Return to Answer

I will start with few observations; along the way, I will correct your question. Let $g\in SO(n,1)$ be an orientation-preserving loxodromic element. Let $U_g\in SO(n)$ denote the rotational part of $g$. Then $U_g$ always has a nonzero fixed vector in ${\mathbb R}^n$. Furthermore, if $n$ is even then it follows that the dimension of the fixed-point set of $U_g$ is $\ge 2$. Define $\delta_n=1$ if $n$ is odd and $\delta_n=2$ if $n$ is even. Let's call $g$ "strictly loxodromic" if dimension of the fixed point set of $U_g$ is exactly $\delta_n$. Clearly, $SO(n,1)$ contains orientation-preserving strictly loxodromic elements.

Let $\rho=\rho_g: {\mathbb Z}\to SO(n,1)$ denote the representation sending the generator to loxodromic $g$. Then dimension of the fixed-point set of $U_g$ is nothing but $b^0_\rho= dim H^0({\mathbb Z}; R^{n,1}_\rho)$. Now, it is a general fact that $$ \{\rho: {\mathbb Z}\to SO(n,1): b^0_\rho\ge m\} $$ is an algebraic subvariety of $Hom({\mathbb Z}, SO(n,1))=SO(n,1)$. The same holds if you consider $$ dim H^i(\Gamma, V_\rho) $$ where $G$ is an algebraic group, $\Gamma$ is a finitely-generated group and $V$ is a finite-dimensional linear algebraic representation of $G$ (just $Hom(\Gamma, G)$ will be more complicated in this case). In our setting, this is especially easy to see since $$ b^0_{\rho_g}= dim Ker (g-I), $$ where I regard $g$ as a matrix in $SO(n,1)$.

In particular, $$ \{g\in SO(n,1): b^0_{\rho_g}\ge \delta_n+1\} $$ is a (proper) algebraic subvariety. In particular, if $\Lambda < SO(n,1)$ is a Zariski dense torsion-free subgroup, then it cannot consist entirely of elements with $b^0_{\rho_g}\ge \delta_n+1$. In other words, such $\Lambda$ always contains a strictly loxodromic element, thus, answering the corrected version of your question.

I will start with few observations; along the way, I will correct your question. Let $g\in SO(n,1)$ be an orientation-preserving loxodromic element. Let $U_g\in SO(n)$ denote the rotational part of $g$. Then $U_g$ always has a nonzero fixed vector in ${\mathbb R}^n$. Furthermore, if $n$ is even then it follows that the dimension of the fixed-point set of $U_g$ is $\ge 2$. Define $\delta_n=1$ if $n$ is odd and $\delta_n=2$ if $n$ is even. Let's call $g$ "strictly loxodromic" if dimension of the fixed point set of $U_g$ is exactly $\delta_n$. Clearly, $SO(n,1)$ contains orientation-preserving strictly loxodromic elements.

Let $\rho=\rho_g: {\mathbb Z}\to SO(n,1)$ denote the representation sending the generator to loxodromic $g$. Then dimension of the fixed-point set of $U_g$ is nothing but $b^0_\rho= dim H^0({\mathbb Z}; R^{n,1}_\rho)$. Now, it is a general fact that $$ \{\rho: {\mathbb Z}\to SO(n,1): b^0_\rho\ge m\} $$ is an algebraic subvariety of $Hom({\mathbb Z}, SO(n,1))=SO(n,1)$. The same holds if you consider $$ dim H^i(\Gamma, V_\rho) $$ where $G$ is an algebraic group, $\Gamma$ is a finitely-generated group and $V$ is a finite-dimensional linear algebraic representation of $G$ (just $Hom(\Gamma, G)$ will be more complicated in this case). In our setting, this is especially easy to see since $$ b^0_{\rho_g}= dim Ker (g-I), $$ where I regard $g$ as a matrix in $SO(n,1)$.

In particular, $$ \{g\in SO(n,1): b^0_{\rho_g}\ge \delta_n+1\} $$ is a (proper) algebraic subvariety. In particular, if $\Lambda < SO(n,1)$ is a Zariski dense torsion-free subgroup, then it cannot consist entirely of elements with $b^0_{\rho_g}\ge \delta_n+1$. In other words, such $\Lambda$ always contains a strictly loxodromic element, thus, answering the corrected version of your question.

Edit: As for your request for a geometric proof of exietence of loxodromic elements, see for instance Lemma 3.24 here. Lastly, the existence of axial isometries of the symmetric space $G/K$ in a Zariski dense subgroup of a (real) semisimple Lie group $G$ is the first step in the proof of the Tits' alternative. Any proof of TA you find (and there are many by now) will contain an explanation generalizing the "unenlightening matrix manipulation".

I will start with few observations; along the way, I will correct your question. Let $g\in SO(n,1)$ be an orientation-preserving loxodromic element. Let $U_g\in SO(n)$ denote the rotational part of $g$. Then $U_g$ always has a nonzero fixed vector in ${\mathbb R}^n$. Furthermore, if $n$ is even then it follows that the dimension of the fixed-point set of $U_g$ is $\ge 2$. Define $\delta_n=1$ if $n$ is odd and $\delta_n=2$ if $n$ is even. Let's call $g$ "strictly loxodromic" if dimension of the fixed point set of $U_g$ is exactly $\delta_n$. Clearly, $SO(n,1)$ contains orientation-preserving strictly loxodromic elements.

Let $\rho=\rho_g: {\mathbb Z}\to SO(n,1)$ denote the representation sending the generator to loxodromic $g$. Then dimension of the fixed-point set of $U_g$ is nothing but $b^0_\rho= dim H^0({\mathbb Z}; R^{n,1}_\rho)$. Now, it is a general fact that $$ \{\rho: {\mathbb Z}\to SO(n,1): b^0_\rho\ge m\} $$ is an algebraic subvariety of $Hom({\mathbb Z}, SO(n,1))=SO(n,1)$. The same holds if you consider $$ dim H^i(\Gamma, V_\rho) $$ where $G$ is an algebraic group, $\Gamma$ is a finitely-generated group and $V$ is a finite-dimensional linear algebraic representation of $G$ (just $Hom(\Gamma, G)$ will be more complicated in this case).

In particular, $$ \{g\in SO(n,1): b^0_{\rho_g}\ge \delta_n+1\} $$ is a (proper) algebraic subvariety. In particular, if $\Lambda < SO(n,1)$ is a Zariski dense torsion-free subgroup, then it cannot consist entirely of elements with $b^0_{\rho_g}\ge \delta_n+1$. In other words, such $\Lambda$ always contains a strictly loxodromic element, thus, answering the corrected version of your question.

I will start with few observations; along the way, I will correct your question. Let $g\in SO(n,1)$ be an orientation-preserving loxodromic element. Let $U_g\in SO(n)$ denote the rotational part of $g$. Then $U_g$ always has a nonzero fixed vector in ${\mathbb R}^n$. Furthermore, if $n$ is even then it follows that the dimension of the fixed-point set of $U_g$ is $\ge 2$. Define $\delta_n=1$ if $n$ is odd and $\delta_n=2$ if $n$ is even. Let's call $g$ "strictly loxodromic" if dimension of the fixed point set of $U_g$ is exactly $\delta_n$. Clearly, $SO(n,1)$ contains orientation-preserving strictly loxodromic elements.

Let $\rho=\rho_g: {\mathbb Z}\to SO(n,1)$ denote the representation sending the generator to loxodromic $g$. Then dimension of the fixed-point set of $U_g$ is nothing but $b^0_\rho= dim H^0({\mathbb Z}; R^{n,1}_\rho)$. Now, it is a general fact that $$ \{\rho: {\mathbb Z}\to SO(n,1): b^0_\rho\ge m\} $$ is an algebraic subvariety of $Hom({\mathbb Z}, SO(n,1))=SO(n,1)$. The same holds if you consider $$ dim H^i(\Gamma, V_\rho) $$ where $G$ is an algebraic group, $\Gamma$ is a finitely-generated group and $V$ is a finite-dimensional linear algebraic representation of $G$ (just $Hom(\Gamma, G)$ will be more complicated in this case). In our setting, this is especially easy to see since $$ b^0_{\rho_g}= dim Ker (g-I), $$ where I regard $g$ as a matrix in $SO(n,1)$.

In particular, $$ \{g\in SO(n,1): b^0_{\rho_g}\ge \delta_n+1\} $$ is a (proper) algebraic subvariety. In particular, if $\Lambda < SO(n,1)$ is a Zariski dense torsion-free subgroup, then it cannot consist entirely of elements with $b^0_{\rho_g}\ge \delta_n+1$. In other words, such $\Lambda$ always contains a strictly loxodromic element, thus, answering the corrected version of your question.

I will start with few observations; along the way, I will correct your question. Let $g\in SO(n,1)$ be an orientation-preserving loxodromic element. Let $U_g\in SO(n)$ denote the rotational part of $g$. Then $U_g$ always has a nonzero fixed vector in ${\mathbb R}^n$. Furthermore, if $n$ is even then it follows that the dimension of the fixed-point set of $U_g$ is $\ge 2$. Define $\delta_n=1$ if $n$ is odd and $\delta_n=2$ if $n$ is even. Let's call $g$ "strictly loxodromic" if dimension of the fixed point set of $U_g$ is exactly $\delta_n$. Clearly, $SO(n,1)$ contains orientation-preserving strictly loxodromic elements.

Let $\rho=\rho_g: {\mathbb Z}\to SO(n,1)$ denote the representation sending the generator to loxodromic $g$. Then dimension of the fixed-point set of $U_g$ is nothing but $b^0_\rho= dim H^0({\mathbb Z}; Ad(\rho))$. Now, it is a general fact that $$ \{\rho: {\mathbb Z}\to SO(n,1): b^0_\rho\ge m\} $$ is an algebraic subvariety of $Hom({\mathbb Z}, SO(n,1))=SO(n,1)$. The same holds if you consider $$ dim H^i(\Gamma, G) $$ where $G$ is an algebraic group and $\Gamma$ is a finitely-generated subgroup (just $Hom(\Gamma, G)$ will be more complicated in this case).

In particular, $$ \{g\in SO(n,1): b^0_{\rho_g}\ge \delta_n+1\} $$ is a (proper) algebraic subvariety. In particular, if $\Lambda < SO(n,1)$ is a Zariski dense torsion-free subgroup, then it cannot consist entirely of elements with $b^0_{\rho_g}\ge \delta_n+1$. In other words, such $\Lambda$ always contains a strictly loxodromic element, thus, answering the corrected version of your question.

I will start with few observations; along the way, I will correct your question. Let $g\in SO(n,1)$ be an orientation-preserving loxodromic element. Let $U_g\in SO(n)$ denote the rotational part of $g$. Then $U_g$ always has a nonzero fixed vector in ${\mathbb R}^n$. Furthermore, if $n$ is even then it follows that the dimension of the fixed-point set of $U_g$ is $\ge 2$. Define $\delta_n=1$ if $n$ is odd and $\delta_n=2$ if $n$ is even. Let's call $g$ "strictly loxodromic" if dimension of the fixed point set of $U_g$ is exactly $\delta_n$. Clearly, $SO(n,1)$ contains orientation-preserving strictly loxodromic elements.

Let $\rho=\rho_g: {\mathbb Z}\to SO(n,1)$ denote the representation sending the generator to loxodromic $g$. Then dimension of the fixed-point set of $U_g$ is nothing but $b^0_\rho= dim H^0({\mathbb Z}; R^{n,1}_\rho)$. Now, it is a general fact that $$ \{\rho: {\mathbb Z}\to SO(n,1): b^0_\rho\ge m\} $$ is an algebraic subvariety of $Hom({\mathbb Z}, SO(n,1))=SO(n,1)$. The same holds if you consider $$ dim H^i(\Gamma, V_\rho) $$ where $G$ is an algebraic group, $\Gamma$ is a finitely-generated group and $V$ is a finite-dimensional linear algebraic representation of $G$ (just $Hom(\Gamma, G)$ will be more complicated in this case).

In particular, $$ \{g\in SO(n,1): b^0_{\rho_g}\ge \delta_n+1\} $$ is a (proper) algebraic subvariety. In particular, if $\Lambda < SO(n,1)$ is a Zariski dense torsion-free subgroup, then it cannot consist entirely of elements with $b^0_{\rho_g}\ge \delta_n+1$. In other words, such $\Lambda$ always contains a strictly loxodromic element, thus, answering the corrected version of your question.