\tilde\omega + e^{-u}\lambda\otimes\tilde\omega = 0

Hence \omega is locally parallel and non-vanishing (e^{-u} > 0), and (M,\nabla) is locally equi-affine. Q.E.D.

Corollary 7
The Ricci tensor for a Galilean manifold is symmetric.

\mathfrak{G}-bundles and holonomy
In the previous section we discussed some facts relating to local parallel volume forms and affine connections on the tangent bundle. In this section we’ll broaden our view to the picture of a general k-dimensional vector bundle E over the base space M. Clearly the tangent bundle case is a subset of our discussion.

As mentioned earlier, to upgrade the local equi-affine property to a global one in the Galilean case, we used the fact that there is some global structure that is invariant under the connection. This leads naturally to considering the group structure of the frame bundle. We have naturally the following

Lemma 8
Let (E,\pi,M) be a k-dimensional vector bundle over M. Let \nabla be a connection on E. Consider the bundle of top forms \Lambda^k(E^*), which is a one-dimensional vector bundle over M, and the induced action of \nabla on it. Then the following are equivalent

  • The frame bundle of E admits an SL(k) structure.
  • There exists a (non-trivial) \nabla-parallel section of \Lambda^k(E^*).

Proof (This is a nice exercise in linear algebra. I will just sketch it here.) There are two methods of going from the first condition to the second. First is via the construction given in my first post on the subject and observe that the the geometric invariant associated to an SL(k) structure on a k-dimensional bundle is precisely a volume form over E. So in this construction by definition there exists a parallel section of \Lambda^k(E^*). The second method is to consider the possible defect in defining such a section by parallel transport. Fix a non-vanishing top form at some point p. Given any other point q and a piece-wise C^2 curve connecting the two points, by using the connection we can parallel transport our given form to a non-vanishing top form at point q (it cannot vanish: by the fundamental existence and uniqueness theorem of ODEs, since the parallel transport equation is first order, linear, and homogeneous, if the solution vanishes at one point it must vanish everywhere). If we try, in the general situation, to define a form this way, we run into one obstruction: parallel transport is path dependent. So the top form may not be well-defined (going through two separate paths to q produce two different answers). This failure is captured in the notion of holonomy. For a general vector bundle E, the difference between parallel transports along two paths is given by an element of the general structure group GL(k). However, if E admits a \mathfrak{G}-structure, then the holonomy is given by an element of \mathfrak{G}. Applying this to our case, we see that the SL(k) structure on E induces a SL(1) structure on \Lambda^k(E^*). Since SL(1) is the trivial group, this means that parallel transports of top forms is path independent.

To go backwards, let \omega be such a parallel section. By the existence and uniqueness theorem of ODE again, \omega is nowhere vanishing, so it is a volume form on E. It is simple linear algebra to check that the sub-bundle of the frame bundle of E which, when acted on by \omega gives the constant value 1, is a principal SL(k) sub-bundle. Q.E.D.

Corollary 9
pseudo-Riemannian manifolds of signature (p,q) and (1+n)-dimensional Galilean manifolds have parallel volume forms.

Proof. The bundle of admissible frames on Riemannian and Galilean manifolds are, respectively, principal sub-bundles of the general frame bundle (of the tangent bundle) with structure groups SO(p,q) and \mathbb{R}^n\rtimes SO(n). The former is a subgroup of SL(p+q), and the latter a subgroup of SL(1+n). Therefore they have parallel volume forms. Q.E.D.

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