# Integration by parts on compact, non-orientable Riemannian manifold with boundary

Let $$(M,g)$$ be a compact Riemannian manifold, not necessarily orientable or without boundary. Let $$\mu$$ be a normalized volume measure on $$M$$ and $$u$$ be a smooth function on $$M$$. In some notes that I received, following equation is claimed by integration by parts:

$$\int_M e^u \Delta u \,d\mu = - \int_M \lvert\nabla u\rvert^2 e^u d\mu$$.

I have some trouble following it. As far as I know, the following "product rule" holds in that case (even if $$M$$ is unorientable):

$$\nabla \cdot (fX) = \nabla f\cdot X + f \nabla \cdot X$$

for any smooth function $$f$$ and smooth vector field $$X$$ on $$M$$. Can anyone confirm this with some reference, as I didn't find a good one? In the above case that would mean,

$$\nabla\cdot (\nabla e^u)=\nabla\cdot (e^u \nabla u)= e^u \lvert \nabla u\rvert^2 + e^u\Delta u.$$

Thus, this yields

$$\int_M e^u \Delta u d\mu = - \int_M \lvert\nabla u\rvert^2 e^u d\mu + \int_M \nabla\cdot (\nabla e^u) d\mu.$$

Now, I do not quite see how to get rid of the second integral on the right hand side. If $$M$$ were orientable, I think the divergence theorem would give

$$\int_M \nabla\cdot (\nabla e^u) d\mu = \int_{\partial M}(\nabla e^u)\cdot dS,$$

which would be zero if the manifold is without boundary. Yet, if $$M$$ is not orientable and does have a boundary, I don't see how to get the initial formula. Can this really hold in the afore-mentioned generality?

• Maybe these functions are compactly supported. Apr 20, 2022 at 7:50

The boundary term won't generally vanish without further conditions on $$u$$. It should be easy to come up with a counterexample on a simple case like $$[0,1]$$.