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Recently I came across the following test question in one of the graduate algebra courses in our school and am trying to solve it:

Let $R: G\rightarrow SL_2(\mathbb{R})$ be a faithful representation of a finite group by real 2$\times$ 2 matricies with determinant 1. Prove that G is a cyclic group.

I am not sure where to begin in this problem. Any help is appreciated

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The proof consists of two steps:

  1. Verify the statement in the case when $R(G)$ is contained in $SO(2)$. I will leave it to you as an exercise.

  2. Reduce the general case to (1): I will explain how to do this. One can think of this reduction as a simple application of Cartan's fixed point theorem: The manifold of negative curvature in your question is the hyperbolic plane (the upper half-plane) on which $SL(2, {\mathbb R})$ acts via linear-fractional transformations; the compact Lie group is the image of your finite group $G$ under $R$. Then Cartan's theorem will imply that $R(G)$ is conjugate (in $SL(2, {\mathbb R})$) to a subgroup of $SO(2)$.

Let me give a direct proof of (2) without using Cartan's theorem: Pick an arbitrary positive-definite bilinear form $b$ on $R^2$ and consider the average $$ B=\frac{1}{|G|}\sum_{g\in G} R(g)^* (b) $$ where $R(g)$ acts on the bilinear form by changing coordinates. (The action on the Gram matrix $M$ of the form is given by the formula $R_g^T M R_g$.) Note that the sum of positive definite bilinear forms is still a positive-definite bilinear form. Furthermore, by construction, $B$ is invariant under the action of $R(G)$. Hence, $B$ is a positive-definite bilinear form invariant under $R(G)$. By changing coordinates again, we can transform $B$ to the standard bilinear form whose Gram matrix is the identity. This amounts to conjugating $R(G)$ to a subgroup of $O(2)$. Since $R(G)$ consisted of matrices with unit determinant, the conjugate group will have this property too. Hence, $R(G)$ is conjugate to a subgroup of $SO(2)$.

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Each matrix of the form $R(g)$ has finite order as $G$ is finite. This excludes the Jordan normal form $\pmatrix{\pm1&1\\0&\pm1}$ so all $R(g)$ must be diagonizable (over $\Bbb C$, the completion of $\Bbb R$). Examining its complex eigenvalues, it will turn out that each $R(g)$ must be actually a rotation.

Finally, let $\alpha_n\in [0,2\pi)$ be the angle of rotation $R(g_n)$ if $G=\{g_0,g_1,g_2,\dots\}$. Then show that all occuring angles are multiples of the minimal nonzero angle among them.

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  • $\begingroup$ I dont follow why it has to be a rotation matrix. could you please elaborate $\endgroup$ – James Bond Dec 2 '13 at 6:12
  • $\begingroup$ The question is actually more subtle: You need to show that if $h_1, h_2$ are two non-commuting elements conjugate to rotations in $SL(2,R)$, then they generate an infinite group. For instance, product of two such elements might still be conjugate to a rotation. The statement itself is true, but not elementary (it can be deduced for instance from Cartan's fixed point theorem). $\endgroup$ – Moishe Kohan Dec 2 '13 at 9:30

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