Isometry group of a compact hyperbolic surface Consider a compact surface $M$ of genus $g \geq 2$ with a metric of constant negative curvature. My question is, is it known under what sorts of sufficient conditions such a metric will have non-trivial isometry group? 
 A: Let me add to this nice answer of Ian Agol, that the isometry group is always finite, and contains at most $84(g-1)$ elements, if we count orientation-preserving (conformal) isometries, where $g\geq 2$ is the genus (Hurwitz, Math Ann, 41 (1893)). For an exposition in English, see the book by Tsuji, Potential theory in modern function theory.
EDIT. Let me also mention the beautiful book, The Eightfold Way (S. Levy, editor) which contains several surveys on the subject, in particular Hurwitz grous and surfaces, by Murray Macbeath, where he describes history and motivation for this theorem. 
A: In genus 2, every surface has a symmetry, namely a hyperelliptic involution. In higher genus, generic surfaces will not have any symmetries. If a surface has a non-trivial symmetry group, then the quotient by the symmetry group will be an orbifold, and the moduli space of hyperbolic structures on this orbifold will have strictly smaller dimension. Hence moduli space will not be covered by these smaller dimensional Teichmüller spaces. 
One possible obstruction: suppose a surface admits a symmetry (not hyperelliptic in genus 2), then there will be simple closed geodesics which get moved by an isometry, and hence there are simple closed curves which have the same lengths. So if a surface has all simple closed curves of different lengths, then it cannot admit any non-trivial symmetries. One could probably also state this for curves of some bounded length depending on the genus and injectivity radius. 
To see that some curve must be sent to a distinct curve, suppose not. Then every non-separating curve intersects another curve in a single point. Then this point must be fixed by the isometry, and in fact the isometry must be an involution fixing this point. Arguing like this, one can conclude that the isometry is a hyperelliptic involution. However, in genus $>2$, there are always curves not fixed by any given hyperelliptic. 
A: I'll just point out an answer based on somewhat different criteria than explicitly knowing features of the hyperbolic metric:  As is well-known, every oriented, compact hyperbolic surface $C$ is canonically a Riemann surface of genus $g\ge2$.  
As Ian points out, if $C$ is hyperelliptic, it always has a symmetry, the hyperelliptic involution $\iota:C\to C$, and it can be written, almost canonically, as an algebraic plane curve
$$
y^2 = (x-\lambda_1)\cdots(x-\lambda_{2g+2}),
$$ 
where the $\lambda_i$ are distinct complex numbers.  Let $\Gamma\subset\mathrm{PSL}(2,\mathbb{C})$ be the (finite) group of linear fractional transformations that preserve the set $\Lambda =\{\lambda_1,\ldots,\lambda_{2g+2}\}$. Then the group of orientation preserving isometries of $C$ will be an extension of $\Gamma$ by a $\mathbb{Z}_2$ (because of the hyperelliptic involution).  If lets $\Gamma_+$ be the (possibly slightly larger) group of $\pm$-holomorphic linear fractional transformations of $\mathbb{P}^1$ that preserve $\Lambda$, then the full group of isometries of $C$ will be a $\mathbb{Z}_2$ extension of $\Gamma_+$.
If $C$ is not hyperelliptic, then its canonical mapping $\kappa:C\to\mathbb{CP}^{g-1}$ is an embedding, and the isometries of $C$ are exactly the $\pm$-holomorphic automorphisms of $\mathbb{CP}^{g-1}$ that preserve the image $\kappa(C)$.  For example, when $g=3$ and $C$ is not hyperelliptic, the image $\kappa(C)\subset\mathbb{CP}^2$ is a nonsingular plane quartic $Q(X,Y,Z)=0$, and every nonsingular plane quartic is a canonical curve of genus $3$.  The orientation-preserving isometries of $C$ are exactly the elements of $\mathrm{PSL}(3,\mathbb{C})$ that preserve the quartic $Q$.  Since one can write down explicit homogeneous nonsingular quartics in $3$ variables that have no nontrivial automorphisms, it follows that the generic curve of genus $3$ has no nontrivial automorphisms, and hence the conformal hyperbolic metric on $C$ has no nontrivial isometries.  
This gives one a way to write down 'explicit' examples, modulo, of course, the fact that there is no known way explicitly to 'write down' the conformal hyperbolic metric on a nonhyperelliptic curve of genus $3$.
