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In Maclachlan-Reid we can read

Let $G$ be a connected semisimple Lie group with trivial centre and no compact factor. Let $\Gamma\subset G$ be a discrete subgroup of finite covolume. Then $\Gamma$ is arithmetic if there exists a semi-simple algebraic group $H$ over $\mathbb{Q}$ and a surjective homomorphism $\phi:H({\mathbb{R}})_0\to G$ with compact kernel such that $\phi(H(\mathbb{Z})\cap H(\mathbb{R})_0)$ and $\Gamma$ are commensurable.

  1. Why would one restrict $\phi$ to the connected component of $H(\mathbb{R})$ containing the identity? How does replacing $H(\mathbb{R})_0$ by $H(\mathbb{R})$ changes the above definition?

  2. Why would one ask for $\Gamma$ to be a lattice? How does changing the sentence "Let $\Gamma\subset G$ be a discrete subgroup of finite covolume" by "Let $\Gamma\subset G$ be a subgroup" changes the above definition?

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    $\begingroup$ In general even if $H$ is connected as algebraic group, then $H(\mathbf{R})$ is not connected as Lie group — simplest example $H=\mathrm{PGL}_2$. This is why one takes the identity component. (One could avoid this anyway by taking $H$ to be the algebraic simply connected covering since then $H(\mathbf{R})$ is a connected Lie group.) $\endgroup$ Commented Aug 19, 2022 at 11:32
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    $\begingroup$ Thanks for your answer. I edited question 2) so that it makes sense. $\endgroup$ Commented Aug 19, 2022 at 12:46
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    $\begingroup$ @YCor, re, I took the question to mean not "why isn't $H(\mathbb R)$ already connected?" but rather "why is it necessary to have a connected group at all?" That is, if I asked not that $\Gamma$ and $\phi(H(\mathbb Z) \cap H(\mathbb R)_0)$, but just $\Gamma$ and $\phi(H(\mathbb Z) \cap H(\mathbb R))$, be commensurable, would I get the same definition? Is the problem that some relevant $\phi$ might not be extendable to $H(\mathbb R)$? $\endgroup$ Commented Aug 19, 2022 at 13:12
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    $\begingroup$ @LSpice the isomorphism $\mathrm{PGL}_2(\mathbf{R})_0\to \mathrm{PSL}_2(\mathbf{R})$ doesn't extend to a homomorphism $\mathrm{PGL}_2(\mathbf{R})\to \mathrm{PSL}_2(\mathbf{R})$ $\endgroup$ Commented Aug 19, 2022 at 13:52
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    $\begingroup$ @Jacques for Q2: the subgroup $\mathrm{SL}_2(\mathbf{Z}[\sqrt{2}])$ is not discrete and hence not an arithmetic lattice in $\mathrm{SL}_2(\mathbf{R})$ (although it embeds as an arithmetic lattice in $\mathrm{SL}_2(\mathbf{R})^2$ through the injection ($i,\bar{i}$) where $i$ is the inclusion and $\bar{i}$ is the inclusion post-composed with Galois automorphism. $\endgroup$ Commented Aug 19, 2022 at 13:58

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