The Turán number of the Cartesian product of a star and an edge
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| Format: | Preprint |
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2026
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| _version_ | 1866918442886496256 |
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| author | Zhao, Xiamiao Cheng, Xin Chi, Cheng Győri, Ervin Tompkins, Casey Wang, Yichen |
| author_facet | Zhao, Xiamiao Cheng, Xin Chi, Cheng Győri, Ervin Tompkins, Casey Wang, Yichen |
| contents | 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 Turán number of $K_2\mathbin{\square} C_{2l}$ is $Θ(n^{\frac{3}{2}})$ for every $l\ge 4$.
In this paper, we obtain upper and lower estimates for the Turán 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 Turán 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 Turán number of $K_2\mathbin{\square}T$ is at most $\frac{\sqrt{t}}{2\sqrt{2}}(1+o(1))n^{\frac{3}{2}}$. |
| format | Preprint |
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arxiv_https___arxiv_org_abs_2604_11366 |
| institution | arXiv |
| publishDate | 2026 |
| record_format | arxiv |
| spellingShingle | The Turán number of the Cartesian product of a star and an edge Zhao, Xiamiao Cheng, Xin Chi, Cheng Győri, Ervin Tompkins, Casey Wang, Yichen Combinatorics 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 Turán number of $K_2\mathbin{\square} C_{2l}$ is $Θ(n^{\frac{3}{2}})$ for every $l\ge 4$. In this paper, we obtain upper and lower estimates for the Turán 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 Turán 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 Turán number of $K_2\mathbin{\square}T$ is at most $\frac{\sqrt{t}}{2\sqrt{2}}(1+o(1))n^{\frac{3}{2}}$. |
| title | The Turán number of the Cartesian product of a star and an edge |
| topic | Combinatorics |
| url | https://arxiv.org/abs/2604.11366 |