# I need help understanding the meaning of the covariant derivative of the dot product of two tangent vector fields along a parameterized curve

In my differential geometry class, I am learning about the covariant derivative of a tangent vector field of an n-surface in R(n+1). We learned that the formula for the covariant derivative of a tangent vector field at a point p is (in English) equal to the regular derivative of the vector field minus the component of the regular derivative normal to the tangent space, thus thereby yielding a covariant derivative that is in the tangent space. One of the theorems we covered is that (XdotY)'=X'dotY+Y'dotX, for two vector fields, X and Y, of length n+1, at a point, p. My problem is, while I feel that I can understand the covariant derivative based on the original definition which involves subtracting out the normal component, I can't follow what it means in the case of the scalar function XdotY. More specifically, how does it relate to the theme of the covariant derivative in n+1 space being an object which subtracts out the component in the direction of the normal? Does this scalar function have a normal direction? Which space are we even talking about?

I presume you're doing the covariant derivative along a curve (or you should think about that case). Consider a curve $\alpha(t)$ and the function $f(t) = X(\alpha(t))\cdot Y(\alpha(t))$. We're differentiating the scalar function $f$. So we apply the product rule to get $f'(t)=(X\circ\alpha)'(t)\cdot Y(\alpha(t)) + X(\alpha(t))\cdot (Y\circ\alpha)'(t))$. Here's the main point: $(X\circ\alpha)'(t)$ is the sum of two terms: the covariant derivative of $X$ in the direction of $\alpha'(t)$ [which is usually written $\nabla_{\alpha'(t)}X$] and a vector $Z$ normal to the surface. But when we compute the dot product $Z\cdot Y$, we of course get $0$, since $Y$ is tangent to the surface. So we get: $$f'(t) = \nabla_{\alpha'(t)}X\cdot Y(\alpha(t)) + X(\alpha(t))\cdot\nabla_{\alpha'(t)}Y.$$