12
$\begingroup$

Let $S_n$ be the symmetric group and $M_n$ the character table of $S_n$ as a matrix (in some order) for $n \geq 2$.

Question: Is it true that the rank of $M_n$ as a matrix modulo $m$ for $m \geq 2$ is equal to the number of partitions of $n$ by numbers that are not divisible by $m$?

Here the matrix modulo $m$ is obtained by replacing each number $l$ by its canonical representative mod $m$ (for example -1 mod 3 =2) and then calculate the rank of the obtained matrix as a matrix with integer entries.

This seems to be true for $m=2,3,4$ by some computer experiments. For $m=2$ the sequence of ranks (for $n \geq 2$) starts with 2,3,4,5,6,8,10,12,15,18,22 , for $m=3$ it starts with 2,4,5,7,9,13,16,22 and for $m=4$ it starts with 2,3,4,6,9,12,16,22,29.

For example for $n=5$ and $m=3$, the matrix $M_5$ looks as follows:

\begin{bmatrix} 1 & 1 & 1 & 1 & 1 \\ 1 & -1 & 1 & 1 & -1 \\ 2 & 0 & 2 & -1 & 0 \\ 3 & -1 & -1 & 0 & 1 \\ 3 & 1 & -1 & 0 & -1 \end{bmatrix}

modulo 3 the matrix is given by

\begin{bmatrix} 1 & 1 & 1 & 1 & 1 \\ 1 & 2 & 1 & 1 & 2 \\ 2 & 0 & 2 & 2 & 0 \\ 0 & 2 & 2 & 0 & 1 \\ 0 & 1 & 2 & 0 & 2 \end{bmatrix}

and this matrix has rank 4.

The paritions of $n=5$ are given by [ [ 1, 1, 1, 1 ], [ 2, 1, 1 ], [ 2, 2 ], [ 3, 1 ], [ 4 ] ] and thus there are 4 partitions of 5 whose parts are not divisble by $m=3$.

$\endgroup$
6
  • 1
    $\begingroup$ How do you define the rank modulo composite number? $\endgroup$ Nov 28, 2021 at 11:00
  • $\begingroup$ @FedorPetrov It is the rank as a matrix with integer entries I think. You just take modulo to obtain this integer matrix. That is what GAP does when it comuputes the rank of a matrix modulo $m$ (I think). $\endgroup$
    – Mare
    Nov 28, 2021 at 11:01
  • 1
    $\begingroup$ so, say, for $m=2$ it is not the same as the rank over $\mathbb{F}_2$? $\endgroup$ Nov 28, 2021 at 11:07
  • $\begingroup$ @FedorPetrov It is probably not in general. I edited what I mean with the matrix modulo $m$ (it is just the integer matrix where each entry is replaced by its canonical representatitive modulo m, for example -1 gets replaced by m-1). $\endgroup$
    – Mare
    Nov 28, 2021 at 11:13
  • $\begingroup$ Please could you check the sequence of ranks for $m=4$: the $2 \times 2$ character table of $S_2$ cannot have rank $4$. It looks like you omitted the initial $2$, since I make the sequence $2, 3, 4, 6, 9, 12, 16, 22, 29, 38, 50, \ldots$. Your conjecture is true when $m=4$ for $n \le 15$. $\endgroup$ Nov 28, 2021 at 11:45

2 Answers 2

21
$\begingroup$

This is true when $m$ is prime and false in general.

Counterexample. Take $S_8$ with $m=6$. Computer calculations show that the $\mathbb{Z}$-rank of the character table of $S_8$ with entries taken modulo $6$ to lie in $\{0,1,2,3,4,5\}$ is $22$. Thus this matrix has full rank, equal to its number of columns. (Its rank as a matrix with entries in $\mathbb{Z}/6\mathbb{Z}$ is $13$.) Since $(6,2)$ and $(6,1,1)$ are the partitions of $8$ with a part divisible by $6$, there are exactly $20$ partitions of $8$ into parts not divisible by $6$.

Proof when $m$ is prime. Let $\chi^\lambda$ be the irreducible character of $S_n$ canonically labelled by the partition $\lambda$. Let $\phi^\mu$ be the $p$-modular Brauer character of $S_n$ labelled by the $p$-regular partition $\mu$. (A partition is $p$-regular if it has at most $p-1$ parts of any given size: such partitions label the simple representations of $S_n$ in characteristic $p$.) There exist decomposition numbers $d_{\lambda\mu} \in \mathbb{N}_0$ such that

$$\chi^\lambda(g) = \sum_{\mu} d_{\lambda\mu} \phi^\mu (g)$$

for all $p$-regular $g \in S_n$. (Here $p$-regular means 'of order not divisible by $p$'.) Hence the span of the columns of the character table labelled by $p$-regular partitions is the same as the span of the columns of the Brauer character table. Since the number of $p$-regular partitions is the number of $p$-regular conjugacy classes (the Brauer character table is square), the rank of this column submatrix is the number of partitions of $n$ into parts coprime to $p$.

Finally the span of the columns of the character table labelled by $p$-singular elements, taken modulo $p$, is contained in the span of the columns labelled by $p$-regular elements, since if $g = g_{p'}h$ is the factorization of $g \in S_n$ into its $p$-regular part $g_{p'}$ and a $p$-element $h$ then $\chi(g) \equiv \chi(g_{p'})$ mod $p$ for any character $\chi$.

Reference for character congruence. The last fact is (12) in this paper of R. Brauer where it's called 'a well known congruence'. It can be seen in Mare's example: the fourth column is for the conjugacy class of non $3$-regular elements with cycle type $(3,1,1)$, i.e. $3$-cycles, and is equal modulo $3$ to the column for the identity element.

$\endgroup$
1
  • 1
    $\begingroup$ Great answer, thanks. This might raise the question, whether it is true when m is a power of a prime (or for which $m$ it is true at all). I will do some tests. $\endgroup$
    – Mare
    Nov 28, 2021 at 12:07
21
$\begingroup$

When $m$ is prime there is a simpler proof. The Smith normal form of the character table of $S_n$ is computed at Problem 14 here (solution here). From this it follows that the rank of the character table mod $m$ equals the number of partitions of $n$ for which every part has multiplicity less than $m$. By a simple generating function argument (generalizing Euler's well-known proof for the case $m=2$) this is also the number of partitions of $n$ for which no part is divisible by $m$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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