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.

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    $\begingroup$ 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 $K\sim K' \iff \tilde G/K \cong \tilde G/K'$. $\endgroup$
    – Lau
    Commented Jun 25, 2022 at 15:08
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    $\begingroup$ Have you considered the basic case of an abelian $\mathfrak g$? $\endgroup$ Commented Jun 25, 2022 at 17:47
  • $\begingroup$ @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 $\mathbb R^n$. $\endgroup$
    – Lau
    Commented Jun 26, 2022 at 6:44
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    $\begingroup$ But it's no longer given by the image of the adjoint (which instead is trivial), right? $\endgroup$ Commented Jun 26, 2022 at 6:59
  • $\begingroup$ @TorstenSchoeneberg Yes the adjoint representation of a commutative Lie group is trivial and its image is not in $R$. $\endgroup$
    – Lau
    Commented Jun 26, 2022 at 7:43


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