Manifold #13. The tangent bundle - an illustration.매니폴드 #13. 탄젠트 번들의 예

These are my notes on XylyXylyX’s What is a Manifold? lectures. This post covers lesson 13 of the playlist, The tangent bundle - an illustration. The previous post set up the definition of a fiber bundle, so this time we build the most classical example, the tangent bundle, from nothing, one piece at a time, to see where the parts of the definition actually sit. We choose the circle $S^1$ as the base. It has one dimension, so everything is written with a single number and only the structure stands out.

Building an atlas on the circle

First we build the manifold that will be the base. The set is the circle $S^1$, and the topology is fixed by taking small open intervals on the circle as a basis. Finite intersections and arbitrary unions of these intervals are all open sets. Now we need an atlas. We cover the circle with about six open intervals $U_1, \dots, U_6$. First pick a few intervals that do not overlap, then lay other intervals over the gaps between them so that neighbors overlap. The overlaps may be large.

An atlas includes not only the open sets but also, for each of them, a homeomorphism onto a subset of the real numbers, that is, a chart $\gamma_i : U_i \to \R$. A point on the circle is pointed to by an angle $\theta$. Once we fix the point where $\theta = 0$, every point has an angle. The charts may be chosen arbitrarily, and the lecture gives two examples. On $U_1$ we use the angle as it is, $\lambda_1 = \theta$, and on $U_2$ we use half the angle, $\lambda_2 = \theta / 2$.

Where two intervals overlap, the map that changes the coordinate of one chart into the coordinate of the other is the transition function of the charts, and in this example it is multiplication by one number. Multiplication is a map from the real numbers to the real numbers, so it can be differentiated, and the circle becomes a differentiable manifold. (Inside an overlap there may in fact be a constant $c$ added, as in $\lambda_j = a \lambda_i + c$, and the only thing we use later is the slope $a$. This explanation is added for understanding and was not covered in the lecture.)

Curves on the circle and the tangent space

Take a point $p$ on the circle and think of the curves through $p$. On the circle there is only one path a curve can take, so all curves follow the same trajectory. Still there are many curves, because the same trajectory can be traversed at different speeds. For example, if $p$ sits at $300^\circ$, then $\sigma(t)$ moves as $\theta = t$ and reaches $p$ at $t = 300$, while $\tau(s)$ moves as $\theta = 2s$ and reaches $p$ at $s = 150$. They are different curves with the same trajectory and twice the speed. The collection of all such curves makes the tangent space $T_pS^1$ at $p$.

If there is a chart $(U_i, \gamma_i)$ containing $p$, that chart fixes a basis. The dimension is one, so there is one basis vector. Writing the coordinate of the chart as $\lambda_i$, the basis vector is $\partial / \partial \lambda_i$, and every vector $X$ of the tangent space is fixed by one component.

Here $i$ is not a spacetime index. It is the label of the chart, so it is not an index to be summed. I put it in parentheses so that it is not confused with the summation rule of general relativity, and it only tells us in which chart the component was read. There is a single component, so no sum is needed either.

Q. Even when we use the angle $\theta$ as $\lambda_i$ as it is, must we still speak of the identity map of a chart?
A. The lecture says so. The angle $\theta$ is a name that points directly at a point of the circle, and for it to become a coordinate it has to pass once through a chart that sends the point to a real number. This chart just happens to be the identity map $\theta \mapsto \theta$. Elementary general relativity often skips this step, but formally it is always there behind the scenes. Taking the whole circle as one interval $U$ with the identity map as the chart gives the simplest atlas.

The bundle made from the tangent spaces

Now we fill in the parts of a fiber bundle one by one. The base is the circle $S^1$. At each point $x$ of the base there is a tangent space $T_xS^1$, and this is the preimage of the projection. We take the standard fiber to be the real numbers $\R$. $T_xS^1$ is a vector space with one basis vector, so it corresponds one to one with the real numbers through a single component, and it is easy to see that it is diffeomorphic to the real numbers. One fiber is attached at each point, and all are copies of $\R$.

For each chart $U_i$ and point $x \in U_i$, we define the map $h_{i,x}$ from the tangent space to the real numbers as follows. It takes a vector and returns the component of that vector.

This map is a smooth isomorphism, and it is the material for a local trivialization. For the total space, in this example we take the simplest choice, the product of the base and the fiber, $E = S^1 \times \R$. The projection sends $(x, X)$ to $x$. The local trivialization that sends the whole preimage of a cover set $U_i$ to a product is built by pairing the projection with $h$. The lecture calls it $\pi h_i$, and across the whole bundle $h_{i,x}$ is used at each point.

Fitting this example onto the picture from the previous post gives the following. The base space is the circle, the total space is $S^1 \times \R$, the preimage of a point $x$ under the projection is the tangent space at that point, and what sends that tangent space to the fiber $\R$ is $h$. The bundle chart is $(U_i, \pi h_i)$. This example is a product as a whole, so one big trivialization would be enough, but we deliberately split it into several charts.

Transition functions on the overlaps

Suppose a point $p$ lies in the overlap of the cover sets $U_1$ and $U_6$. $p$ lies in both charts, so the tangent space at $p$ has two bases. In chart 6 it is $\partial / \partial \lambda_6$ and in chart 1 it is $\partial / \partial \lambda_1$. The tangent space itself is one and the same, and both bases are correct. The same vector $X$ becomes the component $X^{(6)}$ through $h_{6,p}$ and the component $X^{(1)}$ through $h_{1,p}$. What connects the two components is the transition function, and since it is a map from the real numbers to the real numbers, it is multiplication by one number.

$r_{61}$ and $r_{16}$ are reciprocals of each other. This number is fixed by the ratio of the speeds at which the coordinates of the two charts change, that is, the speed at which $\lambda_6$ changes divided by the speed at which $\lambda_1$ changes. Near a point, $\lambda_6$ may grow fast while $\lambda_1$ grows slowly, and one may shrink while the other grows. Then the ratio can be a large number or a negative number.

Q. The words transition function seem to be used in two meanings. How do we tell them apart?
A. The lecture points this out too. One is the map between charts of a differentiable manifold that changes the coordinate $\lambda_6$ into the coordinate $\lambda_1$, and this map must be smooth for the manifold to be differentiable. The other is the transition function of the bundle, a map such as $r_{61}$ that carries the value of one trivialization to the value of another between fibers. This bundle transition function must be an element of the structure group.

How small can the structure group get?

A transition function is a diffeomorphism from fiber to fiber, and every transition function must be an element of the structure group. In this example the fiber is $\R$ and a transition function is multiplication by a nonzero number. So the most generous structure group is all real numbers except 0, $\R \setminus \{0\}$, and this group is closed under multiplication. The lecture puts this group in the place of the diffeomorphism group $\mathrm{Diff}(F)$. (Strictly, $\mathrm{Diff}(\R)$ contains many nonlinear diffeomorphisms as well, and since the fiber is a vector space, the structure group that uses only the linear ones is $\R \setminus \{0\}$. This explanation is added for understanding and was not covered in the lecture.)

But we have to ask whether this structure group is just as large as we need. With six cover sets there are six overlaps, so the transition functions are $r_{12}, r_{23}, r_{34}, r_{45}, r_{56}, r_{61}$ and their reciprocals, twelve in all. This collection of numbers is not a group. We chose the charts freely, so the numbers are arbitrary, and nothing guarantees that the product of two of them lies in the collection, so it is not closed under multiplication. So we need a larger group that contains the whole collection, and one answer is all real numbers except 0.

But if we choose the charts cleverly, a much smaller group is enough. First, if we choose all coordinates to grow in the same direction, the ratio of speeds is always positive, so all transition functions are positive reals, and the structure group shrinks to all positive reals. Second, if we choose all coordinates to grow in the same direction and at the same speed, all transition functions are 1, and the structure group shrinks to the single identity element $\{1\}$. When the structure group shrinks to one identity element, the bundle is a trivial bundle with a product structure over the whole space. We made the total space a product from the start, so this result is no surprise.

Q. Does being able to shrink the structure group to {1} mean the same as the bundle being trivial?
A. In the lecture they are used as the same thing. If we can choose the atlas so that every transition function is the identity, the local trivializations do not disagree and join into one, so the whole bundle is the product of the base and the fiber. The cylinder of the previous post was such a case.

If the tangent space is a larger vector space, as in four-dimensional spacetime, the transition functions become matrices instead of numbers, and the most general structure group is the collection of all $n \times n$ matrices with nonzero determinant, $GL(n)$, the general linear group. In one dimension $GL(1) = \R \setminus \{0\}$.

The Möbius strip that one sign makes

The lecture ends by saying it need not have turned out this way. Above, we defined all the $h$ by the same rule, returning the component as it is. Suppose $h_1, \dots, h_5$ return the component and only $h_6$ returns the component with its sign flipped. The lecture explains that then, however cleverly we choose the charts, the structure group does not shrink below $\{1, -1\}$, that this group is isomorphic to $\mathbb{Z}_2$, and that it corresponds to an infinite Möbius strip. We can no longer make it a product, so the total space becomes a space other than $S^1 \times \R$, and this is a new bundle that is not the tangent bundle. This small change changes everything, and the next post treats the concrete Möbius strip mathematically, without the tangent bundle.

Let me correct one thing here. In the six-interval arrangement above, $U_6$ overlaps its two neighbors $U_5$ and $U_1$. If we flip the sign of $h_6$, the sign changes once at each of the two places $r_{56}$ and $r_{61}$, and the value $r_{12} r_{23} \cdots r_{61}$ multiplied around the loop stays the same. So we can flip $h_6$ back to return all transition functions to 1, and this change alone still gives a trivial bundle. To get a Möbius strip, $-1$ must be multiplied at only one overlap, so that the product around the loop is $-1$. (This correction is added for understanding and was not covered in the lecture. The lecture says that flipping the single $h_6$ already makes the structure group $\{1, -1\}$.) If the product around the loop is positive, we can choose the scales again to make all transition functions 1, and if it is negative, a single $-1$ remains even after we match the sizes to 1, so it cannot be reduced below $\{1, -1\}$. (This criterion is also added for understanding and was not covered in the lecture.)

Try it in the toy below. I cut the six intervals down to three. Drag the point $p$ on the circle to see the cover sets containing $p$ and the components of the tangent vector $X$, and on an overlap the ratio of the two components appears as the transition function. When you change the chart scales with the upper buttons, the structure group shrinks down to $\{1\}$, and with the lower buttons you put in a twist and see how the product around the loop changes. Compare putting $-1$ on a single overlap with flipping the sign of one chart. (This explanation is added for understanding and was not covered in the lecture.)

What this example showed

The tangent bundle is a fiber bundle made by attaching one tangent space at each point. In this example the base was $S^1$, the fiber $\R$ and the total space the product $S^1 \times \R$. The preimage of the projection was the tangent space at a point, and the $h$ that formed the bundle charts was the map returning the component of a vector. The transition function on an overlap was the ratio of the speeds at which the two chart coordinates change, and when we chose the charts cleverly the structure group shrank to $\{1\}$ and the bundle became trivial. The lecture closes by saying that what we covered here is the first example of a bundle, that knowing globally trivial bundles is almost enough, but that talking about nontrivial ones is too tempting, so the next time will treat the Möbius strip a little more.