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Sebastien Palcoux
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First recall the Jordan-Hölder theorem for groups:

Theorem (Jordan-Hölder): Let $G$ be a group, and let $$ G=G_1 \supset G_2 \supset ... \supset G_r = \{ e \} $$$$ G=G_1 \supset G_2 \supset \dots \supset G_r = \{ e \} $$ be a normal tower such that each group $G_i /G_{i+1}$ is simple, and $G_i \neq G_{i+1}$ for $0<i<r$. Then any other normal tower of $G$ having the same properties is equivalent to this one (i.e. the sequence of factor groups in our two towers are the same up to isomorphisms, and a permutation of the indices).

This paper of Kodiyalam-Landau-Sunder contains the basic definition of a planar algebra, of a group (subfactor) planar algebra, the definitions (p16) of planar algebra morphism, planar ideal and quotient.
We note that the planar ideals are precisely the kernel of the planar algebra morphisms !
We call a planar algebra simple if it has no non-trivial planar ideal.

Question 1 : Do the planar ideals of a group planar algebra correspond to the normal subgroups ?

If yes, a group planar algebra is simple iff the group is simple.
If no, how adapt the concept of planar ideal for having a positive answer ?

Question 2 : Let $\mathcal{P}$ be a planar algebra, and let $$ \mathcal{P}=\mathcal{J}_1 \supset \mathcal{J}_2 \supset ... \supset \mathcal{J}_r = (0) $$$$ \mathcal{P}=\mathcal{J}_1 \supset \mathcal{J}_2 \supset \dots \supset \mathcal{J}_r = (0) $$ be an ideal tower such that each quotient $\mathcal{J}_i /\mathcal{J}_{i+1}$ is simple, and $\mathcal{J}_i \neq \mathcal{J}_{i+1}$ for $0<i<r$. Then any other ideal tower of $\mathcal{P}$ having the same properties is equivalent to this one (i.e. the sequence of quotient planar algebras in our two towers are the same up to isomorphisms, and a permutation of the indices) ?

First recall the Jordan-Hölder theorem for groups:

Theorem (Jordan-Hölder): Let $G$ be a group, and let $$ G=G_1 \supset G_2 \supset ... \supset G_r = \{ e \} $$ be a normal tower such that each group $G_i /G_{i+1}$ is simple, and $G_i \neq G_{i+1}$ for $0<i<r$. Then any other normal tower of $G$ having the same properties is equivalent to this one (i.e. the sequence of factor groups in our two towers are the same up to isomorphisms, and a permutation of the indices).

This paper of Kodiyalam-Landau-Sunder contains the basic definition of a planar algebra, of a group (subfactor) planar algebra, the definitions (p16) of planar algebra morphism, planar ideal and quotient.
We note that the planar ideals are precisely the kernel of the planar algebra morphisms !
We call a planar algebra simple if it has no non-trivial planar ideal.

Question 1 : Do the planar ideals of a group planar algebra correspond to the normal subgroups ?

If yes, a group planar algebra is simple iff the group is simple.
If no, how adapt the concept of planar ideal for having a positive answer ?

Question 2 : Let $\mathcal{P}$ be a planar algebra, and let $$ \mathcal{P}=\mathcal{J}_1 \supset \mathcal{J}_2 \supset ... \supset \mathcal{J}_r = (0) $$ be an ideal tower such that each quotient $\mathcal{J}_i /\mathcal{J}_{i+1}$ is simple, and $\mathcal{J}_i \neq \mathcal{J}_{i+1}$ for $0<i<r$. Then any other ideal tower of $\mathcal{P}$ having the same properties is equivalent to this one (i.e. the sequence of quotient planar algebras in our two towers are the same up to isomorphisms, and a permutation of the indices) ?

First recall the Jordan-Hölder theorem for groups:

Theorem (Jordan-Hölder): Let $G$ be a group, and let $$ G=G_1 \supset G_2 \supset \dots \supset G_r = \{ e \} $$ be a normal tower such that each group $G_i /G_{i+1}$ is simple, and $G_i \neq G_{i+1}$ for $0<i<r$. Then any other normal tower of $G$ having the same properties is equivalent to this one (i.e. the sequence of factor groups in our two towers are the same up to isomorphisms, and a permutation of the indices).

This paper of Kodiyalam-Landau-Sunder contains the basic definition of a planar algebra, of a group (subfactor) planar algebra, the definitions (p16) of planar algebra morphism, planar ideal and quotient.
We note that the planar ideals are precisely the kernel of the planar algebra morphisms !
We call a planar algebra simple if it has no non-trivial planar ideal.

Question 1 : Do the planar ideals of a group planar algebra correspond to the normal subgroups ?

If yes, a group planar algebra is simple iff the group is simple.
If no, how adapt the concept of planar ideal for having a positive answer ?

Question 2 : Let $\mathcal{P}$ be a planar algebra, and let $$ \mathcal{P}=\mathcal{J}_1 \supset \mathcal{J}_2 \supset \dots \supset \mathcal{J}_r = (0) $$ be an ideal tower such that each quotient $\mathcal{J}_i /\mathcal{J}_{i+1}$ is simple, and $\mathcal{J}_i \neq \mathcal{J}_{i+1}$ for $0<i<r$. Then any other ideal tower of $\mathcal{P}$ having the same properties is equivalent to this one (i.e. the sequence of quotient planar algebras in our two towers are the same up to isomorphisms, and a permutation of the indices) ?

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Sebastien Palcoux
  • 27k
  • 5
  • 74
  • 186

Jordan-Hölder theorem for planar algebras?

First recall the Jordan-Hölder theorem for groups:

Theorem (Jordan-Hölder): Let $G$ be a group, and let $$ G=G_1 \supset G_2 \supset ... \supset G_r = \{ e \} $$ be a normal tower such that each group $G_i /G_{i+1}$ is simple, and $G_i \neq G_{i+1}$ for $0<i<r$. Then any other normal tower of $G$ having the same properties is equivalent to this one (i.e. the sequence of factor groups in our two towers are the same up to isomorphisms, and a permutation of the indices).

This paper of Kodiyalam-Landau-Sunder contains the basic definition of a planar algebra, of a group (subfactor) planar algebra, the definitions (p16) of planar algebra morphism, planar ideal and quotient.
We note that the planar ideals are precisely the kernel of the planar algebra morphisms !
We call a planar algebra simple if it has no non-trivial planar ideal.

Question 1 : Do the planar ideals of a group planar algebra correspond to the normal subgroups ?

If yes, a group planar algebra is simple iff the group is simple.
If no, how adapt the concept of planar ideal for having a positive answer ?

Question 2 : Let $\mathcal{P}$ be a planar algebra, and let $$ \mathcal{P}=\mathcal{J}_1 \supset \mathcal{J}_2 \supset ... \supset \mathcal{J}_r = (0) $$ be an ideal tower such that each quotient $\mathcal{J}_i /\mathcal{J}_{i+1}$ is simple, and $\mathcal{J}_i \neq \mathcal{J}_{i+1}$ for $0<i<r$. Then any other ideal tower of $\mathcal{P}$ having the same properties is equivalent to this one (i.e. the sequence of quotient planar algebras in our two towers are the same up to isomorphisms, and a permutation of the indices) ?