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.

6 Comments »

  1. I am wondering that why \nabla_n f is a vector?

    Comment by lucky — February 5, 2013 @ 20:14

    • Hi lucky, thanks for your comment, to be exact, I should say that \nabla f is a vector field instead of \nabla_n f. However, as components, we often use \nabla_nf to denote the vector field \nabla f since

      \displaystyle\nabla f = \underbrace {{g^{nj}}\frac{{\partial f}}{{\partial {x^j}}}}_{{\nabla _n}f}\frac{\partial }{{\partial {x^n}}}.

      Comment by Ngô Quốc Anh — February 5, 2013 @ 21:08

  2. Hi Ngo!, To your knowledge,is there an analog relation between the Laplacian and the bi-Laplacian?
    F

    Comment by Fab — September 28, 2013 @ 23:13

    • Dear Fab,

      I have not seen anything like that once. Fortunately, we can derive a similar formula by taking another \Delta to get

      \displaystyle \Delta^2 \nabla_i f = \nabla_i (\Delta^2 f) + \text{Ric}_{ij} \nabla_j f + \text{Ric}_{ij} \nabla_j (\Delta f).

      I hope this could help you a bit😉.

      Comment by Ngô Quốc Anh — September 28, 2013 @ 23:19

  3. Just a curiosity, how can I write down the bi-Laplacian on a manifold?

    Comment by Fab — September 29, 2013 @ 21:28

    • Dear Fab,

      I don’t think we have a very precise formula for \Delta^2 in local coordinates so far as its calculation is quite complicated and possibly useless. Instead, you may use

      \Delta^2 u = g^{ij}\nabla_i \nabla_j (g^{pq}\nabla_p \nabla_q u) = \nabla_i\nabla_i (\nabla_j \nabla_j u)

      to deal with.

      Comment by Ngô Quốc Anh — September 29, 2013 @ 22:30


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