12
$\begingroup$

Edit: I found a serious flaw in the question and my answer, and I had to change a lot. The basic question is still there, but the details are a lot different.

Premodular categories

In braided spherical fusion (=premodular) categories $\mathcal{C}$ with braiding $c_{-,-}$, we can define the double braiding: $$\gamma_{A,X}\colon A \otimes X \to A \otimes X$$ $$\gamma_{A,X} := c_{X,A} \circ c_{A,X}$$

Now we can define encirclings, where we braid an object around another and close the loop on the right hand side:

$$\Delta_{A,X}\colon A \to A$$ $$\Delta_{A,X} = \operatorname{tr}_X(\gamma_{A,X})$$

(The trace is defined via the pivotal structure.)

The "killing" lemma is quite well known. Assume that $A$ is simple. Then:

$$\Omega := \bigoplus_{X \text{ simple}} d(X) X$$ $$\Delta_{A,\Omega} = \left\{ \begin{matrix} d(\Omega) \cdot 1_A && \qquad A \text{ is in the symmetric centre} \\ 0 && \qquad \text{otherwise} \end{matrix}\right.$$ $A$ is in the symmetric centre if braids trivially with every object. One says that $A$ is killed if it's not in the symmetric centre, hence the name of the lemma.

$G$-crossed braided fusion categories

Now consider $G$-crossed braided fusion categories. If you think of fusion categories as some sort of categorified group (algebra), then a $G$-crossed braided fusion category is a categorified crossed module. It is, amongst other things, a $G$-graded fusion category:

$$\mathcal{C} = \bigoplus_{g \in G} \mathcal{C}_g$$

It also has a monoidal $G$-action:

$${}^g(-)\colon \mathcal{C}_h \to \mathcal{C}_{ghg^{-1}} \quad \forall g, h \in G$$

And it has a crossed braiding:

$$ c_{A,X}\colon A \otimes X \to {}^gX \otimes A\quad \forall A \in \mathcal{C}_g$$

This means that the double braiding is now not an endomorphism anymore:

$$ \gamma_{A,X}\colon A \otimes X \to {}^{ghg^{-1}}A \otimes {}^g X \quad\forall A \in \mathcal{C}_g, X \in \mathcal{C}_h$$

You can only take the trace on the right hand side, and thus define encircling, if $X \cong {}^gX$. The only canonical way I can think of is to demand that the grade of $A$ is trivial, i.e. $g = e$, since there is a coherence isomorphism $\epsilon_X\colon {}^eX \to X$. Then we can define:

$$\Omega_h := \bigoplus_{X \text{ simple } \in \mathcal{C}_h} d(X) X$$ $$\Delta_{A, \Omega_h} = \operatorname{tr}_{\Omega_h}\left((1_A \otimes \epsilon_{\Omega_h}) \circ \gamma_{A,\Omega_h} \right) \qquad \forall A \in \mathcal{C}_e$$

But now source and target aren't equal: $$\Delta_{A,\Omega_h}\colon A \to {}^hA$$

In some cases we have $A \cong {}^hA$, and then one can ask whether the endomorphism is 0 or not, which leads me to my question:

Is there a similar lemma to the standard killing lemma in the $G$-crossed case?

$\endgroup$
2
$\begingroup$

Edit: I used to believe that there is a possible generalisation stemming from work of Altschüler and Bruguières. (See Appendix C in Drinfeld, Gelaki, Nikshych, Ostrik - On braided fusion categories I). But that was based on a flawed assumption. (I didn't realise that $A$ must be in the trivial degree for it to work.) With the correct assumptions, the following can be said:

Sliding lemma

Like in braided spherical fusion categories, the sliding lemma is closely related to the killing lemma.

Lemma

Let $Y \in \mathcal{C}_g, A \in \mathcal{C}_e$. Up to coherences in the $G$-crossed fusion category, we have:

$$\Delta_{A,\Omega_h} \otimes 1_Y = c_{Y,A} \circ (\epsilon_Y \otimes \Delta_{A,\Omega_{hg^{-1}}}) \circ c_{A,Y}$$

In words, we can slide the $Y$-strand over the $\Omega_h$-encircling so that it links with the $A$-strand. If you draw a picture, it will become clear.

Killing lemma

Recall that the trivial degree $\mathcal{C}_e$ is a braided spherical fusion category, so the classical killing lemma applies to $\Delta_{A,\Omega_e}$.

Lemma

Let $A \in \mathcal{C}_e$ be a simple object, and $\mathcal{C}_h \not\simeq 0$. Then:

$$\Delta_{A,\Omega_h}\colon A \to {}^hA$$ is nonzero, and thus an isomorphism, iff $A$ is in the symmetric centre of $\mathcal{C}_e$.

Proof

The statement is equivalent to the statement with $- \otimes 1_{\Omega_{h^{-1}}}$ applied to it (recall that $\mathcal{C}_h \not\simeq 0 \implies \Omega_h \neq 0 \implies \Omega_{h^{-1}} \cong \Omega_h^* \neq 0)$. By the sliding lemma, the left hand side is then equal to $c_{\Omega_{h^{-1}}, A} \circ (\epsilon_{\Omega_{h^{-1}}} \otimes \Delta_{A,\Omega_e}) \circ c_{A, \Omega_{h^{-1}}}$. Apply the killing lemma for $\Delta_{A,\Omega_e}$, yielding:

$$\Delta_{A,\Omega_h} \otimes 1_{\Omega_{h^{-1}}} = \left\{ \begin{matrix} d(\Omega_e) \cdot \gamma_{A,\Omega_{h^{-1}}} && \qquad A \text{ is in the symmetric centre of } \mathcal{C}_e \\ 0 && \qquad \text{otherwise} \end{matrix}\right.$$

Observation

This seems to prove that if all degrees of $\mathcal{C}$ are nonzero (that is, the grading is "faithful"), all objects in the symmetric centre of $\mathcal{C}_e$ are automatically equivariant, i.e. possess a family of isomorphisms $A \cong {}^gA, \forall g \in G$. Surely, that result must have been known before?

Edit This has been commented on briefly by Etingof, Nikshych, Ostrik and Meir in Fusion categories and homotopy theory, (39).

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.