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\begin{document}
\begin{center}
{\Large\bf On character tables related to   \\[5pt]
the alternating groups}\\[.1in]

\bigskip \bigskip

{ \large
Christine Bessenrodt} \\[5pt]
Institut f\"ur Mathematik,
Uni\-ver\-si\-t\"at Hannover\\
D-30167 Hannover, Germany \\
Email: bessen@math.uni-hannover.de
\\[3ex]
{\large 

J\o rn B. Olsson\footnote{Partially
    supported by
 The Danish National Research Council.}
} \\[5pt]
Matematisk Afdeling,
University of Copenhagen \\
Copenhagen, Denmark\\
Email: olsson@math.ku.dk
\\[4ex]


{\small 

{\bf Abstract}
\medskip

\parbox{12cm}{
There is a simple formula for the absolute value of the determinant of
the character table of the symmetric group~$S_n$.
It equals  $a_{\mathcal{P}}$, the product of all parts
of all partitions of~$n$ (see \cite[Corollary 6.5]{James}).
In this paper we calculate the absolute values of
the determinants of certain  submatrices of the character table
$\mathcal{X}$ of the alternating group $A_n$, including that of
$\mathcal{X}$ itself (Section~2). We also study explicitly the
powers of 2 occurring in these determinants using generating
functions (Section~3).
}}
\end{center}


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\medskip

\section{Preliminaries}

We fix a positive integer $n$.
We will use the same notation as in \cite{bos}, which we recall here.

If $\mu=(\mu_1,\mu_2, \ldots)$ is a partition of $n$ we write $\mu \in
\P$ and then
$z_{\mu}$ denotes the order of the centralizer of an element of
(conjugacy) type $\mu$ in $S_n$. Suppose $\mu=(1^{m_1(\mu)},
2^{m_2(\mu)},\ldots)$,  is written in exponential notation.
Then we may factor $z_\mu=
a_{\mu}b_{\mu}$,  where
$$a_{\mu}=\prod_{i \ge 1}i^{m_i(\mu)}, \hspace{.1 in}
b_{\mu}=\prod_{i \ge 1}m_i(\mu)!$$


Whenever $\mathcal{Q} \subseteq \mathcal{P}$
we define
$$a_{\mathcal{Q}}=\prod_{\mu \in \mathcal{Q}}a_{\mu}, \hspace{.1 in}
b_{\mathcal{Q}}=\prod_{\mu \in \mathcal{Q}}b_{\mu} \: .$$

\bigskip
We consider the alternating group $A_n$.
We let $\P^+$ denote the even partitions in $\P$,
${\mathcal O}$ the partitions into odd parts, and
${\mathcal D}$  the partitions into distinct parts.

The conjugacy classes in $A_n$ are of two types.
The classes labelled by partitions  $\mu \in \P^+\setminus (\O \cap \D)$
are the non-split classes, which contain all $S_n$-permutations of this type;
we denote a representative by $\sigma_{\mu}$ and note that
the corresponding centralizer is then of order $z_{\mu}'=z_{\mu}/2$.
For the partitions $\mu \in \D \cap \O$, the corresponding $S_n$-class splits
into two conjugacy classes in $A_n$, for which we denote representatives by
$\sigma_{\mu}^+$ and  $\sigma_{\mu}^-$; their centralizers are of order $z_{\mu}'=z_{\mu}$.

We briefly recall some information on the
irreducible $A_n$-characters (see \cite[sect.~2.5]{JK}).

Let $\mu$ be a partition of $n$.
For $\mu \neq \tilde{\mu}$, i.e., $\mu$ non-symmetric,
$[\mu]\downarrow_{A_n}=[\tilde{\mu}]\downarrow_{A_n}$
is  irreducible.
Let $\{\mu\}=\{\tilde{\mu}\}$ denote this irreducible character of $A_n$.\\
For $\mu=\tilde{\mu}$, i.e., $\mu$ symmetric,
$[\mu]\downarrow_{A_n} = \{\mu\}_{+} + \{\mu\}_{-}$
is a sum of two distinct irreducible $A_n$-characters
(which are conjugate in $S_n$).\\
This gives all the irreducible complex characters of $A_n$, i.e.,
$$
\mbox{Irr}(A_n)  =
\{\{\mu\}_{\pm} \mid \mu \vdash n, \mu=\tilde{\mu}\}
\cup \{\{\mu\} \mid \mu \vdash n, \mu \neq \tilde{\mu}\} \: .
$$
The characters $\{\mu\}_{\pm}$, for symmetric $\mu$,
usually have non-rational values on the
corresponding  ``critical'' classes of cycle type
$h(\mu)=(h_{1}, \ldots, h_{l})$, where $h_{1},  \ldots , h_{l}$ are the
principal hook lengths in $\mu$; note that $h(\mu) \in \D \cap \O$,
so the corresponding $S_n$-class splits.
Then we have
$[\mu](\sigma_{h(\mu)}) = (-1)^{\frac{n-l}{2}}=:\epps_{\mu}$
and
$$
\{\mu\}_{+}(\sigma_{h(\mu)}^{\pm}) =
\frac12 \left( \epps_{\mu} \pm \sqrt{\epps_{\mu}\prod_{i=1}^l h_i} \right)
$$
$$
\{\mu\}_{-}(\sigma_{h(\mu)}^{\pm}) =
\frac12 \left( \epps_{\mu} \mp \sqrt{\epps_{\mu}\prod_{i=1}^l h_i} \right)
$$
All other irreducible $A_n$-characters have the same value on these two classes.

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\medskip

For later use, we want to recall  the Jacobi minor
theorem (see \cite[p.\ 21]{gantmacher}).\\
Let $A=(a_{ij})$ be an $n\times n$ matrix.
Let $M_v$ be a $v$-rowed minor of the determinant $\det A$,
corresponding to the rows $i_1, \ldots , i_v$ and the
columns $k_1, \ldots , k_v$.
Then we take the $(n-v)$-rowed complementary minor for $A$ by deleting
all the rows and columns chosen for $M_v$ before,
and define the signed complementary minor $M^{(v)}$ to $M_v$ by
multiplying this complementary minor by the sign
$\pm 1$, depending on $\sum_{j=1}^v i_j + \sum_{j=1}^v k_j$
being even or odd, respectively.
(Note that for principal minors the sign is always $+$.)


Let $A'=(A_{ij})$ be the $n \times n$-matrix of cofactors $A_{ij}$
for $A$, i.e., the adjoint matrix to $A$. Let $M_v$ and $M'_v$
be corresponding $v$-rowed minors of
$A$ and $A'$, respectively, then
$$M'_v = (\det A)^{v-1} M^{(v)} \: .$$

\medskip

\section{Determinants of submatrices of the character table of $A_n$}

We observe that by the Murnaghan-Nakayama formula
we have for any symmetric partition $\mu$ and any $\nu \in \D \cap \O$:
$$\{\mu\}_{\pm}(\sigma_{\nu}^{\pm}) =0 \; \textrm{for all }
\nu > h(\mu) \; \textrm{(in lexicographic order)}$$
Hence, if we order the $k$ (say) partitions in $\D \cap \O$ in
decreasing lexicographic order, and the $k$ symmetric partitions
according to  their principal hook lengths,
then the corresponding $2k \times 2k$ part of the character table
of $A_n$ is almost an upper triangular matrix, except that we have
$2 \times 2$ blocks along the diagonal.
We call this matrix $\X_s$.

Knowing the entries of these diagonal blocks explicitly,
we can easily compute their determinant and hence the
(absolute value of the) determinant of this submatrix of the character table.
A $2\times 2$ block corresponding to the characters
$\{\mu\}_{\pm}$ on the classes $\sigma_{h(\mu)}^{\pm}$ gives a contribution
of absolute value
$$|\epps_{\mu}\sqrt{\epps_{\mu}\prod_i h_i}|=\sqrt{\prod_i h_i}=\sqrt{a_{h(\mu)}} \: ,$$
where $h(\mu)=(h_1, h_2 , \ldots)$.
Hence the absolute value of the determinant of the whole submatrix is given by:

\begin{proposition}
$$|\det \X_s|= \prod_{\nu \in \D \cap \O} \sqrt{a_{\nu}} = \sqrt{a_{\D \cap \O} } \: .$$
\end{proposition}

\medskip

We can also easily determine the (absolute value of the)
determinant for the whole character table $\X$ of $A_n$.
By character orthogonality, we know that
$\bar{\X}^t \X$ is a diagonal matrix with the centralizer orders
as its diagonal entries.
Set $\P^{(+)}=\P^+ \setminus (\D \cap \O)$.
Hence we have
$$\begin{array}{rcl}
|\det \X|^2 & = & \left(\prod_{\mu \in \P^{(+)}} \frac{z_{\mu}}{2}\right)
   \left(\prod_{\mu \in \D \cap \O} z_{\mu}^2\right)\\[8pt]
& = & 2^{-|\P^{(+)}|} z_{\P^+} z_{\D \cap \O} = 2^{-|\P^{(+)}|} a_{\P^+} b_{\P^+} a_{\D \cap \O} \\
\end{array}$$

Now we have $b_{\P^+}=2^{e^+} a_{\P^+}$, for some integer $e^+ \in \Z$.
(This is not hard to prove by a combinatorial argument, see Lemma~\ref{Lemma3}.)

Hence we obtain

\begin{proposition}\label{X}
$$\begin{array}{rcl}
|\det \X|^2 & = & 2^{e^+-|\P^{(+)}|} a_{\P^+}^2 a_{\D \cap \O} \: .
\end{array}$$
\end{proposition}

In the next section we will see that
$e^+=e^+(n) \in \N$, and that there is a nice generating function
for the numbers $e^+(n)$ (Proposition~\ref{Proposition4}).
In particular, an explicit formula for $e^+(n)$ is given by
$$e^+(n)=\sum_{i=1}^{[n/2]} \tau(i) p'(n-2i) \: , $$
where $\tau(i)$ is the number of divisors of~$i$,  and
$p'(j)= |\D(j)\cap \O(j)|$.


\medskip

We are interested in determining the determinant of the integral part
of the character table of $A_n$ corresponding to the non-symmetric
partitions and the non-split conjugacy classes; let us call this
matrix $\X_u$ (with some ordering of rows and columns chosen).
(Note that this is also a submatrix of the character table of $S_n$.)
This part of the character table of $A_n$ is complementary to the submatrix
we have considered above, and we want to compute its determinant
by employing Jacobi's theorem.

\begin{theorem}\label{det(Xu)}
The determinant of the matrix $\X_u$ has absolute value
$$|\det \X_u| = 2^{(e^+-|\P^{(+)}|)/2} a_{\P^{(+)}} \: .  $$
\end {theorem}

\begin{proof}
We assume that the rows of the character table $\X$ of $A_n$ are labelled such
that the rows corresponding to the symmetric partitions come first,
and that the columns are labelled such that the $v=|\P^{(+)}|$ partitions
in $\P^{(+)}$ come first.
Let $\Delta$ be the diagonal matrix with the centralizer orders $z_{\mu}'$ as its
diagonal entries, and let $\Delta^{(+)}$ be the diagonal submatrix corresponding
to the partitions $\mu \in \P^{(+)}$.

As we have $\bar{\X}^t \cdot \X = \Delta$ , we know that
the adjoint matrix to $\X$ is
$$\X'=(\det  \X) \Delta^{-1} \bar{\X}^t\: .$$
We now want to apply Jacobi's minor theorem as it is stated in
Section~1.
We take the $v$-rowed minor $M_v$ corresponding to the
upper left square part in $\X$, i.e., $M_v=\det \X_u$.
The corresponding minor of $\X'$ is then the determinant of
$$(\det \X) (\Delta^{(+)})^{-1} \X_u$$
(remember that $\X_u$ is integral).
The signed complementary minor to $M_v$ in $\X$ is then
just $\det \X_s$.
By Jacobi's theorem we know that
$$(\det \X)^v (\prod_{\mu\in\P^{(+)}} z_{\mu}' ) ^{-1} \det \X_u =
(\det \X)^{v-1} \det \X_s \: .$$
Hence
$$\det \X_u = (\det \X)^{-1} 2^{-v} a_{\P^{(+)}} b_{\P^{(+)}} \det \X_s \: , $$
and thus
$$\begin{array}{rcl}
|\det \X_u| & = & 2^{-(e^+-v)/2} (a_{\P^{+}} \sqrt{a_{\D \cap \O}})^{-1}
2^{-v} a_{\P^{(+)}} b_{\P^{+}} \sqrt{a_{\D \cap\O}}\\
& = & 2^{(e^+-v)/2} a_{\P^{(+)}}
\end{array}  $$
where we have used the relation
$b_{\P^+}=2^{e^+}a_{\P^+}$.
\end{proof}

\medskip

\section{Powers of 2}

We compute the generating functions for the powers of 2 occurring in
the determinants of the previous section.

Let $P(q), P^+(q), P^-(q)$
be the generating function for the number of partitions
(resp. even/odd partitions) of $n$.
The following is well-known:

\medskip

\begin{lemma}\label{Lemma1}
 $P^+(q)-P^-(q)=\Delta(q)$, where
$$\Delta(q)=\prod_{k \ge 0}(1+q^{2k+1})\quad ( \; = \frac{P(q)P(q^4)}{P(q^2)^2})$$
is the generating function
for the number of partitions of $n$ into distinct odd parts.
\end{lemma}

Indeed, using that in
$P(q)=\prod_{k \ge 1} \frac{1}{1-q^k}$
the factor $\frac{1}{1-q^k}$ accounts for the parts equal to $k$ we
see that
$$ P^+(q)-P^-(q)=\prod_{k \ge 1} \frac{1}{1+(-q)^k} \: .$$
Substituting $q \rightarrow -q$ in the Euler identity
$ \prod_{k \ge 1}(1+q^k) = \prod_{k \ge 0} \frac{1}{1-q^{2k+1}}$
and inverting we get
$$\prod_{k \ge 1} \frac{1}{1+(-q)^k}=\prod_{k \ge 0} (1+q^{2k+1}) \: ,$$
proving the Lemma. $\Box$

\medskip

We assume in the following always that $\delta=+$ or $-$ is a sign.

\medskip

\begin{corollary}\label{Corollary2}
We have
$$P^{\delta}(q)=\frac{P(q)+\delta \Delta(q)}{2} \; .$$
\end{corollary}

We let $\mathcal{P}^{\delta}(n)$ be the set
of partitions of $n$ with sign $\delta$.
Then define
$$a^{\delta}(n)=a_{\P^{\delta}(n)}= \prod_{\mu \in \mathcal{P}^{\delta}(n)}a_{\mu},
\hspace{.1 in}
b^{\delta}(n)=b_{\P^{\delta}(n)}= \prod_{\mu \in \mathcal{P}^{\delta}(n)}b_{\mu}\; .$$

We factor each $i \in \mathbb{N}$ as a product $i=i_2i'$, where
$i_2$ is a power of 2 and $i'$ is odd and consider two involutory
bijections $\iota, \iota'$ on the set
$$ \mathcal{T}(n)=\{(\mu,d,k)| \mu \in \mathcal{P}(n), m_d(\mu) \ge k \} \: .$$
Here
$$\iota: (\mu,d,k) \mapsto (\hat{\mu},k,d)$$
where $\hat{\mu}$ is obtained from $\mu$ by replacing $k$ parts equal to
$d$ by $d$ parts equal to $k$ and leaving all other parts unchanged
and
$$\iota': (\mu,d,k) \mapsto (\tilde{\mu},d_2k',k_2d')$$
where $\tilde{\mu}$ is obtained from $\mu$ by replacing $k$ parts equal to
$d$ by $k_2d'$ parts equal to $d_2k'$ and leaving all other parts
unchanged.
Let
$$\mathcal{T}^{\delta}_{d,k}(n)=\{\mu \in \mathcal{P}^{\delta}(n)|
m_d(\mu) \ge k \}\: .$$
Then
\beq
|\mathcal{T}^{\delta}_{d,k}(n)|=p^{(-1)^{(d-1)k}\delta}(n-dk)
\eeq
Indeed removing $k$ parts equal to $d$ from a partition $\mu$ with
sign $\delta$ gives you a
partition with sign $(-1)^{(d-1)k}\delta$ and of cardinality
$|\mu|-dk$.

Note that this means that the partitions $\mu, \hat{\mu}$ in the definition
of $\iota$ have different signs if and only if $(d-1)k$ and $d(k-1)$
have different parities, ie. if and only if $d$ and $k$ have different
parities. Moreover the partitions  $\mu, \tilde{\mu}$ in the definition
of $\iota'$ have the same sign.

Thus
 $\iota$ induces a bijection between
$\mathcal{T}^{\delta}_{d,k}(n)$ and  $\mathcal{T}^{\delta}_{k,d}(n)$
if $d,k$ have the same parity and between
$\mathcal{T}^{\delta}_{d,k}(n)$ and  $\mathcal{T}^{-\delta}_{k,d}(n)$
if $d,k$ have different parities. Moreover the bijection $\iota'$
shows that
\beq
 a^{\delta}(n)'= b^{\delta}(n)',
\eeq
and hence
\medskip

\begin{lemma}\label{Lemma3}
$b^{\delta}(n)/ a^{\delta}(n)=2^{e^{\delta}(n)}$ for some
integer $e^{\delta}(n)$.
\end{lemma}

\medskip

The power of 2 in $a^{\delta}(n)$ is
$$x^{\delta}(n)= \underset{d \: \text{even}}{\prod_{d,k} }
d_2^{|\mathcal{T}^{\delta}_{d,k}(n)| } $$
and the power of 2 in $b^{\delta}(n)$ is
$$y^{\delta}(n)= \underset{ k \: \text{even}}{\prod_{d,k }}
k_2^{|\mathcal{T}^{\delta}_{d,k}(n)| }$$

Let $x_o^{\delta}(n),x_e^{\delta}(n) $ be the product of the factors
in $x^{\delta}(n)$, where $k$ is odd/even and correspondingly
$y_o^{\delta}(n),y_e^{\delta}(n) $ be the product of the factors
in $y^{\delta}(n)$, where $d$ is odd/even. Using the map $\iota$ we
see that
$$x_e^{\delta}(n)=y_e^{\delta}(n), ~~x_o^{\delta}(n)=y_o^{-\delta}(n) $$
Thus the power of 2 in $b^{\delta}(n)/ a^{\delta}(n)$ is
$x_o^{-\delta}(n)/x_o^{\delta}(n)$.
\\
Suppose that $x_o^{\delta}(n)=2^{f_o^{\delta}(n)}$ and
 $x_e^{\delta}(n)=2^{f_e^{\delta}(n)}$.
Then $e^{\delta}(n)=f_o^{-\delta}(n)-f_o^{\delta}(n)$.
\\
We have (since $\nu_2(d)=0$, when $d$ is odd)
$$f_o^{\delta}(n)=
\underset{ k \: \text{odd}}{\sum_{d,k}} \nu_2(d)|\mathcal{T}^{\delta}_{d,k}(n)|=
\underset{ k \: \text{odd}}{\sum_{d,k }} \nu_2(d)p^{-\delta}(n-dk) \: .$$
Let
$\tau_o(n)$ the
number of odd divisors of $n$. Note that $\tau_o(n)\nu_2(n)$ equals
the number $\tau_e(n)$ of {\it even} divisors of $n$.
We then get (substituting $dk=t$ in the above sum and noting that then
$\nu_2(d)=\nu_2(t)$ )
$$ f_o^{\delta}(n)=
\sum_{t=1}^{n}\tau_o(t)\nu_2(t)p^{-\delta}(n-t)=
\sum_{t=1}^{n}\tau_e(t)p^{-\delta}(n-t) \: . $$
Let $T(q)=\sum_{t \ge 1}\frac{q^t}{1-q^t}$
be the generating function for $\tau(n)$. Then $T(q^2)$ is the
generating function for the number $\tau_e(n)$ of even divisors of $n$.
If $F_o^{\delta}(q)$ is the
generating function for  $f_o^{\delta}(n)$ we obtain
\beq
F_o^{\delta}(q)=T(q^2)P^{-\delta}(q).
\eeq
Using Lemmas~\ref{Lemma1} and~\ref{Lemma3} we deduce
\begin{proposition}\label{Proposition4}
The generating function for $e^{\delta}(n)$ is
$$E^{\delta}(n)= F_o^{-\delta}(q)-F_o^{\delta}(q)=\delta T(q^2)\Delta(q) \: .$$
\end{proposition}


\begin{remark}
{\rm
This Proposition was also proved in \cite{mueller} in a different way.
Our approach was partially inspired by an unpublished note of John Graham.\\
Note that the proposition shows that $e^+=e^+(n)$ is always a {\em positive} integer.
}
\end{remark}

\medskip

Let us also consider $F_e^{\delta}(q)$.
We have
$$f_e^{\delta}(n)=\sum_{\{d,k| k~
  even \}}\nu_2(d)|\mathcal{T}^{\delta}_{d,k}(n)|=
\sum_{\{d,k| k~  even \}}\nu_2(d)p^{\delta}(n-dk) \: .$$
We substitute $dk=2t$ in the above and obtain
$$f_e^{\delta}(n)=\sum_{t \ge 1} \tau^*(t)p^{\delta}(n-2t) \: ,$$
where
$\displaystyle \tau^*(t)=\sum_{d \mid t}\nu_2(d)$.
We have
$$\tau^*(t)=\binom{\nu_2(t)+1}{2} \prod_{p~odd}(\nu_p(t)+1) \: .$$
Thus if $T^*(q)$ is the generating function for $\tau^*(t)$ then
$$F_e^{\delta}(q)=T^*(q^2)P^{\delta}(q) \: .$$
It is easily seen that
$$T^*(q)=\sum_{j \ge 1} T(q^{2^j}) \: .$$

\medskip

\begin{proposition}\label{Proposition5}
The exponent of 2 in $a^{\delta}(n)$ has the
generating function
$$F_e^{\delta}(q)+F_o^{\delta}(q)=
T^*(q^2)P^{\delta}(q)+T(q^2)P^{-\delta}(q) \: .$$
\end{proposition}

\bigskip


In Theorem~\ref{det(Xu)} we have seen that
$|\det \X_u| = 2^{(e-|\P^{(+)}|)/2} a_{\P^{(+)}}$.\\
By Proposition~\ref{Proposition4},
$e=e^+(n)$ has generating function
$E^+(q)=\Delta(q)T(q^2)$. \\
Moreover $|\P^{(+)}(n)|$ has  generating function
$P^+(q)-\Delta(q)=P^-(q)$ (Lemma \ref{Lemma1}).
Clearly, $ a_{P^{(+)}}(n)$ is divided by the same power of 2 as
$a^+(n)$, as the removed partitions have only odd
parts. The generating function for the corresponding exponent
is given by Proposition~\ref{Proposition5}.
Hence  the exponent of 2 in  $\det \X_u$ has  the generating function
$$G(q)=\frac{1}{2}\left(T(q^2) \Delta(q)-P^-(q)\right)+T^*(q^2)P^+(q)+T(q^2)P^-(q),$$
and this then yields

\begin{theorem}
The exponent of 2 in  $\det \X_u$ has  the generating function
$$G(q)= \frac{1}{2}\left(T(q^2)P(q)-P^-(q)\right)+T^*(q^2)P^+(q) \: .$$
\end{theorem}

\medskip
According to MAPLE the first values of the coefficients of $G(q)$ are
the following for $n=2, \ldots ,14$:
0 0 2 2 4 6 15 19 30 43 70 94 138.

\medskip

Let us finally remark that the Propositions~\ref{Proposition4}
and~\ref{Proposition5} also allow  to compute the generating function
for the exponent of 2 in $|\det(\X)|$, using Proposition~\ref{X}.

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\bibitem{bos} C.\ Bessenrodt, J.~B.\ Olsson, R.~P.\ Stanley,
Properties of some character tables related to the symmetric groups,
preprint 2003; to appear in: J.\ Algebraic Combinatorics.
\bibitem{gantmacher} F.~R.\ Gantmacher, \emph{The Theory of Matrices},
  vol.\ 1, Chelsea, New York, 1960.
\bibitem{James} G.\ James, The representation theory of the
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  Springer--Verlag, 1978.
\bibitem{JK} G.~James, A.~Kerber, \emph{The Representation Theory of the
  Symmetric Group}, Addison--Wesley, 1981.
\bibitem{mueller} J.\ M\"uller, On a remarkable partition identity,
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\bibitem{OlsReg} J.~B.~Olsson,  Regular character tables of symmetric
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\end{thebibliography}

\end{document}
