MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $g$ be a Riemannian metric on the $d$-dimensional flat space $\mathbb R^d$, and consider the usual Lagrangian $$L(x, \dot x) = \tfrac 1 2 g_{ij}(x) \dot x^i \dot x^j.$$ Let $\hat g := \sqrt g$ denote the square root of the metric $g$, implicitly defined by the formula $\hat g_{ai} \hat g_{bj} \delta^{ab} = g_{ij}$, where $\delta^{ab}$ is the identity 2-tensor. I want to introduce phase-space-type coordinates $$u_i = \hat g_{ij} \dot x^j.$$ In the coordinates $(x,u)$, the metric on $u$ is just the Euclidean metric: $\langle u, u \rangle := \delta^{ij} u_i u_j$.

Let $x = x(t)$ denote a geodesic for the metric $g$, and define $u(t) := \hat g_{ij} \dot x^j$. These coordinates are convenient, because along the geodesic, $u(t)$ remains on the sphere of radius $|u(0)|$: $$\tfrac{d}{dt} \langle u, u \rangle = \tfrac{d}{dt} \delta^{ij} u_i u_j = \tfrac{d}{dt} g_{ij} \dot x^i \dot x^j = 0,$$ since geodesics are parametrized by unit speed. Conveniently, this means that $\langle u, \dot u \rangle \equiv 0$.

(You may wonder why don't I just use Hamiltonian phase-space coordinates $(x,p)$. In my research, I consider $g$ as a parameter, ranging over all possible Riemannian metrics on the plane $\mathbb R^2$. Hamiltonian coordinates have the nice property that for a fixed metric, the energy shells $\{ g^{ij}(x) p_i p_j = \mathrm{constant} \}$ are invariant under the geodesic flow. Unfortunately, these energy shells are not independent of the metric parameter $g$. In the coordinates $(x,u)$, on the other hand, the shells $\{ \langle u, u \rangle = \mathrm{constant} \}$ are just spheres in Euclidean space, and do not depend on $g$. In particular, it is important to me that these spherical shells are invariant under rotations in the phase space $\mathbb R^d \times \mathbb R^d$).

I want to calculate the geodesic equation in the coordinates $(x,u)$, particularly for the case that $d=2$. It is easy to see that $\dot x^j = \hat g^{ji} u_i$, where the superscripts denote the inverse of $\hat g$. When I calculate $\dot u$, though, I get a mess: $$\dot u_a = \big( \hat g_{ab,c} \hat g^{cj} \hat g^{bi} - \hat g_{ab} \Gamma_{uv}^b \hat g^{ui} \hat g^{vj} \big) u_i u_j,$$ where $\Gamma_{uv}^b$ are the Christoffel symbols for the metric $g$. I tried simplifying this expression, to no effect. There is plenty of symmetry around (e.g., $\langle u, \dot u \rangle = 0$), and I'm sure that the formula for $\dot u$ takes a much, much simpler form.

Question: Is there a simple expression in these coordinates for the evolution of $u(t)$?

Let me explain why the above expression is inadequate. For the metric $g$, let $U_g$ denote the vector field given by $U_g(x,u) = (u, \dot u)$ (where $\dot u$ is the expression above), so that solutions to the differential equation $(\dot x, \dot u) = U_g(x,u)$ are geodesics for the metric $g$. I need to calculate the (Euclidean) divergence $\operatorname{div} U$. I am pretty sure that in the end, $\operatorname{div} U$ can be expressed in some simple geometric quantities involving the metric (like the Riemannian divergence $\operatorname{div}_g$ of some vector field, scalar curvature $K_g$, etc.). For the messy $\dot u$ above, though, it is impossible for me to see what the true character of $\operatorname{div} U$ is.

share|cite|improve this question
This is a totally naive remark, but if you select a particular, explicit, easily invertible $g$, and compute $\dot u$, is your hope that it takes a simpler form borne out? – Joseph O'Rourke May 28 '11 at 23:36
A very good question, Joseph. If the metric is in appropriate normal coordinates, then the first derivatives and the Christoffel symbols vanish. Consequently, $\dot u$ vanishes. I have not tried calculating the expression with any fixed, non-trivial metric. – Tom LaGatta May 29 '11 at 8:01
up vote 4 down vote accepted

The answer that you want, namely div $U_g$, is not going to expressible in terms of a geometrically invariant quantity (such as, say, the scalar curvature of $g$) because it depends on the underlying coordinates $x$ in which you have presented the metric. For example, if $\det g(x)$ is constant (which is a coordinate dependent thing), then the divergence of $U_g$ (computed in the $xu$-coordinates, which, I assume, is what you mean by the 'Euclidean divergence') will vanish identically (and conversely, as a matter of fact).

I'll try to explain this in the symplectic formulation, since that's the version I find the clearest, but I think that you can make the translation on your own. You start with a Riemannian metric $g = dx\cdot G(x) dx$, where $G$ is a function on $\mathbb{R}^n$ with values in positive definite symmetric matrices. Let's write $G = F^TF$, where $F$ is invertible (but not necessarily symmetric; you can take $F$ to be the positive definite square root of $G$ if you like, but that's not necessary for my argument). The Lagrangian is then $L = \tfrac12 u\cdot u$, where $u = F(x)\ dx$, regarded as an $\mathbb{R}^n$-valued function on the tangent bundle of $\mathbb{R}^n$ (i.e., an orthonormal coframing of the underlying manifold). Applying the Legendre tranform, the symplectic form on the cotangent bundle pulls back to the tangent bundle to be $$ \Omega = dp\wedge dx = d((F(x)\ u)^T) \wedge dx = (du)^T\ F(x)^T \wedge dx + u^T\ d(F(x)^T)\wedge dx $$ The vector field $U_g$ is the $\Omega$-Hamiltonian vector field associated to $L$, i.e., it satisfies $$ \iota(U_g)\ \Omega = -dL = - u\cdot du $$ (where $\iota(X)$ denotes interior product, what I normally call 'lefthook'). By the standard identity, the divergence of $U_g$ with respect to the Liouville volume form, i.e., $\mu = \tfrac1{n!}\Omega^n$, vanishes identically. Now, you can compare the Liouville volume form with the 'Euclidean' volume form in $xu$-coordinates by noting that, by exterior algebra, we have $$ \tfrac1{n!}\Omega^n = \det(F(x))\ \tfrac1{n!}\bigl((du)^T\wedge dx\bigr)^n $$ It follows that the 'Euclidean' divergence of $U_g$ in the $xu$-coordinates is given by the formula $$ -U_g\bigl(\log|\det(F(x))|\bigr) = -\tfrac12\ U_g\bigl(\log|\det(G(x))|\bigr). $$ Since, as you have already computed, $U_g(x) = F^{-1} u$, it follows that the divergence you want is, as a function, linear in the $u$-coordinates. You might write it as $$ -\tfrac12 \nabla \bigl(\log|\det(G(x))|\bigr)\cdot (F^{-1} u). $$

share|cite|improve this answer
Robert, thank you for your elegant and elucidating answer; very little translation is necessary. I am not surprised that symplectic geometry plays a role in the solution. I was so in the thick of coordinates, that I hadn't grasped that $(x,u)$ really are just coordinates for the cotangent bundle. – Tom LaGatta May 29 '11 at 19:12
@Tom: I'm glad this was helpful. I also meant to remark on your main question about simplifying the formula for $U_g$. The reason your expression is complicated is that it involves the derivatives of both $\hat g$ and $g$ but doesn't take advantage of the relation between them generated by differentiating $g = {\hat g}^2$. If you express everything directly in terms of $\hat g $ and its derivatives (or, more generally, $F$ in my presentation), you'll get a simpler expression. Also, you can figure this formula out directly with a minimum of fuss from the 'lefthook' definition of $U_g$. – Robert Bryant May 30 '11 at 14:33

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.