# Integration by parts for tensor fields on Riemannian manifold

I'm working on the following exercise in my Riemannian manifolds book:

Suppose $$M$$ is a compact, oriented Riemannian manifold with boundary. Show that if $$\omega$$ is any $$k$$-tensor field and $$\eta$$ is any $$k+1$$-tensor field, $$\int_M \langle \nabla \omega, \eta \rangle \, dV = -\int_M \left\langle \omega, \mathrm{tr}_g \nabla \eta \right\rangle \, dV + \int_{\partial M} \left\langle \omega \otimes N, \eta \right\rangle \, d\tilde V,$$ where the trace is on the last two indices of $$\nabla\eta$$ and $$N$$ is the unit normal field along $$\partial M$$.

I really don't know where to start with this. I've been able to show the divergence theorem and the integration by parts formula for the divergence operator $$\int_M \langle \mathrm{grad}f, X \rangle \, dV = -\int_M f\mathrm{div}X\,dV + \int_{\partial M} f\langle X, N \rangle\,d\tilde V$$ and the book has led me to believe this problem is somehow related to divergence, but past this I'm really stuck. I don't even know how to interpret the integrands in the expression, since this is the notation the book uses for the Riemannian metric or the pairing of vectors and covectors, but the arguments in the brackets are $$k+1$$-tensors, $$k$$-tensors, or mixed tensors. Can anyone help?

EDIT: Building on the answer below, I've managed to work out that we can say $$\langle \omega, \eta \rangle$$ is a $$1$$-form via $$X \mapsto \langle \omega, \iota_X \eta \rangle$$, where $$\iota_X$$ is interior multiplication (ie. $$\iota_X \omega$$ is the $$n-1$$-form given by plugging $$X$$ into the first argument of the $$n$$-form $$\omega$$, etc). Then we want its corresponding vector field $$\langle \omega, \eta \rangle^\sharp$$ to satisfy:

1. $$\iota_{\langle \omega, \eta \rangle^\sharp} dV = \langle \omega, \iota_N \eta \rangle d\tilde V$$ restricted to $$T\partial M$$, where $$N$$ is the outer unit normal to $$\partial M$$; and
2. $$d\left(\iota_{\langle \omega, \eta \rangle^\sharp} dV\rangle\right) = \langle \nabla \omega, \eta \rangle dV + \langle \omega, \mathrm{tr}\nabla \eta \rangle dV$$.

First follows from the analogous results for arbitrary vector fields: $$\iota_X dV = \langle X, N \rangle dV$$. Still working on showing the second. Does it even make logical sense if $$\eta$$ is a $$k+1$$-tensor field? Because then $$\nabla \eta$$ is a $$k+2$$-tensor field, and in order to take its trace we need $$\eta$$ to have at least one vector index in order for $$\nabla \eta$$ to have a vector index.

• Can you write the left-hand side in coordinates? – Giuseppe Negro Mar 6 at 21:05
• Well, I don't know what $\langle \nabla \omega, \eta \rangle$ is in the first place, so... not really. I guess it would look something like $\sum \omega_{i_1 \cdots i_k ; j} \eta_{i_1 \cdots i_k j}$? – D Ford Mar 6 at 21:09

Let $$\varphi=\langle \omega, \eta\rangle$$, this is a $$1$$-form, locally defined by $$\varphi_r=g^{i_1j_1}...g^{i_kj_k}\omega_{i_1, ... ,i_k}\eta_{r, j_1, ... ,j_k}$$. Then the integrand $$\langle\nabla\omega, \eta\rangle+\langle \omega, \text{Tr}\nabla\eta\rangle$$ over $$M$$ is just the divegence of $$\varphi$$: that is, $$\nabla_{e_i}\varphi(e_i)$$ with $$e_i$$ any (normal) orthonormal frame at the point where you compute; and the boundary integrand is just $$\varphi(N)$$.
In another word, let $$X$$ be the dual vector of $$\varphi$$, then the $$M$$ integrand above is the divergence of $$X$$, i.e. $$\langle \nabla_{e_i} X, e_i\rangle$$, and the boundary integrand is $$\langle X, N\rangle$$.
• $\langle \omega, \eta \rangle$ is a $1$-form? Your coordinate formula seems to imply that $\phi(X) = \langle \omega, \iota_X \eta \rangle$, where $\iota_X$ denotes the interior product by $X$, and in this case $\langle \cdot, \cdot \rangle$ should be interpreted to be a canonical inner product on the space of $k$-tensors. Am I interpreting your response correctly? – D Ford Mar 7 at 21:01
• Perhaps in the same vein, should we interpret $\langle \omega \otimes N, \eta \rangle = \langle \omega, \iota_N \eta \rangle$? – D Ford Mar 7 at 21:04
• Yes, that is correct. One may want to point out $\iota_V\eta$ is "plug in $V$ as the first variable of $\eta$". – Yu Ding Mar 8 at 1:57