The isometry group of a metric space is a topological group (with the compact open topology). The isometry group of a Riemann Manifold is a Liegroup. (Thm. of Steenrod-Myers)
So, is every topological group isomorphic (in the category of topological groups) to the isometry group of a metric space?
And what about the differentiable version: Is every Liegroup isomorphic (in the category of Liegroups) to a isometry group of a Riemann manifold?
Edit: Benjamin Steinberg gave a reference, which fully answers the question in the topological case.
Ryan Budney gave an idea how to realize at least every compact lie group as the isometry group of a Riemannian manifold. (This is proven in THE ISOMETRY GROUPS OF MANIFOLDS
AND THE AUTOMORPHISM GROUPS OF DOMAINS, by RITA SAERENS and WILLIAM R. ZAME)
So for me, still open is the question about the non-compact case: Is even every non-compact Lie group realizable as the isometrygroup of a riemannian manifold?
Best Answer
This seems to be completely answered in the topological category by Thm 1.4 of http://arxiv.org/pdf/1202.3368v3.pdf.
Edit
If X is a topological space which is not Dieudonne complete (meaning its topology cannot be given by a complete uniformity), then it seems theorem 7.24 of http://books.google.com/books?id=v3_PVdvJek4C&pg=PA35&lpg=PA35&dq=free+topological+group+dieudonne+complete&source=bl&ots=QIt0C2TCjN&sig=d4BJjO2h3zM4jDLoTzijrDVYt3w&hl=en&sa=X&ei=VIw8UeuBJpO30QHek4C4AQ&ved=0CC4Q6AEwAA and Thm 1.4 of the paper above shows that the free topological group on X is not the isometry group of a metric space. Googling shows completely regular spaces exist which are not Dieudonne.
I believe that any polish group or locally compact group is an isometry group of a metric space by the paper I linked.
Also the author of the first paper has shown that every Lie group is the isometry group of another Lie group with respect to some proper metric. http://arxiv.org/pdf/1201.5675v2.pdf