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Given a complex simply-laced Lie algebra $\mathfrak{g}$. Denote the unique simply connected Lie group whose Lie algebra is $\mathfrak{g}$ by $G$. Consider the real Lie algebra generated by the centre of the Lie group $G$, denoted $Lie(Z(G),\mathbb{R})$. What is the relation between $Lie(Z(G),\mathbb{R}$ and the coset space $\Lambda_\mathfrak{g}^*/\Lambda_\mathfrak{g}$?

Conventions used include: length-squared of simple roots is $2$ for simply-laced Lie algebra $\mathfrak{g}$ whose simple roots have equal length squared. The lattice $\Lambda_\mathfrak{g}^*$ is the dual lattice of $\Lambda_\mathfrak{g}$ which contains the root lattice in simply-laced case.

For completeness, given the fundamental root system of $\mathfrak{g}$, denoted $\Pi$, the root lattice of $\mathfrak{g}$, denoted $\Lambda_\mathfrak{g}$ is defined to be the set $\{\sum_an_a\alpha_a|n_a\in\mathbb{Z},\alpha_a\in\Pi\}$.

Rescy_
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    Cf. also https://math.stackexchange.com/q/2596716/96384. – Torsten Schoeneberg Jun 11 '23 at 17:22
  • Note that the "real Lie algebra generated by the centre of the Lie group" is a bit of an odd notion. That's just a vector space of the same dimension $d$ as the centre, it literally catches not more information than the number $d$. – Torsten Schoeneberg Jun 11 '23 at 17:23
  • Thx for pointing out. I was thinking there could be a group whose centre, whilst being an Abelian subgroup, may not necessarily be a vector space over any field. I know this is true for some Abelian group, but I am not certain about centre of simply-connected ones. – Rescy_ Jun 11 '23 at 17:30
  • Well the centres themselves are not vector spaces, of course, but their Lie algebras are, of course. – Torsten Schoeneberg Jun 11 '23 at 18:21
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    The centre of any semisimple Lie group is discrete as shown in the link. Thus it and its Lie algebra have dimension $0$. – Callum Jun 11 '23 at 20:55

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