Let $Y$ be the Fermat quintic, i.e. $Y \subset \mathbb{C}P^4$ is defined by $$ \sum_{i=1}^5 z_i^5 = 0 $$ In Section 5.3 of this paper by Volker Braun the author computes the K-groups of a quotient $X = Y/(\mathbb{Z}/5\mathbb{Z})$, where the generator acts by $$ [z_1: \dots : z_5] \mapsto [z_1: \alpha z_2 :\alpha^2 z_3 : \alpha^3 z_4 : \alpha^4 z_5] $$ with $\alpha = e^{\frac{2 \pi i}{5}}$. This quotient is a non-singular Calabi-Yau manifold and its $K$-groups turn out to be $$ K^0(X) \cong \mathbb{Z}^4 \oplus \mathbb{Z}/5\mathbb{Z}\\ K^1(X) \cong \mathbb{Z}^{44} \oplus \mathbb{Z}/5\mathbb{Z} $$ The computation uses the Atiyah-Hirzebruch spectral sequence, which unfortunately does not tell us much about the multiplication on $K^*(X)$.

Question: What is the ring structure on $K^*(X)$?

  • $\begingroup$ It seems to me that $Y$ would be an honest subspace of a weighted projective space, which nonequivariantly is homeomorphic to $\mathbb{C}P^4$. So, not being an expert, it seems to me that computations of Bahri, Franz and Ray. The equivariant cohomology ring of weighted projective space. Math. Proc. Cambridge Philos. Soc. 146 (2009), no. 2, 395–405 might give some idea on the integral cohomology, which may help to deal with K theory as well. The subject of arxiv.org/pdf/1306.1641.pdf might be relevant and helpful. $\endgroup$
    – user51223
    Mar 5, 2018 at 6:20

1 Answer 1


The Atiyah-Hirzebruch spectral sequence does have a multiplicative structure, and I think this can be used to determine the multiplication on K-theory. From the paper of Braun, it follows that the spectral sequence actually degenerates integrally: the only potentially nontrivial differentials are the ones with target $\mathbb{Z}/5\mathbb{Z}\in H^5(X,K^{2q})$, i.e. the differential $d_3^{2,2q}:H^2(X,K^{2q})\to H^5(X,K^{2q-2})$ on the $E_3$ page and the differential $d_5^{0,2q}:H^0(X,K^{2q})\to H^5(X,K^{2q-4})$. By the computations in Braun's paper, these differentials must be trivial because the $\mathbb{Z}/5\mathbb{Z}$ actually appears in the K-theory of $X$. So the spectral sequence must degenerate integrally at the $E_2$-page. Furthermore the extension problems all split, since otherwise there would be no torsion in K-theory.

Now the multiplicative structure of the Atiyah-Hirzebruch spectral sequence states that

  • the multiplication on the $E_2$-page is induced from the cup-product on cohomology of $X$.

  • the multiplication on K-theory is compatible with the relevant filtration for the spectral sequence, and the induced multiplication on the $E_\infty$-page coincides with the multiplicative structure of the spectral sequence.

Since the filtration on K-theory is split, the multiplication is given by the multiplication on the $E_\infty$-page, which by the degeneration coincides with the multiplication on the $E_2$-page induced from the cup product. All in all, I think the multiplicative structure of the Atiyah-Hirzebruch spectral sequence implies that the multiplication on $K^\ast(X)$ (which actually is just $K^0=H^{ev}$ and $K^1=H^{odd}$) coincides with the cup product on the cohomology of $X$.

Some more explanations on the relation between multiplication on $K^\ast(X)$ and the $E_\infty$-page are in order. It is true that the relation between multiplication on $K^\ast$ and on the $E_\infty$-page is rather weak: there is a filtration $F^\ast$ on $K^\ast(X)$ which satisfies $F^p\times F^q\subset F^{p+q}$. In particular, even if the filtration splits, the product of elements of $H^p$ and $H^q$ will land in $H^{p+q}\oplus H^{p+q+2}\oplus\cdots \oplus H^{max}$.

I claim that this doesn't happen in the specific case at hand. First of all, we can split off $H^0$ from $K^0$, viewed as K-theory of the point, generated by the trivial line bundle. Then $H^0\times K^\ast(X)\to K^\ast(X)$ will always just be the multiplication coming from the $\mathbb{Z}$-algebra structure. Now for degree reasons, the only product where something strange can happen is the one for $H^2\times H^2\to F^4K^0(X)$. (For instance, multiplication of $H^2$ and $H^3$ lands in the filtration step whose only component is $H^5$ for dimension reasons and therefore this is determined by the cup product on the $E_\infty$-page. Similar arguments for all the other cases.) So the product of two classes from $H^2$ lands a priori in $H^4\oplus H^6$, but then we can check using the Chern character that there is no error term in $H^6$ and the product is really the cup product.

  • 3
    $\begingroup$ "Since the filtration on K-theory is split, the multiplication is given by the multiplication on the E∞-page" This is not generally true. Why is it true in this case? $\endgroup$ Mar 5, 2018 at 15:21
  • $\begingroup$ @OscarRandal-Williams: true. I hope the additional explanations show that nothing strange happens in this case; mainly due to dimension reasons. $\endgroup$ Mar 6, 2018 at 9:45
  • $\begingroup$ If $K^0_f(X)$ is the quotient of $K^0(X)$ by its torsion subgroup then the chern character injects this as a maximal-rank lattice: $$ K^0_f(X) \subset H^{ev}(X, \mathbb{Q}). $$ Of course $H^{ev}(X, \mathbb{Q})$ already contains a maximal-rank lattice, namely the quotient of $H^{ev}(X, \mathbb{Z})$ by its torsion subsgorup. Do the arguments here imply that these two lattices coincide? $\endgroup$ Mar 6, 2018 at 12:38
  • 2
    $\begingroup$ @OliverNash: I think so. This would also be an alternative way to determine the ring structure: modulo torsion everything is determined by the Chern character, and the only potentially nontrivial multiplication involving torsion is $H^2\times H^3\to H^5$ which is just the cup product because it lands in the smallest filtration step. $\endgroup$ Mar 7, 2018 at 9:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.