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Let S be a surface whose fundamental group is NOT finitely generated. Does there exist a complete hyperbolic metric on S for which the area is finite? I suspect the answer in general is NO, but I do not see a simple argument for that. Any help is welcome.

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    $\begingroup$ Yes, it's correct (that the answer is no). Every non-simply-connected complete hyperbolic surface has a funnel/pants decomposition; funnels have infinite area; and pants can have cusps. If there are finitely many pants, clearly the $\pi_1$ is finitely generated. Otherwise, since all non-funnels pants have the same nonzero area, the total area is infinite. $\endgroup$
    – YCor
    Commented Apr 13, 2018 at 18:49

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This is a classical theorem: hyperbolic Riemann surfaces of finite hyperbolic area are compact surfaces with finitely many punctures. Tsuji (Theorem XI.12) credits this to Siegel (1945). The proof is a simple computation of the area of the fundamental polygon of the uniformizing Fuchsian group.

Tsuji, Potential theory in modern function theory, Maruzen Ltd, Tokyo 1959. (There is an AMS reprint).

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