# What is the rank of the period lattice of modular forms?

Let $$f$$ be a weight $$2$$ cusp form for the group $$\Gamma_0(N)$$. I was experimenting with integrals of the form $$\int_r^s f(z) \, dz$$ where $$r, s \in \mathbf{P}^1(\mathbf{Q})$$ and the integral above is over the geodesic in the upper-half plane connecting $$r$$ and $$s$$. When I plotted these period integrals, I noticed that the resulting values formed a rank $$2$$ lattice in $$\mathbf{C}$$. I have two questions about this:

1. Why would the above period lattice have rank $$2$$? Since we are integrating $$f(z) \, dz$$ over paths in the relative homology group $$H^1(X_0(N), \{\text{cusps}\}, \mathbf{Z})$$, I would expect the rank of the period lattice to grow with the rank of this homology group. Why is the rank of the period lattice always $$2$$, even if the rank of the homology group is bigger?

2. Inside the period lattice obtained from integrating over $$H^1(X_0(N), \{\text{cusps}\}, \mathbf{Z})$$, we can consider the sublattice from integrating over just the ordinary homology group $$H^1(X_0(N), \mathbf{Z})$$, omitting paths that are not closed loops in $$X_0(N)$$. (Equivalently, the sublattice is formed by only plotting $$\int_r^s f(z) \, dz$$ when $$r$$ and $$s$$ are $$\Gamma_0(N)$$-equivalent.) What is the index of this sublattice inside the full period lattice? Can one answer this in general?

For Q1: I'm assuming your $$f$$ is a normalized Hecke newform of level $$N$$ with coefficients in $$\mathbf{Z}$$. Then the choice of $$f$$ determines a splitting of the homology as

$$H_1(X_0(N), \{cusps\}, \mathbf{Q}) = (f\text{-generalised eigenspace}) \oplus \text{(other stuff)}.$$

By the Eichler--Shimura isomorphism, we see that the $$f$$-generalized eigenspace is semisimple (i.e. it's a genuine eigenspace) and has dimension 2. When we integrate against $$f$$, it kills the "other stuff"; so the map $$H_1(X_0(N), \{cusps\}, \mathbf{Z}) \to \mathbf{C}$$ given by integration against $$f$$ factors through the image of $$H_1(X_0(N), \{cusps\}, \mathbf{Z})$$ in the $$f$$-eigenspace of $$H_1(X_0(N), \{cusps\}, \mathbf{Q})$$, and a little elementary linear algebra shows that this eigenspace quotient has to be free of rank 2.

So the image in $$\mathbf{C}$$ has rank at most 2. Since this lattice has non-trivial intersection with both $$\mathbf{R}$$ and $$i \mathbf{R}$$, its rank must be exactly 2.

For Q2, the comparison of lattices arising from "usual" and "relative" homology: one can show that the cokernel of the natural map from $$H_1(X_0(N)) \to H_1(X_0(N), \{cusps\})$$ is a free $$\mathbf{Z}$$-module of finite rank which is Eisenstein as a Hecke module, i.e. every Hecke eigenvalue system that shows up looks like an Eisenstein series (this is essentially the Manin--Drinfeld theorem). It follows that, for a given cuspidal eigenform $$f$$ as above, the only primes $$p$$ that can divide the index of the two period lattices are those such that $$f$$ is congruent mod $$p$$ to an Eisenstein series. If I remember correctly, you see this concretely with the unique weight 2 newform of level 11: it is congruent to an Eisenstein series mod 5, and the quotient between the two lattices is a cyclic group of order 5.

If you want to learn more, then you should read about modular symbols; the books by John Cremona and William Stein are both excellent references.

• en.wikipedia.org/wiki/Generalized_eigenvector Commented Jan 2, 2022 at 14:15
• Thanks for the reference. Where can I find references for the homology splitting in Q1 and the Eisenstein quotient you mentioned in Q2? I'm not sure if Stein's book contains those two pieces. Commented Jan 2, 2022 at 15:14
• The homology splitting is just a special case of a general result about modules over Artinian rings (math.SE question: math.stackexchange.com/questions/1089183/…). Because the commutative subring of $End_{\mathbf{Q}}(H_1(\dots))$ generated by the Hecke operators is finite-dimensional over $\mathbf{Q}$, it is necessarily Artinian. Commented Jan 2, 2022 at 19:09

My attempt.

$$f$$ is an eigenform with integer coefficients $$\in S_2(\Gamma_0(N))$$, otherwise it is usually not true that its periods form a lattice in $$\Bbb{C}$$. As $$f=\sum_{d|N} c_d \tilde{f}(dz)$$ for some newform of lower level, we can restrict to the case that $$f$$ is a newform.

Following Cremona, taking finitely many $$\alpha\in \Bbb{Q}$$ representing the finitely many cusps, for some prime $$p\equiv 1\bmod N$$ not dividing the numerator and denominator of any $$\alpha$$, $$(1+p-a_p(f))\int_\alpha^{i\infty} f(z)dz=\int_\alpha^{i\infty} (1+p-T_p) f(z)dz$$ $$=\int_\alpha^{p\alpha} f(z)dz+\sum_{k=0}^{p-1} \int_\alpha^{(\alpha+k)/p} f(z)dz\tag{1}$$ $$p\alpha$$ and $$(\alpha+k)/p$$ are $$\Gamma_0(N)$$ equivalent to $$\alpha$$, whence the RHS is equal to a sum of integrals over closed-loops in $$X_0(N)$$.

From the convergence of the Rankin Selberg integral $$\langle f E_2,f\rangle$$ you'll get that $$p+1-a_p(f)\ne 0$$ for some $$p\equiv 1\bmod N$$ not dividing the numerator and denominator of the $$\alpha$$, obtaining that the subgroup of $$\Bbb{C}$$ generated by the $$\int_\alpha^\beta f(z),\alpha,\beta\in \Bbb{Q}\cup i\infty$$ is a lattice iff $$\{\int_\gamma f(z),\gamma$$ closed-loop in $$X_0(N)\}$$ is a lattice.

Then the theorem we need is that with $$g=\dim_\Bbb{C} S_2(\Gamma_0(N))$$ then $$\pi_1(X_0(N))^{ab}\cong \Bbb{Z}^{2g}$$.

$$S_2(\Gamma_0(N))$$ has a $$\Bbb{C}$$-basis $$f_1,\ldots,f_g$$ of modular forms with integer coefficients. We choose it such that $$f_1=f$$. With $$\gamma_1,\ldots,\gamma_{2g}$$ a $$\Bbb{Z}$$-basis of $$\pi_1(X_0(N))^{ab}$$, and $$\lambda_l=(\int_{\gamma_l} f_1(z)dz),\ldots,\int_{\gamma_l} f_g(z)dz))$$, using that a harmonic function on $$X_0(N)$$ must be constant,you'll get that the columns of the $$2g\times 2g$$ matrix with rows $$(\Re(\lambda_l),\Im(\lambda_l))$$ are $$\Bbb{R}$$-linearly independent, whence so are its rows, so the $$\lambda_l$$ generate a lattice $$\Lambda$$ in $$\Bbb{C}^g$$.

Let $$\Bbb{T}$$ be the $$\Bbb{Z}$$-algebra generated by the Hecke operators. Using that $$f_1$$ is a newform we can take an element $$P\in \Bbb{T}$$ such that $$Pf_j=0$$ except for $$Pf_1=cf_1\ne 0$$.

Each $$\int_\gamma Pf_j(z)dz$$ is given by integrating $$f_j$$ on some curves from cusps to cusps, with the same $$p$$ as above and the second equality of $$(1)$$ you'll get that there is some $$d$$ such that for all closed-loop $$\gamma$$ in $$X_0(N)$$, $$(\int_\gamma (1+p-T_p)Pf_1(z)dz,\ldots,\int_\gamma (1+p-T_p)Pf_g(z)dz) \in \frac1d\Lambda$$

This proves that for any $$\lambda\in \Lambda$$, $$((1+p-a_p(f))c\lambda_1,0\ldots,0) \in \frac1d\Lambda$$, and hence $$\{ \int_\gamma f_1(z)dz, \gamma \in \pi_1(X_0(N))\}$$ is a lattice in $$\Bbb{C}$$, from which the $$\int_\alpha^\beta f_1(z)dz,\alpha,\beta\in \Bbb{Q}\cup i\infty$$ generate a lattice in $$\Bbb{C}$$.

• Ah, this is also quite illuminating! It's nice to see an explicit calculation approach as well. Thanks. Commented Jan 3, 2022 at 0:37