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.

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