Example of $omega(G times H) leq min{omega(G), omega(H)}$











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It's written in this paper by Alon and Lubetzky that $omega(G times H) leq min{omega(G), omega(H)}$, where $omega$ denotes the clique number, and $times$ denotes the tensor product on a graph, where $((a, b), (c, d))$ is an edge in $G times H$ iff $(a, c)$ is an edge in $G$ and $(b, d)$ is an edge in $H$.



I am wondering if there are any examples to the "less than" case of this inequality. How come this isn't an equality? Suppose we take the maximum clique in $G$ and in $H$ and consider all 2-tuples representing vertices in $G times H$ composed of vertices from these cliques. Then based on the definition, by connecting all of these vertices, we can find a new clique in $G times H$ has size which is equal to the smaller of the two clique sizes. So why is there a $leq$?










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  • Was the $le$ in the title of your question supposed to be a $lt$?
    – bof
    Nov 15 at 6:06










  • In the paper you linked to, the identity $omega(Gtimes H)=min{omega(G),omega(H)}$ is asserted at the top of page 2, and again a couple of lines below the statement of Conjecture 3.4. I don't have time to read the whole paper; where do they say that $omega(Gtimes H)lemin{omega(G),omega(H)}$?
    – bof
    Nov 15 at 6:14










  • I must be going insane, because I swore it said $leq$. $=$ is consistent with what I believe it to be.
    – Richard
    2 days ago















up vote
2
down vote

favorite












It's written in this paper by Alon and Lubetzky that $omega(G times H) leq min{omega(G), omega(H)}$, where $omega$ denotes the clique number, and $times$ denotes the tensor product on a graph, where $((a, b), (c, d))$ is an edge in $G times H$ iff $(a, c)$ is an edge in $G$ and $(b, d)$ is an edge in $H$.



I am wondering if there are any examples to the "less than" case of this inequality. How come this isn't an equality? Suppose we take the maximum clique in $G$ and in $H$ and consider all 2-tuples representing vertices in $G times H$ composed of vertices from these cliques. Then based on the definition, by connecting all of these vertices, we can find a new clique in $G times H$ has size which is equal to the smaller of the two clique sizes. So why is there a $leq$?










share|cite|improve this question







New contributor




Richard is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • Was the $le$ in the title of your question supposed to be a $lt$?
    – bof
    Nov 15 at 6:06










  • In the paper you linked to, the identity $omega(Gtimes H)=min{omega(G),omega(H)}$ is asserted at the top of page 2, and again a couple of lines below the statement of Conjecture 3.4. I don't have time to read the whole paper; where do they say that $omega(Gtimes H)lemin{omega(G),omega(H)}$?
    – bof
    Nov 15 at 6:14










  • I must be going insane, because I swore it said $leq$. $=$ is consistent with what I believe it to be.
    – Richard
    2 days ago













up vote
2
down vote

favorite









up vote
2
down vote

favorite











It's written in this paper by Alon and Lubetzky that $omega(G times H) leq min{omega(G), omega(H)}$, where $omega$ denotes the clique number, and $times$ denotes the tensor product on a graph, where $((a, b), (c, d))$ is an edge in $G times H$ iff $(a, c)$ is an edge in $G$ and $(b, d)$ is an edge in $H$.



I am wondering if there are any examples to the "less than" case of this inequality. How come this isn't an equality? Suppose we take the maximum clique in $G$ and in $H$ and consider all 2-tuples representing vertices in $G times H$ composed of vertices from these cliques. Then based on the definition, by connecting all of these vertices, we can find a new clique in $G times H$ has size which is equal to the smaller of the two clique sizes. So why is there a $leq$?










share|cite|improve this question







New contributor




Richard is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











It's written in this paper by Alon and Lubetzky that $omega(G times H) leq min{omega(G), omega(H)}$, where $omega$ denotes the clique number, and $times$ denotes the tensor product on a graph, where $((a, b), (c, d))$ is an edge in $G times H$ iff $(a, c)$ is an edge in $G$ and $(b, d)$ is an edge in $H$.



I am wondering if there are any examples to the "less than" case of this inequality. How come this isn't an equality? Suppose we take the maximum clique in $G$ and in $H$ and consider all 2-tuples representing vertices in $G times H$ composed of vertices from these cliques. Then based on the definition, by connecting all of these vertices, we can find a new clique in $G times H$ has size which is equal to the smaller of the two clique sizes. So why is there a $leq$?







graph-theory ramsey-theory






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asked Nov 15 at 5:16









Richard

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133




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Check out our Code of Conduct.












  • Was the $le$ in the title of your question supposed to be a $lt$?
    – bof
    Nov 15 at 6:06










  • In the paper you linked to, the identity $omega(Gtimes H)=min{omega(G),omega(H)}$ is asserted at the top of page 2, and again a couple of lines below the statement of Conjecture 3.4. I don't have time to read the whole paper; where do they say that $omega(Gtimes H)lemin{omega(G),omega(H)}$?
    – bof
    Nov 15 at 6:14










  • I must be going insane, because I swore it said $leq$. $=$ is consistent with what I believe it to be.
    – Richard
    2 days ago


















  • Was the $le$ in the title of your question supposed to be a $lt$?
    – bof
    Nov 15 at 6:06










  • In the paper you linked to, the identity $omega(Gtimes H)=min{omega(G),omega(H)}$ is asserted at the top of page 2, and again a couple of lines below the statement of Conjecture 3.4. I don't have time to read the whole paper; where do they say that $omega(Gtimes H)lemin{omega(G),omega(H)}$?
    – bof
    Nov 15 at 6:14










  • I must be going insane, because I swore it said $leq$. $=$ is consistent with what I believe it to be.
    – Richard
    2 days ago
















Was the $le$ in the title of your question supposed to be a $lt$?
– bof
Nov 15 at 6:06




Was the $le$ in the title of your question supposed to be a $lt$?
– bof
Nov 15 at 6:06












In the paper you linked to, the identity $omega(Gtimes H)=min{omega(G),omega(H)}$ is asserted at the top of page 2, and again a couple of lines below the statement of Conjecture 3.4. I don't have time to read the whole paper; where do they say that $omega(Gtimes H)lemin{omega(G),omega(H)}$?
– bof
Nov 15 at 6:14




In the paper you linked to, the identity $omega(Gtimes H)=min{omega(G),omega(H)}$ is asserted at the top of page 2, and again a couple of lines below the statement of Conjecture 3.4. I don't have time to read the whole paper; where do they say that $omega(Gtimes H)lemin{omega(G),omega(H)}$?
– bof
Nov 15 at 6:14












I must be going insane, because I swore it said $leq$. $=$ is consistent with what I believe it to be.
– Richard
2 days ago




I must be going insane, because I swore it said $leq$. $=$ is consistent with what I believe it to be.
– Richard
2 days ago










1 Answer
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0
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The bound is always tight. To see this, first note that it’s enough to prove it when $G$ and $H$ are complete graphs and, since $K_m$ is an induced subgraph of $K_n$ (when $mle n$), it’s enough to consider the case $K_mtimes K_m$. But the subgraph induced by the diagonal of $Gtimes G$ is isomorphic to $G$, for any graph $G$.






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    1 Answer
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    1 Answer
    1






    active

    oldest

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    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    0
    down vote



    accepted










    The bound is always tight. To see this, first note that it’s enough to prove it when $G$ and $H$ are complete graphs and, since $K_m$ is an induced subgraph of $K_n$ (when $mle n$), it’s enough to consider the case $K_mtimes K_m$. But the subgraph induced by the diagonal of $Gtimes G$ is isomorphic to $G$, for any graph $G$.






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      up vote
      0
      down vote



      accepted










      The bound is always tight. To see this, first note that it’s enough to prove it when $G$ and $H$ are complete graphs and, since $K_m$ is an induced subgraph of $K_n$ (when $mle n$), it’s enough to consider the case $K_mtimes K_m$. But the subgraph induced by the diagonal of $Gtimes G$ is isomorphic to $G$, for any graph $G$.






      share|cite|improve this answer























        up vote
        0
        down vote



        accepted







        up vote
        0
        down vote



        accepted






        The bound is always tight. To see this, first note that it’s enough to prove it when $G$ and $H$ are complete graphs and, since $K_m$ is an induced subgraph of $K_n$ (when $mle n$), it’s enough to consider the case $K_mtimes K_m$. But the subgraph induced by the diagonal of $Gtimes G$ is isomorphic to $G$, for any graph $G$.






        share|cite|improve this answer












        The bound is always tight. To see this, first note that it’s enough to prove it when $G$ and $H$ are complete graphs and, since $K_m$ is an induced subgraph of $K_n$ (when $mle n$), it’s enough to consider the case $K_mtimes K_m$. But the subgraph induced by the diagonal of $Gtimes G$ is isomorphic to $G$, for any graph $G$.







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        share|cite|improve this answer










        answered 2 days ago









        Chris Godsil

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