Commutator of Delta and Gradient on functions

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.

Comments (6)

6 Comments »

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

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

Reply
- 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
  
  Reply
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

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

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

Reply
- 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
  
  Reply

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Ngô Quốc Anh

May 8, 2011