Isoperimetric constant on random graph












3














Show that there is a constant $c=c(p) > 0 $ such that almost all graphs in $mathcal{G}_{n,p}$ verify the following property : for each subset $X in V(G)$ with cardinality $|X|leq n/2$,
$$ e(X, Vbackslash X) geq c |X| $$



Where $ e(X, Vbackslash X)$ denotes the number of edges connecting $X$ and $Vbackslash X$.



I found a "standard" solution, using combinatorial arguments (the constant $c(p)=p$ satisfies the condition), but I have the feeling that a "more elegant" proof should exist.



My intuition is as follow : the problem is equivalent to proving, with $i(G)$ the isoperimetric constant :
$$ forall p, exists c(p)>0, s.t. mathbb{P}[ i(G) > c] rightarrow 1 text{ when }nrightarrow infty $$



And for any graph $G$, we know a lower bound for $i(G)$:
$$ mu_2 /2 leq i(G)$$
With $mu_2$ being the second smallest eigenvalue of the Laplacian matrix of $G$. Now if I can find some bound for $mu_2$ I should be able to do something.



I know that for fixed $p$ the probability that $G$ is connected tends to 1, hence $mu_2 > 0$. But can we find similar argument for $mu_2 > c$ for some constant $c$ ? in term of connectivity surely?
Thanks for the help!






graph-theory eigenvalues-eigenvectors random-graphs







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    3














    Show that there is a constant $c=c(p) > 0 $ such that almost all graphs in $mathcal{G}_{n,p}$ verify the following property : for each subset $X in V(G)$ with cardinality $|X|leq n/2$,
    $$ e(X, Vbackslash X) geq c |X| $$



    Where $ e(X, Vbackslash X)$ denotes the number of edges connecting $X$ and $Vbackslash X$.



    I found a "standard" solution, using combinatorial arguments (the constant $c(p)=p$ satisfies the condition), but I have the feeling that a "more elegant" proof should exist.



    My intuition is as follow : the problem is equivalent to proving, with $i(G)$ the isoperimetric constant :
    $$ forall p, exists c(p)>0, s.t. mathbb{P}[ i(G) > c] rightarrow 1 text{ when }nrightarrow infty $$



    And for any graph $G$, we know a lower bound for $i(G)$:
    $$ mu_2 /2 leq i(G)$$
    With $mu_2$ being the second smallest eigenvalue of the Laplacian matrix of $G$. Now if I can find some bound for $mu_2$ I should be able to do something.



    I know that for fixed $p$ the probability that $G$ is connected tends to 1, hence $mu_2 > 0$. But can we find similar argument for $mu_2 > c$ for some constant $c$ ? in term of connectivity surely?
    Thanks for the help!






    graph-theory eigenvalues-eigenvectors random-graphs







    share|cite|improve this question



























      3












      3








      3


      2





      Show that there is a constant $c=c(p) > 0 $ such that almost all graphs in $mathcal{G}_{n,p}$ verify the following property : for each subset $X in V(G)$ with cardinality $|X|leq n/2$,
      $$ e(X, Vbackslash X) geq c |X| $$



      Where $ e(X, Vbackslash X)$ denotes the number of edges connecting $X$ and $Vbackslash X$.



      I found a "standard" solution, using combinatorial arguments (the constant $c(p)=p$ satisfies the condition), but I have the feeling that a "more elegant" proof should exist.



      My intuition is as follow : the problem is equivalent to proving, with $i(G)$ the isoperimetric constant :
      $$ forall p, exists c(p)>0, s.t. mathbb{P}[ i(G) > c] rightarrow 1 text{ when }nrightarrow infty $$



      And for any graph $G$, we know a lower bound for $i(G)$:
      $$ mu_2 /2 leq i(G)$$
      With $mu_2$ being the second smallest eigenvalue of the Laplacian matrix of $G$. Now if I can find some bound for $mu_2$ I should be able to do something.



      I know that for fixed $p$ the probability that $G$ is connected tends to 1, hence $mu_2 > 0$. But can we find similar argument for $mu_2 > c$ for some constant $c$ ? in term of connectivity surely?
      Thanks for the help!






      graph-theory eigenvalues-eigenvectors random-graphs







      share|cite|improve this question















      Show that there is a constant $c=c(p) > 0 $ such that almost all graphs in $mathcal{G}_{n,p}$ verify the following property : for each subset $X in V(G)$ with cardinality $|X|leq n/2$,
      $$ e(X, Vbackslash X) geq c |X| $$



      Where $ e(X, Vbackslash X)$ denotes the number of edges connecting $X$ and $Vbackslash X$.



      I found a "standard" solution, using combinatorial arguments (the constant $c(p)=p$ satisfies the condition), but I have the feeling that a "more elegant" proof should exist.



      My intuition is as follow : the problem is equivalent to proving, with $i(G)$ the isoperimetric constant :
      $$ forall p, exists c(p)>0, s.t. mathbb{P}[ i(G) > c] rightarrow 1 text{ when }nrightarrow infty $$



      And for any graph $G$, we know a lower bound for $i(G)$:
      $$ mu_2 /2 leq i(G)$$
      With $mu_2$ being the second smallest eigenvalue of the Laplacian matrix of $G$. Now if I can find some bound for $mu_2$ I should be able to do something.



      I know that for fixed $p$ the probability that $G$ is connected tends to 1, hence $mu_2 > 0$. But can we find similar argument for $mu_2 > c$ for some constant $c$ ? in term of connectivity surely?
      Thanks for the help!






      graph-theory eigenvalues-eigenvectors random-graphs




      graph-theory eigenvalues-eigenvectors random-graphs






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      edited 2 days ago









      David G. Stork

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      asked 2 days ago









      Thomas LesgourguesThomas Lesgourgues

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          From this paper we can estimate the normalized Laplacian spectrum of $mathcal G_{n,p}$ from the spectrum of an "average-case Laplacian" $$bar{L} = I - bar{D}^{-1/2} bar{A}bar{D}^{-1/2}$$
          where $bar{D} = (n-1)pI$ is the expected value of $D$, and $bar{A} = p(J-I)$ is the expected value of $A$. (As usual, $J$ is the all-ones matrix.) Since $bar{D}$ commutes with everything, we can rewrite this as $I - bar{D}^{-1}bar{A}$ which simplifies to$$frac{n}{n-1}I - frac{1}{n-1}J.$$ The eigenvalues of $J$ are $n,0,0,dots,0,0$ and so the eigenvalues of $bar{L}$ are $0, frac{n}{n-1}, frac{n}{n-1}, dots, frac{n}{n-1}$.



          This looks like a rather stupid estimate that doesn't depend on $p$, but the dependence on $p$ is hidden in the error bound. For all constant $epsilon>0$, we have $$|lambda_i(L) - lambda_i(bar{L})| le mathcal Oleft(sqrt{frac{log(n/epsilon)}{(n-1)p}}right)$$ provided $(n-1)p > k(epsilon)log n$. In other words, as long as $p$ is not a too-small function of $n$, we have $lambda_2(L) ge 1 - o(1)$ with high probability.



          I guess Theorem 4 in the paper provides a slightly-tighter answer more easily, but then this answer would have been too short. (Also I didn't find Theorem 4 until I wrote all of the above.)






          share|cite|improve this answer





















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            From this paper we can estimate the normalized Laplacian spectrum of $mathcal G_{n,p}$ from the spectrum of an "average-case Laplacian" $$bar{L} = I - bar{D}^{-1/2} bar{A}bar{D}^{-1/2}$$
            where $bar{D} = (n-1)pI$ is the expected value of $D$, and $bar{A} = p(J-I)$ is the expected value of $A$. (As usual, $J$ is the all-ones matrix.) Since $bar{D}$ commutes with everything, we can rewrite this as $I - bar{D}^{-1}bar{A}$ which simplifies to$$frac{n}{n-1}I - frac{1}{n-1}J.$$ The eigenvalues of $J$ are $n,0,0,dots,0,0$ and so the eigenvalues of $bar{L}$ are $0, frac{n}{n-1}, frac{n}{n-1}, dots, frac{n}{n-1}$.



            This looks like a rather stupid estimate that doesn't depend on $p$, but the dependence on $p$ is hidden in the error bound. For all constant $epsilon>0$, we have $$|lambda_i(L) - lambda_i(bar{L})| le mathcal Oleft(sqrt{frac{log(n/epsilon)}{(n-1)p}}right)$$ provided $(n-1)p > k(epsilon)log n$. In other words, as long as $p$ is not a too-small function of $n$, we have $lambda_2(L) ge 1 - o(1)$ with high probability.



            I guess Theorem 4 in the paper provides a slightly-tighter answer more easily, but then this answer would have been too short. (Also I didn't find Theorem 4 until I wrote all of the above.)






            share|cite|improve this answer


























              1














              From this paper we can estimate the normalized Laplacian spectrum of $mathcal G_{n,p}$ from the spectrum of an "average-case Laplacian" $$bar{L} = I - bar{D}^{-1/2} bar{A}bar{D}^{-1/2}$$
              where $bar{D} = (n-1)pI$ is the expected value of $D$, and $bar{A} = p(J-I)$ is the expected value of $A$. (As usual, $J$ is the all-ones matrix.) Since $bar{D}$ commutes with everything, we can rewrite this as $I - bar{D}^{-1}bar{A}$ which simplifies to$$frac{n}{n-1}I - frac{1}{n-1}J.$$ The eigenvalues of $J$ are $n,0,0,dots,0,0$ and so the eigenvalues of $bar{L}$ are $0, frac{n}{n-1}, frac{n}{n-1}, dots, frac{n}{n-1}$.



              This looks like a rather stupid estimate that doesn't depend on $p$, but the dependence on $p$ is hidden in the error bound. For all constant $epsilon>0$, we have $$|lambda_i(L) - lambda_i(bar{L})| le mathcal Oleft(sqrt{frac{log(n/epsilon)}{(n-1)p}}right)$$ provided $(n-1)p > k(epsilon)log n$. In other words, as long as $p$ is not a too-small function of $n$, we have $lambda_2(L) ge 1 - o(1)$ with high probability.



              I guess Theorem 4 in the paper provides a slightly-tighter answer more easily, but then this answer would have been too short. (Also I didn't find Theorem 4 until I wrote all of the above.)






              share|cite|improve this answer
























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                From this paper we can estimate the normalized Laplacian spectrum of $mathcal G_{n,p}$ from the spectrum of an "average-case Laplacian" $$bar{L} = I - bar{D}^{-1/2} bar{A}bar{D}^{-1/2}$$
                where $bar{D} = (n-1)pI$ is the expected value of $D$, and $bar{A} = p(J-I)$ is the expected value of $A$. (As usual, $J$ is the all-ones matrix.) Since $bar{D}$ commutes with everything, we can rewrite this as $I - bar{D}^{-1}bar{A}$ which simplifies to$$frac{n}{n-1}I - frac{1}{n-1}J.$$ The eigenvalues of $J$ are $n,0,0,dots,0,0$ and so the eigenvalues of $bar{L}$ are $0, frac{n}{n-1}, frac{n}{n-1}, dots, frac{n}{n-1}$.



                This looks like a rather stupid estimate that doesn't depend on $p$, but the dependence on $p$ is hidden in the error bound. For all constant $epsilon>0$, we have $$|lambda_i(L) - lambda_i(bar{L})| le mathcal Oleft(sqrt{frac{log(n/epsilon)}{(n-1)p}}right)$$ provided $(n-1)p > k(epsilon)log n$. In other words, as long as $p$ is not a too-small function of $n$, we have $lambda_2(L) ge 1 - o(1)$ with high probability.



                I guess Theorem 4 in the paper provides a slightly-tighter answer more easily, but then this answer would have been too short. (Also I didn't find Theorem 4 until I wrote all of the above.)






                share|cite|improve this answer












                From this paper we can estimate the normalized Laplacian spectrum of $mathcal G_{n,p}$ from the spectrum of an "average-case Laplacian" $$bar{L} = I - bar{D}^{-1/2} bar{A}bar{D}^{-1/2}$$
                where $bar{D} = (n-1)pI$ is the expected value of $D$, and $bar{A} = p(J-I)$ is the expected value of $A$. (As usual, $J$ is the all-ones matrix.) Since $bar{D}$ commutes with everything, we can rewrite this as $I - bar{D}^{-1}bar{A}$ which simplifies to$$frac{n}{n-1}I - frac{1}{n-1}J.$$ The eigenvalues of $J$ are $n,0,0,dots,0,0$ and so the eigenvalues of $bar{L}$ are $0, frac{n}{n-1}, frac{n}{n-1}, dots, frac{n}{n-1}$.



                This looks like a rather stupid estimate that doesn't depend on $p$, but the dependence on $p$ is hidden in the error bound. For all constant $epsilon>0$, we have $$|lambda_i(L) - lambda_i(bar{L})| le mathcal Oleft(sqrt{frac{log(n/epsilon)}{(n-1)p}}right)$$ provided $(n-1)p > k(epsilon)log n$. In other words, as long as $p$ is not a too-small function of $n$, we have $lambda_2(L) ge 1 - o(1)$ with high probability.



                I guess Theorem 4 in the paper provides a slightly-tighter answer more easily, but then this answer would have been too short. (Also I didn't find Theorem 4 until I wrote all of the above.)







                share|cite|improve this answer












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                answered yesterday









                Misha LavrovMisha Lavrov

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