Example of SO(2) being not simply connected

[Hill]

TL;DR

The example from "QFT and SM" by Schwartz

The example goes like this:

The group SO(2) is specified by angles $\theta$. Let's parametrize a path by $0 \leq t \lt 1$ and consider the path $\theta (t) = 2 \pi t$. Then it says, "There is no smooth function $\theta (t,u)$ for $0 \leq u \leq 1$, such that $\theta (t,0) = \theta (t)$ and $\theta (t,1) = 0$."

Why? What is wrong with $\theta (t,u) = \theta (t) (1-u)$? What am I missing? I suspect it has something to do with the path being closed, but where does it appear in this example?

 

[PeterDonis]

The group SO(2) is specified by angles $\theta$.

More precisely, by real numbers $\theta$ such that the numbers $\theta + 2 \pi n$, where $n$ is any integer, all label the same group element of SO(2). The obvious model for this group is the unit circle, where each value of $\theta$ labels a point on the circle by its angle, conventionally measured counterclockwise from the positive $x$ axis. Hence the term "angle" for $\theta$.

Let's parametrize a path by $0 \leq t \lt 1$ and consider the path $\theta (t) = 2 \pi t$.

This would mean "going around the circle" once.

Then it says, "There is no smooth function $\theta (t,u)$ for $0 \leq u \leq 1$, such that $\theta (t,0) = \theta (t)$ and $\theta (t,1) = 0$."

Why? What is wrong with $\theta (t,u) = \theta (t) (1-u)$?

When $u$ is nonzero, you are no longer describing a closed loop within SO(2) (namely SO(2) itself). You're describing an open subset of SO(2). That's not what you need to show that a group is simply connected; you need a smooth function that shrinks a closed loop within the group to a point, while keeping it a closed loop the entire time.

 

[Hill]

More precisely, by real numbers $\theta$ such that the numbers $\theta + 2 \pi n$, where $n$ is any integer, all label the same group element of SO(2). The obvious model for this group is the unit circle, where each value of $\theta$ labels a point on the circle by its angle, conventionally measured counterclockwise from the positive $x$ axis. Hence the term "angle" for $\theta$.


This would mean "going around the circle" once.


When $u$ is nonzero, you are no longer describing a closed loop within SO(2) (namely SO(2) itself). You're describing an open subset of SO(2). That's not what you need to show that a group is simply connected; you need a smooth function that shrinks a closed loop within the group to a point, while keeping it a closed loop the entire time.

Thank you, I understand all this. But none of this is encoded in the example in the text. IOW, the "example" as it is does not demonstrate anything. Maybe it needs some additional condition on the function $\theta (t,u)$ that would keep the path closed for all $u$?

 

[PeterDonis]

none of this is encoded in the example in the text.

That's because the example is leaving it to you to prove the claimed statement. (Unless there is some proof given later on in the text, that you haven't quoted.) Finding a single valid counterexample would disprove the claim, but your proposed counterexample is not a valid one. But proving the statement requires an argument that proves that there is no possible counterexample.

 

[PeterDonis]

Maybe it needs some additional condition on the function $\theta (t,u)$ that would keep the path closed for all $u$?

You're missing the point. There is no possible function $\theta(t, u)$ that would keep the path closed for all $u$. That is why SO(2) is not simply connected.

 

[Hill]

You're missing the point. There is no possible function $\theta(t, u)$ that would keep the path closed for all $u$. That is why SO(2) is not simply connected.

My point is that in their example, the function $\theta (t,u)$ is not defined to keep the path closed. (This is an example, not an exercise.)

 

[PeterDonis]

My point is that in their example, the function $\theta (t,u)$ is not defined to keep the path closed.

Yes, it is, because it's defined to be a smooth function. That requires that if the path is closed for $u = 0$, it must remain closed for all values of $u$ (where we treat the single point at $u = 1$ as an edge case of a "closed path"). Otherwise the function will not be smooth when you transition from a value of $u$ where the path is closed to a value of $u$ where it isn't.

 

[Hill]

How to show that the function $\theta (t,u) = 2 \pi t (1-u)$ is not smooth on $0 \leq t \lt 1, 0 \leq u \leq 1$?

 

[PeroK]

How to show that the function $\theta (t,u) = 2 \pi t (1-u)$ is not smooth on $0 \leq t \lt 1, 0 \leq u \leq 1$?

There seems to be a lot of mathematics missing here. The definition of simply connected is that if we take any two paths with the same start and end points, then we can continuously deform one path into the other.

In this case, to show that $SO(2)$ is not simply connected, we would need two paths that cannot be continuously deformed into one another.

Also, we need to look at what the continuous deformation of one path into another means. To do this we have to consider the space of paths within $SO(2)$, which is the space of continuous functions from $[0,1] \to SO(2)$. Let's call this space $Y$. A continuous deformation of one path into another is a family of functions (i.e. a continuous mapping, $h$, from $[0, 1] \to Y$) such that $h(0)$ is the first path and $h(1)$ is the second path. Note that continuity here depends on the topology defined on $Y$ (the space of continuous maps on $SO(2)$.

If you want to prove that $SO(2)$ is not simply connected, you need:

a) Two paths $p, q: [0,1] \to SO(2)$, with $p(0) = q(0)$ and $p(1) = q(1)$

PS it looks like we only need to consider closed paths, where $p(0) = p(1)$.

b) Show that there is no continuous function $h: [0, 1] \to Y$, such that $h(0) = p$ and $h(1) = q$, and where, for every $h(t)(0) = p(0)$ and $h(t)(1) = p(t)(1)$.

To prove b) from first principles looks difficult to me. I would expect, instead, to use some theorem of simply connected spaces that implies a property that $SO(2)$ lacks. That said, I guess I could go looking for a proof.

 

[PeroK]

TL;DR Summary: The example from "QFT and SM" by Schwartz

The example goes like this:

The group SO(2) is specified by angles $\theta$. Let's parametrize a path by $0 \leq t \lt 1$ and consider the path $\theta (t) = 2 \pi t$. Then it says, "There is no smooth function $\theta (t,u)$ for $0 \leq u \leq 1$, such that $\theta (t,0) = \theta (t)$ and $\theta (t,1) = 0$."

Why? What is wrong with $\theta (t,u) = \theta (t) (1-u)$? What am I missing? I suspect it has something to do with the path being closed, but where does it appear in this example?

In this example, you are trying to deform the path $p: [0,1] \to SO(2)$, s.t. $p(t) = 2\pi t$ into the path $q: [0,1] \to SO(2)$, s.t. $q(t) = 0$.

You need the topology on $Y$ to do this formally. But, the idea is clear that there is no way to continuously deform the loop into the null path, without changing the endpoints of the path.

 

[PeroK]

I've not studied topology for 40 years, so this is hard work!

 

[PeterDonis]

In this case, to show that $SO(2)$ is not simply connected, we would need two paths that cannot be continuously deformed into one another.

I don't think that is sufficient, since we can have two open paths on SO(2) that can be continuously deformed into one another (just "slide" one along the circle until it coincides with the other, stretching or compressing as necessary). The definition I am familiar with is that a topological space is simply connected if every closed path can be continuously deformed into a single point; so finding any closed path that cannot be continuously deformed into a point is sufficient to show that a space is not simply connected.

I would expect, instead, to use some theorem of simply connected spaces that implies a property that $SO(2)$ lacks.

I think the statement of yours quoted below points at a solution along these lines.

there is no way to continuously deform the loop into the null path, without changing the endpoints of the path

Not just changing the endpoints, but disconnecting the endpoints, i.e., changing the closed path into an open path. Heuristically, there is no closed path in SO(2) that is "in between" SO(2) itself (which is a closed path since it is homeomorphic to S1, the circle) and a single point in SO(2).

 

[PeroK]

I don't think that is sufficient, since we can have two open paths on SO(2) that can be continuously deformed into one another (just "slide" one along the circle until it coincides with the other, stretching or compressing as necessary). The definition I am familiar with is that a topological space is simply connected if every closed path can be continuously deformed into a single point; so finding any closed path that cannot be continuously deformed into a point is sufficient to show that a space is not simply connected.

Yes, it looks like the definition only specifies closed paths. It doesn't have to be every pair of points. Although, if we can do it for every path, then we can do it for every loop.

I wonder whether every path is an equivalent condition or genuinely stronger?

PS it's the additional path-connected condition that requires every pair of points to be connected by a path.

 

[PeroK]

Not just changing the endpoints, but disconnecting the endpoints, i.e., changing the closed path into an open path. Heuristically, there is no closed path in SO(2) that is "in between" SO(2) itself (which is a closed path since it is homeomorphic to S1, the circle) and a single point in SO(2).

I was just typing a post to say that the proof is essentially the same as that of the circle in 2D - which is path connected, but not simply connected.

 

[PeterDonis]

if we can do it for every path, then we can do it for every loop.

I wonder whether every path is an equivalent condition or genuinely stronger?

Any open path can obviously be continuously deformed to a single point--just do it by shrinking the path itself, which is already known to be contained in the topological space. So including open paths does not change anything.

 

[PeroK]

Any open path can obviously be continuously deformed to a single point--just do it by shrinking the path itself, which is already known to be contained in the topological space. So including open paths does not change anything.

The question is whether any path between two points can be continuously deformed into any other path between those two points. That's (possibly) a stronger condition than the condition of simple-connectedness. It might, however, be equivalent. Hypothesis:

If $X$ is path connected and if every closed path can be continuously deformed to a single point (usual definition of simple connectedness), then any path between two points can be continuously deformed into any other path between those two points (alternative definition?).

PS this is suggested in several sources, although I haven't found it stated explicitly yet.

 

[jbergman]

TL;DR Summary: The example from "QFT and SM" by Schwartz

The example goes like this:

The group SO(2) is specified by angles $\theta$. Let's parametrize a path by $0 \leq t \lt 1$ and consider the path $\theta (t) = 2 \pi t$. Then it says, "There is no smooth function $\theta (t,u)$ for $0 \leq u \leq 1$, such that $\theta (t,0) = \theta (t)$ and $\theta (t,1) = 0$."

Why? What is wrong with $\theta (t,u) = \theta (t) (1-u)$? What am I missing? I suspect it has something to do with the path being closed, but where does it appear in this example?

What page in Schwartz is this. Clearly as written here the statement is wrong. As you pointed out there is a smooth function with the properties specified.

 

[PeroK]

What page in Schwartz is this. Clearly as written here the statement is wrong. As you pointed out there is a smooth function with the properties specified.

It's not wrong. If we take $u = 1/2$, then the path is:
$$p(t) = \theta(t, 1/2) = \frac 1 2 \theta(t)$$The endpoint for that path is $p(1) = \pi$. Therefore, that is not a mapping to a family of closed loops.

 

[PeterDonis]

The question is whether any path between two points can be continuously deformed into any other path between those two points. That's (possibly) a stronger condition than the condition of simple-connectedness. It might, however, be equivalent.

I believe it's equivalent, because if we have two distinct paths that both connect the same points, the two paths combined are a single closed path. And continuously deforming one of the two distinct paths into the other is then equivalent to continously deforming the closed path into a single point (since I can always continuously deform either of the two paths, which both must be open since they connect two distinct points, into a point, or vice versa).

 

[Hill]

What page in Schwartz is this

P. 177.

 

[jbergman]

What page in Schwartz is this. Clearly as written here the statement is wrong. As you pointed out there is a smooth function with the properties specified.

I looked it up on pg. 177. There is an error. The problem is that the domain of t needs to be the closed interval $t \in [0, 1]$. Then you get a non-contractible loop around the entire circle.

 

[jbergman]

It's not wrong. If we take $u = 1/2$, then the path is:
$$p(t) = \theta(t, 1/2) = \frac 1 2 \theta(t)$$The endpoint for that path is $p(1) = \pi$. Therefore, that is not a mapping to a family of closed loops.

Sure. Even at u=0 it isn't a closed loop. But that wasn't the question. It just said to show there is no smooth function with the stated conditions. It doesn't say anything about paths. There is a problem with the domain of t as I just stated which fixes the exercise.

 

[jbergman]

P. 177.

The domain of t should be $t \in [0, 1]$ on the problem.

 

[PeroK]

Sure. Even at u=0 it isn't a closed loop. But that wasn't the question. It just said to show there is no smooth function with the stated conditions. It doesn't say anything about paths. There is a problem with the domain of t as I just stated which fixes the exercise.

You've missed the point that the mapping is from $[0,1]$ to the set of paths in $SO(2)$. It's not a mapping from $[0,1]$ into $SO(2)$. Look at the definition of simply connected.

 

[jbergman]

You've missed the point that the mapping is from $[0,1]$ to the set of paths in $SO(2)$. It's not a mapping from $[0,1]$ into $SO(2)$. Look at the definition of simply connected.

You've missed the point because you haven't closely read the question. There are two parameters, t and u. The t parameter describes where one is on a path and the u parameter controls which path you are on.

The problem is that even on the path at u=0, as t ranges from $t\in [0,1)$ theta ranges as $\theta(t) \in [0,2\pi)$. In other words the loop isn't closed at u=0.

This is simply fixed by changing the domain of t so that you get a full loop around the circle at u=0.

 

[PeroK]

You've missed the point because you haven't closely read the question. There are two parameters, t and u. The t parameter describes where one is on a path and the u parameter controls which path you are on.

The problem is that even on the path at u=0, as t ranges from $t\in [0,1)$ theta ranges as $\theta(t) \in [0,2\pi)$. In other words the loop isn't closed at u=0.

This is simply fixed by changing the domain of t so that you get a full loop around the circle at u=0.

Is that all it was? Note that, in that case, we are not even dealing with a path in the first place. A path in topology is a mapping from a closed interval.

 

[jbergman]

Is that all it was? Note that, in that case, we are not even dealing with a path in the first place. A path in topology is a mapping from a closed interval.

Schwartz' book is known to have quite a few minor mistakes like this.