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Yesterday I asked about an example in Chevalley's book "Theory of Lie Groups I". Well, the second example on pages 38 and 39 made me work a lot to understand.

http://books.google.com.br/books?id=t-41JV63sRIC&dq=editions%3AOCLC490045337&source=gbs_book_other_versions

But there are still some facts I don't understand. I'll put them below.

1) First paragraph, page 39. He says that $s_{2}$$s_{1}$$r$ $\in$ $g_{1}$ implies that $r$ is in the group generated by $g_{1}$ and $g_{2}$. This must be easy to see, but I'm not convinced yet.

2) Thir paragraph. I'm not convinced that $V$ (with this condition) exists, neither why this fact implies $V$ to be mapped in a continuous univalent way by $\theta$.

3) Still third paragraph. Why A (the complement of $V_{1}$$\cup$$(-e_{0})$$V_{1}$ in $Sp(1)$ is compact?

4) Right after that, what does "run over all compact neighbourhood" mean?

Sorry for ask too many things. You can find the book by clicking on that link above. I would appreciate if you could help me :) Or maybe discuss the example two, if you want.

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up vote 1 down vote accepted

(1) Let $s_3=s_2s_1r$. Then $s_1,s_3\in g_1$ and $s_2\in g_2$, so $s_1^{-1}\in g_1$ and $s_2^{-1}\in g_2$. Finally, $r=s_1^{-1}s_2^{-1}s_3$ is then in the group generated by $g_1$ and $g_2$, since that group certainly contains all products of elements of $g_1$ and $g_2$.

(3) $Sp(1)$ is homeomorphic to $S^3$ (see the bottom of p. $36$), so it’s compact. $A$ is the union of two open sets, so it’s open. $Sp(1)\setminus A$ is therefore a closed subset of the compact set $Sp(1)$, so it must be compact.

(4) He’s looking at the family

$$\{\theta(U)\cap A:U\text{ is a compact neighborhood of the unit matrix in }SO(3)\}\;;$$

Letting $U$ run over all these compact neighborhoods is just letting it become each of them in turn, as a dummy variable.

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