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Adam Harris
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Consider $\tau$ in the upper half plane such that $\tau$ is fixed by some $\alpha \in GL_2^{+}(\mathbb{Q})$ (i.e. positive determinant). Now suppose that there isand $\tau'$ in the upper half plane such that $j(g \tau) = j(g \tau')$ for all $g \in GL_2^{+}(\mathbb{Q})$, where $j$ is the modular $j$-function and $GL_2^{+}(\mathbb{Q})$ acts as Mobius transformations. Then does $\tau = \tau'$?

Consider $\tau$ in the upper half plane such that $\tau$ is fixed by some $\alpha \in GL_2^{+}(\mathbb{Q})$ (i.e. positive determinant). Now suppose that there is $\tau'$ in the upper half plane such that $j(g \tau) = j(g \tau')$ for all $g \in GL_2^{+}(\mathbb{Q})$, where $j$ is the modular $j$-function. Then does $\tau = \tau'$?

Consider $\tau$ and $\tau'$ in the upper half plane such that $j(g \tau) = j(g \tau')$ for all $g \in GL_2^{+}(\mathbb{Q})$, where $j$ is the modular $j$-function and $GL_2^{+}(\mathbb{Q})$ acts as Mobius transformations. Then does $\tau = \tau'$?

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Adam Harris
  • 1.9k
  • 1
  • 17
  • 19

A question regarding the j-function

Consider $\tau$ in the upper half plane such that $\tau$ is fixed by some $\alpha \in GL_2^{+}(\mathbb{Q})$ (i.e. positive determinant). Now suppose that there is $\tau'$ in the upper half plane such that $j(g \tau) = j(g \tau')$ for all $g \in GL_2^{+}(\mathbb{Q})$, where $j$ is the modular $j$-function. Then does $\tau = \tau'$?