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Because the genus 2 surface with hyperbolic metric has constant sectional curvature, its Riemann tensor is covariantly zero. i.e. $\bigtriangledown R \equiv 0$ Therefore, it is locally symmetric space. However, I can't understand why it is not a symmetric space. I guess this is because it has negative sectional curvature and it is compact, but I failed to prove it. Does every symmetric space whose sectional curvature is negative need to be non-compact?

user63310
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A compact surface of genus $g \geq 2$ (viewed as a complex curve uniformized by a constant-curvature metric) admits only finitely many automorphisms by Hurwitz's theorem, hence admits only finitely many isometries (since every isometry is an automorphism or anti-automorphism of the holomorphic structure).

Andrew D. Hwang
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    To be more specific, the local isometry that reverses geodesics at $p$ (which is a global isometry of the hyperbolic plane) does not give a globally-defined map of the surface. – Ted Shifrin Dec 20 '13 at 15:06