# The Unapologetic Mathematician

## Tangent Vectors and Coordinates

Let’s say we have a coordinate patch $(U,x)$ around a point $p$ in an $n$-dimensional manifold $M$. We can use the function $f$ to give us some tangent vectors at $p$ called the “coordinate vectors”.

We define the coordinate vector $\frac{\partial}{\partial x^i}(p)\in\mathcal{T}_pM$ as follows: given a smooth function $f\in\mathcal{O}_p$, we define

$\displaystyle\left[\frac{\partial}{\partial x^i}(p)\right](f)=\left[D_i(f\circ x^{-1})\right](x(p))$

Okay, I know that that’s confusing. But all we mean is this: start with a function $f:U\to\mathbb{R}$. We compose it with the inverse of the coordinate map to get $f\circ x^{-1}:x(U)\to\mathbb{R}$, where $x(U)$ is some open neighborhood of the point $x(p)\in\mathbb{R}^n$. Now we can take that $i$th partial derivative of this function and evaluate it at the point $x(p)$.

The first thing we really should check is that it doesn’t matter which representative $f$ we pick. That is, if $f=g$ in some neighborhood of $p$, do we get the same answer? Indeed, in that case $f\circ x^{-1}=g\circ x^{-1}$ in some neighborhood of $x(p)$, and so their partial derivatives are identical. Thus this operation only depends on the germ $f\in\mathcal{O}_p$.

But is it a tangent vector? It’s easy to see that it’s a linear functional, so we just have to check that it’s a derivation at $p$:

\displaystyle\begin{aligned}\left[\frac{\partial}{\partial x^i}(p)\right](fg)=&\left[D_i((fg)\circ x^{-1})\right](x(p))\\=&\left[D_i((f\circ x^{-1})(g\circ x^{-1}))\right](x(p))\\=&\left[D_i(f\circ x^{-1})\right](x(p))\left[g\circ x^{-1}\right](x(p))\\&+\left[f\circ x^{-1}\right](x(p))\left[D_i(g\circ x^{-1})\right](x(p))\\=&\left[\frac{\partial}{\partial x^i}(p)\right](f)g(p)+f(p)\left[\frac{\partial}{\partial x^i}(p)\right](g)\end{aligned}

And so we have at least these $n$ vectors at each point $p\in M$. We can even tell that they much be distinct — and even linearly independent — since we can calculate

\displaystyle\begin{aligned}\left[\frac{\partial}{\partial x^i}(p)\right](x^j)&=\left[\frac{\partial}{\partial x^i}(p)\right](\pi^j\circ x)\\&=\left[D_i(\pi^j\circ x\circ x^{-1})\right](x(p))\\&=\left[D_i(\pi_j)\right](x(p))\end{aligned}

where $\pi^j$ is the $j$th coordinate projection $\pi^j:\mathbb{R}^n\to\mathbb{R}$. But we know that $D_i(\pi^j)$ is always and everywhere $\delta_i^j$ — it takes the value $1$ if $i=j$ and $0$ otherwise.

Thus $\frac{\partial}{\partial x^i}(p)$ takes a different value on $x^i$ than on all the other $x^j$. Further, any linear combination of the $\frac{\partial}{\partial x^j}(p)$ for $j\neq i$ must take the value $0$ on $x^i$, while $\frac{\partial}{\partial x^i}(p)$ takes the value $1$; we see that none of the coordinate vectors can be written as a linear combination of the rest, and conclude that the dimension of $\mathcal{T}_pM$ is at least $n$.

March 30, 2011