I'm trying to show that the connected component has a finite number of distinct maximal tori. So the group generated by its generators must be dense. But I don't know if it is really true.
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1This proof strategy won't work: if a compact Lie group $G$ has non-abelian identity component, then it has uncountable many distinct maximal tori. This follows, e.g., from the Baire category theorem since each maximal torus is closed nowhere dense subset of the identity component of $G$. – Jason DeVito - on hiatus Jul 18 '18 at 19:04
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Do you have any hint for another strategy? – Andre Gomes Jul 18 '18 at 19:12
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2Here is the approach I am trying, but I don't know if it works: Lemma (which I don't know how to prove, but I think it true): If $H\subseteq G$ are connected Lie groups, with $H\neq G$, then there is an element $g\in G\setminus H$ for which the group generated by $g$ only intersects $H$ at ${e}$. If this lemma is true, I think I can prove your result by induction on $\dim G$. – Jason DeVito - on hiatus Jul 18 '18 at 19:15
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I think I've got it, writing an answer now.... – Jason DeVito - on hiatus Jul 18 '18 at 19:20
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We begin with a Lemma.
Lemma: Suppose $H\subseteq G$ are compact connnected Lie groups with $H\neq G$. Then there is a subgroup $S^1\subseteq G$ for which $S^1\cap H$ is finite.
Proof: Look on the Lie algebra level. We have $\mathfrak{h}\subseteq \mathfrak{g}$, and $\mathfrak{h}\neq \mathfrak{g}$ because $H$ and $G$ are connected and $H\neq G$. Then $\mathfrak{g}\setminus \mathfrak{h}$ is an open dense subset of $\mathfrak{g}$, so contains a vector $v$ for which $\exp(tv)$ closes up. Then $\{\exp(tv)\} = S^1$ is the desired $S^1$.
To see this, note that if $S^1\cap H$ is infinite, then it has an accumulation point (since $G$ is compact). Now, by using the group multiplication, we may assume this accumulation point is the identity. It follows that $\exp(t_n v) \in H$ for a decreasing sequence $t_n\rightarrow 0$. This, then, implies that $v\in\mathfrak{h}$, a contradiction. $\square$
Now, we prove the theorem. Suppose $G$ is any connected Lie group. Let $H = T\subseteq G$ be a maximal torus. Under the identification $T\cong \mathbb{R}^n/\mathbb{Z}^n$, if we pick an element $x=(x_1,...,x_n)\in T$ for which $\operatorname{span}_\mathbb{Q}\{1,x_1, x_2,...,x_n\}$ has dimension $n+1$, it follows that $x$ generates a dense subgroup of $T$.
If $T = G$, we are done. Otherwise, using the lemma, pick $S^1\subseteq G$ with $S^1\cap T$ finite. We pick $y\in S^1$ which generates a dense subgroup of $S^1$.
Let $\langle\langle x,y\rangle\rangle$ denote the closure of the subgroup generated by $x$ and $y$. Then clearly $\langle\langle x,y\rangle\rangle$ contains $H$ and $S^1$. On the Lie algebra level, the Lie algebra of $\langle \langle x,y\rangle\rangle$ is a subspace containing $\mathfrak{t}$ and $v$, so has dimension at least that of $\dim T + 1$.
Now, we induct. If $\langle \langle x,y\rangle\rangle \neq G$, we use the lemma to pick $z\in G\setminus \langle\langle x,y\rangle \rangle$. By the same argument as above, $\langle \langle x,y,z\rangle\rangle$ has larger dimension that $\langle \langle x,y,\rangle \rangle$. Continuing in this way, since the dimension of the generated subset increases at every stage, the process must stop after finitely many steps.
Jason DeVito - on hiatus
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Why is $\mathfrak{g}\setminus\mathfrak{h}$ open and dense in $\mathfrak{g}$? And what do you mean by "closes up"? – Andre Gomes Jul 23 '18 at 20:50
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1Up to isomorphism, $\mathfrak{h}\subseteq \mathfrak{g}$ is just $\mathbb{R}^k\subseteq \mathbb{R}^n$ for some $k < n$. View $\mathbb{R}^k$ as spanned by the first $k$ of the standard basis vectors of $\mathbb{R}^n$. Considering $f:=d(\cdot, \mathbb{R}^k):\mathbb{R}^n\rightarrow \mathbb{R}$, $\mathbb{R}^n\setminus \mathbb{R}^k$ is $f^{-1}( \mathbb{R}\setminus {0})$, so is open. For density, just note that every open set in $\mathbb{R^n}$ contains points whose $k+1$st coordinate is non-zero. "Closes up" means ${\exp(tv):t\in\mathbb{R}}$ is a circle. – Jason DeVito - on hiatus Jul 23 '18 at 21:01
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1The sources I've used for most of my knowledge are: 1. Lecture notes by my advisor, Wolfgang Ziller http://www.math.upenn.edu/~wziller/math650/LieGroupsReps.pdf. 2. Fulton and Harris - Representation theory 3. Brocker and tom Dieck - Representations of Compact Lie groups – Jason DeVito - on hiatus Jul 25 '18 at 19:57
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@JasonDeVito Can you please explain why the fact $\mathfrak{g}\setminus \mathfrak{h}$ is open and dense in $\mathfrak{g}$ implies it contains a vector $v$ for which $\exp(tv)$ closes up? – Asaf Shachar Dec 05 '18 at 14:28
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2@Asaf: Actually, I just need the fact that $\mathfrak{g}\setminus \mathfrak{h}$ is open; density is not required. Let $X = {v\in \mathfrak{g}: v\neq 0\text{ and } \exists t_0 > 0 \text{ with} \exp(t_0 v) = e}$, so $X$ is the set of vectors which exponentiate to circles. Then the claim is that $X$ is dense. The details are a bit annoying. In the Lie algebra of a maximal torus $T$, $X\cap \mathfrak{t} \cong \mathbb{R}^n$ consists of vectors of the form $(v_1,..., v_n)$ with the $v_i \in 2\pi \mathbb{Q}$. This is dense, because it's essentially a copy of $\mathbb{Q}^n \in $\mathbb{R}^n$. – Jason DeVito - on hiatus Dec 05 '18 at 16:31
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2Now, consider the map $f:G\times \mathfrak{g}\rightarrow \mathfrak{t}$ given by $(g,v)\mapsto Ad_g v$. This is a surjective map by the maximal torus theorem, so it maps dense sets to dense sets. Now one verifies that $f(G\times (X\cap \mathfrak{t}) = X$, so $X$ is dense. (There is a probably a much nicer proof - this is just what I thought of spur of the moment.) – Jason DeVito - on hiatus Dec 05 '18 at 16:33
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2Sorry about the texing in the first comment - I got a phone call and missed my 5 minute editing window. When I wrote $X\cap \mathfrak{t} \cong \mathbb{R}^n$, I meant $\mathfrak{t}\cong \mathbb{R}^n$, and we are considering $X\cap \mathfrak{t}$. – Jason DeVito - on hiatus Dec 05 '18 at 16:47