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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

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

2 Answers 2

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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 $G=\mathrm{GL}(3,\mathbf R)$.

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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.

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

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