We can get arithmetic lattices isomorphic to free groups in $\mathrm{SL}\_2\mathbb{R}$, so in general we can’t expect homomorphisms of lattices into semisimple Lie groups to say much about $\mathrm{SL}\_2\mathbb{R}$, even if we require the image to be discrete. In particular, unlike in higher-rank cases, such homomorphisms won’t necessarily extend to homomorphisms of $\mathrm{SL}\_2\mathbb{R}$.


However, if $\Gamma<\mathrm{SL}\_2\mathbb{R}$ is cocompact and we have an discrete embedding (by which I mean a map which is both a homeomorphism and an isomorphism onto its image) of $\Gamma$ into some semisimple Lie group $G$, will we not necessarily get a copy of $\mathrm{SL}\_2\mathbb{R}$ in $G$ containing (the image of) $\Gamma$?


By the considerations above, the image of $\Gamma$ in such a copy of $\mathrm{SL}\_2\mathbb{R}$ might not look like it originally did (meaning the embedding of $\Gamma$ hasn’t necessarily extended to all of $\mathrm{SL}\_2\mathbb{R}$), but part of me imagines that the image of $\Gamma$ should still determine something. Is this naïve?



Do discrete embeddings of surface groups not necessarily carry an embedding of SL_2?
Homomorphisms from surface groups are very flexible, so there are indeed such examples.


For instance, use: Breuillard–Gelander–Souto–Storm (Dense embeddings of surface groups.
Geom. Topol. 10 (2006), 1373–1389) ([DOI link](http://dx.doi.org/10.2140/gt.2006.10.1373))


They proved, for a surface group $\Gamma$ ($\pi\_1$ of oriented closed surface of genus $\ge 2$) that, among others, $\Gamma$ embeds densely into arbitrary nontrivial connected semisimple Lie groups.


To deduce a discrete embedding, argue as follows: choose a dense embedding into $G$ (with $G$ pretty arbitrary, even $G=\mathrm{SL}\_2(\mathbf{R})$ is fine) and consider a standard discrete embedding into $\mathrm{PSL}\_2(\mathbf{R})$. Then the diagonal embedding into $G\times\mathrm{SL}\_2(\mathbf{R})$ has a Zariski-dense image, so its image is not contained in any copy of any subgroup locally isomorphic to $\mathrm{SL}\_2(\mathbf{R})$.




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Variant, **not** using the above reference [BGSS]:


Choose two Fuchsian embeddings $i\_1,i\_2:\Gamma\to H$ where $H=\mathrm{PSL}\_2(\mathbf{R})$, such that the images of $i\_1$ and $i\_2$ are not conjugate under $\mathrm{PGL}\_2(\mathbf{R})$. Then the diagonal embedding $i\_1\times i\_2:\Gamma\to H\times H$ works (its image is Zariski-dense in $H^2$).


