On Integral Cohomology Ring of Symmetric Products
Dmitry V. Gugnin
TL;DR
This work proves that the integral cohomology ring modulo torsion of symmetric products, H^*(Sym^n X;Z)/Tor, is a functor of A^*=H^*(X;Z)/Tor for connected CW complexes with finite homology type, and provides an explicit, computable description via the S^n_Z functor. Central to the result is Nakaoka's Integrality Lemma, which lifts integral classes from the rational setting and addresses torsion complications that thwart naive functoriality. The paper then applies this framework to Macdonald's theorem for symmetric products of compact Riemann surfaces M^2_g, resolving gaps in the original proof and delivering corrected, explicit generators and relations in both the rational and integral settings, including precise unstable-range corrections ( ilde{iii}_A, ilde{iii}_B). As a consequence, it yields a concrete description and algorithm to compute the integral cohomology rings H^*(Sym^n M^2_g;Z), with no torsion in these rings for the cases considered, and demonstrates the broader utility of the functorial approach via explicit basis and multiplication data.
Abstract
We prove that the integral cohomology ring modulo torsion $H^*(\mathrm{Sym}^n X;\mathbb{Z})/\mathrm{Tor}$ for the symmetric product of a connected CW-complex $X$ of finite homology type is a functor of $H^*(X;\mathbb{Z})/\mathrm{Tor}$ (see Theorem 1). Moreover, we give an explicit description of this functor. We also consider the important particular case when $X$ is a compact Riemann surface $M^2_g$ of genus $g$. There is a famous theorem of Macdonald of 1962, which gives an explicit description of the integral cohomology ring $H^*(\mathrm{Sym}^n M^2_g;\mathbb{Z})$. The analysis of the original proof by Macdonald shows that it contains three gaps. All these gaps were filled in by Seroul in 1972, and, therefore, he obtained a complete proof of Macdonald's theorem. Nevertheless, in the unstable case $2\le n\le 2g-2$ Macdonald's theorem has a subsection, that needs a slight correction even over $\mathbb{Q}$ (see Theorem 2).