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The module of vector-valued modular forms is Cohen-Macaulay
Czechoslovak Mathematical Journal ( IF 0.5 ) Pub Date : 2020-10-27 , DOI: 10.21136/cmj.2020.0476-19
Richard Gottesman

Let $H$ denote a finite index subgroup of the modular group $\Gamma$ and let $\rho$ denote a finite-dimensional complex representation of $H.$ Let $M(\rho)$ denote the collection of holomorphic vector-valued modular forms for $\rho$ and let $M(H)$ denote the collection of modular forms on $H$. Then $M(\rho)$ is a $\textbf{Z}$-graded $M(H)$-module. It has been proven that $M(\rho)$ may not be projective as a $M(H)$-module. We prove that $M(\rho)$ is Cohen-Macaulay as a $M(H)$-module. We also explain how to apply this result to prove that if $M(H)$ is a polynomial ring then $M(\rho)$ is a free $M(H)$-module of rank $\textrm{dim } \rho.$

中文翻译:

向量值模形式的模是 Cohen-Macaulay

令$H$表示模群$\Gamma$的有限索引子群,令$\rho$表示$H的有限维复数表示。令$M(\rho)$表示全纯向量的集合- $\rho$ 的值模形式,让 $M(H)$ 表示 $H$ 上模形式的集合。那么 $M(\rho)$ 是一个 $\textbf{Z}$-graded $M(H)$-module。已经证明 $M(\rho)$ 作为 $M(H)$-模可能不是投影的。我们证明 $M(\rho)$ 是作为 $M(H)$-模的 Cohen-Macaulay。我们还解释了如何应用这个结果来证明如果 $M(H)$ 是一个多项式环,那么 $M(\rho)$ 是一个免费的 $M(H)$-秩 $\textrm{dim } \ rho.$
更新日期:2020-10-27
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