Mixing
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
reference-request co.combinatorics combinatorial-identities elementary-proofs
asked 1 hour ago
T. Amdeberhan
16.4k228122
add a comment |Â
up vote
2
down vote
favorite
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
asked 1 hour ago
T. Amdeberhan
16.4k228122
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
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
asked 1 hour ago
T. Amdeberhan
16.4k228122
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
reference-request co.combinatorics combinatorial-identities elementary-proofs
asked 1 hour ago
T. Amdeberhan
16.4k228122
asked 1 hour ago
T. Amdeberhan
16.4k228122
edited 1 hour ago
asked 1 hour ago
T. Amdeberhan
16.4k228122
asked 1 hour ago
T. Amdeberhan
16.4k228122
asked 1 hour ago
T. Amdeberhan
16.4k228122
16.4k228122
add a comment |Â
add a comment |Â
1 Answer
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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
$2n+1+(i-j)^2$.
answered 1 hour ago
Sam Hopkins
4,02312348
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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
$2n+1+(i-j)^2$.
answered 1 hour ago
Sam Hopkins
4,02312348
add a comment |Â
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
$2n+1+(i-j)^2$.
answered 1 hour ago
Sam Hopkins
4,02312348
add a comment |Â
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
$2n+1+(i-j)^2$.
answered 1 hour ago
Sam Hopkins
4,02312348
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
$2n+1+(i-j)^2$.
answered 1 hour ago
Sam Hopkins
4,02312348
edited 53 mins ago
answered 1 hour ago
Sam Hopkins
4,02312348
answered 1 hour ago
Sam Hopkins
4,02312348
answered 1 hour ago
Sam Hopkins
4,02312348
4,02312348
add a comment |Â
add a comment |Â
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