# Is every Lie subgroup of a Lie group isometric to all its conjugates?

Let $$G$$ be a Lie group with a left invariant metric. Assume that $$N$$ is a Lie subgroup of $$G$$.

For a given $$g\in G$$, are $$N$$ and $$g^{-1} N g$$ necessarily isometric Riemannian manifold when they inherit the original metric of $$G$$?

I was inspired by this MSE question:

https://math.stackexchange.com/questions/3121058/almost-normal-subgroup

• It seems you would want a bi-invariant metric for this to be true... Feb 21, 2019 at 21:38
• @AndySanders Not necessarily, as they can be isometric via some map other than $\operatorname{Ad}_g:n\mapsto gng^{-1}$. (E.g. the identity, when $N=G$.) Feb 21, 2019 at 23:30
• @AndySanders Of course, bi-invariant is sufficient but not necessary. For instance it's obviously true when $N$ has dimension $1$, or when $N$ is normal. In particular, for $G$ non-abelian of dimension 2, it's true for all $N$.
– YCor
Feb 22, 2019 at 10:09

Unless I miscomputed, the left-invariant metric $$Q(dg,dg)=\operatorname{Tr}\bigl(\overline{g^{-1}dg}\,g^{-1}dg\bigr)$$ (bar $$=$$ transpose) on $$$$G=\left\{g=\begin{pmatrix}a&b&c\\0&1&e\\0&0&1\end{pmatrix}: \begin{matrix}a>0,\\b,c,e\in\mathbf R\end{matrix}\right\}, \qquad N=\left\{n=\begin{pmatrix}a&b&0\\0&1&0\\0&0&1\end{pmatrix}: \begin{matrix}a>0,\\b\in\mathbf R\end{matrix}\right\}$$$$ provides a counterexample. Indeed, taking $$g=\smash[b]{\begin{pmatrix}1&0&c\\0&1&e\\0&0&1\end{pmatrix}}$$ and following Milnor (1976, pp. 303, 312–314), one finds that the metric $$$$\ \\ (\operatorname{Ad}_g^*Q)(dn,dn)=(a^{-1}da\quad a^{-1}db) \begin{pmatrix}1+c^2&ce\\ce&1+e^2\end{pmatrix} \begin{pmatrix}a^{-1}da\\a^{-1}db\end{pmatrix}$$$$ (restricted to $$N$$) has scalar curvature $$\ -\dfrac{1+e^2}{1+c^2+e^2},\$$ which depends on $$g$$.

Added: For simpler, one could of course let $$e=0$$ throughout, or do this inside $$G=\mathrm{GL}(3,\mathbf R)$$.

As Francois Ziegler pointed out, it is not true in general. The map $$\sigma: N \to gNg^{-1}$$, $$\sigma(n)=gng^{-1}$$, is an isometry if and only if $$\langle Ad(g)\cdot X,Ad(g)\cdot Y\rangle=\langle X,Y\rangle \quad\text{for all X,Y \in\mathfrak n:=\textrm{Lie}(N).}$$ This follows since a left-invariant metric on a Lie group is determined by the inner product on the tangent space at the identity (identified with the Lie algebra), and also because the Lie algebra of $$gNg^{-1}$$ is $$Ad(g)\cdot \mathfrak n$$.

Note that a bi-invariant metric satisfies this condition, but the space of them may be usually larger. I am not sure whether there might be an example of an isometry between $$N$$ and $$gNg^{-1}$$ when $$\sigma$$ is not an isometry.

• In the last sentence of the answer, was "different than $\sigma$" intended to be "when $\sigma$ is not an isometry"? Feb 22, 2019 at 13:28
• Yes there can be an isometry when the conjugating map is not an isometry. One obvious example is when $N$ is the 1-dimensional normal subgroup in a 2-dimensional non-abelian connected Lie group.
– YCor
Feb 22, 2019 at 16:16