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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

flag	asked Aug 29 at 20:25 Riadh 3

This is probably explain in the book, but see my answer below. – Benoît Kloeckner Aug 29 at 21:13

Benoît Kloeckner · Answer 1 · 2012-08-29 21:12:57Z UTC

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

answered Aug 29 at 21:12

Benoît Kloeckner
5,489●18●29

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 – Riadh Aug 30 at 15:04

Laplace-Beltrami operator expression

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