# Ngô Quốc Anh

## May 8, 2011

### Commutator of Delta and Gradient on functions

Filed under: Riemannian geometry — Ngô Quốc Anh @ 0:05

Let us prove following interesting identity between $\Delta$ and $\nabla$

$\Delta \nabla_i f=\nabla_i \Delta f+{\rm Ric}_{ij}\nabla_j f$

for any function $f$.

For any function $f$, using the formula

$\Delta = g^{ij}\nabla_i \nabla_j$

we obtain

$\Delta {\nabla _i}f = {g^{mn}}{\nabla _m}{\nabla _n}({\nabla _i}f).$

Since $f$ is a funtion, we know that

${\nabla _n}({\nabla _i}f) = {\nabla _i}({\nabla _n}f)$

then we obtain

$\Delta {\nabla _i}f = {g^{mn}}{\nabla _m}({\nabla _i}({\nabla _n}f)).$

Now $\nabla_n f$ is a vector, we need to use the Riemmanian curvature tensor, which measures the difference between $\nabla_m \nabla_i$ and $\nabla_i\nabla_m$.

${\nabla _m}({\nabla _i}({\nabla _n}f)) = {\nabla _i}({\nabla _m}({\nabla _n}f)) + {\rm Rm}({e^m},{e^i}){\nabla _n}f.$

Notice that there was an extra term involving Lie bracket. Fortunately, that term vanishes. Thus

$\Delta {\nabla _i}f = {g^{mn}}{\nabla _i}({\nabla _m}({\nabla _n}f)) + {g^{mn}}{\rm Rm}({e^m},{e^i}){\nabla _n}f = {\nabla _i}\Delta f + {\rm Ric}_{ik}{\nabla _k}f.$

The proof follows.