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Let ${\cal O}, {\cal O}'$ be two order in ${\mathrm M}_2({\Bbb R})$ that are sets of all $2 \times 2$ matrices over real number ${\Bbb R}$. Assume that we have the relation ${\cal O}' = a{\cal O}a^{-1}$ with $a \in {\mathrm GL}_2({\Bbb R})$. Then it is seen that both ${\cal O}[\tau,1]^t$ and ${\cal O}'[\tau,1]^t$ are lattices in ${\Bbb C}^2$ of ${\Bbb Z}$-rank $4$.

Question: How can one make a natural isomorphism ${\Bbb C}^2/{\cal O}[\tau,1]^t \cong {\Bbb C}^2/{\cal O}'[\tau,1]^t$ ?

This is concerned on the moduli theoretic viewpoint of Shimura curve $C$ associated quaternion algerba $D$ and its maximal order ${\cal O}$. The above question arose to make up the ambiguity of ${\cal O}$ up to inner automorphisms od $D$.

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Let $\tau^\prime=a^{-1}(\tau)$. Then the linear transformation $$ \binom{z_1}{z_2}\mapsto j(a^{-1},\tau)^{-1}a \binom{z_1}{z_2} $$ tranforms the lattice $\cal O\binom{\tau^\prime}{1}$ into $a\cal Oa^{-1}\binom{\tau}{1}$.


You shouldn't expect to realize the isomorphism with $\tau^\prime=\tau$. To identify $\cal O\otimes\Bbb R$ with $\Bbb C^2$ via action on $\binom\tau1$ amounts to endow $\cal O\otimes\Bbb R$ with a complex structure. In general conjugation on a real space doesn't preserve a complex structure.

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  • $\begingroup$ Thanks. Anyway, is it OK that I think ${\Bbb C}^2/{\cal O}(\tau,1)^t \cong {\Bbb C}^2/{\cal O}(\tau',1)^t$? $\endgroup$ – Pierre MATSUMI Feb 18 '14 at 16:58
  • $\begingroup$ I think in the answer, $j(a^{-1}, \tau)^{-1}$ should be replaced by $j(a^{-1}, \tau)$. I am not sure but I almost got hold of what the answerer is teaching me. Very many thanks. $\endgroup$ – Pierre MATSUMI Feb 18 '14 at 17:54
  • $\begingroup$ $\binom{\tau^\prime}1=\binom{a^{-1}(\tau)}1=j(a^{-1},\tau)^{-1}a\binom\tau1$ $\endgroup$ – Andrea Mori Feb 18 '14 at 18:02

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