# Lie groups admitting flat (bi)invariant metrics.

I would like to see an example of a non-abelian compact lie group admitting a bi/left/right-invariant flat metric.

Is there any non-abelian compact lie group admitting a flat metric that is bi or one-sided invariant?

The motivation is the following: as a manifold a torus does admit a flat metric, but it is also a abelian group. The first impression would be that the triviality of the lie algebra could imply the existence of a flat metric, or even be a necessary condition. Of course it is reasonable to ask for some relationship between the metric and the algebraic structure; thus some invariance would be expected (even though I would be happy to see a less restrictive assumption and a more powerful restriction dictated by the algebraic structure in the compact case).

Thus one could ask for an example of a non-abelian lie group admitting a flat metric.

A easy counter example (flat $\Rightarrow$ abelian) is the group of affine transformations of the line. The connected component of the identity is diffeomorphic to the real plane, so as a manifold (forgetting about the group structure) it does admit a flat metric. But if one considers a one-sided invariant metric one gets the Lobachevsky plane.

Since I could not think of a counter example of a non-abelian compact lie group admitting a flat metric (not necessary related to the group structure), I will also be happy to see a counter example of it. I will keep the invariance of the metric on the question though.

P.S.: Since for bi-invariant metrics $R(X,Y)Z=\frac{1}{4}[[X,Y],Z]$ (see e.g.: do Carmo, Riemannian Geometry), it is true in this case that abelian $\Rightarrow$ flat.

EDITED

Due to Igor's comment I will also include another question:

Is there any non-abelian lie group admitting a flat metric that is bi or one-sided invariant?

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Compact flat manfolds are aspherical, and the only aspherical compact Lie group is a torus. –  Igor Belegradek Jul 10 '11 at 18:07
Igor, is it obvious that there is no nonabelian structure on a torus? –  Marco Golla Jul 10 '11 at 18:43
Nice, so there is a powerful restriction dictated by the algebraic structure in the compact case. Where can I found a proof of it? It would be interesting to see which role the commutativity of the group plays in the proof. –  Romero Solha Jul 10 '11 at 18:46
You might also want to consult - Milnor "Curvatures of Left Invariant Metrics on Lie Groups", which deals with your problem in a broader context. In particular he answers your question in his article. –  Asaf Jul 10 '11 at 21:01

It is a standard fact that the only Lie groups that have contractible universal covers are the products of a torus and Euclidean space (where a factor could be zero dimensional). The proof uses classification of compact Lie groups, and some other Lie theory.

Below I outline a more elementary proof in the case you care about: if a Lie group admits a complete left-invariant flat metric, the group is isomorphic to the product of a torus and Euclidean space.

First by a standard procedure (see Bredon's book on "Compact transformation groups", section I.9) any action of a connected Lie group $G$ action on a manifold $M$ gives rise to an action of the universal cover of $G$ on the universal cover of $M$. Moreover, if the action is free, transitive, and isometric, the same is true for the action on the universal cover. Also the kernel of the surjection $\tilde G\to G$ is a subgroup of the deck-transformation group of the covering.

Since the metric is flat, the universal cover is isometric to $\mathbb R^n$. Thus we have an isometric free transitive action of the universal cover $\tilde G$ of $G$ on $\mathbb R^n$. The only isometries that act freely on $\mathbb R^n$ are translations, and since the action is transitive, we conclude that $\tilde G$ is the group of translations, i.e. $\mathbb R^n$ itself. Thus $\tilde G=\mathbb R^n$, and the kernel of the surjection $\tilde G\to G$ is a discrete subgroup $\Gamma$ that consists of translations. These translations stabilize a subspace $\mathbb R^k$ on which $\Gamma$ acts cocompactly. Thus we have the claimed splitting of $G$ as $T^k\times \mathbb R^{n-k}$.

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The isomorphism here is as manifolds, not as groups, unless I am very mistaken. There are many nonabelian Lie groups (the solvable connected simply-connected ones) that are isomorphic as manifolds to $\mathbb R^n$. Since the tangent bundle to any Lie group is trivializable by right translations, every Lie group admits a left-invariant metric (pick any metric on the underlying vector space of the Lie algebra). In particular, Alain below gives a noncommutative group with left-invariant flat metric, which is as a manifold just $\mathbb R^3$, but not as a group. –  Theo Johnson-Freyd Jul 13 '11 at 12:38

If we are only interested in invariant metrics, I think the answer to your questions readily derives from J. Milnor's paper Milnor, J., Adv. Math. 21 (1976), no. 3, 293-329.

Let G be a connected Lie group of dimension m, with a left invariant Riemannian metric g. We denote again by g the scalar product induced by g on the Lie algebra $\mathcal G$ of G. Then g is flat if and only if its Levi- Civita connection $\nabla$ gives rise to a homomorphism $\rho : x \mapsto \rho(x) := \nabla_x$ from $\mathcal G$ to the Lie algebra $\mathcal O(m)$ consisting of all skew-adjoint linear maps from $\mathcal G$ to itself. This allows Milnor (Theorem 1.5 of Milnor op. cit) to establish that $(G; g)$ is flat if and only if $\mathcal G$ splits as a g-orthogonal sum $\mathcal G = A_1\oplus A_2$ of a commutative ideal $A_1 := ker(\rho)$ and a commutative subalgebra $A_2$ acting on $A_1$ by skew-adjoint transformations obtained by restricting each $\rho(a)$ to $A_1$, for all $a$ in $A_2.$ So as a Lie algebra, this is simply the semi-direct product $\mathcal G = A_2\ltimes_\rho A_1,$ via $\rho.$

The "motion group" discribed above by Alain Valette lies in this picture, as its Lie algebra is the semi-direct product $A_2\ltimes_\rho A_1$, where
$A_2=\mathbb R$ acts on $A_1:= \mathbb R^2 = span (~e_1, ~e_2 ~)$, via $\rho(t) e_1 =t e_2$ and $\rho(t)e_2 =-te_1.$

So obviously, a Lie group with a left (resp. right) invariant flat Riemannian metric must be solvable. Of course, in the above description of Milnor's result, $\mathcal G$ is compact if and only if $\rho =0$ or one of the $A_i's$ is trivial and hence $\mathcal G$ must be abelian. This is an alternative to Igor Belegradek's above proof.

Recall that the Lie algebra of a compact Lie group is either Abelian or the direct sum $\mathcal S\oplus \mathcal R$ (as Lie algebras) of two ideals $\mathcal S$ and $\mathcal R$, where $\mathcal S$ is semisimple and compact and $\mathcal R$ is abelian (its center, actually).

Now, if we drop the "positive definite" property of the metric, allowing for some submanifolds (subgroups) to be "isotropic" and hence for some vectors to be of length zero, so that the metric is no longuer Riemannian, but rather "pseudo-Riemannian", the situation becomes much harder to handle. In this case, the question of existence is an (ongoing) open problem.

For Lie groups with flat Lorentzian metrics, see in F. J. Grunewald and G. A. Margulis,
J. Geom. Phys. 5 (1988), no. 4, 493-531 (1989); and for general pseudo-metrics see e.g. Aubert, A.; Medina, A., Tohoku Math. J. (2) 55 (2003), no. 4, 487–506 and S. Bromberg and A. Medina, arXiv:1103.1356.

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To complement Igor's answer. Consider the semi-direct product $G=\mathbb{C}\rtimes\mathbb{R}$ associated with the action of $\mathbb{R}$ on $\mathbb{C}$ by rotations; so the product in $G$ is given by $(z,s)(z's')=(z+e^{is}z',s+s')$. Note that $G$ is just the universal cover of the motion group of $\mathbb{R}^2$. Take the standard euclidean structure on $\mathbb{C}\times\mathbb{R}$: from the formula for the product, it is clearly left-invariant but not right invariant.

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For semisimple case the answer is no in a more general context. H. Matsushima and K. Okamoto proved a beautiful theorem in 1978: on semisimple Lie groups there are no left-invariant torsion-free flat connections.

I recall the proof is that you would get a non-trivial element on the $H^2({\frak{g}}^*)$ Lie algebra Cartan-cohomology. On a one-page paper in Hiroshima Mathematics Journal! (I just have the result and reference with me now).

Of course there are other non-abelian groups other than the semisimple. But then you may try to use a decomposition theorem. First one has to see the solvable case...

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