1
$\begingroup$

In the book Shape and Shape Theory of Kendall in p.147 I found the following expression of the Laplace-Beltrami operator: $\sum_i\left({v_i^2-\nabla_{v_i}v_i}\right)$ where $v_i$ are orthonormal tangent vectors. So please what does the exponent 2 stands for?

Thank you

$\endgroup$

closed as off-topic by Benoît Kloeckner, RP_, Ryan Budney, Stefan Kohl, Wolfgang Jan 5 '17 at 7:24

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Benoît Kloeckner, RP_, Ryan Budney, Stefan Kohl, Wolfgang
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ This is probably explain in the book, but see my answer below. $\endgroup$ – Benoît Kloeckner Aug 29 '12 at 21:13
7
$\begingroup$

The $v_i$ are vector fields, and as such are derivations. The square usually means that you apply it twice (so, e.g. in the Euclidean space one can take $v_i=\frac{\partial}{\partial x_i}$ and its square is simply $\frac{\partial^2}{\partial x_i^2}$).

$\endgroup$
  • $\begingroup$ Thank you Mr. Benoît for that consideration. Actually in the book shape and shape theory I'am supposed to know this definition of Laplace-Beltrami operator, yet I do not know it though I do know other commun definitions of the $\Delta$ operator. Actually up to now I assimilated $v_i^2$ to the directional derivative of $v_i$ in the direction of $v_i$, the problem is that when $v_i$ is a vector field, both directional derivative $v_i^2$ and the covariant derivative $\nabla_{v_i}v_i$ are identical and then $\sum_i\left(v_i^2-\nabla_{v_i}v_i\right)$ reduces to zero. Thank you again $\endgroup$ – Riadh Aug 30 '12 at 15:04

Not the answer you're looking for? Browse other questions tagged or ask your own question.