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There is a very interesting question How can a group of matrices form a manifold. From the answers it looks more like group of matrices form euclidean space than a general manifold. I understand that euclidean space is a manifold, but manifold is very general and has curvature. My question is what exactly makes a group of matrices a manifold but not simply a euclidean space.

  • I am not a mathematician so please correct me if there is anything wrong with the question or the way I posed it.
Shew
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    Under addition, they can be Euclidean. However, all the interesting math happens multiplicatively, where they usually aren't flat. – Randall Oct 29 '18 at 18:11

2 Answers2

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Take $SO(2,\mathbb{R})$, for instance. This is the group of the $2\times2$ orthogonal matrices whose determinant is $1$. But then$$SO(2,\mathbb{R})=\left\{\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}\,\middle|\,\theta\in\mathbb R\right\}.$$This can be seen as a circle in $\mathbb{R}^2$. Therefore, it is naturally a manifold, but in no way an Euclidean space.

  • circle is a 1-dimensional manifold, right ?. A $n\times n$ matrix group should form an $n$ dimensional manifold right ?. – Shew Nov 11 '18 at 15:30
  • Wrong, as my example shows. – José Carlos Santos Nov 11 '18 at 15:34
  • which part is wrong ?. $n \times n$ matrix group forms $$ dimensional manifold part ?. Is there any relationship between matrix dimension and manifold dimension ? – Shew Nov 11 '18 at 15:58
  • If a group of $n\times n$ matrices forms a manifold, then the dimension of that manifold can take any value from the set ${0,1,2,\ldots,n^2}$. The example that I gave is a case in which $n=2$ and the dimension is $1$. – José Carlos Santos Nov 11 '18 at 17:38
  • great. Thank you. If you got some time, can you please check my other question https://math.stackexchange.com/questions/2994146/group-of-rectangular-matrices – Shew Nov 11 '18 at 17:39
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In the question to which you refer the group of matrices is given a manifold structure by virtue of its embedding in a Euclidean space. So for example you can think of a torus as a two dimensional manifold (with interesting curvature) by using the differentiable structure it gets from the usual embedding as a curved surface in $3$-space. That doesn't make the torus a Euclidean space.

Ethan Bolker
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