Please explain the repeated gradient of a function on manifold.

87 Views Asked by At
1

There are 1 best solutions below

0
On

In general, if $T\colon \mathfrak{X}(M)^k \to \mathcal{C}^\infty(M)$ is a covariant $k$-tensor and $X \in \mathfrak{X}(M)$, one defines $$(\nabla_XT)(X_1,\ldots,X_k) = X(T(X_1,\ldots, X_k)) - \sum_{k=1}^n T(X_1,\ldots, \nabla_XX_i,\ldots,X_k),$$thinking of a product rule. If $u$ is a smooth function, $u$ can be seen as a $0$-tensor and the above definition applies to give $\nabla_Xu = X(u) = {\rm d}u(X)$. Now $(\nabla u)(X) = {\rm d}u(X)$ is a $1$-tensor, and the definition again applies to give $$(\nabla^2u)(X,Y) \doteq (\nabla_X(\nabla u))(Y) = X({\rm d}u(Y))- {\rm d}u(\nabla_XY).$$In coordinates $(x^1,\ldots, x^n)$, take $X = \partial_i$ and $Y = \partial_j$ to obtain $$(\nabla^2u)_{ij} = \partial_i(\partial_ju) - \Gamma_{ij}^k \partial_ku.$$For $\nabla^3u$, you repeat what was done above for the $2$-tensor $\nabla^2u$.