### Newton-Cartan, Part 2

After writing up the previous post on Newton-Cartan theory, I came to realize that it is actually a very nice exercise for myself to dig into the geometry more. So here goes a bit more on the implications of the Galilean geometry and the Newton-Cartan theory.

Symmetries of space-time
We start by making precise what we mean by a symmetry

Definition 1
A symmetry of a Galilean manifold $(M, \epsilon_t,\sigma,\nabla)$ is a diffeomorphism $f:M\to M$ such that it preserves the geometric structures: that the pullback $f^*\epsilon_t = \epsilon_t$, the pushforward $f_*\sigma = \sigma$, and the pullback connection $f_*\nabla = \nabla$.

If we have a one-parameter family of symmetries continuous to the identity, we can define an infinitesimal symmetry

Definition 2
An infinitesimal symmetry of a Galilean manifold is a vector-field $X$ on $M$ such that $\mathcal{L}_X\epsilon_t = 0$, $\mathcal{L}_X\sigma = 0$, and $[ \mathcal{L}_X, \nabla ] = 0$, where $\mathcal{L}_X$ denote the Lie derivative by $X$ and square-brackets denote commutator of operators.

Now, by the Cartan relation, $\mathcal{L}_X\epsilon_t = i_X d\epsilon_t + d(i_X\epsilon_t)$. By assumption $\epsilon_t$ is closed, and this shows that

Equation 3
$\mathcal{L}_X\epsilon_t = 0 \iff \epsilon_t(X) \equiv \mbox{ constant}$

On the other hand, using the definition

$(\mathcal{L}_X K)^{ab} = (\nabla_XK)^{ab} - K^{cb}\nabla_cX^a - K^{ac}\nabla_cX^b$

and that $\sigma$ is parallel and symmetric, this implies

Equation 4
$\mathcal{L}_X\sigma = 0 \iff \sigma \circ \nabla X$ is an anti-symmetric (2,0)-tensor (i.e. a bivector)

Equation 4 is an analogue of Killing’s equation.

Equations 3 and 4 together implies that if $X$ is a symmetry and it is tangent to a spatial hypersurface $\Sigma_\tau$ at one point, then $X$ is always tangent to spatial hypersurfaces and that $X |_{\Sigma_\tau}$ is a Killing vector field. This is also evident from Definition 1, where one immediately sees that by the construction given in the previous post, a symmetry map $f$ will carry hypersurfaces $\Sigma_\tau \to \Sigma_{\tau'}$ to hypersurfaces.

We can also prove a version of the familiar equation in pseudo-Riemannian geometry relating the Riemann curvature tensor to the second derivative of a Killing vector field (see, for example, Appendix C.3 of Robert Wald’s General Relativity). By definition, for arbitrary vector fields $X,U,V$

$\displaystyle [\mathcal{L}_X,\nabla]V(U) = [X,\nabla_UV] - \nabla_{[X,U]}V - \nabla_U[X,V]$

where on the right hand side the square-brackets now denote the Lie bracket. Using that $\nabla$ is torsion free, we re-group the terms and find that

Equation 5
$[\mathcal{L}_X,\nabla] = 0 \iff R(X,U)V = \nabla_{\nabla_UV}X - \nabla_U\nabla_VX =: -\nabla^2_{U,V}X$

and thus, as in the pseudo-Riemannian case, a symmetry vector field is uniquely determined by its value and the value of its covariant derivatives at one point.

Volume form and Stokes theorem
First recall Stokes theorem from differential topology: given an orientable smooth $k$ dimensional manifold $N$ with boundary $\partial N$, and given a $(k-1)$-form $\eta$, Stokes theorem tells us that

$\displaystyle \int_N d\eta = \int_{\partial N} \eta$

where on the right hand side we use the induced orientation on $\partial N$. From this, we can derive a dual-version: the divergence theorem. Now let $\eta$ denote, instead, a volume form on $N$ (remember that the set of volume forms is a torsor over the multiplicative group of positive smooth functions; there are infinitely many to choose from). Consider a vector field $X$ on $N$, then the interior derivative $i_X\eta$ is a $(k-1)$-form. Apply Stokes theorem to this form, and we have

$\displaystyle \int_{\partial N} i_X\eta = \int_N d(i_X\eta) = \int_N(\mathcal{L}_X\eta - i_X d\eta)$

where in the last equality we used the Cartan relation again. Since $\eta$ is a top form, it is automatically closed, so the integrand on the far right simplifies to $\mathcal{L}_X\eta$. Now, since $\Lambda^k(T^*N^k)$ is one-dimensional over the ring of smooth functions, and that the units in this ring corresponds to non-vanishing functions, we have that $\mathcal{L}_X\eta = F\eta$ for some smooth function $F$. This way we define the divergence of the vector field $X$:

Definition 6
Let $N$ be a smooth manifold and $\eta$ a volume form. For a given vector field $X$, the divergence of $X$ relative to $\eta$, written $(\mathop{div}_\eta X)$ is the unique function given by $\mathcal{L}_X\eta = (\mathop{div}_\eta X)\eta$.

Notice that the definition does not depend on a connection! Furthermore it depends on the choice of volume form. Let $\eta' = e^u\eta$ be another volume form. Then a computation shows that

$\displaystyle \mathcal{L}_X\eta' = X(e^u)\eta + e^u\mathcal{L}_X\eta = (Xu + \mathop{div}_\eta X)\eta'$

On the other hand, we can also define the object $\nabla_a X^a = c(\nabla X)$, the tensor contraction of the covariant derivative of $X$. This function, for a fixed vector field, only depends on the choice of connection. Now, in 3D vector calculus, the two definitions are equal. In general, we have that

Lemma 7
Given a torsion-free connection $\nabla$ and a volume form $\eta$. Then the following two conditions are equivalent

1. $\nabla\eta = 0$
2. $c(\nabla X) = \mathop{div}_\eta X$ for all vector fields $X$.

Proof. Let $e_1,\ldots,e_k \in TN$. A direct calculation shows that

$\mathcal{L}_X\eta(e_1,\ldots,e_k) = (\nabla_X\eta)(e_1,\ldots,e_k) + \sum_{j = 1}^k \eta(e_1,\ldots, e_{j-1},\nabla_{e_j}X, e_{j+1},\ldots e_k)$

Now if $\{e_j\}$ are linearly dependent, then both sides evaluate simply to zero. So assume $\{e_j\}$ form a basis. Now $\nabla X\in T^{1,1}M$ can be regarded as a linear map from vector fields to vector fields. In the basis we have, it can be represented by some matrix $A_{ij}$ so that $\nabla_{e_j}X = \sum A_{jm}e_m$. Now, plugging this expression into the above equation, we notice that if $j\neq m$, then by linearity ($A_{jm}$ is just a real number) and anti-symmetry, the form evaluates to zero. So we can simplify

$\mathcal{L}_X\eta(e_1,\ldots,e_k) = (\nabla_X\eta)(e_1,\ldots,e_k) + (\mathop{Tr} A)\eta(e_1,\ldots,e_k)$

Lastly observe that the trace of a linear operator is basis independent, and in fact is, by definition, the tensor contraction, we have that

$(\mathop{div}_\eta X - c(\nabla X))\eta = \nabla_X\eta$

and the lemma follows. Q.E.D.

Observe that this Lemma is why the divergence theorem is stated so simply in pseudo-Riemannian geometry: the metric-induced volume form is by definition parallel relative to the Levi-Civita connection.

Now we specialise to Galilean geometry.

Proposition 8
Let $(M,\epsilon_t,\sigma,\nabla)$ be a Galilean $(n+1)$-dimensional manifold, and assume that $M$ is orientable. Then there exists a preferred volume form $\mathrm{Vol}$ with the property that $\nabla \mathrm{Vol} = 0$.

Proof. $\mathrm{Vol}$ will be, morally speaking, given by $\epsilon_t\wedge \mu_\sigma$, where $\mu_\sigma$ is any $n$-form such that its restriction to the spatial hypersurfaces $\Sigma_\tau$ agrees with the volume form given by the induced Riemannian metric. We show such a volume form exists. Essential uniqueness follows from the fact that $\nabla (f\mathrm{Vol}) = \nabla f \mathrm{Vol} + f \nabla\mathrm{Vol}$, so that any other parallel volume form must be a constant multiple of the one we construct.

It suffices to show that this volume form exists locally. Suppose we defined such forms on two overlapping neighborhoods $U,V$. That the two forms both restrict to $\Sigma_\tau\cap U\cap V$ as the induced volume form by the Riemannian metric means that we have a preferred normalisation by construction, and hence on $U\cap V$ the two forms agree. A partition of unity argument then allows us to patch the local definitions to a global one. Now, fix a local neighborhood $U$ and pick a co-frame $\{\epsilon_0,\epsilon_1,\ldots, \epsilon_n\}$ where $\epsilon_0 = \epsilon_t$ and that $\sigma(\epsilon_i,\epsilon_k) = \delta_{ij}$ if $1\leq i,j\leq n$. The connection can be expressed in the method of moving frames

$\nabla \epsilon_0 = 0$ by definition
$\nabla \epsilon_i = \sum_j \omega^j_i\otimes\epsilon_j + \omega^0_i\otimes\epsilon_0$

where the $\omega^*_*$ are real-valued one-forms called the rotation coefficients. Now, we use the fact that $\sigma$ is parallel

$0 = \nabla[\sigma(\epsilon_i,\epsilon_j)] = \sigma(\epsilon_i,\nabla\epsilon_j) + \sigma(\nabla\epsilon_i,\epsilon_j) = \omega^i_j + \omega^j_i$

so the $\omega^*_*$ are anti-symmetric in the indices. Now consider the top form $\epsilon_0\wedge \cdots \wedge\epsilon_n$. Its covariant derivative is, after substituting in the rotation coefficients,

$\nabla (\epsilon_0\wedge \cdots \wedge\epsilon_n) =$
$\sum_{j = 1}^n (-1)^j [\sum_{k = 1}^n \omega^k_j\otimes (\epsilon_k\wedge \epsilon_0 \wedge \cdots \hat{\epsilon}_j \cdots \wedge \epsilon_n) + \omega^0_j\otimes (\epsilon_0\wedge \epsilon_0 \wedge \cdots \hat{\epsilon}_j \cdots \wedge \epsilon_n) ]$

where the hat above a co-frame component means that the component is omitted. It is simple to see that many of the terms drop-out by anti-symmetry, and in the end we are left with

$\nabla(\epsilon_0\wedge \cdots \wedge\epsilon_n) = \sum_{j=1}^{n} \omega^j_j \otimes (\epsilon_0\wedge \cdots \wedge\epsilon_n)$

Now, using the anti-symmetry of the indices of $\omega^*_*$, we see that the right hand side vanishes identically. Therefore $\mathrm{Vol} = \epsilon_0\wedge \cdots \wedge\epsilon_n$ is parallel. To finish the proof we need to show that $\mu_\sigma = \epsilon_1\wedge \cdots \wedge \epsilon_n$ restricts to the volume form on $\Sigma_\tau$ (without a fixed normailsation, we cannot use the partition of unity argument unless $M$ is simply connected). But this follows from definition: by construction $\sum \epsilon_i\otimes\epsilon_i |\Sigma_\tau$ is precisely the induced Riemannian metric. Q.E.D.

Now, combining Proposition 8, Lemma 7, and the Stokes theorem, we see the following fact (assuming all the integrals converge):

Proposition 9
Let $X$ be a vector field on a Galilean manifold $(M,\epsilon_t,\sigma,\nabla)$, and let $\mathrm{Vol}$ as defined as in Proposition 8. Let $U$ be a region in $M$ such that its boundary is the disjoint union of $\Sigma_\tau$ and $\Sigma_{\tau'}$ with the latter “to the future” of the former. Then the divergence theorem takes on the form
$\displaystyle \int_U c(\nabla X) \mathrm{Vol} = \int_{\Sigma_{\tau'}} \epsilon_t(X) \mu_\sigma - \int_{\Sigma_{\tau}} \epsilon_t(X) \mu_\sigma$
where $\mu_\sigma$ is the volume form associated to the induced Riemannian metric on spatial slices.

The right hand side of the above equation follows because the restriction $\epsilon_t | \Sigma_\tau = 0$.

In particular, the above proposition implies

Corollary 10 (Conservation of energy)
Given a solution of the Newton-Cartan system, the total matter energy $E = \int_\Sigma T(\epsilon_t,\epsilon_t) \mu_\sigma$ is conserved.

Proof. Apply Proposition 9 to $X = T(\epsilon_t, \cdot)$ and note that $c(\nabla X) = 0$ using that $\epsilon_t$ is parallel and that the stress-energy tensor is divergence free. Q.E.D.