# Lie Groups

## Circle as a 2-Manifold

The definition of a smooth manifold is primarily concerned with finding "coordinate patches" that fit specified constraints. Finding coordinate patches for a circle is easy enough using parametrization with cos and sin. The tricky part comes in finding the appropriate domain. The restrictions in the definition are there to ensure that the domains of the patches are "well-defined" and differentiable. The first part of the definition mentions injective functions and open subsets:

An n-dimensional differentiable manifold is a set $$M$$ and a family $${\gamma_1,\gamma_2,\dots}$$ of injective functions from domains in $$\mathbb{R}^n$$ into $$M$$ such that:

1. For each $$i$$ the domain $$U_i$$ of the function $$\gamma_i$$ is an open subset of $$\mathbb{R}^n$$. The domains $$U_i$$ and the functions $$\gamma_i$$ are called coordinate patches.

An injective function, or a one-to-one function, is simply one in which there are no two domain values that when plugged into the function give the same answer. When dealing with sin and cos, the domain has to be restricted so that the values on the circle are not repeated. This can be done by keeping the domain between 0 and 2pi. The definition says that each domain $$U$$ has to be an open subset of n-dimensional real space (1-dimensional in this case). So the coordinate patch has to be $$\left(0,2\pi\right)$$. This does not "cover" the whole circle, because it leaves the value 0/2pi undefined for the circle. The 2nd part of the definition addresses coverage:

1. The union of the coordinate patches cover $$M$$. That is, $$\gamma_1\left(U_1\right)\cup\gamma_2\left(U_2\right)\cdots = M$$.

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If we introduce another coordinate patch that includes 0/2pi in the domain then the union of the two patches will cover all of the circle and criterion 2 will be satisfied. Let's define the two coordinate patches in this way:

\begin{eqnarray} r_1:\left(0,2\pi\right)\rightarrow S^1, t_1\mapsto {\cos t_1\choose\sin t_1} \\ r_2:\left(-\pi,\pi\right)\rightarrow S^1, t_2\mapsto {\cos t_2\choose\sin t_2} \\ \end{eqnarray}

To view the coordinate patches, right-click on the buttons below the plot.

Now we need to verify that the third part of the definition holds in order to prove that the circle is a 2-manifold. The third condition needs a little more unravelling than the first two:

1. The coordinate systems are compatible on overlaps. That is, if we consider the overlaps $$W_{ij} := \gamma_i\left(U_i\right)\cap\gamma_j\left(U_j\right)$$ then the sets $$\gamma_i^{-1}\left(W_{ij}\right)$$ and $$\gamma_j^{-1}\left(W_{ij}\right)$$ must be open in $$\mathbb{R}^n$$ and the transition functions $$\Gamma_{ji}:=\gamma_j^{-1}\circ\gamma_i : \gamma_i^{-1}\left(W_{ij}\right)\rightarrow\gamma_j^{-1}\left(W_{ij}\right)$$ must be infinitely differentiable.

What are the overlaps, the sets, and the transition functions? First replace the terms with what we know about the circle. The overlaps become $$W_{1,2}:=\gamma_1\left(\left(0,2\pi\right)\right)\cap\gamma_2\left(\left(-\pi,\pi\right)\right)$$, which is the intersection of the circle excluding the point (1,0) and the circle excluding the point (-1,0). So, the overlap is the circle without the points (1,0) and (-1,0). The sets $$\gamma_1^{-1}\left(W_{1,2}\right)$$ and $$\gamma_2^{-1}\left(W_{1,2}\right)$$ are the parts of the domains of each of the coordinate patches that corresponds the circle, exluding (1,0) and (-1,0). Following the definition, the sets $$\gamma_1^{-1}\left(W_{1,2}\right):=\left(0,\pi\right)\cup\left(\pi,2\pi\right)$$, and $$\gamma_2^{-1}\left(W_{1,2}\right):=\left(-\pi,0\right)\cup\left(0,\pi\right)$$ are open

Side note: Remember that all the inverse does is give us the domain instead of the range. For example, if x is the domain of a function f, then $$f\left(x\right)$$ is the range defined as $$f\left(x\right)=y$$. The inverse function $$f^{-1}$$ returns the domain, while the range is the independent variable: $$f^{-1}\left(y\right)=x$$. Or, thinking about it another way, if the domain is the raw material that goes into a machine, f, then the range is the product that the machine makes out of the raw material. In this scenario, the inverse function would be a machine which turns product into raw material.

So, the transition function is simply the function that can turn the open set $$\gamma_1^{-1}\left(W_{1,2}\right):=\left(0,\pi\right)\cup\left(\pi,2\pi\right)$$, into the open set $$\gamma_2^{-1}\left(W_{1,2}\right):=\left(-\pi,0\right)\cup\left(0,\pi\right)$$:

\Gamma_{2,1}:=\gamma_2^{-1}\circ\gamma_1:\left(0,\pi\right)\cup\left(\pi,2\pi\right)\rightarrow\left(-\pi,0\right)\cup\left(0,\pi\right)

The function is easy to come up with. The points on the circle between (1,0) and (-1,0) on the top half of the circle are the same in either set, (0,pi). The points on the bottom half are (pi,2pi) in one set and (-pi,0) in the other. So the function can be defined in this way:

\begin{eqnarray} \Gamma_{2,1}:\left(0,\pi\right)\cup\left(\pi,2\pi\right)\rightarrow\left(-\pi,0\right)\cup\left(0,\pi\right) \\ t_1 \mapsto \begin{cases} t_1 & t_1 \in \left(0,\pi\right) \\ t_1 - 2\pi & t_1 \in \left(\pi,2\pi\right) \end{cases} \end{eqnarray}

Thus, the transition function is infinitely differentiable (if the lines on the plot had closed ends, then the function would not be infinitely differentiable), and the circle is a 2-manifold.

This was a detailed look at how a circle fits the definition of a 2-manifold. The next page looks briefly at how one might go about showing that the sphere is a manifold (a 4-manifold, because it takes 4 coordinate patches to cover a sphere)