$\require{AMScd}$For $G$ a compact, connected Lie group, here's an alternate construction of the complexification $G\subset G_{\mathbb{C}}$ such that this inclusion induces $\pi_1(G)=\pi_1(G_{\mathbb{C}})$ (from the book of Brocker-Dieck). I have written this down partially for my own understanding; comments are more than welcome. I believe the argument in Bump to be incorrect; I'm leaving the question open for someone to convince me otherwise.
For $K$ either $\mathbb{R}$ or $\mathbb{C}$, let
$$\mathcal{F}(G,K) := \left\lbrace L(\pi(.)v)\in C(G,K) \mid G\xrightarrow {\pi} GL(N,K) \text{ is a representation }, v\in K^N, L \in (K^N)^{\vee} \right\rbrace.$$
These are the $\textbf{matrix coefficients}$ of $G$ and, by using tensors and direct sums of representations, one can see that this is a $K$-algebra under pointwise multiplication and addition of functions.
$\textbf{Claim 1: (The polynomial structure of the matrix coefficients)}$ If $r$ is a faithful, real representation (existence of $r$ is equivalent to Peter-Weyl), $\mathcal{F}(G,\mathbb{R})$ is generated as a $\mathbb{R}$-algebra over the matrix coefficients $r_{jk}$. Moreover, $\mathcal{F}(G,\mathbb{C})$ is generated as a $\mathbb{C}$ algebra over $r_{jk}$. Thus $\mathcal{F}(G,K)\simeq K[X_{jk}]/I$ for some ideal $I$.
$\textbf{Proof:}$ One can check that, since the real and imaginary parts of matrix coefficients of a complex representation are matrix coefficients of a real representation, we have $\mathcal{F}(G,\mathbb{R})\otimes \mathbb{C} \simeq \mathcal{F}(G,\mathbb{C})$. Hence the second statement will follow from the first.
Using faithfulness, we apply Stone-Weierstrass to see that $\mathbb{R}[r_{jk}]$ is dense in $C(G, \mathbb{R})$ and hence in $\mathcal{F}(G,\mathbb{R})$ with respect to the supremum norm. Being bounded by the supremum norm, density also holds in the $L^2$ norm. By Schur's lemma, we can decompose $F(G,\mathbb{R})$ into an $L^2$ orthogonal direct sum $\bigoplus \mathcal{F}(\rho)$ of matrix coefficients over all finite domensional irreducible real representations $\rho$ (If $f$ is a coefficient orthogonal to all $\mathcal{F}(\rho)$, noticing that $f$ is contained in a finite dimensional $G$ module gives a contradiction). Then note $\mathbb{R}[r_{jk}]=\bigoplus \left(\mathbb{R}[r_{jk}]\cap \mathcal{F}(\rho) \right)$. Using density and finiteness of each summand, we see that $\mathbb{R}[r_{jk}]$ is in fact all of $\mathcal{F}(G,\mathbb{R})$. Hence $\mathbb{C}[r_{jk}]=\mathcal{F}(G,\mathbb{C})$. $\blacksquare$
Note, as a non-example to the above, it is necessary to take a faithful $\textit{real}$ representation $r$. The result is clearly not true if you take a faithful complex representation. Given the standard embedding $S^1\to GL(1,\mathbb{C})$, one can see $\mathcal{F}(S^1,\mathbb{C})=\mathbb{C}[Z,Z^{-1}] \neq \mathbb{C}[Z].$
$\textbf{Definition 2: (Double dual)}$ Let $G_K$ be the set of all $K$-algebra morphisms $\mathcal{F}(G,K)\to K$. Give it the topology induced by the functions $s \mapsto s(f)$ where $f\in\mathcal{F}(G,K)$. For $s,t \in G_K$ define the product $s\cdot t$ by the commutative diagram
$$ \mathcal{F}(G,K) \rightarrow \mathcal{F}(G\times G,K) \xrightarrow{\simeq} \mathcal{F}(G,K)\otimes \mathcal{F}(G,K) \xrightarrow {s\otimes t} K\otimes K \simeq K .$$
Note, the first arrow is induced by multiplication in $G$ i.e. $f(x)\mapsto f(a.b)$. The second arrow is the inverse of the isomorphism $f(.)\otimes h(.)\mapsto f\circ\pi_1(.,.)\cdot h\circ\pi_2(.,.).$ In fact, one can check that, for $\rho$ a representation, the composition $\mathcal{F}(G,K)\to \mathcal{F}(G,K)\otimes \mathcal{F}(G,K)$ is given by $\rho_{jk} \mapsto \sum \rho_{jl} \otimes \rho_{lk}$. One can also check that $s\cdot t$ is indeed a $K$-algebra morphism. Let $\epsilon$ denote the element in $G_K$ given by evaluation at the identity. Define $s^{-1}$ by the composition
$$\mathcal{F}(G,K) \to \mathcal{F}(G,K) \xrightarrow{s} K$$
where the first map is induced by inversion on $G$.
$\textbf{Claim 3: (Topological group structure of $G_{\mathbb{R}}$)}$ Definition $2$ makes sense and $(G_k,\cdot, \epsilon)$ gives $G_K$ the structure of a topological group. $\blacksquare$
$\textbf{Definition 4:}$ For $r$ any $K$-representation to $GL(n,K)$, define $r_K$ by the rule $s\mapsto \left(s(r_{jk})\right)$ and define $i$ by sending $g\in G$ to evaluation at $g$. This gives a commutative diagram
$$
\begin{CD}
G @>{i}>> G_{K} \\
@V{r}VV @V{r_K}VV \\
GL(n,K) @>{id}>> GL(n,K)
\end{CD} \qquad \qquad \qquad (1)$$
$\textbf{Claim 5:}$ $i$ and $r_K$ are continuous group morphisms. $i$ is injective. When $r$ is a faithful real representation so is $r_{\mathbb{R}}$.
$\textbf{Proof:}$ Continuity is obvious. $i$ is injective since the matrix coefficients separate points (Peter-Weyl). If $r$ is faithful, real, then any $s\in G_{\mathbb{R}}$ is determined by its value of the $r_{jk}$ and injectivity follows. $\blacksquare$
$\textbf{Claim 6: (Lie group structure of $G_{\mathbb{R}}$)}$ The map $i:G\to G_{\mathbb{R}}$ is an isomorphism of compact Lie groups.
$\textbf{Proof:}$ For this we assume the representation $r$ in diagram $(1)$ is faithful into $O(n)$.
$r_K$ is injective since the $r_{jk}$ generate $\mathcal{F}(G,\mathbb{R})$. Observe that the topological group $G_{\mathbb{R}}$ is compact as follows. Map $G_{\mathbb{R}}\to \prod\limits_{f\in \mathcal{F}(G,\mathbb{R})} \mathbb{R}$ by the rule $s\mapsto \left(s(f)\right)$. Any one of these $f$ can be written as a polynomial $p(r_{jk})$. Using that $r$ is an orthogonal representation and the boundedness of the coefficients of $p$, we can assume the co-domain of the above map is $\prod\limits_{f}J_f$ where each $J_f$ is a compact interval. With the product topology, this map becomes a continuous homeomorphism (onto image). Moreover, the image is defined in terms of the algebra relations $s(f_1\cdot f_2)=s(f_1)s(f_2)$ etc. making it closed and hence compact by Tychonov.
Hence, through $r_{\mathbb{R}}$, $G_{\mathbb{R}}$ is embedded as a closed Lie subgroup of $O(n)$. This gives it a unique Lie group structure.
It remains to show the map $i$ from diagram $(1)$ is onto (it is injective since the matrix coefficients separate points). For this we consider the matrix coefficients $\mathcal{F}(G_{\mathbb{R}},\mathbb{R})$ and the $\mathbb{R}$-algebra morphism $\lambda:\mathcal{F}(G,\mathbb{R}) \to \mathcal{F}(G_{\mathbb{R}},\mathbb{R})$ defined by the rule $f\mapsto \left(s\mapsto s(f)\right)$. Dualizing $i:G\to G_{\mathbb{R}}$ we also get a $\mathbb{R}$-algebra morphism $i^*:\mathcal{F}(G_{\mathbb{R}},\mathbb{R}) \to \mathcal{F}(G,\mathbb{R})$. Clearly $i^*\circ \lambda = Id$. $\lambda$ is surjective since claim $5$ with claim $1$ imply that the $(r_{\mathbb{R}})_{jk}\left(=\lambda(r_{jk})\right)$ generate $\mathcal{F}(G_{\mathbb{R}},\mathbb{R})$. This shows that $i^*$ is a bijection, in particular injective. Finally injectivity of $i^*$ proves the surjectivity of $i$; the matrix coefficients are dense in continuous functions
\begin{CD} \mathcal{F}(G_{\mathbb{R}},\mathbb{R}) @>{i^*}>{\simeq}> \mathcal{F}(G,\mathbb{R}) \\
@VVV & @VVV \\
C(G_{\mathbb{R}},\mathbb{R}) @>{i^*}>> C(G,\mathbb{R})
\end{CD} .$\blacksquare$
$\textbf{The complexification of $G$.}$
The discussion of the polynomial structure of the matrix coefficient ring and the claim about $G_{\mathbb{R}}$ are crucial to what follows (cf. claim $9$). Assume $r$ is a real, faithful, orthogonal representation. Extend it to a complex representation by extension of scalars and define $r_{\mathbb{C}}$ as before to obtain the following commutative diagram.
$$
\begin{CD}
G @>{i}>> G_{\mathbb{C}} \\
@V{r}VV @V{r_{\mathbb{C}}}VV \\
GL(n,\mathbb{C}) @>{id}>> GL(n,\mathbb{C})
\end{CD}$$
$\textbf{Claim 7:}$ $r_{\mathbb{C}}$ is a homeomorphism onto its image.
$\textbf{Proof:}$ Injectivity follows from claim $1$ which says $\mathcal{F}(G,\mathbb{C})$ is generated by the $r_{jk}$. $r_{\mathbb{C}}$ is continuous by definition. Openness of the map follows again from claim $1$.
$\textbf{Claim 8:}$ The image of $r_{\mathbb{C}}$ in $GL(n,\mathbb{C})$ is the zero set of a collection of polynomials in $n^2$ variables.
$\textbf{Proof:}$ Dualizing the isomorphism $\mathbb{C}[X_{jk}]/I \to \mathcal{F}(G,\mathbb{C})$ from claim $1$, we get a map $\sigma: G_{\mathbb{C}}=Hom_{\mathbb{C}-algebra}\left(\mathcal{F}(G,\mathbb{C}),\mathbb{C}\right) \to V(I)\subset \mathbb{C}^{n\cdot n}$ which is given by the rule $s\mapsto \left(s(r_{jk})\right)$ In particular, $V(I)\subset GL(n,\mathbb{C})$ . The following commutative diagram that arises then proves the claim
$$
\begin{CD}
G_{\mathbb{C}} @>{\sigma}>{\simeq}> V(I) \\
@V{r_{\mathbb{C}}}VV @VV{\text{inclusion}}V \\
GL(n,\mathbb{C}) @>{id}>> GL(n,\mathbb{C})
\end{CD}$$. $\blacksquare$
Claim $7$ and $8$ together show that $G_{\mathbb{C}}$ has the structure of a closed complex analytic subgroup of $GL(n,\mathbb{C})$. It thus has a unique complex analytic group structure.
Let $U(n)$ denote the unitary group, $P(n)$ denote the set of positive definite hermitian matrices. It is well known that multiplication $U(n)\times P(n)\to GL(n,\mathbb{C})$ is a diffeomorphism.
$\textbf{Claim 9: (Computing the Lie algebra and fundamental group of $G_{\mathbb{C}}$)}$ Denote the image of $r_{\mathbb{C}}$ by $\widetilde{G}$. Then
$(1) \text{ }\widetilde{G}\cap U(n) = r(G).$
$(2) \text{ Under the multiplication map, }\widetilde{G}\simeq \left(\widetilde{G}\cap U(n)\right) \times \left(\widetilde{G}\cap P(n)\right).$
$(3)$ If we denote $\mathfrak{g}=Lie(r(G)) \subset \mathfrak{u}(n)$, then there is an isomorphism $\mathfrak{g}\to \widetilde{G}\cap P(n)$ given by the rule $X \mapsto \exp(iX)$. Moreover the Lie algebra of $\widetilde{G}= \mathfrak{g}\oplus i\mathfrak{g}.$
$\textbf{Proof:} (1)$ Let $r_{\mathbb{C}}(s)\in U(n)$. So we have $(s(r_{jk}))\cdot (s(r_{jk}))^* = I_n$
\begin{equation}
I_n = \left(s(r_{jk})\right)\cdot\left(\overline{s(r_{kj})}\right) = \left(s(r_{jk})\right)\left(\overline{s(r_{jk})}\right)^{-1}
\end{equation}
This shows that each $s(r_{jk})$ is real. Hence the restriction of $s$ to $\mathcal{F}(G,\mathbb{R})$ is real i.e. in $G_{\mathbb{R}}$. Claim $6$ then implies that $r_{\mathbb{C}}(s)=r(g)$ for some $g$.
$(2)$ Let $A \in \widetilde{G}$. Say $A = MH$ for some $M$ unitary and $P$ positive-definite, hermitian. We can find $N$ unitary such that $NHN^*=D$ for some diagonal matrix with positive diagonal elements. Further, put $D=\exp(X)$ for $X$ diagonal with real entries. So $$A^*A=H^*M^*MH=H^*H = N^*DNN^*DN=N^*D^2N= N^*(\exp 2X)N.$$
By hypothesis, we also know $A_{jk}=s(r_{jk})$ so that $A^*_{jk}=\overline{s(r_{kj})}$. Since $r_{jk}$ are real-valued functions, we can further write $A^*_{jk}=\overline{s(\overline{r_{jk}})}$. The upshot of this is that the function $\overline{s(\overline{\cdot})}$ is in $G_{\mathbb{C}}$ whence $A^*$ and $A^*A\in \widetilde{G}$.
Thus we have $N^*\exp(2kX)N\in \widetilde{G}$ for each $k\in \mathbb{Z}$. Let $\phi:\mathbb{C}^{n^2}\to \mathbb{C}^{n^2}$ be the polynomial map given by the rule $V\mapsto N^*VN$. It is an isomorphism and the pre-image of $V(I)$ is some $V(J)$. For any polynomial $q\in I$, we know $q(N^*\exp(2kX)N)=0$ for each $k\in \mathbb{Z}$. This in fact shows $q(N^*\exp(tX)N)=0$ for all real $t$. Whence $\exp(tX)\in V(J).$ for all real $t$.
In particular we get $\phi(\exp(X)) \in V(I)$ and so $H\in \widetilde{G}$. So also $M\in \widetilde{G}$ and $(2)$ is proved.
$(3)$ Let $X\in \mathfrak{g} \subset \mathfrak{u}(n).$ Then $\exp(tX)\in \widetilde{G}$ for all real $t$. This means that any polynnomial $q$ in $I$ has $q(\exp(tX)=0$ identically. Viewing the argument $t$ as a complex variable, we see that the analytic function $t\mapsto q(\exp(tX))$ must in fact vanish everywhere. Hence $iX$ is in the Lie algebra of $\widetilde{G}$ and is hermitian. The argument goes both ways and so we see $Lie\left(\widetilde{G}\cap P(n)\right)=i\mathfrak{g}.$ Exponentiation is a diffeomorphism from the vector space of hermitian matrices to positive definite hermitian matrices and so we are done.
- Answering your question in a comment: It is known that every compact homogenous manifold $M$ satisfies $\chi(M)\ge 0$, this was proven by Mostow, in
Mostow, G. D., A structure theorem for homogeneous spaces, Geom. Dedicata 114, 87-102 (2005). ZBL1086.57024.
- This theorem is a consequence of a deeper structural results for homogeneous manifolds that Mostow proves in his paper. In section 4, Mostow proves that every connected homogeneous manifold (compact or not) admits an iterated fibration with fibers that are homogeneous manifolds satisfying further properties. Specializing to 3-manifolds $M$, it follows that:
i. either $M$ itself fibers, or
ii. (The exceptional case.) Either $G$ is solvable or is semisimple and in this case $H$ has finitely many connected components. Unfortunately, this exceptional case is poorly explained in the statement of the theorem in the introduction: In this case Mostow should have allowed one more subgroup in his sequence. In Mostow's notation, $k=1$, causing appearance of a subgroup $F_{-1}$ which has to be equal to $\{1\}$, while
$F_0=H$. Then the fibration in the statement of Theorem C reads:
$$
H/F_{-1}=H \to G/F_{-1}=G\to G/H.
$$
Then $G_k=G_1=G$, $\Gamma_k=\Gamma_1=H$ and we are in Mostow's case 3, where $G=G_1$ is semisimple or solvable and, in the semisimple case, $\Gamma_k$ has finitely many components. Mostow explains this better when he repeats the formulation of Theorem C in the end of section 4 (where this theorem is actually proven).
Case i. If a compact connected 3-manifold fibers, then either the fiber is a surface and the base is a circle or vice-versa.
(a) If the fiber is a surface, it has to be of nonnegative Euler characteristic (see part 1). Hyperbolic 3-manifolds do not admit such fibrations:
Namely, asphericity of $M$ implies that the fiber $F$ cannot be $S^2, P^2$. The long exact sequences of homotopy groups of a fibration implies that we have a short exact sequence
$$
1\to \pi_1(F)\to \pi_1(M)\to \pi_1(S^1)= {\mathbb Z}\to 1.
$$
If $F$ is the torus or the Klein bottle then $\pi_1(M)$ would contain a nontrivial normal virtually abelian subgroup which is known to be impossible.
(b) On the other hand, a compact hyperbolic manifold also cannot fiber over a surface $B$ with circular fibers $F$. This again follows from asphericity of $M$ which results in a short exact sequence
$$
1\to \pi_1(F)= {\mathbb Z}\to \pi_1(M)\to \pi_1(B)\to 1.
$$
Then, again we would obtain a normal infinite cyclic subgroup in $\pi_1(M)$, which is impossible.
Thus, compact hyperbolic 3-manifolds cannot admit transitive actions of Lie groups.
With more work, one can relax the compactness assumption in (2) to finiteness of volume and prove a similar theorem in higher dimensions as well.
ii. Consider now the exceptional case. If $G$ is semisimple and $H$ has finitely many components, $\pi_1(M)=\pi_1(G/H)$ is finite. This is impossible for a hyperbolic manifold. Lastly, if $G$ is solvable, so is $H$, hence, $H/H^c$ is solvable too (I am using Mostow's notation where $H^c$ denotes the identity component with respect to the Lie group topology). But then $\pi_1(M)$ is solvable, which is impossible for a hyperbolic manifold of finite volume.
Best Answer
Yes, every connected Lie group $G$ has a dense torsion-free subgroup.
You can construct this inductively. The key is the following.
Lemma. Let $H \subset G$ be a countable torsion-free subgroup that is not dense ($\overline H \neq G$). Then there exists $g \in G - \overline H$ such that $\langle H, g \rangle$ is torsion-free. In fact, the set of such $g$ is dense in $G$.
To prove the claim using the lemma, start with $H_0 = \{1\}$ and keep adding elements that lie closer and closer to the identity to construct torsion-free subgroups $H_1, H_2, \ldots$. Then either the dimension of $\overline H$ will jump by at least 1 when going from $H_n$ to $H_{n+1}$, or this happens in countably many steps. Either way after reaching at most $\dim(G)$ countable limit ordinals you should have $\overline{H_n} = G$.
Proof of lemma. We use the Baire Category theorem. For that, we show that the following countable many closed sets have empty interior: The set $\overline H$, and for all integers $n,k \geq 1$, all $h_1, \ldots, h_k \in H$ and every word $w : G^{k+1} \to G$, the set $$S = \{g \in G : w(g, h_1,\ldots, h_k)^n = 1 \text{ and } w(g, h_1,\ldots, h_k)\neq 1\} \,.$$ By BCT there then exists an element $g \in G$ not in these sets and such $g$ works.
Clearly $\overline H$ has empty interior because it is a Lie group of lower dimension.
Note that for every $n$, the identity is an isolated point in the set of $n$-torsion elements. When $G$ is linear, this is because torsion elements are semisimple and have eigenvalues equal to $n$th roots of unity. When $G$ is not linear, take a discrete central subgroup $\Gamma \subset G$ such that $G/\Gamma$ is linear, and then it follows similarly. This implies that the set $S$ is indeed closed.
Now if $S$ had nonempty interior, then $w(G,h_1, \ldots, h_k)^n = 1$ by real analyticity of $w$. But $w(1, h_1, \ldots, h_k) = 1$ because $H$ has no torsion, and by connectedness of $G$ and the fact that $1$ is isolated in $n$-torsion, it follows that $w(G,h_1, \ldots, h_k) = 1$, a contradiction to the assumption that $S$ is nonempty. $\square$