Mixing
SYT and contents of a partition
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Let $lambda$ be an integer partition, denote the number of Standard Young Tableaux of shape $lambda$ by $f_lambda$. This number is computed by the formula
$$f_lambda=fracn!prod_uinlambdah_u$$
where $h_u$ is a hook length. It is also well-known that
$$sum_lambdavdash nf_lambda^2=n!tag1$$
Recall the notation for the content of a cell $u=(i,j)$ in a partition is $c_u=j-i$.
Question. Is this true?
$$sum_lambdavdash nf_lambda^2prod_uinlambda(t+c_u)=n!,t^n.$$
NOTE. An affirmative answer would imply (1): divide both sides by $t^n$ and take the limit $trightarrowinfty$.
co.combinatorics rt.representation-theory partitions
asked 5 hours ago
T. Amdeberhan
16.3k228122
add a comment |Â
up vote
5
down vote
favorite
Let $lambda$ be an integer partition, denote the number of Standard Young Tableaux of shape $lambda$ by $f_lambda$. This number is computed by the formula
$$f_lambda=fracn!prod_uinlambdah_u$$
where $h_u$ is a hook length. It is also well-known that
$$sum_lambdavdash nf_lambda^2=n!tag1$$
Recall the notation for the content of a cell $u=(i,j)$ in a partition is $c_u=j-i$.
Question. Is this true?
$$sum_lambdavdash nf_lambda^2prod_uinlambda(t+c_u)=n!,t^n.$$
NOTE. An affirmative answer would imply (1): divide both sides by $t^n$ and take the limit $trightarrowinfty$.
co.combinatorics rt.representation-theory partitions
asked 5 hours ago
T. Amdeberhan
16.3k228122
2
For any cell $u$ of $lambda$, let $h_lambda, u$ denote the hook length of $u$ in $lambda$. Using the formula quoted in mathoverflow.net/a/263658 as Corollary 7.21.4, we can rewrite $prodlimits_u in lambda left(t+c_uright) = s_lambdaleft(1^tright) prodlimits_u in lambda h_lambda, u = s_lambdaleft(1^tright) dfracn!f_lambda$ (by the hook-length formula). Thus, the left hand side of your question simplifies to $n! cdot sumlimits_lambda vdash n f_lambda s_lambdaleft(1^tright)$. The rest ...
– darij grinberg
4 hours ago
2
... is an easy application of RSK: The symmetric function $sumlimits_lambda vdash n f_lambda s_lambda$ is the generating function of all $n$-letter words over the alphabet $leftx_1,x_2,x_3,ldotsright$ (with the $f_lambda$ counting all possible Q-tableaux and the $s_lambda$ being the generating function of all possible P-tableaux) and thus equals $left(x_1+x_2+x_3+cdotsright)^n$ (a fact that also follows by applying the Pieri rule $n$ times); now substitute $1^t$ for the variables and you're done.
– darij grinberg
4 hours ago
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Let $lambda$ be an integer partition, denote the number of Standard Young Tableaux of shape $lambda$ by $f_lambda$. This number is computed by the formula
$$f_lambda=fracn!prod_uinlambdah_u$$
where $h_u$ is a hook length. It is also well-known that
$$sum_lambdavdash nf_lambda^2=n!tag1$$
Recall the notation for the content of a cell $u=(i,j)$ in a partition is $c_u=j-i$.
Question. Is this true?
$$sum_lambdavdash nf_lambda^2prod_uinlambda(t+c_u)=n!,t^n.$$
NOTE. An affirmative answer would imply (1): divide both sides by $t^n$ and take the limit $trightarrowinfty$.
co.combinatorics rt.representation-theory partitions
asked 5 hours ago
T. Amdeberhan
16.3k228122
Let $lambda$ be an integer partition, denote the number of Standard Young Tableaux of shape $lambda$ by $f_lambda$. This number is computed by the formula
$$f_lambda=fracn!prod_uinlambdah_u$$
where $h_u$ is a hook length. It is also well-known that
$$sum_lambdavdash nf_lambda^2=n!tag1$$
Recall the notation for the content of a cell $u=(i,j)$ in a partition is $c_u=j-i$.
Question. Is this true?
$$sum_lambdavdash nf_lambda^2prod_uinlambda(t+c_u)=n!,t^n.$$
NOTE. An affirmative answer would imply (1): divide both sides by $t^n$ and take the limit $trightarrowinfty$.
co.combinatorics rt.representation-theory partitions
co.combinatorics rt.representation-theory partitions
asked 5 hours ago
T. Amdeberhan
16.3k228122
asked 5 hours ago
T. Amdeberhan
16.3k228122
asked 5 hours ago
T. Amdeberhan
16.3k228122
asked 5 hours ago
T. Amdeberhan
16.3k228122
asked 5 hours ago
T. Amdeberhan
16.3k228122
16.3k228122
2
For any cell $u$ of $lambda$, let $h_lambda, u$ denote the hook length of $u$ in $lambda$. Using the formula quoted in mathoverflow.net/a/263658 as Corollary 7.21.4, we can rewrite $prodlimits_u in lambda left(t+c_uright) = s_lambdaleft(1^tright) prodlimits_u in lambda h_lambda, u = s_lambdaleft(1^tright) dfracn!f_lambda$ (by the hook-length formula). Thus, the left hand side of your question simplifies to $n! cdot sumlimits_lambda vdash n f_lambda s_lambdaleft(1^tright)$. The rest ...
– darij grinberg
4 hours ago
2
... is an easy application of RSK: The symmetric function $sumlimits_lambda vdash n f_lambda s_lambda$ is the generating function of all $n$-letter words over the alphabet $leftx_1,x_2,x_3,ldotsright$ (with the $f_lambda$ counting all possible Q-tableaux and the $s_lambda$ being the generating function of all possible P-tableaux) and thus equals $left(x_1+x_2+x_3+cdotsright)^n$ (a fact that also follows by applying the Pieri rule $n$ times); now substitute $1^t$ for the variables and you're done.
– darij grinberg
4 hours ago
add a comment |Â
2
For any cell $u$ of $lambda$, let $h_lambda, u$ denote the hook length of $u$ in $lambda$. Using the formula quoted in mathoverflow.net/a/263658 as Corollary 7.21.4, we can rewrite $prodlimits_u in lambda left(t+c_uright) = s_lambdaleft(1^tright) prodlimits_u in lambda h_lambda, u = s_lambdaleft(1^tright) dfracn!f_lambda$ (by the hook-length formula). Thus, the left hand side of your question simplifies to $n! cdot sumlimits_lambda vdash n f_lambda s_lambdaleft(1^tright)$. The rest ...
– darij grinberg
4 hours ago
2
... is an easy application of RSK: The symmetric function $sumlimits_lambda vdash n f_lambda s_lambda$ is the generating function of all $n$-letter words over the alphabet $leftx_1,x_2,x_3,ldotsright$ (with the $f_lambda$ counting all possible Q-tableaux and the $s_lambda$ being the generating function of all possible P-tableaux) and thus equals $left(x_1+x_2+x_3+cdotsright)^n$ (a fact that also follows by applying the Pieri rule $n$ times); now substitute $1^t$ for the variables and you're done.
– darij grinberg
4 hours ago
2
2
For any cell $u$ of $lambda$, let $h_lambda, u$ denote the hook length of $u$ in $lambda$. Using the formula quoted in mathoverflow.net/a/263658 as Corollary 7.21.4, we can rewrite $prodlimits_u in lambda left(t+c_uright) = s_lambdaleft(1^tright) prodlimits_u in lambda h_lambda, u = s_lambdaleft(1^tright) dfracn!f_lambda$ (by the hook-length formula). Thus, the left hand side of your question simplifies to $n! cdot sumlimits_lambda vdash n f_lambda s_lambdaleft(1^tright)$. The rest ...
– darij grinberg
4 hours ago
For any cell $u$ of $lambda$, let $h_lambda, u$ denote the hook length of $u$ in $lambda$. Using the formula quoted in mathoverflow.net/a/263658 as Corollary 7.21.4, we can rewrite $prodlimits_u in lambda left(t+c_uright) = s_lambdaleft(1^tright) prodlimits_u in lambda h_lambda, u = s_lambdaleft(1^tright) dfracn!f_lambda$ (by the hook-length formula). Thus, the left hand side of your question simplifies to $n! cdot sumlimits_lambda vdash n f_lambda s_lambdaleft(1^tright)$. The rest ...
– darij grinberg
4 hours ago
2
2
... is an easy application of RSK: The symmetric function $sumlimits_lambda vdash n f_lambda s_lambda$ is the generating function of all $n$-letter words over the alphabet $leftx_1,x_2,x_3,ldotsright$ (with the $f_lambda$ counting all possible Q-tableaux and the $s_lambda$ being the generating function of all possible P-tableaux) and thus equals $left(x_1+x_2+x_3+cdotsright)^n$ (a fact that also follows by applying the Pieri rule $n$ times); now substitute $1^t$ for the variables and you're done.
– darij grinberg
4 hours ago
... is an easy application of RSK: The symmetric function $sumlimits_lambda vdash n f_lambda s_lambda$ is the generating function of all $n$-letter words over the alphabet $leftx_1,x_2,x_3,ldotsright$ (with the $f_lambda$ counting all possible Q-tableaux and the $s_lambda$ being the generating function of all possible P-tableaux) and thus equals $left(x_1+x_2+x_3+cdotsright)^n$ (a fact that also follows by applying the Pieri rule $n$ times); now substitute $1^t$ for the variables and you're done.
– darij grinberg
4 hours ago
add a comment |Â
1 Answer
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Notice that $f_lambda$ is the number of standard Young tableaux of shape $lambda$, whereas $f_lambdafracprod_uin lambda (t+c_u)n!$ is the number of semistandard Young tableaux of shape $lambda$ and content $t$. It was mentioned in a previous question that the identity
$$sum_=nleft|textSSYT(lambda)right|left|textSYT(lambda)right|=t^n.$$
can either be proved bijectively from RSK (as Darij explains in the comments) or by appealing to Schur-Weyl duality (as David explains in the link), which gives an isomorphism of $S_ntimes GL(V)$ representations
$$sum_=n Sp_lambda boxtimes S_lambda(V) cong V^otimes n$$
where $V$ is a $t$ dimensional vector space, $S_lambda$ is the Schur functor, and $Sp_lambda$ the corresponding Specht module. Our identity follows by taking the dimension of both sides.
answered 4 hours ago
Gjergji Zaimi
59.7k3154296
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
Notice that $f_lambda$ is the number of standard Young tableaux of shape $lambda$, whereas $f_lambdafracprod_uin lambda (t+c_u)n!$ is the number of semistandard Young tableaux of shape $lambda$ and content $t$. It was mentioned in a previous question that the identity
$$sum_=nleft|textSSYT(lambda)right|left|textSYT(lambda)right|=t^n.$$
can either be proved bijectively from RSK (as Darij explains in the comments) or by appealing to Schur-Weyl duality (as David explains in the link), which gives an isomorphism of $S_ntimes GL(V)$ representations
$$sum_=n Sp_lambda boxtimes S_lambda(V) cong V^otimes n$$
where $V$ is a $t$ dimensional vector space, $S_lambda$ is the Schur functor, and $Sp_lambda$ the corresponding Specht module. Our identity follows by taking the dimension of both sides.
answered 4 hours ago
Gjergji Zaimi
59.7k3154296
add a comment |Â
up vote
4
down vote
Notice that $f_lambda$ is the number of standard Young tableaux of shape $lambda$, whereas $f_lambdafracprod_uin lambda (t+c_u)n!$ is the number of semistandard Young tableaux of shape $lambda$ and content $t$. It was mentioned in a previous question that the identity
$$sum_=nleft|textSSYT(lambda)right|left|textSYT(lambda)right|=t^n.$$
can either be proved bijectively from RSK (as Darij explains in the comments) or by appealing to Schur-Weyl duality (as David explains in the link), which gives an isomorphism of $S_ntimes GL(V)$ representations
$$sum_=n Sp_lambda boxtimes S_lambda(V) cong V^otimes n$$
where $V$ is a $t$ dimensional vector space, $S_lambda$ is the Schur functor, and $Sp_lambda$ the corresponding Specht module. Our identity follows by taking the dimension of both sides.
answered 4 hours ago
Gjergji Zaimi
59.7k3154296
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Notice that $f_lambda$ is the number of standard Young tableaux of shape $lambda$, whereas $f_lambdafracprod_uin lambda (t+c_u)n!$ is the number of semistandard Young tableaux of shape $lambda$ and content $t$. It was mentioned in a previous question that the identity
$$sum_=nleft|textSSYT(lambda)right|left|textSYT(lambda)right|=t^n.$$
can either be proved bijectively from RSK (as Darij explains in the comments) or by appealing to Schur-Weyl duality (as David explains in the link), which gives an isomorphism of $S_ntimes GL(V)$ representations
$$sum_=n Sp_lambda boxtimes S_lambda(V) cong V^otimes n$$
where $V$ is a $t$ dimensional vector space, $S_lambda$ is the Schur functor, and $Sp_lambda$ the corresponding Specht module. Our identity follows by taking the dimension of both sides.
answered 4 hours ago
Gjergji Zaimi
59.7k3154296
Notice that $f_lambda$ is the number of standard Young tableaux of shape $lambda$, whereas $f_lambdafracprod_uin lambda (t+c_u)n!$ is the number of semistandard Young tableaux of shape $lambda$ and content $t$. It was mentioned in a previous question that the identity
$$sum_=nleft|textSSYT(lambda)right|left|textSYT(lambda)right|=t^n.$$
can either be proved bijectively from RSK (as Darij explains in the comments) or by appealing to Schur-Weyl duality (as David explains in the link), which gives an isomorphism of $S_ntimes GL(V)$ representations
$$sum_=n Sp_lambda boxtimes S_lambda(V) cong V^otimes n$$
where $V$ is a $t$ dimensional vector space, $S_lambda$ is the Schur functor, and $Sp_lambda$ the corresponding Specht module. Our identity follows by taking the dimension of both sides.
answered 4 hours ago
Gjergji Zaimi
59.7k3154296
answered 4 hours ago
Gjergji Zaimi
59.7k3154296
answered 4 hours ago
Gjergji Zaimi
59.7k3154296
answered 4 hours ago
Gjergji Zaimi
59.7k3154296
59.7k3154296
add a comment |Â
add a comment |Â
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2
For any cell $u$ of $lambda$, let $h_lambda, u$ denote the hook length of $u$ in $lambda$. Using the formula quoted in mathoverflow.net/a/263658 as Corollary 7.21.4, we can rewrite $prodlimits_u in lambda left(t+c_uright) = s_lambdaleft(1^tright) prodlimits_u in lambda h_lambda, u = s_lambdaleft(1^tright) dfracn!f_lambda$ (by the hook-length formula). Thus, the left hand side of your question simplifies to $n! cdot sumlimits_lambda vdash n f_lambda s_lambdaleft(1^tright)$. The rest ...
– darij grinberg
4 hours ago
2
... is an easy application of RSK: The symmetric function $sumlimits_lambda vdash n f_lambda s_lambda$ is the generating function of all $n$-letter words over the alphabet $leftx_1,x_2,x_3,ldotsright$ (with the $f_lambda$ counting all possible Q-tableaux and the $s_lambda$ being the generating function of all possible P-tableaux) and thus equals $left(x_1+x_2+x_3+cdotsright)^n$ (a fact that also follows by applying the Pieri rule $n$ times); now substitute $1^t$ for the variables and you're done.
– darij grinberg
4 hours ago