I just read the fact $M_0(\Gamma)=\mathbb{C}$ (constant functions) where $M_0(\Gamma)$ is the space of modular forms of weight $0$ on the congruence subgroup $\Gamma$.

But in this article at page four under the picture of the fundamental domain, the function $y(t)$ is said to be in $M_0(\Gamma_1(6))$ but $y$ is not constant. Have I misunderstood something?


This is a little confusing. Let $Y_1(6)$ be the open modular curve $\mathcal{H}/\Gamma_1(6)$. It has a canonical compactification $X_1(6)$ on which the vector space of functions is indeed one-dimensional. The field of modular functions is the function field of $X_1(6)$, which has transcendence degree 1 over the complex numbers (and in particular is certainly not 1-dimensional).

Only the constant functions extend holomorphically to all of $X_1(6)$; however, as is stated in the paper, the function $y$ they write down has a unique zero in the fundamental domain (hence a unique pole somewhere, presumably at a cusp although I didn't read the definition of $y$), so gives an isomorphism with $\mathbb{P}^1$.

Or, if you don't like the algebraic language here, the function $y$ has a pole at $1/2$ (as stated in that paper) which corresponds to its $q$-expansion there failing the cuspidal condition of being a modular form.

  • $\begingroup$ Thanks a lot! I have now nearly the same question: In the same article, same page, the function $t$ is written in terms of the function $\Delta$, which is modular of weight $12$. So $t$ is modular of weight $(12+12-12-12)\cdot 1/2=0$. So $t$ is constant but this seems not true. Where is my mistake? $\endgroup$ – user337073 May 20 '16 at 14:38
  • $\begingroup$ @R.A. same idea. $t$ has a pole at the cusp at 1/2 (as explained in that paper). Your mistake is to assume that automorphic forms of weight 0 are constant; you only know this if they extend holomorphically to the cusps. $\endgroup$ – hunter May 20 '16 at 15:37
  • $\begingroup$ So the point is that $t$ is only equal to the quotient of the $\Delta$'s on the upper half plane and not on the cusps? But why can we say that $t$ is equal to this fraction of the $\Delta$'s with knowing that it's modular, invariant under $\tau\mapsto -1/6\tau$ and the zeroes and poles coincide with those of $t$? And why do the poles coincide since $t$ has one at $1/2$ but the quotient hasn't? $\endgroup$ – user337073 May 20 '16 at 15:56
  • $\begingroup$ @R.A. I don't completely follow. $t$ is expressible as the quotient of $\Delta$ functions everywhere, even at the cusps. A quotient of holomorphic functions can certainly have poles! $\endgroup$ – hunter May 20 '16 at 21:38
  • $\begingroup$ the pole is not in the upper half plane proper, but at the cusp at 1/2 so there is no contradiction. The phrase "modular form" only implies holomorphicity in the upper half plane. $\endgroup$ – hunter May 21 '16 at 6:14

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