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First recall the Jordan-Hölder theorem for groups:

Theorem (Jordan-Hölder): Let $G$ be a group, and let $$ G=G_1 \supset G_2 \supset \dots \supset G_r = \{ e \} $$ 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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The Jordan-Holder theorem is true for finite modular lattices. Is the lattice of ideals in your planar algebra a finite modular lattice? –  Benjamin Steinberg Jan 29 at 19:25
    
@BenjaminSteinberg : I've developed an answer to your question (for subfactors instead of planar algebras) in the post : Jordan-Hölder theorem for subfactors –  Sébastien Palcoux Jan 31 at 19:02

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up vote 3 down vote accepted

The answer is no. In fact, the group planar algebra has no non-trivial planar ideals.

If $P_\bullet$ is a spherical planar algebra with non-zero modulus, and if $P_{0,\pm}$ are one dimensional, then every planar ideal is contained in the planar ideal $N_\bullet$ of negligible elements, i.e., those $x\in P_{n,\pm}$ for which tr$(xy)=0$ for all $y\in P_{n,\pm}$. (See Proposition 3.5 of the extended Haagerup paper of Bigelow-Morrison-Peters-Snyder MR2979509.)

Now the group planar algebra is a subfactor planar algebra. In particular, the negligible ideal is the zero ideal, since the form $\langle x,y\rangle = \text{tr}(y^*x)$ is positive definite.

I'll think about the followup question.

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I see, thank you Dave. So every subfactor planar algebra is simple ! Do you know the concepts, in the planar algebra framework, generalizing the concepts of subgroup and normal subgroup ? –  Sébastien Palcoux Jan 29 at 20:24

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