February 22, 2020

# The Tangent Bundle

UPDATE: I am aware of an issue in this post regarding the definition of the topology on the tangent bundle, namely that it is not Hausdorff and thus the tangent bundle as I've defined it is not even a topological manifold. I will correct this issue when I have time. However, the intuition and most of the details of this post are still valid, so I encourage you to read it anyway even before I correct my mistake.

Before we begin, I need to introduce a few new concepts. The first is that of a module, which is like a vector space but whose scalars come from a ring instead of a field. I guess I should formally define what a ring is before defining a module.

Definition. A ring is a set $R$ of elements called scalars, together with two binary operations:

which assigns to any pair of scalars $a,b\in R$ the scalar $a+b\in R$,

Multiplication
which assigns to any pair of scalars $a,b\in R$ the scalar $ab\in R$.

Any ring and its operations must satisfy the following properties:

There exists $0\in R$ such that $0+a=a$ for every $a\in R$.

For every $a\in R$, there exists $b\in R$ for which $a+b=0$.

For all $a,b\in R$, we have that $a+b=b+a$.

For all $a,b,c\in R$, we have that $(a+b)+c=a+(b+c)$.

Associative Property of Multiplication
For all $a,b,c\in R$, we have that $(ab)c=a(bc)$.

Distributive Property
For all $a,b,c\in R$, we have that $a(b+c)=ab+ac$.

So basically a ring is just like a field, except it does not necessarily have a multiplicative identity or multiplicative inverses, and multiplication is not necessarily commutative. If a ring does have a multiplicative identity, it is called a ring with unity. If it additionally has multiplicative inverses (since this would make no sense without an identity) it is called a division ring. If its multiplication is commutative, it is called a commutative ring.

The definition of a module is exactly the same as that of a vector space, except its scalars come from a ring.

Definition. A module over a ring $R$ is a set $\Gamma$, together with two operations:

which assigns to any pair $u,v\in \Gamma$ the element $u+v\in \Gamma$,

Scalar Multiplication
which assigns to any scalar $a\in R$ and any $v\in \Gamma$ the element $av\in \Gamma$.

Any module and its operations must satisfy the following properties:

Zero Element
There exists $0\in \Gamma$ such that $0+v=v$ for every $v\in \Gamma$.

For every $u\in \Gamma$, there exists $v\in \Gamma$ for which $u+v=0$.

For all $u,v\in \Gamma$, we have that $u+v=v+u$.

For all $u,v,w\in \Gamma$, we have that $(u+v)+w=u+(v+w)$.

Compatibility with Ring Multiplication
For all $a,b\in R$ and $v\in \Gamma$, we have that $(ab)v=a(bv)$.

Scalar Multiplicative Identity
For every $v\in \Gamma$, we have that $1v=v$.

First Distributive Property
For all $a,b\in R$ and $v\in \Gamma$, we have that $(a+b)v=av+bv$.

Second Distributive Property
For all $a\in R$ and $u,v\in \Gamma$, we have that $a(u+v)=au+av$.

The next new concept we need is that of the disjoint union of sets:

Definition. Given an indexing set $I$ and a collection of sets $\{A_i\}_{i\in I}$ indexed by $I$, their disjoint union is the set

$$\bigsqcup_{i\in I}A_i = \bigcup_{i\in I} A_i\times\{i\}.$$

The main point of the disjoint union is to keep an identifier for which set the elements originally came from.

Example. Let $A_1=A_2=\R$. Normally the union of $A_1$ and $A_2$ would just be $\R$ itself, since every element of $A_2$ is already in $A_1$. However, their disjoint union is the set

$$A_1\sqcup A_2 = (\R\times\{1\})\cup(\R\times\{2\}).$$

Each element of this disjoint union is either of the form $(a, 1)$ or $(a, 2)$, where $a\in\R$, so these elements look and feel just like real numbers except that we can tell whether they came from the set $A_1$ or $A_2$.

There is, of course, no reason why all the sets being unioned together need to be the same set. Later we shall see an example where each set is different, but it is still useful to keep the identifying index.

Lastly, we will need the idea of a section. This is sort of the dual concept to that of a retraction, which I introduced when discussing the Brouwer Fixed Point Theorem in my post on connectedness.

Definition. Let $f:A\to B$ be a function between sets. A function $s:B\to A$ is called a section of $f$ if $f\circ s=1_B$. That is, if $f\circ s(b)=b$ for every $b\in B$.

Informally speaking, sections provide a way to pick out representatives of level sets of the function $f$. To see what I mean, let's look at an example.

Example. Let $A$ denote a set of people and let $B$ denote a set of continents. Define $f:A\to B$ as the function which tells you which continent a person was born in. For instance, $f($me$) =$ North America.

In order to define a section $s:B\to A$ for which $f\circ s$ is the identity map on $B$, all we need to do is send each continent to one person who lives there. Here are two possible sections for $f$, $s_1, s_2:B\to A$.

Check for yourself that $f\circ s_1$ and $f\circ s_2$ are both the identity on $B$. And notice that my claim earlier holds true: these sections pick one representative of each continent. This concept will be very important later on.

Now we have the necessary machinery to proceed!

So far we've seen how to construct tangent vectors to a manifold at a point, and tangent spaces to manifolds at a point. But what if we wanted to do this at every point on our manifold? In multivariable calculus, we have the concept of a smooth vector field, which is an assignment of a vector to each point in $\R^n$ in such a way that the vectors vary smoothly as we move around in space. We would like to do the same on smooth manifolds, but we immediately run into some problems.

The first and most obvious problem is that defining smoothness of vector fields is going to be extremely tricky, since smoothness is a property of maps from smooth manifolds to smooth manifolds.

The second problem is even more troubling. We would like to define a vector field $X$ as a map which acts on any point $p$ in a smooth manifold and returns a tangent vector in $T_pM$. However, we don't currently have any set which can act as the codomain of such a map. This is because each point in the manifold has a different tangent space, so a vector at $p$ lives in $T_pM$ while a vector based at $q$ lives in $T_qM$.

So before we can proceed, we need to come up with a set comprised of all possible tangent vectors to all the points on our manifold, so this set can act as the codomain of a vector field. Furthermore, since we want a concept of smoothness for our vector fields, this set will have to be a smooth manifold. This smooth manifold I'm describing will be the tangent bundle.

Definition. Given a smooth manifold $M$, its tangent bundle is the set

$$TM = \bigsqcup_{p\in M} T_pM.$$

The name "bundle" is actually very appropriate since the tangent bundle, as a set, is really just all the tangent spaces bundled together. Every element of the tangent bundle $TM$ is of the form $(X, p)$, where $p\in M$ is a point in our manifold and $X\in T_pM$ is a tangent vector based at $p$. Our goal now is to introduce a smooth manifold structure on the tangent bundle.

A smooth manifold is a second countable, Hausdorff, locally Euclidean topological space. So the first thing we need to do is give $TM$ a topology. To this end, we introduce a simple but extremely important definition:

Definition. Given a smooth manifold $M$ and its tangent bundle $TM$, the bundle projection map is the function $\pi:TM\to M$ defined by $\pi\big((X,p)\big)=p$ for every $(X,p)\in TM$.

It is extremely important to note that the bundle projection map is surjective!

We will equip the tangent bundle $TM$ wth the bare minimum topology required to make the bundle projection map $\pi$ continuous. This is actually quite simple. We define a set $U\subseteq TM$ as open if and only if $U=\pi^{-1}[V]$ for some open set $V\subseteq M$. In this sense, $TM$ inherits its topology from $M$, which is exactly what we want!

Let's verify that this is actually a topology, and that $\pi$ is continuous with this topology, as claimed.

Theorem. The topology on $TM$ defined above is in fact a topology, and $\pi:TM\to M$ is continuous.

Proof. To show that this is a topology on the tangent bundle, we must verify three things: that the empty set and $TM$ itself are open, that the union of any collection of open sets is open, and that the intersection of any finite collection of open sets is open.

To see that the empty set is open in $TM$, we note that $\varnothing\subseteq M$ is open because $M$ is a topological space. Thus, $\varnothing=\pi^{-1}[\varnothing]$ is open by definition. To see that $TM$ is open, we note that $M$ is open in itself, and so $\pi^{-1}[M]=TM$ is open by definition.

Next, we argue that the union of open sets in $TM$ is open. Let $I$ be an indexing set for which $\{U_i\}_{i\in I}$ is a collection of open subsets of $TM$. Then each $U_i=\pi^{-1}[V_i]$ for some open set $V_i\subseteq M$. It follows from the properties of the preimage that

\begin{align} \bigcup_{i\in I}U_i &= \bigcup_{i\in I}\pi^{-1}[V_i] \\ &= \pi^{-1}\Big[\bigcup_{i\in I}V_i\Big]. \end{align}

Since the union of open sets in $M$ is open, it follows that $\bigcup_{i\in I}V_i$ is open in $M$ and thus $\bigcup_{i\in I}U_i$ is open in $TM$.

Lastly, we argue that the intersection of a finite collection of open sets in $TM$ is open. Let $U_1, U_2, \ldots, U_n$ be a collection of open subsets of $TM$. Then each $U_i=\pi^{-1}[V_i]$ for some open set $V_i\subseteq M$. It follows from the properties of the preimage that

\begin{align} \bigcap_{i=1}^n U_i &= \bigcap_{i=1}^n \pi^{-1}[V_i] \\ &= \pi^{-1}\Big[\bigcap_{i=1}^n V_i\Big]. \end{align}

Since the intersection of a finite collection of open sets in $M$ is open, it follows that $\bigcup_{i=1}^n V_i$ is open in $M$ and thus $\bigcup_{i=1}^n U_i$ is open in $TM$. Thus, this is indeed a topology on $TM$.

It is almost trivial to show that $\pi:TM\to M$ is continuous. Consider any open set $V\subseteq M$. Then $\pi^{-1}[V]$ is open in $TM$ by the definition of our topology! Thus, $\pi$ is a continuous function.

So $TM$ is indeed a topological space with the topology induced by $M$ as we described. The next step is to show that it is a topological manifold. To do this, we technically need to check that $TM$ is Hausdorff and second countable. I will not do this here, though, since it is not very informative. It is vital however to show that it is locally Euclidean. To do this, we simply need to demonstrate that every point in $TM$ in contained in some chart domain.

To this end, choose any point $(X, p)$ in $TM$. Then $p\in M$ and $X\in T_pM$. We will use the atlas on $M$ to construct a chart whose domain contains $(X, p)$. Since $M$ is a manifold, there is at least one chart $(U, x)$ for $M$ whose domain contains $p$.

We use this information to define a chart $(U', x')$ for $TM$ whose domain contains $(X, p)$ as follows:

\begin{align} U' &= \bigsqcup_{q\in U} T_q M \\ x'\big((Y, q)\big) &= \big(x^1(q), x^2(q), \ldots, x^n(q), Y^1, Y^2, \ldots, Y^n\big), \end{align}

where $n=\dim M$ and $Y=\displaystyle\sum_{i=1}^n Y_i\frac{\partial}{\partial x^i}\at{q}\in T_qM$.

Notice that $x':TM\to\R^{2n}$, so what I am saying is that $TM$ is a manifold whose dimension is twice the dimension of $M$. We of course need to verify that the set $U'$ as defined above is open in $TM$. Technically we also need to show that $x'$ is an embedding, but I will omit this step since it is not difficult or interesting.

Theorem. The set $U'$ as defined above is open in the tangent bundle $TM$. That is, if $U$ is an open set in $M$ then $U'=\bigsqcup_{q\in U}T_qM$ is an open set in $TM$.

Proof. We need to construct an open set $V\subseteq M$ for which $U'=\pi^{-1}[V]$. It just so happens that this set is $U$ itself, since

\begin{align} \pi^{-1}[U] &= \{(X, q)\in TM \mid \pi\big((X,q)\big)\in U\} \\ &= \bigsqcup_{q\in U} T_qM \\ &= U'. \end{align}

Since $U'$ is the preimage under $\pi$ of an open set in $M$, it is open in the tangent bundle $TM$.

So we've shown that $TM$ is a topological manifold, and it was almost too easy! In fact, most things related to the tangent bundle are. It is defined to be nice and to make our lives easier.

Lastly, we should show that the atlas we've exhibited on $TM$ is a smooth atlas. That is, if $(U', x')$ and $(V', y')$ are charts for $TM$ with nonempty intersection, $U'\cap V'\ne\varnothing$, then we need to show that the transition map $x'\circ(y')^{-1}:\R^{2n}\to\R^{2n}$ is smooth in the sense of ordinary multivariable calculus. It is not too difficult to do so, but it is a bit ugly. To simplify the proof, we'll need a lemma which explains what happens to the components of tangent vectors under a change of chart.

Lemma. Suppose $(U, x)$ and $(V, y)$ are charts on a smooth manifold $M$ of dimension $n$ with $U\cap V\ne\varnothing$. If $p\in U\cap V$ and $Z\in T_p M$ is the tangent vector

$$Z = \sum_{i=1}^n b^i \frac{\partial}{\partial y^i}\at{p}$$

with respect to the chart map $y$, then

$$Z = \sum_{i=1}^n c^i \frac{\partial}{\partial x^i}\at{p}$$

with respect to the chart map $x$, where each component $c^j$ is given by

$$c^j = \sum_{i=1}^n b^i \frac{\partial}{\partial y^i}\at{p} x^j.$$

Proof. Since $\displaystyle\Big(\frac{\partial}{\partial x^i}\at{p}\Big)_{i=1}^n$ is a basis for $T_p M$, we know there exist scalars $c^1, c^2, \ldots, c^n\in\R$ for which

$$\sum_{i=1}^n b^i \frac{\partial}{\partial y^i}\at{p} = \sum_{i=1}^n c^i \frac{\partial}{\partial x^i}\at{p}.$$

If we apply the differential of $x^j$, the cotangent vector $\d x^j\smallat{p}:T_p M\to\R$, to both sides of the above equation, things simplify considerably. Applying it first to the right side, we obtain

\begin{align} \d x^j\smallat{p}\Big(\sum_{i=1}^n c^i \frac{\partial}{\partial x^i}\at{p}\Big) &= \sum_{i=1}^n c^i \d x^j\smallat{p} \Big(\frac{\partial}{\partial x^i}\at{p}\Big) \\ &= \sum_{i=1}^n c^i\delta^j_i \\ &= c^j. \end{align}

Applying $\d x^j\smallat{p}$ to the left side, we see that

\begin{align} \d x^j\smallat{p}\Big(\sum_{i=1}^n b^i \frac{\partial}{\partial y^i}\at{p}\Big) &= \sum_{i=1}^n b^i \d x^j\smallat{p} \Big(\frac{\partial}{\partial y^i}\at{p}\Big) \\ &= \sum_{i=1}^n b^i \frac{\partial}{\partial y^i}\at{p} x^j \end{align}

Equating the two sides, we see that

$$c^j = \sum_{i=1}^n b^i \frac{\partial}{\partial y^i}\at{p} x^j,$$

as desired.

Now that we know how tangent vectors transform under a change of basis, we can prove that the chart maps we've defined on the tangent bundle are smooth maps in the sense of multivariable calculus.

Theorem. Let $M$ be a smooth manifold. If $(U', x')$ and $(V', y')$ are charts for $TM$ with nonempty intersection, then the transition map $x'\circ(y')^{-1}:\R^{2n}\to\R^{2n}$ is smooth.

Proof. Choose any point $(a^1, a^2, \ldots, a^n, b^1, b^2, \ldots, b^n)$ in the image of the transition map $\im x'\circ(y')^{-1}\subseteq\R^{2n}$. Then

\begin{align} x'\circ(y')^{-1}(a^1, a^2, \ldots, a^n, b^1, b^2, \ldots, b^n) &= x'\Big(y^{-1}(a^1, a^2,\ldots,a^n), \sum_{i=1}^n b^i \frac{\partial}{\partial y^i}\at{p}\Big) \\ &= (A^1, A^2, \ldots, A^n, B^1, B^2, \ldots, B^n), \end{align}

where each $A^j$ is given by

\begin{align} A^j &= x^j\circ y^{-1}(a^1, a^2, \ldots, a^n) \\ &= (x\circ y^{-1})^j (a^1, a^2, \ldots, a^n) \end{align}

and each $B^j$ is given by

\begin{align} B^j &= \sum_{i=1}^n b^i \frac{\partial}{\partial y^i}\at{p} x^j \\ &= \sum_{i=1}^n b^i \partial_i(x^j\circ y^{-1})\at{y(p)} \\ &= \sum_{i=1}^n b^i \partial_i(x\circ y^{-1})^j\at{y(p)}. \end{align}

These components are all smooth, since the transition map $x\circ y^{-1}$ on $M$ is smooth. It follows that the transition map $x'\circ(y')^{-1}$ on $TM$ is smooth, completing the proof.

We've arrived where we wanted to: the tangent bundle is a smooth manifold! We have all the machinery we need now to develop vector fields in my next post, as well as covector and tensor fields.