# How to show the space of inverse-invariant metrics on a Lie group is infinite dimensional?

Let $G$ be a Lie group. I am trying to convinve myself there are 'many' Riemannian metrics on $G$ for which the inverse automorphism is an isometry.

Denote the iverse by $i$. For any metric $g$ on $G$, $g + i^*g$ is $i$-invariant.

Intuitively, it is evident from this that the space of such metrics is "infinite dimensional" (since the space of all metrics is inf' dim').

However, a priori it's possible that many different metrics $g_k$ will give rise to identical inverse-invariant metrics (up two scalar multiplication) $g_k + i^*g_k$.

How can I be sure there is no substantial degeneracy here?

(I thought something like looking at perturbations of the metric to achieve different curvatures will do, but the Levi-Civita connection does not behave well under sums of metrics, there are nonlinear parts like inversing the $g_{ij}$ matrix...)

• Pick a large region $U$ of $G$ disjoint from $iU$ such that you can specify the metric arbitrarily on $U$. (I think it suffices for the closure of $U$ to be disjoint from the closure of $iU$.) – Qiaochu Yuan Jan 22 '16 at 16:21
• I think I managed to complete the details of your suggestion. I guess you meant for something like this? (Or is there an easier way to see this?) – Asaf Shachar Jan 23 '16 at 13:03


I am filling details in Qiaochu's suggestion.

First, we prove there exist a neighbourhood $U \subseteq G$ such that $U \cap i(U) =\emptyset$: (Is there an easier way?)

Denote by $\g$ the Lie algebra of $G$. ($\g=T_eG$).

$\exp$ is a local diffeomorphism in a neighbourhood of $0 \in \g$. Let $V \subseteq \g$ be an open neighbourhood where $\exp:V \to \exp(V)$ is a diffeomorphism.

Take some inner product $\langle,\rangle$ on $\mathfrak{g}=T_eG$. Assume $\ep$ is small enough such that $B_{2\ep}(0) \subseteq V$. Let $v \in \g, |v|=\ep$ be some arbitrary vector in the $\ep$-sphere of $\g$. Define $U=\exp(B_\frac{\ep}{2}(v))$. ($U$ is open since $B_\frac{\ep}{2}(v) \subseteq B_{2\ep}(0)$).

Let's prove $U \cap i(U)= \emptyset$. Assume $x \in U \cap i(U)$. Then there exist $u, \til u \in B_\frac{\ep}{2}(v)$ such that: $$x=\exp(u),x^{-1}=\exp(\til u)$$, but then $\exp(-u)=\exp(\til u) , \, u,\til u \in B_{2\ep}(0)$ so ($\exp$ is injective there) this forces $-u = \til u$.

Note that: $||u|-\ep|=||u|-|v|| \le |u-v|<\frac{\ep}{2} \Rightarrow |u| > \frac{\ep}{2}$.

$$\ep<2|u|=d(u,-u)=d(u,\til u)\le d(u,v)+d(v,\til u) < \frac{\ep}{2} + \frac{\ep}{2} =\ep$$
(*I think taking $B_\ep(v)$ should be enough (the $\ep/2$ is not necessary, but somehow it was needed...)
Now we can specify a metric how we want to on $U$. We can complete $\{U,i(U)\}$ to an open cover whose intersection with $U,i(U)$ is "small", so on some open subset of $U$ we can select the metric to be as we like. Then, using a partition of unity argument we can build inverse-invariant metrics (on whole $G$) which can be arbitrary on some open subset of $U$.