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











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












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






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.











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.









share|cite|improve this question




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.









asked Nov 15 at 5:16









Richard

133




133




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.





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.






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


















  • 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
1






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 $G$.






share|cite|improve this answer





















    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });






    Richard is a new contributor. Be nice, and check out our Code of Conduct.










     

    draft saved


    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2999252%2fexample-of-omegag-times-h-leq-min-omegag-omegah%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    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 $G$.






    share|cite|improve this answer

























      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 $G$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 days ago









        Chris Godsil

        11.5k21634




        11.5k21634






















            Richard is a new contributor. Be nice, and check out our Code of Conduct.










             

            draft saved


            draft discarded


















            Richard is a new contributor. Be nice, and check out our Code of Conduct.













            Richard is a new contributor. Be nice, and check out our Code of Conduct.












            Richard is a new contributor. Be nice, and check out our Code of Conduct.















             


            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2999252%2fexample-of-omegag-times-h-leq-min-omegag-omegah%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            AnyDesk - Fatal Program Failure

            How to calibrate 16:9 built-in touch-screen to a 4:3 resolution?

            QoS: MAC-Priority for clients behind a repeater