Stuck on an example of why a Matrix lie algebra needs to be closed with respect to GL(n,C)

Question

Can someone please help me on this example.

$$ G = \left\{ \begin{pmatrix} { e }^{ it } & 0 \\ 0 & { e }^{ 2\pi it } \end{pmatrix},\quad t\in\mathbb{R} \right\} $$

G is a matrix group, but not a matrix Lie group, why not?

I cannot think of a sequence which is in G and converges in GL(n,C) but not in G.

Thank you in advance.

Harch

Do you believe that the set ${e^{2\pi i(2n+1)\pi}\mid n \in \mathbb{N}}$ contains points arbitrarily close to $-1$? (More generally that the set ${e^{\pi i an} \mid n \in \mathbb{N}$ is dense in $S^1$ when $a$ is irrational.) — fuglede, Jan 08 '14 at 14:36

score 2 · Answer 1 · answered Jan 08 '14 at 14:26

2

Hint $-I\notin G$ can you see that????????????????

answered Jan 08 '14 at 14:26

Myshkin

Having trouble with your keyboard? – Joel Jan 08 '14 at 14:28
4

@Joel probably having trouble with the character limit. – Igor Rivin Jan 08 '14 at 14:47
1

east or west igor is correct – Myshkin Jan 08 '14 at 15:27
just finished a 6 hour math marathon, brain is fried. please clarify. – Harch Jan 08 '14 at 18:49

1 Answers1