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Consider the symmetric space $X_\textsf{d}=\textsf{SL}(d,\Bbb R)/\textsf{SO}(d)$. For any $M\in X_d$, by transporting the Killing form on $T_I(X_d)$ (the space of symmetric matrices with trace zero) to $T_MX_d$, we have a Riemannian metric on $X_d$, which in turn gives a metric $d_{X_d}$ on $X_d$. Then $$d_{X_d}(I,M)=\frac12\sqrt{\sum_{i=1}^n\left(\log\sigma_i(M)\right)^2}\text{ for }M\in X_d,$$ where $\sigma_i(M)$ denote a singular value of $M$.

Note that we have an action of $\textsf{SL}(d,\Bbb R)$ on $X_d$ given by $A(M):=AMA^T$ for $A\in \textsf{SL}(d,\Bbb R)$ and $M\in X_d$.

Question: Is it true that $d_{X_d}(I, A(I))\leq d_{X_d}(M,A(M))$ for all $A\in \textsf{SL}(d,\Bbb R)$ and all $M\in X_d$?

Reference: Anosov representations: informal lecture notes by Richard D. Canary.

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    $\begingroup$ Quantifiers are missing in your question. Do you mean "for all $A$ and all $M$"? Then the answer is of course negative. $\endgroup$
    – Moishe Kohan
    Commented Dec 23, 2023 at 23:32
  • $\begingroup$ @MoisheKohan I have edited my question by adding the Quantifiers. Unfortunately, I couldn't find $A\in \textsf{SL}(d,\Bbb R)$ and $M\in X_d$ such that $d_{X_d}(I, A(I))> d_{X_d}(M,A(M))$. $\endgroup$
    – Random
    Commented Dec 26, 2023 at 15:49
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    $\begingroup$ Did you consider the case $d=2$? Then $X$ is the hyperbolic plane. To get an example, find a hyperbolic isometry which in the upper half plane model is given by $z\mapsto z+1$. $\endgroup$
    – Moishe Kohan
    Commented Dec 26, 2023 at 15:57

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