For \(k\ge0\), let \(M_k\) be the space of rational \(k\)-differentials \[ f(z)\,dz^k \] on \(\widehat{\mathbb C}\) whose only poles are at \(0\), \(1\), and \(\infty\), each of order at most \(k\).
Determine \(\dim M_k\) and give a basis for \(M_k\).
Give a finite set of generators for the graded ring \[ \bigoplus_{k\ge0}M_k. \]
This is the ring of modular forms for \(\Gamma(2)\) in the source’s model.
Solution
For every \(k\ge0\), \[ \boxed{\dim M_k=k+1.} \] A basis is \[ \boxed{ \omega_{k,j} =\frac{z^j}{z^k(z-1)^k}\,dz^k, \qquad 0\le j\le k. } \] Moreover, if \[ u=\frac{dz}{z(z-1)}, \qquad v=\frac{dz}{z-1}, \] then \[ \boxed{\bigoplus_{k\ge0}M_k\cong\mathbb C[u,v], \qquad \deg u=\deg v=1.} \]
1. Every element of \(M_k\) has the form\[ \frac{P(z)}{z^k(z-1)^k}\,dz^k \]for a polynomial \(P\).
Let \[ \omega=f(z)\,dz^k\in M_k. \] At the finite points \(0\) and \(1\), the differential \(dz^k\) is nonvanishing, so the pole-order condition says that \(f\) has poles of order at most \(k\) at those two points and no other finite poles. Therefore \[ P(z)=z^k(z-1)^k f(z) \] is a rational function with no finite poles. Hence \(P\) is a polynomial, and \[ \omega =\frac{P(z)}{z^k(z-1)^k}\,dz^k. \]
2. The condition at \(\infty\) is exactly\[ \deg P\le k. \]
Use the local coordinate \[ w=\frac1z \] near \(\infty\). Since \[ dz=-w^{-2}dw, \] we have \[ z^k(z-1)^k =\frac{(1-w)^k}{w^{2k}}. \] Therefore \[ \begin{aligned} \omega &=\frac{P(1/w)}{(1-w)^k/w^{2k}} (-1)^k w^{-2k}dw^k\\ &=(-1)^k\frac{P(1/w)}{(1-w)^k}\,dw^k. \end{aligned} \] The factor \((1-w)^{-k}\) is a unit at \(w=0\). If \(d=\deg P\), then \(P(1/w)\) has a pole of order \(d\) at \(w=0\). Thus the pole order of \(\omega\) at \(\infty\) is exactly \(d\). The requirement that this order be at most \(k\) is therefore equivalent to \[ \deg P\le k. \]
3. This gives the dimension and basis of \(M_k\).
By 1 and 2, the map \[ P(z)\longmapsto \frac{P(z)}{z^k(z-1)^k}\,dz^k \] is a linear isomorphism from the vector space of polynomials of degree at most \(k\) onto \(M_k\). The polynomials \[ 1,z,\ldots,z^k \] form a basis of that space. Hence \[ \dim M_k=k+1 \] and the corresponding basis of \(M_k\) is \[ \omega_{k,j} =\frac{z^j}{z^k(z-1)^k}\,dz^k, \qquad0\le j\le k. \]
4. The two degree-one forms \(u\) and \(v\) generate every homogeneous piece.
Both \[ u=\frac{dz}{z(z-1)}, \qquad v=\frac{dz}{z-1} \] belong to \(M_1\). For \(0\le j\le k\), \[ \begin{aligned} u^{k-j}v^j &=\left(\frac{dz}{z(z-1)}\right)^{k-j} \left(\frac{dz}{z-1}\right)^j\\ &=\frac{z^j}{z^k(z-1)^k}\,dz^k\\ &=\omega_{k,j}. \end{aligned} \] By 3 these monomials form a basis of \(M_k\). Hence every graded piece, and therefore the whole graded ring, is generated by \(u\) and \(v\).
5. There is no algebraic relation between \(u\) and \(v\).
Any homogeneous polynomial relation of degree \(k\) would have the form \[ \sum_{j=0}^k c_j u^{k-j}v^j=0. \] By 4 this is \[ \sum_{j=0}^k c_j\omega_{k,j}=0. \] The basis statement in 3 forces every \(c_j=0\). A relation in the graded direct sum decomposes into homogeneous relations, so there are no nonzero relations at all. Thus \[ \bigoplus_{k\ge0}M_k\cong\mathbb C[u,v] \] as a graded ring, with both generators in degree \(1\).