1
$\begingroup$

Let $(N \subset M)$ be a finite index irreducible subfactor, and $P = P(N \subset M)$ its planar algebra.

We can see $M$ as an algebraic $N$-$N$-bimodule, it decomposes into irreducible algebraic $N$-$N$-bimodules: $$M = \bigoplus_{i \in I} V_i \otimes B_i$$ with $(B_i)_{i \in I}$ the irreducible algebraic sub-$N$-$N$-bimodules of $M$, and $V_i$ the multiplicity space.

Now $P_{2,+} = N' \cap M_1$ is a finite dimensional ${\rm C}^*$-algebra and by the double characterization of the principal graph, we know that:
$$ P_{2,+} \simeq \bigoplus_{i \in I} End(V_i)$$

The one-to-one correspondence between the projections of $p \in P_{2,+}$ and the algebraic sub-$N$-$N$-bimodules $X_p$ of $M$, comes from the one-to-one correspondence between the projections of $P_{2,+}$ and the subspaces of $V=\oplus_{i \in I}V_i$ on one hand (through image and range projection), and the one-to-one correspondence between the subspaces of $V$ and the algebraic sub-$N$-$N$-bimodules of $M$ on the other hand (by definition of the decomposition of $M$). Moreover this correspondence is an isomorphism of poset, which means $$p \le q \Rightarrow X_p \subseteq X_q$$ The set of biprojections is a subposet of the set of projections and the set of intermediate subfactors is also a subposet of the set of algebraic sub-$N$-$N$-bimodules of $M$ and using this paper of D. Bisch, $$p \text{ is a biprojection iff } X_p \text{ is an intermediate subfactor. } $$

Corollary: The biprojection $\langle p \rangle$ generated by the projection $p \in P_{2,+}$ corresponds to the intermediate subfactor $\langle X_p \rangle$ generated by the algebraic sub-$N$-$N$-bimodules $X_p$ of $M$, which can be reformulated by:
$$X_{\langle p \rangle} = \langle X_{ p } \rangle$$ proof: First $\langle p \rangle$ is the minimal biprojection $b \ge p$, and next $\langle X_p \rangle$ is the minimal von Neumann algebra containing $X_p$ in $M$, but $N \subset \langle X_p \rangle$ because $X_p$ is a $N$-$N$-bimodule, and so $\langle X_p \rangle$ is an intermediate subfactor by irreducibility of $(N \subset M)$. The result follows by the poset isomorphism respecting the one-to-one correspondence between biprojections and intermediate subfactors $\square$

Questions:

  • Is there an explicit formula for $X_p$ using $M$, $p$ (and $e^M_N$)?
  • Is it true that $X_p^* = X_{\overline{p}}$ (with $x^*$ the dual of $x \in M$ and $\overline{p}$ the contragredient of $p \in P_{2,+}$)?
  • Is it true that $Vect_{\mathbb{C}}(X_p X_q) = X_{S(p * q)}$ (with $X_p X_q$ the product of $X_p$ and $X_q$ in $M$, $p*q$ the coproduct of $p$ and $q$, and $S(u)$ the support projection of $u \in P_{2,+}$)?

Marcel has pointed out in comment that for $p$ a biprojection $X_p = \{p \}' \cap M$ (theorem 3.2. p213 of this paper).
Note that if $p$ is not a biprojection then $X_p$ has not an algebra (see the algebra objects of a fusion category).

$\endgroup$
12
  • $\begingroup$ That there is a one-to-one correspondence readily follows, for the rest I don't completely understand the notation but to me it seems like an exercise in planar algebras using the results of Jones on algebraic bi modules. $\endgroup$ Jun 14, 2015 at 18:05
  • $\begingroup$ @MarcelBischoff: Which notations you don't understand? Which paper of V. Jones contains the result on algebraic bimodules? $\endgroup$ Jun 15, 2015 at 3:17
  • $\begingroup$ @MarcelBischoff: about the one-to-one correspondence, $M$ decompose into irreducible $N$-$N$-bimodules $(B_i)_i$: $_NM_N = \bigoplus_i V_i \otimes B_i$ (with $V_i$ the multiplicity space). And the $2$-box space $P_{2,+} = N' \cap M_1 = \bigoplus_i End(V_i)$ (with the same $V_i$ as above). I know it's true (in good cases) using the double characterization of the principal graph, but do you see a direct proof for that? $\endgroup$ Jun 16, 2015 at 3:40
  • $\begingroup$ @MarcelBischoff: for any projection $p \in P_{2,+}$ then $pMp$ is a $N$-$N$-bimodule because $p \in N'$, but $p \in N' \cap M_1$ and $pMp \not\subset M$, because (assuming irreducible) $N' \cap M = \mathbb{C}$, so $p \not \in M$ in general. What the relation between $X_p (\subset M$, see the post) and $pMp$? Does $pMp = X_pe^M_N$? $\endgroup$ Jun 16, 2015 at 4:10
  • 1
    $\begingroup$ I mean, in Section 3 projecteuclid.org/download/pdf_1/euclid.pjm/1102622455 it is shown that $p\in \mathrm{IS}(N,M)\subset N'\cap M_1\mapsto \{p\}'\cap M$ is a one-to-one correspondence. $\endgroup$ Jun 17, 2015 at 21:10

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.