# Upper Bounds for Totally Symmetric Sets

Kevin Kordek, Qiao Li, and Caleb Partin

###### Abstract.

Totally symmetric sets are a recently introduced tool for studying
homomorphisms between groups. In this paper, we give full classifications of
totally symmetric sets in certain families of groups and bound their sizes in
others. As a consequence, we derive restrictions on possible homomorphisms
between these groups. One sample application of our results is that any
homomorphism of a braid group to a direct product of solvable groups must have
cyclic image.

##  1\. Introduction

Given any two groups $G$ and $H$, understanding all the possible homomorphisms
$G\xrightarrow[]{}H$ is a naturally important but technically difficult task.
This motivates using a carefully-chosen subset of the information in the given
groups to constrain and eventually classify homomorphisms. As a simple
example, cyclic groups and subgroups map to cyclic images. Analogously, the
motivation for using totally symmetric sets to study homomorphisms follows
this pattern.

######  Definition 1.1.

A totally symmetric set of a group $G$ is a finite subset
$\\{x_{1},x_{2},...,x_{n}\\}$ of $G$ that satisfies two key properties:

  * •

The elements of the subset commute pairwise.

  * •

Any permutation of the subset can be achieved through conjugation: for any
element of the symmetric group on n letters $\sigma\in S_{n}$, there exists
some $g\in G$ such that $x_{\sigma(i)}=gx_{i}g^{-1}$ for all $1\leq i\leq n$.

Loosely speaking, a totally symmetric set is a subset of elements in a group
characterized by commutativity and conjugation properties, and these
properties are preserved under group homomorphisms. Thus, the image of a
totally symmetric set under a homomorphism is also a totally symmetric set. An
even stronger version of this statement holds: the image of a totally
symmetric set under a group homomorphism must be a totally symmetric set of
the same size or a singleton. This fundamental lemma, shown in section 2, is a
major obstruction to the possible homomorphisms between two groups.

Totally symmetric sets are first defined by Kordek and Margalit in the context
of the braid group [KM19]. The fact that odd Artin generators of the braid
group form a totally symmetric set is exploited in characterizing
homomorphisms from certain braid group to other certain braid groups. The same
fact has been used in the work of the authors along with Chudnovsky, where
they derive a lower bound on the size of non-cyclic quotients of the braid
group [CKLP20]. It has also made an appearance in the work of Caplinger and
Kordek [CK20], the work of Scherich and Verberne [SV20], and the work of Chen
and Mukherjea [CM20]. All of these results vitally depend on the properties of
totally symmetric sets.

### Overview

In this paper, we will develop and present some theory about totally symmetric
sets as well as a full classification of them in several families of groups.
In Section 2, we give motivating examples of totally symmetric sets and
present a proof for the fundamental lemma. In Section 3, we give a full
classification of totally symmetric sets in a few special families of groups.
In Section 4, we investigate upper bounds on sizes of totally symmetric sets
in various products of groups. In Section 5, we give constant bounds on sizes
of totally symmetric sets in groups of odd order and solvable groups through
the action of the stabilizer. Finally, in Section 6, we present a table of
sizes of totally symmetric sets and derive classifications of homomorphisms.

### Acknowledgments

The majority of this work was completed during the summer of 2019 during the
Georgia Institute of Technology Mathematics REU under the guidance of Dan
Margalit and Kevin Kordek. The authors would like to thank Dan Margalit for
his mentorship and guidance throughout this project, Alice Chudnovsky for
useful discussions during the Georgia Tech Mathematics REU, and Santana Afton
for helpful conversations on free products.

##  2\. Fundamental Lemma and Examples

Totally symmetric sets are a powerful aid to the study of homomorphisms due to
the following fundamental lemma due to Kordek and Margalit [KM19]:

######  Lemma 2.1 (The Fundamental Lemma of Totally Symmetric Sets).

Let $\varphi:G\xrightarrow{}H$ be a homomorphism between two groups, and
$S\subset G$ a totally symmetric set of size $n$. Then $\varphi(S)$ is a
totally symmetric set in $H$ with $\lvert\varphi(S)\rvert=n$ or
$\lvert\varphi(S)\rvert=1$.

###### Proof.

We will first show that $\varphi(S)$ is a totally symmetric set of $H$ given
that $S$ is a totally symmetric set of $G$. Let
$S=\\{x_{1},x_{2},\dots,x_{n}\\}\subset G$. It is clear that the elements of
the set $\\{\varphi(x_{1}),\varphi(x_{2}),\dots,\varphi(x_{n})\\}$ pairwise
commute, so we just need to show they also satisfy the conjugation
requirement. Given $\sigma\in S_{n}$, let $g_{\sigma}\in G$ be the element
inducing the permutation given by $\sigma$ on $S$, i.e.
$x_{\sigma(i)}=g_{\sigma}x_{i}g_{\sigma}^{-1}$ for all $i$. Applying $\varphi$
gives
$\varphi(x_{\sigma(i)})=\varphi(g_{\sigma})\varphi(x_{i})\varphi(g_{\sigma})^{-1}$.
Thus, $\varphi(g_{\sigma})$ achieves the same conjugation for $\varphi(S)$ as
$g_{\sigma}$ does for $S$.

Now for the second part of the lemma, suppose that
$S=\\{x_{1},x_{2},\dots,x_{n}\\}$, and $\varphi(x_{1})=\varphi(x_{2})$. We
need to show that $\varphi(x_{1})=\varphi(x_{i})$ for all $i$. If
$\varphi(x_{1})=\varphi(x_{2})$, then we know that
$\varphi(x_{1}x_{2}^{-1})=e$, and $x_{1}x_{2}^{-1}\in\ker\varphi$. Since $S$
is a totally symmetric set we know that there exists an element $h\in G$ such
that $hx_{1}h^{-1}=x_{1}$ and $hx_{2}h^{-1}=x_{i}$. Therefore,
$x_{1}x_{i}^{-1}=(hx_{1}h^{-1})(hx_{2}^{-1}h^{-1})=hx_{1}x_{2}^{-1}h^{-1}$.
Since $\ker\varphi$ is a normal subgroup, we know that it is closed under
conjugation and $x_{1}x_{i}^{-1}\in\ker\varphi$, so
$\varphi(x_{1})=\varphi(x_{i})$. Thus if $S$ does not map injectively to a
totally symmetric set of size $n$, then it must map to a set of size 1. ∎

An immediate corollary of this lemma is any totally symmetric set of a
subgroup $H\leqslant G$ is a totally symmetric set in $G$ by the inclusion
map.

Totally symmetric sets or their maximum sizes can provide insight about the
possible homomorphisms between two groups. Going forward, we denote the
maximum size of a totally symmetric set in a group $G$ by $S(G)$. We will
discuss this in much greater detail and rigor in section 6. Before diving deep
into the theory, we present two primary motivating examples to show what a
totally symmetric set looks like in the wild.

######  Example 1.

The motivating example for totally symmetric sets comes from the braid group,
$B_{n}$. The braid group is generated by a collection of half-twists,
$\sigma_{1},\sigma_{2},\dots,\sigma_{n-1}$, the Artin generators. Take the set
of odd Artin generators,

$\displaystyle X=\\{\sigma_{2i-1}\\}_{i=1}^{\lfloor{\frac{n}{2}}\rfloor}.$

One can deduce this is a totally symmetric set from the commutation relations
of the braid group and the change-of-coordinates principle from mapping class
group theory [FM12, Section 1.3]. This example is especially useful, since the
commutator subgroup of the braid group, $B^{\prime}_{n}$, is normally
generated by $\sigma_{1}\sigma_{3}^{-1}$; see [Lin04]. From the fundamental
lemma of totally symmetric sets above, we know that if $B_{n}$ maps to a group
$G$ with $S(G)<\lfloor\frac{n}{2}\rfloor$, then the homomorphism collapses $X$
to a singleton. Thus $B^{\prime}_{n}$ is in the kernel of the homomorphism,
and the map factors through the abelianization of $B_{n}$, which is
$\mathbb{Z}$ for all $n$.

This fact that all homomorphisms from $B_{n}$ to a group $G$ with
$S(G)<\lfloor\frac{n}{2}\rfloor$ are cyclic becomes a powerful tool in
studying homomorphisms from the braid group. This is the driving force behind
the recent results on braid groups from the authors, Chudnovsky, and Kordek
[CKLP20], Kordek and Margalit [KM19] and Caplinger and Kordek [CK20].

######  Example 2.

Another example of large totally symmetric sets comes from the symmetric group
$S_{n}$. Take the standard homomorphism $\varphi:B_{n}\rightarrow S_{n}$ where
$\sigma_{i}\mapsto(i\,\,\,\,\,i+1)$. From the fundamental lemma, we know that
the image of the totally symmetric set in Example 1 consisting of odd Artin
generators will also be a totally symmetric set. Therefore, the set of
disjoint transpositions
$\\{(2i-1\,\,\,\,\,2i)\\}_{i=1}^{\lfloor\frac{n}{2}\rfloor}$ is a totally
symmetric set. We will later see that these two examples are rare instances of
large totally symmetric sets, since in the many classifications following this
section, most families of groups will have a constant upper bound on the size
of their totally symmetric sets.

##  3\. Classifying Totally Symmetric Sets in Specific Groups

In this section, we will provide upper bounds for the sizes of totally
symmetric sets of free groups, dihedral groups and a subset of the
Baumslag–Solitar groups. These results will naturally lead to the bounds given
on general products of groups in section 4.

Recall that one of the defining conditions of a totally symmetric set is that
all of its elements pairwise commute. Thus, a natural place to start when
searching for totally symmetric sets is one where all the elements commute,
i.e. abelian groups.

######  Proposition 3.1.

Let $A$ be an abelian group and $S\subset A$ a totally symmetric set, then
$|S|=1$. Hence, the only totally symmetric sets of $A$ are all the singleton
subsets.

###### Proof.

From the definition of a totally symmetric set, we know that any permutation
of the elements of $S$ must be achieved through conjugation in $A$. Since $A$
is abelian, conjugation is always a trivial action, thus the only achievable
permutation is the identity. Since all permutations must be possible, we can
conclude that $|S|=1$. ∎

It is perhaps disappointing to the reader to start out on such an
uninteresting example, but this should motivate us to search elsewhere for
interesting totally symmetric sets. At the risk of over-correcting, we can
investigate the totally symmetric sets of a highly non-abelian group, the free
group.

######  Theorem 3.2.

Let $F_{2}$ be the free group on two generators. For all $G\leqslant F_{2}$,
$S(G)=1$.

###### Proof.

To prove this theorem, we just need to show that $S(F_{2})=1$, that is,
$F_{2}$ only has trivial totally symmetric sets.

First, we will check when any two elements in $F_{2}$ commute. If two elements
$a,b\in F_{2}$ commute, we know that they must generate an abelian subgroup.
The Nielsen–Schreier theorem states that any subgroup of a free group is free,
and the only abelian free group is $F_{1}\cong\mathbb{Z}$. This implies that
both of these elements are powers of the generator of $\mathbb{Z}$. If
$\mathbb{Z}\cong\langle x\rangle$, then $a=x^{n}$ and $b=x^{m}$.

For $a$ and $b$ to be members of a totally symmetric set, there must exist an
element $h\in F_{2}$ such that $hx^{n}h^{-1}=x^{m}$ and $hx^{m}h^{-1}=x^{n}$.
Take the first of these equalities and raise it to the $m$th power, to obtain
the following:

$\displaystyle(hx^{n}h^{-1})^{m}=(x^{m})^{m}=x^{m^{2}}.$

Using our second conjugation equality, we can manipulate the above expression
to see that:

$\displaystyle(hx^{n}h^{-1})^{m}=hx^{n\cdot
m}h^{-1}=(hx^{m}h^{-1})^{n}=(x^{n})^{n}=x^{n^{2}}.$

Thus, $x^{n^{2}}=x^{m^{2}}$ and $n=\pm m$. This reduces our problem to two
distinct cases.

If $n=m$, then these are the same element and our totally symmetric set
consists of a singleton. On the other hand, if $n=-m$, then
$hx^{n}h^{-1}=x^{-n}$ and $hx^{-n}h^{-1}=x^{n}$. Combining these equalities
shows us that $h^{2}x^{n}h^{-2}=x^{n}$, implying $h^{2}$ commutes with $x^{n}$
and therefore $h^{2}$ and $x^{n}$ can both be expressed as some element $y$ to
a power.

Suppose that $h^{2}=y^{a}$ and $x^{n}=y^{b}$. Moreover, we can assume that $y$
is chosen to be minimal in the sense that it cannot be written as another
element $z^{c}$, where $c>1$. This follows from the residual finiteness of the
free groups. Hence, $a$ must be even and $h=y^{a/2}$. Therefore, $h$ commutes
with $x^{n}$, and the equation $hx^{n}h^{-1}=x^{-n}$ reduces to
$x^{n}=x^{-n}$. This implies that $n=0$, and thus $F_{2}$ has trivial totally
symmetric sets. ∎

Note since every free group on a finite or countably infinite number of
generators is a subgroup of $F_{2}$, it follows that all free groups have
trivial totally symmetric sets. Another consequence is that totally symmetric
sets under quotients will not be well-behaved, as every group is a quotient of
some free group.

From these first two examples, we can make an observation: The two properties
that define a totally symmetric set are inherently at odds with each other. On
the one hand, we want to find subsets of elements that all pairwise commute,
so groups that are “more abelian” seem to be the natural place to find large
sets of this nature. Yet we also need conjugation to be efficacious in our
group in order to achieve the full set of permutations, but conjugation is
trivial when elements commute. So if we want to find examples of large totally
symmetric sets, we need to find groups that intuitively achieve this abelian,
non-abelian balance that large totally symmetric sets would require.

After seeing two important classes of groups that both have trivial totally
symmetric sets, we now present our first example of a family of groups
exhibiting non-trivial totally symmetric sets.

######  Theorem 3.3.

The maximal size of a totally symmetric subset of the dihedral group $D_{2n}$
is 2 for all $n\geq 3$. Furthermore, every totally symmetric set of size 2
must take the form $\\{r^{i},r^{-i}\\}$ or $\\{sr^{i},sr^{i+\frac{n}{2}}\\}$
for $0\leq i\leq n-1$, with the latter only occurring if $n$ is divisible by
4.

###### Proof.

We will use the following presentation for the proof: $D_{2n}=\langle
r,s\,|\,r^{n}=1,s^{2}=1,srs=r^{-1}\rangle$. Hence, any element of $D_{2n}$ can
be written as $s^{\epsilon}r^{i}$ where $\epsilon=0,1$ and $0\leq i\leq n-1$.
We first check the necessary conditions for any two elements of the group to
commute with each other. Any $r^{i}$ and $r^{j}$ commute by definition, but
for elements $sr^{i}$ and $sr^{j}$, we use the following implications to show
that if they commute, $i=j$ or $j=i+\frac{n}{2}$:

$\displaystyle sr^{i}sr^{j}=sr^{j}sr^{i}\implies
s^{2}r^{-i}r^{j}=s^{2}r^{-j}r^{i}\implies r^{j-i}=r^{i-j}.$

Thus, $j-i\equiv i-j\text{ mod n}$, or $2(j-i)\equiv 0\text{ mod n}$. Either
$i=j$, or $j-i=\frac{n}{2}\implies j=i+\frac{n}{2}$, which can only occur when
$n$ is even. Therefore, the only nontrivial case of two elements of this form
commuting is $sr^{i}$ and $sr^{i+\frac{n}{2}}$. From the definition of
conjugation and the fact that $s^{2}=1$, two conjugate elements must have the
same exponents on $s$. Hence, we do not consider any further cases and the two
cases discussed above are disjoint from each other.

We will now check the conditions necessary for conjugation. The following
shows two elements of the form $r^{i}$ and $r^{j}$ are conjugate if and only
if they are inverses:

$\displaystyle sr^{k}r^{i}=r^{j}sr^{k}\implies sr^{k+i}=sr^{k-j}\implies
k+i=k-j\implies i=-j.$

This gives the first totally symmetric set $\\{r^{i},r^{-i}\\}$. For our other
candidates, elements of the form $sr^{i}$ and $sr^{i+\frac{n}{2}}$ are
conjugate if and only if $i=j+\frac{n}{4}$.

$\displaystyle sr^{j}sr^{i}=sr^{i+\frac{n}{2}}sr^{j}\implies
s^{2}r^{i-j}=s^{2}r^{j-i-\frac{n}{2}}\implies i-j=j-i-\frac{n}{2}\implies
2(i-j)\equiv\frac{n}{2}\text{ mod n }.$

This implies that $i-j=\frac{n}{4}$ or $i=j+\frac{n}{4}$, which can only occur
if $n$ is divisible by $4$. If two elements of this form are conjugate, there
are no other elements in their conjugancy class, as
$sr^{(i+\frac{n}{2})+\frac{n}{2}}=sr^{i}$. We have thus obtained a full
classification of the totally symmetric sets of $D_{2n}$, and $S(D_{2n})=2$.

Note: This proof extends to the infinite dihedral group as well
$D_{\infty}=\mathbb{Z}\ltimes\mathbb{Z}_{2}$, showing that its totally
symmetric sets also have a maximal size of 2. ∎

The classification in the above proof relies on the convenient presentation of
the dihedral group. While this direct proof approach will not work in general,
there is another class of groups where we can explicitly compute totally
symmetric sets from the group presentation: Baumslag–Solitar groups. For
nonzero integers $m,n$, the Baumslag–Solitar group $BS(m,n)$ is defined by the
following presentation: $\langle a,b\,|\,ba^{m}b^{-1}=a^{n}\rangle$.

######  Theorem 3.4.

The maximal size of a totally symmetric subset of the Baumslag–Solitar group
$BS(1,n)$ is 1 when $n\neq-1$ and 2 when $n=-1$.

###### Proof.

The group $BS(1,n)$ has the one-relation presentation $\langle
a,b\,|\,bab^{-1}=a^{n}\rangle$. This relation is equivalent to $ba=a^{n}b$,
which allows us to write any element of this group in the form $a^{i}b^{j}$.
We can use this fact to check necessary conditions for different elements of
the group to commute and be conjugates. If elements $a^{i}b^{j}$ and
$a^{x}b^{y}$ commute, we have the following implication:

$\displaystyle a^{i}b^{j}a^{x}b^{y}=a^{x}b^{y}a^{i}b^{j}\implies
a^{i}a^{xn^{j}}b^{j}b^{y}=a^{x}a^{in^{y}}b^{j}b^{y}.$

This gives the condition that $i+xn^{j}=x+in^{y}$. We couple this with
necessary conditions for conjugation. If elements $a^{i}b^{j}$ and
$a^{x}b^{y}$ are conjugated to each other by $a^{x_{0}}b^{y_{0}}$, we have
that

$\displaystyle
a^{x_{0}}b^{y_{0}}a^{i}b^{j}=a^{x}b^{y}a^{x_{0}}b^{y_{0}}\implies
a^{x_{0}}a^{in^{y_{0}}}b^{y_{0}}b^{j}=a^{x}a^{x_{0}n^{y}}b^{y}b^{y_{0}}.$

Hence, conjugation give us two conditions: $x_{0}+in^{y_{0}}=x+x_{0}n^{y}$ and
$y_{0}+j=y+y_{0}\implies j=y$. Using the latter condition, we can combine this
with the conditions for commutativity to obtain the following statement:
elements $a^{x}b^{y}$ and $a^{i}b^{y}$ (as $j=y$) will commute and be
conjugate when $i+xn^{y}=x+in^{y}$. Rearranging and solving for $i$, we obtain
an expression for $i$ in terms of $x,y,n$:

$i=\frac{x(1-n^{y})}{1-n^{y}}.$

This implies that $i=x$, which makes the two elements the same, except when
the denominator is 0. For the denominator to be 0, we have three cases to
check. It must either be the case that $y=0$, $y\neq 0$ and $n=1$, or $y=2m$
for some integer $m$ and $n=-1$.

If $y=0$, we only need to check when two elements $a^{x}$ and $a^{y}$ are
conjugate. By definition of a totally symmetric set there exists an element
$h$ swapping $a^{x}$ and $a^{y}$ through conjugation: $ha^{x}h^{-1}=a^{y}$ and
$ha^{y}h^{-1}=a^{x}$. We derive conditions on the exponents by raising the
former equality to the power of $y$:

$\displaystyle(ha^{x}h^{-1})^{y}=a^{y^{y}}\implies
ha^{xy}h^{-1}=a^{y^{2}}\implies(ha^{y}h^{-1})^{x}=a^{y^{2}}\implies(a^{x})^{x}=a^{y^{2}}.$

So $a^{x^{2}}=a^{y^{2}}$ and $x=\pm y$. If $x=y$, then the two elements are
the same. Otherwise, if $x=-y$, we instead have $ha^{x}h^{-1}=a^{-x}$. As
previously discussed, we can represent our general element $h$ as the element
$b^{i}a^{j}$. Plugging this in gives $b^{i}a^{x}b^{-i}=a^{-x}$, which becomes
$a^{xn^{i}}=a^{-x}$, so $x(1+n^{i})=0$. Either $x=0$ or $n^{i}=-1$. The former
case is trivial and the latter can only occur when $n=-1$ and $i$ is an odd
integer. Thus, we can examine this case by assuming $n=-1$ and writing the
conjugating element as $b^{2m+1}a^{j}$. This gives
$b^{2m+1}a^{x}b^{-(2m+1)}=a^{-x}$. Using the conjugation relation, we can see
that conjugating $a^{x}$ an odd number of times by $b$ gives $a^{-x}$, so the
equation is always true. Hence, when $n=-1$, two elements of the form $a^{x}$,
$a^{-x}$ form a totally symmetric set of size 2.

The other case to consider is when $y\neq 0$. This breaks down into two
subcases, the first of which is $n=1$. When $n=1$, the relation for $BS(1,n)$
becomes the commuting relation for $a$ and $b$, implying that $BS(1,1)$ is
abelian and thus only has trivial totally symmetric sets. The other case is
when $n=-1$ and $y=2m$ for $m\in\mathbb{Z}$. Thus, we need to check when two
elements of the form $a^{x}b^{2m}$ and $a^{y}b^{2m}$ commute and are
conjugate. Using our general commuting condition from above, we substitute
$2m$ for $j$ and $y:i+x(-1)^{2m}=x+i(-1)^{2m}$. This reduces to $i+x=x+i$,
which is always true.

We now must check that for any two elements $a^{x}b^{2m}$ and $a^{y}b^{2m}$,
there exists an element $h$ that swaps them through conjugation, i.e.
$ha^{x}b^{2m}h^{-1}=a^{y}b^{2m}$ and $ha^{y}b^{2m}h^{-1}=a^{x}b^{2m}$. As
above, we raise the former equation to $y$ and simplify:

$(ha^{x}b^{2m}h^{-1})^{y}=(a^{y}b^{2m})^{y}\implies
ha^{xy}b^{2my}h^{-1}=a^{y^{2}}b^{2my}\implies(ha^{y}b^{2m}h^{-1})^{x}=a^{y^{2}}\implies(a^{x})^{x}=a^{y^{2}}.$

Therefore, $x^{2}=y^{2}$ and $x=\pm y$. Again, if $x=y$, then the elements are
equal, otherwise if $x=-y$, we can write $h$ as $b^{i}a^{j}$ to obtain the
following:

$\displaystyle b^{i}a^{j}a^{x}b^{2m}a^{-j}b^{-i}$ $\displaystyle=a^{-x}b^{2m}$
$\displaystyle a^{(x+j)(-1)^{i}}b^{2m+i}a^{-j}b^{-i}$
$\displaystyle=a^{-x}b^{2m}$ $\displaystyle
a^{(x+j)(-1)^{i}+(-j)(-1)^{2m+i}}b^{2m}$ $\displaystyle=a^{-x}b^{2m}$
$\displaystyle a^{x(-1)^{i}}b^{2m}$ $\displaystyle=a^{-x}b^{2m}.$

Thus, the elements $a^{x}b^{2m}$ and $a^{-x}b^{2m}$ form a totally symmetric
set. Any other element in such a totally symmetric set would have to satisfy
the above conditions with $a^{x}b^{2m}$ and $a^{-x}b^{2m}$, which is
impossible for an element not equal to either. ∎

For the groups above, we are able to directly compute the totally symmetric
sets by exploiting the existence of their normal forms. In general, we want to
be able to compute the totally symmetric sets of groups without having to rely
on assumptions such as the existence of normal forms or explicit descriptions
of centralizers and conjugacy classes. For many groups, such a direct
computation will not be possible, so we need to introduce more powerful
techniques.

##  4\. Totally Symmetric Sets in Products of Groups

A natural question that arises in studying totally symmetric sets is how they
behave under various products, such as direct products, free products, and
semi-direct products. To this end, we will prove the following theorem:

######  Theorem 4.1.

Let $G$ and $H$ be groups.

$\displaystyle S(G\times H)=\max\\{S(G),S(H)\\}$ $\displaystyle
S(G*H)=\max\\{S(G),S(H)\\}$

The next step after understanding how totally symmetric sets behave under
direct and free products is to explore semi-direct products or other non-
trivial group extensions. While we will not classify all totally symmetric
sets in semi-direct products, we will generalize our results on dihedral
groups from the previous section.

######  Proposition 4.2.

For $p$ prime and $m$ any integer where $p|m$,
$S(\mathbb{Z}_{p}\ltimes\mathbb{Z}_{m})=2$.

###  4.1. Direct and Free Products

While we may expect that taking the direct product of two groups provides a
way of creating larger totally symmetric sets, we begin by showing that direct
products do not create larger totally symmetric sets.

######  Lemma 4.3.

Let $G$ and $H$ both be groups and $S$ a subset of $G\times H$. If there
exists elements $(x_{1},y_{1}),(x_{1},y_{2}),(x_{2},y)\in S$, with $x_{1}\neq
x_{2}$, $y_{1}\neq y_{2}$, and $y$ any element in $H$, then $S$ cannot be a
totally symmetric subset of $G\times H$.

###### Proof.

Suppose that $S$ is a totally symmetric subset of $G$, and it contains
elements of the form above. Since $S$ is a totally symmetric set, we know that
there is a permutation $\phi$ that sends $(x_{1},y_{1})$ to $(x_{1},y_{2})$
and $(x_{1},y_{2})$ to $(x_{2},y)$, and there is a conjugating element
$(h_{G},h_{H})$ that achieves this permutation. By the fundamental lemma, the
image of $S$ under the projection map $\pi_{G}$ is a totally symmetric subset
of $G$. From the permutation $\phi$, we have
$(h_{G},h_{H})(x_{1},y_{1})(h_{G},h_{H})^{-1}=(x_{1},y_{2})$ and
$(h_{G},h_{H})(x_{1},y_{2})(h_{G},h_{H})^{-1}=(x_{2},y)$. Using these
equations, we derive a contradiction that $h_{G}$ conjugates $x_{1}$ to two
different elements. To do this, we first apply $\pi_{G}$ to both sides of the
first equation. This yields:

$\displaystyle\pi_{G}((h_{G},h_{H})(x_{1},y_{1})(h_{G},h_{H})^{-1})$
$\displaystyle=\pi_{G}((h_{G},h_{H}))\pi_{G}((x_{1},y_{1}))\pi_{G}((h_{G},h_{H})^{-1})$
$\displaystyle=h_{G}x_{1}h_{G}^{-1}$
$\displaystyle=\pi_{G}((x_{1},y_{2}))=x_{1}.$

Similarly, applying $\pi_{G}$ to both sides of the second equation, we find
that:

$\displaystyle\pi_{G}((h_{G},h_{H})(x_{1},y_{2})(h_{G},h_{H})^{-1})$
$\displaystyle=\pi_{G}((h_{G},h_{H}))\pi_{G}((x_{1},y_{2}))\pi_{G}((h_{G},h_{H})^{-1})$
$\displaystyle=h_{G}x_{1}h_{G}^{-1}$ $\displaystyle=\pi_{G}((x_{2},y))=x_{2}.$

Since $h_{G}$ conjugates $x_{1}$ to itself and $x_{2}$, we have a
contradiction. Therefore, $S$ cannot be a totally symmetric subset of $G\times
H$. ∎

######  Corollary 4.4.

Let $G$ and $H$ be groups, and let
$S=\\{(x_{1},y_{1}),\dots,(x_{k},y_{k})\\}\subseteq G\times H$ be a totally
symmetric set. All the $x_{i}$ are equivalent, or they are all distinct. The
same holds for $y_{i}$. Moreover, $S(G\times H)=\max\\{S(G),S(H)\\}$.

###### Proof.

By Lemma 4.3, if a set $S$ is a totally symmetric subset of $G\times H$, it
cannot simultaneously contain elements of the form
$(x_{1},y_{1}),(x_{1},y_{2})$ and $(x_{2},y)$. There are thus three
possibilities for the elements of $S$:

  1. (1)

$S$ contains elements of the form $(x_{1},y_{1})$ and $(x_{2},y)$ but none of
the form $(x_{1},y_{2})$. We also can’t contain elements of the form
$(x_{2},y^{\prime})$ for some $y^{\prime}\neq b$, since the labels could
easily be swapped between the elements. This implies that a totally symmetric
set in this case will take the form

$\displaystyle S=\\{(x_{1},y_{1}),\dots,(x_{n},y_{n})\\}.$

All the $x_{i}$ are distinct elements of $G$ and each $y_{i}$ is any element
of $H$.

  2. (2)

We allow elements of the form $(x_{1},y_{2})$ and $(x_{2},y)$ but none of the
form $(x_{1},y_{1})$. This is equivalent to the previous case by relabeling.

  3. (3)

$S$ contain elements of the form $(x_{1},y_{1})$ and $(x_{1},y_{2})$, but none
of the form $(x_{2},y)$. This implies that a set in this case will take the
form

$\displaystyle S=\\{(x_{1},y_{1}),\dots,(x_{1},y_{n})\\},$

where $x_{1}\in G$ and all the $y_{i}\in H$.

The above implies that a totally symmetric set
$S=\\{(x_{1},y_{1}),\dots,(x_{k},y_{k})\\}$ must have all the $x_{i}$
equivalent or all distinct. The same can be said of the $y_{i}$ by the
symmetry of the direct product.

Now that we know the totally symmetric sets of $G\times H$ must take on these
specific forms, we can first show that $S(G\times H)\leq\max\\{S(G),S(H)\\}$.
Any non-trivial totally symmetric subset of $G\times H$ will have either the
$x_{i}$ or $y_{i}$ elements all be distinct. Suppose that we have a totally
symmetric set $S^{\prime}$ such that all the $x_{i}$ elements are all
distinct. By Lemma 2.1, the set $\pi_{G}(S^{\prime})$ is a totally symmetric
subset of $G$, in this case with the same cardinality as $S^{\prime}$ due to
all the $x_{i}$ being distinct. The same can be said if $S^{\prime}$ had all
the $y_{i}$ distinct and we mapped it to $H$ by $\pi_{H}$. Together these
imply that a totally symmetric subset of $G\times H$ has cardinality bounded
above by $\max\\{S(G),S(H)\\}$.

To see that $S(G\times H)=\max\\{S(G),S(H)\\}$, we will construct a totally
symmetric subset of size $\max\\{S(G),S(H)\\}$ in $G\times H$. Without loss of
generality, suppose that $S(G)\geq S(H)$ and let $T=\\{x_{1},\dots,x_{n}\\}$
be a maximal totally symmetric subset of $G$. Consider the set
$T\times\\{e_{H}\\}\subset G\times H$. This is a totally symmetric subset of
$G\times H$. Any two elements $(x_{i},e_{H})$ and $(x_{j},e_{H})$ commute.
Additionally, any permutation on the elements of $T\times\\{e_{H}\\}$
corresponds to a permutation of $T$. If $g_{\sigma}$ is the element of $G$
that induces a permutation on $T$, then $(g_{\sigma},e_{H})$ will be the
element that induces the same permutation on $T\times\\{e_{H}\\}$.

∎

The following corollary will not be used again in the rest of the paper, but
it describes totally symmetric sets in direct products if we ask all the
elements to be distinct in each coordinate.

######  Corollary 4.5.

Let $G$ and $H$ be groups, and let
$S=\\{(x_{1},y_{1}),\dots,(x_{k},y_{k})\\}\subseteq G\times H$ be a totally
symmetric set. If all the $x_{i}$ are distinct and all they $y_{i}$ are
distinct, then $|S|=\min\\{S(G),S(H)\\}$.

As shown, totally symmetric sets in direct products of groups behave in a
straightforward way. We might now ask whether the same behavior persists in
additional group constructions, such as free products, semi-direct products,
group extensions. While the latter two are much more difficult to answer, it
turns out that we can say the following about totally symmetric sets of the
free product of two groups.

######  Proposition 4.6.

Let $G$ and $H$ be groups, and let $G*H$ denote their free product. Then
$S(G*H)=\max\\{S(G),S(H)\\}$. Moreover, any totally symmetric subset of $G*H$
is of the form $wSw^{-1}$, where $w\in G*H$, and $S$ is a totally symmetric
subset of $G$ or $H$.

###### Proof.

The direction $S(G*H)\geq\max\\{S(G),S(H)\\}$ is almost immediate. To show
this let $S$ be a totally symmetric subset of either $G$ or $H$. Since $G$ and
$H$ both inject into their free product, we have that $S$ is also a totally
symmetric subset of $G*H$. Thus, we see $S(G*H)\geq\max\\{S(G),S(H)\\}$. For
the other direction, we will show that every totally symmetric set in $G*H$
comes from a totally symmetric subset of either $G$ or $H$ in an injective
manner, thus proving $S(G*H)\leq\max\\{S(G),S(H)\\}$.

To begin, let $w_{1}$ and $w_{2}$ be commuting elements of $G*H$. Then $w_{1}$
and $w_{2}$ are both in the same conjugate of a factor of $G*H$ or both powers
of some element $w\in G*H$ [MKS04, Corollary 4.1.6]. The former of these means
that $w_{1}$ and $w_{2}$ are both contained in $xGx^{-1}$ or $xHx^{-1}$ for
some element $x\in G*H$. This corollary proves an even stronger statement: if
we have a set of pairwise commuting elements $\\{x_{1},\dots,x_{n}\\}$ where
only one is in the conjugate of one of the free factors, then all the elements
must be in that same conjugate.

Hence, given a totally symmetric set $S$ in $G*H$, there are three possible
forms of the elements of the set:

  1. (1)

Conjugates of $G$: $S=\\{wg_{1}w^{-1},\dots,wg_{n}w^{-1}\\}$ for $g_{i}\in G$
and $w\in G*H$.

  2. (2)

Conjugates of $H$: $S=\\{wh_{1}w^{-1},\dots,wh_{n}w^{-1}\\}$ for $h_{i}\in H$
and $w\in G*H$.

  3. (3)

Powers of the same element: $S=\\{v^{i_{1}},\dots,v^{i_{n}}\\}$ for $v\in G*H$
and $i_{1}\leq i_{2}\leq\dots\leq i_{n}$.

We will first examine the first two of these three cases. Without loss of
generality, suppose that we have a totally symmetric set $S$ of the first
type, with all its elements in a conjugate of $G$. Since this is a totally
symmetric set, any permutation of its elements can be achieved through
conjugation. Let $\sigma$ be a permutation of the elements of $S$ and let
$w_{\sigma}\in G*H$ be an element of $G*H$ that achieves this permutation by
conjugation. If $\sigma(i)=j$, then
$w_{\sigma}wg_{1}w^{-1}w_{\sigma}^{-1}=wg_{j}w^{-1}$, which implies
$w_{\sigma}=wg_{\sigma}w^{-1}$ and $g_{\sigma}g_{i}g_{\sigma}^{-1}=g_{j}$ for
$g_{\sigma}\in G$ . This combined with commutativity of $g_{i}$ shows
$S^{\prime}=\\{g_{1},\dots,g_{n}\\}$ is a totally symmetric subset of $G$ with
the same size as $S$. Therefore, any totally symmetric subset of the first two
forms above can have size at most $\max\\{S(G),S(H)\\}$.

In the other case, all elements of $S$ are powers of the same element $v\in
G*H$. We can assume $v$ is not an element of a conjugate of one of the
factors, since otherwise, all powers of it would be as well, which would then
reduce to case $(1)$ and $(2)$. Moreover, all elements with torsion in a free
product are conjugates of finite-order elements in one of the factors [MKS04,
Corollary 4.1.4], implying $v$ must have infinite order. Since our set
$S=\\{v^{i_{1}},\dots,v^{i_{n}}\\}$ is totally symmetric, there exists an
element $w\in G*H$ such that $wv^{i_{1}}w^{-1}=v^{i_{2}}$,
$wv^{i_{2}}w^{-1}=v^{i_{1}}$, and $w$ fixes all other element of $S$. This
reduces to a similar situation as in the proof of Theorem 3.2 and it has a
similar solution. We can show that $i_{2}=\pm i_{1}$ by taking
$wv^{i_{1}}w^{-1}=v^{i_{2}}$, and exponentiating both sides to the $i_{2}$th
power:

$\displaystyle(v^{i_{2}})^{i_{2}}=(wv^{i_{1}}w^{-1})^{i_{2}}=wv^{i_{1}i_{2}}w^{-1}=(wv^{i_{2}}w^{-1})^{i_{1}}=(v^{i_{1}})^{i_{1}}.$

This implies $v^{i_{2}^{2}}=v^{i_{1}^{2}}$, so $i_{2}^{2}=i_{1}^{2}$ and
$i_{2}=\pm i_{1}$. In either case we have now reduced our totally symmetric
set $S$ to either a singleton or $\\{v^{i_{1}},v^{-i_{1}}\\}$, as permutations
such as the one above must be possible for all pairs of elements in $S$.

We now check if $\\{v^{i_{1}},v^{-i_{1}}\\}$ is a totally symmetric set.
Again, we must have an element $w\in G*H$ such that
$wv^{i_{1}}w^{-1}=v^{-i_{1}}$, and $wv^{-i_{1}}w^{-1}=v^{i_{1}}$. As we saw in
the proof of Theorem 3.2, this implies that $w^{2}$ commutes with $v^{i_{1}}$.
From above, either $w^{2}$ and $v^{i_{1}}$ are in the same conjugate of one of
the factors of $G*H$, or they are powers of the same element. If $v^{i_{1}}$
was a conjugate of an element in either $G$ or $H$, $v$ is also a conjugate of
an element in $G$ or $H$, contradicting our initial assumption. Therefore,
$v^{i_{1}}=z^{a}$ and $w^{2}=z^{b}$ for some element $z\in G*H$. We can assume
$z$ is primitive, that it can’t be written as a power of another element in
$G*H$.

We now show that $b$ is even and $w=z^{\frac{b}{2}}$. Since $w$ is a freely
reduced word in $G$ and in $H$, it either has even or odd length. If $w$ has
even length, then its starting and ending elements cannot both be from the
same group. Thus, $w^{2}$ is exactly the concatenation of $w$ with itself,
with no cancelling or reducing involved. Since $z$ is reduced and primitive,
$w$ is a concatenation of $\frac{b}{2}$ copies of $z$. Alternatively, suppose
that $w$ has odd length, implying it ends and begins with an element from the
same factor. If cancellation occurs when concatenating, $w$ and $z$ are both
conjugates of the same element $x\in G*H$:
$z=xz^{\prime}x^{-1},w=xw^{\prime}x^{-1}$. This reduces to proving that
$w^{\prime}=z^{\prime\frac{b}{2}}$, where $w^{\prime},z^{\prime}$ cannot be
written as conjugates of some shorter length element. Thus, assume $w$ and $z$
are not conjugates of a smaller length element. When taking powers of both
elements, the only reduction will be happening at the letters where they are
concatenated, so we can conclude that $w=z^{\frac{b}{2}}$.

Since $w$ and $v^{i_{1}}$ are both powers of $z$, $w$ commutes with
$v^{i_{1}}$. This implies $v^{i_{1}}=v^{-i_{1}}$ and thus, $S$ is a singleton.
Hence, there are no non-trivial totally symmetric sets of form $(3)$, and all
totally symmetric sets of a free product come from conjugates of totally
symmetric sets in the factors. ∎

We can now see that totally symmetric sets behave in a reasonably nice way
with regards to these basic product operations and have a full proof of
Theorem 4.1. However, before we rejoice over this predictable behavior of
totally symmetric sets under products, it’s worth pointing out a
disappointment: at this point, we still don’t have a way to construct larger
totally symmetric sets from small ones on the group level: The maximum size of
totally symmetric sets under direct and free products are limited by the
totally symmetric sets within its factors. It has been a theme throughout this
paper that large totally symmetric sets are rare.

###  4.2. Semi-direct Product of Cyclic Groups

######  Lemma 4.7.

Let $G$ be a group and $S\subset G$ a totally symmetric subset of $G$ such
that $g,g^{-1}\in S$ for some $g\in G$. Then $S=\\{g,g^{-1}\\}$.

###### Proof.

Label the elements of the totally symmetric set $S$ as
$S=\\{g,g^{-1},x_{1},\dots,x_{n}\\}$ for some non-negative integer $n$. Since
$S$ is a totally symmetric set, there exists an element $h\in G$ such that
$hgh^{-1}=g$ and $hg^{-1}h^{-1}=x_{i}$ for one of the $x_{i}\in S$. Inverting
both sides of the former of these equalities shows $hg^{-1}h^{-1}=g^{-1}$,
thus we must have that $x_{i}=g^{-1}$. This process can be continued for all
elements of $S$, hence, $S=\\{g,g^{-1}\\}$. ∎

######  Proposition 4.2.

For $p$ prime and $m$ any integer where $p|m$,
$S(\mathbb{Z}_{p}\ltimes\mathbb{Z}_{m})=2$.

###### Proof.

We use the presentation
$\mathbb{Z}_{p}\ltimes\mathbb{Z}_{m}=\\{s,r|r^{p}=e,s^{m}=e,srs^{-1}=r^{k}\\}$
for the proof. A typical element in the group is of the form $r^{a}s^{b}$,
where $a\in\mathbb{Z}/p\mathbb{Z}$ and $b\in\mathbb{Z}/m\mathbb{Z}$. We first
investigate the conditions for two elements, $r^{a}s^{b}$ and $r^{x}s^{y}$, to
commute.:

$\begin{split}r^{a}s^{b}r^{x}s^{y}&=r^{x}s^{y}r^{a}s^{b}\\\
r^{a-x}s^{b}r^{s}&=s^{y}r^{a}s^{b-y}\\\
r^{a-x}s^{b}r^{s}&=s^{y}r^{a}s^{-y}s^{b}\\\
r^{a-x}s^{b}r^{s}&=r^{ak^{y}}s^{b}\\\ r^{a-x-ak^{y}}(s^{b}r^{x}s^{-b})&=e\\\
r^{a-x-ak^{y}}r^{xk^{b}}&=e\\\ r^{a-x-ak^{y}+xk^{b}}&=e.\\\ \end{split}$

By the group presentation, we obtain the following equation:

(1) $a-x-ak^{y}+xk^{b}=0\mod p.$

We put this aside and discuss the conditions for the same two elements,
$r^{a}s^{b}$ and $r^{x}s^{y}$, to also be conjugates of each other. Before we
go through the calculations, we take a detour to discuss $s^{-1}rs$. By
$srs^{-1}=r^{k}$, it follows that $s^{-1}rs=r^{l}$, where $l$ is the
multiplicative inverse of $k$ modulo $p$, which exists because $p$ is prime.
Suppose there exists some element $r^{e}s^{f}$ that conjugates $r^{a}s^{b}$ to
$r^{x}s^{y}$. We can use the group presentation to show that the equation
$b+y=0\mod m$ must hold:

$\begin{split}r^{e}s^{f}r^{a}s^{b}s^{-f}r^{-e}&=r^{x}s^{y}\\\
r^{e}s^{b}s^{f}s^{-b}r^{a}s^{b}s^{-f}r^{-e}&=r^{x}s^{y}\\\
r^{e}s^{b}s^{f}r^{al^{b}}s^{-f}r^{-e}&=r^{x}s^{y}\\\
r^{e}s^{b}r^{al^{b}k^{f}-e}&=r^{x}s^{y}\\\
r^{e}s^{b}r^{al^{b}k^{f}-e}s^{-b}s^{-y}s^{b}&=r^{x}\\\
r^{e+(al^{b}k^{f}-e)k^{b}}s^{b+y}&=r^{x}.\end{split}$

Deciphering the exponents, we now have the following relations: $b+y=0\mod m$,
and $e+(al^{b}k^{f}-e)k^{b}=x\mod p$. Recall $b,y\in\mathbb{Z}/m\mathbb{Z}$
and thus, $0\leq b,y<m$. The relation $b+y=m\mod m$ then implies that either
$b=y=0$, or $y=m-b$.

The case that $b=y=0$ amounts to considering elements of the form $r^{a}$ and
$r^{x}$ as candidate elements of the totally symmetric set. The commuting
relation $(1)$ shows elements of this form always commute, and the second of
the two conjugation relations reduces to $ak^{f}\equiv x\mod p$ given $b=0$
and $k^{b}\equiv 1\mod p$. Therefore, choosing a conjugating element
$r^{e}s^{f}$ determines $x$, and the relation shows the choice of $e$ will not
affect the conjugation. It is important to note that because $f$ is an integer
between $0$ and $m-1$, multiple choices of $f$ could lead to the same $x$. Let
$0\leq f\neq f^{\prime}\leq m-1$ such that $k^{f}\equiv k^{f^{\prime}}\mod p$.
This implies that $k^{f-f^{\prime}}\equiv 1\mod p$ and thus, $f$ and
$f^{\prime}$ must differ by a multiple of $p-1$.

Choose $f$ as above to determine an $x$, and let $f^{\prime}=f+n(p-1)$. We
conclude by the following calculation that $r^{e}s^{f^{\prime}}$ acts on
$r^{x}$ by conjugation in the same way that $r^{e}s^{f}$ does, i.e. different
choices of $f$ that determine the same $x$ will act the same by conjugation:

$\displaystyle
r^{e}s^{f^{\prime}}r^{x}s^{-f^{\prime}}r^{-e}=r^{e}r^{xk^{f^{\prime}}}r^{-e}=r^{xk^{f+n(p-1)}}=r^{xk^{f}}.$

In order for $r^{a}$ and $r^{x}$ to form a totally symmetric set, we need an
element that conjugates $r^{a}$ to $r^{x}$ to also conjugate $r^{x}$ back to
$r^{a}$. Hence, we need $xk^{f}\equiv a\mod p$. Combining this with
$ak^{f}\equiv x\mod p$, we have that $k^{2f}\equiv 1\mod p$ and thus,
$k^{f}\equiv\pm 1\mod p$. This tells us that $x=\pm a$, and one cannot make a
larger totally symmetric set containing the elements $r^{a}$ and $r^{x}$.

For the other case, assume that $y=m-b$, and use $(1)$ to derive an equation
to determine the following equations expressing $x$ in terms of $a$ and $b$:

(2) $\begin{split}a-x-ak^{m-b}+xk^{b}=0\mod p\\\ x(k^{b}-1)=a(k^{m-b}-1)\mod
p.\\\ \end{split}$

If $k^{b}-1\neq 0\mod p$, then given $a,b$, there is a unique solution for
$x$. Otherwise, we consider the case where $k^{b}-1\equiv 0\mod p$.

From above we know that $x=\pm a$ and thus we have the following three
elements to consider for a totally symmetric set, $r^{a}s^{b},r^{a}s^{-b}$,
and $r^{-a}s^{-b}$. We know that we can construct two totally symmetric sets
from these elements, namely $\\{r^{a}s^{b},r^{a}s^{-b}\\}$ and
$\\{r^{a}s^{b},r^{-a}s^{-b}\\}$, but we also need to check if all three
elements can form a totally symmetric set. Under our assumption that
$k^{b}\equiv 1\mod p$, we can show that $r^{a}s^{b}$ and $r^{-a}s^{-b}$ are
inverses:

$\displaystyle
r^{a}s^{b}r^{-a}s^{-b}=r^{a}(s^{b}rs^{-b})^{-a}=r^{a}(r^{k^{b}})^{-a}=r^{a}r^{-a}=e.$

By Lemma 4.7, this inhibits all three elements from being in a totally
symmetric set together, so the maximal size of a totally symmetric set in
$\mathbb{Z}_{p}\ltimes\mathbb{Z}_{m}$ is 2.

∎

##  5\. The Stabilizer

In this section, we introduce the stabilizer of the totally symmetric set and
use it to bound the cardinality of totally symmetric sets in groups. In
particular, we will show that the totally symmetric sets in odd-ordered finite
groups and solvable groups are small.

Let $G$ be a group, $S$ a totally symmetric subset of size $n$ in $G$, and
$\text{Stab}_{G}(S)$ the stabilizer of $S$ in $G$ under conjugation. We claim
that $\text{Stab}_{G}(S)$ surjects onto $S_{n}$, the symmetric group on $n$
letters. The stabilizer acts on $S$ by conjugation, so we can explicitly
construct a homomorphism $\varphi:\text{Stab}_{G}(S)\rightarrow Sym(S)\cong
S_{n}$ described by sending $\gamma\in\text{Stab}_{G}(S)$ to the automorphism
of $S$ given by $s\rightarrow\gamma s\gamma^{-1}$. By definition, every
permutation of $S$ is realized as a conjugation by some
$\gamma\in\text{Stab}_{G}(S)$, so this homomorphism is surjective.
Equivalently, we can write this as a short exact sequence, where $k$ denotes
the kernel of $\varphi$:

$1\xrightarrow{}k\xrightarrow{}\text{Stab}_{G}(S)\xrightarrow{}S_{n}\xrightarrow{}1$

In the case that $G$ is of finite order,
$|G|\geq|\text{Stab}_{G}(S)|=|k||S_{n}|=n!|k|$. The kernel consists of
elements that commute with all elements of the totally symmetric set, or
equivalently the intersection of centralizers of each element. In particular,
the subgroup generated by the totally symmetric set will be a subgroup of the
kernel $k$ by commutativity. We can relate the cardinality of the group $G$ to
the size of its totally symmetric sets through bounding the size of the
kernel. Using a similar technique, Chudnovsky and the authors showed that for
braid groups and their commutator subgroups, the required cardinality of $G$
grows super-exponentially as $n$ increases [CKLP20].

We use the existence of the surjection from the stabilizer to the symmetric
group to bound the sizes of totally symmetric sets in the following groups.

######  Proposition 5.1.

Let $G$ be a finite group with odd order. Then $S(G)=1$.

###### Proof.

Let $|G|=2n+1$ for some $n\in\mathbb{N}$, and let $S\subset G$ be a totally
symmetric set in $G$. From the short exact sequence above, the order of
$Sym(S)$ must divide the order of $\text{Stab}_{G}(S)$. Moreover, since the
order of $\text{Stab}_{G}(S)$ divides the order of the whole group $G$, we
have that $|S|!\,\,|\,\,|G|$ and hence $|S|!\,\,|\,\,2n+1$. Since for $|S|\geq
2$, $|S|!$ is even, we must have that $|S|=1$. ∎

######  Theorem 5.2.

Let $G$ be a solvable group. Then $S(G)\leq 4$.

###### Proof.

Let $S\subset G$ be a totally symmetric subset of $G$. Since
$\text{Stab}_{G}(S)$ is a subgroup of $G$, and $G$ is solvable,
$\text{Stab}_{G}(S)$ is also solvable. Moreover, from the short exact sequence
above, we know that $Sym(S)$ is a quotient of $\text{Stab}_{G}(S)$, so it is
solvable. For $n\geq 5$, $S_{n}$ is not solvable. Thus, $|S|\leq 4$. ∎

We do not know if this bound is sharp. In other words, it’s an open question
whether there are totally symmetric sets in solvable groups of size 3 or 4.

##  6\. Corollaries for Homomorphisms

In this last section, we summarize the results and display various corollaries
on possible homomorphisms between the groups discussed in this paper. The
corollaries on the homomorphisms we present stem from the following two
corollaries of the fundamental lemma of totally symmetric sets:

######  Corollary 6.1.

Let $G$ and $H$ be groups with $S(G)>S(H)$. No homomorphism
$f:G\xrightarrow[]{}H$ can be injective.

###### Proof.

Let $S\subset G$ be a totally symmetric set of size $S(G)$ and
$f:G\xrightarrow[]{}H$ be a homomorphism. From the fundamental lemma, $S$ maps
to a totally symmetric set in $H$ of size $S(G)$ or a singleton. Since
$S(G)>S(H)\geq 1$, there are no totally symmetric sets in $H$ of size $S(G)$,
and $S$ maps to a singleton. Since $S$ is not a singleton, $f$ is not
injective. ∎

This corollary tells us that $S(G)$ serves as an obstruction to $G$ being a
subgroup of another group. A further corollary that has been used frequently
in the recent results on braid groups using totally symmetric sets is the
following:

######  Corollary 6.2.

Let $B_{n}$ be the braid group on $n$ strands for $n\geq 5$ and let $G$ be a
group such that $S(G)<\lfloor\frac{n}{2}\rfloor$, then any homomorphism
$f:B_{n}\xrightarrow[]{}G$ is cyclic, i.e. it factors through $\mathbb{Z}$.

A refinement of this statement using totally symmetric sets by Chudnovsky and
the authors [CKLP20] and improved by Caplinger and Kordek [CK20] says that
this is true if $|G|<\lfloor\frac{n}{2}\rfloor!\cdot
3^{\lfloor\frac{n}{2}\rfloor-1}$. This bound was recently improved upon by
Scherich and Verberne [SV20].

The table below summarizes the bounds on totally symmetric sets derived in
this paper. Applying the above two corollaries gives us a list of groups which
can have no injective homomorphisms between them. For example,
$\mathbb{Z}_{p}\ltimes\mathbb{Z}_{mp}$ cannot be a subgroup of the
Baumslag–Solitar group $BS(1,n)$. Likewise, we obtain many example of groups
that the braid groups only map cyclically to. For example, by Corollary 6.2
any homomorphism from $B_{n}$ where $n\geq 10$ to solvable groups must have
cyclic image. Using the last line, we can take arbitrary products of the
groups in this table, and $B_{n}$ with $n\geq 10$ will map cyclically to it.
For example, any homomorphism from $B_{n}$ into
$(BS(1,420)*(\mathbb{Z}_{2017}\ltimes\mathbb{Z}_{8068}))\times D_{914}$ has
cyclic image.

$S(G)$ | Group |  |  |
---|---|---|---|---
1 | Abelian |  |  |
| Free Group ($F_{n}$) |  |  |
| Odd Order Group |  |  |
|  $BS(1,n)$, $n\neq 1$ |  |  |
2 | Dihedral ($D_{2n}$) |  |  |
| $\mathbb{Z}_{p}\ltimes\mathbb{Z}_{np}$ |  |  |
| $BS(1,-1)$ |  |  |
$\leq 4$ | Solvable Groups |  |  |
Max$\\{S(G),S(H)\\}$ | Direct Product, $G\times H$ |  |  |
| Free Product, $G*H$ |  |  |

## References

  * [CK20] Noah Caplinger and Kevin Kordek.  Small quotients of braid groups, 2020.
  * [CKLP20] Alice Chudnovsky, Kevin Kordek, Qiao Li, and Caleb Partin.  Finite quotients of braid groups.  Geometriae Dedicata, pages 1–8, 2020.
  * [CM20] Lei Chen and Aru Mukherjea.  From braid groups to mapping class groups, 2020.
  * [FM12] B. Farb and D. Margalit.  A Primer on Mapping Class Groups.  Princeton mathematical series. Princeton University Press, 2012.
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# Weak radiative decay $\Lambda_{c}^{+}\to\Sigma^{+}\gamma$ using light-cone
sum rules

Yu-Ji Shi1<EMAIL_ADDRESS>Zhi-Peng Xing 2
<EMAIL_ADDRESS>(corresponding author) and Ulf-G. Meißner 1,3,4
<EMAIL_ADDRESS>1 Helmholtz-Institut für Strahlen- und
Kernphysik and Bethe Center for Theoretical Physics,
Universität Bonn, 53115 Bonn, Germany
2 Tsung-Dao Lee Institute, Shanghai Jiao Tong University, Shanghai 200240,
China
3 Institute for Advanced Simulation, Institut für Kernphysik and Jülich Center
for Hadron Physics, Forschungszentrum Jülich, D-52425 Jülich, Germany
4 Tbilisi State University, 0186 Tbilisi, Georgia

###### Abstract

We calculate the decay width of the $\Lambda_{c}^{+}\to\Sigma^{+}\gamma$ using
light-cone sum rules. For the initial quark radiation an effective Hamiltonian
is constructed, where the internal quark line shrinks to a point. The final
quark radiation is studied within the full theory. The leading twist light-
cone distribution amplitudes of the $\Sigma^{+}$ serve as the non-perturbative
input for the sum rules calculation, and the perturbative kernel is calculated
at leading order. The branching fraction we obtain is ${\cal
B}(\Lambda_{c}^{+}\to\Sigma^{+}\gamma)=1.03\pm 0.36\times 10^{-4}$, which is
below the recent upper limit $<2.6\times 10^{-4}$ given by the Belle
collaboration.

##  I Introduction

Weak radiative decays of charmed hadrons are an ideal platform for
investigating the interplay of the strong and the weak interactions. Unlike
the flavor-changing neutral-current transition of bottom hadrons, the penguin
contribution in such charm decays is highly suppressed. As a result, the weak
radiative decay of charmed hadrons are Cabibbo-favored and dominated by long-
distance non-perturbative effects, where the decay is induced by internal
W-exchange bremsstrahlung processes such as $cd\to us\gamma$. Studying the
weak radiative decays of charmed hadrons both from the experimental and the
theoretical side can help us to understand the strong dynamics inside hadrons.

Over the past few decades, there are several measurements of the weak
radiative decays of charmed meson Belle:2003vsx ; BaBar:2008kjd ;
Belle:2016mtj , and the corresponding theoretical researches Cheng:2010ry ;
Artuso:2008vf ; Fajfer:2006zqn ; Fajfer:2005ke ; Pakvasa:2005pp ;
deBoer:2018zhz ; Dias:2017nwd ; deBoer:2017rzd ; Adolph:2022ujd ;
Adolph:2021ncg ; Gisbert:2020vjx ; Fu:2018yin ; Biswas:2017eyn . However, the
experimental researches in the charmed baryon sector are rare. Recently, the
Belle collaboration announced the first search for the weak radiative decays
$\Lambda_{c}^{+}\to\Sigma^{+}\gamma$ and $\Xi_{c}^{0}\to\Xi^{0}\gamma$
Belle:2022raw , where the upper limits for their absolute branching fractions
are given as:

$\displaystyle{\cal B}_{\rm exp}(\Lambda_{c}^{+}\to\Sigma^{+}\gamma)<2.6\times
10^{-4},~{}~{}~{}~{}{\cal B}_{\rm exp}(\Xi_{c}^{0}\to\Xi^{0}\gamma)<1.7\times
10^{-4}.$ (1)

On the theoretical side, the corresponding branching fractions have been
predicted by various theoretical approaches, which include a modified
nonrelativistic quark model Kamal:1983zt , the constituent quark model
Uppal:1992cc  and the effective Hamiltonian approach combined with the pole
model Cheng:1994kp . The theoretical predictions of the branching fractions of
$\Lambda_{c}^{+}\to\Sigma^{+}\gamma$ and $\Xi_{c}^{0}\to\Xi^{0}\gamma$ are in
the range $(4.5-29.1)\times 10^{-5}$ and $(3.0-19.5)\times 10^{-5}$,
respectively. Most of these predictions are consistent with the experimental
constraints given above, while the one from the constituent quark model are
slightly larger than the upper limits in Eq. (1).

Nowadays, except the model-based theoretical approaches mentioned above, there
is no model-independent calculation for the weak radiative decays of charmed
baryons. In this work, we will fill this gap and calculate the decay width of
the $\Lambda_{c}^{+}\to\Sigma^{+}\gamma$ with the use of light-cone sum rules
(LCSR). In terms of the initial quark radiation, following Refs. Uppal:1992cc
; Cheng:1994kp , we construct an effective Hamiltonian to simplify the
calculation. Since the radiating quark comes from the heavy baryon
$\Lambda_{c}^{+}$, its velocity can be assumed to be parallel to the velocity
of the $\Lambda_{c}^{+}$. This enables us to shrink the internal off-shell
quark line to a point so that the decay amplitude can be effectively induced
by a local Hamiltonian of $cd\to us\gamma$. In terms of the final quark
radiation, since the final state $\Sigma^{+}$ is a light baryon, thus we
cannot make the same assumption on its composite quark velocities. Therefore
we have to treat the final quark radiation in the full theory. The leading
twist light-cone distribution amplitudes (LCDAs) of the $\Sigma^{+}$ will
serve as the non-perturbative input for the sum rules calculation. These LCDAs
are taken from the latest Lattice QCD calculation with $N_{f}=2+1$
RQCD:2019hps . Furthermore, the perturbative kernel will be calculated at
leading order. It should be mentioned that these LCDAs are defined according
to the light-cone expansion, which are most reliable when the quark masses
vanish. Therefore the LCDAs of $\Xi^{0}$ are not as good as those of
$\Sigma^{+}$ which has less massive $s$ quarks, and we will not consider
$\Xi_{c}^{0}\to\Xi^{0}\gamma$ in this work.

This paper is organized as follows. In Sec. II, we construct an effective
Hamiltonian for the initial quark radiation in the
$\Lambda_{c}^{+}\to\Sigma^{+}\gamma$ decay and express the decay amplitude by
several calculable matrix elements. In Sec. III, we define suitable
correlation functions to calculate the decay amplitude at the hadron level. In
Sec. IV, we perform the QCD level calculation for the correlation function
defined above with the use of $\Sigma^{+}$ LCDAs. Sec.V contains the numerical
results on the decay amplitudes and branching fraction. We will also compare
them with those from literature. Sec. VI is a brief summary of this work.

##  II Decay amplitudes for initial and final radiation

The weak effective Hamiltonian contributing to the
$\Lambda_{c}^{+}\to\Sigma^{+}\gamma$ decay reads

$\displaystyle{\cal H}_{\rm eff}$
$\displaystyle=\frac{G_{F}}{\sqrt{2}}V_{cd}V^{*}_{ud}\left(C_{1}{\cal
O}_{1}+C_{2}{\cal O}_{2}\right),$ $\displaystyle{\cal O}_{1}$
$\displaystyle=\bar{s}\gamma^{\mu}(1-\gamma_{5})c\
\bar{u}\gamma_{\mu}(1-\gamma_{5})d,$ $\displaystyle{\cal O}_{2}$
$\displaystyle=\bar{u}\gamma^{\mu}(1-\gamma_{5})c\
\bar{s}\gamma_{\mu}(1-\gamma_{5})d,$ (2)

where the $C_{1,2}$ are the Wilson coefficients. Fig. 1 shows the W-exchange
bremsstrahlung processes $cd\to us\gamma$. In the case of initial radiation
where the photon is emitted by the $c$ or $d$ quark as shown in Fig. 1,
following the approach given in Ref. Cheng:1994kp  we can construct an
effective Hamiltonian to simplify the calculation. Here, we take the $c$ quark
radiation in the ${\cal O}_{1}$ contribution, namely Fig. 1(a) as an example
to illustrate the procedure.

Figure 1: W-exchange bremsstrahlung processes $cd\to us\gamma$ induced by
${\cal O}_{1}$, where the double crossed dots denote ${\cal O}_{1,2}$. The
diagrams for ${\cal O}_{2}$ are similar, just exchanging $u$ and $s$.

The amplitude of Fig.1(a) reads

$\displaystyle{\cal A}_{{\rm Initial},c}^{{\cal
O}_{1}}=i\frac{G_{F}}{\sqrt{2}}V_{cs}V_{ud}^{*}\varepsilon^{*\mu}(k)\bar{s}(p_{s})\gamma^{\nu}(1-\gamma_{5})\frac{\not{p}_{c}-\not{k}+\bar{m}_{c}}{(p_{c}-k)^{2}-\bar{m}_{c}^{2}}\gamma_{\mu}c(p_{c})\bar{u}(p_{u})\gamma_{\nu}(1-\gamma_{5})d(p_{d}),$
(3)

where $p_{c,s,u,d}$ are the on-shell quark momenta, $\bar{m}_{c,d}$ are the
constituent quark masses in the $\Lambda_{c}$ and $k$ satisfies $k^{2}=0$ and
$k\cdot\varepsilon=0$. Since the initial $c,\ d$ quarks are confined in the
heavy baryon $\Lambda_{c}$, we can assume that $c$, $d$ and $\Lambda_{c}$ have
the same velocity, in other words
$p_{c,d}=(\bar{m}_{c,d}/m_{\Lambda_{c}})p_{\Lambda_{c}}$. Thus the denominator
of Eq. (3) becomes

$\displaystyle(p_{c}-k)^{2}-m_{c}^{2}=\frac{\bar{m}_{c}}{m_{\Lambda_{c}}}(m_{\Sigma}^{2}-m_{\Lambda_{c}}^{2}),$
(4)

which implies that effectively the internal off-shell quark line shrinks to a
point. Further, the numerator of Eq. (3) can be simplified by using the
equation of motion of the $c$ quark. For the case of $d$ quark radiation the
derivation is almost the same. Finally, the amplitude in Eq. (3) can be
effectively generated by the following Hamiltonian

$\displaystyle{\cal H}_{{\rm eff}}^{{\cal O}_{1}}$
$\displaystyle=\frac{G_{F}}{\sqrt{2}}V_{cs}V_{ud}^{*}\
C_{1}\sum_{q}\left[A_{\mu}J_{{\cal
O}_{1},q}^{\mu}-\frac{i}{2}F_{\mu\nu}K_{{\cal O}_{1},q}^{\mu\nu}\right],$ (5)

where $q=c,d$ and $J_{{\cal O}_{1},q},K_{{\cal O}_{1},q}$ are the effective
four-quark currents

$\displaystyle J_{{\cal O}_{1},c}^{\mu}$ $\displaystyle=2i\ Q_{c}\lambda_{c}\
\bar{s}\gamma^{\nu}(1-\gamma_{5})\partial^{\mu}c\
\bar{u}\gamma_{\nu}(1-\gamma_{5})d,$ $\displaystyle K_{{\cal
O}_{1},c}^{\mu\nu}$ $\displaystyle=i\ Q_{c}\lambda_{c}\
\bar{s}\gamma^{\alpha}(1-\gamma_{5})\sigma^{\mu\nu}c\
\bar{u}\gamma_{\alpha}(1-\gamma_{5})d,$ $\displaystyle J_{{\cal
O}_{1},d}^{\mu}$ $\displaystyle=2i\ Q_{d}\lambda_{d}\
\bar{s}\gamma^{\nu}(1-\gamma_{5})c\
\bar{u}\gamma_{\nu}(1-\gamma_{5})\partial^{\mu}d,$ $\displaystyle K_{{\cal
O}_{1},d}^{\mu\nu}$ $\displaystyle=i\ Q_{d}\lambda_{d}\
\bar{s}\gamma^{\alpha}(1-\gamma_{5})c\
\bar{u}\gamma_{\alpha}(1-\gamma_{5})\sigma^{\mu\nu}d~{}.$ (6)

Here, $Q_{q}$ is the electric charge and
$\lambda_{q}=\frac{m_{\Lambda_{c}}}{\bar{m}_{q}(m_{\Lambda_{c}}^{2}-m_{\Sigma}^{2})}$.
For the case of ${\cal O}_{2}$ the corresponding operators can be obtained by
just exchanging the $u,s$ fields. Now the initial radiation amplitude induced
by ${\cal O}_{1,2}$ can be expressed as:

$\displaystyle{\cal A}_{{\rm Initial}}^{{\cal
O}_{i}}=-i\frac{G_{F}}{\sqrt{2}}V_{cs}V_{ud}^{*}\ C_{i}\
\varepsilon_{\mu}^{*}(k)\left[\langle\Sigma(p)|J_{{\cal
O}_{i}}^{\mu}(0)|\Lambda_{c}(q)\rangle+\langle\Sigma(p)|k_{\alpha}K_{{\cal
O}_{i}}^{\alpha\mu}(0)|\Lambda_{c}(q)\rangle\right],$ (7)

where $i=1,2$, $J_{{\cal O}_{i}}^{\mu}=J_{{\cal O}_{i,c}}^{\mu}+J_{{\cal
O}_{d}}^{\mu}$ and $K_{{\cal O}_{i}}^{\mu\nu}=K_{{\cal
O}_{i,c}}^{\mu\nu}+K_{{\cal O}_{d}}^{\mu\nu}$. $k=q-p$ is the on-shell photon
momentum.

For the final quark radiation this effective Hamiltonian approach is not
suitable. The reason is that in our case the final baryon $\Sigma^{+}$
contains no heavy quark, and thus we cannot equate its velocity with its
constituent quarks, namely the momentum relation
$p_{u,s}=(\bar{m}_{u,s}/m_{\Sigma})p_{\Sigma}$ cannot be used any more. The
amplitude for the final quark radiation is calculated in the full theory. It
can be written as

$\displaystyle{\cal A}_{{\rm Final}}^{{\cal
O}_{i}}=-i\frac{G_{F}}{\sqrt{2}}V_{cs}V_{ud}^{*}\ C_{i}\
\varepsilon_{\mu}^{*}(k)\int d^{4}x\ \langle\Sigma(p)|T\\{j^{\mu}(0){\cal
O}_{i}(x)\\}|\Lambda_{c}(q)\rangle,$ (8)

where $j^{\mu}=iQ_{u}\bar{u}\gamma^{\mu}u+iQ_{s}\bar{s}\gamma^{\mu}s$ is the
quark electromagnetic current. According to the Ward-identity, the matrix
elements appearing in Eq. (7) and Eq. (8) can be parameterized as

$\displaystyle\langle\Sigma(p)|{\cal J}_{{\cal
O}_{i}}^{\mu}(0)|\Lambda_{c}(q)\rangle$ $\displaystyle=i\
\varepsilon^{*\mu}(k)\bar{u}_{\Sigma}\left(a_{i,{\cal J}}^{+}+b_{i,{\cal
J}}^{+}\gamma_{5}\right)\,\sigma^{\mu\nu}\frac{k_{\nu}}{m_{\Lambda_{c}}}u_{\Lambda_{c}}(q),$
(9)

where ${\cal J}_{{\cal O}_{i}}^{\mu}=J_{{\cal O}_{i}}^{\mu}+k_{\alpha}K_{{\cal
O}_{i}}^{\alpha\mu}$ for the initial radiation and ${\cal J}_{{\cal
O}_{i}}^{\mu}=\int d^{4}xT\\{j^{\mu}{\cal O}_{i}(x)\\}$ for the final
radiation. The amplitudes $a^{+}_{i,\cal J},b^{+}_{i,\cal J}$ will be
calculated using LCSR in the next section.

##  III Hadron level calculation in LCSR

Now we present the calculation of $\Lambda_{c}^{+}\to\Sigma^{+}\gamma$ decay
width within the LCSR approach. To obtain the matrix elements given in Eq.
(9), one has to define a suitable correlation function and calculate it both
at the hadron and the QCD level. Matching these two levels by the quark-hadron
duality enables us to extract the decay amplitudes. Here we define a two-point
correlation function as:

$\displaystyle\Pi_{{\cal O}_{i},{\cal J}}(p,q)=p^{\mu}\int d^{4}x\ e^{-iq\cdot
x}\langle\Sigma(p)|T\\{{{\cal J}_{\mu}^{{\cal
O}_{i}}(0)\bar{J}_{\Lambda_{c}}(x)}\\}|0\rangle,$ (10)

where $\bar{J}_{\Lambda_{c}}$ is a current creating the $\Lambda_{c}$ baryon
and its explicit form will be given later. Here, we have contracted the
correlation function with a momentum vector $p^{\mu}$. Without this
contraction, the correlation function will have 12 independent structures
$\gamma_{\mu},\gamma_{\mu}\gamma_{5},\gamma_{\mu}\not{q},\gamma_{\mu}\not{q}\gamma_{5},p_{\mu},~{}p_{\mu}\gamma_{5},p_{\mu}\not{q},p_{\mu}\not{q}\gamma_{5},q_{\mu},~{}q_{\mu}\gamma_{5},~{}q_{\mu}\not{q},~{}q_{\mu}\not{q}\gamma_{5},1,\gamma_{5},\not{q}$,
and $\not{q}\gamma_{5}$. However, in Eq. (9) there are only two independent
amplitudes. Thus it will become ambiguous which two of the 12 structures
should be chosen to extract the two amplitudes. Contracting the momentum
vector $p^{\mu}$ reduces the number of independent structures to four, namely
$1,\gamma_{5},\not{q},\not{q}\gamma_{5}$, which is still too much. This
mismatch can be solved by doubling the number of amplitudes, as will be
explained next.

At the hadron level, this correlation function is calculated by inserting a
complete set of states between the two currents. The lowest single particle
state should be explicitly kept while the higher excited states will be
attributed to the continuous spectrum. To match the four independent
structures of the correlation function with the number of decay amplitudes, we
have to introduce two extra amplitudes from the decay of the negative parity
state $\Lambda_{c}({1/2}^{-})$. Similarly to Eq. (9), the corresponding
amplitudes are

$\displaystyle i\ \varepsilon^{*\mu}(k)\bar{u}_{\Sigma}\left(a_{i,{\cal
J}}^{-}+b_{i,{\cal
J}}^{-}\gamma_{5}\right)\sigma_{\mu\nu}\frac{k^{\nu}}{m_{\Lambda_{c}}}(i\gamma_{5})u_{\Lambda_{c}}(q).$
(11)

Now we have four amplitudes $a_{i,{\cal J}}^{\pm},b_{i,{\cal J}}^{\pm}$
mapping to the four structures $1,\gamma_{5},\not{q},\not{q}\gamma_{5}$.
Keeping both the two lowest states $\Lambda_{c}({1/2}^{\pm})$ and attributing
higher excited states to the continuous spectrum, we express the hadron level
correlation function of Eq. (10) as

$\displaystyle\Pi_{{\cal O}_{i},{\cal J}}(p,q)_{\rm H}=$
$\displaystyle\frac{\lambda_{+}}{m_{\Lambda_{c}+}^{2}-q^{2}}\bar{u}_{\Sigma}(a_{i,{\cal
J}}^{+}+b_{i,{\cal
J}}^{+}\gamma_{5})\sigma_{\mu\nu}(\not{q}+m_{\Lambda_{c}+})\frac{p^{\mu}k^{\nu}}{m_{\Lambda_{c}+}}$
$\displaystyle+\frac{\lambda_{-}}{m_{\Lambda_{c}-}^{2}-q^{2}}\bar{u}_{\Sigma}(a_{i,{\cal
J}}^{-}+b_{i,{\cal
J}}^{-}\gamma_{5})\sigma_{\mu\nu}(\not{q}-m_{\Lambda_{c}-})\frac{p^{\mu}k^{\nu}}{m_{\Lambda_{c}-}}+\int_{s_{\rm
th}}^{\infty}ds\frac{\rho_{{\cal O}_{i},{\cal J}}(s,p)}{s-q^{2}}~{}.$ (12)

The last term is the continuous spectrum contribution including all the states
above the $\Lambda_{c}({1/2}^{-})$. $s_{\rm th}$ is the threshold parameter of
this continuous spectrum and should be larger than $m_{\Lambda_{c}-}^{2}$.
$\lambda_{\pm}$ are the decay constants of the $\Lambda_{c}({1/2}^{\pm})$
which are defined as

$\displaystyle\langle\Lambda_{c}({1/2}^{+})(q)|\bar{J}_{\Lambda_{c}}(0)|0\rangle=\bar{u}_{\Lambda_{c}}(q)\lambda_{+},$
$\displaystyle\langle\Lambda_{c}({1/2}^{-})(q)|\bar{J}_{\Lambda_{c}}(0)|0\rangle=\bar{u}_{\Lambda_{c}}(q)(i\gamma_{5})\lambda_{-}.$
(13)

The same correlation function should also be calculated at the QCD level,
which can be expressed as a dispersion integral:

$\displaystyle\Pi_{{\cal O}_{i},{\cal J}}(p,q)_{\rm QCD}=\frac{1}{2\pi
i}\int_{m_{c}^{2}}^{\infty}ds\frac{{\rm Disc}\ \Pi_{{\cal O}_{i},{\cal
J}}(p,s)_{\rm QCD}}{s-q^{2}}.$ (14)

The discontinuity part can be parameterized as:

$\displaystyle{\rm Disc}\ \Pi_{{\cal O}_{i},{\cal J}}(s,p)_{\rm
QCD}=F^{(1)}_{{\cal O}_{i},{\cal J}}\ \not{q}\gamma_{5}+F^{(2)}_{{\cal
O}_{i},{\cal J}}\ \not{q}+F^{(3)}_{{\cal O}_{i},{\cal J}}\
\gamma_{5}+F^{(4)}_{{\cal O}_{i},{\cal J}}.$ (15)

In principle, the correlation function calculated at the hadron and the QCD
level should be equivalent. According to the quark-hadron duality, the
continuous spectrum contribution in Eq. (12) is canceled by the corresponding
QCD level dispersion integral in the region $s_{\rm th}<s<\infty$.
Furthermore, since the QCD level calculation can only be explicitly performed
using a light-cone expansion (LCE), one has to perform a Borel transformation
of the correlation function at both levels to improve the LCE convergence.
Finally one can extract the amplitudes as

$\displaystyle a_{i,{\cal J}}^{+}$
$\displaystyle=\frac{1}{\pi}\int_{m_{c}^{2}}^{s_{{\rm th}}}ds\
e^{\frac{m_{\Lambda_{c}+}^{2}-s}{T_{2}}}\frac{m_{\Lambda_{c}+}\big{[}m_{\Lambda_{c}-}F_{{\cal
O}_{i},{\cal J}}^{(2)}+F_{{\cal O}_{i},{\cal
J}}^{(4)}\big{]}}{\lambda_{+}(m_{\Lambda_{c}+}+m_{\Lambda_{c}-})(m_{\Lambda_{c}+}-m_{\Sigma})^{2}},$
$\displaystyle b_{i,{\cal J}}^{+}$
$\displaystyle=\frac{1}{\pi}\int_{m_{c}^{2}}^{s_{{\rm th}}}ds\
e^{\frac{m_{\Lambda_{c}+}^{2}-s}{T_{2}}}\frac{m_{\Lambda_{c}+}\big{[}F_{{\cal
O}_{i},{\cal J}}^{(3)}-m_{\Lambda_{c}-}F_{{\cal O}_{i},{\cal
J}}^{(1)}\big{]}}{\lambda_{+}(m_{\Lambda_{c}+}+m_{\Lambda_{c}-})(m_{\Lambda_{c}+}+m_{\Sigma})^{2}},$
(16)

where $T_{2}$ is the Borel parameter which will be determined during the
numerical calculation. Here, $a_{i,{\cal J}}^{-},b_{i,{\cal J}}^{-}$ are not
shown since we only care about the decay amplitudes of the
$\Lambda_{c}({1/2}^{+})$. The coefficients $F_{{\cal O}_{i},{\cal J}}^{(n)}$
will be explicitly calculated by the LCE at the QCD level.

##  IV QCD level calculation in LCSR

In this section, we will use light-cone expansion to calculate the correlation
function defined in Eq. (10), and extract the coefficients $F_{{\cal
O}_{i},{\cal J}}^{(n)}$. Now the correlation function reads

$\displaystyle\Pi_{{\cal O}_{i},{\cal J}}(p,q)_{\rm QCD}=p^{\mu}\int d^{4}x\
e^{-iq\cdot x}\langle\Sigma(p)|T\\{{{\cal J}_{\mu}^{{\cal
O}_{i}}(0)\bar{J}_{\Lambda_{c}}(x)}\\}|0\rangle,$ $\displaystyle{\rm
with}~{}~{}~{}~{}\bar{J}_{\Lambda_{Q}}=-\epsilon_{abc}\bar{Q}_{c}(\bar{d}_{b}C\gamma_{5}\bar{u}_{a}^{T}),$
(17)

where $a,b,c$ are color indices. Here, $q^{2}\ll 0$ is taken in the deep
Euclidean region to realize the light-cone expansion. Let us take $i=1$ and
${\cal J}=J$ as an example to illustrate the detailed calculation for the $c$
quark radiation.

At leading order the corresponding correlation function becomes

$\displaystyle\Pi_{{\cal O}_{i},{\cal J}}^{c}(p,q)_{\rm QCD}=$
$\displaystyle-2i\ Q_{c}\lambda_{c}\epsilon_{abc}\ p_{\mu}\int d^{4}x\
e^{-iq\cdot x}[\gamma^{\nu}(1-\gamma_{5})\partial_{w}^{\mu}S_{c}(w,x)]_{ig}$
$\displaystyle\times[\gamma_{\nu}(1-\gamma_{5})S_{d}(0,x)C\gamma_{5}]_{kn}\langle\Sigma(p)|\bar{s}_{c}^{i}(0)\bar{u}_{b}^{k}(0)\bar{u}_{a}^{n}(x)|0\rangle,$
(18)

where $S_{q}(x,y)$ is the free propagator of the quark $q$. Fig. 2(a) shows
the corresponding Feynman diagram, where the black dot at coordinate $x$
denotes the $\Lambda_{c}$ current and the white crossed dot at coordinate $0$
denotes the effective current $J_{{\cal O}_{1},c}^{\mu}$.

Figure 2: Diagrams for the QCD level correlation function in Eq. (17). (a) is
the initial quark radiation where the white crossed dot denotes the effective
four-quark currents in Eq. (6). (b) is the final $s$ quark radiation where the
white crossed dot denotes the current $j^{\mu}$. The black dot denotes the
$\Lambda_{c}$ current. The grey ellipse represents the LCDAs of the
$\Sigma^{+}$.

The last matrix element in Eq. (18) is represented by the grey ellipse in Fig.
2, which can be parameterized by three leading twist LCDAs of the
$\Sigma^{+}$Lepage:1980fj ; Efremov:1979qk ; Chernyak:1983ej ; Krankl:2011gch
; RQCD:2019hps

$\displaystyle\langle\Sigma(p)|\bar{s}_{c}^{i}(0)\bar{u}_{b}^{k}(0)\bar{u}_{a}^{n}(x)|0\rangle=$
$\displaystyle-\frac{1}{4}\epsilon_{abc}\int
du_{1}du_{2}du_{3}\delta(1-u_{1}-u_{2}-u_{3})\ e^{iu_{1}p\cdot x}$
$\displaystyle\times\Big{\\{}[\bar{u}^{B}\gamma_{5}]_{i}[C\not{\tilde{n}}]_{kn}V^{B}(u_{1},u_{2},u_{3})+[\bar{u}^{B}]_{i}[C\gamma_{5}\not{\tilde{n}}]_{kn}A^{B}(u_{1},u_{2},u_{3})$
$\displaystyle+\
i\tilde{n}^{\alpha}g_{\perp}^{\beta\rho}[C\sigma_{\beta\alpha}]_{kn}[\bar{u}^{B}\gamma_{\rho}\gamma_{5}]_{i}T^{B}(u_{1},u_{2},u_{3})\Big{\\}},$
(19)

where
$g_{\mu\nu}^{\perp}=g_{\mu\nu}-(\tilde{n}_{\mu}n_{\nu}+\tilde{n}_{\nu}n_{\mu})/(\tilde{n}\cdot
n)$ with $\tilde{n}_{\mu}=p_{\mu}-(m_{\Sigma}^{2}/2p\cdot n)n_{\mu}$ and $n$
is a light-cone vector.
$\bar{u}^{B}=\bar{u}_{\Sigma}\not{n}\not{p}/2m_{\Sigma}$ with $u_{\Sigma}$ the
Dirac spinor of the $\Sigma^{+}$ baryon. The coordinate $x$ of the quark field
is parallel to $n$, $x=(x\cdot p/m_{\Sigma})n$, where we have used
$p=m_{\Sigma}v$, $v=(n+\bar{n})/2$ and $n\cdot\bar{n}=2$. In the chiral limit
$m_{u}=m_{d}=0$ the contribution of $V^{B}$ and $A^{B}$ to the correlation
function vanishes. The explicit form of $T^{B}$ of the $\Sigma^{+}$ baryon now
reads

$\displaystyle
T^{B}(u_{1},u_{2},u_{3})=120u_{1}u_{2}u_{3}\left(\pi_{00}^{B}\mathcal{P}_{00}+\pi_{11}^{B}\mathcal{P}_{11}+\ldots\right),$
(20)

where the $\mathcal{P}_{ij}$ are polynomials,
$\mathcal{P}_{00}=1,~{}~{}\mathcal{P}_{11}=7\left(u_{1}-2u_{3}+u_{2}\right)$
RQCD:2019hps . $\pi_{00}^{B}$ and $\pi_{11}^{B}$ are the shape parameters
which encode all non-perturbative information of the baryon. The ellipsis
denotes terms of higher power polynomials, which are suppressed and omitted
here.

Using the $\Sigma^{+}$ LCDAs given above, we can express the correlation
function in Fig. 2(a) as

$\displaystyle\Pi_{{\cal O}_{i},{\cal J}}^{c}(p,q)_{\rm QCD}=$
$\displaystyle-3iQ_{c}\lambda_{c}\frac{1}{2m_{\Sigma}}\int du_{1}du_{2}\int
d^{4}x\int\frac{d^{4}k_{1}}{(2\pi)^{4}}\frac{d^{4}k_{2}}{(2\pi)^{4}}\
e^{-i(q-k_{1}-k_{2}-u_{1}p)\cdot x}$ $\displaystyle\times
T^{B}(u_{1},u_{2},1-u_{1}-u_{2})\frac{1}{k_{1}^{2}-m_{c}^{2}}\frac{1}{k_{2}^{2}}\
p\cdot k_{1}$
$\displaystyle\times\left[p^{\alpha}g^{\rho\beta}+(\frac{1}{2}m_{\Sigma}g^{\rho\alpha}-\frac{1}{m_{\Sigma}}p^{\rho}p^{\alpha})n^{\beta}+\frac{1}{2}p^{\alpha}n^{\beta}n^{\rho}\right]n^{\kappa}$
$\displaystyle\times\bar{u}_{\Sigma}\gamma_{\kappa}\not{p}\gamma_{\rho}\gamma_{5}\gamma^{\nu}(1-\gamma_{5})(\not{k}_{1}+m_{c})\
{\rm tr}[\gamma_{\nu}(1-\gamma_{5})\not{k}_{2}\sigma_{\alpha\beta}].$ (21)

Here, we have defined
$\tilde{T}^{B}(u_{1},u_{2})=T^{B}(u_{1},u_{2},1-u_{1}-u_{2})$. Note that since
$n=(m_{\Sigma}/x\cdot p)x$, we can use the following trick to remove the $x$
in the denominator:

$\displaystyle\int du_{1}du_{2}\int d^{4}x\ e^{-i(q-k_{1}-k_{2}-u_{1}p)\cdot
x}\ \tilde{T}^{B}(u_{1},u_{2})n_{\kappa}\ \cdots$ $\displaystyle=\ $
$\displaystyle m_{\Sigma}\frac{\partial}{\partial q^{\kappa}}\int
du_{1}du_{2}\int d^{4}x\ e^{-i(q-k_{1}-k_{2}-u_{1}p)\cdot x}\
\tilde{T}_{(1)}^{B}(u_{1},u_{2})\ \cdots,$ (22)

where the ellipses represents all the terms independent of $u_{1},u_{2}$, and

$\displaystyle\tilde{T}_{(i)}^{B}(u_{1},u_{2})=\int_{0}^{u_{1}}dt\
\tilde{T}_{(i-1)}^{B}(t,u_{2})~{}~{}~{}~{}~{}{\rm
with}~{}~{}~{}\tilde{T}_{(0)}^{B}(t,u_{2})=\tilde{T}^{B}(t,u_{2}).$ (23)

From Eq. (22), it follows that for each $n_{\kappa}$ one can equivalently
replace it with an operator $\hat{n}_{\kappa}=m_{\Sigma}\ \partial/\partial
q^{\kappa}$ and simultaneously replace $\tilde{T}^{B}$ with
$\tilde{T}_{(1)}^{B}$. Therefore, the correlation function takes the form

$\displaystyle\Pi_{{\cal O}_{i},{\cal J}}^{c}(p,q)_{\rm QCD}=$
$\displaystyle-3iQ_{c}\lambda_{c}\frac{1}{2m_{\Sigma}}\int du_{1}du_{2}\ {\cal
N}[\hat{n},T_{(i)}^{B}]^{\alpha\beta\rho\kappa}\int\frac{d^{4}k_{1}}{(2\pi)^{4}}\frac{d^{4}k_{2}}{(2\pi)^{4}}$
$\displaystyle\times(2\pi)^{4}\delta^{4}(q-u_{1}p-k_{1}-k_{2})\frac{1}{k_{1}^{2}-m_{c}^{2}}\frac{1}{k_{2}^{2}}\
p\cdot k_{1}$
$\displaystyle\times\bar{u}_{\Sigma}\gamma_{\kappa}\not{p}\gamma_{\rho}\gamma_{5}\gamma^{\nu}(1-\gamma_{5})(\not{k}_{1}+m_{c})\
{\rm tr}[\gamma_{\nu}(1-\gamma_{5})\not{k}_{2}\sigma_{\alpha\beta}],$ (24)

where the operator $\cal N$ is defined as

$\displaystyle{\cal N}[\hat{n},T_{(i)}^{B}]^{\alpha\beta\rho\kappa}$
$\displaystyle=$ $\displaystyle\
\hat{n}^{\kappa}\left[p^{\alpha}g^{\rho\beta}\tilde{T}_{(1)}^{B}(u_{1},u_{2})+\left(\frac{1}{2}m_{\Sigma}g^{\rho\alpha}-\frac{1}{m_{\Sigma}}p^{\rho}p^{\alpha}\right)\hat{n}^{\beta}\tilde{T}_{(2)}^{B}(u_{1},u_{2})+\frac{1}{2}p^{\alpha}\hat{n}^{\beta}\hat{n}^{\rho}\tilde{T}_{(3)}^{B}(u_{1},u_{2})\right].$
(25)

Now we have to express the QCD level correlation function as a dispersive
integral. The discontinuity part can be extracted from the cutting rules:

$\displaystyle{\rm Disc}\ \Pi_{{\cal O}_{i},{\cal J}}^{c}(p,q)_{\rm QCD}=$
$\displaystyle-3iQ_{c}\lambda_{c}\frac{(2\pi)^{2}}{2m_{\Sigma}}{\cal
N}[\hat{n},T_{(i)}^{B}]^{\alpha\beta\rho\kappa}\int du_{1}du_{2}\int
d\Phi_{2}[(q-u_{1}p)^{2}]\ p\cdot k_{1}$
$\displaystyle\times\bar{u}_{\Sigma}\gamma_{\kappa}\not{p}\gamma_{\rho}\gamma_{5}\gamma^{\nu}(1-\gamma_{5})(\not{k}_{1}+m_{c})\
{\rm tr}[\gamma_{\nu}(1-\gamma_{5})\not{k}_{2}\sigma_{\alpha\beta}],$ (26)

where

$\displaystyle
d\Phi_{2}[(q-u_{1}p)^{2}]=\int\frac{d^{3}k_{1}}{(2\pi)^{3}}\frac{1}{2E_{k_{1}}}\frac{d^{3}k_{1}}{(2\pi)^{3}}\frac{1}{2E_{k_{1}}}\delta^{4}(q-u_{1}p-k_{1}-k_{2})$
(27)

is the two-body phase space integration, which corresponds to cutting off the
$c,d$ quark loop in Fig. 2(a). Further, $\Pi_{{\cal O}_{2},{\cal
J}}^{c}(p,q)_{\rm QCD}=-\Pi_{{\cal O}_{1},{\cal J}}^{c}(p,q)_{\rm QCD}$ so
that we only have to calculate the amplitudes induced by ${\cal O}_{1}$. The
integration in Eq. (26) is involved but straightforward, so we will not
present further calculational details here.

For the case of the final quark radiation, the corresponding diagram is shown
in Fig. 2(b), where we take the $s$ quark radiation as an example. The
calculation for this diagram is similar to Fig. 2(a) and the only difference
is that now we have an extra $s$ quark propagator:

$\displaystyle\frac{1}{(q-(u_{1}+u_{2})p)^{2}}=\frac{1}{u_{3}(s-(u_{1}+u_{2})m_{\Sigma}^{2})}.$
(28)

It should be mentioned that for the final quark radiation ${\cal
J}_{\mu}^{{\cal O}_{i}}(0)$ is an composite operator of $j_{q^{\prime}\mu}$
and ${\cal O}_{i}$, so that the hadron level correlation function in Eq. (10)
is actually induced by three operators. However, since we only insert a
complete set of states between ${\cal J}_{\mu}^{{\cal O}_{i}}(0)$ and
$\bar{J}_{\Lambda_{c}}(x)$, the composite operator ${\cal J}_{\mu}^{{\cal
O}_{i}}(0)$ is not disconnected. Therefore, at the QCD level when extracting
the discontinuity part, we only have to cut off the $c,d$ quark loop in Fig.
2(b) and keep the $s$ quark propagator unchanged.

##  V Numerical Results

We first give the input parameters. We use the $\overline{\rm MS}$ masses for
the quarks, $m_{c}(\mu)=1.27$ GeV and $m_{s}(\mu)=0.103$ GeV with $\mu=1.27$
GeV ParticleDataGroup:2020ssz . The masses of $u,d$ quarks are omitted. The
composite masses of the $c,d$ quarks are taken as $\bar{m}_{c}=1.6$ GeV and
$\bar{m}_{d}=0.32$ GeV Cheng:1994kp . The masses of the baryons are
$m_{\Sigma}=1.19$ GeV, $m_{\Lambda_{c}+}=2.286$ GeV and $m_{\Lambda_{c}-}=2.6$
GeV ParticleDataGroup:2020ssz . The decay constant of the
$\Lambda_{c}({1/2}^{+})$ is taken as $\lambda_{+}=0.01\pm 0.001$ Zhao:2020mod
. From Eq. (16), it can be seen that the amplitudes are proportional to the
inverse of $\lambda_{+}$ so that its uncertainty may affect the result a lot.
Therefore, we will include the uncertainty of $\lambda_{+}$ when evaluating
the uncertainty of the decay amplitudes. The shape parameters of the
$\Sigma^{+}$ LCDAs are taken from a lattice calculation with $N_{f}=2+1$ and
vanishing lattice spacing limit $a\to 0$ : $\pi_{00}^{B}=5.14\times 10^{-3}$
GeV2 and $\pi_{11}^{B}=-0.09\times 10^{-3}$ GeV2RQCD:2019hps .

Further, the LCSR contains two kinds of extra parameters, namely the threshold
parameter $s_{\rm th}$ and the Borel parameter $T_{2}$. The threshold
parameter should in principle be process independent and only related to the
corresponding hadron state. Here, $s_{\rm th}$ is taken from a QCD sum rules
study on the decay constant of the $\Lambda_{c}$ Zhao:2020mod : $s_{\rm
th}=2.85^{2}$ GeV2. Generally, the sum rules results are sensitive to the
threshold parameter, thus here we consider a small uncertainty $\pm 0.5$ GeV2
near this value to evaluate the uncertainty from the threshold parameter on
the decay amplitudes.

Figure 3: Decay amplitudes $a_{\cal J}^{+}$ and $b_{\cal J}^{+}$ (in unit
$10^{-3}$ GeV2) as functions of the Borel parameter $T^{2}$. In each diagram,
the blue band denotes the error from the uncertainty of the threshold $s_{\rm
th}=2.85^{2}\pm 0.5$ GeV2. The upper and lower red bands denote the error from
the uncertainty of $\lambda_{+}$.

Generally, the Borel parameter $T_{2}$ is chosen to satisfy three
requirements. First, $T_{2}$ cannot be too large so that the continuous
spectrum contribution is suppressed. Second, $T_{2}$ must be large enough to
ensure the light-cone expansion to convergence. Finally, the result must be
stable in a window of $T_{2}$. The first and the second requirement can
determine the upper and lower bound of the $T_{2}$ window, respectively. Fig.
3 shows the amplitudes $a_{i,{\cal J}}^{+}$ and $b_{i,{\cal J}}^{+}$ as
functions of $T_{2}$. To determine the upper bound, we require that the pole
contribution must be larger than the continuous spectrum contribution, namely:

$\displaystyle\displaystyle\frac{\displaystyle\int_{m_{c}^{2}}^{s_{\rm th}}ds\
e^{-s/T^{2}}{\rm Disc}\ \Pi_{{\cal O}_{i},{\cal J}}(p,s)_{\rm
QCD}}{\displaystyle\int_{m_{c}^{2}}^{\infty}ds\ e^{-s/T^{2}}{\rm Disc}\
\Pi_{{\cal O}_{i},{\cal J}}(p,s)_{\rm QCD}}>0.5.$ (29)

The numerator is the pole contribution, which represents the integral on the
right-hand side of Eq. (16). The denominator is the same integral but the
upper limit of $s$ is extended to infinity, which contains both pole and
continuous spectrum contributions. Note that although the value for this
fraction is derived from experience, as long as the third requirement for
stability is satisfied, the result will be insensitive to this fraction, and
its uncertainty can be attributed to choosing the window of $T_{2}$.

On the other hand, in principle, the lower bound of the $T_{2}$ is determined
by the ratio between the contribution from the leading order and next-to-
leading order QCD corrections to the perturbative kernel. However, in this
work only the leading order contribution is considered so that this method
cannot be used. Following our previous work Shi:2022kfa , to get the window of
$T_{2}$, we can set the center value of $T_{2}$ as its upper bound, and find a
range $\pm 1$ GeV2 around this center value. The amplitudes and the
corresponding errors from the uncertainties of $s_{\rm th}$, $T_{2}$ and
$\lambda_{+}$ are listed in Table 1. Note that the center value of the $T_{2}$
is already in a relatively stable region as shown in Fig. 3, thus the
procedure given above is sufficient for determining the errors of the
amplitudes. Generally, the Borel parameters are close to the corresponding
mass square of hadrons. From Table 1, the $T_{2}$ s for initial-state
radiation are close to $m_{\Lambda_{c}}^{2}$ which is as expected. However,
the $T_{2}$ s for final-state radiation are much smaller. The reason is that
in Fig. 2(b) the extra propagator as shown in Eq. (28) provides a lighter mass
scale $m_{\Sigma}$. Now the $s$ dominates around $m_{\Sigma}^{2}$, which
reduces the optimal value of $T_{2}$.

Table 1: Decay amplitudes $a_{\cal J}^{+}$ and $b_{\cal J}^{+}$ (in unit $10^{-3}$ GeV2) and the corresponding Borel parameters (GeV2) for the initial- and the final-state radiation. $a_{{\rm Ini}}^{+}$ | $T_{2}$ | $b_{{\rm Ini}}^{+}$ | $T_{2}$ | $a_{{\rm Fin}}^{+}$ | $T_{2}$ | $b_{{\rm Fin}}^{+}$ | $T_{2}$
---|---|---|---|---|---|---|---
$-6.03\pm 1.22$ | $5.5\pm 1.0$ | $0.37\pm 0.11$ | $4.7\pm 1.0$ | $-1.56\pm 0.17$ | $2.3\pm 0.5$ | $0.13\pm 0.05$ | $1.45\pm 0.5$

Using the amplitudes given in Table 1, we can obtain the decay width of the
$\Lambda_{c}^{+}\to\Sigma^{+}\gamma$ from the formula

$\displaystyle\Gamma\left(\Lambda_{c}^{+}\to\Sigma^{+}\gamma\right)=\frac{1}{8\pi\
m_{\Lambda_{c}+}^{2}}\left(\frac{m_{\Lambda_{c}+}^{2}-m_{\Sigma}^{2}}{m_{\Lambda_{c}+}}\right)^{3}\frac{G_{F}^{2}}{2}|V_{cs}V_{ud}|^{2}(C_{1}-C_{2})^{2}\left(|a|^{2}+|b|^{2}\right)$
(30)

with $a=a_{{\rm Ini}}^{+}+a_{{\rm Fin}}^{+}$ and $b=b_{{\rm Ini}}^{+}+b_{{\rm
Fin}}^{+}$. The Wilson coefficients are taken as $C_{1}=1.22$ and
$C_{2}=-0.43$ at $\mu=m_{c}$ Li:2012cfa . The CKM matrix elements are
$|V_{cs}|=0.975$ and $|V_{ud}|=0.973$ ParticleDataGroup:2020ssz . Using the
$\Lambda_{c}^{+}$ lifetime $\tau(\Lambda_{c}^{+})=2.01\times 10^{-13}$ s
ParticleDataGroup:2020ssz , we can obtain the branching fraction

$\displaystyle{\cal B}(\Lambda_{c}^{+}\to\Sigma^{+}\gamma)=1.03\pm 0.36\times
10^{-4},$ (31)

which is below the experimental upper limit given recently by the Belle
Collaboration Belle:2022raw :

$\displaystyle{\cal B}_{\rm
expr}(\Lambda_{c}^{+}\to\Sigma^{+}\gamma)<2.6\times 10^{-4}.$ (32)

Table 2 gives a comparison of the $\Lambda_{c}^{+}\to\Sigma^{+}\gamma$
branching fraction from this work, the result from the Belle Collaboration,
the modified nonrelativistic quark model (NRQM) Kamal:1983zt , the constituent
quark model (CQM) Uppal:1992cc  and the effective Hamiltonian approach (EHA)
Cheng:1994kp . The branching fraction from the CQM is slightly larger than the
experimental upper limit, while the branching fractions from other theoretical
methods are nearly one order smaller than the upper limit. Our result is
between these theoretical predictions and the experimental upper limit. Due to
the limitation on the data sample and resolution, an extremely small branching
fraction is difficult to be measured. However, the relatively larger branching
fraction predicted in this work is more likely to be tested by future
experiments.

Table 2: Comparison of the branching fraction ${\cal B}(\Lambda_{c}^{+}\to\Sigma^{+}\gamma)$ from this work with those from the literature and the Belle experiment. Method | ${\cal B}(\Lambda_{c}^{+}\to\Sigma^{+}\gamma)$
---|---
This Work | $1.03\pm 0.36\times 10^{-4}$
NRQM Kamal:1983zt  | $3.2\times 10^{-5}$
CQM Uppal:1992cc  | $2.8\pm 0.6\times 10^{-4}$
EHA Cheng:1994kp  | $4.9\times 10^{-5}$
Exp. Belle:2022raw  | $<2.6\times 10^{-4}$

##  VI Conclusion

We have calculated the decay width of $\Lambda_{c}^{+}\to\Sigma^{+}\gamma$
using light-cone sum rules. For the initial quark radiation we constructed an
effective Hamiltonian to simplify the calculation, where the internal quark
line shrinks to a point. The final quark radiation is studied utilizing the
full theory. The leading twist light-cone distribution amplitudes of the
$\Sigma^{+}$ serve as the non-perturbative input for the sum rule calculation,
and the perturbative kernel is calculated at leading order. The branching
fraction we obtain is ${\cal B}(\Lambda_{c}^{+}\to\Sigma^{+}\gamma)=1.03\pm
0.36\times 10^{-4}$, which is between previous theoretical predictions and the
experimental upper limit. Considering the data sample and resolution of the
experiment, we believe that our prediction can be tested in the near future.

## Acknowledgements

The authors are grateful to Ji-Bo He, Chengping Shen, Wei Wang, Zhen-Xing Zhao
and Chien-Yeah Seng for useful discussions. This work is supported in part by
the NSFC under Grant No.12147147 and the Deutsche Forschungsgemeinschaft (DFG,
German Research Foundation) through the funds provided to the Sino-German
Collaborative Research Center TRR110 “Symmetries and the Emergence of
Structure in QCD” (NSFC Grant No. 12070131001, DFG Project-ID 196253076 - TRR
110). The work of UGM was supported in part by the Chinese Academy of Sciences
(CAS) President’s International Fellowship Initiative (PIFI) (Grant No.
2018DM0034) and by VolkswagenStiftung (Grant No. 93562).

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