# Covariant Derivative Along Arbitrary Vector

The inspiration for this question is section 1.3 of "A Course in Minimal Surfaces," by Colding and Minicozzi. This section has to do with deriving the first variation formula. We are dealing with an $n$-dimensional Riemannian manifold $M$ and a $k$-dimensional submanifold $\Sigma \subset M$. $M$ has a covariant derivative $\nabla$, and the induced covariant derivative on $\Sigma$ is given by $\nabla_{\Sigma} = (\nabla)^{T}$, where the superscript $T$ means that we take only the part that is tangent to $\Sigma$.

Here are two sentences from the very beginning of section 1.3:

• Let $F: \Sigma \times (-\epsilon, \epsilon) \to M$ be a variation of $\Sigma$ with compact support and fixed boundary.
• The vector field $F_{t}$ restricted to $\Sigma$ is often called the variational vector field.

I'm not entirely sure what $F_{t}$ is supposed to be, and I couldn't find any information about variational vector fields, so I came up with the following definition: for $x \in \Sigma$, $$F_{t}(x) := F_{x}'(0),$$ where $F_{x}'(0) \in T_{x} M$ is the equivalence class of the curve in $M$ that we obtain by fixing $x \in \Sigma$ and considering $F$ as a function of the single variable $t \in (-\epsilon, \epsilon)$.

On the next page, the authors write:

• $\dots = \sum_{i=1}^{k} \langle \nabla_{F_{t}} F_{x_{i}}, F_{x_{i}} \rangle = \dots$

This is an issue because, with my definition, nothing guarantees that $F_{t}$ is tangent to $\Sigma$. (Also, I'm assuming that $\nabla$ is the induced covariant derivative on $\Sigma$.)

If $Y$ is a vector field on $M$ and $X$ is a vector that is not tangent to $\Sigma$, then what is meant by $\nabla_{X} Y$? If this is not defined, then $\nabla_{F_{t}} F_{x_{i}}$ is not defined, so my definition of $F_{t}$ is not defined. In that case, what is the proper definition of $F_{t}$?

• Your definition of the variational vector field is correct. $\nabla$ is the connection on $M$ (not that of $\Sigma$) so $F_t$ don't have to be tangent to $\Sigma$. – user99914 Sep 14 '15 at 1:41