$r_{(a,b)}=\begin{cases}\left\lfloor{\frac{a-b}{6}}\right\rfloor+1&\text{if
}a\equiv_{2}b\equiv_{2}1\\\ \left\lfloor{\frac{a-b}{6}}\right\rfloor&\text{if
}a\equiv_{2}b\equiv_{2}0\\\ 0&\text{if }a\not\equiv_{2}b.\end{cases}$

###### Proof.

The group $\mathfrak{S}_{2}\times\mathfrak{S}_{3}$ is included in
$\operatorname{GL}_{2}(\mathbb{Z})$ by sending the transposition
$(12)\in\mathfrak{S}_{2}$ to $-I_{2}$, transpositions in $\mathfrak{S}_{3}$
map to matrices with eigenvalues $(1,-1)$ and its $3$-cycles map to matrices
with eigenvalues $(\zeta_{3},\zeta_{3}^{-1})$ where $\zeta_{3}$ is a primitive
$3$rd root of unity.

The character of the Schur functor $\mathbb{S}_{(a,b)}(\mathbb{Q}^{2})$ is
given by sending a matrix $A\in\operatorname{GL}_{2}(\mathbb{Z})$ with
eigenvalues $(x,y)$ to the Schur polynomial
$s_{(a,b)}(x,y)=\sum_{i=b}^{a}x^{i}y^{a+b-i}$. One readily checks that
transpositions in $\mathfrak{S}_{3}$ acts with trace equal to $(-1)^{b}$ times
the residue of $(a-b+1)$ mod $2$, and $3$-cycles act with trace $(a-b+1)$ mod
$3$ (taking $2$ mod $3$ to have residue $-1$). With these character values the
scalar product with $\operatorname{sgn}_{3}$ is as claimed. ∎

######  Example 6.9 (Multiplicity of $\chi_{(2,1^{n-2})}$).

Let us use calculations of bead representations to see that the
$\mathfrak{S}_{n}$-representation
$\chi_{(2,1^{n-2})}\cong\operatorname{Std}_{n}\otimes\operatorname{sgn}_{n}$
never occurs in $\operatorname{gr}^{W}_{0}H^{n+2}_{c}(\mathcal{M}_{2,n})$, as
conjectured in [BCGY21, Conj. 3.8]. Indeed, in Example 4.16 we explained that
Powell-Vespa effectively compute $\chi_{(2,1^{n-2})}$ to never occur in the
cohomlogy $H^{n-1}_{c}(F(X,n))$. In light of Proposition 1.11, this
immediately implies that this irreducible does not appear in
$\operatorname{gr}^{W}_{0}H^{n+2}_{c}(\mathcal{M}_{2,n})$.

The other half of the conjecture involves the standard representation
$\chi_{(n-1,1)}$. The multiplicity of these are at the moment out of reach,
but see the first point of Conjecture 6.17 below.

Together with our computations of $\Phi[n,\lambda]$ for $n\leq 10$ below,
Proposition 1.11 recovers the $\mathfrak{S}_{n}$-character of
$\operatorname{gr}_{0}^{W}H_{c}^{*}(\mathcal{M}_{2,n})$ as computed in
[BCGY21]. For $n\geq 11$ we give a new lower bound. For example, we find the
following lower bound for $n=11$:

$\displaystyle\operatorname{gr}_{0}^{W}H_{c}^{13}($
$\displaystyle\mathcal{M}_{2,11},\mathbb{Q})\geq
2\chi_{(9,1^{2})}+2\chi_{(8,3)}+4\chi_{(8,2,1)}+2\chi_{(8,1^{3})}+\underline{2\chi_{(7,4)}}+$
$\displaystyle
12\chi_{(7,3,1)}+\underline{5\chi_{(7,2^{2})}}+12\chi_{(7,2,1^{2})}+\underline{2\chi_{(7,1^{4})}}+3\chi_{(6,5)}+12\chi_{(6,4,1)}+$
$\displaystyle
18\chi_{(6,3,2)}+23\chi_{(6,3,1^{2})}+20\chi_{(6,2^{2},1)}+18\chi_{(6,2,1^{3})}+5\chi_{(6,1^{5})}+8\chi_{(5^{2},1)}+$
$\displaystyle
16\chi_{(5,4,2)}+22\chi_{(5,4,1^{2})}+15\chi_{(5,3^{2})}+42\chi_{(5,3,2,1)}+25\chi_{(5,3,1^{3})}+12\chi_{(5,2^{3})}+$
$\displaystyle
32\chi_{(5,2^{2},1^{2})}+17\chi_{(5,2,1^{4})}+\underline{7\chi_{(5,1^{6})}}+7\chi_{(4^{2},3)}+24\chi_{(4^{2},2,1)}+17\chi_{(4^{2},1^{3})}+$
$\displaystyle
21\chi_{(4,3^{2},1)}+23\chi_{(4,3,2^{2})}+42\chi_{(4,3,2,1^{2})}+21\chi_{(4,3,1^{4})}+19\chi_{(4,2^{3},1)}+23\chi_{(4,2^{2},1^{3})}+$
$\displaystyle\underline{11\chi_{(4,2,1^{5})}}+\underline{2\chi_{(4,1^{7})}}+10\chi_{(3^{3},2)}+8\chi_{(3^{3},1^{2})}+19\chi_{(3^{2},2^{2},1)}+18\chi_{(3^{2},2,1^{3})}+$
$\displaystyle\underline{11\chi_{(3^{2},1^{5})}}+4\chi_{(3,2^{4})}+13\chi_{(3,2^{3},1^{2})}+\underline{8\chi_{(3,2^{2},1^{4})}}+\underline{6\chi_{(3,2,1^{6})}}+2\chi_{(2^{5},1)}+$
$\displaystyle\underline{4\chi_{(2^{4},1^{3})}}+\underline{2\chi_{(2^{3},1^{5})}}+\underline{2\chi_{(2^{2},1^{7})}}+\underline{\chi_{(1^{11})}},$

which thus has dimension at least $727840$. The underlined terms are the ones
for which our lower bound agrees with the exact multiplicity computed in
[BCGY21]. The other terms are either known to have higher multiplicity, or
their exact multiplicity is still unknown.

For number $n$ larger still, the super-exponential multiplicity of every Schur
functor discussed in Example 6.4 implies similar growth in
$\operatorname{gr}^{W}_{0}H^{*}_{c}(\mathcal{M}_{2,n})$. The reader can find
our equivariant lower bound for
$\operatorname{gr}_{0}^{W}H_{c}^{*}(\mathcal{M}_{2,n})$ presented on this
webpage101010https://louishainaut.github.io/GH-ConfSpace/. Forgetting the
$\mathfrak{S}_{n}$-action, a lower bound on the dimension of the cohomology
$\operatorname{gr}^{W}_{0}H^{n+2}_{c}(\mathcal{M}_{2,n})$ for all $n\leq 17$
are listed in Table 2.

$n$ | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ | $9$ | $10$
---|---|---|---|---|---|---|---|---|---|---|---
$\dim$ | $0$ | $0$ | $0$ | $0$ | $1$ | $5$ | $26$ | $155$ | $1066$ | $8666$ | $81012$

$n$ | $11$ | $12$ | $13$ | $14$
---|---|---|---|---
$\dim$ | $\geq 727\,840$ | $\geq 7\,920\,155$ | $\geq 94\,325\,925$ | $\geq 1\,220\,494\,146$

$n$ | $15$ | $16$ | $17$
---|---|---|---
$\dim$ | $\geq 17\,048\,375\,436$ | $\geq 255\,669\,776\,040$ | $\geq 4\,096\,729\,778\,379$

Table 2: Dimension of
$\operatorname{gr}_{0}^{W}H_{c}^{n+2}(\mathcal{M}_{2,n},\mathbb{Q})$

###  6.3 Highly connected complexes of $\mathfrak{S}_{n}$-representations

It is well known that the Whitney homology of the partition lattice admits a
differential making the homology into an acyclic complex – this fact was used
by Sundaram [Sun94] to give a calculation of the poset homology, going by the
term _the Whitney homology trick_. This acyclic complex in fact shows up as
the bottom row of the $E_{1}$-page of the collision spectral sequence.

Looking at the other rows, one thus find analogous highly connected complexes
of $\mathfrak{S}_{n}$-representations: one attached to every Schur functor
$\mathbb{S}_{\lambda}$; with Sundaram’s complex attached to the trivial
functor $\mathbb{S}_{()}$. These complexes are no longer acyclic, instead we
have the following.

######  Corollary 6.10.

For every partition $\lambda\vdash q\leq n-1$, there is an $n-q-2$-connected
complex of $\mathfrak{S}_{n}$-representations

$V^{\lambda}_{0}\to V^{\lambda}_{1}\to\ldots\to V^{\lambda}_{n-q}$ (54)

with terms $V^{\lambda}_{i}$ given by the $\mathbb{S}_{\lambda}$-isotypic
component of the Chevalley-Eilenberg complex

$C^{CE}_{i-n}(A_{V}\otimes\operatorname{S}\operatorname{Lie})(n)$ (55)

and the differential is the usual $\partial^{CE}$. These complexes have only
their top two cohomology groups $\neq 0$, and they are isomorphic to
$\Phi^{n-q-1}[n,\lambda]$ and $\Phi^{n-q}[n,\lambda]$ respectively.

The complexes can be directly expressed in terms of plethysms, and they give
highly nontrivial relations between such. Two new families of complexes are
afforded by Theorem 1.6.

######  Example 6.11.

For $\lambda=(m)$ one gets an $(n-m-2)$-connected complex of
$\mathfrak{S}_{n}$-representations of length $n-m$, whose top two cohomology
representations are the Whitehouse modules
$H_{2(n-m-1)}(F(\mathbb{R}^{3},n-1))$ and $H_{2(n-m)}(F(\mathbb{R}^{3},n-1))$
respectively111111Really the Chevalley-Eilenberg complex terminates with a
sign twist of these modules. But then the sign twist of that complex is the
one discussed in the example..

######  Example 6.12.

For $\lambda=(1^{m})$ one gets an $(n-m-2)$-connected complex of
$\mathfrak{S}_{n}$-representations of length $n-m$ whose top two cohomology
representations are Getzler’s modules $H_{n-m-2}(\mathcal{M}_{0,n})$ and
$H_{n-m}(\mathcal{M}_{0,n})$.

###### Proof of Corollary 6.10.

The $q$-th row of the Chevalley-Eilenberg bicomplex for a wedge of circles
decomposes as a sum of Schur functors of degree $q$. By Proposition 5.5 its
homology vanishes in degrees $\leq n-q-2$. Decomposing into
$\mathbb{S}_{\lambda}$-isotypic components gives the proposed complex of
$\mathfrak{S}_{n}$-representations. ∎

###  6.4 Patterns and conjectures

We start this section by proving the fourth statement of Proposition 5.5,

######  Proposition 6.13.

The coefficient $\Phi^{0}[n,n-1]$ of the polynomial decomposition is

$\Phi^{0}[n,n-1]=\chi_{(1^{n})}\boxtimes\chi_{{(1^{n-1})}}.$

###### Proof.

From Proposition 5.15 the equivariant multiplicity of the alternating power
$\Lambda^{n-1}$ in $\Psi^{0}[n,n-1]$ is

$\Phi^{0}[n,{(1^{n-1})}]=H_{0}(F(\mathbb{R}^{3},n-1))\otimes\operatorname{sgn}_{n}=\operatorname{sgn}_{n},$

so $\Phi^{0}[n,n-1]$ is at least as large as indicated, and it remains to
prove that it does not contain any other terms.

For this we look closely at the collision spectral sequence. Once again, it is
enough to prove the statement when $X=\bigvee_{i=1}^{g}{S^{1}}$ is a wedge of
sufficiently many circles. In fact $g\geq n-1$ is enough, but we may as well
consider all $g$’s simultaneously. The polynomial representation
$\Phi^{0}[n,n-1]\otimes_{\mathfrak{S}_{n-1}}{\mathbb{Q}^{g}[-1]}^{\otimes
n-1}$ is by definition the kernel of the differential $d_{1}\colon
E_{1}^{0,n-1}\to E_{1}^{1,n-1}$. The multiplicity of alternating powers
implies that this kernel contains at least one copy of
$\operatorname{sgn}_{n}\boxtimes\operatorname{Sym}^{n-1}(\mathbb{Q}^{g})[1-n]$,
which has dimension $\binom{n+g-2}{n-1}$. We claim that the kernel of this
differential cannot have larger dimension.

A basis for
$E_{1}^{0,n-1}=\Lambda^{n}\left(H^{*}(X)\otimes\operatorname{S}\operatorname{Lie}(1)\right)^{n-1}$
is given by tuples

$\langle(x_{1},\alpha_{i_{1}}),\ldots,(x_{k-1},\alpha_{i_{k-1}}),(x_{k},1),(x_{k+1},\alpha_{i_{k+1}}),\ldots,(x_{n},\alpha_{i_{n}})\rangle$

where the symbols $x_{i}$ are copies of the ‘Lie variable’ in
$\operatorname{S}\operatorname{Lie}(1)\cong\mathbb{Q}x$, and the vectors
$\alpha_{1},\ldots,\alpha_{g}\in\tilde{H}^{1}(X)$ form a basis. Here $1$
refers to the unit $1\in H^{0}(X)$, and the indices are such that $1\leq k\leq
n$ along with $1\leq i_{1},\ldots,\widehat{i_{k}},\ldots,i_{n}\leq g$. We
denote this basis element by
$\mathcal{B}_{(i_{1},\ldots,\widehat{i_{k}},\ldots,i_{n})}$.

For
$E_{1}^{1,n-1}\cong\bigoplus_{\binom{n}{2}}(H^{*}(X)\otimes\operatorname{S}\operatorname{Lie}(2))\bigotimes\Lambda^{n-2}\left(H^{*}(X)\otimes\operatorname{S}\operatorname{Lie}(1)\right)^{n-2}$
we work with the basis of tuples

$\langle([x_{j},x_{k}],\alpha_{i_{0}}),(x_{1},\alpha_{i_{1}}),\ldots,\widehat{(x_{j},\alpha_{i_{j}})},\ldots,\widehat{(x_{k},\alpha_{i_{k}})},\ldots,(x_{n},\alpha_{i_{n}})\rangle$

for every pair $\\{j,k\\}$ and sequence of indices
$(i_{0},i_{1},\ldots,\widehat{i_{j}},\ldots,\widehat{i_{k}},\ldots,i_{n})$.
Note that now all terms in the tuple contain a cohomology class
$\alpha_{i}\in\tilde{H}^{1}(X)$. We denote this basis element by
$\mathcal{C}_{(i_{0},i_{1},\ldots,\widehat{i}_{j},\ldots,\widehat{i}_{k},\ldots,i_{n})}$.

Now, the differential $d_{1}$ acts on our basis by

$d_{1}:\mathcal{B}_{(i_{1},\ldots,\widehat{i_{k}},\ldots,i_{n})}\longmapsto\sum_{j\neq
k}\pm\mathcal{C}_{(i_{j},i_{1},\ldots,\widehat{i}_{j},\ldots,\widehat{i}_{k},\ldots,n)}$
(56)

and in particular all resulting basis elements appear with coefficient $\pm
1$. Let us consider the dual problem: which $\mathcal{B}_{\mathbf{i}}$’s are
sent to a linear combination containing a nonzero multiple a particular
$c=\mathcal{C}_{(i_{0},\ldots,\widehat{i}_{j},\ldots,\widehat{i}_{k},\ldots,i_{n})}$?

From the description of $d_{1}$ there are exactly two such basis elements
$b_{1}=\mathcal{B}_{(i_{1},\ldots,i_{0},\ldots,\widehat{i_{k}},\ldots,i_{n})}$
and
$b_{2}=\mathcal{B}_{(i_{1},\ldots,\widehat{i}_{j},\ldots,i_{0},\ldots,i_{n})}$,
with $i_{0}$ inserted in the $j$-th, resp. $k$-th, empty slot. It follows that
every element of $\ker(d_{1})$ that contains a nontrivial multiple of $b_{1}$
must also contain one of $b_{2}$, and the coefficient of one uniquely
determines the coefficient of the other.

Define an equivalence relation of the $\mathcal{B}_{\mathbb{i}}$’s, generated
by the relation that $b_{1}\sim b_{2}$ if their indexing tuples differ by
moving one $i_{j}$ of $b_{1}$ to the empty slot of $b_{2}$. Then the
discussion in the previous paragraph shows that $\ker(d_{1})$ is spanned by
$\sim$-equivalence classes, and it remains to count these classes.

It is straightforward to see that equivalence classes are in bijection with
multisets $\langle i_{1},\ldots,i_{n-1}\rangle$ with $1\leq i_{j}\leq g$ for
all $j$. Thus there are exactly $\binom{n+g-2}{n-1}$ many equivalence classes,
and the kernel has at most this dimension. This is what we wanted to show. ∎

######  Remark 6.14.

Using Euler characteristic, one can obtain from the previous proposition a
description of $\Phi^{1}[n,n-1]$, equivalent to the one in [PV18, Corollary
3].

Calculations for small values of $n$ also suggest the following pattern.

######  Conjecture 6.15.

For $n\geq 4$, the coefficients $\Phi^{1}[n,n-2]$ are given by

$\Phi^{1}[n,n-2]=(\chi_{(3,1^{n-3})}\boxtimes\chi_{(1^{n-2})})\oplus(\chi_{(n)}\boxtimes\chi_{(n-2)}).$
(57)

Up to this point we mainly discussed the $\mathfrak{S}_{n}$-equivariant
multiplicity of individual Schur functors in
$\operatorname{gr}H^{*}_{c}(F(X,n))$, equivalently the coefficients
$\Phi[-,\mu]$. Let us now shift perspective and instead consider for fixed
$\mathfrak{S}_{n}$-irreducible $\chi_{\lambda}$ the corresponding isotypic
component as a sum of Schur functors. These are equivalently given by the
coefficient $\Phi[\lambda,-]:=\oplus_{m=0}^{|\lambda|}\Phi[\lambda,m]$.

Through our calculations we have identified several patterns in these
coefficients – see below. We present our conjectural formulas only for
multiplicities in codimension 1 cohomology of $F(\vee S^{1},n)$, but using the
spectral sequence to compute the Euler characteristic, one can readily compute
the corresponding multiplicity in codimension 0 as well.

In some of the conjectural patterns below we use the notation

$M_{n}=\sum_{2a+b=n}{\mathbb{S}_{{(a,1^{b})}^{\ast}}},$

where the sum is over $a\geq 1$ and $b\geq 0$. The following can be deduced
from [PV18, Examples 3-5] using the dictionary in §4.4.

######  Proposition 6.16.

For partition $\lambda\vdash n$, the $\chi_{\lambda}$-isotypic component of
$\operatorname{gr}H^{n-1}_{c}(F(X,n))$, equivalently the bead representation
$U_{\lambda}^{II}$ from [TW19, §2.5], is the following polynomial functor.

  * •

For $\lambda=(n)$, the trivial representation it is $M_{n-1}$.

  * •

For $\lambda=(1^{n})$ the sign representation, it is
$\mathbb{S}_{{(1^{n-1})}^{\ast}}$.

  * •

For $\lambda=(2,1^{n-2})$ it is $0$.

We observed the following further patterns, verified computationally for up to
$n\leq 11$ and always matching the estimate given by the lower bound in §6.1.
Any multiplicity involving $\mathbb{S}_{{(k)}^{\ast}}$ and
$\mathbb{S}_{{(1^{k})}^{\ast}}$ is completely determined by Theorem 1.6.

######  Conjecture 6.17.

The $\chi_{\lambda}$-isotypic components of $\Psi(n,\bullet)$ continue as
follows.

  * •

For $\lambda=(n-1,1)$ it is $\begin{cases}\mathbb{S}_{{(n/2)}^{\ast}}&\text{if
}n\text{ is even}\\\ 0&\text{if }n\text{ is odd}\end{cases}.$

  * •

For $\lambda=(n-2,2)$ it is the sum
$\sum_{k=0}^{\lfloor{\frac{n}{4}}\rfloor-1}{M_{n-2-4k}}+\sum_{k=0}^{\lfloor{\frac{n-7}{4}}\rfloor}{M_{n-5-4k}}$

$+\left\\{\begin{array}[]{ll}\sum_{k=0}^{\lfloor{\frac{n}{4}}\rfloor-2}\mathbb{S}_{{\left(\frac{n-4}{2}-2k\right)}^{\ast}}&\text{if
}n\text{ is even}\\\
\sum_{k=0}^{\lfloor{\frac{n-5}{4}}\rfloor}\mathbb{S}_{{\left(\frac{n-1}{2}-2k\right)}^{\ast}}&\text{if
}n\text{ is odd}\end{array}\right.$

  * •

For $\lambda=(n-2,1,1)$ it is the sum
$\sum_{k=0}^{\lfloor{\frac{n-5}{4}}\rfloor}{M_{n-3-4k}}+\sum_{k=0}^{\lfloor{\frac{n-6}{4}}\rfloor}{M_{n-4-4k}}$

$+\left\\{\begin{array}[]{ll}\sum_{k=0}^{\lfloor{\frac{n-6}{4}}\rfloor}\mathbb{S}_{{\left(\frac{n-2}{2}-2k\right)}^{\ast}}&\text{if
}n\text{ is even}\\\
\sum_{k=0}^{\lfloor{\frac{n-3}{4}}\rfloor}\mathbb{S}_{{\left(\frac{n+1}{2}-2k\right)}^{\ast}}&\text{if
}n\text{ is odd}\end{array}\right.$

  * •

For $\lambda={(n-3,3)}^{\ast}$ it is
$\sum_{k=0}^{n-6}{\lfloor{\frac{k+3}{3}}\rfloor\mathbb{S}_{{(n-4-k)}^{\ast}}}+a_{n}\mathbb{S}_{{(1,1)}^{\ast}}+b_{n}\mathbb{S}_{{(1)}^{\ast}}$

  * •

For $\lambda={(n-3,2,1)}^{\ast}$ it is
$\sum_{k=0}^{n-5}{\left\lfloor{\frac{2k+3}{3}}\right\rfloor\mathbb{S}_{{(n-3-k)}^{\ast}}}+\left\lfloor{\frac{n-2}{3}}\right\rfloor\mathbb{S}_{{(1,1)}^{\ast}}+\left\lfloor{\frac{n-4}{3}}\right\rfloor\mathbb{S}_{{(1)}^{\ast}}$

  * •

For $\lambda={(n-3,1,1,1)}^{\ast}$ it is
$\sum_{k=0}^{n-6}{\left\lfloor{\frac{k+3}{3}}\right\rfloor\mathbb{S}_{{(n-4-k)}^{\ast}}}+c_{n}\mathbb{S}_{{(1,1)}^{\ast}}+d_{n}\mathbb{S}_{{(1)}^{\ast}}$

  * •

For $\lambda={(n-2,2)}^{\ast}$ it is
$\sum_{k=0}^{\lfloor{n/2}\rfloor-2}{\mathbb{S}_{{(n-3-2k)}^{\ast}}}$

  * •

For $\lambda={(n-2,1,1)}^{\ast}$ it is
$\sum_{k=0}^{\lfloor{(n-3)/2}\rfloor}{\mathbb{S}_{{(n-2-2k)}^{\ast}}}$

where the numbers $a_{n},b_{n},c_{n}$ and $d_{n}$ are $\lfloor{n/6}\rfloor$
plus $6$-periodic functions of $n$. Explicitly,

$\displaystyle a_{n}=\lfloor{n/6}\rfloor$ $\displaystyle+$
$\displaystyle(-1,0,-1,0,0,0)_{n}$ $\displaystyle b_{n}=\lfloor{n/6}\rfloor$
$\displaystyle+$ $\displaystyle(-1,-1,-1,0,-1,0)_{n}$ $\displaystyle
c_{n}=\lfloor{n/6}\rfloor$ $\displaystyle+$ $\displaystyle(0,0,0,0,1,0)_{n}$
$\displaystyle d_{n}=\lfloor{n/6}\rfloor$ $\displaystyle+$
$\displaystyle(0,-1,0,0,0,0)_{n}$

where $(-)_{n}$ returns the entry in position residue of $n$ modulo $6$.

######  Remark 6.18.

Interestingly, the complicated patterns for $\lambda=(n-2,2)$ and $(n-2,1,1)$
become far simpler when _summed together_ ; and the same is true for their
conjugate partitions. For the partitions $\lambda={(n-3,3)}^{\ast}$ and
${(n-3,1,1,1)}^{\ast}$ it is now their _difference_ that has a significantly
simpler form: one can observe that $c_{n}-a_{n}$ and $d_{n}-b_{n}$ both take
the value $1$ if $n$ is even and $0$ if $n$ is odd. We have no guess as to why
this should be the case.

These conjectural multiplicities give conjectural ranks for the corresponding
isotypical components of $H_{c}^{*}(\mathcal{M}_{2,n},\mathbb{Q})$ through
Proposition 1.11.

##  Appendix A Full decomposition for $10$ particles

We explain here how to combine insights from the different presentations of §4
to compute the full cohomology of $F(X,10)$ for $X$ any wedge of circles.
Conceptually, the $2$-term complex of §4.3 gives access to individual
$\mathfrak{S}_{n}$-isotypic components, most effective for Schur functors
$\mathbb{S}_{\lambda}$ for $\lambda$ with few parts. On the other hand, the
Chevalley-Eilenberg complex of §4.1 accesses individual
$\operatorname{GL}$-isotypic components, most effective for
$\mathbb{S}_{\lambda}$ of large total degree $|\lambda|$. Between these two,
one can uniquely determine the complete representation.

To illustrate our approach we focus on the multiplicity of a few specific
$\mathfrak{S}_{10}$-irreducibles in the bottom cohomology $H^{9}(F(X,10))$.
Note that by the bound on polynomial degree in Proposition 5.5 and since the
highest degree term $\Phi^{0}[10,9]$ is also computed there, the only
remaining multiplicities that need to be computed are those of Schur functors
of degree $\leq 8$. In the following we will show how to compute the
equivariant multiplicity of the $\mathfrak{S}_{10}$-representations
$\chi_{(8,2)}$, $\chi_{(6,3,1)}$, $\chi_{(4^{2},2)}$ and $\chi_{(3,2,1^{5})}$.

### Step 1: Lower bound

First compute the lower bounds for these $\mathfrak{S}_{10}$-irreducible as in
§6.1, corrected with the knowledge of the exact multiplicity of the symmetric
and alternating powers $\mathbb{S}_{(k)}$ and $\mathbb{S}_{(1^{k})}$ from
§5.3:

$\mathfrak{S}_{10}$-irrep |  $\operatorname{GL}$-equivariant multiplicities
---|---
$(8,2)$ |  $\geq\mathbb{S}_{(4)}+\mathbb{S}_{(3)}+\mathbb{S}_{(2,1)}+\mathbb{S}_{(2)}+\mathbb{S}_{(1^{7})}+\mathbb{S}_{(1^{4})}+\mathbb{S}_{(1^{3})}$
$(6,3,1)$ |  $\geq\mathbb{S}_{(5)}+6\mathbb{S}_{(4)}+11\mathbb{S}_{(3)}+2\mathbb{S}_{(2,1)}+10\mathbb{S}_{(2)}+\mathbb{S}_{(1^{6})}+2\mathbb{S}_{(1^{5})}+3\mathbb{S}_{(1^{4})}+5\mathbb{S}_{(1^{3})}+7\mathbb{S}_{(1^{2})}+4\mathbb{S}_{(1)}$
$(4^{2},2)$ |  $\geq\mathbb{S}_{(5)}+6\mathbb{S}_{(4)}+6\mathbb{S}_{(3)}+\mathbb{S}_{(2,1)}+10\mathbb{S}_{(2)}+2\mathbb{S}_{(1^{4})}+6\mathbb{S}_{(1^{3})}+5\mathbb{S}_{(1^{2})}+\mathbb{S}_{(1)}$
$(3,2,1^{5})$ |  $\geq\mathbb{S}_{(7)}+\mathbb{S}_{(6)}+2\mathbb{S}_{(5)}+3\mathbb{S}_{(4)}+3\mathbb{S}_{(3)}+4\mathbb{S}_{(2)}+2\mathbb{S}_{(1^{2})}+2\mathbb{S}_{(1)}$
Table 3: Lower bounds for $n=10$

We already know the multiplicity of all symmetric powers $\mathbb{S}_{(m)}$,
so we proceed to determine the multiplicities of Schur functors with two parts
$\mathbb{S}_{(a,b)}$.

### Step 2: Traces on Schurs with two parts

To gain understanding of the $\operatorname{End}(\mathbb{Z}^{2})$-action on
each of the above multiplicity spaces, we use the 2-step complex from 4.3 with
genus $g=2$ to compute its character. Start by computing the trace of the
action induced by specific diagonal matrices on each multiplicity space:
$diag(1,1),diag(1,2)$ and $diag(2,2)$ – these calculations are not too
demanding as they only involve one or two doubling maps on $S^{1}\vee S^{1}$.

Once these traces are obtained, we subtract from them the traces of the Schur
functors appearing in the lower bound. We thus obtain the traces of terms
missing from our lower bound. In our example, the traces corresponding to the
lower bounds, as well as the true computed traces of our $3$ diagonal matrices
and the difference between them are as in Table 4

$\mathfrak{S}_{10}$-irreducible | Lower bound | Computed traces | Difference
---|---|---|---
$(8,2)$ | $[14,59,140]$ | $[14,59,140]$ | $[0,0,0]$
$(6,3,1)$ | $[129,522,1220]$ | $[145,580,1396]$ | $[16,58,176]$
$(4^{2},2)$ | $[99,428,1024]$ | $[109,464,1136]$ | $[10,36,112]$
$(3,2,1^{5})$ | $[72,684,2256]$ | $[72,684,2256]$ | $[0,0,0]$
Table 4: Traces for the matrices diag(1,1), diag(1,2) and diag(2,2)

One thus immediately sees that for the first and last rows the lower bound is
sharp, i.e. it accounts for all Schur functors $\mathbb{S}_{(a,b)}$. The
remaining rows are missing a $16$-dimensional and a $10$-dimensional
representation respectively, with known character values at $diag(1,2)$ and
$diag(2,2)$.

The problem thus reduces to finding a sum of Schur functors that produces the
traces seen in the last column of Table 4. Note that while the complete traces
are rather large, their difference from our lower bound is significantly
smaller. Moreover, as stated at the outset of this calculation, only Schur
functors of degree $\leq 8$ may appear. The corresponding traces on these
Schur functors are given in Table 5.

Schur functor | Traces | Schur functor | Traces
---|---|---|---
$(2,1)$ | $[2,6,16]$ | $(3,1)$ | $[3,14,48]$
$(2^{2})$ | $[1,4,16]$ | $(4,1)$ | $[4,30,128]$
$(3,2)$ | $[2,12,64]$ | $(5,1)$ | $[5,62,320]$
$(4,2)$ | $[3,28,192]$ | $(3^{2})$ | $[1,8,64]$
$(6,1)$ | $[6,126,768]$ | $(5,2)$ | $[4,60,512]$
$(4,3)$ | $[2,24,256]$ | $(7,1)$ | $[7,254,1792]$
$(6,2)$ | $[5,124,1280]$ | $(5,3)$ | $[3,56,768]$
$(4^{2})$ | $[1,16,256]$ |  |
Table 5: Traces of Schur functors at diag(1,1), diag(2,1) and diag(2,2)

It is useful to note that $s_{(a,b)}(2,2)=2^{a+b}s_{(a,b)}(1,1)$ for any Schur
polynomial $s_{(a,b)}$, i.e. the third trace in the table is always $2^{a+b}$
larger than the first. Therefore since $\mathbb{S}_{(2,1)}$ is the only Schur
functor in our table for which $2^{a+b}<16$, any sum of other Schur functors
must have its third trace at least $16$ times larger than than the first.
Since the missing Schur functors for $\chi_{(6,3,1)}$ and $\chi_{(4^{2},2)}$
have their third trace $<16$ times their first, one immediately concludes that
$\mathbb{S}_{(2,1)}$ occurs several times.

Let us therefore account for these copies of $\mathbb{S}_{(2,1)}$ and remove
them from the trace until the third column is $\geq 16$ times the first. For
example, from the traces $[10,36,112]$ corresponding to $\chi_{(4^{2},2)}$ we
subtract $m\times[2,6,16]$ with $m$ such that $16\cdot(10-2m)\leq 112-16m$.
That is, $m\geq 3$, leaving traces $[4,18,64]$ to be accounted for. In general
the $\mathbb{S}_{(2,1)}$-multiplicity is $\geq c_{1}-c_{3}/16$ where $c_{i}$
is the $i$-th trace in our table.

At this point the remaining Schur functors can be uniquely determined. The
Schur functors $\mathbb{S}_{(a,b)}$ that need to be added can be found in
Table 6; the interested reader is invited to verify that no other combination
of Schur functors gives the desired traces.

$\mathfrak{S}_{10}$-irreducible | Missing Schur functors
---|---
$(8,2)$ | $0$
$(6,3,1)$ | $5\mathbb{S}_{(2,1)}+2\mathbb{S}_{(3,1)}$
$(4^{2},2)$ | $3\mathbb{S}_{(2,1)}+\mathbb{S}_{(3,1)}+\mathbb{S}_{(2,2)}$
$(3,2,1^{5})$ | $0$
Table 6: Corrections to the lower bounds in genus $2$

We therefore add these Schur functors to our lower bound estimate and proceed
to find Schur functors with $3$ or more parts.

### Step 3: Ranks in genus 3 and 4

One could attempt the same game in higher genus, i.e. finding multiplicities
of $\mathbb{S}_{(a,b,c)}$ and so on, but it turns out that one can (almost
always) make due with only the nonequivariant ranks in genus $3$ and $4$. For
example, Table 7 shows the ranks in genus $3$ and $4$ of the revised lower
bound compared with the true rank computed using the $2$-step complex in §4.3.

$\mathfrak{S}_{10}$-irreducible | Lower bound | Actual ranks | Difference
---|---|---|---
$(8,2)$ | $[40,90]$ | $[46,126]$ | $[6,36]$
$(6,3,1)$ | $[405,897]$ | $[411,931]$ | $[6,34]$
$(4^{2},2)$ | $[308,691]$ | $[308,691]$ | $[0,0]$
$(3,2,1^{5})$ | $[217,541]$ | $[217,541]$ | $[0,0]$
Table 7: Ranks in genus 3 and 4

From Table 7 one sees that only rather small Schur functors may appear. Table
8 lists the ranks of all Schur functors of degree $\leq 8$ with $3$ or $4$
parts. For most Specht modules of $\mathfrak{S}_{10}$ this table is already
sufficient for uniquely determining the multiplicity of every Schur functor
with $3$ or $4$ parts.

Schur functor | Ranks in genus $3,4$ | Schur functor | Ranks in genus $3,4$
---|---|---|---
$(2,1^{2})$ | $[3,15]$ | $(3,2^{2})$ | $[3,36]$
$(3,1^{2})$ | $[6,36]$ | $(3,2,1^{2})$ | $[0,20]$
$(2^{2},1)$ | $[3,20]$ | $(2^{3},1)$ | $[0,4]$
$(2,1^{3})$ | $[0,4]$ | $(6,1^{2})$ | $[21,189]$
$(4,1^{2})$ | $[10,70]$ | $(5,2,1)$ | $[24,256]$
$(3,2,1)$ | $[8,64]$ | $(5,1^{3})$ | $[0,35]$
$(3,1^{3})$ | $[0,10]$ | $(4,3,1)$ | $[15,175]$
$(2^{3})$ | $[1,10]$ | $(4,2^{2})$ | $[6,84]$
$(2^{2},1^{2})$ | $[0,6]$ | $(4,2,1^{2})$ | $[0,45]$
$(5,1^{2})$ | $[15,120]$ | $(3^{2},2)$ | $[3,45]$
$(4,2,1)$ | $[15,140]$ | $(3^{2},1^{2})$ | $[0,20]$
$(4,1^{3})$ | $[0,20]$ | $(3,2^{2},1)$ | $[0,15]$
$(3^{2},1)$ | $[6,60]$ | $(2^{4})$ | $[0,1]$
Table 8: Ranks of Schur functors $\mathfrak{S}_{10}$-irreducible | Missing Schur functors
---|---
$(8,2)$ | $\mathbb{S}_{(3,1^{2})}+\mathbb{S}_{(2,1^{4})}$
$(6,3,1)$ | $2\mathbb{S}_{(2,1^{2})}+\mathbb{S}_{(2,1^{3})}$
$(4^{2},2)$ | $0$
$(3,2,1^{5})$ | $0$
Table 9: Remaining corrections to the lower bounds

However, this is not always the case, e.g. for partition $(8,2)$ in our
example. Let us explain why the remaining combination of Schurs occuring in
our examples are as presented in Table 9. To gain more insight one can take
the following steps:

  * •

As before, with $\mathbb{S}_{(2,1)}$ having the smallest ratio between its
first and third traces, now it is $\mathbb{S}_{(2,1^{2})}$ that has the
smallest ratio between its ranks in genus $3$ and $4$ (at least among the
Schurs that are small enough to contribute nontrivially). Thus
$\mathbb{S}_{(2,1^{2})}$ must appear with multiplicity high enough to make the
second column $\leq 6$ times the first. E.g. it appears at least once for
partition $(6,3,1)$ leaving unknown ranks $[3,19]$, at which point it is clear
that it must appear with multiplicity $2$ to account for the rank in genus
$3$. The remaining ranks $[0,4]$ do not uniquely determine the multiplicities
of Schur functors with $4$ parts.

  * •

For partition $(8,2)$ ranks alone are not enough to uniquely determine the
multiplicity of $\mathbb{S}_{(3,1^{2})}$. However, it is feasible to compute
the trace of the matrix $diag(2,1,1)$, and this additional information is
sufficient.

  * •

The second term of partition $(8,2)$, the functor $\mathbb{S}_{(2,1^{4})}$,
does not even show up until one considers wedges of $\geq 5$ circles. However,
since the representation $\chi_{(8,2)}$ has relatively small dimension, it is
feasible to compute the nonequivariant rank of the isotypic component
$H^{9}_{c}(F(X,10))_{(8,2)}$ in genus $5$ using the $2$-term complex. This is
sufficient for determining the $\mathbb{S}_{(3,1^{2})}$ term from the previous
point, and further detects that there exists some $\mathbb{S}_{\lambda}$ for
$\lambda$ with $5$ parts (though it can not decide which).

With these additional considerations we have completely accounted for all
Schur functors with $\leq 3$ parts and reduced the ambiguity regarding
functors with $4$ parts to a small number of cases of very small dimension:
deciding between $\mathbb{S}_{(2,1^{3})},\mathbb{S}_{(2^{3},1)}$ and
$\mathbb{S}_{(2^{4})}$.

Looking further to $\mathbb{S}_{\lambda}$ for $\lambda$ with $\geq 5$ parts we
reach the computational limits of the 2-term complex, where it is no longer
feasible to compute further ranks or traces. Instead, observe that any yet
undetected Schur functor must have relatively large polynomial degree. More
precisely, any Schur functor with $\geq 5$ parts is either an exterior power
or it has total polynomial degree $\geq 6$, as are the degrees of
$\mathbb{S}_{(2^{3},1)}$ and $\mathbb{S}_{(2^{4})}$.

### Step 4: CE-complex in high polynomial degree

Fortunately, at high polynomial degree a new tool becomes available: high
polynomial degrees appear as $\Phi^{p-1}[10,10-p]$ for $p$ small, so the
Chevalley-Eilenberg complex more readily gives access to the nonequivariant
multiplicity of individual Schur functors. Although this complex forgets the
data as to which $\mathfrak{S}_{n}$-isotypical component they belong, and
further simplification is needed.

This couples with the crucial observation that by plugging wedges of even
spheres into the CE-complex, the partitions of $\mathbb{S}_{\lambda}$ are the
conjugate of the ones appearing for wedges of circles. For example the functor
$\mathbb{S}_{{(2,1^{6})}^{\ast}}$, which would have been extremely difficult
to access on a wedge of $7$ circles, contributes $\mathbb{S}_{(7,1)}$ on a
wedge of only two even spheres! And its high polynomial degree makes it even
easier to access.

With this, the last step in our calculation is to compute the nonequivariant
multiplicity of Schur functors of polynomial degree $\geq 6$ and compare them
with the total multiplicity of those functors in our lower bound. If they are
found to agree, the bound is sharp and all Schur functors have been counted.
Fortunately, this always turned out to be the case

To make the last computations feasible we use one more trick: inputting wedges
of spheres of different dimensions. Let us explain one such computation in
detail, and for the other ones we will only specify what the analogous
computation produces.

### Step 4’: Mixing spheres of different dimensions

Say we want to compute the nonequivariant multiplicity of the Schur functor
$\mathbb{S}_{{(2,1^{4})}^{\ast}}$, which on even spheres gives
$\mathbb{S}_{(5,1)}(\tilde{H}^{*}(X))$ so two spheres suffice. Consider the
space $X=S^{2}\bigvee S^{4}$.

The CE-complex for this $X$ admits a multigrading where the multidegree
$(a,b)$ consists of tensors

$(H^{2}(S^{2})^{\otimes a}\otimes H^{4}(S^{4})^{\otimes
b})^{\oplus\binom{a+b}{a}}\leq\tilde{H}^{*}(X)^{\otimes(a+b)}.$

This multigrading is preserved by the CE-differential, so one may decompose
the complex and consider one multidegree at a time. The $(a,b)$-multigraded
subcomplex is substantially smaller than most $\mathbb{S}_{\lambda}$-isotypic
components, and furthermore it is spanned by a basis of pure tensors. These
facts make calculations feasible, and we are able to determine the
$(a,b)$-multigraded part of the CE-homology.

At the same time, $\operatorname{map}({X},{X})$ acts on these subcomplexes,
giving them the structure of polynomial
$\operatorname{GL}(1)\times\operatorname{GL}(1)$-representations, with the
$(a,b)$-multigraded component coinciding with the $Sym^{a}(H^{2}(X))\otimes
Sym^{b}(H^{4}(X))$-isotypic component of this action. Knowing the multiplicity
of each of these representations in homology, one is able to deduce the
multiplicity of every $\mathbb{S}_{\lambda}$ for $\lambda$ with $2$ parts: one
needs only to count how often every $Sym^{a}\otimes Sym^{b}$ occurs in each
$\mathbb{S}_{\lambda}$.

Let us consider the $(5,1)$-multigraded component. One can compute121212In
general the multiplicity with which $Sym^{a_{1}}\otimes\ldots\otimes
Sym^{a_{r}}$ appears in $\mathbb{S}_{\lambda}$ is equal to the number of
semistandard Young tableax of shape $\lambda$ filled with $a_{i}$ many $i$’s.
that $Sym^{5}\otimes Sym^{1}$ only occurs once in $\mathbb{S}_{(5,1)}$ and
once in $\mathbb{S}_{(6)}$. It follows that its total multiplicity in CE-
homology is $\Phi^{3}[10,(5,1)]\oplus\Phi^{3}[10,(6)]$. Given that we already
know the multiplicities of all symmetric powers, this determines the rank of
$\Phi^{3}[10,(5,1)]$.

One then compares this rank with the lower bound, and find that they agree. It
follows that the bound accounts for the entire multiplicity of
$\mathbb{S}_{(2,1^{4})}$ on wedges of circles, which is thus known
equivariantly. Note further that approaching this calculation with the
multigraded components has the added benefit that it is sensitive to multiple
Schur functors at a time: if the lower bounds of all of them sum to the
calculated multiplicity of the multigraded component, then all their lower
bounds in fact exhaust the whole multiplicity space.

Other cases of this approach needed to completely determine
$H^{9}_{c}(F(X,10))$ are:

  * •

$X=S^{2}\bigvee S^{4}$, multidegree $(4,3)$: computing rank of
$\Phi^{2}[10,(4,3)]+\Phi^{2}[10,(5,2)]+\Phi^{2}[10,(6,1)]+\Phi^{2}[10,(7)]$.

  * •

$X=S^{1}\bigvee S^{2}$, multidegree $(2,5)$: computing rank of
$\Phi^{2}[10,(5,1^{2})]+\Phi^{2}[10,(6,1)]$.

  * •

$X=S^{2}\bigvee S^{4}$, multidegree $(4,4)$: computing rank of
$\Phi^{1}[10,(4,4)]+\Phi^{1}[10,(5,3)]+\Phi^{1}[10,(6,2)]+\Phi^{1}[10,(7,1)]+\Phi^{1}[10,(8)]$.

  * •

$X=S^{2}\bigvee S^{4}\bigvee S^{6}$, multidegree $(5,2,1)$: computing rank of
$\Phi^{1}[10,(5,2,1)]+\Phi^{1}[10,(5,3)]+\Phi^{1}[10,(6,1,1)]+2\cdot\Phi^{1}[10,(6,2)]+2\cdot\Phi^{1}[10,(7,1)]+\Phi^{1}[10,(8)]$.

  * •

$X=S^{1}\bigvee S^{2}$, multidegree $(3,5)$: computing rank of
$\Phi^{1}[10,(5,1^{3})]+\Phi^{1}[10,(6,1^{2})]$.

Note that in some of these cases we mix even and odd dimensional spheres. The
partition $(5,1^{3})$ has $4$ parts, and its transpose has $5$ parts. So it
would take $\geq 4$ even spheres to access its multiplicity. But mixing even
and odd spheres makes $\mathbb{S}_{(5,1^{3})}$ contribute nontrivially already
with only two spheres.

In all these cases the rank obtained from the CE-homology calculation agrees
with the rank obtained from Table 10. Therefore all Schur functors
$\mathbb{S}_{\lambda}$ for which $\Phi[10,{\lambda}^{\ast}]$ appeared in one
of the cases above cannot occur with higher multiplicity in any
$\mathfrak{S}_{10}$-isotypical component.

Careful bookkeeping shows that at this point every multiplicity of every Schur
functor has been uniquely determined, therefore Table 10 contains the full
decomposition of every isotypical component.

$\mathfrak{S}_{10}$-irrep |  $\operatorname{GL}$-equivariant multiplicities in $\operatorname{gr}H^{9}_{c}(F(X,10))$
---|---
$(10)$ |  $\mathbb{S}_{(4,1)}+\mathbb{S}_{(3,1^{3})}+\mathbb{S}_{(2,1^{5})}+\mathbb{S}_{(1^{8})}$
$(9,1)$ |  $\mathbb{S}_{(5)}$
$(8,2)$ |  $\mathbb{S}_{(4)}+\mathbb{S}_{(3,1^{2})}+\mathbb{S}_{(3)}+\mathbb{S}_{(2,1^{4})}+\mathbb{S}_{(2,1)}+\mathbb{S}_{(2)}+\mathbb{S}_{(1^{7})}+\mathbb{S}_{(1^{4})}+\mathbb{S}_{(1^{3})}$
$(8,1^{2})$ |  $\mathbb{S}_{(4)}+\mathbb{S}_{(3,1)}+\mathbb{S}_{(3)}+\mathbb{S}_{(2,1^{3})}+\mathbb{S}_{(2,1^{2})}+\mathbb{S}_{(2)}+\mathbb{S}_{(1^{6})}+\mathbb{S}_{(1^{5})}+\mathbb{S}_{(1^{2})}+\mathbb{S}_{(1)}$
$(7,3)$ |  $\mathbb{S}_{(5)}+4\mathbb{S}_{(3)}+\mathbb{S}_{(2,1^{2})}+2\mathbb{S}_{(2,1)}+2\mathbb{S}_{(2)}+\mathbb{S}_{(1^{5})}+\mathbb{S}_{(1^{4})}+\mathbb{S}_{(1^{3})}+2\mathbb{S}_{(1^{2})}+\mathbb{S}_{(1)}$
$(7,2,1)$ |  $\mathbb{S}_{(5)}+3\mathbb{S}_{(4)}+\mathbb{S}_{(3,1)}+5\mathbb{S}_{(3)}+\mathbb{S}_{(2,1^{3})}+\mathbb{S}_{(2,1^{2})}+4\mathbb{S}_{(2,1)}+5\mathbb{S}_{(2)}+\mathbb{S}_{(1^{6})}+\mathbb{S}_{(1^{5})}+2\mathbb{S}_{(1^{4})}+3\mathbb{S}_{(1^{3})}+3\mathbb{S}_{(1^{2})}+2\mathbb{S}_{(1)}$
$(7,1^{3})$ |  $\mathbb{S}_{(6)}+2\mathbb{S}_{(4)}+2\mathbb{S}_{(3)}+\mathbb{S}_{(2,1^{2})}+\mathbb{S}_{(2,1)}+3\mathbb{S}_{(2)}+\mathbb{S}_{(1^{5})}+\mathbb{S}_{(1^{4})}+\mathbb{S}_{(1^{3})}+2\mathbb{S}_{(1^{2})}$
$(6,4)$ |  $2\mathbb{S}_{(4)}+2\mathbb{S}_{(3)}+2\mathbb{S}_{(2,1)}+4\mathbb{S}_{(2)}+\mathbb{S}_{(1^{4})}+3\mathbb{S}_{(1^{3})}+2\mathbb{S}_{(1^{2})}$
$(6,3,1)$ |  $\mathbb{S}_{(5)}+6\mathbb{S}_{(4)}+2\mathbb{S}_{(3,1)}+11\mathbb{S}_{(3)}+\mathbb{S}_{(2,1^{3})}+2\mathbb{S}_{(2,1^{2})}+7\mathbb{S}_{(2,1)}+10\mathbb{S}_{(2)}+\mathbb{S}_{(1^{6})}+2\mathbb{S}_{(1^{5})}+3\mathbb{S}_{(1^{4})}+5\mathbb{S}_{(1^{3})}+7\mathbb{S}_{(1^{2})}+4\mathbb{S}_{(1)}$
$(6,2^{2})$ |  $\mathbb{S}_{(5)}+4\mathbb{S}_{(4)}+7\mathbb{S}_{(3)}+6\mathbb{S}_{(2,1)}+8\mathbb{S}_{(2)}+3\mathbb{S}_{(1^{4})}+6\mathbb{S}_{(1^{3})}+4\mathbb{S}_{(1^{2})}+\mathbb{S}_{(1)}$
$(6,2,1^{2})$ |  $2\mathbb{S}_{(5)}+6\mathbb{S}_{(4)}+\mathbb{S}_{(3,1)}+13\mathbb{S}_{(3)}+2\mathbb{S}_{(2,1^{2})}+8\mathbb{S}_{(2,1)}+10\mathbb{S}_{(2)}+2\mathbb{S}_{(1^{5})}+3\mathbb{S}_{(1^{4})}+5\mathbb{S}_{(1^{3})}+7\mathbb{S}_{(1^{2})}+5\mathbb{S}_{(1)}$
$(6,1^{4})$ |  $\mathbb{S}_{(5)}+2\mathbb{S}_{(4)}+4\mathbb{S}_{(3)}+3\mathbb{S}_{(2,1)}+4\mathbb{S}_{(2)}+2\mathbb{S}_{(1^{4})}+3\mathbb{S}_{(1^{3})}+2\mathbb{S}_{(1^{2})}+\mathbb{S}_{(1)}$
$(5^{2})$ |  $\mathbb{S}_{(5)}+3\mathbb{S}_{(3)}+3\mathbb{S}_{(2,1)}+\mathbb{S}_{(1^{4})}+\mathbb{S}_{(1^{3})}+2\mathbb{S}_{(1)}$
$(5,4,1)$ |  $\mathbb{S}_{(5)}+5\mathbb{S}_{(4)}+\mathbb{S}_{(3,1)}+10\mathbb{S}_{(3)}+\mathbb{S}_{(2,1^{2})}+7\mathbb{S}_{(2,1)}+9\mathbb{S}_{(2)}+\mathbb{S}_{(1^{5})}+3\mathbb{S}_{(1^{4})}+5\mathbb{S}_{(1^{3})}+6\mathbb{S}_{(1^{2})}+3\mathbb{S}_{(1)}$
$(5,3,2)$ |  $2\mathbb{S}_{(5)}+7\mathbb{S}_{(4)}+\mathbb{S}_{(3,1)}+18\mathbb{S}_{(3)}+\mathbb{S}_{(2,1^{2})}+12\mathbb{S}_{(2,1)}+13\mathbb{S}_{(2)}+\mathbb{S}_{(1^{5})}+4\mathbb{S}_{(1^{4})}+7\mathbb{S}_{(1^{3})}+9\mathbb{S}_{(1^{2})}+6\mathbb{S}_{(1)}$
$(5,3,1^{2})$ |  $\mathbb{S}_{(6)}+\mathbb{S}_{(5)}+13\mathbb{S}_{(4)}+2\mathbb{S}_{(3,1)}+16\mathbb{S}_{(3)}+\mathbb{S}_{(2,1^{2})}+10\mathbb{S}_{(2,1)}+20\mathbb{S}_{(2)}+\mathbb{S}_{(1^{5})}+4\mathbb{S}_{(1^{4})}+10\mathbb{S}_{(1^{3})}+12\mathbb{S}_{(1^{2})}+5\mathbb{S}_{(1)}$
$(5,2^{2},1)$ |  $4\mathbb{S}_{(5)}+8\mathbb{S}_{(4)}+\mathbb{S}_{(3,1)}+19\mathbb{S}_{(3)}+\mathbb{S}_{(2,1^{2})}+11\mathbb{S}_{(2,1)}+16\mathbb{S}_{(2)}+\mathbb{S}_{(1^{5})}+3\mathbb{S}_{(1^{4})}+8\mathbb{S}_{(1^{3})}+11\mathbb{S}_{(1^{2})}+6\mathbb{S}_{(1)}$
$(5,2,1^{3})$ |  $\mathbb{S}_{(6)}+3\mathbb{S}_{(5)}+8\mathbb{S}_{(4)}+14\mathbb{S}_{(3)}+9\mathbb{S}_{(2,1)}+13\mathbb{S}_{(2)}+3\mathbb{S}_{(1^{4})}+6\mathbb{S}_{(1^{3})}+8\mathbb{S}_{(1^{2})}+5\mathbb{S}_{(1)}$
$(5,1^{5})$ |  $\mathbb{S}_{(7)}+2\mathbb{S}_{(5)}+\mathbb{S}_{(4)}+5\mathbb{S}_{(3)}+2\mathbb{S}_{(2,1)}+3\mathbb{S}_{(2)}+\mathbb{S}_{(1^{3})}+2\mathbb{S}_{(1^{2})}+2\mathbb{S}_{(1)}$
$(4^{2},2)$ |  $\mathbb{S}_{(5)}+6\mathbb{S}_{(4)}+\mathbb{S}_{(3,1)}+6\mathbb{S}_{(3)}+\mathbb{S}_{(2^{2})}+4\mathbb{S}_{(2,1)}+10\mathbb{S}_{(2)}+2\mathbb{S}_{(1^{4})}+6\mathbb{S}_{(1^{3})}+5\mathbb{S}_{(1^{2})}+\mathbb{S}_{(1)}$
$(4^{2},1^{2})$ |  $3\mathbb{S}_{(5)}+3\mathbb{S}_{(4)}+14\mathbb{S}_{(3)}+\mathbb{S}_{(2,1^{2})}+8\mathbb{S}_{(2,1)}+7\mathbb{S}_{(2)}+\mathbb{S}_{(1^{5})}+2\mathbb{S}_{(1^{4})}+3\mathbb{S}_{(1^{3})}+6\mathbb{S}_{(1^{2})}+5\mathbb{S}_{(1)}$
$(4,3^{2})$ |  $\mathbb{S}_{(5)}+4\mathbb{S}_{(4)}+\mathbb{S}_{(3,1)}+8\mathbb{S}_{(3)}+\mathbb{S}_{(2,1^{2})}+4\mathbb{S}_{(2,1)}+6\mathbb{S}_{(2)}+\mathbb{S}_{(1^{5})}+\mathbb{S}_{(1^{4})}+2\mathbb{S}_{(1^{3})}+5\mathbb{S}_{(1^{2})}+3\mathbb{S}_{(1)}$
$(4,3,2,1)$ |  $4\mathbb{S}_{(5)}+15\mathbb{S}_{(4)}+2\mathbb{S}_{(3,1)}+25\mathbb{S}_{(3)}+16\mathbb{S}_{(2,1)}+23\mathbb{S}_{(2)}+4\mathbb{S}_{(1^{4})}+11\mathbb{S}_{(1^{3})}+14\mathbb{S}_{(1^{2})}+9\mathbb{S}_{(1)}$
$(4,3,1^{3})$ |  $\mathbb{S}_{(6)}+4\mathbb{S}_{(5)}+9\mathbb{S}_{(4)}+17\mathbb{S}_{(3)}+9\mathbb{S}_{(2,1)}+16\mathbb{S}_{(2)}+2\mathbb{S}_{(1^{4})}+7\mathbb{S}_{(1^{3})}+10\mathbb{S}_{(1^{2})}+5\mathbb{S}_{(1)}$
$(4,2^{3})$ |  $\mathbb{S}_{(5)}+7\mathbb{S}_{(4)}+\mathbb{S}_{(3,1)}+8\mathbb{S}_{(3)}+4\mathbb{S}_{(2,1)}+11\mathbb{S}_{(2)}+\mathbb{S}_{(1^{4})}+5\mathbb{S}_{(1^{3})}+6\mathbb{S}_{(1^{2})}+\mathbb{S}_{(1)}$
$(4,2^{2},1^{2})$ |  $\mathbb{S}_{(6)}+4\mathbb{S}_{(5)}+11\mathbb{S}_{(4)}+\mathbb{S}_{(3,1)}+19\mathbb{S}_{(3)}+9\mathbb{S}_{(2,1)}+16\mathbb{S}_{(2)}+\mathbb{S}_{(1^{4})}+5\mathbb{S}_{(1^{3})}+11\mathbb{S}_{(1^{2})}+8\mathbb{S}_{(1)}$
$(4,2,1^{4})$ |  $2\mathbb{S}_{(6)}+3\mathbb{S}_{(5)}+7\mathbb{S}_{(4)}+8\mathbb{S}_{(3)}+3\mathbb{S}_{(2,1)}+11\mathbb{S}_{(2)}+4\mathbb{S}_{(1^{3})}+6\mathbb{S}_{(1^{2})}+3\mathbb{S}_{(1)}$
$(4,1^{6})$ |  $\mathbb{S}_{(6)}+\mathbb{S}_{(5)}+\mathbb{S}_{(4)}+2\mathbb{S}_{(3)}+2\mathbb{S}_{(2)}+2\mathbb{S}_{(1^{2})}+\mathbb{S}_{(1)}$
$(3^{3},1)$ |  $\mathbb{S}_{(6)}+5\mathbb{S}_{(4)}+5\mathbb{S}_{(3)}+2\mathbb{S}_{(2,1)}+8\mathbb{S}_{(2)}+3\mathbb{S}_{(1^{3})}+5\mathbb{S}_{(1^{2})}+\mathbb{S}_{(1)}$
$(3^{2},2^{2})$ |  $3\mathbb{S}_{(5)}+3\mathbb{S}_{(4)}+12\mathbb{S}_{(3)}+7\mathbb{S}_{(2,1)}+5\mathbb{S}_{(2)}+\mathbb{S}_{(1^{4})}+2\mathbb{S}_{(1^{3})}+4\mathbb{S}_{(1^{2})}+5\mathbb{S}_{(1)}$
$(3^{2},2,1^{2})$ |  $\mathbb{S}_{(6)}+3\mathbb{S}_{(5)}+9\mathbb{S}_{(4)}+\mathbb{S}_{(3,1)}+13\mathbb{S}_{(3)}+6\mathbb{S}_{(2,1)}+14\mathbb{S}_{(2)}+\mathbb{S}_{(1^{4})}+5\mathbb{S}_{(1^{3})}+8\mathbb{S}_{(1^{2})}+4\mathbb{S}_{(1)}$
$(3^{2},1^{4})$ |  $\mathbb{S}_{(7)}+4\mathbb{S}_{(5)}+2\mathbb{S}_{(4)}+9\mathbb{S}_{(3)}+4\mathbb{S}_{(2,1)}+4\mathbb{S}_{(2)}+\mathbb{S}_{(1^{3})}+3\mathbb{S}_{(1^{2})}+4\mathbb{S}_{(1)}$
$(3,2^{3},1)$ |  $3\mathbb{S}_{(5)}+6\mathbb{S}_{(4)}+\mathbb{S}_{(3,1)}+8\mathbb{S}_{(3)}+3\mathbb{S}_{(2,1)}+8\mathbb{S}_{(2)}+2\mathbb{S}_{(1^{3})}+5\mathbb{S}_{(1^{2})}+3\mathbb{S}_{(1)}$
$(3,2^{2},1^{3})$ |  $2\mathbb{S}_{(6)}+2\mathbb{S}_{(5)}+7\mathbb{S}_{(4)}+8\mathbb{S}_{(3)}+3\mathbb{S}_{(2,1)}+9\mathbb{S}_{(2)}+2\mathbb{S}_{(1^{3})}+5\mathbb{S}_{(1^{2})}+3\mathbb{S}_{(1)}$
$(3,2,1^{5})$ |  $\mathbb{S}_{(7)}+\mathbb{S}_{(6)}+2\mathbb{S}_{(5)}+3\mathbb{S}_{(4)}+3\mathbb{S}_{(3)}+4\mathbb{S}_{(2)}+2\mathbb{S}_{(1^{2})}+2\mathbb{S}_{(1)}$
$(3,1^{7})$ |  $\mathbb{S}_{(8)}+\mathbb{S}_{(6)}+\mathbb{S}_{(4)}+\mathbb{S}_{(2)}$
$(2^{5})$ |  $2\mathbb{S}_{(4)}+2\mathbb{S}_{(2)}+\mathbb{S}_{(1^{3})}$
$(2^{4},1^{2})$ |  $2\mathbb{S}_{(5)}+\mathbb{S}_{(4)}+4\mathbb{S}_{(3)}+2\mathbb{S}_{(2,1)}+\mathbb{S}_{(2)}+\mathbb{S}_{(1^{2})}+2\mathbb{S}_{(1)}$
$(2^{3},1^{4})$ |  $\mathbb{S}_{(6)}+\mathbb{S}_{(5)}+\mathbb{S}_{(4)}+2\mathbb{S}_{(3)}+2\mathbb{S}_{(2)}+\mathbb{S}_{(1^{2})}$
$(2^{2},1^{6})$ |  $\mathbb{S}_{(7)}+\mathbb{S}_{(5)}+\mathbb{S}_{(3)}+\mathbb{S}_{(1)}$
$(2,1^{8})$ |  0
$(1^{10})$ |  $\mathbb{S}_{(9)}$
Table 10: Equivariant multiplicity of each isotypic component in
$\operatorname{gr}H_{c}^{9}(F(X,10))$

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# Kings in Multipartite Hypertournaments

Jiangdong Ai Department of Computer Science. Royal Holloway University of
London<EMAIL_ADDRESS>Stefanie Gerke Department of
Mathematics. Royal Holloway University of London<EMAIL_ADDRESS>Gregory Gutin Department of Computer Science. Royal Holloway University of
London<EMAIL_ADDRESS>
###### Abstract

In his paper “Kings in Bipartite Hypertournaments” (Graphs $\&$ Combinatorics
35, 2019), Petrovic stated two conjectures on 4-kings in multipartite
hypertournaments. We prove one of these conjectures and give counterexamples
for the other.

##  1 Introduction

Given two integers $n$ and $k$, $n\geq k>1$, a $k$-hypertournament $T$ on $n$
vertices is a pair $(V,A)$, where $V$ is a set of vertices, $|V|=n$ and $A$ is
a set of $k$-tuples of vertices, called arcs, so that for any $k$-subset $S$
of $V$, $A$ contains exactly one of the $k!$ tuples whose entries belong to
$S$. For an arc $x_{1}x_{2}\ldots x_{k}$, we say that $x_{i}$ precedes $x_{j}$
if $i<j.$ A 2-hypertournament is merely an (ordinary) tournament.
Hypertournaments have been studied in a large number of papers, see e.g. [1,
2, 3, 4, 5, 8, 9, 11, 12].

Recently, Petrovic [10] introduced multipartite hypertournaments in a similar
way. Let $n$ and $k$ be integers such that $n>k\geq 2.$ Let $V$ be a set of
$n$ vertices and $V=V_{1}\uplus V_{2}\uplus\dots\uplus V_{p}$ be a partition
of $V$ into $p\geq 2$ non-empty subsets. A $p$-partite $k$-hypertournament
(or, multipartite hypertournament) $H$ can be obtained from a
$k$-hypertournament $T$ on vertex set $V$ by deleting all arcs
$x_{1}x_{2}\ldots x_{k}$ such that $\\{x_{1},x_{2},\ldots,x_{k}\\}\subseteq
V_{i}$ for some $i\in[p].$ We call $V_{i}$’s partite sets of $H.$ The set of
arcs of $H=(V,A)$ will be denoted by $A(H),$ i.e., $A(H)=A.$ A $p$-partite
2-hypertournament is a $p$-partite tournament.

For $u\in V_{i},w\in V_{j}$ with $i\neq j,$ $A_{H}(u,w)$ is the set of arcs of
$H$ which contain $u$ and $w$ and where $u$ precedes $w.$ We will write $xey$
if $e\in A_{H}(x,y).$ We let $A_{H}(x,y)=\emptyset$ if either $x$ and $y$
belong to the same partite set of $H.$ A path in $H$ is an alternating
sequence $P=x_{1}a_{1}x_{2}a_{2}\dots x_{q-1}a_{q-1}x_{q}$ of distinct
vertices $x_{i}$ and distinct arcs $a_{j}$ such that $x_{j}a_{j}x_{j+1}$ for
every $j\in[q-1].$ We will call $P$ an $(x_{1},x_{q})$-path of length $q-1.$

Let $q\geq 1$ be a natural number. A vertex $x$ of $H$ is a $q$-king if for
every $y\in V$, $H$ has an $(x,y)$-path of length at most $q.$ Generalizing a
well-known theorem of Landau that every tournament has a 2-king (see e.g.
[6]), Brcanov et al. [4] showed that every hypertournament has a 2-king. A
vertex $v$ of $H$ is a transmitter if for every vertex $u$ from a different
partite set than $v,$ $A_{H}(u,v)=\emptyset.$

Note that for every $u\in V_{i},w\in V_{j}$ $(i\neq j)$, we have
$|A_{H}(u,w)|+|A_{H}(w,u)|=\binom{n-2}{k-2}.$ A majority multipartite
tournament $M(H)$ of $H$ has the same partite sets as $H$ and for every $u\in
V_{i}$ and $w\in V_{j}$ with $i\neq j$, $uw\in M(H)$ if
$|A_{H}(u,w)|>\frac{1}{2}\binom{n-2}{k-2}.$ If
$|A_{H}(u,w)|=\frac{1}{2}\binom{n-2}{k-2}$ then we can choose either $uw$ or
$wu$ for $M(H).$

For a graph $G=(V,E)$ and $U\subseteq V$, let $N_{G}(U)=\\{v\in V\setminus U:\
uv\in E,u\in U\\}.$

Gutin [7] and independently Petrovic and Thomassen [11] proved the following:

######  Theorem 1.

[7, 11] Every multipartite tournament with at most one transmitter contains a
4-king.

Petrovic [10] proved that the same result holds for bipartite
$k$-hypertournaments:

######  Theorem 2.

[10] Every bipartite $k$-hypertournaments ($k\geq 2$) with at most one
transmitter contains a 4-king.

In the same paper he conjectured the following:

######  Conjecture 3.

[10] Every multipartite $k$-hypertournament ($k\geq 2$) with at most one
transmitter contains a 4-king.

In this short paper, we will solve this conjecture in the affirmative.

The next conjecture of Petrovic [10] is motivated by the fact that Petrovic
and Thomassen [11] proved that the assertion of the conjecture holds for
bipartite tournaments.

######  Theorem 4.

[11] Every bipartite tournament $B$ without transmitters has at least two
4-kings in each partite set of $B.$

######  Conjecture 5.

[10] Every bipartite $k$-hypertournament $B$ ($k\geq 2$) without transmitters
has at least two 4-kings in each partite set of $B.$

In this paper, we will first show a couterexample to Conjecture 5 and then
exhibit a wide family of bipartite hypertournaments for which the conclusion
of the conjecture holds.

The paper is organized as follows. In the next section, we prove a lemma
(Lemma 7) which we call the Majority Lemma, and which is used to show the
positive above-mentioned results. In Section 3, we provide the counterexample
and positive results. The terminology not introduced in this paper can be
found in [6].

##  2 The Majority Lemma

The Majority Lemma, Lemma 7, is the main technical result of this paper. To
prove Lemma 7, we will use the following simple lemma.

######  Lemma 6.

Let $G$ be a bipartite graph with partite sets $U$ and $W$ and let every
vertex in $U$ have degree at least $p\geq 1$ and every vertex in $W$ have
degree at most $p$, except for one vertex which has degree at most $2p-1.$
Then $G$ has a matching saturating $U.$

###### Proof.

By Hall’s theorem, if for every $S\subseteq U$, $|S|\leq|N_{G}(S)|$ then $G$
has a matching saturating $U.$ Suppose that there is a subset $S$ of $U$ such
that $|S|\geq|N_{G}(S)|+1.$ Let $e$ be the number of edges in the subgraph of
$G$ induced by $S\cup N_{G}(S)$ and observe that

$p|S|\leq e\leq(|N(S)|-1)p+(2p-1)\leq(|S|-2)p+(2p-1)=|S|p-1,$

a contradiction. ∎

Proposition 14 proved in the next section shows that Lemma 7 cannot be
extended to $n=4$ and $p=2.$

######  Lemma 7.

Let $H$ be a $p$-partite $k$-hypertournament with $p\geq 2.$ Let $n\geq 5$ and
$n>k\geq 3.$ If a majority $p$-partite tournament $M(H)$ has an $(x,y)$-path
$P$ of length at most 4, then $H$ has such a path of length at most 4.

###### Proof.

It suffices to prove this lemma for the case when $P$ is of length 4 as the
other cases are simpler and similar. Thus, assume that
$P=x_{1}x_{2}x_{3}x_{4}x_{5}.$ By definition of a path, for every $i\in[4],$
$x_{i}$ and $x_{i+1}$ belong to different partite sets of $H.$ Now consider
the following cases covering all possibilities.

Case 1: $n\geq 9$ and $3\leq k<n$ or $n\geq 7$ and $4\leq k<n-1$. Observe that
if for every $i\in\\{1,2,3,4\\}$,

$|A_{H}(x_{i},x_{i+1})|>3$ (1)

then we can choose distinct arcs $a_{i}\in A_{H}(x_{i},x_{i+1})$ such that
$x_{1}a_{1}x_{2}a_{2}x_{3}a_{3}x_{4}a_{4}x_{5}$ is the required path in $H$.
In particular, inequalities (1) will hold if $\frac{1}{2}\binom{n-2}{k-2}>3.$

If $n\geq 9$ and $3\leq k<n,$ we have

$\frac{1}{2}\binom{n-2}{k-2}\geq\frac{n-2}{2}>3$

and hence inequalities (1) hold. If $n\geq 7$ and $4\leq k<n-1,$ we have

$\frac{1}{2}\binom{n-2}{k-2}\geq\frac{(n-2)(n-3)}{4}>3.$

Case 2: $k=3$ and $5\leq n\leq 8$. Then

$|A_{H}(x_{i},x_{i+1})|\geq\frac{1}{2}\binom{n-2}{k-2}\geq\frac{1}{2}\binom{3}{1}=\frac{3}{2}$
(2)

for $i=1,2,3,4.$ Consider a bipartite graph $G$ with partite sets
$Z=\\{z_{1},z_{2},z_{3},z_{4}\\}$ and $A(H).$ We have an edge $z_{i}a_{j}$ if
$a_{j}\in A_{H}(x_{i},x_{i+1}).$ By (2), each vertex in $Z$ has degree at
least two. Since $k=3$, vertices $z_{i}$ and $z_{j}$ in $G$ have no common
neighbor unless $|i-j|=1.$ Thus, every vertex of $G$ in $A(H)$ has degree at
most 2. Thus, by Lemma 6, $G$ has a matching saturating $Z$. In other words,
there are distinct $a_{1},a_{2},a_{3},a_{4}\in A(H)$ such that
$x_{1}a_{1}x_{2}a_{2}x_{3}a_{3}x_{4}a_{4}x_{5}$ is a path in $H.$

Case 3: $k=4$ and $5\leq n\leq 6$. Consider the bipartite graph $G$
constructed as in the previous case. Using the computations analogous to those
in (2), we see that the minimum degree of a vertex in $Z$ is at least $3$ when
$n=6$ and at least $2$ when $n=5.$ Since $k=4$, there is no common neighbor of
all vertices in $Z.$ Thus, every vertex of $G$ in $A(H)$ has degree at most
$3$. Now consider two subcases.

Subcase 1: $n=6$. Since every vertex of $G$ in $A(H)$ has degree at most $3$
and every vertex of $G$ in $Z$ has degree at least $3$, by Lemma 6, $G$ has a
matching saturating $Z$ and we are done as in Case 2.

Subcase 2: $n=5$. Recall that the minimum degree of a vertex in $Z$ is at
least $2$. Suppose that there are two vertices of $G$ in $A(H)$ of degree $3$.
This means that

$N_{G}(z_{i})\cap N_{G}(z_{i+1})\cap N_{G}(z_{i+2})\neq\emptyset$ (3)

for $i=1$ or $2$. Indeed, since $k=4$, $N_{G}(z_{1})\cap N_{G}(z_{j})\cap
N_{G}(z_{4})=\emptyset$ when either $j=2$ or $3$. Without loss of generality,
we assume that (3) holds when $i=1$ and let $e_{1}\in N_{G}(z_{1})\cap
N_{G}(z_{2})\cap N_{G}(z_{3}).$ Thus, $e_{1}=x_{1}x_{2}x_{3}x_{4}.$

If $x_{1}$ and $x_{4}$ are in different partite sets of $H$, then
$x_{1}{e_{1}}{x_{4}}$. Since $e_{1}$ does not contain $x_{5}$, we can choose
an arc $e_{2}$ of $H$ which is different from $e_{1}$ such that
$x_{4}{e_{2}}{x_{5}}$. Then $x_{1}{e_{1}}{x_{4}}{e_{2}}{x_{5}}$ is a path in
$H$. Now we assume that $x_{1}$ and $x_{4}$ are in the same partite set of
$H.$ Then there is an arc $e_{1}$ of $H$ such that $x_{1}{e_{1}}{x_{3}}$.
Since the degree of $z_{3}$ in $G$ is at least $2$, we can choose an arc
$e_{2}$ of $H$ which is different from $e_{1}$ such that
$x_{3}{e_{2}}{x_{4}}.$ We can also choose an arc $e_{3}$ of $H$ which is
different from $e_{1}$ and $e_{2}$ such that $x_{4}{e_{3}}{x_{5}}$. Indeed,
$e_{3}\neq e_{1}$ since $e_{1}$ does not contain $x_{5}$ and $e_{3}\neq e_{2}$
since the degree of $z_{4}$ in $G$ is at least $2$. Then
$x_{1}{e_{1}}{x_{3}}{e_{2}}{x_{4}}{e_{3}}{x_{5}}$ is a path in $H$. Thus, we
may assume that every vertex of $G$ in $A(H)$ has degree at most $2$, except
for one vertex which has degree at most $3$. Then we can use Lemma 6 and thus
we are done as above.

Case 4: $k\in\\{5,6,7\\}$ and $n=k+1$. Consider the bipartite graph $G$
constructed as in Case $2$.

Subcase 1: $k\in\\{6,7\\}$. Using the computations analogous to those in (2),
we see that the minimum degree of a vertex in $Z$ is at least $3$. If there is
a vertex with degree $4$ in $A(H)$, then it means
$\\{x_{1},x_{2},x_{3},x_{4},x_{5}\\}$ is a subset of a vertex set of an arc
$e_{1}$ and the relative order is $x_{1},x_{2},x_{3},x_{4},x_{5}$. If $x_{1}$
and $x_{5}$ are in different partite sets, then $x_{1}{e_{1}}x_{5}$ is a path
in $H$. Otherwise $x_{1}$ and $x_{4}$ are in different partite sets, so
$x_{1}{e_{1}}x_{4}.$ There is an arc $e_{2}$ different from $e_{1}$ such that
$x_{4}{e_{2}}x_{5}$ (since the degree of $z_{4}$ is at least $3$). Now
$x_{1}{e_{1}}x_{4}{e_{2}}x_{5}$ is a path in $H.$ Thus, we assume each vertex
in $A(H)$ has degree at most $3$, and we are done by Lemma 6.

Subcase 2: $k=5$. Suppose that the lemma does not hold in this case. Using the
computations analogous to those in (2), we see that the minimum degree of a
vertex in $Z$ is at least $2$. To obtain a contradiction, it suffices to show
that $G$ has at most one vertex of degree at least $3$ in $A(H).$ Suppose that
$G$ has at least two vertices of degree at least $3$ in $A(H).$ This means
that (3) holds for $i=1$ or $2$. Since $H$ can have only one arc with vertex
set $\\{x_{1},x_{2},x_{3},x_{4},x_{5}\\},$ we have

$\sum^{3}_{j=2}|N_{G}(z_{1})\cap N_{G}(z_{j})\cap N_{G}(z_{4})|\leq 1$ (4)

Without loss of generality, we assume that (3) holds when $i=1$ and let
$e_{1}\in N_{G}(z_{1})\cap N_{G}(z_{2})\cap N_{G}(z_{3}).$ If we restrict
$e_{1}$ to the vertices $\\{x_{1},x_{2},x_{3},x_{4}\\},$ we obtain
$e^{\prime}_{3}=x_{1}x_{2}x_{3}x_{4}.$

If $x_{1}$ and $x_{4}$ are in the different partite sets, then
$x_{1}{e_{1}}{x_{4}}$. Since the degree of $z_{4}$ in $G$ is at least 2, we
can choose an arc $e_{2}$ of $H$ which is different from $e_{1}$ such that
$x_{4}{e_{2}}{x_{5}}$. Then $x_{1}{e_{1}}{x_{4}}{e_{2}}{x_{5}}$ is a path in
$H$, a contradiction. Now we assume $x_{1}$ and $x_{4}$ are in the same
partite set. Then $x_{1}{e_{1}}{x_{3}}$. Since the degree of $z_{3}$ in $G$ is
at least 2, we can choose an arc $e_{2}$ of $H$ which is different from
$e_{1}$ such that $x_{3}{e_{2}}{x_{4}}.$ Since the degree of $z_{4}$ in $G$ is
at least $2$, we can choose an arc $e_{3}$ of $H$ such that
$x_{4}{e_{3}}{x_{5}}$ and $e_{3}\neq e_{2}.$ Suppose $e_{3}=e_{1}.$ Then
$e_{1}=x_{1}x_{2}x_{3}x_{4}x_{5}$ and $x_{1}e_{1}x_{5},$ a contradiction.
Thus, $e_{3}\neq e_{1}$ and $x_{1}{e_{1}}{x_{3}}{e_{2}}{x_{4}}{e_{3}}{x_{5}}$
is a path in $H$, a contradiction. ∎

##  3 Main Results

In Section 3.1, using the Majority Lemma and other results, we solve
Conjecture 3 in affirmative. In Section 3.2, we describe a family of
couterexamples to Conjecture 5 and prove a sufficient condition of when the
statement of Conjecture 5 holds.

###  3.1 Results on Conjecture 3

######  Lemma 8.

Let $H=(V,A)$ be a multipartite $k$-hypertournament with at most one
transmitter and let $M(H)$ be a majority multipartite tournament of $H$. Let
$n\geq 5$ and $n>k\geq 3.$ If $M(H)$ has at least one transmitter, then $H$
has a 2-king.

###### Proof.

Let $V_{1}$ be the partite vertex set containing all transmitters of $M(H).$
Let $v$ be the transmitter of $H,$ if $H$ has a transmitter, and an arbitrary
transmitter of $M(H)$, otherwise. Clearly, $v\in V_{1}.$ Observe that for
every $u\in V\setminus V_{1}$, there is an arc $a\in A_{H}(v,u)$ implying that
$vau.$ Note that for every $w\in V_{1}\setminus\\{v\\},$ there are a vertex
$u\in V\setminus V_{1}$ and an arc $e$ of $H$ such that $uew.$ As in Lemma 7,
it is easy to see that $|A_{H}(v,u)|\geq 2.$ Thus, there is an arc $a\in
A_{H}(v,u)$ distinct from $e$ implying that $vauew$ is a path. ∎

######  Lemma 9.

Let $H=(V,A)$ be a multipartite $k$-hypertournament and let $n\geq 5$ and
$n>k\geq 3.$ If $H$ has at most one transmitter then $H$ has a 4-king.

###### Proof.

Let $M(H)$ be a majority multipartite tournament of $H$. If $M(H)$ has no
transmitters, then by Theorem 1, $M(H)$ has a 4-king $x$. By Lemma 7, $x$ is a
4-king of $H.$ If $M(H)$ has transmitters, then we apply Lemma 8. ∎

######  Lemma 10.

Let $H=(V,A)$ be a $p$-partite $k$-hypertournament with $k=3,$ $n=4$ and
$p\geq 2.$ If $H$ has at most one transmitter then $H$ has a 4-king.

###### Proof.

By Theorem 2, this lemma holds for $p=2$ and so we may assume that $p\geq 3.$
It is well known that every $k$-hypertournament with more than $k$ vertices
has a Hamilton path [8]. Observe that for $p=4$ the first vertex of a Hamilton
path in $H$ is a 3-king. Now we may assume that $p=3.$ Let $V=V_{1}\cup
V_{2}\cup V_{3}$ be a partition of vertices of $H$. Without loss of
generality, we may assume that $V_{1}=\\{x_{1},x_{2}\\}$, $V_{2}=\\{x_{3}\\}$
and $V_{3}=\\{x_{4}\\}$.

First assume that $H$ has the unique transmitter $v.$ If $v=x_{3}$ or
$v=x_{4}$, then $v$ is a 1-king of $H$. Thus, we assume without loss of
generality that $v=x_{1}.$ Since $v$ is a transmitter, $va_{1}x_{3}$ and
$va_{2}x_{4}$ for some arcs $a_{1}$ and $a_{2}$ of $H.$ Since $x_{2}$ is not a
transmitter, there is an arc $e_{1}$ such that $y{e_{1}}x_{2}$, where $y\in
V_{2}\cup V_{3}$. By the definition of a transmitter, $v$ precedes $y$ in
every arc containing $v$ and $y$. Consequently, there is an arc $e_{2}$
different from $e_{1}$ such that $v{e_{2}}y$. Hence $v{e_{2}}y{e_{1}}{x_{2}}$
is a path from $v$ to $x_{2}.$ So $v$ is a 2-king.

Now assume that $T$ has no transmitter. Consider the arc $e_{1}$ containing
$x_{1}$, $x_{3}$, and $x_{4}.$ If $x_{1}$ is in the first position of $e_{1}$,
since $x_{2}$ is not a transmitter, there is an arc $e_{2}$ different from
$e_{1}$ such that $x_{3}{e_{2}}x_{2}$ or $x_{4}{e_{2}}x_{2}$. Hence
$x_{1}{e_{1}}x_{3}{e_{2}}{x_{2}}$ or $x_{1}{e_{1}}x_{4}{e_{2}}{x_{2}}$ is a
path from $x_{1}$ to $x_{2}$, implying that $x_{1}$ is a 2-king. Without loss
of generality, we now assume that $x_{3}$ is in the first position of $e_{1}$.
Since $x_{2}$ is not a transmitter, there is an arc $e_{2}$, where $x_{3}$ or
$x_{4}$ preceds $x_{2}$. Hence $x_{3}$ is a 2-king. ∎

Lemmas 9 and 10 imply the following result solving Conjecture 3 in
affirmative.

######  Theorem 11.

Every multipartite hypertournament with at most one transmitter has a 4-king.

###  3.2 Results on Conjecture 5

The next result describes a family of counterexamples to Conjecture 5.

######  Proposition 12.

For every $k\geq 3,$ there is a bipartite $k$-hypertournament $B$ without
transmitters which has at most one 4-king in each of its partite sets.

###### Proof.

Let $U$ and $W$ be partite sets of $B.$ Choose a vertex $u$ in $U$ and a
vertex $w$ in $W.$ Let every arc of $B$ with both $u$ and $w$ have both of
them in the first and second position such that in at least one such arc $u$
is the first and in at least one such arc $w$ is the first. Let every arc of
$B$ containing $u$ but not $w$ have $u$ in the first position and let every
arc of $B$ containing $w$ but not $u$ have $w$ in the first position. Clearly,
$B$ has no transmitters, but no vertex $v$ in $(U\cup W)\setminus\\{u,w\\}$
can be a 4-king as there is no path from $v$ to either $u$ or $w.$ ∎

The next result is a sufficient condition of when the conclusion of Conjecture
5 holds. It follows directly from Theorem 4 and the Majority Lemma.

######  Theorem 13.

Let $B$ be a bipartite hypertournament with partite sets $U$ and $W$ and with
at least 5 vertices. If a majority bipartite tournament $M(B)$ has no
transmitters, then $B$ has at least two 4-kings in each $U$ and $W.$

Our final result shows that the Majority Lemma cannot be extended to $n=4$ and
$p=2.$ The proof provides another counterexample to Conjecture 5.

######  Proposition 14.

For $k=3$ and $n=4,$ there is a bipartite hypertournament $H$ with partite
sets $U$ and $W$ such that (i) $|U|=|W|=2$, (ii) a majority bipartite
tournament $M(H)$ has no transmitters, (iii) $M(H)$ has an $(x,y)$-path of
length 3, but $H$ has no $(x,y)$-path, (iv) $H$ has only one 4-king in $U.$

###### Proof.

Let $H$ be a bipartite hypertournament with partite sets $U=\\{x_{1},x_{3}\\}$
and $W=\\{x_{2},x_{4}\\}$, arc set $\\{a_{1},a_{2},a_{3},a_{4}\\}$ where

$a_{1}=x_{4}x_{1}x_{2},a_{2}=x_{2}x_{3}x_{4},a_{3}=x_{3}x_{2}x_{1},a_{4}=x_{4}x_{3}x_{1}.$

Let the arcs of $M(H)$ be $x_{4}x_{1},x_{1}x_{2},x_{2}x_{3},x_{3}x_{4}$ (see
Fig. 1). Clearly, (i) and (ii) hold and $x_{1}x_{2}x_{3}x_{4}$ is an
$(x_{1},x_{4})$-path in $M(H)$.

$x_{1}$$x_{2}$$x_{3}$$x_{4}$ Figure 1: $M(H)$

Now consider $H.$ Suppose that $H$ has an $(x_{1},x_{4})$-path $P.$ Since
$A_{B}(x_{1},x_{4})=\emptyset$, $P=x_{1}b_{1}x_{2}b_{2}x_{3}b_{3}x_{4}$ for
some distinct arcs $b_{1},b_{2},b_{3}$ of $H.$ By inspection of the arcs of
$H$, we conclude that $b_{1}=a_{1},b_{2}=a_{2},b_{3}=a_{2},$ which is
impossible since $b_{1},b_{2},b_{3}$ must be distinct. So $H$ has no
$(x_{1},x_{4})$-path and (iii) holds. Observe that $x_{3}$ is a 4-king of $H$
since $x_{3}a_{3}x_{2},$ $x_{3}a_{2}x_{4}$ and $x_{3}a_{2}x_{4}a_{1}x_{1}$ is
an $(x_{3,}x_{1})$-path of length 2. Moreover, $x_{1}$ cannot be a 4-king by
the discussion in (iii), so (iv) holds.∎

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