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%references
\def\StanAI{6}
\def\ProcAE{5}
\def\MiRRAD{4}
\def\KratBD{3}
\def\KratBG{2}
\def\FuKrAA{1}

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\document


\subhead Case 5: Self-complementary plane partitions\endsubhead
The enumeration is done in \cite{\StanAI}. There are 3 cases
depending on the parity of the side-lengths of the box (see
\cite{\StanAI, (3a)-(3c)}). Basically, what Stanley shows is that the
number of self-complementary plane partitions in a given box is equal
to the product of two Schur functions, both of rectangular shape,
evaluated at $x_1=x_2=\cdots=1$ (see \cite{\StanAI, Theorem~3.4}).
However, he does actually a weighted enumeration. If this is
specialized to $q$-enumeration, then, roughly, he is able to count
self-complementary plane partitions $\pi$ with respect to
$q^{w(\pi)}$, where $w(\pi)$ is the sum of the entries of the plane
partition $\pi$ (now, when interpreted as planar array according to the
original definition by MacMahon) which are larger or equal to half of
the maximal possible size of an entry (see \cite{\StanAI,
next-to-last paragraph on p.~112} for one of the three cases). 

If I am correct, in your $(-1)$-enumeration you
would exactly consider just this part of the plane partition (now in
the geometric interpretation as heap of unit boxes) which is
larger than half of the height of the box, and count each box in the
upper half by $-1$. Therefore, all we have to do is to set $q=-1$ in
Stanley's formulas (equivalently, $x_i=(-1)^i$ in \cite{\StanAI, 
Theorem~3.4}). The result is, up to some sign, the product of two
hook-content formulas evaluated in the limit $q\to-1$.


\midinsert
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\subhead Case 6: Transpose-complementary plane partitions\endsubhead
In a trans\-pose-comp\-le\-men\-tary plane partition, viewed as rhombus
tiling, there are forced rhombi on the symmetry axis and along two
sides of the hexagon, see
Figure~1 (only one half of a hexagon with side-lengths
$a+1,a+1,2n,a+1,a+1,2n$ is shown, the other half is determined by
symmetry,
the forced rhombi are shaded). The figure shows a particular
tiling. The rhombi which are drawn by bold lines
determine all the others of the tiling.
Then introduce nonintersecting lattice paths in the usual
way (see Figure~2). What you get is a family
$(P_1,P_2,\dots,P_n)$ of nonintersecting lattice paths,
$P_i:(-i+1,i-1)\to (a-i+1,a+i-1)$, which do not cross the diagonal
$y=x-1$
(see Figure~2). 

\vskip10pt
\vbox{
$$
\Einheit.64cm
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\Diagonale(-1,-2)5
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\Pfad(0,0),122211\endPfad
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\Label\u{y=x-1}(-1,-1)
\hskip1.5cm
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\centerline{\eightpoint Figure 2}
}
\vskip10pt
Now, if you are just interested in plain
enumeration, then it is easy to see (see e.g\. \cite{\FuKrAA, Sec.~5,
proof of the symplectic dual Jacobi--Trudi identity}) that
what you want to know is the symplectic character
$\sp_{(a^n)}(x_1^{\pm1},x_2^{\pm1},\dots,x_a^{\pm1})$ evaluated at
$x_i=1$, and this is a special case of the well-known dimension formula
for irreducible representations of the symplectic group.
This is, in the language of rhombus tilings, the argument
of Proctor's original solution of that case.

I know at least two ways of doing the $(-1)$-enumeration (in addition
to your solution in ``Exploration \dots"). Here is the
one which bases on Proctor's observations from plain enumeration.
Clearly, we have to introduce weights. A
possible way of doing that is by saying that a horizontal step
$(u-1,v)\to (u,v)$ has weight $(-1)^{u+v}$. An even better way of
introducing weights (which allows to connect with Proctor's analysis) 
is by saying that a horizontal step
$(u-1,v)\to (u,v)$ has weight $-\sqrt{-1}(-1)^{u+v}$. 
This just amounts to multiplying the total $(-1)$-count by a
constant. (To be precise, the constant is
$(-\sqrt{-1})^{na}$.)

For convenience, let us say that $u+v$ is the {\it label\/} of the
edge $(u-1,v)\to (u,v)$.
In \cite{\FuKrAA, Sec.~5,
proof of the symplectic dual Jacobi--Trudi identity} we explain that
if you give weight $x_{e}$ to an edge with label $2e-1$, and weight
$x_e^{-1}$ to an edge with label $2e$, then the weighted count of all
families of nonintersecting paths that we are considering here equals
exactly the symplectic character from above, 
$\sp_{(a^n)}(x_1^{\pm1},x_2^{\pm1},\dots,x_a^{\pm1})$.

Therefore, if we specialize in this symplectic characters 
$x_i$ to $\sqrt{ -1}$ for all $i$, then, in terms of weights of
horizontal edges, we obtain exactly the above described weights, and
therefore this gives, apart
from the above mentioned constant, exactly the desired
$(-1)$-enumeration.

To get an expression for 
$\sp_{(a^n)}((\sqrt{-1})^{\pm1},(\sqrt{-1})^{\pm1},\dots,(\sqrt{-1})^{\pm1})$,
you can proceed as follows:
First compute 
$\sp_{(a^n)}((q\sqrt{-1})^{\pm1},
(q^2\sqrt{-1})^{\pm1},
\dots,(q^a\sqrt{-1})^{\pm1})$. The Weyl character formula gives
$$\multline 
\frac {\det_{1\le i,j,\le n}(\(q^j\sqrt{-1}\)^{a+n-i+1}-\(q^j\sqrt{-1}\)^{-(a+n-i+1)})} 
{\det_{1\le i,j,\le n}(\(q^j\sqrt{-1}\)^{n-i+1}-\(q^j\sqrt{-1}\)^{-(n-i+1)})}\\
=(\sqrt{-1})^{an}
\frac {\det_{1\le i,j,\le n}(\(q^j\)^{n-i+1}+(-1)^{n-i}\(q^j\)^{-(n-i+1)})} 
{\det_{1\le i,j,\le n}(\(q^j\)^{n-i+1}+(-1)^{n-i}\(q^j\)^{-(n-i+1)})}.
\endmultline$$
To evaluate the determinants make use of the identity
$$\multline 
\det_{1\le i,j,\le n}\(x_j^{n-i+1}+(-1)^{n-i}x_j^{-(n-i+1)}\)\\
=(x_1\cdots x_n)^{-n}\prod
_{1\le i<j\le n} ^{}\big((x_i-x_j)(1+x_ix_j)\big)
\ \prod _{i=1} ^{n}(x_i^2+1).
\endmultline$$
Finally let $q$ tend to 1 in the result.

\medskip
A second way to approach the problem is to introduce nonintersecting
lattice paths which run from south-west to east, and handle the
resulting determinant in the way of the solution for
$(-1)$-enumeration in Case~8 which I describe below.


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\endinsert


\subhead Case 7: Symmetric self-complementary plane partitions\endsubhead
Proctor's approach from \cite{\ProcAE} to plain enumeration, in the
language of rhombus tilings, is as follows. Figure~3 shows just one
``quarter" of a hexagon with side-lengths $a+1,a+1,2n,a+1,a+1,2n$.
The rhombi which are taken on the vertical axis vary. 
The tiling of the quarter determines everything else by symmetry.
Again, introduce nonintersecting lattice paths, see Figure~4. 


\vskip10pt
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$$
\Einheit.64cm
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\Label\u{y=x-1}(-1,-1)
\hskip1.5cm
$$
\centerline{\eightpoint Figure 4}
}
\vskip10pt
In difference to Case~6, now the end
points of the lattice paths may have any position on the line
$x+y=a$. 
But if you
fix end points, then again, by \cite{\FuKrAA, Sec.~5,
proof of the symplectic dual Jacobi--Trudi identity}, the number of
nonintersecting lattice paths is a symplectic character of some
shape, specialized at $x_1=x_2=\cdots=1$.
If $a$ is even, then what you get in total is the sum of all
specialized symplectic characters whose shape is contained in
$((a/2)^n)$. What Proctor shows is that the sum of the {\it
unspecialized\/} symplectic characters equals a Schur function of
rectangular shape.

If $a$ is odd, then he has to play a certain trick, so that
everything reduces to computing a sum of symplectic characters whose
shape is contained in a rectangle, but where in addition all the columns have
even length.

Anyway, it seems that my prediction was wrong in this case.
I do not know a way to build on Proctor's approach. If I try the same
trick as in Case~6, then I would run into trouble because for each
symplectic character the constant (remember $\sqrt{-1}^{an}$ ?) 
would be a different one, depending on the size of the 
shape of the characters.
So, probably one has to look for something else. (On the character
level, there does not seem to be a nice expression for this sum of
symplectic characters multiplied by some power of $\sqrt{-1}$.)


\subhead Case 8: Cyclically
symmetric transpose-complementary plane partitions\endsubhead
The enumeration was solved in \cite{\MiRRAD}. It is very easy to
solve the $(-1)$-enumeration as well, by relying on the crucial
determinant evaluation from \cite{\MiRRAD},
$$\multline
\det_{0\le i,j\le n-1}\(\binom {x+i+j}{2i-j} \)\\
=(-1)^{\chi(n\equiv 3\mod 4)}2^{\binom {n-1}2}\prod _{i=1} ^{n-1}
\frac {\(x+i+1\)_{\fl{(i+1)/2}}\,
\(-x-3n+i+\frac {3} {2}\)_{\fl{i/2}}} {(i)_i}.
\endmultline
\tag1
$$

\midinsert
\vskip15pt
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  \rhombus \rhombus  \rhombus \rhombus \rhombus \rhombus \rhombus
  \rhombus \ldreieck  
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  \rdreieck \rhombus \rhombus \rhombus
  \rhombus 
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%\rtext td:0 (4.3 -2.7){$\sideset {} \and n\to 
%    {\left.\vbox{\vskip1cm}\right\}}$}
%\rtext td:60 (2 -.5){$\sideset {} \and {} \to 
%    {\left.\vbox{\vskip2.5cm}\right\}}$}
\rtext td:120 (-1.2 -.8){$\sideset {} \and {} \to 
    {\left.\vbox{\vskip2.2cm}\right\}}$}
%\rtext td:0 (-2.3 -3.6){$\sideset {c} \and {}\to 
%    {\left\{\vbox{\vskip2.2cm}\right.}$}
%\rtext td:240 (0 -7){$\sideset {} \and {} \to 
%    {\left.\vbox{\vskip1.7cm}\right\}}$}
%\rtext td:300 (3.03 -7.25){$\sideset {} \and {} \to 
%    {\left.\vbox{\vskip1.4cm}\right\}}$}
\htext (-1.7 -.4){$n$}
%\htext (2.8 -.1){$a+1$}
%\htext (3.2 -7.8){$a$}
%\htext (-0.3 -7.65){$b$}
}
\centerline{\eightpoint Figure 5}
}
\vskip10pt
\endinsert


\vskip10pt
\vbox{
$$
\Einheit.64cm
\Gitter(3,8)(-6,-1)
%\Koordinatenachsen(3,8)(-6,-1)
\Kreis(0,0)
\Kreis(-2,1)
\Kreis(-4,2)
\Kreis(-6,3)
\Kreis(1,1)
\Kreis(0,3)
\Kreis(-1,5)
\Kreis(-2,7)
\Kreis(2,-1)
\Pfad(0,0),12\endPfad
\Pfad(-2,1),1221\endPfad
\Pfad(-4,2),222111\endPfad
\Pfad(-6,3),22211211\endPfad
\hskip-2cm
$$
\centerline{\eightpoint Figure 6}
}
\vskip10pt
In terms of rhombus tilings, what Mills, Robbins and Rumsey do is to
introduce nonintersecting lattice paths in the way as depicted in
Figures~5 and 6. (In Figure~5 only a fundamental region of the
hexagon is shown.) What is obtained is a bijection of these rhombus
tilings with families $(P_0,P_1,\dots,P_{n-1})$ of nonintersecting
lattice paths, $P_i$ running from $(-2i+1,i-1)$ to $(-i+1,2i-1)$,
$i=0,1,\dots,n-1$.
The corresponding Gessel--Viennot determinant is the
determinant in (1) with $x=0$. 

If we want to do $(-1)$-enumeration, then we obtain again a
Gessel--Viennot determinant,
only that now paths are counted with weight. An appropriate way to
introduce weights is by saying that a horizontal step at height $h$
contributes weight $(-1)^h$.

Now, to be able to continue, we should figure out what the
$(-1)$-count for paths from one point to another is. 
What we need is to set $q=-1$ in a $q$-binomial coefficient.
As is easy to figure out this is
$$\bmatrix n\\k\endbmatrix_{-1}=\cases 0&\text {if $n$ is even and $k$
is odd}\\
\binom {\fl{n/2}}{\fl{k/2}}&\text {otherwise.}\endcases$$
Therefore, what we want to know is 
$M:=\det_{0\le i,j\le n-1}\(m_{ij}\)$, where
$$m_{ij}=\cases 0&\text {if $i$ and $j$ are odd}\\
\binom {\fl{(i+j)/2}}{i-\cl{j/2}}&\text {otherwise.}\endcases$$

First let $n$ be even. Because of the many entries which are zero
(arranged in a checkerboard fashion), 
the determinant $M$ decomposes, up to some sign, into the
product of the determinant containing all the entries from
even-numbered rows and odd-numbered columns times 
the determinant containing all the entries from
odd-numbered rows and even-numbered columns. 
The first of these two determinants is zero since the first row
consists of zeroes only.

Now let $n$ be odd. In the determinant $M$ the first row consists of a $1$
as the left-most entry and 0's otherwise. Hence, the determinant $M$
is equal to $\det_{1\le i,j\le n-1}\(m_{ij}\)$. Similarly to before,
this determinant decomposes into the product of the determinant
containing all the entries from
even-numbered rows and odd-numbered columns times 
the determinant containing all the entries from
odd-numbered rows and even-numbered columns.
Both of these determinants are special cases of the
Mills--Robbins--Rumsey determinant evaluation (1).

\subhead Case 10: Totally
symmetric self-complementary plane partitions\endsubhead

I was wrong, the $q$-results from \cite{\KratBG} are of no use
because there is a certain extra term in the determinant, which
disappears if $q=1$, but does not disappear is $q=-1$. I expect that
an anlysis similar to the one in \cite{\KratBD} could lead to a
solution. However, that may be hard work.


\Refs

\ref\no \FuKrAA\by M.    Fulmek and C. Krattenthaler \yr 1997 
\paper Lattice path proofs for determinant formulas for symplectic and orthogonal characters
\jour J. Combin\. Theory Ser.~A\vol 77
\pages 3--50\endref

\ref\no \KratBG\by C.    Krattenthaler \yr 1996 
\paper Some $q$-analogues of determinant identities which arose in plane partition enumeration
\jour S\'eminaire Lotharingien Combin\.\vol 36
\pages paper~B36e, 23~pp\endref

\ref\no \KratBD\by C.    Krattenthaler \yr 1997 \paper Determinant 
identities and a generalization of the number of totally symmetric 
self-complementary plane partitions\jour Electron\. J. Combin\.\vol 
4\rm(1)\pages \#R27, 62~pp\endref

\ref\no \MiRRAD\by W. H. Mills, D. H. Robbins and H. Rumsey \yr 1987 
\paper Enumeration of a symmetry class of plane partitions
\jour Discrete Math.\vol 67
\pages 43--55 \endref

\ref\no \ProcAE\by R. A. Proctor \yr 1983 
\paper Shifted plane partitions of trapezoidal shape
\jour Proc\. Amer\. Math\. Soc\.\vol 89
\pages 553--559\endref

\ref\no \StanAI\by R. P. Stanley \yr 1986 
\paper Symmetries of plane partitions
\jour J. Combin. Theory A\vol 43
\pages 103--113 \moreref 
\jour Erratum \vol44 \yr 1987\page310 \endref

\endRefs
\enddocument

