# Fake degrees of classical Weyl groups

William M. McGovern

###### Abstract.

We compute the fake degrees of representations of classical Weyl groups in
terms of major indices of domino tableaux.

###### Key words and phrases:

fake degrees, hyperoctahedral groups, major index, domino tableau

###### 1991 Mathematics Subject Classification:

22E47,22E46

##  1\. Introduction

Let $W$ be the complex reflection group $C_{d}\wr S_{n}$, where $C_{d}$ is the
cyclic group of order $d$. The action of $W$ on $\mathbb{C}^{n}$ by coordinate
permutations and scalar multiplications by complex $d$th roots of unity then
extends to the coordinate ring $S$ of $\mathbb{C}^{n}$, preserving the natural
grading of $S$. Let $I$ be the ideal of $S$ generated by $W$-invariant
polynomials of positive degree. The coinvariant algebra $C=S/I$ is then well
known to be isomorphic to the regular representation of $W$; like $S$ it has a
graded structure preserved by $W$. Given an irreducible representation $\tau$
of $W$ of degree $d_{\tau}$ its so-called fake degree (polynomial) is the
palindromic polynomial $f_{\tau}(q)=\sum_{i=1}^{d_{\tau}}q^{d_{i}}$, where the
exponents $d_{i}$ are the degrees in which $\tau$ occurs in $C$, each listed
according to its multiplicity. There are well-known formulas for these degrees
as powers of $q$ times ratios of products of differences $q^{m}-1$ for various
$m$ (see [Stei51, L77]). More recently these formulas have been rewritten in
terms of major indices of standard Young tableaux [Sta71, Ste89]. Here we give
new formulas for these degrees for hyperoctahedral groups and Weyl groups of
type $D$, using major indices of domino tableaux. Such tableaux were first
introduced in [G90] to study primitive ideals in enveloping algebras of
classical complex Lie algebras (see also [G92, G93]). They were used to study
orbital subvarieties of nilpotent orbits in classical complex Lie algebras
[M21, M21’]. We remark also that the notion of the major index of a domino
tableau has been generalized to that of a descent of a border strip tableau in
[P21].

##  2\. Types $B$ and $C$

We begin with a quick review of the $q$-analogues of integers, factorials, and
multinomial coefficients. For $n$ a nonnegative integer, $k$ a positive
integer at most equal to $n$, and $\alpha=(\alpha_{1},\ldots,\alpha_{m})$ a
partition of $n$, set

$[n]_{q}=1+q+\cdots+q^{n-1}=\frac{q^{n}-1}{q-1}\text{ for }n\geq 1,[0]_{q}=1$
$[n]_{q}!=[n]_{q}[n-1]_{q}\cdots[1]_{q},{n\choose
k}_{q}=\frac{[n]_{q}!}{[k]_{q}![n-k]_{q}!}$
${n\choose\alpha}_{q}=\frac{[n]_{q}!}{[\alpha_{1}]_{q}!\cdots[\alpha_{m}]_{q}!}$

Identifying $\alpha$ with the Young diagram of the corresponding shape, so
that $\alpha_{i}$ is the length of the $i$th row of this diagram, denote by
$h_{c}$ the length of the hook of the cell $c\in\alpha$. Set
$b(\alpha)=\sum_{i=1}^{m}(i-1)\alpha_{i}$.

Recall that a standard Young tableau $T$ of shape $\alpha$ is a bijective
filling of the cells of $\alpha$ by the numbers from 1 to the sum $|\alpha|$
of the parts of $\alpha$ such that labels increase to the right in rows and
down columns. The major index maj$(T)$ of $T$, sometimes just called the index
of $T$, is the sum of the labels $i$ such that $i+1$ appears in a lower row
than $i$ in $T$. Denoting by SYT$(\alpha)$ the set of standard Young tableaux
of shape $\alpha$, we have the generating function

$\text{SYT}(\alpha)^{\text{maj}}(q):=\sum_{T\in\text{SYT}(\alpha)}q^{\text{maj}(T)}$

It is well known that irreducible representations of $W$ are parametrized by
ordered $d$-tuples $\lambda=(\lambda^{(1)}\ldots,\lambda^{(d)})$ of partitions
$\lambda^{(i)}$ such that $\sum_{i}|\lambda^{(i)}|=n$ [Ste89, Thm. 4.1].
Denote by $V_{\lambda}$ the representation corresponding to $\lambda$ and
write $b(\lambda)=\sum_{i=1}^{d}(i-1)|\lambda^{(i)}|$. A standard (Young)
tableau $T$ of shape $\lambda$ is a $d$-tuple $(T^{(1)},\ldots,T^{(d)})$ of
fillings of shapes $\lambda^{(1)},\ldots,\lambda^{(d)}$ such that the labels
$1,\ldots,n$ are each used exactly once overall and labels increase across
rows and down columns of each $T^{(i)}$. The major index maj$(T)$ of $T$ is
the sum of the labels $i$ such that either $i$ appears in a higher row than
$i+1$ in the same filling $T^{(j)}$, or $i,i+1$ appear in the fillings
$T^{(j)},T^{(k)}$, respectively, with $j<k$. Then Stanley and Stembridge have
derived the following formula for the fake degree $f_{\lambda}$ corresponding
to $\lambda$ [Sta71, Sta79],[Ste89, Thm. 5.3]. Denote by SYT$(\lambda)$ the
generating function $\sum_{T}q^{\text{maj}(T)}$, where the sum runs over
standard tableaux of shape $\lambda$.

######  Theorem 1.

The fake degree $f_{\lambda}$ corresponding to $\lambda$ is given by

$f_{\lambda}=q^{b(\lambda)}\text{SYT}(\lambda)(q^{d})=q^{b(\lambda)}{n\choose{|\lambda^{(1)}|,\ldots,|\lambda^{(d)}|}}_{q}\cdot\prod_{i=1}^{d}\text{SYT}(\lambda^{(i)})^{\text{maj}}(q^{d})$

where

$\text{SYT}(\alpha)^{\text{maj}}(q)=\frac{q^{b(\alpha)}[r]_{q}!}{\prod_{c\in\alpha}[h_{c}]_{q}}$

for a partition $\alpha=(\alpha_{1},\alpha_{2},\ldots)$ of $r$ and
$f_{\lambda}$ denotes the fake degree of the representation $V_{\lambda}$
corresponding to $\lambda$. Equivalently, the multiplicity of $V_{\lambda}$ in
the $k$-th graded piece of the coinvariant algebra $C$ is the number of
standard tableaux $T$ of shape $\lambda$ with $k=b(\lambda)+d\,\text{maj}(T)$.

We now specialize down to the case $d=2$. Given an ordered pair
$(\lambda^{(1)},\lambda^{(2)})$ of partitions with
$|\lambda^{(1)}|+|\lambda^{(2)}|=n$, we follow Lusztig [L77, §3] to produce a
single partition $\rho_{1}$ of $2n$, as follows (see also [C85]). Add zeroes
to the parts of $\lambda^{(1)},\lambda^{(2)}$ as necessary to make
$\lambda^{(1)}=(\alpha_{1},\ldots,\alpha_{m+1})$ have exactly one more part
than $\lambda^{(2)}=(\beta_{1},\ldots,\beta_{m})$. For $1\leq i\leq m+1$, put
$\alpha_{i}^{*}=\alpha_{i}+m+1-i$; similarly for $1\leq j\leq m$ put
$\beta_{j}^{*}=\beta_{j}+m-j$. Then the $\alpha_{i}^{*}$ and the
$\beta_{j}^{*}$ are distinct. Now set
$\gamma_{i}=2\alpha_{i}^{*},\delta_{i}=2\beta_{i}^{*}+1$, and combine and
rearrange the $\gamma_{i},\delta_{i}$ to make a partition
$\rho_{1}^{\prime}=(p_{1}^{\prime},\ldots,p_{r}^{\prime})$. Then for $1\leq
i\leq r$ set $p_{i}=p_{i}^{\prime}-r+i$, thereby obtaining
$\rho_{1}=(p_{1},\ldots,p_{r})$. In a similar way we also use the
$\alpha_{i}^{*}$ and $\beta_{i}^{*}$ to produce a single partition $\rho_{2}$
of $2n+1$, by putting
$\gamma_{i}^{\prime}=2\alpha_{i}^{*}+1,\delta_{i}^{\prime}=2\beta_{i}^{*}$ and
combining and rearranging the $\gamma_{i}^{\prime},\delta_{i}^{\prime}$ to
make $\rho_{2}^{\prime}=(q_{1}^{\prime},\ldots,q_{r}^{\prime})$, finally
setting $q_{i}=q_{i}^{\prime}-r+i$ to obtain $\rho_{2}=(q_{1},\ldots,q_{r})$.
The partitions $\rho,\rho_{2}$ that arise in this way are exactly those
supporting a standard domino tableau of that shape.

Let $\alpha$ be a partition of $2n$. Recall from [G90] that a domino tableau
$T$ of shape $\alpha$ is an arrangement with shape $\alpha$ of $n$
nonoverlapping dominos, each horizontal or vertical. Such a tableau becomes
standard if each domino is labelled by an integer between 1 and $n$ such that
labels increase across rows and down columns and that every integer between 1
and $n$ occurs exactly once as a label. If instead $\alpha$ is a partition of
$2n+1$, then a domino tableau of shape $\alpha$ is an an arrangement with
shape $\alpha$ of $n$ dominos together with a single square in the upper left
corner. It becomes standard if the dominos are labelled $1,\ldots,n$ obeying
the same rules and the square is labelled 0. The major index maj$(T)$ of a
standard domino tableau $T$ is defined to be the sum of the labels $i$ such
that both squares of the domino labelled $i$ in $T$ lie strictly above both
squares of the domino labelled $i+1$. Denote by SDT$(\alpha)$ the set of
standard domino tableaux of shape $\alpha$ and by
SDT$(\alpha)^{\text{maj}}(q)$ the generating function
$\sum_{T\in\text{SDT}(\alpha)}q^{\text{maj}(T)}$.

######  Theorem 2.

Take $d=2$ and let the partition pair $\lambda=(\lambda^{(1)},\lambda^{(2)})$
correspond as above to the partitions $\rho_{1},\rho_{2}$ of $2n,2n+1$,
respectively. Then we have

$f_{\lambda}=q^{b(\lambda)}\text{SDT}(\rho_{1})^{\text{maj}}(q^{2})=q^{b(\lambda)}\text{SDT}(\rho_{2})^{\text{maj}}(q^{2})$

###### Proof.

We construct bijections $\pi_{C},\pi_{B}$ from the sets of standard domino
tableaux of shapes $\rho_{1},\rho_{2}$, respectively to the set of tableau
pairs of shape $\lambda$ and then modify these to bijections
$\pi_{C}^{\prime},\pi_{B}^{\prime}$ preserving major indices.

First we define $\pi_{C}$. A standard domino tableau $T$ is built from the
empty tableau in stages, at the $i$th of which a domino labelled $i$ is added
to a standard tableau $T_{i-1}$ with $i-1$ dominos to make a new domino
tableau $T_{i}$. Assuming inductively that the pair $(Y_{1},Y_{2})$ of Young
tableaux corresponding to $T_{i-1}$ has already been constructed, we will show
how to add a single cell $c_{i}$ labelled $i$ to one of the $Y_{i}$ to make a
new tableau pair.

Suppose first that the domino $D_{i}$ labelled $i$ in $T_{i}$ is horizontal.

  1. (1)

If $D_{i}$ lies in row $2m$ with its rightmost square in an even column then
$c_{i}$ is added to the (end of the) $m$th row of $Y_{2}$.

  2. (2)

If $D_{i}$ lies in row $2m$ with its rightmost square in an odd column then
$c_{i}$ is added to the $m$th row of $Y_{1}$.

  3. (3)

If $D_{i}$ lies in row $2m+1$ with its its rightmost square in an even column
then $c_{i}$ is added to the $(m+1)$st row in $Y_{1}$.

  4. (4)

If $D_{i}$ lies in row $2m+1$ with its rightmost square in an odd column then
$c_{i}$ is added to the $m$th row of $Y_{2}$ (or the first row, if $m=0$).

Similarly, if instead $D_{i}$ is vertical, then

  1. (1)

If $D_{i}$ lies in an even column $2m$ with its lowest square in an even row,
then $c_{i}$ is added to the $m$th column of $Y_{1}$.

  2. (2)

If $D_{i}$ lies in an even column $2m$ with its lowest square in an odd row,
then $c_{i}$ is added to the $m$th column of $Y_{2}$.

  3. (3)

If $D_{i}$ lies in an odd column $2m+1$ with its lowest square in an even row,
then $c_{i}$ is added to the $(m+1)$st column of $Y_{2}$.

  4. (4)

If $D_{i}$ lies in an odd column $2m+1$ with its lowest square in an odd row,
then $c_{i}$ is added to the $(m+1)$st column of $Y_{1}$,

Next we define $\pi_{B}$, again proceeding inductively. A domino tableau is
constructed as before, but this time starting with a single square labelled 0.
Defining $T_{i-1},T_{i}$ as above and again letting $D_{i}$ be the domino
labelled $i$ in $T_{i}$, assume first that $D_{i}$ is horizontal.

  1. (1)

If $D_{i}$ lies in an even row $2m$ with its rightmost square in an even
column, then $c_{i}$ is added to the $(m+1)$st row of $Y_{1}$.

  2. (2)

If $D_{i}$ lies in an even row $2m$ with its rightmost square in an odd
column, then $c_{i}$ is added to the $m$th row of $Y_{2}$.

  3. (3)

If $D_{i}$ lies in an odd row $2m+1$ with its rightmost square in an even
column, then $c_{i}$ is added to the $m$th row of $Y_{2}$ (or to the first
row, if $m=0$).

  4. (4)

If $D_{i}$ lies in an odd row $2m+1$ with its rightmost square in an odd
column, then $c_{i}$ is added to the $(m+1)$st row of $Y_{1}$.

If instead $D_{i}$ is vertical then

  1. (1)

If $D_{i}$ lies in an even column $2m$ with its lower square in an even row,
then $c_{i}$ is added to the $m$th column of $Y_{1}$.

  2. (2)

If $D_{i}$ lies in an even column $2m$ with its lower square in an odd row,
then $c_{i}$ is added to the $(m+1)$st column of $Y_{2}$.

  3. (3)

If $D_{i}$ lies in an odd column $2m+1$ with its lower square in an even row,
then $c_{i}$ is added to the $m$th column of $Y_{1}$ (or the first column, if
$m=0$).

  4. (4)

If $D_{i}$ lies in an odd column $2m+1$ with its lower square in an odd row,
then $c_{i}$ is added to the $(m+1)$st column of $Y_{2}$.

Let $\rho_{1}$ be a partition of $2n$ whose shape supports a domino tableau.
it is straightforward to check that if $T$ is a standard domino tableau of
this shape, then the image $\pi_{C}(T)$ is a (Young) tableau pair
$(Y_{1},Y_{2})$ such that the respective shapes $\lambda^{(1)},\lambda^{(2)}$
of $Y_{1},Y_{2}$ form a pair corresponding to $\rho_{1}$ by the above recipe.
Similarly if $\rho_{2}$ is a partition of $2n+1$ whose shape supports a domino
tableau and $T$ is a standard domino tableau of this shape, then $\pi_{B}(T)$
is a pair $(Y_{1},Y_{2})$ whose shapes $(\lambda^{(1)},\lambda^{(2)})$
correspond to $\rho_{2}$.

But now the major indices of $\pi_{C}(T),\pi_{B}(T)$ do not generally match
that of $T$. Instead, in type $C,m=\text{maj}(\pi_{C}(T)$ is given by the
following rule: it is the sum of the indices $i$ such that $i$ lies in a
strictly higher row within its tableau than $i+1$, or in the same row of their
tableaux with the column of $i+1$ strictly to the left of that of $i$, or else
$i,i+1$ lie in the same row and column of their tableaux with $i$ in
$Y_{1},i+1$ in $Y_{2}$. Call this last condition $(\ast)$. Running through the
indices $i=1,\ldots,n-1$ in turn, we then produce a new tableau pair
$(Y_{1}^{\prime},Y_{2}^{\prime})$ by flipping the labels $i$ and $i+1$
whenever either the indices $i,i+1$ satisfy $(\ast),i$ lies in $Y_{2}$, and
$i$ in $Y_{1}$, or else $i,i+1$ do not satisfy $(1),i$ lies in $Y_{1}$, and
$i+1$ lies in $Y_{2}$. (One can check that, had the indices $i,i+1$ originally
been in their current positions, then they would have been flipped, so that no
two tableau pairs $(Y_{1},Y_{2})$ can yield the same pair
$(Y_{1}^{\prime},Y_{2}^{\prime})$.) Having run through the indices once, we
then run through them again, flipping pairs of adjacent indices as before,
except that we do not flip a pair of indices that was flipped previously. We
repeat this procedure until we get a pair $(Z_{1},Z_{2})$ of tableaux whose
major index is exactly the sum of the indices contributing to the major index
of $T$, so that maj$(Z_{1},Z_{2})=\text{maj}(T)$. The map sending
$(Y_{1},Y_{2})$ to $(Z_{1},Z_{2})$ is then a bijection. The result follows in
type $C$, setting $\pi_{C}^{\prime}(T)=(Z_{1},Z_{2})$.

For example, if

$(Y_{1},Y_{2})=\bigg{(}\begin{pmatrix}4\\\
6\end{pmatrix}\,,\,\begin{pmatrix}1&3\\\ 2&5\end{pmatrix}\bigg{)}$

then we interchange first the 3 and the 4, then the 5 and the 6, obtaining

$(Y_{1}^{\prime},Y_{2}^{\prime})=\bigg{(}\begin{pmatrix}3\\\
5\end{pmatrix}\,,\,\begin{pmatrix}1&4\\\ 2&6\end{pmatrix}\bigg{)}$

and then we interchange the 4 and 5, obtaining finally

$(Z_{1},Z_{2})=\bigg{(}\begin{pmatrix}3\\\
4\end{pmatrix}\,,\,\begin{pmatrix}1&5\\\ 2&6\end{pmatrix}\bigg{)}$

If

$(Y_{1},Y_{2})=\Bigg{(}\begin{pmatrix}1\\\ 3\\\
4\end{pmatrix}\,,\,\begin{pmatrix}2\end{pmatrix}\Bigg{)}$

then we interchange first the 2 and the 3, then the 3 and the 4, to obtain

$(Z_{1},Z_{2})=\Bigg{(}\,\begin{pmatrix}1\\\ 2\\\
3\end{pmatrix}\,\,,\,\,\begin{pmatrix}4\end{pmatrix}\Bigg{)}$

Similarly, given a pair $(Y_{1},Y_{2})=\pi_{B}(T)$, we now find that
$m=\text{maj}(T)$ is the sum of the indices $i$ such that $i,i+1$ lie in the
same tableau with $i$ strictly higher in this tableau, or $i$ lies in
$Y_{1},i+1$ in $Y_{2}$, with the row of $i$ higher than or equal to that of
$i+1$, or else they lie in the same rows of their respective tableaux with the
column of $i$ weakly to the left of that of $i+1$. Call this last condition
$(\ast\ast)$. Running through the indices $1,\ldots,n-1$ in order, as in type
$C$, we then flip the indices $i$ and $i+1$ whenever either $i,i+1$ satisfy
$(\ast\ast),i$ is in $Y_{2}$, and $i+1$ is in $Y_{1}$, or else $i,i+1$ do not
satisfy $(\ast\ast),i$ lies in $Y_{1}$, and $i+1$ lies in $Y_{2}$. This time
it is only necessary to run through the indices once, obtaining a tableau pair
$(Z_{1},Z_{2})$ whose major index agrees with that of $T$. The map sending
$(Y_{1},Y_{2})$ to $(Z_{1},Z_{2})$ is again a bijection and the result follows
in type $B$, setting $\pi_{B}^{\prime}(T)=(Z_{1},Z_{2})$. ∎

Recall from [L82, L86] that given any irreducible representation $V$ of $W$
there is a unique special representation $S$ occurring in the unique double
cell of $W$ having $V$ as a subrepresentation.

######  Corollary 1.

With notation as above, assume that $\mu=(\mu^{(1)},\mu^{(2)})$ is the
partition pair corresponding to the special representation corresponding to
$V_{\lambda}$. Then the exponents $d_{1},\ldots,d_{r}$ of $q$ in
$f_{\lambda}$, counting multiplicities, are up to a uniform shift a subset of
the corresponding exponents $e_{1},\ldots,e_{s}$ for $V_{\mu}$.

###### Proof.

The exponents $e_{i}$ are up to a uniform shift twice the major indices of the
standard domino tableaux of shape $\rho_{1}$ or $\rho_{2}$, the partition of
$2n$ or $2n+1$ corresponding as above to $\mu$. A standard domino tableau $T$
of shape $\rho_{1}$ or $\rho_{2}$ can be moved through open cycles in the
sense of [G92] to have shape $\rho_{1}^{\prime}$ or $\rho_{2}^{\prime}$, the
partition corresponding to $\lambda$. Moving through open cycles in this way
preserves the $\tau$-invariant of $T$ in the sense of [G92], which determines
its major index. More precisely, the index $i$ lies in the major index if and
only if the difference $e_{i}-e_{i+1}$ of the $i$th and $(i+1)$st unit
coordinate vectors in $\mathbb{C}^{n}$, regarded as a simple root in the
standard root system of type $B_{n}$ or $C_{n}$, lies in the $\tau$-invariant
of $T$. Finally, the $\tau$-invariant of $T$ is an invariant of the Kazhdan-
Lusztig left cell corresponding to $T$; this left cell $L$ is also the left
cell corresponding to a suitable domino tableau of shape $\rho_{1}^{\prime}$
or $\rho_{2}^{\prime}$ [G93]. Hence the major indices of tableaux of shape
$\rho_{1}$ or $\rho_{2}$, counting multiplicities, are also major indices of
some tableau of shape $\rho_{1}^{\prime}$ or $\rho_{2}^{\prime}$. The result
follows. ∎

A weaker version of this result holds in type $D$; there the multiset of
exponents is the union of two submultisets, each of them up to a uniform shift
a subset of multiset of exponents for $\mu$ (but the shifts can be different
for the two submultisets).

##  3\. Type $D$

Let $W^{\prime}$ be the subgroup of $W=C_{2}\wr S_{n}$ generated by coordinate
permutations and evenly many sign changes. Recall that irreducible
representations of $W$ are parametrized by pairs
$((\lambda^{(1)},\lambda^{(2)}),c)$, where $(\lambda^{(1)},\lambda^{(2)})$ is
an unordered pair of partitions with $|\lambda^{(1)}|+|\lambda^{(2)}|=n$ and
$c=1$ if $\lambda^{(1)}\neq\lambda^{(2)}$ while $c=1$ or $2$ if
$\lambda^{(1)}=\lambda^{(2)}$ [Ste89, Remark after Prop. 6.1]. Given an
unordered pair $\lambda=(\lambda^{(1)},\lambda^{(2)})$ with
$\lambda^{(1)}\neq\lambda^{(2)}$, denote by
$\lambda^{\prime},\lambda^{\prime\prime}$ the respective ordered pairs
$(\lambda^{(1)},\lambda^{(2)}),(\lambda^{(2)},\lambda^{(1)})$. Write
SYT’$(\lambda^{\prime}),SYT^{\prime\prime}(\lambda^{\prime\prime})$ for the
respective generating functions $\sum_{T}q^{\text{maj}(T)}$ where the sum now
ranges respectively over standard tableaux $T=(T^{(1)},T^{(2)})$ of shapes
$\lambda^{\prime},\lambda^{\prime\prime}$ such that in both cases such that
the largest label occurs in $T^{(1)}$. Then Stembridge has shown [Ste89, Cor.
6.4] (cf. also [BKS20, Thm. 2.35]) that

######  Theorem 3.

With notation as above the fake degree $f_{\lambda}$ corresponding to
$\lambda$ is given by

$f_{\lambda}(q)=q^{b(\lambda^{\prime})}\text{SYT'}(\lambda^{\prime})+q^{b(\lambda^{\prime\prime})}\text{SYT}(\lambda^{\prime\prime})$

If instead $\lambda=\lambda^{(1)}=\lambda^{(2)}$, then we have

$f_{\lambda}(q)=q^{b(\lambda)}\text{SYT}^{\prime}(\lambda)$

for either of the representations corresponding to $(\lambda,\lambda)$,
summing as above over standard tableaux $(T^{(1)},T^{(2)})$ with $n$ occurring
in $T^{(1)}$ to define SYT’$(\lambda)$.

Alternatively, a simple calculation leads to the following formula. Instead of
summing over standard tableaux $T$ of shape either $\lambda^{\prime}$ or
$\lambda"$, one can sum over standard tableaux of shape $\lambda^{\prime}$
only, attaching the term $q^{b(\lambda^{\prime})+\text{maj}(T)}$ to $T$ if the
largest label $n$ occurs in $T^{(1)}$ and the term
$q^{b(\lambda^{\prime})+\text{maj}(T)-n}$ to $T$. Thus the fake degrees
attached to $\lambda$ in type $D$ are obtained from those in type $C$ attached
to the ordered pair $\lambda^{\prime}$ by subtracting $n$ from some of them.

Now let $\rho^{\prime},\rho^{\prime\prime}$ be the partitions of $2n$
corresponding as above to $\lambda^{\prime},\lambda^{\prime\prime}$. As an
immediate consequence of this theorem and the proof of the preceding one we
get

######  Theorem 4.

With notation as above we have

$f_{\lambda}(q)=q^{b(\lambda)^{\prime}}\text{SDT}^{\prime}(\lambda^{\prime})(q^{2})+q^{b(\lambda^{\prime\prime})}\text{SDT}^{\prime}(\lambda^{\prime\prime})(q^{2})$

where
$\text{SDT}^{\prime}(\lambda^{\prime}),\text{SDT}^{\prime}(\lambda^{\prime\prime})$
denote the generating functions for standard domino tableaux $T$ of the
respective shapes $\rho^{\prime},\rho^{\prime\prime}$, weighted as above by
their major indices, such that in both cases the pair
$(Z_{1},Z_{2})=\pi_{C}^{\prime}(T)$ has the largest label $n$ occurring in
$Z_{1}$. If instead $\lambda=\lambda^{(1)}=\lambda^{(2)}$, then the right side
is replaced by $q^{b(\lambda)}\text{SDT}^{\prime}(\lambda)(q^{2})$, again
defining $\text{SDT}^{\prime}(\lambda)$ by summing over domino tableaux $T$
such that $n$ occurs in the first coordinate $Z_{1}$ of the pair
$\pi_{C}^{\prime}(T)=(Z_{1},Z_{2})$.

For example, take $\lambda=(\lambda^{(1)},\lambda^{(2)})=((1,1),1)$. This pair
corresponds to the partition $(2,2,2)$ of $6$; the complementary pair
$((1),(1,1))$ corresponds to the partition $(2,2,1,1)$. There are three
standard domino tableaux of shape $(2,2,2)$, having major indices $1,2,3$. The
first two of these contribute to the sum in the theorem, leading to the terms
$q^{3},q^{5}$ in $f_{\lambda}$, given the shift by $q$ in this theorem. There
are three standard domino tableaux of shape $(2,2,1,1)$, of which only the one
with major index 1 contributes to $f_{\lambda}$; since the shift is now by
$q^{2}$, we get $f_{\lambda}=q^{3}+q^{4}+q^{5}$. If
$\lambda=(\lambda^{(1)},\lambda^{(2)})=((2),(2))$, then the corresponding
partition is $(4,4)$; of the six standard domino tableaux of this shape, just
three contribute to $f_{\lambda}$ and they have major indices $0,1,2$. Here
$f_{\lambda}=q^{2}+q^{4}+q^{6}$.

In our first example above, where $\lambda^{(1)}=(1,1),\lambda^{(2)}=1$,
applying the alternative formula using pairs of Young tableaux gives the
degrees $d_{i}$ are 3,5, and $7-3=4$. Alternatively, taking the ordered pair
$((1),(1,1))$ we get that the $d_{i}$ are $6-3=3,4$, and $8-3=5$. In the
second example, taking $(\lambda^{(1)},\lambda^{(2)})=((2),(2))$, the $d_{i}$
are $2,4,6,6-4=2,8-4=4,10-4=6$. Cutting all multiplicities in half (in
accordance with Theorem 4), we get that the $e_{i}$ are $2,4,6$.

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