A New Proof of Wei's Theorem

Introduction

Wei’s theorem [3 Theorem 1.2] concerns the realization of a tuple of partitions $\pi_1,\ldots,\pi_k \vdash d$ as the ramification data of a rational function. The purpose of this note is to give a new proof of this theorem, based on two operations on partitions and spherical coverings: two-cut partitions and marked gluing.

The proof is inductive on

Assuming the base case $\ell\le2$, we show that any counterexample with $\ell\ge3$ can be reduced to auxiliary data with either fewer factors or smaller shortest length. The reduction uses a two-cut partition theorem (Theorem 3.1) and a marked gluing theorem for spherical coverings (Theorem 4.1), together with an additive-defect merging lemma (Lemma 5.1). The final gluing uses common letters in the marked cycles; Riemann–Hurwitz and the Riemann existence theorem then yield the required rational function. Note that the realizability for the case $l = 1$ is well known [1]. The realizability in the case $l = 2$ is proved by Pakovich [2].

Notation and the first lemma

For a partition $\pi\vdash n$, let $|\pi|$ be its number of parts and define the defect

The partition $[1^n]$ is called trivial. The symbol $\sqcup$ denotes multiset union. An allowed cut selects several complete parts of a partition and allows one further part to be shortened; if the shortening is zero, no actual cut occurs.

Proof. Write $r=|\pi|$. A partition $\mu$ with $|\mu|=q$ is attainable iff the sum of the largest $q$ parts of $\pi$ is at least $b$, and the sum of the smallest $q-1$ parts is at most $b-1$. Necessity is clear. For sufficiency, first take the smallest $q$ parts. If their sum reaches $b$, shorten the largest of them. Otherwise, successively replace chosen parts by larger ones until the total first reaches $b$, then shorten the newly chosen part. Before the last replacement the total is less than $b$, so the sum of the remaining $q-1$ chosen parts is less than $b$; hence the shortened part is positive.

In the above criterion, the first condition is monotone increasing in $q$, while the second is monotone decreasing in $q$. Hence the attainable $q$, and therefore the defects $b-q$, form an integer interval.

For the upper bound on $m(\pi,b)$, take

Then the sum of the largest $q$ parts is at least

and the sum of the smallest $q-1$ parts is at most

Thus this $q$ is attainable, and consequently

where we used $r=d-\nu(\pi)$ and the identity $b-\lceil x\rceil=\lfloor b-x\rfloor$ for $b\in\mathbb{Z}$.

For the lower bound on $M(\pi,b)$, we distinguish two cases. If the sum of all nontrivial parts of $\pi$ is less than $b$, take all of them and fill up with $1$'s; the resulting defect is exactly $\nu(\pi)$. Otherwise, select nontrivial parts successively until the sum reaches $b$, and shorten the last selected part. Since each nontrivial part is at least $2$, at most

parts are needed. Hence the defect is at least

Therefore

Finally, since $b<d$ and $\nu(\pi)>0$,

so the first estimate gives

This completes the proof.

The two-cut partition theorem

Proof. Put $S=d+\ell-2$. Any two distinct indices $p,q$ satisfy

since the remaining $\ell-1$ rows have defect at least 1.

Case $b=1$. A partition without part 1 has defect at least $d/2$; by (1) there is at most one such row. Choose it and one further row as $I,J$, split a part $a\ge2$ in each into $1+(a-1)$, and take an existing 1 from every other row. Then all $\mu_i=[1]$, and

The remaining rows keep positive large-side defect.

Case $b\ge2$. Let $e_i$ be the number of 1’s in $\pi_i$. If $e_i<b$, then $\nu_i\ge(d-e_i)/2>(d-b)/2$. Since $d\ge3b$, three such rows would give

@page 3/10 starts_mid=1 ends_mid=1 Therefore

Finally, since $b<d$ and $\nu(\pi)>0$,

so the first estimate gives

This completes the proof.

The two-cut partition theorem

Proof. Put $S=d+\ell-2$. Any two distinct indices $p,q$ satisfy

since the remaining $\ell-1$ rows have defect at least 1.

Case $b=1$. A partition without part 1 has defect at least $d/2$; by (1) there is at most one such row. Choose it and one further row as $I,J$, split a part $a\ge2$ in each into $1+(a-1)$, and take an existing 1 from every other row. Then all $\mu_i=[1]$, and

The remaining rows keep positive large-side defect.

Case $b\ge2$. Let $e_i$ be the number of 1’s in $\pi_i$. If $e_i<b$, then $\nu_i\ge(d-e_i)/2>(d-b)/2$. Since $d\ge3b$, three such rows would give

a contradiction. Hence at most two rows have $e_i < b$, and for such rows

Choose the two rows with $e_i < b$ if there are two; otherwise choose one such row together with a row of maximal defect; if there is none, choose two rows of maximal defect. Call the chosen rows $P,Q$. All other rows contain at least $b$ ones.

For $i=P,Q$, set

Lemma 2.1 and (1) give

We claim

If both $\nu_P$ and $\nu_Q$ are at least $\lfloor b/2\rfloor+1$, then Lemma 2.1 gives $U_i\ge\lfloor b/2\rfloor$ for $i=P,Q$, and (6) follows. Otherwise, after swapping names, let $t=\nu_Q\le\lfloor b/2\rfloor$. We claim that $\nu_i\le t$ for all $i\ne P$. Indeed, by (2) and $t\le\lfloor b/2\rfloor<b+1$, the row $Q$ cannot satisfy $e_Q<b$; hence $e_Q\ge b$. If there is a row with $e_i<b$, then it must be $P$, and $Q$ was chosen as a row of maximal defect among the remaining rows, so every $i\ne P,Q$ satisfies $\nu_i\le\nu_Q=t$. If no row has $e_i<b$, then $P,Q$ are the two rows of maximal defect, so again every $i\ne P,Q$ satisfies $\nu_i\le\nu_Q=t$. Thus $\nu_i\le t$ for all $i\ne P$.

Then

If $t\le|\pi_P|$, the largest $t$ parts of $\pi_P$ sum to at least $td/|\pi_P|>d/\ell\ge b$; if $t>|\pi_P|$, all parts of $\pi_P$ already sum to at least $b$. Hence at most $t$ parts suffice to reach $b$, so $M_P\ge b-t$. Also

so $U_P\ge b-t$. On the other hand, Lemma 2.1 gives

hence $U_Q\ge t-1$. Thus (6) holds.

By (5), (6) and the interval property, choose

Take the corresponding $\mu_P,\mu_Q$, and set $\mu_i=[1^b]$ for all other rows. At this stage at most two actual cuts have been made and the total small-side defect is $b-1$.

It remains to normalize the cuts. If fewer than two cuts have been made, choose the required number of rows outside $P,Q$. Return one original 1 from the small side to the large side, split some $a\ge2$ on the large side into $1+(a-1)$, and put the new 1 on the small side. The small partition remains $[1^b]$, while the large-side defect becomes $\nu_i-1\ge\ell-1>0$.

Finally, suppose two actual cuts have been made and both small-side marked parts are 1. Define

Let $e_i$ be the number of 1’s in $\pi_i$, and let $r_i$ be the number of parts of $\pi_i$ different from 1. Then $|\pi_i|=e_i+r_i$, $\nu_i=d-e_i-r_i$, and

Hence

so

For any two distinct indices $p,q$, (1) gives $\nu_p+\nu_q\le d-1$. Taking $p=P,q=Q$ and using (6), we obtain

After cancellation,

This proves (5):

Now let $f_i$ be the number of original 1’s of $\pi_i$ used on the small side of $\mu_i$, excluding the marked 1’s produced by cuts, and let $r_i'$ be the number of non-1 parts of $\mu_i$. Since each cut row has exactly one marked 1 produced by the cut on the small side,

On the other hand,

so

Because the total small-side defect is $\sum_i\nu(\mu_i)=b-1$, and the ordinary rows $\mu_i=[1^b]$ have defect 0, we have

Combining this with (7),

while

Thus

Since an allowed cut can only decrease or preserve the value of $c$, we have

Using (8) and (10), we get

Therefore at least one of the rows $P$, $Q$ still has an original 1 that has not been used on the small side.

Choose such a row and replace the cut-produced marked 1 on its small side by this unused original 1. The multiset of the small-side partition of that row is unchanged. Cancel the cut in that row and perform a new cut outside $P$, $Q$. The small partition types remain unchanged. After this adjustment, at least one of the two marked small partitions, say $\mu_I$, is $[1^b]$. If the other one $\mu_J$ is also $[1^b]$, then

Otherwise, $\mu_J$ contains the marked part 1 together with at least one further part; hence $|\mu_J| \ge2$ and therefore

Consequently,

Thus assertion (5) holds.

For the large side, uncut rows have positive large-side defect. If a cut row lies outside $P$, $Q$, its large-side defect is at least $\ell- 1$, so the sum of the two cut-row large-side defects is positive. If the two cuts remain at $P$, $Q$, then

If one of $P$, $Q$ satisfies $e_i < b$, this is positive by (2) and $\nu_j \ge\ell$ for the other. Otherwise $P$, $Q$ are the two rows of maximal defect, and

where the last inequality follows from

Each actual cut lowers the total defect by 1, so

Since the $\ell- 1$ uncut rows have defect at least 1,

This proves all assertions.

Marked gluing of spherical coverings

Proof. Let $\gamma_i=\alpha_i\oplus\beta_i$ and $\tau=(x\,y)$. Define

Then

Multiplication by a transposition joining two distinct cycles merges them, so the first two positions perform the required merges. The new cycles contain both old supports, so transitivity is preserved on $A\sqcup B$. The total defect is

Riemann–Hurwitz and the Riemann existence theorem now give the required rational function. $\square$

Remark 4.2. The common-letter hypothesis is essential: it allows the same cross-side transposition to be used at both gluing positions. It cannot be inferred from the separate realizability of the two sides.

Marked merging lemma and the induction

The content of the following lemma is taken from [1]; the marked-part refinement is added for the present application.

Proof. We first construct a bipartite forest $F$ with vertex sets $U,V$, where the degrees on $U$ are given by $\eta$, the degrees on $V$ are given by $\theta$, and the edge set $E$ has cardinality $N$. Such a forest exists by induction on $N$. For $N=1$ it is a single edge. For $N \ge2$, put

If every vertex on both sides had degree at least 2, then

hence $|\eta|+|\theta| \le N$, contradicting $c \ge1$. Thus at least one side has a vertex of degree 1. Suppose $u \in U$ has degree 1. If some $v \in V$ has degree at least 2, delete $u$ and the edge $uv$, and decrease the degree of $v$ by 1. The new degree sequences $\eta',\theta'$ have total $N-1$, and

By induction there is a forest for $\eta',\theta'$; adding back the edge $uv$ gives the required forest. If every vertex of $V$ has degree 1, then $\theta=(1^N)$. In this case each $u \in U$ is joined to $\deg(u)$ distinct vertices of $V$; since all vertices of $V$ have degree 1, these stars are disjoint, and we obtain the required forest directly. The case where a vertex of $V$ has degree 1 is symmetric. This proves the existence of $F$.

At each vertex of $F$, order the incident edges cyclically. Let $a$ be the product of the cycles at the vertices of $U$, and $b$ the product of the cycles at the vertices of $V$, where each edge is regarded as a letter. Deleting leaf edges successively shows that each tree of $F$ contributes exactly one cycle to $ab$. Hence $ab$ has exactly

cycles, so

It remains to treat the marked parts. Let $u \in U$ and $v \in V$ be the marked vertices. We modify $F$ so that $uv$ becomes an edge while preserving the degree sequences and the forest property. If $u$ and $v$ lie in different trees, choose an edge $uu'$ and an edge $vv'$, delete them, and add $uv$ and $u'v'$. The two original trees are split into four subtrees; the new edges join them in pairs, so the result is again a forest with the same degrees, and now $uv$ is an edge.

If $u$ and $v$ lie in the same tree and at least one of them, say $u$, has degree at least 2, choose an edge $uu'$ not on the $u$-$v$ path, and choose the edge $vv'$ on the $u$-$v$ path incident with $v$. Delete $uu'$ and $vv'$. Since two edges of a tree are removed, the original tree is split into three subtrees. Add the edges $uv$ and $u'v'$. These two new edges reconnect the three subtrees into a single tree: the edge $uv$ joins the component containing $u$ to the component containing $v$, while the edge $u'v'$ joins the remaining component containing $u'$ to the component containing $v'$, which lies in the same component as $u$. No cycle is created, the degree sequences are preserved, and $uv$ is an edge of the resulting forest.

If both $u$ and $v$ have degree 1 and $c \ge2$, remove the marked parts 1 from $\eta$ and $\theta$, i.e. set

Then $\eta',\theta' \vdash N-1$, and

By the existence part applied to $\eta',\theta'$, the remaining edges can be completed to a bipartite forest. Together with the isolated edge $uv$, this gives the required forest with marked edge $uv$.

The only excluded case is $c=1$ with both marked vertices of degree $1$; then a connected tree cannot contain the edge $uv$ without collapsing to a single edge. This is precisely the case excluded in the statement.

Finally, suppose the merged data are realized by a transitive permutation group. Conjugate the whole tuple so that the factor corresponding to $ab$ becomes $ab$ itself; this does not change transitivity. Replace that factor by the two factors $a$ and $b$. The product is unchanged, and the new generated group contains the old one, so transitivity is preserved. The marked cycles still share the letter corresponding to the edge $uv$. This completes the proof.

Closing the induction for Wei’s theorem

Let $\pi_1,\ldots,\pi_k\vdash d$ be nontrivial and satisfy

Assume the base case $\ell\le2$ is known.

Suppose a counterexample exists, and choose one with $\ell$ minimal and then $k$ minimal. Then $\ell\ge3$ and

If $k\ge\ell+3$, the two smallest defects among $\pi_1,\ldots,\pi_{k-1}$ sum to at most

Lemma 5.1 merges them, producing data with shortest length at most $\ell$, $k-1\ge\ell+2$ factors, and the same total defect. By minimality these data are realizable, and splitting back realizes the original data, a contradiction. Hence $k=\ell+2$.

If some $i<k$ has $\nu(\pi_i)\le\ell-1$, merging it with $\pi_k$ gives a partition of length $\ell-\nu(\pi_i)\le\ell-1$. The new data have $\ell+1$ nontrivial factors and still satisfy the strict inequality, so minimality gives a contradiction. Thus

Choose a smallest part $b$ of $\pi_k$, and put $D=d-b$. Then $\ell b\le d$. Apply Theorem 3.1 to $\pi_1,\ldots,\pi_{k-1}$, obtaining $\mu_i,\lambda_i$ and indices $I,J$. Set

Then $\lambda_k$ is nontrivial. On the small side,

so the total small-side defect is $2b-2$. Merge the two marked factors first using Lemma 5.1; Theorem 3.1(5) excludes the marked exception. Merge the remaining factors successively and take the inverse of the final $b$-cycle as the last factor. This gives a transitive permutation tuple with total defect $2b-2$, hence a spherical covering with the required markings.

On the large side,

so the total defect is $2D - 2$. Theorem 3.1 gives

Merge these two factors using Lemma 5.1, keeping the marked cycles intersecting. The auxiliary data have $\ell+ 1$ nontrivial factors, including $\lambda_k$ of length $\ell- 1$, so their shortest length is at most $\ell- 1$. Since

minimality realizes the auxiliary data. Splitting back gives the large side with the required markings.

Finally, order the branch values on both sides so that $I,J$ are adjacent, construct the marked factor pairs first, and apply Theorem 4.1. The two cut parts are restored as $u_I + v_I$ and $u_J + v_J$, all other parts are restored by multiset union, and the last row is restored as $[b] \sqcup\lambda_k = \pi_k$. Thus the original data are realized by a rational function, contradicting the counterexample assumption.

Declarations

Generative AI disclosure. The authors used OpenAI’s ChatGPT as an assisting tool in developing this paper. The authors takes full responsibility for its contents.

Data availability. This manuscript has no associated data.

Competing interests. The authors have not disclosed any competing interests.

References

Author information

Zhiqiang Wei, Mengyan Xing, Xingya Zhao

School of Mathematics and Statistics, Henan University, Zhengzhou, China

E-mail: [email protected] (Zhiqiang Wei); [email protected] (Mengyan Xing); [email protected] (Xingya Zhao).