Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|cite|improve this question
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

4 Answers 4

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.

share|cite|improve this answer
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 at 0:39

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.

share|cite|improve this answer

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.)

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.