Heat Invariants on a one - dimensional Riemannian manifold

I am trying to understand the asymptotic heat trace expansion $$\text{Tr}(e^{-t\triangle_g}) \backsim \sum_{k \geq 0} t^{k - \frac{n}{2}}c_{2k} \quad (t \to 0^+)$$ that is associated with the Laplace - Beltrami operator $$\triangle_g = \frac{1}{\sqrt{\text{det}(g)}} \sum^n_{i,j = 1} \frac{\partial}{\partial x_i} g^{ij} \sqrt{\text{det}(g)} \frac{\partial}{\partial x_j}$$ where $g$ denotes the metric on the n - dimensional Riemannian Manifold $(M,g)$ that we consider.

Now, I understand that the first invariant $c_0$ is equal to the Volume of $M$ and the next coefficient is associated with the Scalar curvature (is that correct?). If I take $M$ to be 1-dimensional compact and boundaryless, then this means the second coefficient should evaluate to zero, is that correct ?

Many thanks for your help!

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1 Answer

There is only one closed (i.e. compact and boundaryless) 1-manifold, the sphere $S^1$. Furthermore, the circumference is in that case the only metric invariant, i.e. two 1-dimensional spheres $(S^1, g)$ and $(S^1, h)$ are isometric if and only if they have the same circumference.

So restricting to the case of $M = S^1$ with circumference $l$, the asymptotic of the trace of the heat operator is given by $$\operatorname{Tr} e^{-t\Delta} \sim \frac{l}{\sqrt{4\pi}} t^{-\frac{1}{2}}.$$

So you are right, in this case all higher heat invariants vanish and the only information stored in the asymptotic of the trace of the heat operator are the dimension and the circumference, i.e. the heat invariants give in this case a complete set of isometry invariants.

All this can be found, e.g., in Chapter 1.1 of

Rosenberg: The Laplacian on a Riemannian manifold, London Mathematical Society Student Texts 31, Cambridge University Press, 1997.

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1 dimensional spheres are often called circles:) –  Thomas Rot Jan 16 '12 at 11:26
thanks a lot! that helped a great deal! –  harlekin Jan 16 '12 at 11:28