# Note on generating a divergence identity

Special thanks to Ben Mares for coming up with this identity. $\newcommand{\cd}{\nabla}$

## Motivation

The motivation for this calculation comes from Bochner’s formula for harmonic functions. On a manifold $M$ with metric $g$, a function $u: M\to \mathbb{R}$ is harmonic if $\Delta u = 0$, where $\Delta \equiv \cd^a \cd_a$ is the Laplacian and $\cd$ is the Levi-Civita connection compatible with $g$. Let’s define $X_a =\cd_a u$, which is divergence-free, $\cd_a X^a = 0$. Then Bochner’s formula says

\begin{align} \label{eq:bochner} \Delta( \frac{1}{2} X^a X_a ) = (\cd_a X_b)(\cd^a X^b) + R_{ab} X^a X^b \,. \end{align}

Why is this formula interesting? Well, the left hand side is a total divergence, so it will vanish if integrated over a manifold without boundary (or if the boundary term vanishes). Meanwhile, if the metric is Riemannian, then the first term on the right hand side is non-negative; and the sign of the final term is determined by the eigenvalues of the Ricci tensor. This formula is useful for estimating energy bounds.

Another viewpoint on this formula is to rearrange it as an identity for $|\cd X|^2$, which looks like a kinetic term. Then Bochner’s formula says: $|\cd X|^2$ is a total divergence, modulo lower-derivative curvature terms.

A natural question is: does this divergence identity generalize?

## Generating the divergence identity

From here forward let’s drop the divergence-free condition, so $\cd_a X^a$ may be non-zero (though the identities hold just as well for vanishing $\cd_a X^a$).

It turns out that we can generalize the above identity by constructing carefully anti-symmetrized tensor products of the tensor $M_a{}^b \equiv \cd_a X^b$. These expressions arise in the expansion of the characteristic polynomial of $M_a{}^b$ (the characteristic polynomial made an appearance in an earlier note that I wrote). The characteristic polynomial of an $n\times n$ matrix is given by $p_M(t) \equiv \det(tI - M)$, where $I$ is the $n\times n$ identity matrix.

The term proportional to $t^{n-k}$ is homogeneous of degree $k$ in the coefficients of $M$. Wikipedia writes this as $(-1)^k \mathrm{tr}(\Lambda^k M)$, which is a bit abstruse. Let’s clarify by first forming the $k$-th tensor product,

\begin{align} M_{i_1}{}^{j_1} M_{i_2}{}^{j_2} \cdots M_{i_k}{}^{j_k} \,. \end{align}

Now antisymmetrize on all the i and/or j indices, and completely contract the upper indices with all of the lower indices. Let’s call this $p_{(k)}$,

\begin{align} p_{(k)} = (\cd_{i_1}X^{[i_1}) (\cd_{i_2}X^{i_2})\cdots (\cd_{i_k}X^{i_k]}) \,. \end{align}

It is important to remember that the antisymmetrization happens first, before contraction.

This question is now: is $p_{(k)}$ a total divergence, modulo lower-derivative curvature terms? The answer is yes (why else would I have written these notes).

To see this, let’s define the vector

\begin{align} D_{(k)}^i \equiv X^{[i} (\cd_{i_2}X^{i_2}) \cdots (\cd_{i_k}X^{i_k]}) \,. \end{align}

What happens when we take the divergence of this vector? Using the product rule, we will generate $k$ terms (each of which is actually antisymmetrized over the $k$ upper indices, so it’s really $k\cdot k!$ terms).

The first of these terms, when $\cd_i$ hits $X^i$, is precisely $p_{(k)}$. What about the remaining $k-1$ terms, when $\cd_i$ hits $\cd_{i_r}X^{i_r}$, $2\le r \le k$? That is, a term of the form

\begin{align} X^{[i} (\cd_{i_2}X^{i_2}) \cdots (\cd_i \cd_{i_r}X^{i_r}) \cdots (\cd_{i_k}X^{i_k]}) \,. \end{align}

The crucial observation is that the two derivative indices $i\,i_r$ are contracted with the completely anti-symmetric block of indices $[i\,i_1\,\ldots\,i_k]$. This means that we are free to transfer over the antisymmetry onto the double derivative indices $[i\,i_r]$; and this makes it a commutator of derivatives and thus a curvature term.

Thus we arrive at the following divergence identity,

\begin{align} \cd_i D_{(k)}^i - p_{(k)} = \sum_{r=2}^k \frac{1}{2} X^{[i_1} (\cd_{i_2}X^{i_2}) \cdots R_{i_1 i_r}{}^{i_r}{}_j X^{|j|} \cdots (\cd_{i_k}X^{i_k]}) \,, \end{align}

where the vertical bars around $j$ mean that we do not antisymmetrize that index. Again it is important to antisymmetrize over the upper indices first, and trace second; so the Riemann does not automatically become a Ricci.

Thus we have established that $p_{(k)}$ is a total divergence, modulo lower-derivative curvature terms.

We can give a few examples, but the expressions get very large very quickly. For $k=2$, we have

\begin{align} \cd_i D_{(2)}^i - p_{(2)} = -\frac{1}{2} R_{ab}X^a X^b \,, \end{align}

which is a way of re-writing Bochner’s formula Eq. \eqref{eq:bochner}. For $k=3$, we have

\begin{align} \cd_i D_{(3)}^i - p_{(3)} = \frac{1}{3} X^a X^b \left( R_{bc}\cd_a X^c -R_{ab}\cd_c X^c +R_{acb}{}^d\cd_d X^c \right) \,. \end{align}

For $k=4$, we have the intimidating

\begin{align} \cd_i D_{(4)}^i - p_{(4)} = -\frac{1}{8} X^{a} X^{b} \Big[& 2 \cd_{a}X^{c} (R_{bd} \cd_{c}X^{d} - R_{bc} \cd_{d}X^{d}) \\ & {}+ R_{ab} (\cd_{c}X^{c} \cd_{d}X^{d} - \cd_{c}X^{d} \cd_{d}X^{c}) \nonumber\\ & {}+ 2 \cd_{e}X^{d} (R_{acb}{}^{e} \cd_{d}X^{c} + R_{cdb}{}^{e} \cd_{a}X^{c} - R_{adb}{}^{e} \cd_{c}X^{c}) \Big]\,.\nonumber \end{align}

Obviously I would recommend using xTensor to do these computations. My notebook to do these calculations is available upon request.

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