Lie groups must be smooth manifolds, so intuition would suggest a Lie group must have an infinite number of continuously parameterized elements.

But on the other hand, I am studying Lie groups from Brian C. Hall's "Lie Groups, Lie Algebras, and Representations," in which he focuses on matrix Lie groups instead of general Lie groups. By his definition, the group $O(1)$ is a matrix Lie group. It consists of only the $1\times 1$ matrices $\{[1],[-1]\}$. Yet Hall has a theorem (Theorem 1.19) which states,

Every matrix Lie group is a smooth embedded submanifold of $M_n(\mathbb C)$ and is thus a Lie group.

So it seems either this theorem is incorrect as stated (and needs to exclude these discrete cases), or my intuition about smooth manifolds needing to be infinite is incorrect. Which of these is the case?

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    $\begingroup$ Discrete groups are the zero-dimensional Lie groups. $\endgroup$ Dec 17, 2019 at 4:18

2 Answers 2


Any (countable) set $X$ admits a unique smooth $0$-manifold structure: just give $X$ the discrete topology, and take the unique map $\{x\}\to\mathbb{R}^0$ as a chart for each $x\in X$, since $\mathbb{R}^0$ is just a single point. Compatibility of the charts is trivial since no two distinct charts overlap. Moreover, any map from a $0$-dimensional smooth manifold to any smooth manifold is automatically smooth (in local charts, you just have a map $\mathbb{R}^0\to\mathbb{R}^n$ which is always smooth, since there are no partial derivatives that need to exist). In particular, any group structure on $X$ makes $X$ a Lie group.

(The countability requirement is if you require manifolds to be second-countable, which is sometimes not done but generally is in the context of Lie groups.)

  • $\begingroup$ Thanks, this is the articulation I was looking for. Wasn't sure what to make of 0-dimensional manifolds. $\endgroup$
    – WillG
    Dec 17, 2019 at 5:12

As mentioned in the comments discrete groups are Lie groups, but to give info on the theorem consider that every finite group $G$ acts by automorphism on itself, i.e $G$ can be embedded in some permutation group $S_n$ ($n$ can be taken as the order of $G$) and these admit a representation on $k^n$ by permuting coordinates, where $k$ is your favorite field, so we have $G \to S_n \to GL(n, k)$ so in fact $G$ is embedded as a closed subgroup of $GL(n,k)$. Discrete groups are even complex Lie groups. The only thing to have in mind is that a discrete group is algebraic if and only if it is finite.

  • $\begingroup$ Thanks for this answer, but the part I'm confused about is how these discrete groups are represented as manifolds. $\endgroup$
    – WillG
    Dec 17, 2019 at 4:32
  • $\begingroup$ @WillG : Smooth relative to what topology. Zero dimensional manifolds with finitely many connected components invite the use of the discrete topology and discrete metric. So they are smooth... $\endgroup$ Dec 17, 2019 at 4:38
  • $\begingroup$ Some definitions admit 0-dimentional manifolds, i.e. arbitrary collections of points. Well you need an open cover $U_i$ (just take singletons) of your discrete set with charts isomorphism $U_i \to \mathbb{R}^0 = \{0\}$. Depending on the definition you might want to ask if you admit non numerable discrete sets. Note that the maps defined are unique so they satisfy the manifold axiom, as the map $\{0\} \to \{0\}$ is smooth. $\endgroup$ Dec 17, 2019 at 4:39
  • $\begingroup$ The topology is the discrete one as it is the only one locally homeomorphic to $\mathbb{R}^0$. $\endgroup$ Dec 17, 2019 at 4:42
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    $\begingroup$ @WillG It's a finite collection of isolated points. All tangent spaces are trivial. They're rather boring manifolds. $\endgroup$ Dec 17, 2019 at 4:55

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