# 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

## The Tangent Bundle of a Euclidean Space

Let’s look at the tangent bundle to a Euclidean space. That is, we let $V$ be a finite-dimensional real vector space with its standard differentiable structure and see what these constructions look like.

Well, if $V$ has dimension $n$, then we know $V$ is an $n$-manifold, and thus we know that $\mathcal{T}_vV$ is an $n$-dimensional vector space for every point $v\in V$. Now, of course all $n$-dimensional vector spaces are equivalent, but I say that $\mathcal{T}_vV$ can be made canonically equivalent to $V$ itself.

First of all, let’s set up an isomorphism between $V$ and $\mathcal{T}_0V$. Given a vector $v\in V$, we can set up the curve $c(t)=tv$. Then we just define a vector in $\mathcal{T}_0V$ by picking the tangent vector to $c$ at $c(0)=0\in V$. This correspondence is clearly linear. For instance, if $v$ and $w$ are two vectors in $V$, then $tv+tw = t(v+w)$. The linearity of the derivative shows that the tangent vector to $t(v+w)$ is the sum of the tangent vectors to $tv$ and $tw$.

In terms of components, pick a basis $\{e_i\}$ of $V$ and use it to get a coordinate map on all of $V$. We also get a basis of coordinate vectors for $\mathcal{T}_vV$ at each point $v\in V$, and in particular at $v=0$. What does this isomorphism look in terms of these coordinates?

Well, the $i$th component of the tangent vector at $c(0)=0$ is the derivative of the $i$th component of the curve written out in terms of coordinates. And this component is $tv^i$, where $v^i$ is the $i$th component of $v$ in the $\{e_i\}$ basis, so the derivative in question is just $v^i$. That is, if we use a particular basis of $V$ and the basis of coordinate vectors it induces on $\mathcal{T}_0V$, we get the exact same components for each vector $v\in V$ and its corresponding vector in $\mathcal{T}_0V$.

In fact, there’s nothing particularly special about $0\in V$ here. We can do pretty much the exact same thing at any other point $p\in V$; just replace the curve $c(v)=tv$ with the curve $c(v)=p+tv$. That is, we use the same line translated (slid around) parallel to itself by adding an offset of $p$ to every point. This is such a fundamental technique in Euclidean spaces that we have a name for this method of comparing vectors in $\mathcal{T}_pV$ with vectors in $\mathcal{T}_0V$ — and thus with vectors in any other $\mathcal{T}_qV$ — “parallel translation”.

Parallel translation works as simply as this for two fundamental reasons: we can cover the entire manifold $V$ by a single coordinate patch, and the vector space structure allows us to sensibly define an operation of “add $p$ to each point”. In more general manifolds, this parallel translation operation is not possible, or at least not so straightforwardly. But Euclidean spaces are, of course, special.

In fact, they’re so special that they’re basically all that most people study in multivariable calculus. And in that situation everything works so smoothly that people often conflate many vector spaces which are really only equivalent. It’s common to automatically compare tangent vectors rooted at different points, parallel translating them with impunity. It’s even common to cavalierly identify the tangent space $\mathcal{T}_vV$ with the space $V$ itself, identifying a point $p$ with its position vector — that which points from $0$ to $p$.

But to move forward in differential geometry it is absolutely essential to unlearn this identification. When working in a Euclidean space, it’s okay to use it to simplify matters, but in general it pays off to be careful whether we’re talking about a point in a manifold, a tangent vector at one point in a manifold, or a tangent vector at a different point in the manifold. They’re just not all the same thing.

April 11, 2011

## 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

## Functoriality of the Derivative

We’ve said that the tangent bundle construction is a functor with the derivative as the action on morphisms. But we haven’t actually verified that it obeys the conditions of functoriality.

First off, if $f=1_M:M\to M$ is the identity map on a smooth manifold $M$, then $f_*:\mathcal{T}M\to\mathcal{T}M$ should be the identity map between the tangent bundles. That is, at each point $p$ we should have $f_{*p}:\mathcal{T}_pM\to\mathcal{T}_pM$ the identity map on this vector space. And indeed, if we let $(U,x)$ be any coordinate patch around $p$ we know that the matrix of $f_{*p}$ with respect to these local coordinates is the Jacobian of the coordinate function $x\circ1_M\circ x^{-1}=1_{x(U)}$. But this Jacobian is clearly the identity matrix, proving our claim.

More importantly, if $f:L\to M$ and $g:M\to N$ are two smooth maps, then their composition $g\circ f$ is also smooth. Given a point $p\in L$ we can define the derivatives $f_{*p}:\mathcal{T}_pL\to\mathcal{T}_{f(p)}M$, $g_{*f(p)}:\mathcal{T}_{f(p)}M\to\mathcal{T}_{g(f(p))}N$, and $(g\circ f)_{*p}\mathcal{T}_pL\to\mathcal{T}_{g(f(p))}N$. I say that $g_{*f(p)}\circ f_{*p}=(g\circ f)_{*p}$. And since this holds at every point we can write $(g\circ f)_*=g_*\circ f_*$, proving functoriality.

So, let’s take a vector $v\in\mathcal{T}_pL$ and see what happens. Taking a test function $\phi:N\to\mathbb{R}$ we calculate

\displaystyle\begin{aligned}\left[(g\circ f)_{*p}(v)\right](\phi)&=v\left(\phi\circ(g\circ f)\right)\\&=v\left((\phi\circ g)\circ f\right)\\&=\left[f_{*p}(v)\right](\phi\circ g)\\&=\left[g_{*f(p)}\left(f_{*p}(v)\right)\right](\phi)\\&=\left[\left[g_{*f(p)}\circ f_{*p}\right](v)\right](\phi)\end{aligned}

And so $(g\circ f)_{*p}=g_{*f(p)}\circ f_{*p}$, just as we claimed.

We should note, here, how this recalls the Newtonian notation for the chain rule, where we wrote $\left[g\circ f\right]'(p)=g'(f(p))f'(p)$. Of course, multiplication is changed into composition of linear maps, but that little detail will be cleared up soon (if you don’t already see it).

April 7, 2011

## 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

## The Tangent Bundle

So far we’ve talked about tangent spaces one at a time. For each $p\in M$ we get a tangent space $\mathcal{T}_pM$ at $p$. But things get really interesting when we start to sew them all together.

So, let’s take the (disjoint) union of all the tangent spaces $\mathcal{T}_pM$ at once. It’s important here that we never identify any of these with each other; a tangent vector at $p$ is never the same as another tangent vector at $q$. So far this is just a bunch of $n$-dimensional vector spaces parameterized by the $n$-dimensional manifold $M$, but I say that we can actually give this whole set the structure of a $2n$-dimensional smooth manifold!

Indeed, all we need is to give a collection of patches covering the whole space, and we can do this starting from any atlas on $M$. Given an open coordinate patch $(U,x)$ in $M$, we will make a coordinate patch that covers all of the $\mathcal{T}_pM$ with $p\in U$. Given a $v\in\mathcal{T}_pM$ we need to come up with a point $\bar{x}(v)\in\mathbb{R}^{2n}$. We write

$\displaystyle\bar{x}(v)=\left(x^1(p),\dots,x^n(p),v(x^1),v(x^2),\dots,v(x^n)\right)$

As $v$ varies within $\mathcal{T}_pM$, the first $n$ components of the image — $x(p)$ — stay fixed, and the rest of them give a linear isomorphism from $\mathcal{T}_pM$ to $\mathbb{R}^n$.

On the other hand, how do things change as we vary the base point $p$? This, it turns out, is where the interesting stuff is happening. Normally we wouldn’t be able to tell anything about how $\mathcal{T}_pM$ and $\mathcal{T}_qM$ are related for $p\neq q$, but within a single coordinate patch we can use the coordinate map $x$ to define coordinate vectors at every single point $p\in U$, and this lets us compare vectors at different points by comparing their components with respect to these coordinate vector bases.

The catch is, this only works for points in the same coordinate patch, and different coordinate patches give us different ways of comparing tangent vectors at nearby points. So we can’t really say much about them, but it’s enough to define a coordinate patch.

So we have to check about where patches overlap. If we have $(U,x)$ and $(V,y)$ as two patches on $M$ then we already know that $y\circ x^{-1}$ is a smooth transition function. This handles the smoothness of the first $n$ components of the transition function $\bar{y}\circ\bar{x}^{-1}$. For the other components, we know that the transition function is a linear isomorphism, which is clearly smooth.

We call this manifold the “tangent bundle” of $M$ and write $\mathcal{T}M$. It comes equipped with a map $\pi:\mathcal{T}M\to M$, which I say is smooth. Indeed, we just need to check this on a single pair of coordinate patches. We can pick any patch $(U,x)$ on $M$ and the corresponding patch on $\mathcal{T}M$. Then given $(a,b)\in x(U)\times\mathbb{R}^n$ we get a vector $\bar{x}^{-1}(a,b)=v\in\mathcal{T}_{x^{-1}(a)}M$ by writing

$\displaystyle v=\sum\limits_{i=1}^nb^i\left(\frac{\partial}{\partial x^i}(x^{-1}(a))\right)$

Our projection then sends $v$ to its base-point $x^{-1}(a)$, which $x$ sends to $a$. That is, if we write our projection $\pi$ out in terms of coordinates $\bar{x}$ on $\mathcal{T}M$ and $x$ on $M$, it’s just the projection $x(U)\times\mathbb{R}^n\to x(U)$, which is obviously smooth.

April 4, 2011

## Coordinate Transforms on Tangent Vectors

Given a coordinate patch $(U,x)$ in a neighborhood of a point $p$ in an $n$-dimensional manifold $M$, we get $n$ coordinate vectors $\frac{\partial}{\partial x^i}(p)$ which form a basis for the tangent space $\mathcal{T}_pM$. But this is true of any coordinate patch! If we have another patch $(V,y)$, we can get another basis $\frac{\partial}{\partial y^j}(p)$. Today we’ll examine how these bases are related.

As is usual, we can use the two bases to come up with a change of basis matrix. This can be used to take the components of any vector written out in terms of the $\frac{\partial}{\partial x^i}(p)$ basis and get the components of the same vector written out in terms of the $\frac{\partial}{\partial y^j}(p)$ basis. And we come up with the matrix by asking how to write out each of the latter basis vectors in terms of the former.

So, how did we find the components of the tangent vector $v$ in terms of the $\frac{\partial}{\partial x^i}(p)$ basis? We evaluated $v(x^i)$. Let’s stick $x^i$ into $\frac{\partial}{\partial y^j}(p)$ and see what we get:

$\displaystyle\left[\frac{\partial}{\partial y^j}(p)\right](x^i)=\left[D_j(x^i\circ y^{-1})\right](y(p))=\left[D_j(u^i\circ(x\circ y^{-1}))\right](y(p))$

But all this means is that we take the transition function $x\circ y^{-1}:\mathbb{R}^n\to\mathbb{R}^n$, take the $i$th component, and take the $j$th partial derivative of that function. And this is precisely the definition of the Jacobian of this transition function!

This basic fact starts showing us how everything we did when talking about multivariable calculus is on the one hand a special case of the concepts of differential geometry we’re coming up with now, while on the other hand they’re exactly the groundwork we need to build up the more general tools. What we did before in the simple Euclidean spaces is the model for all the more complicated manifolds we study.

April 1, 2011