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I am trying to understand the following definition of matrix Lie group:

A matrix Lie group is a subgroup $G$ of a $GL(n;\mathbb{C})$ such that if $A_m$ is any sequence of matrices in $G$ and $A_m$ converges to some matrix $A$, then either $A$ is in $G$ or $A$ is not invertibile.

Consider that the set of matrices with complex entries $M_n(\mathbb{C})$ can be thought as $C^{n^2}$ that again can be thought as $\mathbb{R}^{2n^2}$ and hence has a natural structure that makes $M_n(\mathbb{C})$ a manifold and then a Lie group.
$GL(n;\mathbb{C})$ can be seen through the determinant map as $GL(n;\mathbb{C})=det^{-1}(\mathbb{R}/0)$ and hence is open in $M_n(\mathbb{C})$ because is the preimage of an open set through a continuos map. This shows that as $GL(n;\mathbb{C})$ is open in $M_n(\mathbb{C})$ it can inherit the structure of manifold and is a Lie group.
Then I guess that the definition of a matrix Lie group given before use the theorem that states that a closed subgroup (in the algebraic meaning) of a Lie group is a Lie subgroup. This should show that the definition it makes sense, if I am not wrong.

My question is why we define matrix Lie groups to be closed subgroups with respect to $GL(n;\mathbb{C})$ and not respect to $M_n(\mathbb{C})$ in the following way:

A matrix Lie group is a subgroup G of a $M_n(\mathbb{C})$ such that if $A_m$ is any sequnce of matrices in G and $A_m$ converges to some matrix A, then A is in G.

I think the last definition, excluding the case in which the series converge to an invertible matrix is more restrictive, so maybe it could exclude some interesting groups. If this is the reason could you provide a meaningful example? or there are just patological cases?. There are even other motivations (assumed that mine is correct which I am not sure)?
thanks a lot for the help

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    $\begingroup$ Would Mathematics be a better home for this question? $\endgroup$ – Qmechanic May 12 '20 at 20:29
  • $\begingroup$ I posted it here as the meaningful example I was looking for it was an example with a physical meaning, if you think I should move this to Mathematics there is no problem for me $\endgroup$ – Ratman May 12 '20 at 20:34
  • $\begingroup$ Never think in general terms without considering examples: if $G=GL(n,\mathbb{C})$ itself, then your definition says it is not a matrix Lie group. That would be a bummer. $\endgroup$ – Abdelmalek Abdesselam May 12 '20 at 20:38
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Note that $M_n(\mathbb C)$ (equipped with standard matrix multiplication) is not a group, because many of its elements do not have inverses. The largest possible subset of $M_n(\mathbb C)$ which can constitute a group under matrix multiplication is $GL(n,\mathbb C)$.

That being the case, your definition reduces to

A matrix Lie group is a subgroup $G$ of $GL(n,\mathbb C)$ such that if $A_m$ is any sequence of matrices in $G$ and $A_m$ converges to some matrix $A$, then $A$ is in $G$.

This definition is problematic because it excludes some non-compact matrix groups. If $G$ is non-compact, then could contain a sequence $A_m$ whose determinants are unbounded. However, because $G$ is a group, it would also contain the sequence $A^{-1}_m$, whose determinants go to zero - implying that $\lim_{m\rightarrow\infty} A^{-1}_m = B$ for some non-invertible matrix $B$. Since $B\notin GL(n,\mathbb C)$, it's certainly not in $G$, meaning that $G$ is not a Lie group by your definition.

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    $\begingroup$ I fell stupid, didn't think to check if $M_n(\mathbb{C})$ is a group. Thanks a lot even for the other part of the answer, for now I don't think it is problem as we are not considering non compact group (even if it not specified) but anyway it is useful to know $\endgroup$ – Ratman May 12 '20 at 21:10

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