# integrating over a geodesic

Suppose $\mathcal{M}$ is a Riemannian manifold with the volume measure $d\mu$(induced by the Riemannian metric ) and $\gamma$ is a geodesic on $M$. Is it true that $$\int_{\gamma}fd\nu\leq\int _{\mathcal{M}}fd\mu,$$ if yes what replaces $d\nu$ exactly? Edit:

$$\int_{\gamma}fd\nu\int _{\mathcal{M}}d\mu\leq\int _{\mathcal{M}}fd\mu\int_{\gamma}d\nu,$$

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## 1 Answer

What is $d\nu$? Do you mean the measure induced by the induced metric on $\gamma$ as a submanifold? In that case the answer is no.

Let $\mathcal{M} = S^1\times S^1$ with coordinates $(x,y)\in [0,2\pi)^2$, with the line element $ds^2 = dx^2 + \epsilon dy^2$. And let $\gamma$ be the geodesic $y = 0$. The induced Riemannian metric on the geodesic is $d\ell^2 = dx^2$.

Let $f = 1$ the constant function. Then $\int_\gamma f d\nu = 2\pi$, while $\int_{\mathcal{M}}fd\mu = 4\pi^2 \epsilon$. By choosing $\epsilon$ sufficiently small you invalidate the proposed inequality.

More generally, purely by dimensional considerations you see that it is unreasonable to expect any expressions of the form you wrote to hold, except possibly in the case that $\gamma$ is treated as a (measure zero) subset of $\mathcal{M}$, and $d\nu$ is the measure $d\mu$, and $f$ is non-negative. Then the LHS is trivially zero and the RHS is non-negative.

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@ Willie Wong I changed the inequality. – timhortons Aug 15 '11 at 19:26
Edit is still false. Let $M$ be the manifold given by the disjoint union of what I wrote in my above example, and $\mathbb{S}^2_r$, the standard sphere of radius $r$. Let $f = 0$ on the sphere. Now take $r\to\infty$. The LHS blows up, the RHS remains the fixed. – Willie Wong♦ Aug 15 '11 at 19:53
If you want a connected manifold, you can consider the case where $f$ has compact support in a small tubular neighborhood of the geodesic. – Willie Wong♦ Aug 15 '11 at 19:54
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