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In Margulis' book, there are actually two definitions of arithmetic lattices:

  • If $\mathbf{G}$ is a connected semisimple algebraic $\mathbb{R}$-group, then a lattice $\Gamma \subset \mathbf{G}(\mathbb{R})$ is arithmetic if there exists a connected almost $\mathbb{Q}$-simple $\mathbb{Q}$-group $\mathbf{F}$ and an $\mathbb{R}$-epimorphism $\tau \colon \mathbf{F} \to \mathbf{G}$ such that $(\ker\tau)(\mathbb{R})$ is compact and $\tau(\mathbf{F}(\mathbb{Z}))$ is commensurable with $\Gamma$.
  • If $H$ is a connected semisimple real Lie group, then a lattice $\Gamma \subset H$ is arithmetic if $\text{Ad}\,\Gamma$ is arithmetic in $(\overline{\text{Ad}\,H})(\mathbb{R})$ in the above sense.

As far as I understand it, the second definition is necessary because $H$ itself may not be of the form $\mathbf{G}(\mathbb{R})$. But if this is the case, i.e. if $H =\mathbf{G}(\mathbb{R})$ for some connected semisimple linear $\mathbb{R}$-group $\mathbf{G}$, is then the second definition equivalent to the first one?

Assuming the answer is yes, it would be interesting to know for which Lie groups $H$ the second definition is actually neccessary. If $H$ is not linear, then it clearly is. But if $H$ has finite center, can we always find a group $\mathbf{G}$ with $H = \mathbf{G}(\mathbb{R})$?

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  • $\begingroup$ I'm not sure what you mean by "the second definition is necessary". It would be better rename the definitions to make things meaningful. Also, for semisimple connected Lie groups, linear implies finite center, but the converse is false. $\endgroup$
    – YCor
    Commented Feb 3, 2019 at 16:45

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