## § Derivatives in diffgeo

- A function of the form $f: \mathbb R^i \to \mathbb R^o$ has derivative specified by an $(o \times i)$ matrix, one which says how each output varies with each input.
- Now consider a vector field $V$ on the surface of the sphere, and another vector field $D$. Why is $W \equiv \nabla_D V$another vector field? Aren't we differentiating a thing with 3 coordinates with another thing with 3 coordinates?
- Well, suppose we consider the previous function $f: \mathbb R^i \to \mathbb R^o$, and we then consider a curve $c: (-1, 1) \to \mathbb R^i$. Then the combined function $(f \circ c): (-1, 1) \to \mathbb R^o$ needs only $o$ numbers to specify the derivative, since there's only one parameter to the curve (time).
- So what's going on in the above example? Well, though the full function we're defining is from $\mathbb R^i$ to $\mathbb R^o$, composing with $c$ "limits our attention" to a 1D input slice. In this 1D input slice, the output is also a vector.
- This should be intuitive, since for example, we draw a circle parameterized by arc length, and then draw its tangents as vectors, and then
*we draw the normal as vectors * to the tangents! Why does *that * work? In both cases (position -> vel, vel -> accel) we have a single parameter, time. So in both cases, we get vector fields! - That's somehow magical, that the derivative of a thing needs the same "degrees of freedom" as the thing in itself. Or is it magical? Well, we're used to it working for functions from $\mathbb R$ to $\mathbb R$. It's a little disconcerting to see it work for functions from $\mathbb R$to $\mathbb R^n$.
- But how does this make sense in the case of diffgeo? We start with a manifold $M$. We take some curve $c: (-1, 1) \to M$. It's derivative must live as $c': (-1, 1) \to TM$. Now what about $c''$? According to our earlier explanation, this too should be a vector! Well... it is and it isn't, right? but how? I don't understand this well.
- Looping back to the original question, $W \equiv \nabla_D V$ is a vector field because the value of $W(p)$ is defined as taking $D(p) \in T_p M$, treating it as a curve $d_p: [-1, 1] \to M$ such that $d_p(0) = p$ and $d_p'(0) = D(p)$, and then finally taking $V()$.