# Help needed to prove that SO(2) is a Lie group

1. Nov 7, 2013

### mnb96

Hello,

I want to prove that the set SO(2) of orthogonal 2x2 matrices with det=1 is a Lie group.
The group operation is of course assumed to be the ordinary matrix multiplication $\times$:SO(2)→SO(2).

I made the following attempt but then got stuck at one point.

We basically have to prove that SO(2) is a differentiable manifold.
In order to do so I consider the group isomorphism $$\left( SO(2), \times \right) \cong \left( [0,2\pi), +_{\mathrm{mod\, 2\pi}} \right)$$ where +mod 2∏ means addition modulo 2∏.

At this point I should prove that $f(x,y) = (x+y)\, \mathrm{mod}\, 2\pi$ is smooth, and that $f^{-1}(x) = (2\pi -x)\, \mathrm{mod}\, 2\pi$ is also smooth.

This is where I get stuck.
For instance, $f^{-1}$ is certainly smooth in the open interval (0,2∏), but it is not differentiable at x=0, because at x=0 we don't have an open neighborhood where the $f^{-1}$ would be defined.
How do we address this issue?

2. Nov 7, 2013

### fzero

The parameter space is a circle. It is impossible to cover the circle with a single chart. Around $x=0$, we can use the chart on the interval $(-\pi, \pi)$.

3. Nov 8, 2013

### mnb96

thanks fzero!

That is also what I thought, and it was the approach I wanted to follow, initially.
But then something went wrong, because I crashed against a big problem:

If I want to define a coordinate chart $\xi:U\subseteq SO(2) \rightarrow V\subseteq \mathbb{R}$, then I must define $\xi$ so that both U and V are open subsets. But unfortunately I have no clue of what could be an "open subset U of SO(2)".
As far as I know, open subsets are defined for metric spaces in terms of open-balls at some point. Here, we don't have metric, so it is impossible to tell how close two elements of SO(2) are to each other.

Perhaps I should just use the isomorphism $\left( SO(2), \times \right) \cong \left( [0,2\pi), +_{\mathrm{mod\, 2\pi}} \right)$ and just work on that, and forget about SO(2).

4. Nov 8, 2013

### fzero

You can take $U_1=(0,2\pi)$ and $U_2=(-\pi,\pi)$. These are both open intervals, which are open in both the topological and metric sense.

5. Nov 8, 2013

### jgens

Another way of doing this by the way is identifying what the Lie algebra of SO(2) ought to be. Then you can use the exponential map on matrices to give yourself the desired charts.

6. Nov 8, 2013

### mnb96

uh...actually the approach of using two charts does not seem to work either.
I tried to do the following:

1) Applied the isomorphism that I mentioned: so now we have to prove that $G = \left( [0,2\pi), +_{\mathrm{mod\, 2\pi}} \right)$ is a Lie group.

2) The group operation for any $x,y\in [0,2\pi)$ is given by: $f(x,y)=(x+y) \, \mathrm{mod}\, 2\pi$

3) We have to prove that the two variable function f is smooth. But it is clearly not smooth (not even continuous), and I don't see how multiple charts could help.

Any hint?

7. Nov 8, 2013

### fzero

$\left( [0,2\pi), +_{\mathrm{mod\, 2\pi}} \right)$ is topologically the circle. The interval together with modular addition should be understood as the quotient space $\mathbb{R}/(2\pi\mathbb{Z})$. We start with smooth functions on $\mathbb{R}$. Smooth functions on $\mathbb{R}/(2\pi\mathbb{Z})$ are the subset that satisfy the appropriate periodicity conditions.

I'm assuming that your issue with continuity is still related to the idea that something strange is happening as $x\rightarrow 2\pi$. But this is precisely the issue that we address by using a 2nd chart to cover the circle. This is a good illustration of how a manifold differs from Euclidean space. Locally, meaning in any neighborhood of a point, we can find Euclidean coordinates. But these coordinates will not necessarily extend to the neighborhood of a distant point.

8. Nov 8, 2013

### mnb96

thanks fzero!

I think that I previously misunderstood the definition of Lie Group. That was probably the source of my confusion. Now I am re-reading all your posts and I am starting to get what you wanted to say.
Basically, if I am not wrong I should do the following steps:

1) First of all: prove that SO(2) is a manifold! (I have been skipping this step so far!)

1a) In order to do prove point 1) I mapped each matrix of SO(2) onto one point of the circle S1. Actually it is more convenient to map onto the unit complex numbers.

2) Prove that S1 is a manifold (for this I obviously need two charts: the ones you mentioned)

3) Use the isomorphism between SO(2) and the group of unit complex numbers with multiplication in order to obtain the maps:
mult: S1 × S1S1 and
inv: S1S1

4) prove that mult and inv are smooth.

I think I am almost done with points 1) and 2), so I'll let you know how it goes...

9. Nov 8, 2013

### mnb96

Ok...I think I convinced myself that SO(2) is a manifold isomorphic to the unit circle.
As you said, I had to use two coordinate charts to prove that there exists an atlas that covers the circle. So far so good.

The charts were:

$\xi_1(\theta) = \left( \cos\theta, \, \sin\theta \right) \quad \quad;\; \theta\in(0,2\pi)$
$\xi_2(\theta) = \left( \cos\theta, \, \sin\theta \right) \quad \quad;\; \theta\in(-\pi,\pi)$

The transition maps:

$\tau_{21}(\theta) = \tau_{12}(\theta) = \theta \quad \quad ;\; \theta\in(0,\pi)$

Unfortunately all this does not seem to help. In fact, if I try to express the group operation mult: S1×S1→S1 in terms of the parametrization above, we still have:

$mult(\theta,\alpha) = (\theta + \alpha)\mod 2\pi$

This function is not continuous, and it is not going to be smooth.
Can anyone please explain how to prove smoothness of mult? It should be a trivial issue, since all textbooks on Lie groups I have seen do not even spend a single word on this.

Last edited: Nov 8, 2013
10. Nov 8, 2013

### fzero

What you're calling charts here are defined in a backwards way. I would phrase them as

$$\xi_1 : \begin{pmatrix} \cos\theta_1 & \sin\theta_1 \\ - \sin\theta_1 & \cos\theta_1 \end{pmatrix} \longmapsto \theta_1$$

$$\xi_2 : \begin{pmatrix} \cos\theta_2 & \sin\theta_2 \\ - \sin\theta_2 & \cos\theta_2 \end{pmatrix} \longmapsto \theta_2,$$

with appropriate redefinitions of the transition functions.

Can you explain why you believe this is not a continuous function? It's obviously continuous if you fix, say $\alpha$, and choose appropriate open sets for for the domain and image.

11. Nov 9, 2013

### mnb96

Perhaps I should have called them "coordinate patches" rather than charts. The books from where I am studying (M. Do Carmo's "Differential Geometry", and B O'Neill's "Elementary Differential Geometry") use functions like the one I defined, and call them respectively "parametrizations", and "coordinate patches".
I noticed that other texts instead use coordinate charts in the same way as you mentioned.

I guess it does not make much difference here, since they are 1-to-1 functions?

Well, then let me fix $\alpha = \pi /2$, for instance.
Now we have got a function $(\theta + \pi/2)\mod 2\pi$.
At $\theta=\frac{3}{2}\pi$ we clearly have a discontinuity. Please have a look at this plot to see what I mean.

12. Nov 9, 2013

### fzero

That's because $f(x) =2\pi \equiv 0$ on the circle. You've just come around to another point on the circle. This is perfectly continuous since the points are identified.

13. Nov 9, 2013

### mnb96

Ok! So my misunderstanding probably originated from the fact that I was trying to prove smoothness exclusively on the domain of the coordinate patches, i.e. (0,2∏) and (-∏,+∏).
From your response I understood that, loosely speaking, one should instead prove smoothness on the circle (i.e. the points of the manifold).

14. Nov 9, 2013

### fzero

Well this sort of issue at the end of a single patch is why we need two patches for the circle. We could also show continuity by moving onto the 2nd patch on the overlap before we get to the end of the first patch. We move to good local coordinates for the point where the original coordinates would have broken down.

15. Nov 9, 2013

### D H

Staff Emeritus
I think you're going about this all wrong. The connection between the unit circle and SO(2) is a consequence of the fact that SO(2) is a Lie group. The Lie algebra is so(2), and this is isomorphic to the unit circle. By assuming an isomorphism from SO(2) to the unit circle you are implicitly assuming that SO(2) is a Lie group. You are, without knowing it, assuming what you are trying to prove!

As fzero mentioned, you need to prove that SO(2) is a differentiable manifold, show that a group operator exists, and show that this is consistent with the smooth structure of the manifold. In fact, it's not a big jump to proving those three things for any SO(n). That SO(2) is a Lie group is direct consequence of every SO(n) being a Lie group.

16. Nov 9, 2013

### mnb96

Oh no! I am not really assuming it, although from the previous posts it might seem so.
What I did was the following (Note: I don't claim it's correct).

1) I proved that the set SO(2) with the operation of matrix multiplication is isomorphic to the group U(1) of unit complex numbers with complex multiplication.

2) I proved there is a 1-to-1 correspondence between unit complex numbers and points of the unit circle (kind of trivial).

3) Point 2) implies that I can construct a group G whose elements are the points (x,y) on the unit circle, and the group operation is now given by mapping the points into complex numbers, performing complex multiplication, and converting back the result to a point of the circle. Now we have a group isomorphic to SO(2) whose elements are points of the circle, but this is still not sufficient to prove that G is a Lie group.

3) I proved that the set of points of G (the unit circle) is a manifold (fzero helped me with this by suggesting to cover the circle with two different parametrizations).

4) I finally proved that the group operation of G is a smooth map (fzero helped me again, and I had many troubles with this; I am not sure yet I fully understood this point)

17. Nov 9, 2013

### jgens

If you know that SO(2) is a topological group, then the obvious group isomorphism SO(2)→S1 is also a homeomorphism. You can then use this map to pull the smooth structure on S1 back to SO(2) and you end up with a Lie group. This works and does not assume a priori that SO(2) is a Lie group.