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

$\endgroup$
1
  • $\begingroup$ This is probably explain in the book, but see my answer below. $\endgroup$ Aug 29, 2012 at 21:13

1 Answer 1

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 $\frac{\partial^2}{\partial x_i^2}$).

$\endgroup$
1
  • $\begingroup$ 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, 2012 at 15:04

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