# Examples of familiar, easy-to-visualize manifolds that admit Lie group structures

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 $2$D 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

Some important examples include:

• Any finite-dimensional vector space $(\mathbb{V}, +, \cdot)$ over $\mathbb{R}$ or $\mathbb{C}$ is a Lie group under addition $+$; in particular this includes the familiar spaces $\mathbb{R}^n$.

• Any Euclidean space $\mathbb{R}^n$, $n > 1$ admits more than one group structure up to isomorphism. For example, $$(x, y) \ast (x', y') := (x + x',e^{x'} y + y')$$ defines a group structure on $\mathbb{R}^2$, isomorphic to the group $\mathbb{R}^* \rtimes \mathbb{R}$ of affine transformations $t \mapsto a t + b$ with positive slope $a$ under composition. The operation $$(x, y, z) \star (x', y', z') = (x + x', y + y', z + z' + xy')$$ defines a group structure on $\mathbb{R}^3$, and this is usually called the Heisenberg $3$-group. It's easy to verify that both $\ast$ and $\star$ are nonabelian, and so define group structures nonisomorphic to the familiar structures $(\mathbb{R}^n, +)$. In particular, this shows that while we can certainly ask whether a given manifold has a group structure, in general there can be more than one.

• Any torus $\mathbb{S}^1 \times \cdots \times \mathbb{S}^1$ is a Lie group under the direct product multiplication; this is just a quotient of $(\mathbb{R}^n, +)$ by $\mathbb{Z}^n$.

• Isaac has pointed out in another answer (which I include here for some limited semblance of completeness) that $\mathbb{S}^3$ admits a Lie group structure, which we can think of as the group of unit quaternions under quaternionic multiplication. We often call this group $SU(2)$, which we can think of the group preserving a Hermitian form and a (compatible) complex volume form on $\mathbb{R}^4 \cong \mathbb{C}^2 \cong \mathbb{H}$, which we can thus identify explicitly with the group of matrices $A \in GL(2, \mathbb{C})$ satisfying $A^* A = I_2$ and $\det A = 1$.

• The real projective space $\mathbb{RP}^3$ admits a natural group structure, namely that of the orthogonal group $SO(3, \mathbb{R})$. One can see this from the previous example if one identifies $\mathbb{S}^3$ with the double cover $Spin(3, \mathbb{R})$ of $SO(3, \mathbb{R})$.

• By the above example the product $\mathbb{S}^3 \times \mathbb{S}^3$ can be given the group structure $SU(2) \times SU(2)$ (or $Spin(3) \times Spin(3)$), but (in a feature exclusive to this dimension), this group is isomorphic to $Spin(4)$; so, we can regard $SO(4, \mathbb{R})$ (which is double-covered by $Spin(4, \mathbb{R})$) as a quotient of $\mathbb{S}^3 \times \mathbb{S}^3$, namely the one given by identifying a pair $(x, y)$ with its "double antipodes" $(-x, -y)$.

• The punctured spaces $\mathbb{R}^*$, $\mathbb{C}^*$, $\mathbb{H}^*$ are all groups under multiplication; in particular, $\mathbb{C}^*$ is sometimes called $CSO(2, \mathbb{R})$ and is the group of oriented conformal linear transformations of $\mathbb{R}^2$, that is, those preserving angles.

• We can naturally give the interior $\mathbb{D} \times \mathbb{S}^1$ of a solid torus (here $\mathbb{D}$ is the open unit disk) a group structure by identifying a pair $(a, \zeta)$ with the conformal isomorphism $\mathbb{D} \to \mathbb{D}$ defined by $$z \mapsto \zeta \frac{z - a}{1 - \bar{a} z};$$ these exhaust such isomorphisms, and so this group (under composition) is isomorphic to $PSL(2, \mathbb{R})$.

• Trivially, any group structure on the underlying set of a $0$-manifold is automatically smooth and hence defines a Lie group structure.

• An important nonexample: The sphere $\mathbb{S}^2$ admits no Lie group structure; in fact, the only spheres that admit Lie groups are $\mathbb{S}^n$ with $n = 0, 1, 3$. (The $7$-sphere $\mathbb{S}^7$ comes close---it admits a loop structure with a smooth operation given by octonionic multiplication, but this operation is nonassociative and so cannot define a group structure.)

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I find $\mathbb{R} - \lbrace 0 \rbrace$ a helpful example for thinking about Lie groups which aren't connected.

We can see that $(0,\infty)$ is a normal subgroup. We also see that any neighborhood of the identity can generate this subgroup.

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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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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
It's perhaps worth saying that this Lie group is usually called $SU(2)$, which is the group preserving a Hermitian structure and a compatible complex volume form on $\mathbb{R}^4 \cong \mathbb{C}^2 \cong \mathbb{H}$, which can in turn be regarded as the group of matrices $A \in GL(2, \mathbb{C})$ satisfying $A^* A = I_2$ and $\det A = 1$. This group is also called $Spin(3, \mathbb{R})$, the double cover of the orthogonal group $SO(3, \mathbb{R})$ (which is hence diffeomorphic to $\mathbb{RP}^2$, furnishing another answer to the original question). – Travis Mar 16 '15 at 0:39