# Do discrete embeddings of surface groups not necessarily carry an embedding of SL_2?

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?

• Homomorphisms of a surface group into a lie group $G$ form a nice space, the character variety (with $2g$ coordinates in $G$ and one equation). The ones whose image lands in a copy of $SL_2(\mathbb R)$ are a high codimension subset (their closure should just consists of themselves plus maps that factor through a degneration of $SL_2(\mathbb R)$). Your condition that the embedding be homeomorphic to its image is not quite an open condition but it seems more generic than not. Commented Sep 21, 2020 at 1:30
• There are vast number of counterexamples. For instance, see the following paper for surface subgroups of $SL(3,\mathbb{R})$ that are Zariski dense in the entire group: web.math.ucsb.edu/~long/pubpdf/CoCompact_IJM.pdf
– Tara
Commented Sep 21, 2020 at 2:21
• Up to conjugation, there only finitely many homomorphisms $SL(2,R)\to G$ (where $G$ is, say, connected and semisimple). At the same time, the dimension of the space of homomorphisms $\Gamma\to G$, up to conjugation, is $(2p-2)dim(G)$, where $p$ is the genus, provided $p\ge 2$. Commented Sep 21, 2020 at 2:27
• @Tara That’s exactly the kind of thing I was looking for. If you’d like to post that as an answer, then I’ll accept it. Commented Sep 21, 2020 at 3:55

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)

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 $$\mathrm{SL}_2(\mathbf{R})$$.

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 $$H^2$$).

• Sadly, this does not answer the posed question since it is about discrete and faithful representation (an embedding in question is required to be a homeomorphism to its image). Commented Sep 21, 2020 at 9:41
• @MoisheKohan indeed, right... the word "discrete" is indeed useful to emphasize. I'm leaving the (cw) answer anyway; hopefully the discrete example in the comment will be posted too. I tend not to read (or read too quickly) a definition of a term if I'm already familiar with the term, so can miss an unusual additional assumption.
– YCor
Commented Sep 21, 2020 at 9:47
• @MoisheKohan oh, it's straightforward to deduce discrete embeddings too: I added the missing argument.
– YCor
Commented Sep 21, 2020 at 9:56
• Yes, it is correct now. The stronger result however is that every semisimple Lie group without compact factors and of dimension $>3$ contains a discrete subgroup isomorphic to a surface group and not contained in any subgroup locally isomorphic to $SL(2,R)$. Commented Sep 21, 2020 at 10:01
• @MoisheKohan sure it's not the best result (probably it's even true with "Zariski dense"). I was rather looking at the easiest approach. Note that the very elementary argument I give at the end carries over all $G$ containing a subgroup locally isomorphic to $\mathrm{SL}_2\times\mathrm{SL}_2$ (this covers $SL_4$ and $Sp_4$ but not $SL_3$ or rank one groups). Also one of the most important and studied examples is that of quasi-Fuchsian groups.
– YCor
Commented Sep 21, 2020 at 10:04