# Opposite of the universal covering group

Let $$\mathfrak g$$ be a finite dimensional real Lie algebra and let $$\tilde G$$ be the unique simply connected Lie group with Lie algebra $$\mathfrak g$$.

I think that the set $$R$$ of connected Lie groups $$G$$ (modulo isomorphisms) whose lie algebras is also $$\mathfrak g$$ (or isomorphic to $$\mathfrak g$$) is partially ordered by $$r \preceq r'$$ if $$G'$$ is a covering group of $$G$$ for some representatives $$G$$ and $$G'$$ of $$r$$ and $$r'$$, respectively, i.e., if there exists a surjective Lie group homomorphism $$\pi : G' \to G$$ such that the induced map on the Lie algebra $$\mathfrak g$$ is the identity (an isomorphism).

From the uniqueness of the universal covering group it follows that $$R$$ has a unique maximal element $$r_\text{max}$$, namely the isomorphism class of $$\tilde G$$. I think that if $$\mathfrak g$$ has trivial center that there is also a minimal element, namely the (isomorphism class of the image of the) adjoint representation $$\mathrm{Ad}: \tilde G \to \mathrm{Gl}(\mathfrak g)$$, $$\mathrm{Ad}(g)(X) = g X g^{-1}$$.

My question is if there always is a minimal element, even if $$\mathfrak g$$ has a center, and if so whether it is unique (up to isomorphism of course).

I am not too familiar with the subject, I hope the question makes sense.

• I found an equivalent way to write $R$. Maybe this helps. If $G$ is a connected Lie group with Lie algebra $\mathfrak g$, then $K =\pi^{-1}(e_G)$ is a closed discrete central subgroup of $\tilde G$ and such a $K$ determines a Lie group $G$ with the same Lie algebra via $G = \tilde G / K$. Two closed discrete central subgroups $K,K'$ of $\tilde G$ determine the same element of $R$ if $T=\tilde G/K \cong \tilde G/K'=G'$. So $R$ is isomorphic to the set to discrete subgroups of the center $Z(\tilde G)$ modulo the equivalence relation $K\sim K' \iff \tilde G/K \cong \tilde G/K'$.
– Lau
Commented Jun 25, 2022 at 15:08
• Have you considered the basic case of an abelian $\mathfrak g$? Commented Jun 25, 2022 at 17:47
• @TorstenSchoeneberg Yes it holds for abelian Lie algebras. The $n$-dimensional abelian Lie groups are of the form $\mathbb R^k \times U(1)^{n-k}$, $0\leq k \leq n$, and the minimal one is $U(1)^n$ while the maximal one is $\mathbb R^n$.
– Lau
Commented Jun 26, 2022 at 6:44
• But it's no longer given by the image of the adjoint (which instead is trivial), right? Commented Jun 26, 2022 at 6:59
• @TorstenSchoeneberg Yes the adjoint representation of a commutative Lie group is trivial and its image is not in $R$.
– Lau
Commented Jun 26, 2022 at 7:43