### Parallel volume forms

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.