Definition of matrix Lie group

Question

In Lie Groups, Lie Algebras and Representations by Brian C. Hall, one finds the following on page 3:

Definition 1.4 A matrix Lie Group is any subgroup $G$ of $GL(n;\mathbb{C})$ with the following property: If $A_m$ is any sequence of matrices in $G$, and $A_m$ converges to some matrix $A$ then either $A\in G$, or $A$ is not invertible.

Why do we have "or $A$ is not invertible?". I was expecting: "If $A_m$ is any sequence of matrices in $G$, and $A_m$ converges to some matrix $A$ then $A\in G$.", without an "or" clause, to express the closure property of $G$. It seems that the "or" clause defeats the closure?

I guess my problem is the difference between "closed" and "complete".

In the definition of a "complete" metric space, we stipulate: "a metric space $M$ is called complete (or a Cauchy space) if every Cauchy sequence of points in $M$ has a limit that is also in $M$". But we also have "In a topological space, a closed set can be defined as a set which contains all its limit points".

I naively drew an analogy between the two concepts, but that "or" clause seems to say that some sequences $A_m$ can have limit points that are not in $G$? Can we contrast "closed" and "complete" so that this becomes more intuitive?

Answer 1

$G$ may be "closed under multiplication", but that doesn't make it a "closed set" topologically. Consider $GL(n; \Bbb C)$ itself, which is surely a Lie Group. Consider the case $n = 1$, and the matrices $M_k = [ \frac{1}{k}]$. The limit is the zero matrix, which is not in $GL(1; \Bbb C)$.

Answer 2

If $A$ is not invertible then it in fact cannot be a member of $G$, since that defies the condition that every group element must have an inverse.

This is similar to saying a closed subset of the rationals $X\subset\mathbb{Q}$ is a set such that for any sequence $x_n$ of numbers in $X$ which converges to some number $x$, either $x\in X$ or $x$ is irrational.

Answer 3

$\operatorname{GL}(n,\mathbb C)$ is the preimage $\det^{-1}\{\mathbb C\setminus\{0\}\}$ and thus is open in the set of all $n$ by $n$ matrices, $M_n(\mathbb C)$. Not only that, $\operatorname{GL}(n,\mathbb C)$ is dense open subset, meaning that its closure is $M_n(\mathbb C)$. Consequently, for any singular matrix $A$, you can find a sequence $(A_n)$ of invertible matrices converging to $A$.

So, how do we describe closed subsets of $\operatorname{GL}(n,\mathbb C)$? We can say that $C$ is closed in $\operatorname{GL}(n,\mathbb C)$ if for all sequences $(A_n)$ in $C$ that converge in $\operatorname{GL}(n,\mathbb C)$ the limit $A=\lim_n A_n$ is in $C$. Notice that this way we are ignoring that $\operatorname{GL}(n,\mathbb C)$ is naturally embedded in $M_n(\mathbb C)$.

However, in practice we don't really ignore the fact that topology on $\operatorname{GL}(n,\mathbb C)$ is the relative topology inherited from $M_n(\mathbb C)$. In this way $C$ is closed in $\operatorname{GL}(n,\mathbb C)$ if there exists closed set $C'$ in $M_n(\mathbb C)$ such that $C = C'\cap \operatorname{GL}(n,\mathbb C)$.

Can you see how these descriptions relate to the definition you were given?