7
$\begingroup$

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

$\endgroup$
3
  • 1
    $\begingroup$ It seems you would want a bi-invariant metric for this to be true... $\endgroup$ Feb 21, 2019 at 21:38
  • 1
    $\begingroup$ @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 $N=G$.) $\endgroup$ Feb 21, 2019 at 23:30
  • 1
    $\begingroup$ @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 $N$. $\endgroup$
    – YCor
    Feb 22, 2019 at 10:09

2 Answers 2

7
$\begingroup$

Unless I miscomputed, the left-invariant metric $Q(dg,dg)=\operatorname{Tr}\bigl(\overline{g^{-1}dg}\,g^{-1}dg\bigr)$ (bar $=$ transpose) on \begin{equation} 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\} \end{equation} 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 \begin{equation} \ \\ (\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} \end{equation} (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 $G=\mathrm{GL}(3,\mathbf R)$.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
2
  • 2
    $\begingroup$ In the last sentence of the answer, was "different than $\sigma$" intended to be "when $\sigma$ is not an isometry"? $\endgroup$ Feb 22, 2019 at 13:28
  • 2
    $\begingroup$ 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. $\endgroup$
    – YCor
    Feb 22, 2019 at 16:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.