# The Unapologetic Mathematician

## Tangent Vectors Geometrically

Now we’re in a position to tie our notion of tangent vectors back into geometry. Let $M$ be a manifold containing a point $p$. Now consider a smooth curve $c$ in $M$ that passes through this point. Without loss of generality, we can let $c:(-1,1)\to M$, and let $c(0)=p$. This will simplify a lot of our discussion by standardizing some of the details.

We know that $c$ gives us a tangent vector $c'(0)\in\mathcal{T}_pM$. The important thing for our purposes is that this tangent vector only depends on the germ of $c$ at $0$. That is, it is a very local property of the curve at $0$ and not in some particular neighborhood of $0$.

Indeed, if we pick some coordinate patch $(U,x)$ around $p$ we can write out $c'(0)$ in components. If $d$ is another smooth curve whose derivative has the same components at $0$, then it makes sense to say that $c$ and $d$ have the same tangent vector at $0$. Tangent vectors, then, are equivalence classes of curves under this relation.

We have to be careful, though. Does this definition depend on the coordinates we use? No, and our algebraic approach makes it easy to see why: if $y$ is another set of coordinates around $p$ we use the Jacobian of the transition function $y\circ x^{-1}$ to transform tangent vectors from one coordinate basis to another. Thus if $c'(0)$ and $d'(0)$ have the same components with respect to one coordinate system the same must be true with respect to all coordinate systems. Proving this from the geometric definition gets hairier.

Now it’s clear that every geometric tangent vector — every equivalence class of curves — gives rise to a unique algebraic tangent vector — a certain linear functional on $\mathcal{O}_p$. Indeed, we can turn around our calculation of the derivative $c'(0)$ and use it as a definition: $\displaystyle\left[c'(0)\right](f)=\frac{d}{dt}(f\circ c)\bigg\vert_0$

Given a coordinate map $x$ we can write this out \displaystyle\begin{aligned}\left[c'(0)\right](f)&=\frac{d}{dt}\left((f\circ x^{-1})\circ(x\circ c)\right)\bigg\vert_0\\&=\sum\limits_iD_i(f\circ x^{-1})\frac{d}{dt}(x^i\circ c)\bigg\vert_0\\&=\sum\limits_ic'(0)^i\left[\frac{\partial}{\partial x^i}(p)\right](f)\end{aligned}

where we have used the multivariable chain rule to pass to the second line. Thus if two curves have the same components with respect to some local coordinate map (and thus with respect to all of them) they define the same operator on germs $f$.

The flip side is where it gets a little hairier. Just because a geometric tangent vector gives a well-defined algebraic tangent vector, do all algebraic tangent vectors arise in this way? That is, given a vector $v\in\mathcal{T}_pM$, is there guaranteed to be some smooth curve $c$ passing through $c(0)=p$ with tangent vector $c'(0)=v$?

Given such a vector $v$ and local coordinates $x$ at $p$ — without loss of generality we can pick $x(p)=0\in\mathbb{R}^n$ — we get components $v^i$. Now we just define a curve $\gamma$ in $\mathbb{R}^n$ whose $i$th component is $\gamma^i(t)=tv^i$, and define $c=x^{-1}\circ\gamma$. Now it’s clear that the component $c'(0)^i=v^i$, so $c'(0)=v$, as desired.

Thus defining tangent vectors algebraically as we have done gives the same result as defining them geometrically. The geometric intuition had to wait, but it made establishing our desired results significantly easier.

April 12, 2011 - Posted by | Differential Topology, Topology