# Is every topological (resp. Lie-) group the isometrygroup of a metric space (resp. Riemannian manifold)?

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?

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

• By "completely answered in the topological category" means that a certain analogous question is answered. In the setting here, it shows that any Lie group (with an arbitrary number of components) is isometry group of a complete metric space (btw I don't really call this the "topological category"). But it does not say if it can be chosen as a Riemannian manifold.
– YCor
Mar 8, 2013 at 6:36
• By the topological category I meant topological groups not Lie groups. So as a topological group we just expect a metric space not a Riemannian manifold. Mar 8, 2013 at 11:42
• The OP asks 2 questions. One involves metric spaces and topological groups and the other Lie groups and manifolds. The first question is answered in the paper I linked. Mar 8, 2013 at 11:44
• Not all groups are G_\delta complete. The paper later shows that such groups are not Dieudonne complete and esoteric examples can be found by googling. Locally compact groups and Polish groups are G_'\delta complete if I understood correctly. Mar 10, 2013 at 13:49
• @archipelago: if you look at example 4.5 in the first paper cited in Benjamin's answer (page 14), you will see an explicit example. Particularly, let $H$ be a non-trivial complete group, and let $G$ be the product of $\aleph_1$ copies of $H$. The subgroup of $G$ consisting of tuples where at most countably many components are non-identity elements furnishes an explicit counterexamople. Mar 10, 2013 at 15:47

The answer is yes for finite groups -- you can even ensure your space is a hyperbolic surface, or hyperbolic 3-manifold. The result for 3-manifolds is Sadayoshi Kojima's. For hyperbolic surfaces I forget who its due to, but I think the reference is in Kojima's paper.

For compact Lie groups I think you can just make your space some variant of the Lie group itself (with a left-invariant metric). The idea would be to take the Lie group $G$ with its left-invariant metric. If that has a bigger isometry group than $G$ itself, take the product of $G$ with a ball (with some metric), and perturb the metric on $G \times B$, $G$-equivariantly. Some generic perturbation of the metric should kill all isometries other than the ones coming from the $G$-action.

Andre raises the question of whether or not this construction could be performed for $S^1 \equiv SO_2$. The issue being that with its left-invariant metric the isometry group is $O_2$. I don't see any reason why it shouldn't work. For example, take $S^1 \times D^2$. Put a metric on this space which is locally a product, but where the fibre is a disc whose isometry group is $\mathbb Z_3$, orientation-preserving isometries of a triangle. We make the holonomy of the bundle around the base circle the generator of this isometry group. So there can be no isometry of this bundle that reverses the direction of the base space, since that would mean the holonomy is equal to its inverse, which in $\mathbb Z_3$ can not happen. So the only symmetries of this bundle act as orientation-preserving isometries of the base, and orientation-preserving on the fibre, and this group is $SO_2$.

• Does this only apply to Lie groups with semisimple Lie algebra? <=> the Killing form is nondegenerate. Otherwise the manifold is not Riemannian, correct? Mar 7, 2013 at 19:44
• Any connected Lie group admits a left-invariant Riemannian metric (compact allows to find a bi-invariant one but this is probably useless here).
– YCor
Mar 7, 2013 at 20:46
• Ryan: can you make the last step of your construction explicit in the case $G=S^1$? Mar 7, 2013 at 22:30
• I think that's manageable. I haven't given it as much thought as I should have but I've edited in a sketch of how to address your concern. Mar 7, 2013 at 23:13
• Sorry, I fail to understand how your "thickened moebius band" construction cuts down the isometry group from $O(2)$ to $SO(2)$. Mar 7, 2013 at 23:48