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](https://arxiv.org/pdf/hep-th/0005103.pdf) 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 $K^\*(X)$. 



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Ring structure on K-theory of a quotient of the Fermat quintic
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 $X$. 




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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. 


