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This post was inspired by an exchange with the indian woman mathematician Ajit Iqbal Singh.

The chinese remainder theorem can be stated as follows:
Let $n_1, \dots, n_r \ge 2$ be positive integers such that $n_i \wedge n_j = 1$ iff $i \neq j$, then the system of equations $$ x \equiv a_i [n_i] \text{, for } i=1, \dots , r$$ admits a solution $x$ which is unique modulo $\prod_i n_i$.

There is a natural way to generalize this statement to the cyclic subfactor planar algebras, as follows
(I warn the reader who don't know the subfactor planar algebra theory, that I've reformulated the statement in the case of finite index inclusions of groups, just after).

Let $P$ be a finite index irreducible subfactor planar algebra, and assume that $P$ is cyclic, i.e. admits a distributive biprojections lattice. The statement is the following:
Let $b_1, \dots , b_r \not \in \{ e_1, id \}$ be a subset of biprojections satisfying $ b_i \vee b_j = id$ iff $i \neq j$, then the system: $$ b_i * x * b_i \sim b_i * a_i * b_i \text{, for } i=1, \dots ,r $$ (with $a_i, x$ positive operators) admits a solution $x$ which is unique (up to $\sim$) as $B * x * B$, with $B = \bigwedge_i b_i$.

Now in the group framework: let $(H \subset G)$ be a finite index inclusion of groups, such that the interval lattice $[H,G]$ is distributive, then the statement is the following:
Let $K_1, \dots , K_r$ be strict intermediate subgroups (i.e. $H \subsetneq K_i \subsetneq G$, $\forall i$), such that $\langle K_i , K_j \rangle = G$ iff $i \neq j$, then the system of double-coset equations: $$ K_ixK_i = K_ig_iK_i \text{, for } i=1, \dots ,r $$ admits a solution $x$ which is unique as $KxK$ with $K = \bigcap_i K_i$.

Remark: This group statement above with $G = \mathbb{Z}$, $K_i = n_i\mathbb{Z}$ and $H = K = \prod_i n_i \mathbb{Z}$ is equivalent to the classical statement of the chinese remainder theorem.

Question: Is the chinese remainder statement true for the cyclic subfactor planar algebras?

Remark: A counter-example for an inclusion of finite groups gives a counter-example for the subfactors planar algebras. I will edit a ckeck by GAP soon.

Remark: The distributivity assumption is very useful for generalizing the beginning of the proof here.

Motivation: Try to better understand in what sense the cyclic subfactor theory is a quantum arithmetic.

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    $\begingroup$ @DanielLoughran: when a question has been answered, the usual process is not to close it, but to accept the answer. I'm sorry, I just forgot to accept my answer, now it's done, thanks for your comment! $\endgroup$ – Sebastien Palcoux Sep 28 '15 at 9:11
  • $\begingroup$ The answer below shows that this generalization is not working. Nevertheless, there is an other possible generalization (which could works), by using right cosets $Kg$ instead of double cosets $KgK$ (and at the planar algebra level, $b∗u$ instead of $b∗u∗b$). $\endgroup$ – Sebastien Palcoux Oct 15 '16 at 2:40
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No, we show above an inclusion of finite groups counter-example.

First of all, the statement for an inclusion of finite groups $(H \subset G)$ can be reformulate as follows:

Let $K_1, \dots , K_r$ be strict intermediate subgroups (i.e. $H \subsetneq K_i \subsetneq G$, $\forall i$), such that $\langle K_i , K_j \rangle = G$ iff $i \neq j$ (note that $r$ is assumed $\ge 2$). Without lose of generality, we can assume that $K:=\bigcap_{i}K_i = H$ (otherwise we replace $H$ by $K$). Let $\{ K_i g_{i,1} K_i, \dots , K_i g_{i,r_i} K_i \}$ be the set of all the double cosets for $K_i$ in $G$.
Then the Chinese remainder statement is equivalent to: $$\forall g_1, g_2, \dots , g_r \in G \ , \ \exists g \in G \ ; \ \bigcap_{i}(K_i g_i K_i) \subset HgH$$

The first counter-example having the interval lattice $[H,G]$ distributive is the following:

gap> G:=TransitiveGroup(10,4);
1/2[F(5)]2
gap> H:=Stabilizer(G,1);
Group([ (2,10)(3,9)(4,8)(5,7) ])

First the lattice is distributive because:

gap> IntermediateSubgroups(G,H).inclusions;
[ [ 0, 1 ], [ 0, 2 ], [ 1, 3 ], [ 2, 3 ] ]

Let $K_1$ and $K_2$ the two strict intermediate subgroups:

gap> K1:=IntermediateSubgroups(G,H).subgroups[1];
Group([ (2,10)(3,9)(4,8)(5,7), (1,6)(2,9,10,3)(4,5,8,7) ])
gap> K2:=IntermediateSubgroups(G,H).subgroups[2];
Group([ (2,10)(3,9)(4,8)(5,7), (1,9,7,5,3)(2,10,8,6,4) ])

Obviously $\langle K_1, K_2 \rangle = G$ and $K_1 \cap K_2 = H$.
Let $D$ be the intersection $(K_1 g_{1,2} K_1) \cap (K_2 g_{2,2} K_2)$:

gap> D:=Intersection(DoubleCosets(G,K1,K1)[2], DoubleCosets(G,K2,K2)[2]);
[ (1,2,5,4)(3,8)(6,7,10,9), (1,2,9,8)(3,6,7,4)(5,10), (1,4,5,2)(3,8)(6,9,10,7), (1,4,3,10)(2,7)(5,6,9,8), (1,8,9,2)(3,4,7,6)(5,10), (1,8,7,10)(2,5,6,3)(4,9), 
  (1,10,7,8)(2,3,6,5)(4,9), (1,10,3,4)(2,7)(5,8,9,6) ]

Then $\forall g \in G$, $D \not \subset HgH$:

gap> List(DoubleCosets(G,H,H), S->IsSubset(S,D));
[ false, false, false, false, false, false ]

So $(H \subset G)$ is a counter-example with a distributive lattice.

Remark: up to equivalent there are $166$ index $\le 10$ inclusions of finite groups:

gap> Sum(List([1..10], d->NrTransitiveGroups(d)));
166

Among them (with distributive lattice or not), only $6$ do not check the Chinese remainder statement:

gap> G:=TransitiveGroup(d,r);
gap> H:=Stabilizer(G,1);

with $(d,r) = ( 6, 2 ), ( 8, 3 ), ( 8, 4 ), ( 9, 5 ), ( 10, 2 ), ( 10, 4 )$.
(The last one, i.e. $(d,r) = ( 10, 4 )$, is the first distributive lattice counter-example)

It follows that in this case, more than $96 \%$ check the Chinese remainder statement!

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