The Tur\'{a}n number of the Cartesian product of a star and an edge

Abstract

Let $C_k$ denote the cycle of length $k$, $S_t$ be a star with $t$ edges. And let $B_t$ be the graph consisting of $t$ copies of $C_4$ sharing one fixed edge. Equivalently, $B_t=K_2 \mathbin{\square} S_t$, which is the Cartesian product of a star with $t$ edges and an edge. Recently, Gao, Janzer, Liu and Xu [\textit{Israel J. Math. 269(2025)}] proved that the Tur\'an number of $K_2\mathbin{\square} C_{2l}$ is $\Theta(n^{\frac{3}{2}})$ for every $l\ge 4$. In this paper, we obtain upper and lower estimates for the Tur\'an number of $B_t$ in both the general and bipartite settings for every $t\geq 2$. For the lower bound, we use random construction based on the extremal structure of $C_4$. These results imply that $\frac{1}{2\sqrt{2}}\leq \lim_{t\to \infty} \frac{\mathrm{ex}(n,B_t)}{\sqrt{t}}\leq \frac{1}{2}$, and $\frac{1}{4}\leq \lim_{t\to \infty} \frac{\mathrm{ex}_{bip}(n,B_t)}{\sqrt{t}}\leq \frac{1}{2\sqrt{2}}.$ In the case of $B_2$, we obtain sharper estimates. We show that the Tur\'an number of $B_2$ is approximately between $(0.518+o(1))n^{\frac{3}{2}}$ and $(0.603+o(1))n^{\frac{3}{2}}$. And in the bipartite setting, it is approximately between $(0.385+o(1))n^{\frac{3}{2}}$ and $(0.468+o(1))n^{\frac{3}{2}}$. Moreover, in the bipartite setting, we give a more general result, which shows that for every tree $T$ with $t$ edges, the bipartite Tur\'an number of $K_2\mathbin{\square}T$ is at most $\frac{\sqrt{t}}{2\sqrt{2}}(1+o(1))n^{\frac{3}{2}}$.