A link between hooks and contents: Part II

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This is a question in the spirit of an earlier problem.



Let $lambda$ be an integer partition: $lambda=(lambda_1geqlambda_2geqdotsgeq0)$.



Recall also the notation for the content of a cell $u=(i,j)$ in a partition: $c_u=j−i$. Further, let $h_u$ denote the hook-length of the cell $u$.



The below identity is rather cute for which I don't remember a reference or a proof, so here I ask.




Identity. Given a partition $lambdavdash n$, it holds that
$$sum_uinlambdah_u^2=n^2+sum_uinlambdac_u^2.$$




For example, if $lambda=(4,3,1)vdash 8$ then the hooks are $6,4,3,1,4,2,1,1$ and the contents are $0,1,2,3,-1,0,1,-2$. Hence
beginalign
LHS=6^2+4^2+3^2+1^2+4^2+2^2+1^2+1^2&=84 \
RHS=mathbf8^2+0^2+1^2+2^2+3^2+(-1)^2+0^2+1^2+(-2)^2&=84.
endalign










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    This is a question in the spirit of an earlier problem.



    Let $lambda$ be an integer partition: $lambda=(lambda_1geqlambda_2geqdotsgeq0)$.



    Recall also the notation for the content of a cell $u=(i,j)$ in a partition: $c_u=j−i$. Further, let $h_u$ denote the hook-length of the cell $u$.



    The below identity is rather cute for which I don't remember a reference or a proof, so here I ask.




    Identity. Given a partition $lambdavdash n$, it holds that
    $$sum_uinlambdah_u^2=n^2+sum_uinlambdac_u^2.$$




    For example, if $lambda=(4,3,1)vdash 8$ then the hooks are $6,4,3,1,4,2,1,1$ and the contents are $0,1,2,3,-1,0,1,-2$. Hence
    beginalign
    LHS=6^2+4^2+3^2+1^2+4^2+2^2+1^2+1^2&=84 \
    RHS=mathbf8^2+0^2+1^2+2^2+3^2+(-1)^2+0^2+1^2+(-2)^2&=84.
    endalign










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      up vote
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      up vote
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      This is a question in the spirit of an earlier problem.



      Let $lambda$ be an integer partition: $lambda=(lambda_1geqlambda_2geqdotsgeq0)$.



      Recall also the notation for the content of a cell $u=(i,j)$ in a partition: $c_u=j−i$. Further, let $h_u$ denote the hook-length of the cell $u$.



      The below identity is rather cute for which I don't remember a reference or a proof, so here I ask.




      Identity. Given a partition $lambdavdash n$, it holds that
      $$sum_uinlambdah_u^2=n^2+sum_uinlambdac_u^2.$$




      For example, if $lambda=(4,3,1)vdash 8$ then the hooks are $6,4,3,1,4,2,1,1$ and the contents are $0,1,2,3,-1,0,1,-2$. Hence
      beginalign
      LHS=6^2+4^2+3^2+1^2+4^2+2^2+1^2+1^2&=84 \
      RHS=mathbf8^2+0^2+1^2+2^2+3^2+(-1)^2+0^2+1^2+(-2)^2&=84.
      endalign










      share|cite|improve this question















      This is a question in the spirit of an earlier problem.



      Let $lambda$ be an integer partition: $lambda=(lambda_1geqlambda_2geqdotsgeq0)$.



      Recall also the notation for the content of a cell $u=(i,j)$ in a partition: $c_u=j−i$. Further, let $h_u$ denote the hook-length of the cell $u$.



      The below identity is rather cute for which I don't remember a reference or a proof, so here I ask.




      Identity. Given a partition $lambdavdash n$, it holds that
      $$sum_uinlambdah_u^2=n^2+sum_uinlambdac_u^2.$$




      For example, if $lambda=(4,3,1)vdash 8$ then the hooks are $6,4,3,1,4,2,1,1$ and the contents are $0,1,2,3,-1,0,1,-2$. Hence
      beginalign
      LHS=6^2+4^2+3^2+1^2+4^2+2^2+1^2+1^2&=84 \
      RHS=mathbf8^2+0^2+1^2+2^2+3^2+(-1)^2+0^2+1^2+(-2)^2&=84.
      endalign







      reference-request co.combinatorics combinatorial-identities elementary-proofs






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      edited 1 hour ago

























      asked 1 hour ago









      T. Amdeberhan

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          You can also prove this inductively by adding boxes to outside corners of $lambda$.



          Suppose the results is true for $lambdavdash n$ and then we add a new box $(i,j)$ in an outside corner.



          The RHS increases by $(n+1)^2-n^2 + (i-j)^2=2n+1+(i-j)^2$



          How much does the LHS increase by? Let me use $h_u$ for the hook length of $u$ in $lambda$ (as opposed to the hook length in $lambda + (i,j)$.)



          We increase the hook length by one for every $u$ in the same row or same column as $(i,j)$. So for each of these boxes $u$ we increase the squared hook length by $2h_u+1$.



          Note that
          $$ sum_substacku textrm in same row \textrmor column as $(i,j)$h_u = n-(i-1)times(j-1) +(0+1+...+i-2)+(0+1+...+j-2)$$
          because every box not in the $(i-1)times(j-1)$ northwest rectangle of $lambda$ is covered exactly once by the hooks in this row or column, except that the boxes in this row or column are covered many extra times as accounted for by those arithmetic progressions.



          Thus the LHS increases by
          $$ 2(n-(i-1)(j-1)+(i-1)(i-2)/2+(j-1)(j-2)/2)+(i-1)+(j-1)+1$$
          (where that last $+1$ is for the hook length of $(i,j)$), and this indeed equals
          $2n+1+(i-j)^2$.






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            up vote
            3
            down vote













            You can also prove this inductively by adding boxes to outside corners of $lambda$.



            Suppose the results is true for $lambdavdash n$ and then we add a new box $(i,j)$ in an outside corner.



            The RHS increases by $(n+1)^2-n^2 + (i-j)^2=2n+1+(i-j)^2$



            How much does the LHS increase by? Let me use $h_u$ for the hook length of $u$ in $lambda$ (as opposed to the hook length in $lambda + (i,j)$.)



            We increase the hook length by one for every $u$ in the same row or same column as $(i,j)$. So for each of these boxes $u$ we increase the squared hook length by $2h_u+1$.



            Note that
            $$ sum_substacku textrm in same row \textrmor column as $(i,j)$h_u = n-(i-1)times(j-1) +(0+1+...+i-2)+(0+1+...+j-2)$$
            because every box not in the $(i-1)times(j-1)$ northwest rectangle of $lambda$ is covered exactly once by the hooks in this row or column, except that the boxes in this row or column are covered many extra times as accounted for by those arithmetic progressions.



            Thus the LHS increases by
            $$ 2(n-(i-1)(j-1)+(i-1)(i-2)/2+(j-1)(j-2)/2)+(i-1)+(j-1)+1$$
            (where that last $+1$ is for the hook length of $(i,j)$), and this indeed equals
            $2n+1+(i-j)^2$.






            share|cite|improve this answer


























              up vote
              3
              down vote













              You can also prove this inductively by adding boxes to outside corners of $lambda$.



              Suppose the results is true for $lambdavdash n$ and then we add a new box $(i,j)$ in an outside corner.



              The RHS increases by $(n+1)^2-n^2 + (i-j)^2=2n+1+(i-j)^2$



              How much does the LHS increase by? Let me use $h_u$ for the hook length of $u$ in $lambda$ (as opposed to the hook length in $lambda + (i,j)$.)



              We increase the hook length by one for every $u$ in the same row or same column as $(i,j)$. So for each of these boxes $u$ we increase the squared hook length by $2h_u+1$.



              Note that
              $$ sum_substacku textrm in same row \textrmor column as $(i,j)$h_u = n-(i-1)times(j-1) +(0+1+...+i-2)+(0+1+...+j-2)$$
              because every box not in the $(i-1)times(j-1)$ northwest rectangle of $lambda$ is covered exactly once by the hooks in this row or column, except that the boxes in this row or column are covered many extra times as accounted for by those arithmetic progressions.



              Thus the LHS increases by
              $$ 2(n-(i-1)(j-1)+(i-1)(i-2)/2+(j-1)(j-2)/2)+(i-1)+(j-1)+1$$
              (where that last $+1$ is for the hook length of $(i,j)$), and this indeed equals
              $2n+1+(i-j)^2$.






              share|cite|improve this answer
























                up vote
                3
                down vote










                up vote
                3
                down vote









                You can also prove this inductively by adding boxes to outside corners of $lambda$.



                Suppose the results is true for $lambdavdash n$ and then we add a new box $(i,j)$ in an outside corner.



                The RHS increases by $(n+1)^2-n^2 + (i-j)^2=2n+1+(i-j)^2$



                How much does the LHS increase by? Let me use $h_u$ for the hook length of $u$ in $lambda$ (as opposed to the hook length in $lambda + (i,j)$.)



                We increase the hook length by one for every $u$ in the same row or same column as $(i,j)$. So for each of these boxes $u$ we increase the squared hook length by $2h_u+1$.



                Note that
                $$ sum_substacku textrm in same row \textrmor column as $(i,j)$h_u = n-(i-1)times(j-1) +(0+1+...+i-2)+(0+1+...+j-2)$$
                because every box not in the $(i-1)times(j-1)$ northwest rectangle of $lambda$ is covered exactly once by the hooks in this row or column, except that the boxes in this row or column are covered many extra times as accounted for by those arithmetic progressions.



                Thus the LHS increases by
                $$ 2(n-(i-1)(j-1)+(i-1)(i-2)/2+(j-1)(j-2)/2)+(i-1)+(j-1)+1$$
                (where that last $+1$ is for the hook length of $(i,j)$), and this indeed equals
                $2n+1+(i-j)^2$.






                share|cite|improve this answer














                You can also prove this inductively by adding boxes to outside corners of $lambda$.



                Suppose the results is true for $lambdavdash n$ and then we add a new box $(i,j)$ in an outside corner.



                The RHS increases by $(n+1)^2-n^2 + (i-j)^2=2n+1+(i-j)^2$



                How much does the LHS increase by? Let me use $h_u$ for the hook length of $u$ in $lambda$ (as opposed to the hook length in $lambda + (i,j)$.)



                We increase the hook length by one for every $u$ in the same row or same column as $(i,j)$. So for each of these boxes $u$ we increase the squared hook length by $2h_u+1$.



                Note that
                $$ sum_substacku textrm in same row \textrmor column as $(i,j)$h_u = n-(i-1)times(j-1) +(0+1+...+i-2)+(0+1+...+j-2)$$
                because every box not in the $(i-1)times(j-1)$ northwest rectangle of $lambda$ is covered exactly once by the hooks in this row or column, except that the boxes in this row or column are covered many extra times as accounted for by those arithmetic progressions.



                Thus the LHS increases by
                $$ 2(n-(i-1)(j-1)+(i-1)(i-2)/2+(j-1)(j-2)/2)+(i-1)+(j-1)+1$$
                (where that last $+1$ is for the hook length of $(i,j)$), and this indeed equals
                $2n+1+(i-j)^2$.







                share|cite|improve this answer














                share|cite|improve this answer



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                edited 53 mins ago

























                answered 1 hour ago









                Sam Hopkins

                4,02312348




                4,02312348



























                     

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