# What does $\nabla^j$ mean in Riemannian geometry?

I was learning about Holder space from this paper page 8, and it uses the notation $$\nabla^j u$$ for a function $$u$$ on a Riemannian manifold $$M$$.

I am confused about the meaning of it. I think it might mean $$\nabla_{\frac{\partial}{\partial x_j}}u = \frac{\partial}{\partial x_j }u$$ locally, since this description would match the definition of the Holder space on the Euclidean space. But it is not well defined in my opinion, since it is not invariant under coordinate change.

After recalling the definition of a Holder space on on an Euclidean space, I am now convinced the author should mean something like $$\nabla^j u = \nabla_{\frac{\partial}{\partial x_{l_1}}}\ldots \nabla_{\frac{\partial}{\partial x_{l_j}}}u$$. However as I said before, this seems to not be independent of the coordinate choice.

• Typically $\nabla^j$ means $\nabla \circ \cdots \circ \nabla$ ($j$-times). Jul 10 '19 at 4:46
• But how to read off from this notation in which direction are we taking the covariant derivative in this notation? @AmorFati Jul 10 '19 at 4:51

As stated in the comments, in this context $$\nabla^j u$$ denotes the $$j^\rm{th}$$ covariant derivative of $$u,$$ i.e. $$\underbrace{\nabla\cdots\nabla}_ju,$$ which (when $$u$$ is a scalar function) is a $$(0,j)$$-tensor field.
(Remember that for any $$(a,b)$$-tensor $$T,$$ its derivatives in all directions can be bundled up into the $$(a,b+1)$$-tensor $$\nabla T$$ defined by $$(\nabla T)(v,\cdots) = (\nabla_v T)(\cdots).$$)
• Thank you but I am still confused. The $C^k$ norm of a smooth function $u$ on $M$ is defined to be $\sum_{j=0}^k sup_{x \in M}|\nabla^j u(x)|$. Without specifying the direction how do I compute $\nabla^j u$? Jul 10 '19 at 14:12
• @penny For example, $\nabla u$ is a vector field, so at each point $x$ you can calculate the length of this vector. Of course you need the Riemannian metric to do that. In general the Riemannian metric also defines the length of a $(0,j)$-tensor. Jul 11 '19 at 3:22