five above categories. Mathematica files containing these fusion rings and
twists can be found attached to the arXiv submission of this paper. We compute
the group of fusion ring automorphisms which preserve the twists. We will
refer to these groups as the braided fusion ring symmetries.

For
$\mathcal{C}(\mathfrak{sl}_{4},8)^{0}_{\operatorname{Rep}(\mathbb{Z}_{2})}$ we
see that every braided fusion ring symmetry is non-exceptional, thus
$\operatorname{EqBr}(\mathcal{C}(\mathfrak{sl}_{4},8)^{0}_{\operatorname{Rep}(\mathbb{Z}_{2})})=D_{2}$
by Theorem 3.2.

For the case of
$\mathcal{C}(\mathfrak{sl}_{2},16)^{0}_{\operatorname{Rep}(\mathbb{Z}_{2})}$
we have that the braided fusion rings symmetries form a group isomorphic to
$S_{3}$, with generators

$(8\Lambda_{1},\chi_{0})\leftrightarrow(8\Lambda_{1},\chi_{1}),$

and

$\Omega\leftrightarrow(8\Lambda_{1},\chi_{0}).$

From Theorem 3.2 we know that the first generator is realised as a braided
auto-equivalence of
$\mathcal{C}(\mathfrak{sl}_{2},16)^{0}_{\operatorname{Rep}(\mathbb{Z}_{2})}$.
Thus
$\operatorname{EqBr}(\mathcal{C}(\mathfrak{sl}_{2},16)^{0}_{\operatorname{Rep}(\mathbb{Z}_{2})})$
is an intermediate subgroup of $\mathbb{Z}_{2}$ and $S_{3}$. There are only
two such intermediate subgroups which are the end points.

For the case of
$\mathcal{C}(\mathfrak{sl}_{3},9)^{0}_{\operatorname{Rep}(\mathbb{Z}_{3})}$ we
have that the braided fusion rings symmetries form a group isomorphic to
$\mathbb{Z}_{2}\times S_{4}$, with generators

$(3\Lambda_{1},1)\leftrightarrow(3\Lambda_{2},1)\quad,\quad(3\Lambda_{1}+3\Lambda_{2},\chi_{1})\leftrightarrow(3\Lambda_{1}+3\Lambda_{2},\chi_{2}),$

and

$(3\Lambda_{1}+3\Lambda_{2},\chi_{0})\mapsto(3\Lambda_{1}+3\Lambda_{2},\chi_{1})\mapsto(3\Lambda_{1}+3\Lambda_{2},\chi_{2}),$

and

$\Omega\mapsto(3\Lambda_{1}+3\Lambda_{2},\chi_{0})\mapsto(3\Lambda_{1}+3\Lambda_{2},\chi_{1})\mapsto(3\Lambda_{1}+3\Lambda_{2},\chi_{2}).$

From Theorem 3.2 we know that the first two generators are realised as braided
auto-equivalences of
$\mathcal{C}(\mathfrak{sl}_{3},9)^{0}_{\operatorname{Rep}(\mathbb{Z}_{3})}$,
and form a group isomorphic to $D_{3}$. Further, this theorem tells us that
any braided auto-equivalence which fixes $\Omega$ must be in the subgroup
generated by the first two generators. Thus
$\operatorname{EqBr}(\mathcal{C}(\mathfrak{sl}_{3},9)^{0}_{\operatorname{Rep}(\mathbb{Z}_{3})})$
is an intermediate subgroup of $D_{3}\subset\mathbb{Z}_{2}\times S_{4}$ with
the property that if $\Omega$ is fixed by an auto-equivalence, then this auto-
equivalence lives in the $D_{3}$ subgroup. With this knowledge, we can study
the intermediate subgroups of $D_{3}\subset\mathbb{Z}_{2}\times S_{4}$ to see
that there are only two such subgroups with this property. These are $D_{3}$
with the first two generators, and $S_{4}$ with the first two generators, and
the new generator

$\Omega\leftrightarrow(3\Lambda_{1}+3\Lambda_{2},\chi_{0})\quad,\quad(3\Lambda_{1}+3\Lambda_{2},\chi_{1})\leftrightarrow(3\Lambda_{1}+3\Lambda_{2},\chi_{2}).$

Thus
$\operatorname{EqBr}(\mathcal{C}(\mathfrak{sl}_{3},9)^{0}_{\operatorname{Rep}(\mathbb{Z}_{3})})$
is isomorphic to either $D_{3}$ or $S_{4}$.

The remaining two cases fall to the same argument. For
$\mathcal{C}(\mathfrak{sl}_{4},8)^{0}_{\operatorname{Rep}(\mathbb{Z}_{4})}$
the group of braided fusion ring symmetries is $\mathbb{Z}_{2}\times S_{4}$.
We have that the generators

$(4\Lambda_{2},\chi_{0})\leftrightarrow(4\Lambda_{2},\chi_{1})\quad,\quad(2\Lambda_{1}+2\Lambda_{2}+2\Lambda_{3},\chi_{i})\mapsto(2\Lambda_{1}+2\Lambda_{2}+2\Lambda_{3},\chi_{i+1})$

and

$(2\Lambda_{1}+\Lambda_{2},1)\leftrightarrow(\Lambda_{2}+2\Lambda_{3},1)\quad,\quad(2\Lambda_{1}+2\Lambda_{2}+2\Lambda_{3},\chi_{1})\leftrightarrow(2\Lambda_{1}+2\Lambda_{2}+2\Lambda_{3},\chi_{3})$

form a $D_{4}$ subgroup of
$\operatorname{EqBr}(\mathcal{C}(\mathfrak{sl}_{4},8)^{0}_{\operatorname{Rep}(\mathbb{Z}_{4})})$,
and that any braided auto-equivalence which fixes $\Omega$ must be in this
subgroup. Analysing the subgroup structure between $D_{4}$ and
$\mathbb{Z}_{2}\times S_{4}$ shows that at most there can be one more
generator

$\Omega\leftrightarrow(4\Lambda_{2},\chi_{0})\quad,\quad(2\Lambda_{1}+\Lambda_{2},1)\mapsto(2\Lambda_{1}+2\Lambda_{2}+2\Lambda_{3},\chi_{0})\mapsto(\Lambda_{2}+2\Lambda_{3},1)\mapsto(2\Lambda_{1}+2\Lambda_{2}+2\Lambda_{3},\chi_{2}),$

in the braided auto-equivalence group, which would form a group isomorphic to
$S_{4}$. Thus
$\operatorname{EqBr}(\mathcal{C}(\mathfrak{sl}_{4},8)^{0}_{\operatorname{Rep}(\mathbb{Z}_{4})})$
is either isomorphic to $D_{4}$ or $S_{4}$.

Finally for the case
$\mathcal{C}(\mathfrak{sl}_{5},5)^{0}_{\operatorname{Rep}(\mathbb{Z}_{5})}$ we
have that the group of braided fusion symmetries is $S_{6}$. We have that the
generators

$(\Lambda_{1}+\Lambda_{2}+\Lambda_{3}+\Lambda_{4},\chi_{i})\mapsto(\Lambda_{1}+\Lambda_{2}+\Lambda_{3}+\Lambda_{4},\chi_{i+1})$

and

$(\Lambda_{1}+\Lambda_{2}+\Lambda_{3}+\Lambda_{4},\chi_{i})\mapsto(\Lambda_{1}+\Lambda_{2}+\Lambda_{3}+\Lambda_{4},\chi_{-i})$

form a $D_{5}$ subgroup of
$\operatorname{EqBr}(\mathcal{C}(\mathfrak{sl}_{5},5)^{0}_{\operatorname{Rep}(\mathbb{Z}_{5})})$,
and that any braided auto-equivalence which fixes $\Omega$ must be in this
subgroup. Analysing the subgroup structure between $D_{5}$ and $S_{6}$ shows
that at most there can be one more generator

$\displaystyle\Omega\mapsto(\Lambda_{1}+\Lambda_{2}+\Lambda_{3}+\Lambda_{4},\chi_{1})$
$\displaystyle\mapsto(\Lambda_{1}+\Lambda_{2}+\Lambda_{3}+\Lambda_{4},\chi_{0}),$
$\displaystyle(\Lambda_{1}+\Lambda_{2}+\Lambda_{3}+\Lambda_{4},\chi_{2})\mapsto(\Lambda_{1}+\Lambda_{2}+\Lambda_{3}+\Lambda_{4},\chi_{3})$
$\displaystyle\mapsto(\Lambda_{1}+\Lambda_{2}+\Lambda_{3}+\Lambda_{4},\chi_{4})$

in the braided auto-equivalence group, which would form a group isomorphic to
$A_{5}$. Thus
$\operatorname{EqBr}(\mathcal{C}(\mathfrak{sl}_{5},5)^{0}_{\operatorname{Rep}(\mathbb{Z}_{5})})$
is either isomorphic to $D_{5}$ or $A_{5}$. ∎

To finish off the proof of the main theorem of this section, we need to deal
with the remaining four cases. These can be dealt with easily by a level-rank
duality argument.

######  Proposition 4.9.

We have the following braided equivalences:

$\displaystyle\mathcal{C}(\mathfrak{sl}_{16},2)^{0}_{\operatorname{Rep}(\mathbb{Z}_{4})}$
$\displaystyle\simeq(\mathcal{C}(\mathfrak{sl}_{2},16)^{0}_{\operatorname{Rep}(\mathbb{Z}_{2})})^{\text{rev}}\boxtimes\operatorname{Vec}\left(\mathbb{Z}_{2},\\{1,e^{2\pi
i\frac{3}{4}}\\}\right),$
$\displaystyle\mathcal{C}(\mathfrak{sl}_{16},2)^{0}_{\operatorname{Rep}(\mathbb{Z}_{2})}$
$\displaystyle\simeq(\mathcal{C}(\mathfrak{sl}_{2},16)^{0}_{\operatorname{Rep}(\mathbb{Z}_{2})})^{\text{rev}}\boxtimes\operatorname{Vec}\left(\mathbb{Z}_{8},\\{1,e^{2\pi
i\frac{15}{16}},e^{2\pi i\frac{3}{4}},e^{2\pi i\frac{7}{16}},1,e^{2\pi
i\frac{7}{16}},e^{2\pi i\frac{3}{4}},e^{2\pi i\frac{15}{16}}\\}\right),$
$\displaystyle\mathcal{C}(\mathfrak{sl}_{9},3)^{0}_{\operatorname{Rep}(\mathbb{Z}_{3})}$
$\displaystyle\simeq(\mathcal{C}(\mathfrak{sl}_{3},9)^{0}_{\operatorname{Rep}(\mathbb{Z}_{3})})^{\text{rev}}\boxtimes\operatorname{Vec}\left(\mathbb{Z}_{3},\\{1,e^{2\pi
i\frac{1}{3}},e^{2\pi i\frac{1}{3}}\\}\right),\text{ and }$
$\displaystyle\mathcal{C}(\mathfrak{sl}_{8},4)^{0}_{\operatorname{Rep}(\mathbb{Z}_{4})}$
$\displaystyle\simeq(\mathcal{C}(\mathfrak{sl}_{4},8)^{0}_{\operatorname{Rep}(\mathbb{Z}_{4})})^{\text{rev}}\boxtimes\operatorname{Vec}\left(\mathbb{Z}_{2},\\{1,e^{2\pi
i\frac{3}{4}}\\}\right).$

###### Proof.

Let us do the computation for
$\mathcal{C}(\mathfrak{sl}_{16},2)^{0}_{\operatorname{Rep}(\mathbb{Z}_{4})}$,
as the other cases follow in a similar fashion.

From Lemma 2.18 we know that
$\mathcal{C}(\mathfrak{sl}_{16},2)^{0}_{\operatorname{Rep}(\mathbb{Z}_{4})}$
has a subcategory braided equivalent to
$(\mathcal{C}(\mathfrak{sl}_{16},2)^{\text{ad}})_{\operatorname{Rep}(\mathbb{Z}_{2})}$.
Via level-rank duality [51], we have a braided equivalence

$\mathcal{C}(\mathfrak{sl}_{16},2)^{\text{ad}}\to(\mathcal{C}(\mathfrak{sl}_{2},16)^{\text{ad}})^{\text{rev}}.$

Thus together we have a braided subcategory of
$\mathcal{C}(\mathfrak{sl}_{16},2)^{0}_{\operatorname{Rep}(\mathbb{Z}_{4})}$
equivalent to
$((\mathcal{C}(\mathfrak{sl}_{2},16)^{\text{ad}})^{\text{rev}})_{\operatorname{Rep}(\mathbb{Z}_{2})}$.
Taking the reverse braiding on the category commutes with taking the de-
equivariantization (as
$\operatorname{Rep}(\mathbb{Z}_{2})^{\text{rev}}=\operatorname{Rep}(\mathbb{Z}_{2})$),
thus we have a subcategory braided equivalent to
$(\mathcal{C}(\mathfrak{sl}_{2},16)^{0}_{\operatorname{Rep}(\mathbb{Z}_{2})})^{\text{rev}}$.
As
$(\mathcal{C}(\mathfrak{sl}_{2},16)^{0}_{\operatorname{Rep}(\mathbb{Z}_{2})})^{\text{rev}}$
is modular, Mugers theorem [46, Theorem 4.2 ] give us that the category
$\mathcal{C}(\mathfrak{sl}_{16},2)^{0}_{\operatorname{Rep}(\mathbb{Z}_{4})}$
factors as

$\mathcal{C}(\mathfrak{sl}_{16},2)^{0}_{\operatorname{Rep}(\mathbb{Z}_{4})}\simeq(\mathcal{C}(\mathfrak{sl}_{2},16)^{0}_{\operatorname{Rep}(\mathbb{Z}_{2})})^{\text{rev}}\boxtimes\mathcal{D},$

where $\mathcal{D}$ is a modular category with global dimension 2. To identify
$\mathcal{D}$ we study the object
$([2\Lambda_{2}],1)\in\mathcal{C}(\mathfrak{sl}_{16},2)^{0}_{\operatorname{Rep}(\mathbb{Z}_{4})}$.
The twist of this object is $e^{2\pi i\frac{3}{4}}$, and the order of the
object is $2$. Thus $([2\Lambda_{2}],1)$ generates a modular subcategory
equivalent to $\operatorname{Vec}\left(\mathbb{Z}_{2},e^{2\pi
i\frac{3}{4}}\right)$. As the category
$\mathcal{C}(\mathfrak{sl}_{2},16)^{0}_{\operatorname{Rep}(\mathbb{Z}_{2})}$
has no non-trivial invertible objects, we must have that $([2\Lambda_{2}],1)$
generates $\mathcal{D}$, which completes the claim.

∎

######  Corollary 4.10.

We have

$\displaystyle\operatorname{EqBr}(\mathcal{C}(\mathfrak{sl}_{16},2)^{0}_{\operatorname{Rep}(\mathbb{Z}_{4})})$
$\displaystyle=\operatorname{EqBr}(\mathcal{C}(\mathfrak{sl}_{2},16)^{0}_{\operatorname{Rep}(\mathbb{Z}_{2})}),$
$\displaystyle\operatorname{EqBr}(\mathcal{C}(\mathfrak{sl}_{16},2)^{0}_{\operatorname{Rep}(\mathbb{Z}_{2})})$
$\displaystyle=\operatorname{EqBr}(\mathcal{C}(\mathfrak{sl}_{2},16)^{0}_{\operatorname{Rep}(\mathbb{Z}_{2})})\times\mathbb{Z}_{2},$
$\displaystyle\operatorname{EqBr}(\mathcal{C}(\mathfrak{sl}_{9},3)^{0}_{\operatorname{Rep}(\mathbb{Z}_{3})})$
$\displaystyle=\operatorname{EqBr}(\mathcal{C}(\mathfrak{sl}_{3},9)^{0}_{\operatorname{Rep}(\mathbb{Z}_{3})})\times\mathbb{Z}_{2},\text{
and }$
$\displaystyle\operatorname{EqBr}(\mathcal{C}(\mathfrak{sl}_{8},4)^{0}_{\operatorname{Rep}(\mathbb{Z}_{4})})$
$\displaystyle=\operatorname{EqBr}(\mathcal{C}(\mathfrak{sl}_{4},8)^{0}_{\operatorname{Rep}(\mathbb{Z}_{4})}).$

###### Proof.

We use the canonical embedding

$\operatorname{EqBr}(\mathcal{C})\times\operatorname{EqBr}(\mathcal{D})\to\operatorname{EqBr}(\mathcal{C}\boxtimes\mathcal{D}).$

By analysing the fusion rings and twists of the the Deligne products from the
previous proposition, we see that this embedding is an isomorphism. ∎

##  5\. Realisation of the exceptionals

In the previous section we identified a finite list of the categories
$\mathcal{C}(\mathfrak{sl}_{r+1},k)^{0}_{\operatorname{Rep}(\mathbb{Z}_{m})}$
which may have an exceptional braided auto-equivalence, and furthermore, gave
upper bounds for the number of such auto-equivalences that may exist. In this
section, our goal is to construct all the exceptional braided auto-
equivalences of these finite number of categories. That is, we want to compute
the braided auto-equivalence groups of the categories

$\mathcal{C}(\mathfrak{sl}_{2},16)^{0}_{\operatorname{Rep}(\mathbb{Z}_{2})},\qquad\mathcal{C}(\mathfrak{sl}_{3},9)^{0}_{\operatorname{Rep}(\mathbb{Z}_{3})},\qquad\mathcal{C}(\mathfrak{sl}_{4},8)^{0}_{\operatorname{Rep}(\mathbb{Z}_{4})},\quad\text{and}\quad\mathcal{C}(\mathfrak{sl}_{5},5)^{0}_{\operatorname{Rep}(\mathbb{Z}_{5})}.$

While it is not immediate from the above list, there are two situations in
play here. The first is for the category
$\mathcal{C}(\mathfrak{sl}_{2},16)^{0}_{\operatorname{Rep}(\mathbb{Z}_{2})}$,
where the exceptional auto-equivalences come from the coincidence of
categories

$\mathcal{C}(\mathfrak{so}_{8},3)\simeq(\mathcal{C}(\mathfrak{sl}_{2},16)^{0}_{\operatorname{Rep}(\mathbb{Z}_{2})})^{\text{rev}}\boxtimes\mathcal{C}(\mathfrak{so}_{8},1).$

The $S_{3}$ worth of braided exceptional auto-equivalences of
$\mathcal{C}(\mathfrak{sl}_{2},16)^{0}_{\operatorname{Rep}(\mathbb{Z}_{2})}$
is then naturally seen due to the triality of the Dynkin diagram $D_{4}$. This
connection was initially discovered in [44].

######  Lemma 5.1.

[44, Theorem 4.3] We have

$\operatorname{EqBr}\left(\mathcal{C}(\mathfrak{sl}_{2},16)^{0}_{\operatorname{Rep}(\mathbb{Z}_{2})}\right)=S_{3}.$

The far more interesting situation occurs with the other three categories.
Studying the dimensions of these three examples, one notices that they are all
defined over small quadratic fields. For
$\mathcal{C}(\mathfrak{sl}_{3},9)^{0}_{\operatorname{Rep}(\mathbb{Z}_{3})}$
the dimensions live in $\mathbb{Q}[\sqrt{3}]$, for
$\mathcal{C}(\mathfrak{sl}_{4},8)^{0}_{\operatorname{Rep}(\mathbb{Z}_{4})}$
the dimensions live in $\mathbb{Q}[\sqrt{3}]$, and for
$\mathcal{C}(\mathfrak{sl}_{5},5)^{0}_{\operatorname{Rep}(\mathbb{Z}_{5})}$
the dimensions live in $\mathbb{Q}[\sqrt{5}]$. Such behaviour with the
dimensions does not hold for general
$\mathcal{C}(\mathfrak{sl}_{r+1},k)^{0}_{\operatorname{Rep}(\mathbb{Z}_{m})}$,
and suggests that these dimensions may have something to do with the potential
exceptional auto-equivalences of these categories.

A large class of categories with objects living in quadratic fields are the
quadratic categories, where the simple objects consist of the group of
invertibles, and an object $\rho$, along with the orbit of $\rho$ under the
action of the invertibles. The natural suspicion to draw, is that the three
categories
$\mathcal{C}(\mathfrak{sl}_{3},9)^{0}_{\operatorname{Rep}(\mathbb{Z}_{3})}$,
$\mathcal{C}(\mathfrak{sl}_{4},8)^{0}_{\operatorname{Rep}(\mathbb{Z}_{4})}$,
and
$\mathcal{C}(\mathfrak{sl}_{5},5)^{0}_{\operatorname{Rep}(\mathbb{Z}_{5})}$
should in some way be connected to quadratic categories. The naive guess, that
these three categories are quadratic categories on the nose, is immediately
thwarted by fact that the dimensions in these examples take on more than two
values. Further, quadratic categories are almost never modular, where as our
three examples are. However, this lack of modularity suggests the next place
to look for a connection.

Taking the Drinfeld centre of a quadratic category gives a modular category
whose dimensions lie in the same field as the quadratic category. While in the
quadratic category, the dimensions of the simples can only have two possible
values, the dimensions of the simples in the centre can be any integer
combination of $1$ and the dimension of the non-invertible of the quadratic.
This provides strong evidence that the three categories
$\mathcal{C}(\mathfrak{sl}_{3},9)^{0}_{\operatorname{Rep}(\mathbb{Z}_{3})}$,
$\mathcal{C}(\mathfrak{sl}_{4},8)^{0}_{\operatorname{Rep}(\mathbb{Z}_{4})}$,
and
$\mathcal{C}(\mathfrak{sl}_{5},5)^{0}_{\operatorname{Rep}(\mathbb{Z}_{5})}$
may be Drinfeld centers of quadratic categories.

Now that we have an idea of what to look for, we can make educated guesses as
to the identity of the quadratic categories. For example, the dimensions of
$\mathcal{C}(\mathfrak{sl}_{3},9)^{0}_{\operatorname{Rep}(\mathbb{Z}_{3})}$
are

$1,\quad 7+4\sqrt{3},\quad 8+4\sqrt{3},\quad\text{and}\quad
3+2\sqrt{3}\quad\text{(6 times)}.$

If $\mathcal{C}(\mathfrak{sl}_{3},9)^{0}_{\operatorname{Rep}(\mathbb{Z}_{3})}$
were the Drinfeld centre of a quadratic category, then a natural guess for the
dimension of the non-invertible would be $3+2\sqrt{3}$, as all the above
dimensions can be constructed as integer combinations of $1$ and
$3+2\sqrt{3}$. There is a known quadratic category with an object of this
dimension, which is a near-group category with group of invertibles
$\mathbb{Z}_{3}=\\{\mathbf{1},g,g^{2}\\}$, and a single non-invertible with
fusion

$\rho\otimes\rho\cong\mathbf{1}\oplus g\oplus g^{2}\oplus 6\rho.$

The global dimension of this category is $24+12\sqrt{3}$, so the global
dimension of its Drinfeld centre is $1008+576\sqrt{3}$. Where as the global
dimension of
$\mathcal{C}(\mathfrak{sl}_{3},9)^{0}_{\operatorname{Rep}(\mathbb{Z}_{3})}$ is
$336+192\sqrt{3}$. These global dimensions are off by a factor of three, which
suggests a $\mathbb{Z}_{3}$ factor is involved. From all this we conjecture
that there is a quadratic category $\mathcal{C}_{3,9,3}$ with fusion as above
such that

$\mathcal{Z}(\mathcal{C}_{3,9,3})\simeq\mathcal{C}(\mathfrak{sl}_{3},9)^{0}_{\operatorname{Rep}(\mathbb{Z}_{3})}\boxtimes\operatorname{Vec}(\mathbb{Z}_{3},\\{1,e^{2i\pi\frac{1}{3}},e^{2i\pi\frac{1}{3}}\\}).$

Using similar reasoning we conjecture the existence of a fusion category
$\mathcal{C}_{4,8,4}$ with invertibles
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}=\\{\mathbf{1},e,m,em\\}$ and non-
invertibles $\\{\rho,m\rho\\}$ with fusion

$\rho\otimes\rho\cong\mathbf{1}\oplus e\oplus 6\rho\oplus 4m\rho,$

such that

$\mathcal{Z}(\mathcal{C}_{4,8,4})\simeq\mathcal{C}(\mathfrak{sl}_{4},8)^{0}_{\operatorname{Rep}(\mathbb{Z}_{4})}\boxtimes\operatorname{Vec}(\mathbb{Z}_{2}\times\mathbb{Z}_{2},\\{1,e^{2i\pi\frac{3}{4}},-1,e^{2i\pi\frac{3}{4}}\\}),$

and a fusion category $\mathcal{C}_{5,5,5}$ with invertibles
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}=\\{\mathbf{1},e,m,em\\}$ and non-
invertibles $\\{\rho,e\rho,m\rho,em\rho\\}$ with fusion

$\rho\otimes\rho\cong\mathbf{1}\oplus\rho\oplus e\rho\oplus m\rho\oplus
em\rho,$

such that

$\mathcal{Z}(\mathcal{C}_{5,5,5})\simeq\mathcal{C}(\mathfrak{sl}_{5},5)^{0}_{\operatorname{Rep}(\mathbb{Z}_{5})}\boxtimes\operatorname{Vec}(\mathbb{Z}_{2}\times\mathbb{Z}_{2},\\{1,-1,-1,-1\\}).$

We make all this precise with the following Theorem.

######  Theorem 5.2.

There exist unitary fusion categories $\mathcal{C}_{3,9,3}$,
$\mathcal{C}_{4,8,3}$, and $\mathcal{C}_{5,5,5}$ with combinatorics as above,
such that

$\displaystyle\mathcal{Z}(\mathcal{C}_{3,9,3})$
$\displaystyle\simeq\mathcal{C}(\mathfrak{sl}_{3},9)^{0}_{\operatorname{Rep}(\mathbb{Z}_{3})}\boxtimes\operatorname{Vec}(\mathbb{Z}_{3},\\{1,e^{2i\pi\frac{1}{3}},e^{2i\pi\frac{1}{3}}\\})$
$\displaystyle\mathcal{Z}(\mathcal{C}_{4,8,4})$
$\displaystyle\simeq\mathcal{C}(\mathfrak{sl}_{4},8)^{0}_{\operatorname{Rep}(\mathbb{Z}_{4})}\boxtimes\operatorname{Vec}(\mathbb{Z}_{2}\times\mathbb{Z}_{2},\\{1,e^{2i\pi\frac{3}{4}},-1,e^{2i\pi\frac{3}{4}}\\}),\quad\text{
and }$ $\displaystyle\mathcal{Z}(\mathcal{C}_{5,5,5})$
$\displaystyle\simeq\mathcal{C}(\mathfrak{sl}_{5},5)^{0}_{\operatorname{Rep}(\mathbb{Z}_{5})}\boxtimes\operatorname{Vec}(\mathbb{Z}_{2}\times\mathbb{Z}_{2},\\{1,-1,-1,-1\\}).$

###### Proof.

Let us begin with the case
$\mathcal{C}(\mathfrak{sl}_{3},9)^{0}_{\operatorname{Rep}(\mathbb{Z}_{3})}$.
It is well known that there is a conformal inclusion
$SU(3)_{9}\subset(E_{6})_{1}$ which induces a commutative algebra object
$A\in\mathcal{C}(\mathfrak{sl}_{3},9)$. The category
$\operatorname{Mod}(\mathcal{C}(\mathfrak{sl}_{3},9),A)$ is then a unitary
fusion category, as described in [6, Figure 4]. The category
$\operatorname{Mod}(\mathcal{C}(\mathfrak{sl}_{3},9),A)$ is
$\mathbb{Z}_{3}$-graded. Let $\mathcal{C}_{3,9,3}$ be the trivially graded
subcategory of this grading.

By [2, Corollary 4.8] we have that

$\mathcal{Z}(\operatorname{Mod}(\mathcal{C}(\mathfrak{sl}_{3},9),A))\simeq\mathcal{C}(\mathfrak{sl}_{3},9)\boxtimes(\operatorname{Mod}(\mathcal{C}(\mathfrak{sl}_{3},9),A)^{0})^{\text{rev}}.$

For this particular case, we have that
$\operatorname{Mod}(\mathcal{C}(\mathfrak{sl}_{3},9),A)^{0}\simeq\mathcal{C}(\mathfrak{e}_{6},1)\simeq\operatorname{Vec}(\mathbb{Z}_{3},\\{1,e^{2i\pi\frac{2}{3}},e^{2i\pi\frac{2}{3}}\\})$.

Now the results of [29] allow us to compute the Drinfeld centre of the
subcategory $\mathcal{C}_{3,9,3}$ in terms of the Drinfeld centre of
$\operatorname{Mod}(\mathcal{C}(\mathfrak{sl}_{3},9),A)$. This gives that

$\mathcal{Z}(\mathcal{C}_{3,9,3})\simeq\mathcal{C}(\mathfrak{sl}_{3},9)^{0}_{\operatorname{Rep}(\mathbb{Z}_{3})}\boxtimes\operatorname{Vec}(\mathbb{Z}_{3},\\{1,e^{2i\pi\frac{1}{3}},e^{2i\pi\frac{1}{3}}\\})$

as desired.

The case of
$\mathcal{C}(\mathfrak{sl}_{4},8)^{0}_{\operatorname{Rep}(\mathbb{Z}_{4})}$ is
almost identical, except now we use the conformal inclusion $SU(4)_{8}\subset
Spin(20)_{1}$. The structure of the associated algebra object, and category of
modules, can be found in [8, Section 2.6].

The case of
$\mathcal{C}(\mathfrak{sl}_{5},5)^{0}_{\operatorname{Rep}(\mathbb{Z}_{5})}$
should follow in the same manner, where now we work with the conformal
inclusion $SU(5)_{5}\subset Spin(24)_{1}$. However, the author was unable to
find a suitable description of the category of modules in this case. Instead
we have [58, Theorem 3.2] which proves the precise statement in this case.
They show that $\mathcal{C}_{5,5,5}$ is the even part of the
$3^{\mathbb{Z}_{2}\times\mathbb{Z}_{2}}$ subfactor. ∎

With these alternate identifications of the categories
$\mathcal{C}(\mathfrak{sl}_{3},9)^{0}_{\operatorname{Rep}(\mathbb{Z}_{3})}$,
$\mathcal{C}(\mathfrak{sl}_{4},8)^{0}_{\operatorname{Rep}(\mathbb{Z}_{4})}$,
and
$\mathcal{C}(\mathfrak{sl}_{5},5)^{0}_{\operatorname{Rep}(\mathbb{Z}_{5})}$
identified, we now have the tools to construct their exceptional braided auto-
equivalences, and hence determine their braided auto-equivalence groups. We
are able to complete this for the cases
$\mathcal{C}(\mathfrak{sl}_{3},9)^{0}_{\operatorname{Rep}(\mathbb{Z}_{3})}$
and
$\mathcal{C}(\mathfrak{sl}_{5},5)^{0}_{\operatorname{Rep}(\mathbb{Z}_{5})}$ in
this paper. Let us begin with
$\mathcal{C}(\mathfrak{sl}_{3},9)^{0}_{\operatorname{Rep}(\mathbb{Z}_{3})}$.

######  Lemma 5.3.

We have

$\operatorname{EqBr}(\mathcal{C}(\mathfrak{sl}_{3},9)^{0}_{\operatorname{Rep}(\mathbb{Z}_{3})})=S_{4}$

###### Proof.

From Theorem 4.2 we have that
$\operatorname{EqBr}(\mathcal{C}(\mathfrak{sl}_{3},9)^{0}_{\operatorname{Rep}(\mathbb{Z}_{3})})$
is either $D_{3}$ or $S_{4}$. By analysing the fusion rings and twists of

$\mathcal{Z}(\mathcal{C}_{3,9,3})\simeq\mathcal{C}(\mathfrak{sl}_{3},9)^{0}_{\operatorname{Rep}(\mathbb{Z}_{3})}\boxtimes\operatorname{Vec}(\mathbb{Z}_{3},\\{1,e^{2i\pi\frac{1}{3}},e^{2i\pi\frac{1}{3}}\\})$

we see that

$\operatorname{EqBr}(\mathcal{Z}(\mathcal{C}_{3,9,3}))=\operatorname{EqBr}(\mathcal{C}(\mathfrak{sl}_{3},9)^{0}_{\operatorname{Rep}(\mathbb{Z}_{3})})\times\mathbb{Z}_{2}.$

It is proven in [35, Section 10.6] that
$\operatorname{Out}(\mathcal{C}_{3,9,3})=D_{4}$. From [42] we have an
embedding

$\operatorname{Out}(\mathcal{C}_{3,9,3})\to\operatorname{EqBr}(\mathcal{Z}(\mathcal{C}_{3,9,3})).$

Thus
$\operatorname{EqBr}(\mathcal{C}(\mathfrak{sl}_{3},9)^{0}_{\operatorname{Rep}(\mathbb{Z}_{3})})\times\mathbb{Z}_{2}$
has a subgroup isomorphic to $D_{4}$. This is only possible if
$\operatorname{EqBr}(\mathcal{C}(\mathfrak{sl}_{3},9)^{0}_{\operatorname{Rep}(\mathbb{Z}_{3})})=S_{4}$.
∎

We now deal with the case of
$\mathcal{C}(\mathfrak{sl}_{5},5)^{0}_{\operatorname{Rep}(\mathbb{Z}_{5})}$.
This case has been examined in the literature previously [58, 30].

######  Lemma 5.4.

We have

$\operatorname{EqBr}(\mathcal{C}(\mathfrak{sl}_{5},5)^{0}_{\operatorname{Rep}(\mathbb{Z}_{5})})=A_{5}.$

###### Proof.

From Theorem 4.2 we have that
$\operatorname{EqBr}(\mathcal{C}(\mathfrak{sl}_{5},5)^{0}_{\operatorname{Rep}(\mathbb{Z}_{5})})$
is either $D_{5}$ or $A_{5}$. By analysing the fusion rings and twists of

$\mathcal{Z}(\mathcal{C}_{5,5,5})\simeq\mathcal{C}(\mathfrak{sl}_{5},5)^{0}_{\operatorname{Rep}(\mathbb{Z}_{5})}\boxtimes\operatorname{Vec}(\mathbb{Z}_{2}\times\mathbb{Z}_{2},\\{1,-1,-1,-1\\}).$

we see that

$\operatorname{EqBr}(\mathcal{Z}(\mathcal{C}_{5,5,5}))=\operatorname{EqBr}(\mathcal{C}(\mathfrak{sl}_{5},5)^{0}_{\operatorname{Rep}(\mathbb{Z}_{5})})\times
S_{3}.$

It is proven in [36, Theorem 9.4] that
$\operatorname{Out}(\mathcal{C}_{5,5,5})=A_{4}$. Thus
$\operatorname{EqBr}(\mathcal{C}(\mathfrak{sl}_{5},5)^{0}_{\operatorname{Rep}(\mathbb{Z}_{5})})\times
S_{3}$ has a subgroup isomorphic to $A_{4}$. This is only possible if
$\operatorname{EqBr}(\mathcal{C}(\mathfrak{sl}_{5},5)^{0}_{\operatorname{Rep}(\mathbb{Z}_{5})})=A_{5}$.
∎

Unfortunately we are unable to use the quadratic category
$\mathcal{C}_{4,8,4}$ to construct the exceptional braided auto-equivalence of
$\mathcal{C}(\mathfrak{sl}_{4},8)^{0}_{\operatorname{Rep}(\mathbb{Z}_{4})}$.
This is because an explicit construction of the quadratic category
$\mathcal{C}_{4,8,4}$ has yet to be given. One way to construct this category
is via the Cuntz algebra method, where it will be realised as endomorphisms on
the $C^{*}$-algebra $O_{12}\rtimes\mathbb{Z}_{2}$. The large multiplicity
spaces of the quadratic category $\mathcal{C}_{4,8,4}$ means that this method
required solving for roughly $1700$ complex variables in $20000$ polynomial
equations. This makes the problem too complex, even for modern computer
algebra programs.

Instead we construct the exceptional braided auto-equivalence of
$\mathcal{C}(\mathfrak{sl}_{4},8)^{0}_{\operatorname{Rep}(\mathbb{Z}_{4})}$
using a coincidence of categories to connect it to $\mathfrak{so}_{8}$,
similar to the
$\mathcal{C}(\mathfrak{sl}_{2},16)^{0}_{\operatorname{Rep}(\mathbb{Z}_{2})}$
case. Recall that $\mathfrak{sl}_{4}$ is isomorphic to $\mathfrak{so}_{6}$.
Hence the category $\mathcal{C}(\mathfrak{sl}_{4},8)$ can be viewed as
$\mathcal{C}(\mathfrak{so}_{6},8)$. Now level-rank duality will give a (non-
trivial) connection between $\mathcal{C}(\mathfrak{so}_{6},8)$ and
$\mathcal{C}(\mathfrak{so}_{8},6)$. Via this connection we can pull back the
triality of $\mathfrak{so}_{8}$ in order to obtain an order 3 auto-equivalence
of
$\mathcal{C}(\mathfrak{sl}_{4},8)^{0}_{\operatorname{Rep}(\mathbb{Z}_{4})}$,
which forces
$\operatorname{EqBr}(\mathcal{C}(\mathfrak{sl}_{4},8)^{0}_{\operatorname{Rep}(\mathbb{Z}_{4})})=S_{4}$.

Let us expand more on this connection between
$\mathcal{C}(\mathfrak{so}_{6},8)$ and $\mathcal{C}(\mathfrak{so}_{8},6)$.
These categories have different ranks, so even as abelian categories they are
not equivalent. Thus there are several steps we must take to get some sort of
equivalence. First let $\mathcal{C}^{\text{Vec}}(\mathfrak{so}_{N},k)$ be the
$\mathbb{Z}_{2}$-graded subcategory of $\mathcal{C}(\mathfrak{so}_{N},k)$
generated by the “vector representation” $\Lambda_{1}$. To get “orthogonal
categories”, where level-rank duality applies we have to ‘’add in” the
determinant representation by taking the $\mathbb{Z}_{2}$-equivariantisation
via the $D_{n}$ Dynkin diagram symmetry. This gives us the braided equivalence

$\mathcal{C}^{\text{Vec}}(\mathfrak{so}_{6},8)^{\mathbb{Z}_{2}}\simeq[\mathcal{C}^{\text{Vec}}(\mathfrak{so}_{8},6)^{\mathbb{Z}_{2}}]^{-\text{rev}}.$

Where $-\text{rev}$ means to take the reverse the braiding, and to negate it
on the non-trivial piece of the grading. However this equivalence doesn’t
preserve the determinant representations, so we can’t de-equivariantise by a
single $\operatorname{Rep}(\mathbb{Z}_{2})$ subcategory to obtain a braided
equivalence. Instead we must de-equivariantise by the maximal Tannakian
subcategory. This gives us a braided equivalence

$\mathcal{C}(\mathfrak{so}_{6},8)^{0}_{\operatorname{Rep}(\mathbb{Z}_{2})}=\mathcal{C}^{\text{Vec}}(\mathfrak{so}_{6},8)_{\langle
8\Lambda_{1}\rangle}\simeq[\mathcal{C}^{\text{Vec}}(\mathfrak{so}_{8},6)_{\langle
6\Lambda_{1}\rangle}]^{-\text{rev}}=[\mathcal{C}(\mathfrak{so}_{8},6)^{0}_{\operatorname{Rep}(\mathbb{Z}_{2})}]^{-\text{rev}}.$

However triality doesn’t preserve the $\langle 8\Lambda_{1}\rangle$
subcategory of $\mathcal{C}^{\text{Vec}}(\mathfrak{so}_{8},6)$, so it won’t
descend to
$[\mathcal{C}(\mathfrak{so}_{8},6)^{0}_{\operatorname{Rep}(\mathbb{Z}_{2})}]^{-\text{rev}}$.
Thus we have to take local modules with respect to the remaining
$\operatorname{Rep}(\mathbb{Z}_{2})$ subcategory to get

$\mathcal{C}(\mathfrak{so}_{6},8)^{0}_{\operatorname{Rep}(\mathbb{Z}_{4})}\simeq[\mathcal{C}(\mathfrak{so}_{8},6)^{0}_{\operatorname{Rep}(\mathbb{Z}_{2}\times\mathbb{Z}_{2})}]^{\text{rev}}.$

Triality now preserves the
$\operatorname{Rep}(\mathbb{Z}_{2}\times\mathbb{Z}_{2})$ subcategory of
$\mathcal{C}(\mathfrak{so}_{8},6)$ and hence descends to give us an order 3
auto-equivalence of
$\mathcal{C}(\mathfrak{so}_{6},8)^{0}_{\operatorname{Rep}(\mathbb{Z}_{4})}=\mathcal{C}(\mathfrak{sl}_{3},8)^{0}_{\operatorname{Rep}(\mathbb{Z}_{4})}$.

To make this all precise we begin the following lemma, which formalises type
$D$-$D$ level-rank duality.

######  Lemma 5.5.

Let $N$ and $k$ be even integers. We have a braided equivalence

$\mathcal{C}(\mathfrak{so}_{N},k)^{0}_{\langle
k\Lambda_{1}\rangle}\simeq[\mathcal{C}(\mathfrak{so}_{k},N)^{0}_{\langle
N\Lambda_{1}\rangle}]^{-\text{rev}}$

###### Proof.

Let

$X_{N,k}:=(\Lambda_{1},+)\in\mathcal{C}^{\text{Vec}}(\mathfrak{so}_{N},k)^{\mathbb{Z}_{2}},$

and

$X_{k,N}:=(\Lambda_{1},+)\in\mathcal{C}^{\text{Vec}}(\mathfrak{so}_{k},N)^{\mathbb{Z}_{2}}.$

From [55] we have that

$\mathcal{P}_{X_{N,k}}=\overline{\text{BMW}\left(e^{2\pi
i\frac{1}{2(N+k-2)}},e^{2\pi i\frac{N-1}{2(N+k-2)}}\right)},$

and

$\mathcal{P}_{X_{k,N}}=\overline{\text{BMW}\left(e^{2\pi
i\frac{1}{2(N+k-2)}},e^{2\pi i\frac{k-1}{2(N+k-2)}}\right)},$

where $\overline{\text{BMW}(q,r)}$ is the semi-simplified planar algebra
generated by a single crossing and the Kauffman relation (see [55, Section 7]
[14, Definition 2.12]). It is routine to verify that

$\raisebox{-0.5pt}{
\includegraphics[scale={.5}]{BMWpositivecrossing}}\mapsto-\raisebox{-0.5pt}{
\includegraphics[scale={.5}]{BMWnegitivecrossing}}$

gives an isomorphism of planar algebras

$\mathcal{P}_{X_{N,k}}\to\mathcal{P}_{X_{k,N}}.$

Via [32, Theorem A] this induces a braided equivalence

$\mathcal{C}^{\text{Vec}}(\mathfrak{so}_{N},k)^{\mathbb{Z}_{2}}\to[\mathcal{C}^{\text{Vec}}(\mathfrak{so}_{k},N)^{\mathbb{Z}_{2}}]^{-\text{rev}}.$

Lets define the two groups

$G_{N,k}:=\operatorname{Inv}\left(\mathcal{C}^{\text{Vec}}(\mathfrak{so}_{N},k)^{\mathbb{Z}_{2}}\right)=\\{(\mathbf{1},+),\quad(\mathbf{1},-),\quad(k\Lambda_{1},+),\quad\text{and}\quad(k\Lambda_{1},-)\\},$

and

$G_{k,N}:=\operatorname{Inv}\left([\mathcal{C}^{\text{Vec}}(\mathfrak{so}_{k},N)^{\mathbb{Z}_{2}}]^{-\text{rev}}\right)=\\{(\mathbf{1},+),\quad(\mathbf{1},-),\quad(N\Lambda_{1},+),\quad\text{and}\quad(N\Lambda_{1},-)\\}.$

The above braided equivalence will preserve the group of invertibles, so we
have that it maps $G_{N,k}$ to $G_{k,N}$.

As $k\Lambda_{1}$ and $N\Lambda_{1}$ are symmetric objects in their respective
categories and have trivial twist (these facts requires both $N$ and $k$
even), we get that

$\operatorname{Rep}(G_{N,k})\subseteq
Z_{2}(\mathcal{C}^{\text{Vec}}(\mathfrak{so}_{N},k)^{\mathbb{Z}_{2}}),$

and

$\operatorname{Rep}(G_{k,N})\subseteq
Z_{2}([\mathcal{C}^{\text{Vec}}(\mathfrak{so}_{k},N)^{\mathbb{Z}_{2}}]^{-\text{rev}}).$

Thus we can de-equivariantise to obtain a braided isomorphism

$\mathcal{C}^{\text{Vec}}(\mathfrak{so}_{N},k)^{\mathbb{Z}_{2}}_{\operatorname{Rep}(G_{N,k})}\to[\mathcal{C}^{\text{Vec}}(\mathfrak{so}_{k},N)^{\mathbb{Z}_{2}}]_{\operatorname{Rep}(G_{k,N})}^{-\text{rev}}.$

From [11, Theorem 4.9] we have that $\langle(\mathbf{1},-)\rangle$ generates a
copy of
$\operatorname{Rep}(\mathbb{Z}_{2})\subseteq\mathcal{C}^{\text{Vec}}(\mathfrak{so}_{N},k)$
(resp.
$[\mathcal{C}^{\text{Vec}}(\mathfrak{so}_{k},N)^{\mathbb{Z}_{2}}]^{-\text{rev}}$),
and that de-equivaraiantising
$\mathcal{C}^{\text{Vec}}(\mathfrak{so}_{N},k)^{\mathbb{Z}_{2}}$ (resp.
$[\mathcal{C}^{\text{Vec}}(\mathfrak{so}_{k},N)^{\mathbb{Z}_{2}}]^{-\text{rev}}$)
by this copy of $\operatorname{Rep}(\mathbb{Z}_{2})$ recovers
$\mathcal{C}^{\text{Vec}}(\mathfrak{so}_{N},k)$ (resp.
$[\mathcal{C}^{\text{Vec}}(\mathfrak{so}_{k},N)]^{-\text{rev}})$ up to braided
equivalence. As $\langle(\mathbf{1},-)\rangle$ is normal in $G_{N,k}$ and
$G_{k,N}$ respectively we can use the above fact, along with [45, Proposition
4.12] to get braided equivalences

$\mathcal{C}^{\text{Vec}}(\mathfrak{so}_{N},k)^{\mathbb{Z}_{2}}_{\operatorname{Rep}(G_{N,k})}\simeq\mathcal{C}^{\text{Vec}}(\mathfrak{so}_{N},k)_{\langle
k\Lambda_{1}\rangle},$

and

$[\mathcal{C}^{\text{Vec}}(\mathfrak{so}_{k},N)^{\mathbb{Z}_{2}}]_{\operatorname{Rep}(G_{k,N})}^{-\text{rev}}\simeq[\mathcal{C}^{\text{Vec}}(\mathfrak{so}_{k},N)]_{\langle
N\Lambda_{1}\rangle}^{-\text{rev}}.$

To finish the proof we observe that $k\Lambda_{1}$ is symmetric in
$\mathcal{C}^{\text{Vec}}(\mathfrak{so}_{N},k)$, yet the full category
$\mathcal{C}(\mathfrak{so}_{N},k)$ is modular. Thus the subcategory of objects
which symmetrise $k\Lambda_{1}$ in $\mathcal{C}(\mathfrak{so}_{N},k)$ is
precisely $\mathcal{C}^{\text{Vec}}(\mathfrak{so}_{N},k)$. This implies that

$\mathcal{C}(\mathfrak{so}_{N},k)^{0}_{\langle
k\Lambda_{1}\rangle}=\mathcal{C}^{\text{Vec}}(\mathfrak{so}_{N},k)_{\langle
k\Lambda_{1}\rangle}.$

The same argument works to show that

$\mathcal{C}(\mathfrak{so}_{k},N)^{0}_{\langle
N\Lambda_{1}\rangle}=\mathcal{C}^{\text{Vec}}(\mathfrak{so}_{k},N)_{\langle
N\Lambda_{1}\rangle},$

which completes the proof. ∎

As a corollary we get our desired exceptional braided auto-equivalence.

######  Corollary 5.6.

We have

$\operatorname{EqBr}(\mathcal{C}(\mathfrak{sl}_{4},8)^{0}_{\operatorname{Rep}(\mathbb{Z}_{4})})=S_{4}.$

###### Proof.

Via a coincidence of Dynkin diagrams we have

$\mathcal{C}(\mathfrak{sl}_{4},8)\simeq\mathcal{C}(\mathfrak{so}_{6},8).$

Thus the above Lemma gives a braided equivalence

$\mathcal{C}(\mathfrak{sl}_{4},8)^{0}_{\operatorname{Rep}(\mathbb{Z}_{2})}\simeq[\mathcal{C}(\mathfrak{so}_{8},6)^{0}_{\operatorname{Rep}(\mathbb{Z}_{2})}]^{-\text{rev}}.$

By taking local-modules with respect to the remaining
$\operatorname{Rep}(\mathbb{Z}_{2})$ subcategory we obtain

$\mathcal{C}(\mathfrak{sl}_{4},8)^{0}_{\operatorname{Rep}(\mathbb{Z}_{4})}\simeq[\mathcal{C}(\mathfrak{so}_{8},6)^{0}_{\operatorname{Rep}(\mathbb{Z}_{2}\times\mathbb{Z}_{2})}]^{-\text{rev}}=[\mathcal{C}(\mathfrak{so}_{8},6)^{0}_{\operatorname{Rep}(\mathbb{Z}_{2}\times\mathbb{Z}_{2})}]^{\text{rev}}.$

The triality of the $D_{4}$ Dynkin diagram induces an order 3 braided auto-
equivalence of $\mathcal{C}(\mathfrak{so}_{8},6)^{\text{- rev}}$ by [14,
Corollary 2.7]. This auto-equivalence preserves the
$\operatorname{Rep}(\mathbb{Z}_{2}\times\mathbb{Z}_{2})$ subcategory, and thus
descends to
$[\mathcal{C}(\mathfrak{so}_{8},6)^{0}_{\operatorname{Rep}(\mathbb{Z}_{2}\times\mathbb{Z}_{2})}]^{\text{rev}}$.
To see that this braided auto-equivalence is non-trivial, we observe that
triality will send

$2\Lambda_{1}\to 2\Lambda_{3}$

in $\mathcal{C}(\mathfrak{so}_{8},6)^{\text{- rev}}$, which implies that the
induced braided auto-equivalence will map

$\mathcal{F}_{\mathbb{Z}_{2}\times\mathbb{Z}_{2}}(2\Lambda_{1})\mapsto\mathcal{F}_{\mathbb{Z}_{2}\times\mathbb{Z}_{2}}(2\Lambda_{3})$

in
$[\mathcal{C}(\mathfrak{so}_{8},6)^{0}_{\operatorname{Rep}(\mathbb{Z}_{2}\times\mathbb{Z}_{2})}]^{\text{rev}}$.
However the two objects $2\Lambda_{1}$ and $2\Lambda_{3}$ are not in the same
orbit under the simple currents of
$\mathcal{C}(\mathfrak{so}_{8},6)^{\text{-rev}}$ which implies

$\mathcal{F}_{\mathbb{Z}_{2}\times\mathbb{Z}_{2}}(2\Lambda_{1})\ncong\mathcal{F}_{\mathbb{Z}_{2}\times\mathbb{Z}_{2}}(2\Lambda_{3}).$

Thus the induced braided auto-equivalence of
$[\mathcal{C}(\mathfrak{so}_{8},6)^{0}_{\operatorname{Rep}(\mathbb{Z}_{2}\times\mathbb{Z}_{2})}]^{\text{rev}}$
is non-trivial.

Recall from Theorem 4.2 that
$\operatorname{EqBr}(\mathcal{C}(\mathfrak{sl}_{4},8)^{0}_{\operatorname{Rep}(\mathbb{Z}_{4})})$
is either $D_{4}$ or $S_{4}$. As we know there exists an order 3 braided auto-
equivalence of
$\mathcal{C}(\mathfrak{sl}_{4},8)^{0}_{\operatorname{Rep}(\mathbb{Z}_{4})}$,
we can only have the latter option. ∎

##  Appendix A Coincidences of Small Dimensions

By Terry Gannon

In this appendix, we prove the following result regarding coincidences of
dimensions in the categories $\mathcal{C}(\mathfrak{sl}_{r+1},k)$.

######  Proposition A.1.

Let $X$ be a simple object of $\mathcal{C}(\mathfrak{sl}_{r+1},k)$ such that
$\dim(X)=\dim(\Lambda_{1}+\Lambda_{r})$. Then either
$X\in[\Lambda_{1}+\Lambda_{r}]$, or

$\displaystyle(r,k)=(8,3),(8,15)\quad$ $\displaystyle\text{and}\quad
X\in[\Lambda_{3}],$ or $\displaystyle(r,k)=(2,9),(14,9)\quad$
$\displaystyle\text{and}\quad X\in[3\Lambda_{1}],$ or
$\displaystyle(r,k)=(3,6),(5,4)\quad$ $\displaystyle\text{and}\quad
X\in[2\Lambda_{2}],$ or $\displaystyle(r,k)=(7,4),(7,6)\quad$
$\displaystyle\text{and}\quad X\in[\Lambda_{4}],$ or
$\displaystyle(r,k)=(3,8),(5,8)\quad$ $\displaystyle\text{and}\quad
X\in[4\Lambda_{1}].$

The main technical tool we use to prove this result is the following lemma,
which allows us to shuffle around the Dynkin labels of a simple object $X$ to
decrease its dimension.

######  Lemma A.2.

Let $X=\sum_{i=0}^{r}\lambda_{i}\Lambda_{i}$ be a simple object of
$\mathcal{C}(\mathfrak{sl}_{r+1},k)$. For any two indices $0\leq j,l\leq r$,
and integers $0\leq c_{j}\leq\lambda_{j}$ and $0\leq c_{l}\leq\lambda_{l}$ we
have that

$\dim(X)\geq\min(\dim(X-c_{j}\Lambda_{j}+c_{j}\Lambda_{l}),\dim(X+c_{l}\Lambda_{j}-c_{l}\Lambda_{l})),$

with equality if and only if $c_{j}=0$ or $c_{l}=0$.

###### Proof.

Note that we can write

$X=\frac{c_{l}}{c_{j}+c_{l}}(X-c_{j}\Lambda_{j}+c_{j}\Lambda_{l})+\frac{c_{j}}{c_{j}+c_{l}}(X+c_{l}\Lambda_{j}-c_{l}\Lambda_{l}).$

Thus the result follows immediately from Lemma 2.3. ∎

With this tool we can now prove Proposition A.1.

###### Proof.

Let $X$ be a simple object of $\mathcal{C}(\mathfrak{sl}_{r+1},k)$ such that
$\dim(X)=\dim(\Lambda_{1}+\Lambda_{r})$. First assume that more than three of
the labels $\lambda_{i}$ are non-zero. Pick two of these non-zero labels
$\lambda_{j},\lambda_{l}$. Then Lemma A.2 tells us that either
$X-\lambda_{j}\Lambda_{j}+\lambda_{j}\Lambda_{l}$ or
$X+\lambda_{l}\Lambda_{j}-\lambda_{l}\Lambda_{l}$ has dimension less than or
equal to $X$. Hence we get an object
$X^{\prime}\in\mathcal{C}(\mathfrak{sl}_{r+1},k)$ with
$\dim(X^{\prime})\leq\dim(\Lambda_{1}+\Lambda_{r})$ and with one less non-zero
label than $X$. By repeating this process we can assume that
$\dim(X)\leq\dim(\Lambda_{1}+\Lambda_{r})$ and $X$ has at most three non-zero
labels.

Now suppose $X$ has exactly three non-zero labels. By applying a simple
current symmetry we can assume that
$X=\lambda_{0}\Lambda_{0}+\lambda_{a}\Lambda_{a}+\lambda_{b}\Lambda_{b}$. By
applying Lemma A.2 with $c_{0}=\lambda_{0}-1$ and $c_{a}=\lambda_{a}-1$ we get
that

$\dim(X)\geq\min(\Lambda_{a}+\lambda_{b}\Lambda_{b},(\lambda_{0}+\lambda_{a}-1)\Lambda_{a}+\lambda_{b}\Lambda_{b}).$

By repeating this process with the $\lambda_{b}$ label, and applying a simple
current symmetry we get that
$\dim(X)\geq\Lambda_{a^{\prime}}+\Lambda_{b^{\prime}}$ for some
$0<a^{\prime}<b^{\prime}$, with equality if and only if
$X\in[\Lambda_{a^{\prime}}+\Lambda_{b^{\prime}}]$. By level-rank duality we
have that
$\dim(\Lambda_{a^{\prime}}+\Lambda_{b^{\prime}})=\dim((b^{\prime}-a^{\prime})\Lambda_{1}+a^{\prime}\Lambda_{2})$.
By applying Lemma A.2 several times we obtain

$\displaystyle\dim((b^{\prime}-a^{\prime})\Lambda_{1}+a^{\prime}\Lambda_{2})$
$\displaystyle\geq\min(\dim(\Lambda_{1}+\Lambda_{2}),\dim(\Lambda_{1}+(k-2)\Lambda_{2}),\dim((k-2)\Lambda_{1}+\Lambda_{2}))$
$\displaystyle=\min(\dim(\Lambda_{1}+\Lambda_{2}),\dim(\Lambda_{1}+\Lambda_{r})).$

Hence all together we have

$\dim(\Lambda_{1}+\Lambda_{r})\geq\min(\dim(\Lambda_{1}+\Lambda_{2}),\dim(\Lambda_{1}+\Lambda_{r}))$

with equality only if $X\in[\Lambda_{a^{\prime}}+\Lambda_{b^{\prime}}]$.

Using [28, Equation (2.1c) ] we compute that

$\frac{\dim(\Lambda_{1}+\Lambda_{2})}{\dim(\Lambda_{1}+\Lambda_{r})}=\frac{\sin\left(\frac{\pi(r+1)}{(k+r+1)}\right)}{\sin\left(\frac{3\pi}{(k+r+1)}\right)}.$

This is always $\geq 1$, with equality only if $r=2$ or $k=3$. Hence if $X$
has exactly three non-zero labels then $r=2$ or $k=3$ and
$X\in[\Lambda_{1}+\Lambda_{2}]$. However in these cases we get that
$\Lambda_{1}+\Lambda_{2}\in[\Lambda_{1}+\Lambda_{r}]$ and so
$X\in[\Lambda_{1}+\Lambda_{r}]$.

Finally suppose $X$ has exactly two non-zero labels (if $X$ has one non-zero
label, then $X=\mathbf{1}$). Then we can write $X=a\Lambda_{b}$ with $a,b\geq
2$. We can assume that $r\geq 3$ and hence $k\geq 4$ by level-rank duality. By
applying Lemma A.2 with $c_{0}=k-a-2$ and $c_{b}=a-2$ to get

$\dim(a\Lambda_{b})\geq\min(\dim(2\Lambda_{b}),\dim((k-2)\Lambda_{b}))=\dim(2\Lambda_{b}).$

By applying level-rank duality and using the same trick we find that
$\dim(a\Lambda_{b})\geq\dim(2\Lambda_{2})$. We compute

$1\geq\frac{\dim(2\Lambda_{2})}{\dim(\Lambda_{1}+\Lambda_{r})}=\frac{\sin\left(\frac{\pi}{k+r+1}\right)\sin\left(\frac{(r+1)\pi}{k+r+1}\right)^{2}}{\sin\left(\frac{2\pi}{k+r+1}\right)^{2}\sin\left(\frac{3\pi}{k+r+1}\right)}\geq\frac{\sin\left(\frac{\pi}{k+r+1}\right)\sin\left(\frac{4\pi}{k+r+1}\right)^{2}}{\sin\left(\frac{2\pi}{k+r+1}\right)^{2}\sin\left(\frac{3\pi}{k+r+1}\right)},$

as $r\geq 3$ and $k\geq 4$. By level-rank duality we can assume that $r+1\leq
k$. Simple calculus shows that
$\frac{\sin(x)\sin^{2}(4x)}{\sin^{2}(4x)\sin(3x)}>1$ when $x<\frac{\pi}{10}$.
Hence $r+k+1\leq 10$. For these finite possible cases, we can directly search
to find when $\dim(a\Lambda_{b})=\dim(\Lambda_{1}+\Lambda_{r})$. The only
solutions are $(r,k)=(3,6),(5,4)$ where $X=2\Lambda_{2}$.

Finally (by level-rank duality) it suffices to consider $X=\Lambda_{b}$ for
$b\leq\frac{r+1}{2}$. We have

$\frac{\dim(\Lambda_{b})}{\dim(\Lambda_{1}+\Lambda_{r})}=\frac{\sin\left(\frac{\pi(k+1)}{1+r+k}\right)\sin\left(\frac{\pi(k-1)}{1+r+k}\right)}{\sin\left(\frac{\pi}{1+r+k}\right)^{2}}\prod_{j=1}^{b}\frac{\sin\left(\frac{\pi(r+2-j)}{1+r+k}\right)}{\sin\left(\frac{\pi
j}{1+r+k}\right)}.$

In particular this shows

$\dim(\Lambda_{1})<\dim(\Lambda_{2})<\cdots<\dim(\Lambda_{\frac{r+1}{2}}).$

Let’s now study when the terms
$\frac{\dim(\Lambda_{b})}{\dim(\Lambda_{1}+\Lambda_{r})}$ are equal to $1$. We
will start by studying the case $b=1$, and will increase $b$ until be can show
that this term is always strictly bigger than $1$.

For the case of $b=1$, we find that
$\frac{\dim(\Lambda_{1})}{\dim(\Lambda_{1}+\Lambda_{r})}=1$ if and only if
$\Lambda_{1}\in[\Lambda_{1}+\Lambda_{r}]$. With the $b=1$ case done, we can
now assume $r\geq 3$.

For the case of $b=2$, we can use the inequality coming from the concavity of
$\ln|\sin(x)|$:

(A.1) $\sin(a)\sin(b)<\sin(a-x)\sin(b+x)$

for $0<b<a<\pi$ and $0<x\leq\frac{a-b}{2}$, to get

$\frac{\dim(\Lambda_{2})}{\dim(\Lambda_{1}+\Lambda_{r})}<1.$

With the $b=2$ case done, we can now assume $r\geq 5$.

For the case of $b=3$ we have to consider several subcases. If
$r\in\\{5,6,7\\}$ and $k>r+1$, then we get
$\frac{\dim(\Lambda_{3})}{\dim(\Lambda_{1}+\Lambda_{r})}<1$ from
$\sin(\pi(r+1)/(1+r+k))<\sin(\pi(r+2)/(1+r+k))$ and the fact that
$\frac{\sin(x)\sin((r-1)x)}{\sin(2x)\sin(3x)}$ is decreasing for
$0<x<\frac{\pi}{4}$. If $r\in\\{5,6,7\\}$ and $k\leq r+1$ then there are just
a small number of cases to check. If $r=8$ then
$\frac{\dim(\Lambda_{3})}{\dim(\Lambda_{1}+\Lambda_{r})}$ is a strictly
increasing function of $k$, which equals $1$ at $k=15$. When $r\geq 9$, $k\geq
4$, and $1+r+k\geq 19$ we can use Equation (A.1) again to obtain

$\frac{\dim(\Lambda_{3})}{\dim(\Lambda_{1}+\Lambda_{r})}>1.$

When $r\geq 9$ and $k=3$ we get

$\frac{\dim(\Lambda_{3})}{\dim(\Lambda_{1}+\Lambda_{r})}=\frac{\sin\left(\frac{\pi}{1+r+k}\right)\sin\left(\frac{5\pi}{1+r+k}\right)}{\sin\left(\frac{2\pi}{1+r+k}\right)^{2}}>1.$

With the $b=3$ case done we can now assume $r\geq 7$.

Finally for the case of $b=4$ we have that when $r=7$ and $k\geq 3$, we have
$\frac{\dim(\Lambda_{4})}{\dim(\Lambda_{1}+\Lambda_{r})}>1$.

Thus all we have remaining is a finite list of pairs $(r,k)$ where we could
possibly have $\dim(\Lambda_{b})=\dim(\Lambda_{1}+\Lambda_{r})$. Searching
these pairs, and applying level-rank duality gives the statement of the
proposition. ∎

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# A Model for Nonthermal Particle Acceleration in Relativistic Magnetic
Reconnection

Xiaocan Li Dartmouth College, Hanover, NH 03750 USA  Fan Guo Los Alamos
National Laboratory, Los Alamos, NM 87545, USA  Yi-Hsin Liu Dartmouth College,
Hanover, NH 03750 USA  Hui Li Los Alamos National Laboratory, Los Alamos, NM
87545, USA

(Received ****; Revised ****; Accepted ****)

###### Abstract

The past decade has seen an outstanding development of nonthermal particle
acceleration in magnetic reconnection in magnetically-dominated systems, with
clear signatures of power-law energy distributions as a common outcome of
first-principles kinetic simulations. Here we propose a semi-analytical model
for systematically investigating nonthermal particle acceleration in
reconnection. We show particle energy distributions are well determined by
particle injection, acceleration, and escape processes. Using a series of
kinetic simulations, we accurately evaluate the energy- and time-dependent
model coefficients. The resulting spectral characteristics, including the
spectral index and lower and upper bounds of the power-law distribution, agree
well with the simulation results. Finally, we apply the model to predict the
power-law indices and break energies in astrophysical reconnection systems.

High energy astrophysics (739); Plasma astrophysics (1261)

††journal: ApJL

##  1 Introduction

Nonthermal particle acceleration processes during magnetic reconnection are
plausible mechanisms responsible for high-energy emissions observed in
magnetically dominated systems, such as relativistic jets from gamma-ray
bursts (GRBs) (Zhang & Yan, 2011; McKinney & Uzdensky, 2012) and active
galactic nuclei (AGN) (Giannios et al., 2009; Zhang et al., 2015, 2018),
pulsar wind nebulae (PWNe) (Uzdensky & Spitkovsky, 2014), and solar flares
(Lin et al., 2003). Recently, interest has surged in studying particle
acceleration and resulting particle energy spectra, mainly via particle-in-
cell simulations (Guo et al., 2014; Sironi & Spitkovsky, 2014). These studies
have successfully obtained power-law energy spectra
$f(\varepsilon)\propto\varepsilon^{-p}$ with spectral index $p$ decreasing
with the plasma magnetization $\sigma$ and approaching 1 when $\sigma\gtrsim
10$ (Sironi & Spitkovsky, 2014; Guo et al., 2014, 2016; Werner et al., 2016).
However, the origin of the nonthermal energy spectrum and how it varies in
different situations is still undergoing active debate.

Despite the advances in kinetic simulations (Hoshino et al., 2001; Zenitani &
Hoshino, 2001; Sironi & Spitkovsky, 2014; Liu et al., 2011; Dahlin et al.,
2014; Guo et al., 2014, 2016; Li et al., 2015, 2018), theoretical explorations
on particle acceleration have limited success in predicting the resulting
energetic particle spectrum. Although a simple derivation shows $p\sim 1$ when
particle escape is ignored (Guo et al., 2014, 2015), numerous simulations have
consistently obtained $p>2$, indicating some key physics is still missing
(Werner et al., 2018; Ball et al., 2018; Uzdensky, 2022). In addition, how the
resulting energy spectra vary with the guide field, and how the results from
kinetic simulations can be extrapolated to astrophysical scales are not clear.

In this Letter, we present a model for particle acceleration in magnetic
reconnection. In contrast to previous studies, we consider particle injection,
acceleration, and escape together as key ingredients for understanding the
energy spectra and evaluate them in a reconnection system. In a series of
fully kinetic simulations, we quantify the acceleration (both first order and
second order) and escape processes with their time and energy dependence. We
use the model coefficients to predict the nonthermal power-law characteristics
(spectral indices, lower and higher bounds of the power-laws) based on a
Fokker-Planck approach and find that the prediction matches the simulations
very well. While our model is capable of explaining nonthermal acceleration in
kinetic simulations in general, it also reveals essential physical factors
that determine energetic particle spectrum, making a crucial step toward a
comprehensive understanding of those processes in large-scale astrophysical
systems.

##  2 A Model

Fig. 1(a) illustrates our model. The reconnection inflow continuously brings
thermal particles into the reconnection layer, where a fraction of them can be
accelerated out of the thermal pool through particle injection processes (Ball
et al., 2019; Kilian et al., 2020; Sironi, 2022; Guo et al., 2023; French et
al., 2023). A primary acceleration phase then leads to the formation of power-
law distribution. While most reconnection studies focus on the first-order
acceleration (Guo et al., 2014, 2015), our model includes the second-order
Fermi acceleration, which is found to be comparable to the first-order
mechanism (see below). This second-order acceleration can be due to
inhomogeneous distribution of island acceleration, or turbulent like behavior,
similar to relativistic turbulence (Wong et al., 2020; Comisso & Sironi, 2018,
2019). In addition, we find that particle escape plays an important role in
determining the spectral shape. Since particles do not participate in the
acceleration once they are trapped by the largest magnetic islands/flux ropes,
they are considered separately as an effective “escape” process. Together, we
consider these processes in a Fokker-Planck approach (Blandford & Eichler,
1987) describing the evolution of the particle energy distribution
$f(\varepsilon)$, according to

$\partial_{t}f+\partial_{\varepsilon}(\alpha_{\text{acc}}\varepsilon
f)=\partial_{\varepsilon}^{2}(D_{\varepsilon\varepsilon}f)-\alpha_{\text{esc}}f+\frac{f_{\text{inj}}}{\tau_{\text{inj}}},$
(1)

where $\varepsilon=(\gamma-1)m_{s}c^{2}$ is the kinetic energy,
$\alpha_{\text{acc}}$ is the acceleration rate,
$D_{\varepsilon\varepsilon}\coloneqq D_{0}\varepsilon^{2}$ is the energy
diffusion coefficient, $\alpha_{\text{esc}}\equiv\tau_{\text{esc}}^{-1}$ is
the escape rate, $f_{\text{inj}}$ is the injected thermal particle
distribution, and $\tau_{\text{inj}}$ is particle injection time scale.
$\alpha_{\text{acc}}\coloneqq(\partial_{t}\varepsilon+\partial_{\varepsilon}D_{\varepsilon\varepsilon})\varepsilon^{-1}$
describes a combination of the first-order Fermi processes and the
accompanying first-order term associated with second-order Fermi mechanisms.
Instead of separating the injection and nonthermal acceleration, we will treat
them as a continuous process and determine $\alpha_{\text{acc}}$ and $D_{0}$
due to parallel or perpendicular electric fields ($\bm{E}_{\parallel}$ and
$\bm{E}_{\perp}$).

We assume that reconnection starts from a single elongated current sheet (with
a length of $L_{x}$), which breaks into a series of magnetic islands (or flux
ropes in 3D) (Loureiro et al., 2007; Bhattacharjee et al., 2009). These
islands tend to merge to form larger islands, and secondary islands are
continuously generated. The entire process lasts a few Alfvén-crossing times
$\tau_{A}\coloneqq L_{x}/V_{Ax}$ until most magnetic flux is reconnected,
where $V_{Ax}$ is the reconnection outflow speed. The total number of
particles in the reconnection layer due to the reconnection inflow
$dN/dt\propto RV_{Ax}$ (where the reconnection rate $R\sim 0.1$ (Liu et al.,
2017; Goodbred & Liu, 2022)). The largest islands grow in the size of
$L_{\text{O}}=d_{i}+RV_{Ax}t$, assuming they are initially in the $d_{i}$ (ion
inertial length) scale. For the rest of the discussion,
$n_{\text{acc}}(\varepsilon)$ and $n_{\text{esc}}(\varepsilon)$ are the
numbers of the accelerating and escaped particles in different energy bins,
respectively.

Figure 1:  (a) A cartoon illustrating the main processes. The blue and orange
regions represent the acceleration and escape regions, respectively. The black
curve shows one typical particle trajectory. (b) Two electron trajectories in
the kinetic simulations. The orange crosses indicate when particles reach
their final energies. The background is at $t\approx 2.44\tau_{lc}$, where
$\tau_{lc}=L_{x}/c$. (c) Time evolution of the electron energies. The colored
lines are the smoothed data that removed the gyromotion effects.

For a single electron, the acceleration rate due to the parallel electric
field is $-e\bm{E}_{\parallel}\cdot\bm{v}/\varepsilon$. Since
$E_{\parallel}\approx E_{R}=RV_{Ax}B_{0}/c$ near the X-lines and $\approx 0$
rest of the reconnection layer, the acceleration rate due to
$\bm{E}_{\parallel}$ is

$\displaystyle\alpha_{\text{acc}\parallel}$
$\displaystyle=\left<-e\bm{E}_{\parallel}\cdot\bm{v}/\varepsilon\right>,$ (2)
$\displaystyle\approx
ecE_{R}\left<\mu_{E}\right>_{\text{X}}\varepsilon^{-1}F_{\text{X}},$ (3)

where $\left<\cdots\right>$ indicates the ensemble average over
$n_{\text{acc}}(\varepsilon)$ particles, $\left<\cdots\right>_{\text{X}}$ is
the average over $n_{\text{X}}(\varepsilon)$ particles near the X-lines, and
$\mu_{E}\coloneqq-\bm{v}\cdot\bm{E}_{\parallel}/(vE_{\parallel})$. The filling
factor of these electrons $F_{\text{X}}\sim d_{e}/L_{\text{O}}\sim t^{-1}$ for
large $t$. $\left<\mu_{E}\right>_{\text{X}}\approx 0$ for low-energy thermal
electrons with nearly isotropic distributions and increases when they are
accelerated by $\bm{E}_{\parallel}$ or the Fermi mechanism (Drake et al.,
2006). We will determine the exact energy dependence using the simulation
results. The energy diffusion rate due to $\bm{E}_{\parallel}$ is
$D_{0\parallel}\coloneqq\left<(\delta\alpha_{\text{acc}\parallel})^{2}\right>\tau_{\text{dec}\parallel}\approx\left(e^{2}c^{2}E_{R}^{2}\left<\mu_{E}^{2}\right>_{\text{X}}\varepsilon^{-2}F_{\text{X}}-\alpha_{\text{acc}\parallel}^{2}\right)\tau_{\text{dec}\parallel}$,
where
$\delta\alpha_{\text{acc}\parallel}=-e\bm{E}_{\parallel}\cdot\bm{v}/\varepsilon-\alpha_{\text{acc}\parallel}$
is the fluctuation of $\alpha_{\text{acc}\parallel}$ and
$\tau_{\text{dec}\parallel}\sim d_{i}/c$ is the decorrelation time for
$\bm{E}_{\parallel}$ (the timescale on which particles see decorrelated
$\bm{E}_{\parallel}$, le Roux et al., 2015). It indicates how fast the
broadening of the energy distribution is due to $\bm{E}_{\parallel}$. Since
$\alpha_{\text{acc}\parallel}^{2}\propto F_{\text{X}}^{2}\propto t^{-2}$ for
large $t$, the first term dominates. For particles with an anisotropy along
the magnetic field, $\left<\mu_{E}^{2}\right>_{\text{X}}$ changes slowly
between 0.5 and 1. Therefore,

$D_{0\parallel}\approx
e^{2}c^{2}E_{R}^{2}\varepsilon^{-2}F_{\text{X}}\tau_{\text{dec}\parallel}.$
(4)

The acceleration rate due to $\bm{E}_{\perp}$ is

$\alpha_{\text{acc}\perp}=\left<-e\bm{E}_{\perp}\cdot\bm{v}/\varepsilon\right>,$
(5)

as particles gaining energy through the Fermi mechanism when colliding with
the reconnection outflow (Guo et al., 2014). The energy gain of each collision
is about
$\Delta\varepsilon=\left(\Gamma_{Ax}^{2}\left(1+2V_{Ax}v_{x}/c^{2}+V_{Ax}^{2}/c^{2}\right)-1\right)\varepsilon$,
and the collision time scale is about $L_{\text{O}}/v_{x}$, where
$\Gamma_{Ax}=(1-V_{Ax}^{2}/c^{2})^{-1/2}$, and $v_{x}$ is the particle
velocity along the reconnection outflow direction. When there is a guide field
$B_{g}=b_{g}B_{0}$, $v_{x}\sim c/(1+b_{g})^{1/2}$ on average, and
$V_{Ax}=c\sqrt{\sigma_{x}/(\sigma+1)}$ (Liu et al., 2015).
$\sigma_{x}=B_{0}^{2}/4\pi w$ is plasma magnetization based on the
reconnection magnetic field $B_{0}$ (Liu et al., 2015), where $w$ is the
enthalpy density. $\sigma=B^{2}/4\pi w$ is total plasma magnetization, where
$B=(B_{0}^{2}+B_{g}^{2})^{1/2}$. Thus, $V_{Ax}$ decreases with $b_{g}$ and
increases with $\sigma_{x}$. We obtain

$\displaystyle\alpha_{\text{acc}\perp}\sim
c\frac{\Gamma_{Ax}^{2}\left(1+2\beta_{Ax}(1+b_{g}^{2})^{-1/2}+\beta_{Ax}^{2}\right)-1}{L_{\text{O}}(1+b_{g}^{2})^{1/2}},$
(6)

where $\beta_{Ax}=V_{Ax}/c$. When the power-law index $p<2$ in highly
magnetized plasmas, most of the kinetic energy is in the high-energy tail
(Sironi & Spitkovsky, 2014; Guo et al., 2014), and the enthalpy density $w$
keeps increasing. The resulting momentum flux density
$w\Gamma_{Ax}^{2}V_{Ax}^{2}$ will keep increasing if $V_{Ax}$ stays constant,
which is not sustainable. Thus, we expect the reconnection outflow to decrease
gradually in highly magnetized plasmas [See Supplemental Material A],
resulting in an approximately constant momentum flux density. The energy
diffusion rate due to the Fermi mechanism is
$D_{0\perp}\sim\Gamma_{Ax}^{2}\beta_{Ax}^{2}c\lambda_{\text{mfp}}^{-1}/3$
(Lemoine, 2019; Comisso & Sironi, 2019), where $\lambda_{\text{mfp}}\sim
L_{\text{O}}(1+b_{g}^{2})^{1/2}$ is scattering mean-free-path. Thus,

$\displaystyle
D_{0\perp}\sim\frac{1}{3}\Gamma_{Ax}^{2}\beta_{Ax}^{2}L_{\text{O}}^{-1}(1+b_{g}^{2})^{-1/2}c.$
(7)

Since energetic particles escape the acceleration region when advected by the
reconnection outflow, the escape rate is

$\displaystyle\alpha_{\text{esc}}$
$\displaystyle=(dn_{\text{esc}}/dt)/n_{\text{acc}},$ (8) $\displaystyle\sim
V_{Ax}/L_{\text{O}}.$ (9)

##  3 Numerical Simulations

We test the proposed model through a series of 2D particle-in-cell kinetic
simulations of magnetic reconnection in a proton-electron plasma in the
$xz$–plane. The simulations all start from a forcefree current sheet (Guo et
al., 2014). The boundaries along $x$ are periodic for both fields and
particles. The boundaries along $z$ are reflective for particles and perfectly
conducting for fields. A long-wavelength perturbation with $\delta
B_{z}=0.03B_{0}$ is included to initiate reconnection (Birn et al., 2001). We
performed the simulations in proton-electron plasmas ($m_{i}/m_{e}=1836$)
similar to earlier studies (Werner et al., 2018; Kilian et al., 2020; Ball et
al., 2019). The cold ion magnetization parameters $\sigma_{ic}\coloneqq
B_{0}^{2}/(4\pi n_{i}m_{i}c^{2})\in[0.4,1.6,6.4,25.6,102.4]$. The guide fields
$b_{g}\in[0.0,0.5,1.0,2.0]$. The plasmas are initially uniform with density
$n_{0}$ and follow the Maxwell–Jüttner distributions with a normalized
electron temperature $\theta_{e}=kT_{e}/m_{e}c^{2}=10\sigma_{ic}$ and
$T_{i}=T_{e}$, resulting in $\sigma_{x}=0.39,1.51,5.2,13.0,19.7$ for the runs
with different $\sigma_{ic}$. The domain sizes are $L_{x}\times
L_{z}=L_{0}\times L_{0}/2$, where $L_{0}=4096\sqrt{\theta_{e}}d_{e0}$ and
$d_{e0}\coloneqq c\sqrt{m_{e}/4\pi n_{0}e^{2}}$ is the nonrelativistic
electron inertial length. The domain is resolved with a grid $n_{x}\times
n_{z}=8192\times 4096$, and the resulting cell sizes $\Delta x=\Delta
z=0.5\sqrt{\theta_{e}}d_{e0}=0.5d_{e}$, where $d_{e}$ is the electron inertial
length including the relativistic correction. With $b_{g}=0.0$, we performed
three additional simulations with $L_{x}\in[L_{0}/4,L_{0}/2,2L_{0}]$ and the
same cell sizes for each $\sigma_{ic}$ to examine the system size dependence.
We use 100 particles/cell/species and track about 1 million tracer particles
at a high cadence to calculate the model coefficients. These tracer particles
are a small, uniformly selected subset of all electrons, and they evolve self-
consistently along with the rest of the particles.

##  4 Model-Simulation Comparison

In all simulations, electron energy spectra develop power-law tails with
various power-law indices. Fig. 2(a) shows a sustainable power-law spectrum
with a stable $p=2.5$ and cutoff energy $\bar{\varepsilon}_{b}$ linearly
increasing with time. This result suggests that the cutoff energy increases
with the box sizes, confirmed in Fig. 2(b). As the magnetization increases,
the reconnection outflow and the acceleration rate due to the motional
electric field will become faster, resulting in a harder spectrum (see Fig.
5). Similarly, the guide field will slow down the outflow and the Fermi
acceleration, leading to a softer spectrum (see Fig. 5). We will demonstrate
that the model can capture these dependences.

Figure 2:  Electron energy spectra in the runs with $\sigma_{ic}=6.4$. (a)
Time evolution of the spectra when $b_{g}=0.0$. The insets show the time
evolution of the cutoff energy and the power-law index. (b) The spectra in
runs with $b_{g}=0.0$ but different $L_{x}$. The inset shows the cutoff energy
changing with $L_{x}$. $L_{0}=4096d_{e}$ for these runs.

To evaluate the model coefficients in simulations, we must separate particles
into the accelerating and the escaped populations. Fig. 1(b) & (c) shows two
electron trajectories to illustrate the escape mechanism. The two electrons
gain substantial energies before being confined inside the large magnetic
island. Their kinetic energies fluctuate without much change after that. Thus,
they are treated as escaped particles after their acceleration stops. After
separating the two populations, one can calculate acceleration and escape
rates using Eqs. (2), (5), and (8). The process is repeated at every step of
the tracer particles, allowing us to measure how the rates change over time.
We evaluate $D_{0}$ by tracking the energy spread of the particles in
different energy bins (Comisso & Sironi, 2019; Wong et al., 2020). Fig. 3
shows the energy dependence of the rates in the runs with $\sigma_{ic}=102.4$.
$\alpha_{\text{acc}\parallel}$ peaks around $\bar{\varepsilon}=10$ and
decreases when $\bar{\varepsilon}>10$, following the scaling in Eq. (3). Fig.
3(b) shows that $\alpha_{\text{acc}\parallel}$ peaks at higher energies when
$b_{g}>0$, due to the increasing $\left<\mu_{E}\right>_{\text{X}}$ with
$b_{g}$ (Li et al., 2018). The transition energy
$\bar{\varepsilon}_{t}=\varepsilon_{t}/(kT_{e})\approx 12+10\tanh(2b_{g})$
according to the simulations, and the dashed lines in Fig. 3(b) follow the
energy dependence

$\displaystyle\alpha_{\text{acc}\parallel}\sim 2.5\times
10^{-5}\frac{\bar{\varepsilon}_{t}^{3}(\bar{\varepsilon}/\bar{\varepsilon}_{t})^{2}}{\left(1+(\bar{\varepsilon}/\bar{\varepsilon}_{t})^{2}\right)^{3/2}},$
(10)

which is $\sim\bar{\varepsilon}^{2}$ for low-energy particles and
$\sim\bar{\varepsilon}^{-1}$ for high-energy particles (Eq. 3). This energy
dependence is determined by fitting the energy-dependent
$\alpha_{\text{acc}\parallel}$ to achieve a smooth transition between the two
parts. As the rates exhibit significant temporal fluctuations, as illustrated
in Fig. 4, the coefficient should be treated as an approximation rather than
an exact value. $\alpha_{\text{acc}\perp}$ is nearly constant, as expected
from Eq. (6). In Fig. 3(a), we adopt a
$\beta_{Ax}=0.8<\sqrt{\sigma_{x}/(\sigma+1)}$ due to the decelerating
reconnection outflow. Fig. 3(a) shows that while the primary acceleration
mechanism at high energies is the Fermi process, the direct and Fermi
mechanisms are comparable at low energies, consistent with prior findings by
Guo et al. (2019). Regarding the escape, $\alpha_{\text{esc}}$ increases with
particle energy until it saturates at a constant $\sim V_{Ax}/L_{\text{O}}$,
consistent with Eq. (9). When $b_{g}$ is finite, the larger anisotropy
indicates that more particles can escape from the acceleration region by
streaming along the magnetic field lines. Consequently, $\alpha_{\text{esc}}$
peaks around $\varepsilon_{t}$ (Fig. 3(c)), at which the anisotropy is the
largest. We use the following function to model the energy dependency of
$\alpha_{\text{esc}}$ from low to high energies.

$\displaystyle\alpha_{\text{esc}}\sim 6\times
10^{-4}\frac{\bar{\varepsilon}^{2}\bar{\varepsilon}_{t}^{2}}{\bar{\varepsilon}^{2}+\bar{\varepsilon}_{t}^{2}}\left[\left(1-\frac{\bar{\varepsilon}_{t0}^{2}}{\bar{\varepsilon}_{t}^{2}}\right)\frac{10\bar{\varepsilon}_{t}^{2}}{10\bar{\varepsilon}_{t}^{2}+\bar{\varepsilon}^{2}}+\frac{\bar{\varepsilon}_{t0}^{2}}{\bar{\varepsilon}_{t}^{2}}\right],$
(11)

where $\bar{\varepsilon}_{t0}$ is $\bar{\varepsilon}_{t}(b_{g}=0)$.
$\alpha_{\text{esc}}\propto\varepsilon^{2}$ for low-energy particles and
approaches $\sim V_{Ax}/L_{\text{O}}$ for high-energy particles (Eq. 9).
Similar to Eq. 10, the energy dependence is established by fitting energy-
dependent $\alpha_{\text{esc}}$ to achieve a smooth transition between the
low-energy and high-energy parts. The complex form accounts for the peak that
arises near the transition energy $\varepsilon_{t}$. Fig. 3 (c) shows some
discrepancies at low energies, especially for $b_{g}=2$ (purple curve) case.
Such discrepancies are caused by the strong temporal fluctuations in the rates
(Fig. 4), likely due to the initial condition or boundary conditions in the
simulations. Fig. 3(a) shows that $D_{0}$ is dominated by
$D_{0\parallel}\propto\varepsilon^{-2}$ (Eq. 4) at low energies and
$D_{0\perp}\sim$ constant (Eq. 7) at high energies.

Figure 3:  The energy dependence of the rates in the runs with
$\sigma_{ic}=102.4$. (a) The rates in the run with $b_{g}=0$. The dashed lines
are based on the model scalings. (b) $\alpha_{\text{acc}\parallel}$ for the
runs with different guide field. The dashed lines follow the scaling. (c)
$\alpha_{\text{esc}}$ for these runs.

Fig. 4 shows how the rates change with time. Fig. 4(a) shows that
$\alpha_{\text{acc}\parallel}\propto t^{-1}$, following the scaling in Eq.
(3). It is larger when there is a finite guide field, agreeing with Fig. 3(b).
Although $\alpha_{\text{acc}\perp}$ strongly fluctuates (Fig. 4(b)), it
decreases with time, as expected from Eq. (6). $\alpha_{\text{acc}\perp}$ can
be $\propto t^{-2}$ when $\sigma$ is large enough due to the decelerating
reconnection outflow [See Supplemental Material A]. Fig. 4(c) shows that
$\alpha_{\text{esc}}\sim t^{-1}$, consistent with Eq. (9). For more model-
simulation comparison of the time dependence, see Supplement Material B.

Figure 4:  (a) The time dependence of the rates within the selected energy bin
for runs with different $b_{g}$. (b) and (c) The acceleration and escape rates
for runs with different $\sigma_{ic}$. The dashed lines follow the predicted
scalings in Eqs. (3), (6), (9) except in (b), where
$\alpha_{\text{acc}\perp}\propto t^{-2}$ when $\sigma_{ic}=102.4$.

We then solve the Fokker-Planck equation using the energy- and time-dependent
rates [See Supplemental Material C for details].

Figure 5:  (a) Comparing the electron spectra obtained from the model (thick
lines) and those in the simulations (thin lines) with $b_{g}=0.0$ and
different $\sigma_{ic}$. (b) Comparing the spectra in the runs with
$\sigma_{ic}=102.4$ and different $b_{g}$. The insets of panels (a) and (b)
show the power-law indices and break energies obtained from the models for
systems of different sizes. The vertical black line indicate the size of the
PIC simulations ($L_{0}$). (c) Power-law indices in astrophysically relevant
systems with different $\sigma_{x}$ and $b_{g}$. The dashed are the fittings.
(d) The gyroradius of the electrons with the break energies.

Fig.5 (a) and (b) show that the modeled spectra (thick lines) agree well with
those in the simulations (thin lines). The model can capture the spectra’
$\sigma$\- and $b_{g}$\- dependences and the power-law break energies
$\varepsilon_{b}$. Despite some notable agreements between the simulation and
model results, as depicted by the overlaying orange curves in Fig. 5 (a) and
the green and blue curves in Fig. 5 (b), there exist discrepancies between the
simulation results and model results. Such discrepancies can be attributed to
the settings of the simulations, such as the initial and boundary conditions,
as well as the constrained system sizes. Consequently, the model results are
anticipated to reveal the overall patterns but not necessarily replicate the
simulation results exactly.

Since the model is simple enough, we can predict the spectra for much larger
systems and study their long-term evolution. The insets of panels (a) and (b)
show that $p$ gradually increases until saturation at a constant value in
large enough systems. Since the guide field plays a more important role in the
high-$b_{g}$ reconnection, it takes a larger box (up to $10^{3}L_{0}$ when
$b_{g}=2$) for the spectra to saturate. The model also shows that the break
energy linearly increases with the system size up to $10^{6}$ times the PIC
simulation sizes. Thus, the model can predict the spectra in astrophysical
reconnection sites. Fig. 5 (c) and (d) show the spectral indices and the
gyroradius of the break energies, normalized by the system sizes $L$. When
$b_{g}=0$, $p$ approaches 2.1 as the $\sigma_{x}$ increases, indicating that
the power-law extension can keep growing without causing the kinetic energy to
diverge. When $b_{g}$ is finite, the spectra are the hardest when the
$\sigma_{x}$ is between 1 and 10. $p$ can be approximated as

$\displaystyle
p\approx\frac{1}{\sigma_{x}+0.2(1+\tanh(b_{g}))}+0.04\tanh(b_{g})\sigma_{x}+1.7b_{g}+2.1,$
(12)

where the first two terms capture the variation with $\sigma_{x}$ when the
guide field is finite. When $b_{g}=0$, $p\approx(\sigma_{x}+0.2)^{-1}+2.1$.
For a fixed magnetization $\sigma_{x}$, the spectrum becomes softer as the
guide field becomes stronger. Fig. 5 (d) shows the gyroradius of the electrons
with the break energies
($\rho_{b}\coloneqq(\varepsilon_{b}/m_{e}c^{2}+1)m_{e}c/eB_{0}$) is a fraction
of the system size $L$. When $b_{g}=0$, $\rho_{b}$ increases with $\sigma_{x}$
and approaches 0.007 times the system sizes at $2\tau_{A}$. When $b_{g}$ is
finite, $\rho_{b}$ peaks when $\sigma_{x}$ is a few, suggesting the
acceleration is most efficient when the $\sigma_{x}$ is between 1 and 10.
$\rho_{b}$ can be approximated as

$\displaystyle\frac{\rho_{b}}{L}=0.007(1-e^{-0.6\sigma_{x}})e^{-0.06\tanh(b_{g})\sigma_{x}}(1+b_{g}^{2})^{-1}$
(13)

at $2\tau_{A}$. When $b_{g}=0$, $\rho_{b}/L=0.007(1-e^{-0.6\sigma_{x}})$. When
$b_{g}$ is finite, $\rho_{b}$ peaks when the $\sigma_{x}$ is between 1 and 10.
For a fixed $\sigma_{x}$, $\rho_{b}$ monotonically decreases with the guide
field. Note that Eqs. 12 and 13 rely only on the hot magnetization parameter
$\sigma_{x}$ and the guide field strength $b_{g}$. Nonetheless, the equations
are applicable to proton-electron plasmas under the conditions where
$0.1<\sigma_{x}<100$ and electrons are relativistic. Although the equations
can be potentially used for pair plasmas as they do not depend on the ion-to-
electron mass ratio, they have not been thoroughly verified against fully
kinetic simulations using pair plasmas.

##  5 Discussion and Conclusions

The present study builds upon the prior research by Guo et al. (2014, 2015)
and makes improvements in the following aspects: firstly, instead of
separating the injection and the nonthermal acceleration, it treats them as a
continuous process and incorporates the energy-dependence of acceleration
rates; secondly, it incorporates 2nd-order acceleration, including those due
to the parallel electric field and the Fermi mechanism; thirdly, it accurately
evaluates particle escape rate in the simulations and compares it with the
model results; fourthly, it extends the model to proton-electron plasmas; and
lastly, besides plasma magnetization, it considers the effect of guide fields
on particle acceleration. While these processes are more or less studied in
previous studies individually, this study is the first attempt to put these
processes together in the same model.

Some important physics is still missing in the current model and could be the
key to addressing the particle acceleration in the astrophysical reconnection
sites. First, previous research has demonstrated that radiative cooling, such
as synchrotron or inverse Compton cooling, significantly impacts the
acceleration mechanisms and the dynamics of the current layer, even when
cooling is only marginally important (Zhang et al., 2018; Werner et al., 2019;
Hakobyan et al., 2019; Sironi & Beloborodov, 2020; Sridhar et al., 2021).
Second, 3D reconnection can self-generate plasma turbulence (Daughton et al.,
2011; Guo et al., 2014), which could affect particle transport and
acceleration (Dahlin et al., 2015; Li et al., 2019; Zhang et al., 2021b), and
change particle acceleration and escape mechanisms (Zhang et al., 2021a).
However, other studies have shown that 3D physics is not as crucial in the
relativistic regime as in the non-relativistic regime (Guo et al., 2021). Our
preliminary results indicate that the spectra are not saturated to the end of
the simulation due to the limited system sizes of the 3D simulations. Thus,
incorporating 3D effects in the model requires more studies using massive 3D
kinetic simulations, which are currently not affordable for revealing the
scalings in system sizes. Third, the simulation setup, including boundary
conditions and initialization (Ball et al., 2018), may impact the energy
dependence and the time evolution of the rates, as shown in Figs. 3 and 4.
These could lead to the deviation of the modeling predictions from the
simulation results, given that the semi-analytical model still needs empirical
fitting of the kinetic simulation results. These simulations are essential in
determining the energy dependencies at low energies and quantifying the
constants in each rate for different simulation setups. Addressing all these
factors using fully kinetic simulations is critical to making the model more
reliable for extending the kinetic simulation results to astrophysical
reconnection systems.

In this Letter, we presented a model for determining the main characteristics
of power-law particle energy spectra in magnetic reconnection in magnetically
dominated plasmas. The power-law spectrum produced by this model is controlled
by particle injection, acceleration, energy diffusion, and escape processes.
Using a series of first-principles 2D kinetic simulations, we evaluate these
transport coefficients and obtained nice agreements with simulation results.
By solving the Fokker-Planck equation describing these processes using the
modeled transport coefficients, we demonstrate that the modeled spectra agree
well with the simulation results for a broad range of magnetization and guide
fields. We then use the simulation-verified model to predict the power-law
spectral indices and break energies for astrophysically relevant reconnection
systems and provide their empirical expressions for different magnetization
and guide fields. Our results have strong implications to understanding
nonthermal particle acceleration and emissions in high-energy astrophysical
systems.

We thank the anonymous referee for a constructive review. We gratefully
acknowledge our discussions with William Daughton and Patrick Kilian. X.L.
acknowledges the support from NASA through Grant 80NSSC21K1313, National
Science Foundation Grant No. AST-2107745, and Los Alamos National Laboratory
through subcontract No. 622828. The simulations used resources provided by the
National Energy Research Scientific Computing Center (NERSC, a U.S. Department
of Energy Office of Science User Facility at Lawrence Berkeley National
Laboratory) and the Texas Advanced Computing Center (TACC, at the University
of Texas at Austin).

##  Appendix A Decelerating Reconnection Outflow

Fig. 6 shows that the reconnection outflow is decelerating as reconnection
evolves. The deceleration is the strongest when the guide field is weak and
the particle acceleration is the most efficient.

Figure 6:  Time evolution of $V_{x-\text{max}}\Gamma(V_{x-\text{max}})$ in
simulations with $\sigma_{ic}=102.4$, where $V_{x-\text{max}}$ is the maximum
outflow speed at the mid-plane and $\Gamma(V_{x-\text{max}})$ is the
corresponding Lorentz factor. $\tau_{lc}$ is the light crossing time.

##  Appendix B Time Dependence of the Rates

Fig. 7 shows how the rates change with time for runs with different $b_{g}$.
Although $\alpha_{\text{acc}\perp}$ strongly fluctuates (Fig. 7(a)), it
decreases with time and $b_{g}$, as expected from Eq. (6) in the main text.
When $b_{g}=0$, due to the decelerating reconnection outflow [See Supplemental
Material C], $\alpha_{\text{acc}\perp}$ decreases faster than $t^{-1}$. Fig.
7(b) shows that $\alpha_{\text{esc}}$ is higher when there is a finite guide
field. $\alpha_{\text{esc}}\sim t^{-1}$ (consistent with Eq. (9) in the main
text) even as the reconnection outflow slows down when $b_{g}=0$ and $\sigma$
is large (Fig. 7(f)) because $V_{Ax}$ can be close to $c$ while $\Gamma_{Ax}$
decreases significantly. The energy diffusion rate $D_{0}$ strongly fluctuates
but gradually decreases with $t$. When $b_{g}=0$, it decreases faster than
$t^{-1}$ due to the decelerating outflow.

Figure 7:  The time dependence of the rates within the selected energy bin for
runs with different $b_{g}$. The dashed lines follow the predicted scalings
$\sim L_{\text{O}}^{-1}\sim t^{-1}$ in Eqs. (6), (9), and (7).

##  Appendix C Solving the Fokker-Planck Equation

We solve the Fokker-Planck equation (1) using the energy- and time-dependent
rates. The acceleration part of Eq. (1) is equivalent to the stochastic
differential equation (SDE) of the Itô type,

$\displaystyle
d\varepsilon=(\alpha_{\text{acc}\parallel}+\alpha_{\text{acc}\perp})\varepsilon
dt+\sqrt{2(D_{0\parallel}+D_{0\perp})\varepsilon^{2}}dW_{t},$ (C1)

where $W_{t}$ is the standard Wiener process, and $dW_{t}$ is the normalized
distributed random number with mean 0 and variance $\Delta t$. The SDE can be
solved using pseudo particles. We apply a taper function
$\varepsilon_{\text{O}}/(\varepsilon_{\text{O}}+\varepsilon)$ to
$\alpha_{\text{acc}\perp}$ and $D_{0\perp}$ when solving the equation to model
the reduction of the rates with particle energies when their gyroradius is
close to the largest island size. Specifically, we use

$\displaystyle\varepsilon_{\text{O}}=eB_{0}\sqrt{1+b_{g}^{2}}L_{\text{O}}/(10m_{e}c\Gamma_{Ax}),$
(C2)

which includes the correction when the outflow is relativistic. Motivated by
the simulation results, we model the reconnection rate $R\approx
0.1V_{Ax}/V_{A0}$, where $V_{A0}=\sqrt{\sigma_{x}/(\sigma_{x}+1)}$. As a
result, the number of injected particles and the parallel electric field have
a strong dependence on the guide field. After updating the energies of the
pseudo particles at each time step according to Eq. (C1), we determine whether
the particles escape according to the escape rate and inject new pseudo
particles. Before the next cycle, we evaluate the enthalpy density of the
particles and update the outflow speed $V_{Ax}$ assuming a constant momentum
flux density.

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