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

There seems to be something curious about the so-called Van-Vleck-Morette determinant, as I cannot find any source that properly defines it in terms of expressions previously defined in that source and that actually proves properties.

To my understanding, a possible definition is $$ \triangle(p, q) = (\det(g_{ij}(x(p)))\det(g_{ij}(y(q)))^{-1/2} \det \left(- \frac{\partial^2\sigma}{\partial x^i \partial y^j}\right),$$ where $\sigma = d(p, q)^2/2$ and the expression means the following: Choose charts $x$ and $y$ such that $p$ and $q$ are in the respective domains of these charts, then we denote by $g_{ij}(x)$ is the matrix of the metric in the chart $x$ at the point $p$ and $g_{ij}(y)$ the corresponding matrix with respect to $y$ at $q$. One can check easily be the usual transformation laws under coordinate changes that this expression is well-defined.

Define on the other hand $$ \mu(p, q) = \det_g (d \exp_p|_X),$$ where $X$ is such that $\exp_p(X) = q$. Now if $x$ is a geodesic chart about $p$, then $\mu(p, q) = \det(g_{ij}(x))^{1/2}$.

Now my question is: These two functions should be related, if one trusts the literature, if I am not mistaken, $\triangle = \mu^{-1}$. However, I cannot find a proof anywhere and I couldn't figure out a proof myself. This should be a straightforward calculation of Riemannian geometry; however, the definition of $\triangle$ might be wrong. Does anybody know how to show this?

share|cite|improve this question
You might be able to use this: – Deane Yang Jul 4 '13 at 3:30
up vote 5 down vote accepted

The Van-Vleck Morette determinant often arises in semiclassical analysis, for example in computations of asymptotic expansions of heat kernels or Feynman path integrals. In these applications it is not easy to trace its origin as a transformation Jacobian.

The space of initial data of the geodesic initial data problem is the tangent bundle $TM$. (Equivalently the cotangent bundle $T^{*}M$ due to the existence of a metric. The space of boundary data of the geodesic boundary value problem is $M \times M$. Trading the dependence on the initial momenta by the dependence on the final positions defines a map $T^{*}M \rightarrow M \times M$ whose Jacobian as will be shown below is the Van-Vleck Morette deterinant. From the derivation it will be explicit that it is the inverse of the Jacobian of the exponential map.

In the following, the initial $t=0$ and final $ t=1$ positions along the geodesic will be denoted by $x(0)$ and $x(1)$, and the corresponding momenta by $p(0)$ and $p(1)$.

The Liouville volume element of $T{*}M$ is given by:

$$d\mu_L = \wedge dx^{i}(0)\wedge_idp_i(0)$$

The Hamilton-Jacobi function is equal to the action integral along the geodesic between the initial and final points;

$$ \sigma(x(0), x(1)) = \int_{x(0), t=0}^{x(1), t=1} dt \sum_{ij} g_{ij}(x(t)) \dot{x}^{i}(t)\dot{x}^{i}(t)$$

$\sigma$ satisfies the Hamilton-Jacobi equation:

$$ g^{ij}(x(0)) \frac{\partial\sigma}{\partial x^{i}(0)} \frac{\partial\sigma}{\partial x^{j}(0)} = E$$

Where $E = \sum_{ij} g^{ij}(x(0)) p_i(0){p_j(0)}$, is the ( invariant) energy.

The initial momentum along the geodesic can be derived from the Hamilton-Jacobi function by:

$$p_i(0) = \frac{\partial\sigma}{\partial x^{i}(0)}$$

Thus the transformed Liouville measure on $M\times M$ is given by:

$$ \wedge_i dx^{i}(0)\wedge_i dp_i(0) = \det\big ( \frac{\partial p(0)}{\partial x(1)} \big) \wedge_i dx^{i}(0) \wedge_i dx^{i}(1) $$

$$ = \frac{\det( \frac{\partial^2 \sigma}{\partial x(0)^i\partial x(1)^j}) }{ \sqrt{\det(g(x(0)) \det(g(x(1))}} \sqrt{\det(g(x(0))} \wedge_i dx^{i}(0) \sqrt{\det(g(x(1))}\wedge_i dx^{i}(1) $$

$$ = \Delta(x(0), x(1)) \det(g(x(0))) \sqrt{\det(g(x(0))} \wedge_i dx^{i}(0) \sqrt{\det(g(x(1))}\wedge_i dx^{i}(1).$$

Now by definition:

$$x(1) = \mathrm{Exp}_{x(0)}(v(0))$$

Where the velocity: $v_0 \in T_{x(0)}$ is given by:

$$v^i(0)= \sum_j g^{ij}(x(0))p_j(0)$$


$$\det_g( d \mathrm{Exp}_{x(0)}) = \sqrt{\det(g(x(0))}^{-1} \sqrt{\det(g(x(1))} \det(\frac{\partial x(1)}{\partial v(0)}) = \sqrt{\det(g(x(0))}^{-1} \sqrt{\det(g(x(1))} \det(g(x(0)) \det(\frac{\partial x(1)}{\partial p(0)}) = \Delta(x(0), x(1))^{-1}$$

Using the intermediate step of the evaluation of the Van-Vleck Morette determinant.

share|cite|improve this answer

For Lorentzian geometry (which uses the same formulas as Riemannian geometry up to a few signs) you can find a pretty concise definition of this determinant and the equivalence of these two formulas in Section 7 of this very nice review article:

Eric Poisson and Adam Pound and Ian Vega, "The Motion of Point Particles in Curved Spacetime", Living Rev. Relativity 14, (2011), 7.

share|cite|improve this answer

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.