How could a group be a manifold?

Question

For example a Lie group is defined as a certain differentiable manifold, but what does this mean geometrically, and what is gained by viewing something abstract and algebraic as a manifold?

First, I know there are severel quite abstract definitions of a manifold, but what I know from my analysis courses, a manifold is something that could be defined by equations (i.e. something like $f^{-1}(0)$ for a regular function) or for example as defined in Munkres: Analysis on Manifolds, p. 109:

A subseteq M of $\mathbb R^n$ is called a $k$-dimensional manifold (in $\mathbb R^n)$ if for every point $x \in M$ the following condition is satisfied:

(M) There is an open set $U$ containing $x$, an open set $V \subseteq \mathbb R^n$, and a diffeomorphism $h : U \to V$ such that $$ h(U\cap M) = V \cap (\mathbb R^k \times \{0\}) = \{ y \in V : y^{k+1} = \ldots = y^n = 0 \}. $$

So a manifold is something concrete, something that I can think of sitting in $\mathbb R^n$ (I know vaguely there are some intrinsic definitions saying something like a set $X$ is manifold if it has a topology and to each point there exists a diffeomorphism on $\mathbb R^n$).

So I am used to think of a manifold as a geomtric object, and in some sense this are the explanations I find everywhere, but in what sense could a group be something "geometrically concrete", for example $SL(n, K)$ is also a manifold (this could be seen by noting that it is the inverse image $\det^{-1}(1)$), but again what does this mean geometrically and what is gained by seeing for example $SL(n,K)$ as a manifold?

Answer 1

I'll address two points here:

• how can a group be a geometric object, and
• what is gained by the geometric viewpoint

Let's start with an easy example. Consider $SO_2$, the group of rotations in $\Bbb R^2$. We can identify this group with the set $S_1 = \{z \in \Bbb C: |z| = 1\}$. Geometrically, this set is a circle in $\Bbb C$ (which, geometrically, is just $\Bbb R^2$). The idea encapsulated here is that we associate every point on the circle with a rotation by the corresponding angle.

The geometry adds a differential topology to the group, so that we can now take limits and derivatives in our group. So, for example, if we have a sequence $(A_n) \subset SO_2$ with $A_n \to A$, then if $A$ acts on $\Bbb R^2$, we'll have $A_n x \to Ax$ for any $x \in \Bbb R^2$. We can also make sense of a derivative: take $$ A(t) = e^{iat}, \quad a \in \Bbb R $$ then we have $$ A'(0) = \lim_{t \to 0} \frac{e^{iat} - 1}{t} = ia $$ So that $ia \in i\Bbb R \subset \Bbb C$ is an element of the tangent space of this group at the identity.

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By far, the most useful thing that comes out of manifold structure to the group is the notion of a tangent space. In particular, we define the tangent space at the identity in a Lie Group $G$ by $$ g = T_I(G) = \{A'(0): A:[-1,1] \to G \text{ is differentiable with } A(0) = I\} $$ This gives us the Lie algebra associated with $G$. The Lie algebra $g$ is a vector space with a natural operation given by "Lie brackets", defined by $$ [X,Y] = XY - YX \quad X,Y \in g $$ It turns out that Lie groups can (often) be completely understood in terms of their Lie algebras, which are (perhaps surprisingly) often easier to work with.

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Tangent space of $SL_n$:

Define the matrix exponential by $$ \exp[X] = \sum_{k=0}^\infty \frac 1{k!}X^k $$ where we define $X^0 = I$. Consider $SL_n$ as a subset of $\Bbb R^{n \times n}$.

For any matrix $X$, we can consider the map $A:[-1,1] \to \Bbb R^{n \times n}$ defined by $$ A(t) = \exp[tX] $$ And we have $A'(t) = Xe^{tX}$, so that $A'(0) = X$.

We note that for any matrix $M$, $\det(\exp[M]) = \exp[\operatorname{trace}(M)]$. It follows that $A(t)$ will only be a path in $SL_n$ if $X$ has trace $0$.

So, the tangent space of $SL_n$ will include all trace-zero matrices. In fact, there's a theorem that allows us to state that this is the entire tangent space to $SL_n$, so that the trace zero matrices form the Lie algebra of $SL_n$.