This post originally appeared between the second and third installments of my write-up of Newton-Cartan theory, and should be regarded as an extended footnote to that article. I must thank my friend and collaborator Pin Yu for bringing some of the literature to my attention.
Like my post on Newton-Cartan theory, I've made many changes on the presentation, as well as the content, of this post since 10 years ago.
In this post we will discuss the presence/availability of a preferred/compatible volume form for manifolds equipped with affine connections.
The notion of manifolds equipped with affine connections1 relaxes the definition of a (pseudo)Riemannian manifold by keeping the geodesic structure (hence connection) but discarding the metric. This post began by introducing the notion of equiaffine manifolds and follows with descriptions of some of the consequences of the definition.
Equiaffine manifolds
(Reference: Affine Differential Geometry by Katsumi Nomizu and Takeshi Sasaki, around page 14)
- The pair $(M,\nabla)$ is said to be 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.
- An equi-affine manifold is a triple $ (M,\nabla,\omega)$ where $ \omega$ is a smooth volume form satisfying $ \nabla\omega \equiv 0$. Alternatively, given a manifold $M$, an equi-affine structure on $M$ is a pair $(\nabla,\omega)$ consisting of a torsion free affine connection $\nabla$ and a volume form $\omega$ with the compatibility condition $\nabla \omega \equiv 0$.
Notice that given two non-vanishing (local) top forms $\omega, \omega'$, there is a scalar function $f$ such that $\omega = f\omega'$ (on the overlap of their domains). If they are both compatible with $\nabla$, then $f$ is a constant. Then by a standard partition of unity argument we see that whenever $(M,\nabla)$ is locally equi-affine and simply connected, we can patch together the local top forms to form a parallel volume form $\omega$ making $ (M,\nabla,\omega)$ an equi-affine manifold.
The simply-connected assumption is crucial, though: consider the circle $\mathbb{S}^1$ and let $v$ denote a non-vanishing tangent vector field. A connection $\nabla$ is entirely specified by the scalar $\Gamma$ defined through \[ \nabla_v v = \Gamma v. \] Let $\omega'$ denote the dual of $v$: the unique one form satisfying $\omega'(v) = 1$. Then we have \[ \nabla_v \omega' = - \Gamma\omega'. \] On a small coordinate chart (an interval), we can always find a function $f$ such that $\nabla_v f = \Gamma$. Then \[ \nabla_v (e^f \omega') = e^f \nabla_v f \omega' + e^f \nabla_v \omega' = 0 \] and we see that $\mathbb{S}^1$ equipped with any affine connection is locally equi-affine. However, to globalize this construction we need to find a function $f: \mathbb{S}^1 \to \mathbb{R}$ such that $\nabla_v f = \Gamma$. This is impossible whenever (for example) $\Gamma$ is everywhere positive. And here we see an example where the topology intervenes and a locally equi-affine manifold is not globally equi-affine.
The importance of local equi-affine structures lies in its relation to symmetries of the Ricci tensor. First, for an arbitrary pair $(M,\nabla)$ (even when $\nabla$ is, as assumed, torsion-free), the associated Ricci curvature $\mathrm{Ric}(X,Y) = c(R(\cdot, X)Y)$ 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 $TM \leftrightarrow T^\star M$ 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
- The Ricci tensor is symmetric.
- The manifold is locally equi-affine.
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$ for some one form $\lambda$. 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 \begin{equation} \label{eq:4} (\nabla\wedge\nabla)\tilde\omega = \underbrace{(\nabla\wedge \lambda)}_{= \mathrm{d}\lambda} \otimes\tilde\omega + \underbrace{(\lambda\wedge\lambda)}_{= 0}\otimes\tilde\omega \end{equation} where we used that $\nabla$ is torsion free. 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 \[ \bigl(\tilde\omega\cdot R(Y,X)\bigr)(Z_1,\ldots Z_n) = \sum_{i = 1}^n (-1)^{i-1}\tilde\omega(R(Y,X)Z_i,Z_1,\ldots\hat{Z}_i \ldots,Z_n) \] 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 \begin{equation}\label{eq:5} \mathrm{d}\lambda(X,Y) = c(R(Y,X)) \end{equation}
Now, observe that the first Bianchi identity (which holds because $\nabla$ is torsion free) gives \[ R(\cdot,X)Y - R(\cdot,Y)X + R(X,Y) = 0. \] Taking the contraction, we have \[ c(R(Y,X)) = \mathrm{Ric}(X,Y) - \mathrm{Ric}(Y,X). \] And so \eqref{eq:5} gives \begin{equation}\label{eq:6} \mathrm{d}\lambda = \bigwedge(\mathrm{Ric}) \end{equation} where the right hand side denotes the antisymmetric part of Ricci tensor.
With these computations performed, we can now prove the lemma.
First, suppose $(M,\nabla)$ is equi-affine, and $\tilde\omega = \omega$ is the parallel volume form that defines the local equi-affine structure. Then $\lambda = 0 \implies d\lambda = 0$. By \eqref{eq:6} this implies that Ricci must be symmetric. For the opposite direction, suppose Ricci is symmetric, and let $\tilde\omega$ be an arbitrary volume form. Equation \eqref{eq:6} implies that $\lambda$ is closed. On a possibly smaller, simply-connected neighborhood, $\lambda$ is exact, so there exists some function $u$ such that $\mathrm{d}u = \lambda$. 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.
Now, in the companion post, we defined the notion of a Galilean manifold (Definition 2 in that post). The proof of Proposition 10 there can be extended2 to state that any Galilean manifold is locally equi-affine, and hence, as a corollary,
$\mathfrak{G}$-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.
Now, in Proposition 10 of the companion post, we were able to upgrade the equi-affine property to a global one in the Galilean case (assuming orientability). There 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.
- The frame bundle of $E$ admits an $SL(k)$ structure.
- There exists a (non-trivial) $\nabla$-parallel section of $\Lambda^k(E^\star)$.
(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 the companion post and observe that 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 $\Lambda^k(E^\star)$. 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 $\Lambda^k(E^\star)$. 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 top-order 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.
Volume preserving diffeomorphisms
A well-known theorem in Riemannian geometry is
It turns out that the same theorem holds more generally for equi-affine manifolds $(M,\nabla,\omega)$.
We begin by first discussing the geometry of $TM$. Since $TM$ is itself a vector bundle over $M$, the connection $\nabla$ specifies a decomposition of $T_{p,V}(TM)$ (which is a $2n$-dimensional vector space) into horizontal and vertical subspaces3 $H$ and $V$. By definition $H$ is canonically isomorphic to $T_pM$, and since $TM$ is a vector bundle, we also have that $V$ is canonically isomorphic to $T_p M$. This means that given a top form $\omega$ on $T_pM$, we can define $\omega_H$ and $\omega_V$ two $n$-forms on $T_{p,v}(TM)$. Their wedge product $\varpi:= \omega_H\wedge \omega_V$ is then a natural volume form on $TM$.
It is simplest to just perform the computation in local coordinates.
Let $U$ denote a small coordinate neighborhood in $M$, on this neighborhood we take coordinates $x^1, \ldots, x^n$. Then $TU$ can be coordinated by $x^1, \ldots, x^n, y^1, \ldots, y^n$, where $(x,y)\in \mathbb{R}^{2n}$ represents the vector $y^i \partial_{x^i}$ in $T_x U$.
The connection $\nabla$ is represented by its Christoffel symbols $\Gamma^i_{jk}$ with the indices running from $1, \ldots, n$, given by $\nabla_{\partial_{x^i}} \partial_{x^j} = \Gamma_{ij}^k \partial_{x^k}$. Now fix $(q, V)\in TU$, the horizontal lift $H_{q,V} z$, where $z\in T_q M$ is given by $z^i \partial_{x^i}$ is the vector $z^i \partial_{x^i} - V^j z^i \Gamma_{ij}^k \partial_{y^k}$. The vertical subspace $V$ is spanned by $\partial_{y^k}$.
By an abuse of notation we will write our volume form $\omega$ on $M$ as \[ \omega ~dx^1 \wedge \cdots \wedge dx^n \] where $\omega$ also stands for the scalar function coefficient. On $TM$, the $n$ form $\omega_H$ acts trivially on the vertical subspace, and via the horizontal lift diffeomorphism we see that \[ \omega_H = \omega ~ dx^1\wedge \cdots \wedge dx^n .\] The vertical $n$ form $\omega_V$ on the other hand acts trivially on the horizontal subspace; this means that there exists coefficients $A_j^k(x,y)$ such that \[ \omega_V = \omega ~ (dy^1 + A_j^1 dx^j) \wedge \cdots (dy^n +A_k^n dx^n) \] with $A$ chosen so that \[ (dy^i + A^i_j dx^j)(\partial_{x^k} - y^\ell \Gamma_{k\ell}^m \partial_{y^m}) = 0 \] which we can solve by $A^i_k(x,y) = y^\ell \Gamma^i_{k\ell}(x)$.
This implies that in local coordinates \[ \varpi = \omega^2 ~dx^1 \wedge \cdots\wedge dx^n \wedge dy^1 \wedge \cdots \wedge dy^n .\]
The geodesic flow vector field is given by \[X = y^k \partial_{x^k} - y^k y^j \Gamma^i_{jk} \partial_{y^i}. \] By Cartan's magic formula, since $\varpi$ is a top form, \begin{aligned} \mathcal{L}_X \varpi &= \mathrm{d} i_X \varpi \newline &= \mathrm{d} \left( (-1)^{k-1} y^k \omega^2 dx^1 \wedge \hat{\cdots}^k \wedge dx^n \wedge dy + (-1)^{n+i} \omega^2 y^k y^j \Gamma^i_{jk} dx \wedge dy^1 \wedge \hat{\cdots}^i \wedge dy^n\right). \end{aligned} Note that both $\Gamma$ and $\omega$ only depend on the $x$ coordinates.
A direct computation shows then \[ \mathcal{L}_X \varpi = y^k \partial_k (\omega^2) ~dx\wedge dy - 2 \omega^2 y^k \Gamma^i_{ki}~ dx \wedge dy. \] On the other hand, for the volume form $\omega$ in local coordinates \[ 0 = \nabla_{y^i \partial_{x^i}} \omega = y^k \partial_k \omega ~ dx - \omega y^k \Gamma^i_{ik} ~dx.\] And hence we see that $\mathcal{L}_X \varpi = 0$ as claimed.
- One may ask why we don't call these "affine manifolds". Unfortunately, in the literature the phrase "affine manifold" primarily refers to a smooth manifold equipped with a flat, torsion-free affine connection on its tangent bundle, which is much more restrictive than what we want to consider. (There's a good reason for this name too: these are manifolds whose transition maps between charts in its atlas are affine mappings.) ^
- The only difference is that in Proposition 10 of the companion post, $M$ is assumed to be orientable. This only is needed to guarantee the existence of a global top form. We can drop that assumption and localize the proof. ^
- Let $\pi$ denote the canonical projection $TM \to M$. Its pushforward $d\pi$ maps from $T_{p,V}(TM) \to T_p M$. The kernel is the vertical subspace. Given $X \in T_p M$, let $\gamma$ denote a path through $p$ with derivative $X$ at $p$. The parallel transport of $V$ using the connection $\nabla$ gives a vector field along $\gamma$, and defines a curve $c$ in $TM$. The derivative of $c$ at $(p,V)$ turns out to be independent of the extension $\gamma$, and depends only on $X$ (linearly). The mapping $X \mapsto \dot{c}$ gives a linear mapping $H_{p,V}$ of $T_pM \to T_{p,V}(TM)$ with $d\pi(\dot{c}) = X$. The image of this mapping is the horizontal subspace. For more about this, see my notes on bundles and derivatives. Note that $(p,V) \mapsto H_{p,V}V$ is a vector field on $TM$, and this is exactly the geodesic spray (or geodesic flow vector field). ^
