# An example of a Lie group

I have a trouble learning Lie groups --- I have no canonical example to imagine while thinking of a Lie group. When I imagine a manifold it is usually some kind of a 2D blanket or a circle/curve or a sphere, a torus etc.

However I have a problem visualizing a Lie group. The best one I thought is SO(2) which as far as I understand just a circle. But a circle apparently lacks distinguished points so I guess there is no way to canonically prescribe a neutral element to turn a circle into a group SO(2).

Examples I saw so far start from a group, describe it as a group of matrices to show that the group is endowed with the structure of a manifold. I would appreciate the other way --- given a manifold show that it is naturally a group. And such a manifold should be easily imaginable.

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We use ℝ^n all the time –  pyCthon Jan 17 '13 at 18:26
R\Z (reals modulo 1) is also a good one. –  Jan Dvorak Jan 17 '13 at 18:27
Concerning "a circle ... lacks distinguished points": That's necessary, for any Lie group (or more generally any topological group). Any two points look alike, topologically, because there is a homeomorphism of the space sending one to the other. Specifically, you van send $x$ to $y$ by multiplying on the left by $yx^{-1}$, and this left-multiplication operation is a homeomorphism. –  Andreas Blass Jan 17 '13 at 18:30
The torus is $\mathbb{S}^1\times\mathbb{S}^1$, a product of Lie groups. For the circle group, perhaps better to think of it as $U(1)$, the unit complex numbers. –  Neal Jan 17 '13 at 19:34
@AndreasBlass so since we can't topologically distinguish points we had to mark a certain (arbitrary) point with a pencil and call it $\text{id}$ (one, zero)? –  Yrogirg Jan 18 '13 at 4:19

A good example is the $3$-sphere $\mathbb{S}^3$. In my mind, I think of this first and foremost as a geometric entity, and certainly people were considering spheres before they started to think about groups. As you may know, $\mathbb{S}^3$ can be made into a group by identifying it with the set of unit quaternions

$$\{ q \in \mathbb{H}: \lVert q \rVert = 1\}$$

Then, the group structure on $\mathbb{S}^3$ is inherited from the group structure of the quaternions, making $\mathbb{S}^3$ a Lie group.

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But where is zero ( quarternion neutral element) on $\mathbb S^3$ ? –  Yrogirg Jan 18 '13 at 3:43
When you represent $\mathbb{S}^3$ as the set of unit quaternions, the one corresponding to $1$ will be the unit element. –  Isaac Solomon Jan 20 '13 at 2:37
Yrogirg, just to make it clear: We consider the unit quaternions as a multiplicative group. So there is no zero-element (additive neutral element), but a one-element (multiplicative neutral element). –  Hauke Strasdat Jan 21 '13 at 22:34

Think of $SO_2$ as the group of $2\times 2$ rotation matrices:

$$\left[\begin{array}{cc} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{array} \right]$$

or the group of complex numbers of unit length $e^{i\theta}$.

You can convince yourself directly from definitions that either of these objects is a group under the appropriate multiplication, and that they are isomorphic to each other.

This group (which is presented in two ways) is a 1-manifold because it admits smooth parametrization in one variable ($\theta$ here).

Why does this group represent the circle $S^1$? Well, the matrices of the above form are symmetries of circles about the origin in $\mathbb{R}^2$, and the trace of $e^{i\theta}$ is the unit circle in the complex plane. I think it is best conceptually to think of Lie groups first as groups, and then to develop geometric intuition to "flavor" your algebraic construction.

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