### Newton-Cartan, Part 2

#### by Willie Wong

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

Asymmetryof a Galilean manifold is a diffeomorphism such that it preserves the geometric structures: that the pullback , the pushforward , and the pullback connection .

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

Definition 2

Aninfinitesimal symmetryof a Galilean manifold is a vector-field on such that , , and , where denote the Lie derivative by and square-brackets denote commutator of operators.

Now, by the Cartan relation, . By assumption is closed, and this shows that

Equation 3

On the other hand, using the definition

and that is parallel and symmetric, this implies

Equation 4

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 is a symmetry and it is tangent to a spatial hypersurface at one point, then is always tangent to spatial hypersurfaces and that 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 will carry hypersurfaces 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

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

Equation 5

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 dimensional manifold with boundary , and given a -form , Stokes theorem tells us that

where on the right hand side we use the induced orientation on . From this, we can derive a dual-version: the divergence theorem. Now let denote, instead, *a* volume form on (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 on , then the interior derivative is a -form. Apply Stokes theorem to this form, and we have

where in the last equality we used the Cartan relation again. Since is a top form, it is automatically closed, so the integrand on the far right simplifies to . Now, since is one-dimensional over the ring of smooth functions, and that the units in this ring corresponds to non-vanishing functions, we have that for some smooth function . This way we *define* the divergence of the vector field :

Definition 6

Let be a smooth manifold and a volume form. For a given vector field , thedivergence of relative to, written is the unique function given by .

Notice that the definition does not depend on a connection! Furthermore it depends on the choice of volume form. Let be another volume form. Then a computation shows that

On the other hand, we can also define the object , the tensor contraction of the covariant derivative of . 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 and a volume form . Then the following two conditions are equivalent

- for all vector fields .

*Proof*. Let . A direct calculation shows that

Now if are linearly dependent, then both sides evaluate simply to zero. So assume form a basis. Now 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 so that . Now, plugging this expression into the above equation, we notice that if , then by linearity ( is just a real number) and anti-symmetry, the form evaluates to zero. So we can simplify

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

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 be a Galilean -dimensional manifold, and assume that is orientable. Then there exists a preferred volume form with the property that .

*Proof*. will be, morally speaking, given by , where is any -form such that its restriction to the spatial hypersurfaces 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 , 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 . That the two forms both restrict to as the induced volume form by the Riemannian metric means that we have a preferred normalisation by construction, and hence on 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 and pick a co-frame where and that if . The connection can be expressed in the method of moving frames

by definition

where the are real-valued one-forms called the rotation coefficients. Now, we use the fact that is parallel

so the are anti-symmetric in the indices. Now consider the top form . Its covariant derivative is, after substituting in the rotation coefficients,

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

Now, using the anti-symmetry of the indices of , we see that the right hand side vanishes identically. Therefore is parallel. To finish the proof we need to show that restricts to the volume form on (without a fixed normailsation, we cannot use the partition of unity argument unless is simply connected). But this follows from definition: by construction 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 be a vector field on a Galilean manifold , and let as defined as in Proposition 8. Let be a region in such that its boundary is the disjoint union of and with the latter “to the future” of the former. Then the divergence theorem takes on the form

where 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 .

In particular, the above proposition implies

Corollary 10 (Conservation of energy)

Given a solution of the Newton-Cartan system, the total matter energy is conserved.

*Proof*. Apply Proposition 9 to and note that using that is parallel and that the stress-energy tensor is divergence free. Q.E.D.

Made a correction to the proof of Proposition 8. It is important to use the fact that in the construction the “trace” of on the spatial slices is precisely the volume form associated to the induced Riemannian metric. Else there may be a problem with topology when the manifold is not simply connected. See also the next post on Parallel Volume Forms for another point of view.

[…] university maths, differential/pseudo-Riemannian geometry — Willie Wong @ 15:33 In the previous post on Newton-Cartan theory and Galilean geometry, I showed that the Galilean manifolds admit a […]