# explicit linear representations of fundamental groups of surfaces

I am looking for an explicit representation of the fundamental group of a closed orientable surface of genus >1. I guess they should be abundant in degree 2. Did anyone see the explicit matrix construction of such a representation? Are there any integral ones? Maybe in higher degrees?

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Are you looking for faithful representations? If not, the group presentation of the fundamental group gives you easy relations to find among matrices. – John Wiltshire-Gordon Jan 29 '11 at 3:42
What does "degree 2" means in this context? Does it mean $2\times2$ matrices? – Igor Rivin Jan 29 '11 at 3:44
@John: Yes, I meant faithful representations. @Igor: Yes, degree 2 means $2 \times 2$ matrices – mathreader Jan 29 '11 at 4:22
I seem to recall that some interesting representations were written down in papers of Narasimhan and Seshadri. My recollection is that they gave some explicit irreducible representations of arbitrary degree. I don't know if they were integral representations though; I suspect not. – Dan Ramras Jan 29 '11 at 6:54

Yes, this has been done.

MR1292919 (96f:30045) Maskit, Bernard(1-SUNYS) Explicit matrices for Fuchsian groups. (English summary) The mathematical legacy of Wilhelm Magnus: groups, geometry and special functions (Brooklyn, NY, 1992), 451–466, Contemp. Math., 169, Amer. Math. Soc., Providence, RI, 1994.

You can find the paper with Google Books, if you don't have easy access to the library.

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Thank you for the link, Igor! As far as I see, the author considers projective representations. I hope they lift to the linear ones. – mathreader Jan 30 '11 at 1:13
The question of when things lift is considered at great length by Bill Goldman in his paper Goldman, William M.(1-MD) Topological components of spaces of representations. Invent. Math. 93 (1988), no. 3, 557–607. 57M05 (22E40 32G15) The basic principle is: keep the traces negative... – Igor Rivin Jan 30 '11 at 2:41
Discrete subgroups of PSL_2 lift to SL_2 provided there's no 2-torsion, see "Lifting representations to covering groups" by Marc Culler. (mathlab.snu.ac.kr/~top/articles/…) – Richard Kent Feb 1 '11 at 21:33
@Richard: thanks for the Culler reference, I did not know this. – Igor Rivin Feb 1 '11 at 21:37
@Igor Sure thing. It comes in handy. Of course, everyone should look at Goldman's paper regardless. :) – Richard Kent Feb 1 '11 at 21:41

Here are the examples from Narasimhan and Seshadri that I mentioned in my comment above. (I could have the reference wrong; I'm actually taking this out of notes from a talk I gave many years ago.)
These examples are really not that exciting, but at least they give irreducible unitary representations of each degree. Writing the generators of $\pi_1 M^g$ as $a_i$ and $b_i$ ($i=1, \ldots, g$) we define a representation $\rho$ by sending $a_1$ to the diagonal matrix $$A_1 = \left[\begin{array}{rrr} z_1 \\ & \ddots\\ && z_n \end{array}\right],$$ where the $z_i$ are distinct, and sending $a_2$ to the permutation matrix $$A_2 = \left[\begin{array}{rrrrrr} 0 & 0 & \cdots & 0 &1 \\ 1 & 0 & \cdots && 0 \\ 0 & \ddots & 0 & \cdots & 0\\ 0 & \cdots & 1& 0 & 0\\ 0 & \cdots & 0 & 1 & 0\end{array}\right].$$ All the other generators are sent to the identity matrix. Since the $b_i$ all go to the identity, all the commutators $[\rho(a_i), \rho(b_i)]$ are trivial and we have a representation. It's irreducible because the invariant subspaces of $A_1$ are just sums of eigenspaces, and these eigenspaces are permuted transitively by $A_2$.

One can vary this example in a number of ways to get other interesting examples.

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Oh, and "not that exciting" includes "not faithful"... – Dan Ramras Feb 1 '11 at 20:58
@Dan: these permutation representations probably predate Narasimhan and Seshadri's birth by several decades (since I am reasonably sure that Hurrwitz knew this, but have no evidence). In this connection, you can take a look at M. Droste and I. Rivin On extension of coverings – Igor Rivin Feb 1 '11 at 21:40
Yeah, these are pretty simple representations, so I'm sure they've been known a long time. I was just trying to remember where I had found them. – Dan Ramras Feb 2 '11 at 5:15
More generally, one can send the $a_i$ to whatever you like and the $b_i$ to the identity (or vice versa). The point is simply that $\pi_1 M_g$ surjects a free group of rank $g$. – HJRW Feb 2 '11 at 5:46
Yes, good point! – Dan Ramras Feb 2 '11 at 5:48