On symmetric functions
Let $wedge^m(X)$ be the set of homogeneous symmetric function on X such that $X$ is finite i.e. $X={x_1,x_2,x_3, cdots, x_n}$. I am able to show that the dimension of $wedge^m(X)$ is equal to the partition of $m$ with at most $n$ parts.
I am interested in the conditions on the partition that will make
monomial, complete, and elementary symmetric function on partitions
that will make them bases.
Thus, the restriction on partitions that will make the monomial symmetric function $m_lambda$ a basis of $wedge^m(X)$ should be:
${m_lambda: lambda in Par(m; le n)}$ will be a bases for $wedge^m(X)$, with $m_lambda= sum_{alpha} x^{alpha}$, where $alpha$ runs over all distinct permutation of $lambda$.
For Complete symmetric functions $h_m$. Since $h_m= sum_{mu in Par(m; le n)} m_mu$, the set ${h_lambda: lambda in Par(m; le n;le m)}$ will be a bases for $wedge^m(X)$.
Lastly, for elementary symmetric functions $e_lambda$. I think that the set ${e_lambda: lambda^* in Par(m; le n)}$ will be a bases for $wedge^m(X)$ since $e_{lambda}= sum_{mu in Par(m; le n) }M_{lambda mu}m_{mu}$ and $M_{lambda mu}$ is only positive when $mu le lambda^*$
Are my answers correct?
combinatorics proof-verification symmetric-functions
asked Nov 29 '18 at 1:29
Jaynot
487515
add a comment |
Let $wedge^m(X)$ be the set of homogeneous symmetric function on X such that $X$ is finite i.e. $X={x_1,x_2,x_3, cdots, x_n}$. I am able to show that the dimension of $wedge^m(X)$ is equal to the partition of $m$ with at most $n$ parts.
I am interested in the conditions on the partition that will make
monomial, complete, and elementary symmetric function on partitions
that will make them bases.
Thus, the restriction on partitions that will make the monomial symmetric function $m_lambda$ a basis of $wedge^m(X)$ should be:
${m_lambda: lambda in Par(m; le n)}$ will be a bases for $wedge^m(X)$, with $m_lambda= sum_{alpha} x^{alpha}$, where $alpha$ runs over all distinct permutation of $lambda$.
For Complete symmetric functions $h_m$. Since $h_m= sum_{mu in Par(m; le n)} m_mu$, the set ${h_lambda: lambda in Par(m; le n;le m)}$ will be a bases for $wedge^m(X)$.
Lastly, for elementary symmetric functions $e_lambda$. I think that the set ${e_lambda: lambda^* in Par(m; le n)}$ will be a bases for $wedge^m(X)$ since $e_{lambda}= sum_{mu in Par(m; le n) }M_{lambda mu}m_{mu}$ and $M_{lambda mu}$ is only positive when $mu le lambda^*$
Are my answers correct?
combinatorics proof-verification symmetric-functions
asked Nov 29 '18 at 1:29
Jaynot
487515
Your notation is unclear and nonstandard. What is the role of $m$ -- are you considering a fixed degree component? $wedge^m$ usually stands for an $m$-th exterior power. See Remark 2.3.9 in arXiv:1409.8356v5 for a few basic results, but I suspect you know them already (sadly, I never got around to write the proofs up, but they are mostly simple variants of the proofs given for the case of infinitely many variables).
– darij grinberg
Dec 3 '18 at 19:15
add a comment |
Let $wedge^m(X)$ be the set of homogeneous symmetric function on X such that $X$ is finite i.e. $X={x_1,x_2,x_3, cdots, x_n}$. I am able to show that the dimension of $wedge^m(X)$ is equal to the partition of $m$ with at most $n$ parts.
I am interested in the conditions on the partition that will make
monomial, complete, and elementary symmetric function on partitions
that will make them bases.
Thus, the restriction on partitions that will make the monomial symmetric function $m_lambda$ a basis of $wedge^m(X)$ should be:
${m_lambda: lambda in Par(m; le n)}$ will be a bases for $wedge^m(X)$, with $m_lambda= sum_{alpha} x^{alpha}$, where $alpha$ runs over all distinct permutation of $lambda$.
For Complete symmetric functions $h_m$. Since $h_m= sum_{mu in Par(m; le n)} m_mu$, the set ${h_lambda: lambda in Par(m; le n;le m)}$ will be a bases for $wedge^m(X)$.
Lastly, for elementary symmetric functions $e_lambda$. I think that the set ${e_lambda: lambda^* in Par(m; le n)}$ will be a bases for $wedge^m(X)$ since $e_{lambda}= sum_{mu in Par(m; le n) }M_{lambda mu}m_{mu}$ and $M_{lambda mu}$ is only positive when $mu le lambda^*$
Are my answers correct?
combinatorics proof-verification symmetric-functions
asked Nov 29 '18 at 1:29
Jaynot
487515
Let $wedge^m(X)$ be the set of homogeneous symmetric function on X such that $X$ is finite i.e. $X={x_1,x_2,x_3, cdots, x_n}$. I am able to show that the dimension of $wedge^m(X)$ is equal to the partition of $m$ with at most $n$ parts.
I am interested in the conditions on the partition that will make
monomial, complete, and elementary symmetric function on partitions
that will make them bases.
Thus, the restriction on partitions that will make the monomial symmetric function $m_lambda$ a basis of $wedge^m(X)$ should be:
${m_lambda: lambda in Par(m; le n)}$ will be a bases for $wedge^m(X)$, with $m_lambda= sum_{alpha} x^{alpha}$, where $alpha$ runs over all distinct permutation of $lambda$.
For Complete symmetric functions $h_m$. Since $h_m= sum_{mu in Par(m; le n)} m_mu$, the set ${h_lambda: lambda in Par(m; le n;le m)}$ will be a bases for $wedge^m(X)$.
Lastly, for elementary symmetric functions $e_lambda$. I think that the set ${e_lambda: lambda^* in Par(m; le n)}$ will be a bases for $wedge^m(X)$ since $e_{lambda}= sum_{mu in Par(m; le n) }M_{lambda mu}m_{mu}$ and $M_{lambda mu}$ is only positive when $mu le lambda^*$
Are my answers correct?
combinatorics proof-verification symmetric-functions
combinatorics proof-verification symmetric-functions
asked Nov 29 '18 at 1:29
Jaynot
487515
asked Nov 29 '18 at 1:29
Jaynot
487515
asked Nov 29 '18 at 1:29
Jaynot
487515
asked Nov 29 '18 at 1:29
Jaynot
487515
asked Nov 29 '18 at 1:29
Jaynot
487515
487515
Your notation is unclear and nonstandard. What is the role of $m$ -- are you considering a fixed degree component? $wedge^m$ usually stands for an $m$-th exterior power. See Remark 2.3.9 in arXiv:1409.8356v5 for a few basic results, but I suspect you know them already (sadly, I never got around to write the proofs up, but they are mostly simple variants of the proofs given for the case of infinitely many variables).
– darij grinberg
Dec 3 '18 at 19:15
add a comment |
Your notation is unclear and nonstandard. What is the role of $m$ -- are you considering a fixed degree component? $wedge^m$ usually stands for an $m$-th exterior power. See Remark 2.3.9 in arXiv:1409.8356v5 for a few basic results, but I suspect you know them already (sadly, I never got around to write the proofs up, but they are mostly simple variants of the proofs given for the case of infinitely many variables).
– darij grinberg
Dec 3 '18 at 19:15
Your notation is unclear and nonstandard. What is the role of $m$ -- are you considering a fixed degree component? $wedge^m$ usually stands for an $m$-th exterior power. See Remark 2.3.9 in arXiv:1409.8356v5 for a few basic results, but I suspect you know them already (sadly, I never got around to write the proofs up, but they are mostly simple variants of the proofs given for the case of infinitely many variables).
– darij grinberg
Dec 3 '18 at 19:15
Your notation is unclear and nonstandard. What is the role of $m$ -- are you considering a fixed degree component? $wedge^m$ usually stands for an $m$-th exterior power. See Remark 2.3.9 in arXiv:1409.8356v5 for a few basic results, but I suspect you know them already (sadly, I never got around to write the proofs up, but they are mostly simple variants of the proofs given for the case of infinitely many variables).
– darij grinberg
Dec 3 '18 at 19:15
add a comment |
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Your notation is unclear and nonstandard. What is the role of $m$ -- are you considering a fixed degree component? $wedge^m$ usually stands for an $m$-th exterior power. See Remark 2.3.9 in arXiv:1409.8356v5 for a few basic results, but I suspect you know them already (sadly, I never got around to write the proofs up, but they are mostly simple variants of the proofs given for the case of infinitely many variables).
– darij grinberg
Dec 3 '18 at 19:15