# The Unapologetic Mathematician

## Derivatives in Coordinates

Let’s take the derivative and see what it looks like in terms of coordinates. Say we have a smooth manifold $M$ and a smooth map $f:U\to N$ from an open subset of $M$ to another smooth manifold $N$. If $p\in U$ is any point, we define the derivative $f_{*p}:\mathcal{T}_pM\to\mathcal{T}_{f(p)}N$ as before.

Now, if $(U,x)$ is a coordinate patch — even if there isn’t a single coordinate patch on the whole domain of $f$ we can restrict $f$ down to a coordinate patch containing $p$ — we get a basis of coordinate vectors at $p$. Similarly, if $(V,y)$ is a coordinate patch around $f(p)$ we get a basis of coordinate vectors at $f(p)$. We want to write down the matrix of $f_{*p}$ in terms of these two bases.

So, the obvious path is to take one of the coordinate vectors at $p$, hit it with $f_{*p}$, and write the result out in terms of the coordinate vectors at $f(p)$. The generic problem, then, is to calculate the $j$th component — the one corresponding to $\frac{\partial}{\partial y^j}(f(p))$ — of $f_{*p}\left(\frac{\partial}{\partial x^i}(p)\right)$. But we know that this coefficient comes from sticking $y^j$ into this vector and seeing what pops out!

\displaystyle\begin{aligned}\left[f_{*p}\left(\frac{\partial}{\partial x^i}(p)\right)\right](y^j)&=\left[\frac{\partial}{\partial x^i}(p)\right](y^j\circ f)\\&=D_i\left(y^j\circ f\circ x^{-1}\right)\\&=D_i\left(u^j\circ(y\circ f\circ x^{-1})\right)\end{aligned}

We’re taking the $i$th partial derivative of the $j$th component of the function $y\circ f\circ x^{-1}$, which goes from the open set $x(U)\in\mathbb{R}^m$ into $\mathbb{R}^n$, where $m$ and $n$ are the dimensions of $M$ and $N$, respectively. Like we saw for coordinate transforms in place, this is just the Jacobian again.

So if we want to write out the derivative $f_{*p}$ in terms of local coordinates, we first write out our local coordinate version of $f$ as a function from one Euclidean space to another, and then we take the Jacobian of that function at the appropriate point.

April 6, 2011

## The Derivative

It turns out that the tangent bundle construction is actually a functor. Given a smooth map $f:M\to N$ between smooth manifolds, we will get a smooth map $f_*:\mathcal{T}M\to\mathcal{T}N$. Yes, we’d usually write $\mathcal{T}f$ for a functor’s action on a map, but the $f_*$ notation is pretty classical.

So if we’re given a tangent vector $v\in\mathcal{T}_pM$ we want to get a tangent vector $f_*(v)\in\mathcal{T}_qN$. And since we already have $f$ sending points of $M$ to points of $N$, it only makes sense to ask that $q=f(p)$. That is, in terms of the tangent bundle projection functions, we can write $f(\pi(v))=\pi(f_*(v))$. In other words, the projection $\pi:\mathcal{T}M\to M$ will be a natural transformation from the tangent bundle functor to the identity functor.

Anyway, this means that for each $p\in M$ we’ll get a map $f_{*p}:\mathcal{T}_pM\to\mathcal{T}_{f(p)}N$. Since these are both vector spaces, it only stands to reason that we’d have a linear map. We haven’t yet established the connection between our “tangent vectors” and the geometric notion, but we do have a notion from multivariable calculus of a linear map that takes tangent vectors to tangent vectors: the Jacobian, which we saw as a certain extension of the notion of the derivative. We will find that our map $f_*$ is the analogue of the same concept on manifolds, and so we will call it the derivative of $f$.

So here’s our definition: if $f:U\to N$ is a differentiable map in some open set $U\subseteq M$ and if $p\in U$, then we define our map $f_{*p}:\mathcal{T}_pM\to\mathcal{T}_{f(p)}N$ by

$\displaystyle\left[f_{*p}(v)\right](\phi)=v(\phi\circ f)$

where $\phi\in\mathcal{O}(V)$ is any smooth function on a neighborhood of $f(p)\in N$. That is, $f_{*p}(v)$ is a linear functional on $\mathcal{O}_{f(p)}$; if $\phi$ represents a germ at $f(p)$ we can compose it with $f$ to represent a germ at $p$, and then we can apply $v$ itself to this germ. It should be immediately clear that this construction is linear in $v$.

April 6, 2011