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?

  • $\begingroup$ Maybe these functions are compactly supported. $\endgroup$
    – IAmNoOne
    Apr 20, 2022 at 7:50

1 Answer 1


The divergence theorem works in the nonorientable case as well, it just requires the use of densities rather than differential forms. See for instance theorem 16.48 of Lee's Introduction to Smooth Manifolds.

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 $[0,1]$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .