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I am trying to evaluate the following integral: \begin{equation} \int_{M} \frac{(g')^2}{ g^{5/2}} - \frac{(g'')}{g^{3/2}} \ dx \end{equation} where $(M,g)$ is a one - dimensional closed and compact Riemannian manifold with metric g. So $g' = \frac{dg}{dx}$ and likewise $g'' = \frac{d^2g}{dx^2}$ (locally).

I suspect the integral to be zero but I am not sure how to proceed. In particular I'm not even sure whether the integral above makes sense .. I am totally new to Riemannian Geometry so in case the above is ill - defined please say so. If anyone could help that would be great, many thanks!

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up vote 1 down vote accepted

The integral is not zero in general. To see this, note that $$d\Big(\frac{g'}{g^{\frac{3}{2}}}\Big)=\left(\frac{g''}{g^{\frac{3}{2}}}-\frac{3}{2}\cdot\frac{(g')^2}{g^{\frac{5}{2}}}\right)dx.$$ Integrate it over $M$, we get $$\tag{1}\int_Md\Big(\frac{g'}{g^{\frac{3}{2}}}\Big)=\int_M\left(\frac{g''}{g^{\frac{3}{2}}}-\frac{3}{2}\cdot\frac{(g')^2}{g^{\frac{5}{2}}}\right)dx.$$ Since $M$ is a $1$-dimensional closed and compact Riemannian manifold, the boundary of $M$ is empty, i.e. $\partial M=\emptyset$. By Stokes' Theorem, the left hand side of $(1)$ is equal to $0$. So $(1)$ becomes $$\int_M\left(\frac{g''}{g^{\frac{3}{2}}}-\frac{(g')^2}{g^{\frac{5}{2}}}\right)dx=\int_M\frac{(g')^2}{g^{\frac{5}{2}}}dx.$$ Therefore, the integral is not zero except $g'=0$, i.e. $g$ is constant.

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Ok thanks very much! I will try to find the bug in the calculation, it looks like I missed something to get to the closed form you mention above. Thks again! – harlekin Jan 15 '12 at 22:10

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