# The Unapologetic Mathematician

## Curves

Now we can start coming back down to geometric earth. A smooth curve in a smooth manifold $M$ is nothing but a smooth map $c:I\to M$, where $I$ is some interval in the real line with its standard differentiable structure. The interval $I$ can, in principle, be half-infinite or infinite, but commonly we just consider finite open intervals like $(0,1)$.

At any point $t_0$ of an open interval, the tangent space $\mathcal{T}_{t_0}I$ is one-dimensional. And, in fact, it comes equipped with a canonical vector to use as a basis: $\frac{d}{dt}(t_0)$, the derivative operator at the point itself! Any other linear functions on germs at $t_0$ that satisfies a product rule must be a scalar multiple of this one.

Since we have a canonical tangent vector in $\mathcal{T}_{t_0}I$, we can hit it with the derivative $c_{*t_0}$ and see what happens. We get a tangent vector

$\displaystyle c_{*t_0}\left(\frac{d}{dt}(t_0)\right)\in\mathcal{T}_{c(t_0)}M$

which we call the tangent vector of $c$ at $t_0$, and we write it as $c'(t_0)$.

Let’s say that $c(t_0)=p\in M$ and let $f$ be a germ at $p$. What does $c'(t_0)$ do to $f$? We can calculate:

\displaystyle\begin{aligned}\left[c'(t_0)\right](f)&=\left[c_{*t_0}\left(\frac{d}{dt}(t_0)\right)\right](f)\\&=\left[\left(\frac{d}{dt}(t_0)\right)\right](f\circ c)\\&=\frac{d}{dt}\left(f\circ c\right)\bigg\vert_{t_0}\end{aligned}

That is, we pull the function $f$ back along $c$ to define a smooth real-valued function on the interval $I$ itself, then we hit it with the derivative operator and evaluate at $t_0$.

If our curve lies with a coordinate patch $(U,x)$ — or if we cut out a segment of the curve that does — then we have a curve $x\circ c:I\to\mathbb{R}^n$. We can also use $x$ to define a coordinate basis on $\mathcal{T}_pM$, and thus get components of $c'(t_0)$ in those coordinates. As usual, we calculate the $i$th component by

\displaystyle\begin{aligned}\left[c'(t_0)\right](x^i)&=\frac{d}{dt}\left(x^i\circ c\right)\bigg\vert_{t_0}\\&=\frac{d}{dt}\left(u^i\circ(x\circ c)\right)\bigg\vert_{t_0}\end{aligned}

But this is just the derivative of the $i$th component of the function $x\circ c$. That is, when we’re working in local coordinates we get $i$th coefficient of the tangent vector $c'(t_0)$ by taking the derivative of the $i$th component function of the curve.

If you remember calculations like this in multivariable calculus, this is almost exactly why it works. There’s one other little caveat, though, that we’ll get to next time.

April 8, 2011