A question about special linear group












2












$begingroup$


Is there any way to find all matrices $G in SL(n,mathbb Z)$ such that there exists a matrix $A in GL(n,mathbb R)$ satisfying
$$
AGA^{-1} in SO(n,mathbb R)?
$$






linear-algebra matrices orthogonal-matrices







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  • $begingroup$
    Yes, these are matrices with finite order. It's an exercise, not really research level.
    $endgroup$
    – YCor
    57 mins ago
















2












$begingroup$


Is there any way to find all matrices $G in SL(n,mathbb Z)$ such that there exists a matrix $A in GL(n,mathbb R)$ satisfying
$$
AGA^{-1} in SO(n,mathbb R)?
$$






linear-algebra matrices orthogonal-matrices







share|cite|improve this question









$endgroup$












  • $begingroup$
    Yes, these are matrices with finite order. It's an exercise, not really research level.
    $endgroup$
    – YCor
    57 mins ago














2












2








2





$begingroup$


Is there any way to find all matrices $G in SL(n,mathbb Z)$ such that there exists a matrix $A in GL(n,mathbb R)$ satisfying
$$
AGA^{-1} in SO(n,mathbb R)?
$$






linear-algebra matrices orthogonal-matrices







share|cite|improve this question









$endgroup$




Is there any way to find all matrices $G in SL(n,mathbb Z)$ such that there exists a matrix $A in GL(n,mathbb R)$ satisfying
$$
AGA^{-1} in SO(n,mathbb R)?
$$






linear-algebra matrices orthogonal-matrices




linear-algebra matrices orthogonal-matrices






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 1 hour ago









TotoroTotoro

1417




1417












  • $begingroup$
    Yes, these are matrices with finite order. It's an exercise, not really research level.
    $endgroup$
    – YCor
    57 mins ago


















  • $begingroup$
    Yes, these are matrices with finite order. It's an exercise, not really research level.
    $endgroup$
    – YCor
    57 mins ago
















$begingroup$
Yes, these are matrices with finite order. It's an exercise, not really research level.
$endgroup$
– YCor
57 mins ago




$begingroup$
Yes, these are matrices with finite order. It's an exercise, not really research level.
$endgroup$
– YCor
57 mins ago










1 Answer
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$begingroup$

These are exactly the finite-order elements:




  1. If $G in text{SL}(n,mathbb{Z})$ has finite order, then there exists an inner product preserved by $G$ (take an arbitrary inner product and add up its images under all powers of $G$). Changing bases to an orthonormal basis for this invariant inner product has the effect of conjugating $G$ into the orthogonal group.


  2. Conversely, if $G in text{SL}(n,mathbb{Z})$ is such that there exists some $A in text{GL}(n,mathbb{R})$ with $A G A^{-1} in text{SO}(n,mathbb{R})$, then since $A cdot text{SL}(n,mathbb{Z}) cdot A^{-1}$ is a discrete subgroup of $text{GL}(n,mathbb{R})$, its intersection with the compact group $text{SO}(n,mathbb{R})$ is a finite group, and thus $G$ has finite order.







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    $begingroup$

    These are exactly the finite-order elements:




    1. If $G in text{SL}(n,mathbb{Z})$ has finite order, then there exists an inner product preserved by $G$ (take an arbitrary inner product and add up its images under all powers of $G$). Changing bases to an orthonormal basis for this invariant inner product has the effect of conjugating $G$ into the orthogonal group.


    2. Conversely, if $G in text{SL}(n,mathbb{Z})$ is such that there exists some $A in text{GL}(n,mathbb{R})$ with $A G A^{-1} in text{SO}(n,mathbb{R})$, then since $A cdot text{SL}(n,mathbb{Z}) cdot A^{-1}$ is a discrete subgroup of $text{GL}(n,mathbb{R})$, its intersection with the compact group $text{SO}(n,mathbb{R})$ is a finite group, and thus $G$ has finite order.







    share|cite|improve this answer









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      3












      $begingroup$

      These are exactly the finite-order elements:




      1. If $G in text{SL}(n,mathbb{Z})$ has finite order, then there exists an inner product preserved by $G$ (take an arbitrary inner product and add up its images under all powers of $G$). Changing bases to an orthonormal basis for this invariant inner product has the effect of conjugating $G$ into the orthogonal group.


      2. Conversely, if $G in text{SL}(n,mathbb{Z})$ is such that there exists some $A in text{GL}(n,mathbb{R})$ with $A G A^{-1} in text{SO}(n,mathbb{R})$, then since $A cdot text{SL}(n,mathbb{Z}) cdot A^{-1}$ is a discrete subgroup of $text{GL}(n,mathbb{R})$, its intersection with the compact group $text{SO}(n,mathbb{R})$ is a finite group, and thus $G$ has finite order.







      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        These are exactly the finite-order elements:




        1. If $G in text{SL}(n,mathbb{Z})$ has finite order, then there exists an inner product preserved by $G$ (take an arbitrary inner product and add up its images under all powers of $G$). Changing bases to an orthonormal basis for this invariant inner product has the effect of conjugating $G$ into the orthogonal group.


        2. Conversely, if $G in text{SL}(n,mathbb{Z})$ is such that there exists some $A in text{GL}(n,mathbb{R})$ with $A G A^{-1} in text{SO}(n,mathbb{R})$, then since $A cdot text{SL}(n,mathbb{Z}) cdot A^{-1}$ is a discrete subgroup of $text{GL}(n,mathbb{R})$, its intersection with the compact group $text{SO}(n,mathbb{R})$ is a finite group, and thus $G$ has finite order.







        share|cite|improve this answer









        $endgroup$



        These are exactly the finite-order elements:




        1. If $G in text{SL}(n,mathbb{Z})$ has finite order, then there exists an inner product preserved by $G$ (take an arbitrary inner product and add up its images under all powers of $G$). Changing bases to an orthonormal basis for this invariant inner product has the effect of conjugating $G$ into the orthogonal group.


        2. Conversely, if $G in text{SL}(n,mathbb{Z})$ is such that there exists some $A in text{GL}(n,mathbb{R})$ with $A G A^{-1} in text{SO}(n,mathbb{R})$, then since $A cdot text{SL}(n,mathbb{Z}) cdot A^{-1}$ is a discrete subgroup of $text{GL}(n,mathbb{R})$, its intersection with the compact group $text{SO}(n,mathbb{R})$ is a finite group, and thus $G$ has finite order.








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        share|cite|improve this answer



        share|cite|improve this answer










        answered 1 hour ago









        Andy PutmanAndy Putman

        31.6k7134214




        31.6k7134214






























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