Signless Laplacian characterization of cones over disjoint unions of cycles, edges and isolated vertices
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| Formato: | Preprint |
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2025
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| _version_ | 1866918151536508928 |
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| author | Ye, Jiachang Qian, Jianguo Stanic, Zoran |
| author_facet | Ye, Jiachang Qian, Jianguo Stanic, Zoran |
| contents | Two graphs are said to be $Q$-cospectral if they share the same signless Laplacian spectrum. A simple graph is said to be determined by its signless Laplacian spectrum (abbreviated as DQS) if there exists no other non-isomorphic simple graph with the same signless Laplacian spectrum. In this paper, we establish the following results:
(1) Let$G \cong K_{1} \vee \bigl(C_{k} \cup qK_{2} \cup sK_{1}\bigr),$ with $q,s \geq 1$, $k \geq 4$, and at least $21$ vertices. If $k$ is odd, then $G$ is DQS. Moreover, if $k$ is even and $F$ is $Q$-cospectral with $G$, then $$F \cong G \quad \text{or} \quad F \cong K_{1} \vee \bigl(C_{4} \cup P_{k-3} \cup P_{3} \cup (q-2)K_{2} \cup sK_{1}\bigr).$$
(2) Let $G\cong K_1\vee (C_{k_1}\cup C_{k_2}\cup\cdots \cup C_{k_t}\cup qK_2\cup sK_1)$ with $t\ge 2$, $q,s\ge 1$, $k_i\ge 4$ and at least $33$ vertices. If each $k_i$ is odd, then $G$ is DQS.
(3) The graph $K_{1} \vee \bigl(C_{3} \cup C_{k_{1}} \cup C_{k_{2}} \cup \cdots \cup C_{k_{t-1}} \cup qK_{2} \cup sK_{1}\bigr),$ with $t,q,s \geq 1$ and $k_{i} \geq 3$, is not DQS. Moreover, it is $Q$-cospectral with $K_{1} \vee \bigl(K_{1,3} \cup C_{k_{1}} \cup C_{k_{2}} \cup \cdots \cup C_{k_{t-1}} \cup qK_{2} \cup (s-1)K_{1}\bigr).$
Here $P_{n}$, $C_{n}$, $K_{n}$ and $K_{n-r,r}$ denote the path, the cycle, the complete graph and the complete bipartite graph on $n$ vertices, while $\cup$ and $\vee$ represent the disjoint union and the join of two graphs, respectively. Furthermore, the signless Laplacian spectrum of the graphs under consideration is computed explicitly. |
| format | Preprint |
| id |
arxiv_https___arxiv_org_abs_2509_25925 |
| institution | arXiv |
| publishDate | 2025 |
| record_format | arxiv |
| spellingShingle | Signless Laplacian characterization of cones over disjoint unions of cycles, edges and isolated vertices Ye, Jiachang Qian, Jianguo Stanic, Zoran Combinatorics Two graphs are said to be $Q$-cospectral if they share the same signless Laplacian spectrum. A simple graph is said to be determined by its signless Laplacian spectrum (abbreviated as DQS) if there exists no other non-isomorphic simple graph with the same signless Laplacian spectrum. In this paper, we establish the following results: (1) Let$G \cong K_{1} \vee \bigl(C_{k} \cup qK_{2} \cup sK_{1}\bigr),$ with $q,s \geq 1$, $k \geq 4$, and at least $21$ vertices. If $k$ is odd, then $G$ is DQS. Moreover, if $k$ is even and $F$ is $Q$-cospectral with $G$, then $$F \cong G \quad \text{or} \quad F \cong K_{1} \vee \bigl(C_{4} \cup P_{k-3} \cup P_{3} \cup (q-2)K_{2} \cup sK_{1}\bigr).$$ (2) Let $G\cong K_1\vee (C_{k_1}\cup C_{k_2}\cup\cdots \cup C_{k_t}\cup qK_2\cup sK_1)$ with $t\ge 2$, $q,s\ge 1$, $k_i\ge 4$ and at least $33$ vertices. If each $k_i$ is odd, then $G$ is DQS. (3) The graph $K_{1} \vee \bigl(C_{3} \cup C_{k_{1}} \cup C_{k_{2}} \cup \cdots \cup C_{k_{t-1}} \cup qK_{2} \cup sK_{1}\bigr),$ with $t,q,s \geq 1$ and $k_{i} \geq 3$, is not DQS. Moreover, it is $Q$-cospectral with $K_{1} \vee \bigl(K_{1,3} \cup C_{k_{1}} \cup C_{k_{2}} \cup \cdots \cup C_{k_{t-1}} \cup qK_{2} \cup (s-1)K_{1}\bigr).$ Here $P_{n}$, $C_{n}$, $K_{n}$ and $K_{n-r,r}$ denote the path, the cycle, the complete graph and the complete bipartite graph on $n$ vertices, while $\cup$ and $\vee$ represent the disjoint union and the join of two graphs, respectively. Furthermore, the signless Laplacian spectrum of the graphs under consideration is computed explicitly. |
| title | Signless Laplacian characterization of cones over disjoint unions of cycles, edges and isolated vertices |
| topic | Combinatorics |
| url | https://arxiv.org/abs/2509.25925 |