\input amstex
\input amsppt.sty

%\userunningheads

\font\Bf=cmbx12
\font\Rm=cmr12
\def\LL{\leavevmode\setbox0=\hbox{L}\hbox to\wd0{\hss\char'40L}}
\def\al{\alpha}
\def\be{\beta}
\def\ga{\gamma}
\def\de{\delta}
\def\ep{\varepsilon}
\def\ze{\zeta}
\def\et{\eta}
\def\th{\theta}
\def\vt{\vartheta}
\def\io{\iota}
\def\ka{\kappa}
\def\la{\lambda}
\def\rh{\rho}
\def\si{\sigma}
\def\ta{\tau}
\def\ph{\varphi}
\def\ch{\chi}
\def\ps{\psi}
\def\om{\omega}
\def\Ga{\Gamma}
\def\De{\Delta}
\def\Th{\Theta}
\def\La{\Lambda}
\def\Si{\Sigma}
\def\Ph{\Phi}
\def\Ps{\Psi}
\def\Om{\Omega}
\def\row#1#2#3{#1_{#2},\ldots,#1_{#3}}
\def\rowup#1#2#3{#1^{#2},\ldots,#1^{#3}}
\def\x{\times}
\def\crf{}            %used for crossreferencing, Tex should ignore.
\def\rf{}             %used for refencing (section-numbers)
\def\rfnew{}          %used for new-section numbers
\def\P{{\Bbb P}}
\def\R{{\Bbb R}}
\def\X{{\Cal X}}
\def\C{{\Bbb C}}
\def\Mf{{\Cal Mf}}
\def\FM{{\Cal F\Cal M}}
\def\F{{\Cal F}}
\def\G{{\Cal G}}
\def\V{{\Cal V}}
\def\T{{\Cal T}}
\def\A{{\Cal A}}
\def\N{{\Bbb N}}
\def\Z{{\Bbb Z}}
\def\Q{{\Bbb Q}}
\def\ddt{\left.\tfrac \partial{\partial t}\right\vert_0}
\def\dd#1{\tfrac \partial{\partial #1}}
\def\today{\ifcase\month\or
 January\or February\or March\or April\or May\or June\or
 July\or August\or September\or October\or November\or December\fi
 \space\number\day, \number\year}
\def\nmb#1#2{#2} %zum Nummerieren
\def\dfrac#1#2{{\displaystyle{#1\over#2}}}
\def\tfrac#1#2{{\textstyle{#1\over#2}}}
\def\iprod#1#2{\langle#1,#2\rangle}
\def\pder#1#2{\frac{\partial #1}{\partial #2}}
\def\iint{\int\!\!\int}
\def\({\left(}
\def\){\right)}
\def\[{\left[}
\def\]{\right]}
\def\supp{\operatorname{supp}}
\def\Df{\operatorname{Df}}
\def\dom{\operatorname{dom}}
\def\Ker{\operatorname{Ker}}
\def\Tr{\operatorname{Tr}}
\def\Res{\operatorname{Res}}
\def\Aut{\operatorname{Aut}}
\def\kgV{\operatorname{kgV}}
\def\ggT{\operatorname{ggT}}
\def\diam{\operatorname{diam}}
\def\Im{\operatorname{Im}}
\def\Re{\operatorname{Re}}
\def\ord{\operatorname{ord}}
\def\rang{\operatorname{rang}}
\def\rng{\operatorname{rng}}
\def\grd{\operatorname{grd}}
\def\inv{\operatorname{inv}}
\def\maj{\operatorname{maj}}
\def\des{\operatorname{des}}
\def\varmaj{\operatorname{\overline{maj}}}
\def\vardes{\operatorname{\overline{des}}}
\def\pvarmaj{\operatorname{\overline{maj}'}}
\def\pmaj{\operatorname{maj'}}
\def\ln{\operatorname{ln}}
\def\der{\operatorname{der}}
\def\Hom{\operatorname{Hom}}
\def\tr{\operatorname{tr}}
\def\Span{\operatorname{Span}}
\def\grad{\operatorname{grad}}
\def\div{\operatorname{div}}
\def\rot{\operatorname{rot}}
\def\Sp{\operatorname{Sp}}
\def\sgn{\operatorname{sgn}}
\def\liml{\lim\limits}
\def\supl{\sup\limits}
\def\bigcupl{\bigcup\limits}
\def\bigcapl{\bigcap\limits}
\def\limsupl{\limsup\limits}
\def\liminfl{\liminf\limits}
\def\intl{\int\limits}
\def\suml{\sum\limits}
\def\maxl{\max\limits}
\def\minl{\min\limits}
\def\prodl{\prod\limits}
\def\tg{\operatorname{tan}}
\def\ctg{\operatorname{cot}}
\def\arctg{\operatorname{arctan}}
\def\arccot{\operatorname{arccot}}
\def\arcctg{\operatorname{arccot}}
\def\tgh{\operatorname{tanh}}
\def\ctgh{\operatorname{coth}}
\def\arcsinh{\operatorname{arcsinh}}
\def\arccosh{\operatorname{arccosh}}
\def\arctgh{\operatorname{arctanh}}
\def\arcctgh{\operatorname{arccoth}}
\def\3{\ss}
\catcode`\@=11

\font@\fourteenrm=cmr10 scaled\magstep2
\font@\fourteenit=cmti10 scaled\magstep2
\font@\fourteensl=cmsl10 scaled\magstep2
\font@\fourteensmc=cmcsc10 scaled\magstep2
\font@\fourteentt=cmtt10 scaled\magstep2
\font@\fourteenbf=cmbx10 scaled\magstep2
\font@\fourteeni=cmmi10 scaled\magstep2
\font@\fourteensy=cmsy10 scaled\magstep2
\font@\fourteenex=cmex10 scaled\magstep2
\font@\fourteenmsa=msam10 scaled\magstep2
\font@\fourteeneufm=eufm10 scaled\magstep2
\font@\fourteenmsb=msbm10 scaled\magstep2
\newtoks\fourteenpoint@
\def\fourteenpoint{\normalbaselineskip15\p@
 \abovedisplayskip18\p@ plus4.3\p@ minus12.9\p@
 \belowdisplayskip\abovedisplayskip
 \abovedisplayshortskip\z@ plus4.3\p@
 \belowdisplayshortskip10.1\p@ plus4.3\p@ minus5.8\p@
 \textonlyfont@\rm\fourteenrm \textonlyfont@\it\fourteenit
 \textonlyfont@\sl\fourteensl \textonlyfont@\bf\fourteenbf
 \textonlyfont@\smc\fourteensmc \textonlyfont@\tt\fourteentt
%Erg„nzung des fetten Small-Capitals-Fonts:
 \textonlyfont@\bsmc\fourteenbsmc
%
 \ifsyntax@ \def\big##1{{\hbox{$\left##1\right.$}}}%
  \let\Big\big \let\bigg\big \let\Bigg\big
 \else
  \textfont\z@=\fourteenrm  \scriptfont\z@=\twelverm  \scriptscriptfont\z@=\tenrm
  \textfont\@ne=\fourteeni  \scriptfont\@ne=\twelvei  \scriptscriptfont\@ne=\teni
  \textfont\tw@=\fourteensy \scriptfont\tw@=\twelvesy \scriptscriptfont\tw@=\tensy
  \textfont\thr@@=\fourteenex \scriptfont\thr@@=\twelveex
        \scriptscriptfont\thr@@=\twelveex
  \textfont\itfam=\fourteenit \scriptfont\itfam=\twelveit
        \scriptscriptfont\itfam=\twelveit
  \textfont\bffam=\fourteenbf \scriptfont\bffam=\twelvebf
        \scriptscriptfont\bffam=\tenbf
  \setbox\strutbox\hbox{\vrule height12.2\p@ depth5\p@ width\z@}%
  \setbox\strutbox@\hbox{\lower.72\normallineskiplimit\vbox{%
        \kern-\normallineskiplimit\copy\strutbox}}%
 \setbox\z@\vbox{\hbox{$($}\kern\z@}\bigsize@=1.7\ht\z@
 \fi
 \normalbaselines\rm\ex@.2326ex\jot4.3\ex@\the\fourteenpoint@}

\font@\twelverm=cmr10 scaled\magstep1
\font@\twelveit=cmti10 scaled\magstep1
\font@\twelvesl=cmsl10 scaled\magstep1
\font@\twelvesmc=cmcsc10 scaled\magstep1
\font@\twelvett=cmtt10 scaled\magstep1
\font@\twelvebf=cmbx10 scaled\magstep1
\font@\twelvei=cmmi10 scaled\magstep1
\font@\twelvesy=cmsy10 scaled\magstep1
\font@\twelveex=cmex10 scaled\magstep1
\font@\twelvemsa=msam10 scaled\magstep1
\font@\twelveeufm=eufm10 scaled\magstep1
\font@\twelvemsb=msbm10 scaled\magstep1
\newtoks\twelvepoint@
\def\twelvepoint{\normalbaselineskip15\p@
 \abovedisplayskip15\p@ plus3.6\p@ minus10.8\p@
 \belowdisplayskip\abovedisplayskip
 \abovedisplayshortskip\z@ plus3.6\p@
 \belowdisplayshortskip8.4\p@ plus3.6\p@ minus4.8\p@
 \textonlyfont@\rm\twelverm \textonlyfont@\it\twelveit
 \textonlyfont@\sl\twelvesl \textonlyfont@\bf\twelvebf
 \textonlyfont@\smc\twelvesmc \textonlyfont@\tt\twelvett
%Erg„nzung des fetten Small-Capitals-Fonts:
 \textonlyfont@\bsmc\twelvebsmc
%
 \ifsyntax@ \def\big##1{{\hbox{$\left##1\right.$}}}%
  \let\Big\big \let\bigg\big \let\Bigg\big
 \else
  \textfont\z@=\twelverm  \scriptfont\z@=\tenrm  \scriptscriptfont\z@=\sevenrm
  \textfont\@ne=\twelvei  \scriptfont\@ne=\teni  \scriptscriptfont\@ne=\seveni
  \textfont\tw@=\twelvesy \scriptfont\tw@=\tensy \scriptscriptfont\tw@=\sevensy
  \textfont\thr@@=\twelveex \scriptfont\thr@@=\tenex
        \scriptscriptfont\thr@@=\tenex
  \textfont\itfam=\twelveit \scriptfont\itfam=\tenit
        \scriptscriptfont\itfam=\tenit
  \textfont\bffam=\twelvebf \scriptfont\bffam=\tenbf
        \scriptscriptfont\bffam=\sevenbf
  \setbox\strutbox\hbox{\vrule height10.2\p@ depth4.2\p@ width\z@}%
  \setbox\strutbox@\hbox{\lower.6\normallineskiplimit\vbox{%
        \kern-\normallineskiplimit\copy\strutbox}}%
 \setbox\z@\vbox{\hbox{$($}\kern\z@}\bigsize@=1.4\ht\z@
 \fi
 \normalbaselines\rm\ex@.2326ex\jot3.6\ex@\the\twelvepoint@}


\def\dddot#1{\vbox{\ialign{##\crcr
      .\hskip-.5pt.\hskip-.5pt.\crcr\noalign{\kern1.5\p@\nointerlineskip}
      $\hfil\displaystyle{#1}\hfil$\crcr}}}

\newif\iftab@\tab@false
\newif\ifvtab@\vtab@false
\def\tab{\bgroup\tab@true\vtab@false\vst@bfalse\Strich@false%
   \def\\{\global\hline@@false%
     \ifhline@\global\hline@false\global\hline@@true\fi\cr}
   \edef\l@{\the\leftskip}\ialign\bgroup\hskip\l@##\hfil&&##\hfil\cr}
\def\endtab{\cr\egroup\egroup}
\def\vtab{\vtop\bgroup\vst@bfalse\vtab@true\tab@true\Strich@false%
   \bgroup\def\\{\cr}\ialign\bgroup&##\hfil\cr}
\def\endvtab{\cr\egroup\egroup\egroup}
\def\stab{\D@cke0.5pt\null 
 \bgroup\tab@true\vtab@false\vst@bfalse\Strich@true\Let@@\vspace@
 \normalbaselines\offinterlineskip
  \openup\spreadmlines@
 \edef\l@{\the\leftskip}\ialign
 \bgroup\hskip\l@##\hfil&&##\hfil\crcr}
\def\endstab{\crcr\egroup
 \egroup}
\newif\ifvst@b\vst@bfalse
\def\vstab{\D@cke0.5pt\null
 \vtop\bgroup\tab@true\vtab@false\vst@btrue\Strich@true\bgroup\Let@@\vspace@
 \normalbaselines\offinterlineskip
  \openup\spreadmlines@\bgroup}
\def\endvstab{\crcr\egroup\egroup
 \egroup\tab@false\Strich@false}

\newdimen\htstrut@
\htstrut@8.5\p@
\newdimen\htStrut@
\htStrut@12\p@
\newdimen\dpstrut@
\dpstrut@3.5\p@
\newdimen\dpStrut@
\dpStrut@3.5\p@
\def\openup{\afterassignment\@penup\dimen@=}
\def\@penup{\advance\lineskip\dimen@
  \advance\baselineskip\dimen@
  \advance\lineskiplimit\dimen@
  \divide\dimen@ by2
  \advance\htstrut@\dimen@
  \advance\htStrut@\dimen@
  \advance\dpstrut@\dimen@
  \advance\dpStrut@\dimen@}
\def\Let@@{\relax\iffalse{\fi%
    \def\\{\global\hline@@false%
     \ifhline@\global\hline@false\global\hline@@true\fi\cr}%
    \iffalse}\fi}
\def\matrix{\null\,\vcenter\bgroup
 \tab@false\vtab@false\vst@bfalse\Strich@false\Let@@\vspace@
 \normalbaselines\openup\spreadmlines@\ialign
 \bgroup\hfil$\m@th##$\hfil&&\quad\hfil$\m@th##$\hfil\crcr
 \Mathstrut@\crcr\noalign{\kern-\baselineskip}}
\def\endmatrix{\crcr\Mathstrut@\crcr\noalign{\kern-\baselineskip}\egroup
 \egroup\,}
\def\smatrix{\D@cke0.5pt\null\,
 \vcenter\bgroup\tab@false\vtab@false\vst@bfalse\Strich@true\Let@@\vspace@
 \normalbaselines\offinterlineskip
  \openup\spreadmlines@\ialign
 \bgroup\hfil$\m@th##$\hfil&&\quad\hfil$\m@th##$\hfil\crcr}
\def\endsmatrix{\crcr\egroup
 \egroup\,\Strich@false}
\newdimen\D@cke
\def\Dicke#1{\global\D@cke#1}
\newtoks\tabs@\tabs@{&}
\newif\ifStrich@\Strich@false
\newif\iff@rst

\def\Stricherr@{\iftab@\ifvtab@\errmessage{\noexpand\s not allowed
     here. Use \noexpand\vstab!}%
  \else\errmessage{\noexpand\s not allowed here. Use \noexpand\stab!}%
  \fi\else\errmessage{\noexpand\s not allowed
     here. Use \noexpand\smatrix!}\fi}
\def\format{\ifvst@b\else\crcr\fi\egroup\iffalse{\fi\ifnum`}=0 \fi\format@}
\def\format@#1\\{\def\preamble@{#1}%
 \def\Str@chfehlt##1{\ifx##1\s\Stricherr@\fi\ifx##1\\\let\Next\relax%
   \else\let\Next\Str@chfehlt\fi\Next}%
 \def\c{\hfil\noexpand\ifhline@@\hbox{\vrule height\htStrut@%
   depth\dpstrut@ width\z@}\noexpand\fi%
   \ifStrich@\hbox{\vrule height\htstrut@ depth\dpstrut@ width\z@}%
   \fi\iftab@\else$\m@th\fi\the\hashtoks@\iftab@\else$\fi\hfil}%
 \def\r{\hfil\noexpand\ifhline@@\hbox{\vrule height\htStrut@%
   depth\dpstrut@ width\z@}\noexpand\fi%
   \ifStrich@\hbox{\vrule height\htstrut@ depth\dpstrut@ width\z@}%
   \fi\iftab@\else$\m@th\fi\the\hashtoks@\iftab@\else$\fi}%
 \def\l{\noexpand\ifhline@@\hbox{\vrule height\htStrut@%
   depth\dpstrut@ width\z@}\noexpand\fi%
   \ifStrich@\hbox{\vrule height\htstrut@ depth\dpstrut@ width\z@}%
   \fi\iftab@\else$\m@th\fi\the\hashtoks@\iftab@\else$\fi\hfil}%
 \def\s{\ifStrich@\ \the\tabs@\vrule width\D@cke\the\hashtoks@%
          \fi\the\tabs@\ }%
 \def\sa{\ifStrich@\vrule width\D@cke\the\hashtoks@%
            \the\tabs@\ %
            \fi}%
 \def\se{\ifStrich@\ \the\tabs@\vrule width\D@cke\the\hashtoks@\fi}%
 \def\cd{\hfil\noexpand\ifhline@@\hbox{\vrule height\htStrut@%
   depth\dpstrut@ width\z@}\noexpand\fi%
   \ifStrich@\hbox{\vrule height\htstrut@ depth\dpstrut@ width\z@}%
   \fi$\dsize\m@th\the\hashtoks@$\hfil}%
 \def\rd{\hfil\noexpand\ifhline@@\hbox{\vrule height\htStrut@%
   depth\dpstrut@ width\z@}\noexpand\fi%
   \ifStrich@\hbox{\vrule height\htstrut@ depth\dpstrut@ width\z@}%
   \fi$\dsize\m@th\the\hashtoks@$}%
 \def\ld{\noexpand\ifhline@@\hbox{\vrule height\htStrut@%
   depth\dpstrut@ width\z@}\noexpand\fi%
   \ifStrich@\hbox{\vrule height\htstrut@ depth\dpstrut@ width\z@}%
   \fi$\dsize\m@th\the\hashtoks@$\hfil}%
 \ifStrich@\else\Str@chfehlt#1\\\fi%
 \setbox\z@\hbox{\xdef\Preamble@{\preamble@}}\ifnum`{=0 \fi\iffalse}\fi
 \ialign\bgroup\span\Preamble@\crcr}
\newif\ifhline@\hline@false
\newif\ifhline@@\hline@@false
\def\hlinefor#1{\multispan@{\strip@#1 }\leaders\hrule height\D@cke\hfill%
    \global\hline@true\ignorespaces}
\def\Item "#1"{\par\noindent\hangindent2\parindent%
  \hangafter1\setbox0\hbox{\rm#1\enspace}\ifdim\wd0>2\parindent%
  \box0\else\hbox to 2\parindent{\rm#1\hfil}\fi\ignorespaces}
\def\ITEM #1"#2"{\par\noindent\hangafter1\hangindent#1%
  \setbox0\hbox{\rm#2\enspace}\ifdim\wd0>#1%
  \box0\else\hbox to 0pt{\rm#2\hss}\hskip#1\fi\ignorespaces}
\def\item"#1"{\par\noindent\hang%
  \setbox0=\hbox{\rm#1\enspace}\ifdim\wd0>\the\parindent%
  \box0\else\hbox to \parindent{\rm#1\hfil}\enspace\fi\ignorespaces}
\let\plainitem@\item
\catcode`\@=13

\magnification1200

\hsize13cm
\vsize19cm

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%Input-Datei zum Erzeugen von Gitterpunktwegen.%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%entering a path: 
%  \Pfad(x-coordinate of starting point,y-coordinate of starting point),path
%  as 1-2-3-4-word\endPfad
%(1=step in x-direction, 2=step in y-direction, 3=upward diagonal step, 4=downward diagonal step)
%
%entering a dotted path: 
%  \SPfad(x-coordinate of starting point,y-coordinate of starting point),path
%  as 1-2-3-4-word\endSPfad
%(1=step in x-direction, 2=step in y-direction, 3=upward diagonal step, 4=downward diagonal step)
%
%coordinate-axes:
%  \Koordinatenachsen(length of positive x-axes, length of positive y-axes)(length of negative x-axes, length of negative y-axes)
%The length of negative axes are entered as negative numbers.
%
%lattice:
%  \Gitter(number of points in positive x-direction, number of points in positive y-direction)(number of points in negative x-direction, number of points in negative y-direction)
%
%diagonal lines:
%  \Diagonale(x-coordinate of SW-most point,y-coordinate of SW-most point)length of the projection on the x-axes
%
%antidiagonal lines:
%  \AntiDiagonale(x-coordinate of NW-most point,y-coordinate of NW-most point)length of the projection on the x-axes
%
%vectors:
%  \Vektor(x-coordinate of incline, y-coordinate of incline)length(x-coordinate of starting point, y-coordinate of starting point)
%
%labelling of points:
%  \Label[location?]{[label]}(x-coordinate,y-coordinate)
%where:
%  [location?]=\l,\lo,\lu,\r,\ro,\ru,\o,\u
%and l=left, r=right, u=bottom, o=top.
%In addition, if by \Einheit?cm the basic unit is changed, there exist
%\llo,\loo,\llu,\luu,\rro,\roo,\rru,\ruu.
%
%The basic unit can be changed by entering
%  \Einheit=?cm
%The default is \Einheit=0.5cm.
%
%The thickness of the paths can be changed by entering
%     \PfadDicke{?cm}
%The default is \PfadDicke=1pt.
%
%The following point sizes are available:
%\DuennPunkt, \NormalPunkt, \DickPunkt. Syntax:
%  \DickPunkt(x-coordinate,y-coordinate), etc.
%
%Besides, a circle is available by \Kreis. Syntax:
%  \Kreis(x-coordinate,y-coordinate)
%
\catcode`\@=11
\font\tenln    = line10
\font\tenlnw   = linew10

\newskip\Einheit \Einheit=0.5cm
\newcount\xcoord \newcount\ycoord
\newdimen\xdim \newdimen\ydim \newdimen\PfadD@cke \newdimen\Pfadd@cke

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%LaTeX counters, dimensions, variables for lines%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\newcount\@tempcnta
\newcount\@tempcntb

\newdimen\@tempdima
\newdimen\@tempdimb


\newdimen\@wholewidth
\newdimen\@halfwidth

\newcount\@xarg
\newcount\@yarg
\newcount\@yyarg
\newbox\@linechar
\newbox\@tempboxa
\newdimen\@linelen
\newdimen\@clnwd
\newdimen\@clnht

\newif\if@negarg

\def\@whilenoop#1{}
\def\@whiledim#1\do #2{\ifdim #1\relax#2\@iwhiledim{#1\relax#2}\fi}
\def\@iwhiledim#1{\ifdim #1\let\@nextwhile=\@iwhiledim
        \else\let\@nextwhile=\@whilenoop\fi\@nextwhile{#1}}

\def\@whileswnoop#1\fi{}
\def\@whilesw#1\fi#2{#1#2\@iwhilesw{#1#2}\fi\fi}
\def\@iwhilesw#1\fi{#1\let\@nextwhile=\@iwhilesw
         \else\let\@nextwhile=\@whileswnoop\fi\@nextwhile{#1}\fi}


\def\thinlines{\let\@linefnt\tenln \let\@circlefnt\tencirc
  \@wholewidth\fontdimen8\tenln \@halfwidth .5\@wholewidth}
\def\thicklines{\let\@linefnt\tenlnw \let\@circlefnt\tencircw
  \@wholewidth\fontdimen8\tenlnw \@halfwidth .5\@wholewidth}
\thinlines
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\PfadD@cke1pt \Pfadd@cke0.5pt
\def\PfadDicke#1{\PfadD@cke#1 \divide\PfadD@cke by2 \Pfadd@cke\PfadD@cke \multiply\PfadD@cke by2}
\long\def\LOOP#1\REPEAT{\def\BODY{#1}\ITERATE}
\def\ITERATE{\BODY \let\next\ITERATE \else\let\next\relax\fi \next}
\let\REPEAT=\fi
\def\Punkt{\hbox{\raise-2pt\hbox to0pt{\hss$\ssize\bullet$\hss}}}
\def\DuennPunkt(#1,#2){\unskip
  \raise#2 \Einheit\hbox to0pt{\hskip#1 \Einheit
          \raise-2.5pt\hbox to0pt{\hss$\bullet$\hss}\hss}}
\def\NormalPunkt(#1,#2){\unskip
  \raise#2 \Einheit\hbox to0pt{\hskip#1 \Einheit
          \raise-3pt\hbox to0pt{\hss\twelvepoint$\bullet$\hss}\hss}}
\def\DickPunkt(#1,#2){\unskip
  \raise#2 \Einheit\hbox to0pt{\hskip#1 \Einheit
          \raise-4pt\hbox to0pt{\hss\fourteenpoint$\bullet$\hss}\hss}}
\def\Kreis(#1,#2){\unskip
  \raise#2 \Einheit\hbox to0pt{\hskip#1 \Einheit
          \raise-4pt\hbox to0pt{\hss\fourteenpoint$\circ$\hss}\hss}}

%%%%%%%%%%%%%%%%%%%%%
%LaTeX line macros%
%%%%%%%%%%%%%%%%%%%%%
\def\Line@(#1,#2)#3{\@xarg #1\relax \@yarg #2\relax
\@linelen=#3\Einheit
\ifnum\@xarg =0 \@vline
  \else \ifnum\@yarg =0 \@hline \else \@sline\fi
\fi}

\def\@sline{\ifnum\@xarg< 0 \@negargtrue \@xarg -\@xarg \@yyarg -\@yarg
  \else \@negargfalse \@yyarg \@yarg \fi
\ifnum \@yyarg >0 \@tempcnta\@yyarg \else \@tempcnta -\@yyarg \fi
\ifnum\@tempcnta>6 \@badlinearg\@tempcnta0 \fi
\ifnum\@xarg>6 \@badlinearg\@xarg 1 \fi
\setbox\@linechar\hbox{\@linefnt\@getlinechar(\@xarg,\@yyarg)}%
\ifnum \@yarg >0 \let\@upordown\raise \@clnht\z@
   \else\let\@upordown\lower \@clnht \ht\@linechar\fi
\@clnwd=\wd\@linechar
\if@negarg \hskip -\wd\@linechar \def\@tempa{\hskip -2\wd\@linechar}\else
     \let\@tempa\relax \fi
\@whiledim \@clnwd <\@linelen \do
  {\@upordown\@clnht\copy\@linechar
   \@tempa
   \advance\@clnht \ht\@linechar
   \advance\@clnwd \wd\@linechar}%
\advance\@clnht -\ht\@linechar
\advance\@clnwd -\wd\@linechar
\@tempdima\@linelen\advance\@tempdima -\@clnwd
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\def\ZeilAS{26}
\def\WiZeAC{25}
\def\StemAG{24}
\def\StemAE{23}
\def\StanAI{22}
\def\StanAH{21}
\def\RahmAC{20}
\def\MiRRAD{19}
\def\MiRRAC{18}
\def\MiRRAA{17}
\def\KupeAA{16}
\def\KratBH{15}
\def\KratBF{14}
\def\KratBD{13}
\def\KratAM{12}
\def\KratAK{11}
\def\KoorAD{10}
\def\GeViAB{9}
\def\GaRaAA{8}
\def\AnStAA{7}
\def\AnBuAA{6}
\def\AndrAW{5}
\def\AndrAS{4}
\def\AndrAN{3}
\def\AndrAK{2}
\def\AndrAJ{1}

\TagsOnRight

\def\fl#1{\left\lfloor#1\right\rfloor}
\def\cl#1{\left\lceil#1\right\rceil}
\def\DetBb{D_{2}}
\def\DetBbb{\overline D_{2}}
\def\DetBt{\tilde D_{2}}

\topmatter 
\title Some $q$-analogues of determinant identities which arose
in plane partition enumeration
\endtitle 
\author C.~Krattenthaler\footnote"$^\dagger$"{Supported in part by EC's Human
Capital and Mobility Program, grant CHRX-CT93-0400 and the\linebreak 
\hbox{Austrian Science Foundation FWF, grant P10191-MAT}}
\endauthor 
\affil 
Institut f\"ur Mathematik der Universit\"at Wien,\\
Strudlhofgasse 4, A-1090 Wien, Austria.\\
e-mail: KRATT\@Pap.Univie.Ac.At\\
WWW: \tt http://radon.mat.univie.ac.at/People/kratt
\endaffil
\address Institut f\"ur Mathematik der Universit\"at Wien,
Strudlhofgasse 4, A-1090 Wien, Austria.
\endaddress
%\email KRATT\@Pap.Univie.Ac.At
%WWW: \tt http://radon.mat.univie.ac.at/People/kratt\endemail
%\dedicatory \enddedicatory
%\date \enddate
%\thanks \endthanks
\subjclass Primary 15A15;
 Secondary 05A15, 05A17, 05A30, 33D20.
\endsubjclass
\keywords determinant evaluations, 
totally symmetric self-complementary plane partitions, basic hypergeometric
series\endkeywords
\abstract We prove $q$-analogues of two determinant identities of a
previous paper of the author. These determinant identities are
related to the enumeration of totally symmetric self-complementary
plane partitions.
\endabstract
\endtopmatter

\rightheadtext{S\'eminaire Lotharingien Combinatoire, B36e (1996), 23 pp.}
\leftheadtext{S\'eminaire Lotharingien Combinatoire, B36e (1996), 23 pp.}

\document

\subhead 1. Introduction\endsubhead
Enumeration of plane partitions almost always leads to the problem of
evaluating some determinant, see \cite{\AndrAJ, \AndrAK, \AndrAN,
\AndrAS,
\AndrAW, \AnBuAA, \AnStAA,
\GeViAB, \KratAK, \KratAM, \KupeAA, \MiRRAA, \MiRRAC, \MiRRAD, \StanAH, \StanAI,
\StemAE, \StemAG}. In a recent paper \cite{\KratBD}, we evaluated 
three determinants in order to prove a conjecture of
Robbins and Zeilberger \cite{\ZeilAS, Conjecture {\bf C'=B'}} which
generalizes the enumeration of totally symmetric self-complementary
plane partitions. Two of these three determinant evaluations
\cite{\KratBD, Theorems~8 and 10} read as follows,
using the usual notation $(a)_k:=a(a+1)\cdots(a+k-1)$,
$k\ge1$, $(a)_0:=1$, for shifted factorials:

{\it For any nonnegative integer $n$ there hold
$$\multline \det_{0\le i,j\le n-1}\(\frac {(x+y+i+j-1)!}
{(x+2i-j)!\,(y+2j-i)!}\)\\
=\prod _{i=0} ^{n-1}\frac {i!\,(x+y+i-1)!\,(2x+y+2i)_i\,(x+2y+2i)_i}
{(x+2i)!\,(y+2i)!}
\endmultline\tag1.1$$
and}
$$\align 
\det_{0\le i,j\le n-1}&\(
\frac
{(x+y+i+j-1)!\,(y-x+3j-3i)}
{(x+2i-j+1)!\,(y+2j-i+1)!}\)\\
&=\prod _{i=0} ^{n-1}\(\frac {i!\,(x+y+i-1)!\,(2x+y+2i+1)_i\,(x+2y+2i+1)_i}
{(x+2i+1)!\,(y+2i+1)!}\)\\
&\hskip2cm\cdot \sum _{k=0} ^{n}(-1)^k\binom nk (x)_k\,(y)_{n-k}.
\tag1.2\endalign$$

The purpose of this paper is to provide $q$-analogues for these two
determinant evaluations. In the statements of our $q$-analogues we 
use the standard ``$q$-notations" $(a;q)_\infty:=\prod _{i=0}
^{\infty}(1-aq^{i})$ and $(a;q)_\be:=(a;q)_\infty/(aq^\be;q)_\infty$
for shifted $q$-factorials,
so that in particular for any nonnegative integer we have
$(a;q)_n=(1-a)(1-aq)\cdots(1-aq^{n-1})$, and 
$$\bmatrix n\\k\endbmatrix_q:=\frac {(1-q^n)(1-q^{n-1})\cdots
(1-q^{n-k+1})} {(1-q^k)(1-q^{k-1})\cdots (1-q)}$$
for the $q$-binomials.

Our $q$-analogue of (1.1) reads as follows.
\proclaim{Theorem 1}For any nonnegative integer there holds
$$
\multline
\det_{0\le i,j\le n-1} \bigg(
{{(q;q)_{x+y+i+j-1}}\over {(q;q)_{x+2i-j} \,(q;q)_{y+2j-i}}}
{{q^{-2 i j}}\over {(-q^{x+y+1};q)_{i+j}}}
\bigg)\\
=
\prod _{i=0} ^{n-1}q^{-2 i^2} 
{{(q^2;q^2)_i \,(q;q)_{x+y+i-1} \,(q^{2x+y+2i};q)_i \,(q^{x+2y+2i};q)_i}
\over
{(q;q)_{x+2i} \,(q;q)_{y+2i} \,(-q^{x+y+1};q)_{n-1+i}}}
.
\endmultline\tag1.3
$$
\endproclaim
Our $q$-analogue of (1.2) is the following.
\proclaim{Theorem 2}For any nonnegative integer $n$ there holds
$$\multline
\det_{0\le i,j\le n-1} \bigg(
{{(q;q)_{x+y+i+j-1}\,
(1-q^{y+2j-i}-q^{y+2j-i+1}+q^{x+y+i+j+1})}
\over {(q;q)_{x+2i-j+1} \,(q;q)_{y+2j-i+1}}}\\
\hskip5cm \cdot{{q^{-2 i j}}\over {(-q^{x+y+2};q)_{i+j}}}
\bigg)\\
=
\prod _{i=0} ^{n-1}\(q^{-2 i^2} 
{{(q^2;q^2)_i \,(q;q)_{x+y+i-1} \,(q^{2x+y+2i+1};q)_i \,(q^{x+2y+2i+1};q)_i}
\over
{(q;q)_{x+2i+1} \,(q;q)_{y+2i+1} \,(-q^{x+y+2};q)_{n-1+i}}}\)\\
\times
\sum _{k=0} ^{n} (-1)^k q^{n k} {n\brack k}_q q^{y k}\, (q^x;q)_{k}
\,(q^y;q)_{n-k}
.
\endmultline\tag1.4
$$
\endproclaim
We prove the (easier) Theorem~1 in section~2, and subsequently
Theorem~2 in section~3. The method that we use is also applied
successfully in \cite{\KratBD, \KratBF, \KratBH}.

\medskip
The reader should observe that the $q$-analogues (1.3) and (1.4)
when specialized to $q=1$ slightly differ from the identities (1.1)
respectively (1.2) which they generalize in that they contain powers
of 2 on both sides, which however cancel as is easily seen. This fact
makes it unclear what the combinatorial significance of (1.3) or (1.4)
could be, while there is definitely a combinatorial meaning for (1.1) and
(1.2), at least in special cases, see \cite{\KratBD, Theorem~1; \AnBuAA,
sec.~5}.

Identity (1.3) is a generalization of a determinant evaluation
of Andrews and Stanton \cite{\AnStAA, Cor.~3} to which it reduces on
setting $y=1$. The paper \cite{\AnStAA}
contains another generalization \cite{\AnStAA, Theorem~8} which is different
from ours.

As mentioned at the very beginning, there is a third determinant
evaluation in \cite{\KratBD, Theorem~2; cf\. Corollary~3}. However, I
was not able to find a $q$-analogue for this determinant evaluation.
Finding such a $q$-analogue could be a challenging problem.

\subhead 2. Proof of Theorem~1\endsubhead
First we rewrite the statement (1.3). 
We take as many common factors out
of the rows and columns of the determinant in (1.3) as possible, 
such that the
entries become polynomials in $q^x$ and $q^y$. 
To be precise, we take
$$\prod _{i=0} ^{n-1}\frac {(q;q)_{x+y+i-1}}
{(q;q)_{x+2i}\,(q;q)_{y+2n-2}\,(-q^{x+y+1};q)_{n-1+i}}
$$
out of the $i$-the row, $i=0,1,\dots,n-1$, and we take 
$(q^{y+2j+1};q)_{2n-2j-2}$, $j=0,1,\dots,n-1$, out of the $j$-th row.
Thus the determinant
in (1.3) becomes
$$\multline
\prod _{i=0} ^{n-1}\frac {(q;q)_{x+y+i-1}}
{(q;q)_{x+2i}\,(q;q)_{y+2i}\,(-q^{x+y+1};q)_{n-1+i}}\\
\times\det_{0\le i,j\le
n-1}\big(q^{-2ij}\, (q^{x+y+i};q)_{j}\,(q^{x+2i-j+1};q)_j\\
\cdot (q^{y+2j-i+1};q)_{i}
\,(-q^{x+y+i+j+1};q)_{n-j-1}\big).
\endmultline$$
By comparing with (1.3) we see that we have to prove
$$\multline \det_{0\le i,j\le
n-1}\big(q^{-2ij}\, (q^{x+y+i};q)_{j}\,(q^{x+2i-j+1};q)_j\\
\cdot (q^{y+2j-i+1};q)_{i}
\,(-q^{x+y+i+j+1};q)_{n-j-1}\big)
\\
=\prod _{i=0} ^{n-1}\Big(q^{-2 i^2} \,
{{(q^2;q^2)_i \,(q^{2x+y+2i};q)_i \,(q^{x+2y+2i};q)_i}
}\Big)
,
\endmultline$$
or, if we replace $q^x$ by $x$ and $q^y$ by $y$, equivalently 
$$\multline \det_{0\le i,j\le
n-1}\big(q^{-2ij}\,(xyq^{i};q)_{j}\,(xq^{2i-j+1};q)_j\,(yq^{2j-i+1};q)_{i}\,
(-xyq^{i+j+1};q)_{n-j-1}
\big)
\\
=\prod _{i=0} ^{n-1}\Big(q^{-2 i^2} \,
{{(q^2;q^2)_i \,(x^2yq^{2i};q)_i \,(xy^2q^{2i};q)_i}
}\Big)
.
\endmultline\tag2.1$$
For convenience, let us denote the determinant on the left-hand side
of (2.1) by $D_1(x,y;n)$.

Our proof of (2.1) is divided into four steps. In steps~1 and 2 we show 
that the right-hand side of (2.1) divides $D_1(x,y;n)$ as a
polynomial in $x$ and $y$. Then, 
in Step~3 we show that the (total) degree in $x$ and $y$ of $D_1(x,y;n)$
is $6\binom n2$, which is exactly the degree of the right-hand side, 
so that $D_1(x,y;n)$ is a constant multiple of the
right-hand side. Finally we show in step~4 
that this constant equals 1.

\smallskip
{\it Step 1. $\prod _{i=0}
^{n-1}(x^2yq^{2i};q)_{i}$
is a factor of $D_1(x,y;n)$.}
Let us first concentrate on a typical factor
$(1-x^2yq^{2i+l})$, $0\le i\le n-1$, $0\le l<i$, of the product
$\prod _{i=1} ^{n-1}(x^2yq^{2i};q)_{i}$.
We claim that 
for each
such factor there is a linear combination of the rows that vanishes
if the factor vanishes. More precisely, we claim that for any $i,l$
with $0\le i\le n-1$, $0\le l<i$ there holds
$$\multline \hskip-10pt\sum _{s=l} ^{\fl{(i+l)/2}}
{q^{ {{\left(s -l \right)  
             \left( 5s+3l+3  \right) }\over 2}-(i-l)(2i+2l+1)}}
{x^{2 \left( s-i \right) }}
\frac
{(1-q^{2i-3s+l})} {(1-q^{i-s})}
\frac {(q^{i-2s+l+1};q)_{s-l}} {(q;q)_{s-l}}
\\
\cdot \frac
{(xq^{2s+1};q)_{2i-2s}}
{(q^{-2i-l+s}/x;q)_{i-s}}
\frac {({\let \over / -{{{q^{-2 i - l + n + s}}}\over x}}; q) _{i -
s}} {(-q;q)_{i-s}}
\,\cdot\,(\text {row $s$ of
$D_1(x,q^{-2i-l}/x^2;n)$})\\
= (\text {row $i$ of $D_1(x,q^{-2i-l}/x^2;n)$}).
\endmultline\tag2.2$$
Restricting to the $j$-th column, it is seen that this means to
check
$$\multline \hskip-15pt\sum _{s=l} ^{\fl{(i+l)/2}}
{q^{ {{\left(s -l \right)  
             \left( 5s+3l+3  \right) }\over 2}-(i-l)(2i+2l+1)}}
{x^{2 \left( s-i \right) }}
\frac
{(1-q^{2i-3s+l})} {(1-q^{i-s})}
\frac {(q^{i-2s+l+1};q)_{s-l}} {(q;q)_{s-l}}
\\
\cdot \frac
{(xq^{2s+1};q)_{2i-2s}}
{(q^{-2i-l+s}/x;q)_{i-s}}
\frac {({\let \over / -{{{q^{-2 i - l + n + s}}}\over x}}; q) _{i -
s}} {(-q;q)_{i-s}}\,
\cdot q^{-2sj}\,(q^{-2i-l+s}/x;q)_{j}\,(xq^{2s-j+1};q)_{j}\\
\cdot (q^{-2i-l+2j-s+1}/x^2;q)_{s}\,
({\let \over / -{{{q^{1 - 2 i + j - l + s}}}\over x}}; q) _{n-j-1 }
\\
=q^{-2ij}\,(q^{-i-l}/x;q)_{j}\,(xq^{2i-j+1};q)_{j}\\
\times (q^{-3i-l+2j+1}/x^2;q)_{i}\,
(-q^{1 - i + j - l}/x;q)_{n-j-1}.
\endmultline\tag2.3$$
Of course, this identity can be proven routinely by means of the
$q$-version of Zeilberger's algorithm (see the description in
\cite{\KoorAD}; see also \cite{\WiZeAC}). However, it is certainly
more interesting to find which basic hypergeometric identity is ``behind"
(2.3). Mizan Rahman has kindly informed me that it is in fact a
special case of a transformation formula of his \cite{\RahmAC,
(3.12); \GaRaAA, (3.8.13)}. Namely, the
left-hand side of (2.3) can be rewritten in the form
$$\multline 
q^{-2jl}\,
 ({\let \over / {{{q^{1 - 2 i + 2 j - 2 l}}}\over {{x^2}}}}; q)
      _{l}  \,
     ({\let \over / -{{{q^{1 - 2 i + j}}}\over x}}; q) _{-1 + i - j - l + n}
\\
  \times     ({\let \over / {{{q^{-i - l}}}\over x}}; q) _{-i + j + l}  
     ({\let \over / {q^{1 - j + 2 l}} x}; q) _{2 i + j - 2 l} \\
\times \sum _{k=0} ^{\fl{(i-l)/2}}
\frac {(1-{q^{3 k - 2 i + 2 l}})} {(1-{q^{-2 i + 2 l}})}
\frac {{\let \over /  (-{q^{-i + l}};q)_k\, ({q^{2 i - 2 j +
      2 l}} {x^2};q)_k\, ({{{q^{-2 i + j}}}\over x} ;q)_k}} 
{{\let \over /  (q;q)_k\, ({q^{1 - i + l}};q)_k\, (-{{{q^{1 -
      2 i + j}}}\over x} ;q)_k }}\\
\cdot
\frac {{\let \over / ({q^{-i + l}};q^2)_k\, ({q^{1 - i +
      l}} ;q^2)_k}} 
{{\let \over /  ({q^{1 - j + 2 l}} x;q^2)_k\, ({q^{2 - j + 2 l}}
x;q^2)_k}}\,  q^k,
\endmultline$$
and so can be summed by using Lemma~A1 with $n=i-l$,
$b=q^{2i-2j+2l}x^2$, and $c=q^{-2i+j}/x$. Lemma~A1 indeed follows
from the aforementioned transformation formula of Rahman, as is shown
in the Appendix.

This establishes the claim
that the determinant $D_1(x,y;n)$ vanishes if a factor 
$(1-x^2yq^{2i+l})$, $0\le i\le n-1$, $0\le l<i$, vanishes. Since for
equal factors the corresponding linear combinations of the rows are
linearly independent, the complete product $\prod _{i=0}
^{n-1}(x^2yq^{2i};q)_{i}$ must divide
the determinant $D_1(x,y;n)$.

\smallskip
{\it Step 2. $\prod _{i=0}
^{n-1}(xy^2q^{2i};q)_{i}$
is a factor of $D_1(x,y;n)$.}
The reasoning that\linebreak $\prod _{i=0}
^{n-1}(xy^2q^{2i})_{i}$ is a factor of $D_1(x,y;n)$ is
similar. Also here, let us concentrate on a typical factor
$(1-xy^2q^{2j+l})$, $0\le j\le n-1$, $0\le l<j$. This time
we claim that for each
such factor there is a linear combination of the columns that vanishes
if the factor vanishes. More precisely, we claim that for any $j,l$
with $0\le j\le n-1$, $0\le l<j$ there holds
$$\multline \sum _{s=l} ^{\fl{(j+l)/2}}
{q^{{{\left( s-l \right)  \left( 5s+3l+3 \right) }\over 2}
-(j-l)(2j+2l+1)}}
{y^{2(s- j) }}
\frac
{(1-q^{2j-3s+l})} {(1-q^{j-s})}\frac {(q^{j-2s+l+1};q)_{s-l}}
{(q;q)_{s-l}}\\
\cdot\frac {(yq^{2s+1};q)_{2j-2s}} {(-q;q)_{j-s}}\,\cdot(\text {column $s$ of
$D_1(q^{-2j-l}/y^2,y;n)$})\\
= (\text {column $j$ of $D_1(q^{-2j-l}/y^2,y;n)$}).
\endmultline$$
Restricting to the $i$-th row, we see that this means to check
$$\multline \sum _{s=l} ^{\fl{(j+l)/2}}
{q^{{{\left( s-l \right)  \left( 5s+3l+3 \right) }\over 2}
-(j-l)(2j+2l+1)}}
{y^{2(s- j) }}
\frac
{(1-q^{2j-3s+l})} {(1-q^{j-s})}\frac {(q^{j-2s+l+1};q)_{s-l}}
{(q;q)_{s-l}}\\
\cdot \frac {(yq^{2s+1};q)_{2j-2s}} {(-q;q)_{j-s}}\,\cdot
q^{-2is}\,(q^{-2j-l+i}/y;q)_{s}\,(q^{-2j-l+2i-s+1}/y^2;q)_{s}
\hskip1cm\\
\cdot
(yq^{2s-i+1};q)_{i}
({\let \over / -{{{q^{1 + i - 2 j - l + s}}}\over y}}; q) _{n-s-1}\\
=
 q^{-2ij}\,
(q^{-2j-l+i}/y;q)_{j}\,(q^{-3j-l+2i+1}/y^2;q)_{j}\\
\cdot (yq^{2j-i+1};q)_{i}\,
  ({\let \over / -{{{q^{1 + i - j - l}}}\over y}}; q) _{n-j-1} .
\endmultline$$
The observation that this summation is equivalent to (2.3) with
$x$ replaced by $y$ and with $i$ and $j$ interchanged
establishes the claim. Similarly to as before, this 
shows that the complete product $\prod _{i=0}
^{n-1}(xy^2q^{2i};q)_{i}$ divides $D_1(x,y;n)$. 

Altogether, this implies that $\prod _{i=0}
^{n-1}\big((x^2yq^{2i};q)_{i}\,
(xy^2q^{2i};q)_{i}\big)$, and hence the
right-hand side of (2.1), divides $D_1(x,y;n)$, as desired.


\smallskip
{\it Step 3. $D_1(x,y;n)$ is a polynomial in $x$ and $y$ of degree
$6\binom n2$.} This is because each term in the defining expansion
of the determinant $D_1(x,y;n)$ (the determinant on the left-hand side of
(2.1)) has degree $6\binom n2$ as a
polynomial in $x$ and $y$. Since the right-hand side of (2.1), which
by steps~1 and 2 divides $D_1(x,y;n)$ as a polynomial in $x$ and $y$,
also has degree $6\binom n2$, $D_1(x,y;n)$ and the right-hand side of
(2.1) differ only by a multiplicative constant.

{\it Step 4. 
The evaluation of the multiplicative constant.}
To show that the multiplicative constant, which according to step~3 
is between $D_1(x,y;n)$ (the left-hand side of (2.1)) 
and the right-hand side of (2.1), is indeed
1, we compare the constant coefficient on both sides of (2.1).

The constant term of $D_1(x,y;n)$
equals $\det_{0\le i,j,\le n-1}(q^{-2ij})$. This is a
Vandermonde determinant and hence equals 
$$\prod _{1\le i<j\le n} ^{}(q^{-2j}-q^{-2i})=q^{\sum _{j=1}
^{n}(-2j^2)}\prod _{1\le i<j\le n} ^{}(1-q^{2j-2i})=
q^{\sum _{j=1} ^{n}(-2j^2)}\prod _{i=0} ^{n-1}(q^2;q^2)_i.$$
This is exactly the constant term of the right-hand side of (2.1). 
So indeed, the left-hand and right-hand side of (2.1) are equal,
which completes the proof of the Theorem.\quad \quad \qed



\subhead 3. Proof of Theorem~2\endsubhead
Proving Theorem~2 is more difficult. The reader may take the fact
that the
determinant in (1.4) does not factor completely into ``cyclotomic"
factors (unlike the determinant in (1.3)) as an indication why this
is the case. 

We begin by manipulating the determinant on the left-hand side of (1.4), 
quite analogously 
as at the beginning of the proof of (1.3). Namely, we
take 
$$\prod _{i=0} ^{n-1}\frac {(q;q)_{x+y+i-1}}
{(q;q)_{x+2i+1}\,(q;q)_{y+2n-1}\,
(-q^{x+y+2};q)_{n-1+i}}$$
out of the $i$-th row, $i=0,1,\dots,n-1$, and we take
$(q^{y+2j+2};q)_{2n-2j-2}$ out of the $j$-th column, $j=0,1,\dots,n-1$.
Thus the determinant in (1.4) becomes
$$\multline \prod _{i=0} ^{n-1}\frac {(q;q)_{x+y+i-1}}
{(q;q)_{x+2i+1}\,(q;q)_{y+2i+1}\,(-q^{x+y+2};q)_{n-1+i}}\\
\times\det_{0\le i,j\le
n-1}\big(q^{-2ij}\,(q^{x+y+i};q)_{j}\,(q^{x+2i-j+2};q)_j\,
(q^{y+2j-i+2};q)_{i}\,(-q^{x+y+i+j+2};q)_{n-j-1}\\
\cdot(1-q^{y+2j-i}-q^{y+2j-i+1}+q^{x+y+i+j+1})\big).
\endmultline$$
By comparing with (1.4), and replacing $q^x$ by $x$ and $q^y$ by $y$,
we see that Theorem~2
is equivalent to the statement: 
$$\multline \det_{0\le i,j\le
n-1}\big(q^{-2ij}\,(xyq^i;q)_{j}\,(xq^{2i-j+2};q)_j\,
(yq^{2j-i+2};q)_{i}\,(-xyq^{i+j+2};q)_{n-j-1}\\
\cdot(1-yq^{2j-i}-yq^{2j-i+1}+xyq^{i+j+1})\big)\\
=\prod _{i=0} ^{n-1}\(q^{-2 i^2} \,
{{(q^2;q^2)_i \,(x^2yq^{2i+1};q)_i \,(xy^2q^{2i+1};q)_i}
}\)\\
\times\sum _{k=0} ^{n} (-1)^k q^{n k} {n\brack k}_q {y^k} \,(x;q)_{k}
\,(y;q)_{n-k}
.
\endmultline\tag3.1$$
For convenience, let us denote the determinant in (3.1) by
$D_2(x,y;n)$.

In order to be able to finally prove (3.1), we have to go through a
sequence of three Lemmas. As a first approximation, we identify most 
of the factors of $D_2(x,y;n)$.
\proclaim{Lemma 1}For any nonnegative integer $n$ there holds
$$\multline D_2(x,y;n)=\det_{0\le i,j\le
n-1}\big(q^{-2ij}\,(xyq^i;q)_{j}\,(xq^{2i-j+2};q)_j\,
(yq^{2j-i+2};q)_{i}\\
\hskip3cm\cdot(-xyq^{i+j+2};q)_{n-j-1}\,
(1-yq^{2j-i}-yq^{2j-i+1}+xyq^{i+j+1})\big)\\
=\prod _{i=0} ^{n-1}
\Big((x^2yq^{2i+1};q)_i \,(xy^2q^{2i+1};q)_i\Big)
\cdot P(x,y;n),
\endmultline\tag3.2
$$
where $P(x,y;n)$ is a polynomial in $x$ and $y$, of degree
$n$ in $x$, and also of degree $n$ in $y$.
\endproclaim
\demo{Proof} What we have to prove is that
$$\prod _{i=0} ^{n-1}
\Big((x^2yq^{2i+1};q)_i \,(xy^2q^{2i+1};q)_i\Big)
\tag3.3$$
divides $D_2(x,y;n)$ as a polynomial in $x$ and $y$. Once this is
done, it follows immediately that the
remaining factor $P(x,y;n)$ then must have degree $n$ in $x$ and also
in $y$. For, in the expansion
of the determinant $D_2(x,y;n)$ each term has degree $3\binom n2+n$
in $x$, and the same holds for the degree in $y$. On the other hand,
the degree in $x$ of the product (3.3) is $3\binom n2$, the same
being true for the degree in $y$. Therefore $P(x,y;n)$ must be a
polynomial with degree $n$ in $x$ and degree $n$ in $y$.

In order to show that indeed the product (3.3) divides $D_2(x,y;n)$, 
we first consider just one half of this product, $\prod _{i=0} ^{n-1}
(x^2yq^{2i+1};q)_i$.
Let us first concentrate on a typical factor
$(1-x^2yq^{2i+l+1})$, $0\le i\le n-1$, $0\le l<i$.
Analogously to the proof of Theorem~1, we claim that for each
such factor there is a linear combination of the rows that vanishes
if the factor vanishes. More precisely, we claim that for any $i,l$
with $0\le i\le n-1$, $0\le l<i$ there holds
$$\multline \hskip-10pt\sum _{s=l} ^{\fl{(i+l)/2}}
{q^{ {{\left(s -l \right)  
             \left( 5s+3l+7  \right) }\over 2}-(i-l)(2i+2l+3)}}
{x^{2 \left( s-i \right) }}
\frac
{(1-q^{2i-3s+l})} {(1-q^{i-s})}
\frac {(q^{i-2s+l+1};q)_{s-l}} {(q;q)_{s-l}}
\\
\cdot \frac
{(xq^{2s+2};q)_{2i-2s}}
{(q^{-2i-l+s-1}/x;q)_{i-s}}
\frac {({\let \over / -{{{q^{-2 i - l + n + s}}}\over x}}; q) _{i -
s}} {(-q;q)_{i-s}}
\,\cdot\,(\text {row $s$ of
$D_2(x,q^{-2i-l-1}/x^2;n)$})\\
=
(\text {row $i$ of $D_2(x,q^{-2i-l-1}/x^2;n)$}).
\endmultline\tag3.4$$
Restricting (3.4) to the $j$-th column, it is seen that this means to
check
$$\multline \hskip-10pt\sum _{s=l} ^{\fl{(i+l)/2}}
{q^{ {{\left(s -l \right)  
             \left( 5s+3l+7  \right) }\over 2}-(i-l)(2i+2l+3)}}
{x^{2 \left( s-i \right) }}
\frac
{(1-q^{2i-3s+l})} {(1-q^{i-s})}
\frac {(q^{i-2s+l+1};q)_{s-l}} {(q;q)_{s-l}}
\\
\cdot \frac
{(xq^{2s+2};q)_{2i-2s}}
{(q^{-2i-l+s-1}/x;q)_{i-s}}
\frac {({\let \over / -{{{q^{-2 i - l + n + s}}}\over x}}; q) _{i -
s}} {(-q;q)_{i-s}}
\,\cdot\,q^{-2sj}\,(q^{-2i-l+s-1}/x;q)_{j}\,(xq^{2s-j+2};q)_{j}\\
\cdot(q^{-2i-l+2j-s+1}/x^2;q)_{s}\,(-q^{s+j+1-2i-l}/x;q)_{n-j-1}\\
\cdot\left(1-\frac {q^{2j-s-2i-l-1}} {x^2}-\frac {q^{2j-s-2i-l}} {x^2}+
\frac {q^{s+j-2i-l}} {x}\right)\\
=q^{-2ij} \, ({\let \over / {{{q^{ - i - l-1}}}\over x}}; q) _{j} 
\,   ({\let \over / {q^{2 i - j+2}} x}; q) _{j} \,
 ({\let \over / {{{q^{- 3 i + 2 j - l+1}}}\over {{x^2}}}};
      q) _{i} \hskip3cm\\  
\times({\let \over / -{{{q^{1 - i + j - l}}}\over x}}; q) _{n-j-1}  
\left( 1 - {{{q^{- 3 i + 2 j - l -1}}}\over {{x^2}}} - 
       {{{q^{-3 i + 2 j - l}}}\over {{x^2}}} + {{{q^{-i + j - l}}}\over x}
        \right).
\endmultline\tag3.5$$
We may rewrite the left-hand side sum as
$$\multline -\frac {{q^{2 {l^2}-2i^2 -2jl +2j - 5 i + l -1}}{x^{2l-2i-2 }}}
{{   {({\let \over / -q}; q) _{i - l-1} }}}
 ({\let \over / {{{q^{1 - 2 i + 2 j - 2 l}}}\over {{x^2}}}};
      q) _{l}  \, ({\let \over / -{{{q^{1 - 2 i + j}}}\over x}}; q)
_{n + i - j - l -1} \\
\times ({\let \over / {{{q^{-1 - i - l}}}\over x}}; q) _{-i + j +
      l} \, ({\let \over / {q^{2 - j + 2 l}} x}; q) _{2 i + j - 2 l} 
  \\
\times \sum _{k=0} ^{\fl{(i-l)/2}} 
(1 + {q} -
       {q^{1 - j + 2 l + 2 k}} x - {q^{1 + 2 i + 2 l -2j+ k}} {x^2})
\frac {{\let \over / (1-{q^{3k - 2 i + 2 l}})}} 
   {{\let \over / (1-{q^{-2 i + 2 l}})}}\hskip2cm
\\  \cdot
\frac {{\let \over / (-{q^{-i + l}};q)_k\, 
   ({q^{2 i - 2 j + 2 l}} {x^2};q)_k\, ({{{q^{-1 - 2 i +
      j}}}\over x};q)_k}} 
   {{\let \over / (-{{{q^{1 -
      2 i + j}}}\over x};q)_k\, ({q^{1 - i + l}};q)_k\,  (q;q)_k}}
\frac {{\let \over / ({q^{-i + l}};q^2)_k\, ({q^{1 - i + l}};q^2)_k}} 
   {{\let \over / ({q^{2 - j + 2 l}} x;q^2)_k\, ({q^{3 - j + 2 l}}
x;q^2)_k}}q^{2k}.
\endmultline$$
The series can be summed by Lemma~A2 with $n=i-l$ and $B=xq^{2l-j}$.
After some manipulation one arrives at the right-hand side of (3.5).

This establishes the claim
that the determinant $D_2(x,y;n)$ vanishes if a factor 
$(1-x^2yq^{2i+l+1})$, $0\le i\le n-1$, $0\le l<i$, vanishes.
Again, since for
equal factors the corresponding linear combinations of the rows are
linearly independent, the complete product $\prod _{i=0}
^{n-1}(x^2yq^{2i+1};q)_{i}$ divides $D_2(x,y;n)$.


\smallskip

The reasoning that $\prod _{i=0}
^{n-1}(xy^2q^{2i+1};q)_{i}$ is a factor of $D_2(x,y;n)$ is
similar. Also here, let us concentrate on a typical factor
$(1-xy^2q^{2j+l+1})$, $0\le j\le n-1$, $0\le l<j$. This time
we claim that for each
such factor there is a linear combination of the columns that vanishes
if the factor vanishes. More precisely, we claim that for any $j,l$
with $0\le j\le n-1$, $0\le l<j$ there holds
$$\multline \sum _{s=l} ^{\fl{(j+l)/2}}
{q^{{\let\over/ {{\left( s-l \right)  \left( 5s+3l+3 \right) }\over 2}-
  \left( \left( j - l \right)  \left( 2 j + 2 l + 1\right)  \right) 
           }}}{y^{2 \left( s-j  \right) }}
\frac {\left( 1 - {q^{2 j + l - 3 s}} \right)  } 
      {\left( 1 - {q^{j - s}} \right)}
\frac {{\let \over / ({\let \over / {q^{1 + j + l - 2 s }}}; q) _{s-l}}} 
      {{\let \over /  ({\let \over / -q}; q) _{j - s}  \,
     ({\let \over / q}; q) _{s-l}}}\\
\cdot({\let \over / {yq^{ 2 s+2}} }; q) _{2 j - 2 s}
\,\cdot\,(\text {column $s$ of
$D_2(q^{-2j-l-1}/y^2,y;n)$})\\
=(\text {column $j$ of $D_2(q^{-2j-l-1}/y^2,y;n)$}).
\endmultline\tag3.6$$
Restricting to the $i$-th row, we see that this means to check
$$\multline \sum _{s=l} ^{\fl{(j+l)/2}}
{q^{{\let\over/ {{\left( s-l \right)  \left( 5s+3l+3 \right) }\over 2}-
  \left( \left( j - l \right)  \left( 2 j + 2 l +1 \right)  \right) 
           }}}{y^{2 \left( s-j  \right) }}
\frac {\left( 1 - {q^{2 j + l - 3 s}} \right)  } 
      {\left( 1 - {q^{j - s}} \right)}
\frac {{\let \over / ({\let \over / {q^{1 + j + l - 2 s }}}; q) _{s-l}}} 
      {{\let \over /  ({\let \over / -q}; q) _{j - s}  \,
     ({\let \over / q}; q) _{s-l}}}\\
\cdot({\let \over / {yq^{ 2 s+2}} }; q) _{2 j - 2 s}
\,\cdot\,
q^{-2is}     ({\let \over / {{{q^{-1 + i - 2 j - l}}}\over y}}; q) _{s}  \,
({\let \over / {{{q^{1 + 2 i - 2 j - l - s}}}\over {{y^2}}}}; q) _{s}
\\
\cdot
    ({\let \over / {yq^{2s - i + 2 }} }; q) _{i} \,
     ({\let \over / -{{{q^{1 + i - 2 j - l + s}}}\over y}}; q) 
_{n -s-1}\,
\left( 1  - y{q^{2s-i}}  - 
       y{q^{2s - i +1}}  + {{{q^{i - 2 j - l + s}}}\over y}\right)\\
={{q^{-2 i j}}} \,
  ({\let \over / {{{q^{-1 + i - 2 j - l}}}\over y}}; q) _{j} 
     ({\let \over / {{{q^{1 + 2 i - 3 j - l}}}\over {{y^2}}}}; q) _{j} \,
     ({\let \over / y{q^{2j - i + 2}} }; q) _{i}\\
\times
 ({\let \over / -{{{q^{1 + i - j - l}}}\over y}}; q)
_{n-j-1}\,\left( 1  - y{q^{2 j-i}}  - 
       y{q^{2j - i + 1}}  + {{{q^{i - j - l}}}\over y}\right) .
\endmultline\tag3.7$$
Again, we rewrite the left-hand side series,
$$\multline -\frac {{q^{1+2j-i - 2 i j }} y}
    {{ ({\let \over / -q}; q) _{j - l-1} }}
 ({\let \over / {{{q^{1 + 2 i - 2 j - 2 l}}}\over
      {{y^2}}}}; q) _{l} \,({\let \over / {{{q^{-1 - 2 j}}}\over y}}; q) _{i
      + 2 j - 2 l} \,({\let \over / -{{{q^{1 + i - 2 j}}}\over y}}; q)
   _{n-l-1} \\
\times({\let \over / {{{q^{-1 + i - 2 j - l}}}\over y}}; q) _{l-i }
       \,({\let \over / {q^{2 - i + 2 j}} y}; q) _{i} \\
\times\sum _{k=0} ^{\fl{(j-l)/2}}
\left(1+\frac {1} {q}-
\frac {q^{2i-2j-2l - k-1}} {y^2} - \frac {q^{i -2l-2k-1}} y
\right)
\frac {{\let\over/ (1-{q^{-3k +2j-2l}})}} 
      {{\let\over/ (1-q^{2j-2l})}}\hskip2cm\\
\cdot\frac {{\let\over/ (-{q^{j - l}};q^{-1})_k\, 
 ({q^{2 i - 2 j - 2 l}}/ {y^2};q^{-1})_k \,
({{{q^{2j- i +  1}}} y};q^{-1})_k}} 
      {{\let\over/ (-{{{q^{-1 - i +
      2 j}}} y};q^{-1})_k\, ({q^{-1 + j - l}};q^{-1})_k \,
      (q^{-1};q^{-1})_k}}\\
\cdot\frac {{\let\over/ ({q^{j - l}};q^{-2})_k\, ({q^{-1 +
      j - l}};q^{-2})_k }} 
      {{\let\over/ ({q^{-2 + i - 2 l}}/ y;q^{-2})_k\, ({q^{-3
+ i -  2 l}}/ y;q^{-2})_k }} q^{-2k}.
\endmultline$$
The series can be summed by Lemma~A2 with $q$ replaced by $1/q$,
$n=j-l$, $B=q^{-2l+i}/y$. Similarly to as before, this eventually
shows that the complete product $\prod _{i=0}
^{n-1}(xy^2q^{2i+1};q)_{i}$ divides $D_2(x,y;n)$. 

Altogether, this implies that $\prod _{i=0}
^{n-1}\big((x^2yq^{2i+1};q)_{i}\,
(xy^2q^{2i+1};q)_{i}\big)$ divides\linebreak $D_2(x,y;n)$,
as desired. This completes the proof of Lemma~1.
\quad \quad \qed


\enddemo

Next, we locates several zeros of the polynomial factor
$P(x,y;n)$ of $D_2(x,y;n)$ (recall (3.2) for the definition of
$P(x,y;n)$ and $D_2(x,y;n)$).
\proclaim{Lemma 2}If $u,v$ are nonnegative integers with $u+v\le
n-1$, then\linebreak 
$P(q^{-u},q^{-v};n)=0$, with $P(x,y;n)$ the polynomial in
{\rm(3.2)}.
\endproclaim
\demo{Proof}Let $u,v$ be nonnegative integers with $u+v\le n-1$. 
The polynomial
$P(x,y;n)$ is defined by (3.2), 
$$D_2(x,y;n)
=\prod _{i=0} ^{n-1}
\Big((x^2yq^{2i+1};q)_i \,(xy^2q^{2i+1};q)_i\Big)
\cdot P(x,y;n),
\tag3.8
$$
where $D_2(x,y;n)$ is the determinant in (3.1), respectively (3.2). What 
we would like to do is to set $x=q^{-u}$ and $y=q^{-v}$ in (3.8), prove that
$D_2(q^{-u},q^{-v};n)$ equals 0, that the product on the right-hand side of
(3.8) is nonzero, and conclude that therefore $P(q^{-u},q^{-v};n)$ must
be 0. However, the product on the right-hand side of (3.8)
unfortunately (usually) {\it is\/} 0 for $x=q^{-u}$ and $y=q^{-v}$. 
Therefore we have to find a way around this difficulty.

To begin with, we set $y=q^{-v}$ in (3.8). Before setting $x=q^{-u}$, we
have to cancel all factors of the form $(1-xq^u)$ that occur in the
product on the right-hand side of (3.8).
To accomplish this, we have to ``generate" these factors on the
left-hand side. This is done by reading through the proof
of Lemma~1 with $y=q^{-v}$. To make this more precise, observe that
$(1-xq^u)$ divides a typical factor $1-x^2q^{-v+2i+l+1}$, $0\le i\le n-1$, $0\le
l<i$, of the first half of the product in (3.8) if and only if
$2u=-v+2i+l+1$. Therefore, if we recall (3.4), for each solution $(i,l)$ of
$$2u=-v+2i+l+1,\quad  \text {with }0\le i\le n-1,\ 0\le l<i,\tag3.9$$
we subtract the linear combination
$$\multline \hskip-10pt\sum _{s=l} ^{\fl{(i+l)/2}}
{q^{ {{\left(s -l \right)  
             \left( 5s+3l+7  \right) }\over 2}-(i-l)(2i+2l+3)}}
{x^{2 \left( s-i \right) }}
\frac
{(1-q^{2i-3s+l})} {(1-q^{i-s})}
\frac {(q^{i-2s+l+1};q)_{s-l}} {(q;q)_{s-l}}
\\
\cdot \frac
{(xq^{2s+2};q)_{2i-2s}}
{(q^{-2i-l+s-1}/x;q)_{i-s}}
\frac {({\let \over / -{{{q^{-2 i - l + n + s}}}}/x}; q) _{i -
s}} {(-q;q)_{i-s}}
\,\cdot\,(\text {row $s$ of
$D_2(x,q^{-v};n)$})\\
\endmultline\tag3.10$$
of rows of $D_2(x,q^{-v};n)$ from row $i$ of $D_2(x,q^{-v};n)$. Let
us denote the resulting determinant by $\DetBt(x,q^{-v};n)$.
By (3.4), the
effect is that $(1-x^2q^{-v+2i+l+1})=(1-x^2q^{2u})$ (the equality being due to
(3.9)), is a factor of each
entry of the $i$-th row of $\DetBt(x,q^{-v};n)$,  
for each solution $(i,l)$
of (3.9), in particular
$(1-xq^u)$ is a factor of each
entry of the $i$-th row of $\DetBt(x,q^{-v};n)$.
For later use we record that the $(i,j)$-entry of
$\DetBt(x,q^{-v};n)$, $(i,l)$ a solution of (3.9), reads
$$\multline q^{-2ij}\,(xq^{-v+i};q)_j\,
(xq^{2i-j+2};q)_j\,(q^{-v+2j-i+2};q)_i\, (-xq^{-v+i+j+2};q)_{n-j-1}\\
\hskip4cm\times\(1-q^{-v+2j-i}-q^{-v+2j-i+1}+xq^{-v+i+j+1}\)\\
-\sum _{s=l} ^{\fl{(i+l)/2}}
{q^{ {{\left(s -l \right)  
             \left( 5s+3l+7  \right) }\over 2}-(i-l)(2i+2l+3)}}
{x^{2 \left( s-i \right) }}
\frac
{(1-q^{2i-3s+l})} {(1-q^{i-s})}
\frac {(q^{i-2s+l+1};q)_{s-l}} {(q;q)_{s-l}}
\hskip1cm\\
\cdot \frac
{(xq^{2s+2};q)_{2i-2s}}
{(q^{-2i-l+s-1}/x;q)_{i-s}}
\frac {({\let \over / -{{{q^{-2 i - l + n + s}}}}/x}; q) _{i -
s}} {(-q;q)_{i-s}}\\
\,\cdot\,
q^{-2sj}\,(xq^{-v+s};q)_{j}\,(xq^{2s-j+2};q)_j\,
(q^{-v+2j-s+2};q)_{s}\\
\cdot(-xq^{-v+s+j+2};q)_{n-j-1}\,
(1-q^{-v+2j-s}-q^{-v+2j-s+1}+xq^{-v+s+j+1}).
\endmultline\tag3.11$$
Similar considerations concern the second half of the product in
(3.8). Omitting the details, for each solution $(j,l)$ of
$$u=-2v+2j+l+1,\quad  \text {with }0\le j\le n-1,\ 0\le l<j,\tag3.12$$
we subtract the linear combination
$$\multline \sum _{s=l} ^{\fl{(j+l)/2}}
{q^{{\let\over/ {{\left( s-l \right)  \left( 5s+3l+3 \right) }\over 2}-
  \left( \left( j - l \right)  \left( 2 j + 2 l + 1\right)  \right) 
           }}}{q^{-2 v\left( s-j  \right) }}\\
\cdot\frac {\left( 1 - {q^{2 j + l - 3 s}} \right)  } 
      {\left( 1 - {q^{j - s}} \right)}
\frac {{\let \over / ({\let \over / {q^{1 + j + l - 2 s }}}; q) _{s-l}}} 
      {{\let \over /  ({\let \over / -q}; q) _{j - s}  \,
     ({\let \over / q}; q) _{s-l}}}
({\let \over / {q^{ -v+2 s+2}} }; q) _{2 j - 2 s}
\,\cdot\,(\text {column $s$ of
$\DetBt(x,q^{-v};n)$})
\endmultline$$
of columns of $\DetBt(x,q^{-v};n)$ (we definitely mean $\DetBt(x,q^{-v};n)$,
and not $\DetBb(x,q^{-v};n)$) from column $j$ of $\DetBt(x,q^{-v};n)$. By
(3.6), each entry of the $j$-th column of the new determinant will
have $(1-xq^u)$ as a factor. We remark that entries that were changed by
a row {\it and\/} column operations will now have $(1-xq^u)^2$ as a
factor. Now we take $(1-xq^u)$ out of the $i$-th row, for each solution
$(i,l)$ of (3.9), and we take $(1-xq^u)$ out of the $j$-th column, 
for each solution $(j,l)$ of (3.12). We denote the resulting
determinant by $\DetBbb(x,q^{-v};n)$. Thus, we have
$$\DetBb(x,q^{-v};n)=(1-xq^u)^{\#\text {(solutions $(i,l)$ of (3.9))}+
\#\text {(solutions $(j,l)$ of (3.12))}}\,\DetBbb(x,q^{-v};n).$$
Plugging this into (3.8), we see that now all factors $(1-xq^u)$ can be
cancelled on both sides, so that we obtain
$$\DetBbb(x,q^{-v};n)=C(x,q^{-v};n)\,P(x,q^{-v};n),$$
for some $C(x,q^{-v};n)$ that does not vanish for $x=q^{-u}$. Hence, if we are
able to prove that $\DetBbb(q^{-u},q^{-v};n)=0$, it would follow that
$P(q^{-u},q^{-v};n)=0$, which is what we want to establish.

So we are left with showing that $\DetBbb(q^{-u},q^{-v};n)=0$. This will be
implied by the following two claims: The matrix of which
$\DetBbb(q^{-u},q^{-v};n)$ is the
determinant has a block form (see (3.13)), where 

Claim 1. the
upper-right block, consisting of the entries that are in one of the
rows $0,1,\dots,u+v$ and one of the columns
$u+v+1,u+v+2,\dots,n-1$, is a zero matrix, and where 

Claim 2. the
determinant of the upper-left block, $\Cal N$, 
consisting of the entries that are in one of the
rows $0,1,\dots,u+v$ and one of the columns
$0,1,\dots,u+v$, equals 0. (Note that it is at this point that we
need the assumption $u+v\le n-1$ in the Lemma. It guarantees that the
picture (3.13) makes sense, meaning that
row $u+v$ and column $u+v$ are really a row and a column of the
matrix; recall that the rows and columns are numbered from 0 to
$n-1$.)
\vskip3pt
$$
\PfadDicke{.3pt}
\Pfad(0,-3),11111\endPfad
\Pfad(0,0),11111\endPfad
\Pfad(0,2),11111\endPfad
\Pfad(0,-3),22222\endPfad
\Pfad(2,-3),22222\endPfad
\Pfad(5,-3),22222\endPfad
\Label\r{\Cal N\hskip15pt}(1,1)
\Label\r{$\fourteenpoint$0\hskip5pt}(3,1)
\Label\o{$\fourteenpoint$*}(1,-2)
\Label\ro{$\fourteenpoint$*}(3,-2)
\Label\r{\eightpoint\hskip4pt\leftarrow i\!=\!u\!+\!v}(6,0)
\hbox{\hskip3.7cm}
\Label\r{\hskip-18pt \eightpoint j\!=\!u\!+\!v}(-5,3)
\Label\ru{\hskip-25pt \eightpoint \downarrow}(-5,3)
\tag3.13
$$
\vskip3pt
Indeed, Claim~1 and Claim~2 imply $\DetBbb(q^{-u},q^{-v};n)=0$.
For, the determinant of a block matrix of the form (3.13) equals the
product of the determinants of the upper-left block and the
lower-right block, the first determinant being equal to $0$ by
Claim~2. 

Claim~1 is most obvious for all the entries that did not change in
the transition from $\DetBb(x,q^{-v};n)$ to $\DetBbb(x,q^{-v};n)$. For, the
$(i,j)$-entry of $\DetBb(x,q^{-v};n)$, by its definition in (3.1), is
$$\multline q^{-2ij}\,(q^{-u-v+i};q)_{j}\,(q^{-u+2i-j+2};q)_j\,
(q^{-v+2j-i+2};q)_{i}\,(-q^{-u-v+i+j+2};q)_{n-j-1}\\
\cdot(1-q^{-v+2j-i}-q^{-v+2j-i+1}+q^{-u-v+i+j+1}).
\endmultline\tag3.14$$
Clearly, if $0\le i\le u+v$ and $u+v+1\le j\le n-1$, we have
$(q^{-u-v+i};q)_j=0$, and so the complete expression in (3.14) is 0. 

On the
other hand, let us consider an $(i,j)$-entry of $\DetBbb(x,q^{-v};n)$
that changed in
the transition from $\DetBb(x,q^{-v};n)$ to $\DetBbb(x,q^{-v};n)$. First we
want to know, where such an entry could be located. If it changed
under a row operation, then $(i,l)$ is a solution of (3.9), for some
$l$. By (3.9) we have
$$-v+2i+1\le -v+2i+l+1=2u\quad \text {and}\quad 2u=-v+2i+l+1\le -v+3i,$$
and so,
$$\frac {2u+v} {3}\le i\le \frac {2u+v-1} {2}.\tag3.15$$
If the $(i,j)$-entry 
changed
under a column operation, then $(j,l)$ is a solution of (3.12), for some
$l$. Similar arguments then give, using (3.12), that
$$\frac {u+2v} {3}\le j\le \frac {u+2v-1} {2}.\tag3.16$$
In particular we have $j<u+v$, so an $(i,j)$-entry that is located
in the upper-right block, which we are currently interested in, did
not change under a column operation.  

But it could have changed under a row operation. Such an
$(i,j)$-entry is given by (3.11) divided by $(1-xq^u)$. (Recall that
(3.11) was the expression for an $(i,j)$-entry that changed under a
row operation {\it before\/} we factored $(1-xq^u)$ out of the $i$-th row.)
Thus, it can be written as
$$\multline \frac {(xq^{-v+i};q)_{u+v-i+1}} {(1-xq^u)}
\bigg(q^{-2ij}\,(xq^{u+1};q)_{i+j-u-v-1}\,
(xq^{2i-j+2};q)_j\,(q^{-v+2j-i+2};q)_i\\
\hskip1cm\times(-xq^{-v+i+j+2};q)_{n-j-1}\,
\(1-q^{-v+2j-i}-q^{-v+2j-i+1}+xq^{-v+i+j+1}\)\\
-\sum _{s=l} ^{\fl{(i+l)/2}}
{q^{ {{\left(s -l \right)  
             \left( 5s+3l+7  \right) }\over 2}-(i-l)(2i+2l+3)}}
{x^{2 \left( s-i \right) }}
\frac
{(1-q^{2i-3s+l})} {(1-q^{i-s})}
\frac {(q^{i-2s+l+1};q)_{s-l}} {(q;q)_{s-l}}
\hskip1cm\\
\cdot \frac
{(xq^{2s+2};q)_{2i-2s}}
{(q^{-2i-l+s-1}/x;q)_{i-s}}
\frac {({\let \over / -{{{q^{-2 i - l + n + s}}}}/x}; q) _{i -
s}} {(-q;q)_{i-s}}\\
\,\cdot\,q^{-2sj}\,(xq^{-v+s};q)_{i-s}\,(xq^{u+1};q)_{j+s-u-v-1}\,
(xq^{2s-j+2};q)_j\,(q^{-v+2j-s+2};q)_{s}\\
\cdot(-xq^{-v+s+j+2};q)_{n-j-1}\,
(1-q^{-v+2j-s}-q^{-v+2j-s+1}+xq^{-v+s+j+1})\bigg).
\endmultline\tag3.17$$
We have to show that (3.17) vanishes for $x\to q^{-u}$. Because of the
denominators, it is not even evident that (3.17) is well-defined when
$x\to q^{-u}$. However, by (3.15) we have $u+v-i\ge(v+1)/2\ge0$.
Hence,
$$\frac {(xq^{-v+i})_{u+v-i+1}} {(1-xq^u)}=(xq^{-v+i})_{u+v-i},$$
and so the first term in (3.17) is well-defined when $x\to q^{-u}$.
Furthermore, the denominator in the sum in (3.17) (neglecting the
terms that do not depend on $x$) when $x\to q^{-u}$
becomes
$$(q^{u-2i-l+s-1};q)_{i-s}=(1-q^{u-2i-l+s-1})\cdots(1-q^{u-i-l-2}).\tag3.18$$
By (3.9) and (3.15) we have
$u-i-l-2=-u-v+i-1\le \frac {1} {2}(-v-3)<0.$
Therefore, all the terms in (3.18) are nonzero, which means that the
denominator in the sum in (3.17) is nonzero when $x\to q^{-u}$. Hence,
(3.17) is well-defined for $x\to q^{-u}$. To demonstrate that it actually
vanishes for $x\to q^{-u}$, we show that the second term in (3.17) (the
term in big parentheses) equals 0 for $x=q^{-u}$. 

To see this, set $x=q^{-u}$, and by (3.9) replace $l$ by $2u+v-2i-1$ in
the sum (3.17), and then convert it into hypergeometric notation, to
obtain
%Achtung: Muss $q^{4ij+2j-4ju-2jv}=q^{-2jl}$ dazumultiplizieren!
$$\multline -{q^{4ij+4j-{{i}\over 2} + {{3 {i^2}}\over 2} - 
 u - 4 i u -4ju + 2 {u^2} - 
      {{5 v}\over 2} - 5 i v -2jv + 4 u v + {{3 {v^2}}\over 2}}}\\
\times  ({\let \over / -1}; q) _{-3 i + 2 u + v} \,
  ({\let \over / q}; q) _{-2 - 2 i + j + u} \,
  ({\let \over / -{q^{1 - 2 i + j + u}}}; q) _{3 i - j + n - 2 u - v}
\\
\times  ({\let \over / {q^{3 + 2 i + 2 j - 2 u - 2 v}}}; q) _{-1 - 2 i + 2 u +
   v} \,({\let \over / {q^{-4 i - j + 3 u + 2 v}}}; q) _{2 + 6 i + j -
   4 u - 2 v} \\
\times \sum _{k=0} ^{\fl{(-2u-v+3i+1)/2}}
(+1 + {q} -
    {q^{-1 - 2 i -2j+ k + 2 u + 2 v}} - {q^{-1 - 4 i - j + 2 k + 3 u +
2v}})\\
\cdot\frac {{(1-{q^{1 - 6 i + 4 u + 2 v}})}} {{(1-{q^{-2 - 6 i + 4 u +
   2 v}})}}
\frac {{\let\over/ (-{q^{-1 - 3 i + 2 u + v}};q)_k\, ({q^{-2
   - 2 i - 2 j + 2 u + 2 v}};q)_k\, ({q^{-1 - 2 i + j + u}};q)_k }} 
    {{\let\over/ ({q^{-3 i + 2 u + v}};q)_k\, -{q^{1 - 2 i + j +
    u}};q)_k\, (q;q)_k}}\\
\cdot\frac {{\let\over/ ({q^{-1 - 3 i +
   2 u + v}};q^2)_k\, ({q^{-3 i + 2 u + v}};q^2)_k }} 
    {{\let\over/ ({q^{-4 i - j +
   3 u + 2 v}};q^2)_k\, ({q^{1 - 4 i - j + 3 u + 2 v}};q^2)_k
}}q^{2k}.
\endmultline\tag3.19$$
The series can be summed by means of Lemma~A2 with
$n=-2u-v+3i+1$ and $B=q^{3u+2v-4i-j-2}$. Then, after
simplification, (3.19) becomes
$$\multline {{q^{-2 i j}}}
     ({\let \over / q}; q) _{i + j - u - v-1} \,
     ({\let \over / {q^{-u+ 2 i - j + 2}}}; q) _{j} \,
     ({\let \over / {q^{-v+2j-i +2}}}; q) _{i} \\
\times     ({\let \over / -{q^{-u-v +i + j +2}}}; q) _{n-j-1} 
\left(1 -q^{-v+2j-i} - q^{-v+2j-i} + {q^{-u-v+ i + j +1}}  \right),
\endmultline$$
which is exactly the first term in big parentheses in (3.17) for
$x=q^{-u}$.
Therefore, the term in big parentheses in (3.17) vanishes for $x=q^{-u}$.
This settles Claim~1.

Next we turn to Claim~2. We have to prove that the determinant of the
matrix $\Cal N$, consisting of the entries of $\DetBbb(q^{-u},q^{-v};n)$ that
are in one of the rows $0,1,\dots,u+v$ and one of the columns
$0,1,\dots,u+v$ (recall (3.13)), equals 0. We do this by locating
enough zeros in the matrix $\Cal N$.

We concentrate on the entries that did not change in the transition
from\linebreak
$\DetBb(x,q^{-v};n)$ to $\DetBbb(x,q^{-v};n)$. 
For the location of the various regions in the matrix $\Cal N$
that we are going to describe, always consult Figure~1 which 
gives a rough sketch.
\vskip10pt
\vbox{
$$\Cal N\ =\ 
\raise2.25cm\hbox{$
\Einheit.25cm
\thinlines
\PfadDicke{.35pt}
\Pfad(0,0),111111111111111111\endPfad
\Pfad(0,-6),111111111111111111\endPfad
\Pfad(0,-10),111111111111111111\endPfad
\Pfad(0,-15),111111111111111111\endPfad
\Pfad(0,-18),111111111111111111\endPfad
\Pfad(0,-18),222222222222222222\endPfad
\Pfad(3,-18),222222222222222222\endPfad
\Pfad(8,-18),222222222222222222\endPfad
\Pfad(12,-18),222222222222222222\endPfad
\Pfad(18,-18),222222222222222222\endPfad
\catcode`\@=11
\hskip3\Einheit\hbox to0pt{\Line@(1,-2)9\hss}\hskip-3\Einheit
\raise-6\Einheit\hbox to0pt{\Line@(2,-1){18}\hss}
\catcode`\@=13
\Diagonale(0,-18){18}
\Vektor(-1,0)8(23,-3)
\Vektor(-1,0)9(23,-13)
\Vektor(1,0)6(-2,-2)
\Label\r{\eightpoint i=0}(19,0)
\Label\r{\eightpoint i=\cl{\frac {u-1} {2}}}(20,-6)
\Label\r{\eightpoint\hskip8pt i=\cl{\frac {2u+v} {3}}-2+\chi(u\equiv
v(3))}(25,-10)
\Label\r{\eightpoint\hskip17pt i=\fl{\frac {2u+v-1} {2}}+1}(21,-15)
\Label\r{\eightpoint\hskip3pt i=u+v}(20,-18)
\Label\o{\eightpoint 0}(0,0)
\Label\o{\eightpoint \raise3pt\hbox{$j=$}}(0,1)
\Label\u{\eightpoint j=}(3,-18)
\Label\u{\eightpoint \raise-10pt\hbox{$\cl{\frac {v-1} {2}}$}}(3,-19)
\Label\o{\eightpoint \cl{\frac {u+2v} {3}}-1}(8,0)
\Label\o{\eightpoint \raise10pt\hbox{$j=$}}(8,1)
\Label\u{\eightpoint j=}(12,-18)
\Label\u{\eightpoint \raise-10pt\hbox{$\fl{\frac {u+2v-1} {2}}+1$}}(12,-19)
\Label\o{\eightpoint u+v}(18,0)
\Label\o{\eightpoint \raise5pt\hbox{$j=$}}(18,1)
\Label\r{\eightpoint\hskip-12pt i=\frac {u+j-2} {2}}(26,-13)
\Label\r{\eightpoint j=\frac {v+i-2} {2}}(-7,-2)
\Label\r{\eightpoint\hskip5pt i+j=u+v}(26,-3)
\Label\r{$\hskip8pt I$}(4,-8)
\Label\r{$II\hskip2pt $}(14,-8)
\Label\ro{$\hskip10pt IV$}(14,-17)
\Label\ro{$III$}(5,-17)
\hskip7cm
$}
$$
\vskip6pt
\centerline{\eightpoint Figure 1}
}
\vskip10pt
By earlier
considerations, an $(i,j)$-entry did not change if $i$ is outside
the range (3.15), i.e.,
$$0\le i\le \cl{\frac {2u+v} {3}}-1\quad \text {or}\quad \fl{\frac
{2u+v-1} {2}}+1\le i\le n-1,\tag3.20$$
and if $j$ is outside
the range (3.16), i.e.,
$$0\le j\le \cl{\frac {u+2v} {3}}-1\quad \text {or}\quad \fl{\frac
{u+2v-1} {2}}+1\le j\le n-1.\tag3.21$$
As we already noted, such an $(i,j)$-entry is given by (3.14).
The first term in (3.14) vanishes if and only if
$$i\le u+v\quad \text {and}\quad i+j>u+v.\tag3.22$$
The second term in (3.14) vanishes if and only if
$$\cl{\frac {u-1} {2}}\le i\le \frac {u+j-2} {2}.\tag3.23$$
The third term in (3.14) vanishes if and only if
$$\cl{\frac {v-1} {2}}\le j\le \frac {v+i-2} {2}.\tag3.24$$
Obviously, the fourth term in (3.14) never vanishes.
Finally,
we need the following sufficient conditions in order the fifth
term to vanish: 
The fifth term in (3.14) vanishes if 
$$u-v=3(i-j)\quad \text {and}\quad u+v-i-j=0\text { or }2.\tag3.25$$

Now we claim that in the following four regions of $\Cal N$ all the
entries are 0, except for the cases $u=0$, $v=1$, and $u=1$, $v=0$, 
which we treat
separately. Again, to get an idea of the location of these regions,
consult Figure~1.

\smallskip
Region I: All $(i,j)$-entries with
$$\multline 
\cl{\frac {u-1} {2}}\le i\le \cl{\frac {2u+v} {3}}-2+\chi(u\equiv
v\pmod 3)\\
\quad \text {and}\quad \cl{\frac {v-1} {2}}\le j\le 
\cl{\frac {u+2v} {3}}-1,
\endmultline\tag3.26$$
where $\chi(\Cal A)$=1 if $\Cal A$ is
true and $\chi(\Cal A)$=0 otherwise.

Region II: All $(i,j)$-entries with
$$\multline
\cl{\frac {u-1} {2}}\le i\le \cl{\frac {2u+v} {3}}-2+\chi(u\equiv
v\pmod 3)\\
\quad \text {and}\quad \fl{\frac {u+2v-1} {2}}+1\le j\le
u+v.
\endmultline\tag3.27$$

Region III: All $(i,j)$-entries with
$$\fl{\frac {2u+v-1} {2}}+1\le i\le u+v\quad \text {and}\quad 
\cl{\frac {v-1} {2}}\le j\le 
\cl{\frac {u+2v} {3}}-1.\tag3.28$$

Region IV: All $(i,j)$-entries with
$$\fl{\frac {2u+v-1} {2}}+1\le i\le u+v\quad \text {and}\quad 
\fl{\frac {u+2v-1} {2}}+1\le j\le
u+v.\tag3.29$$

\smallskip
Instantly we observe that all four regions satisfy (3.20) and (3.21).
So, all the entries in these regions are given by (3.14). Hence, to
verify that all these entries are 0 we have to show that for each
entry one of (3.22)--(3.25) is true. Of course, we treat the four
regions separately.

\smallskip
ad Region I. First let $i\le \cl{(2u+v)/3}-2$. In case that $i\le
j+(u-v)/3$, we have
$$\multline
i\le \frac {i+j+\frac {u-v} {3}} {2}\le \frac {\cl{\frac {2u+v}
{3}}-2+j+\frac {u-v} {3}} {2}\\
\le \frac {\frac {2u+v} {3}+\frac {2}
{3}-2+j+\frac {u-v} {3}} {2}=\frac {u+j-2} {2}+\frac {1} {3}.
\endmultline$$
Combined with (3.26), this implies that (3.23) is satisfied. On the
other hand, in case that $i>j+(u-v)/3$, or equivalently, 
$$i\ge j+\frac {u-v} {3}+\frac {1} {3},\tag3.30$$
we have, using the last inequality in (3.26),
$$\multline
j\le \frac {i+j-\frac {u-v} {3}-\frac {1} {3}} {2}\le 
\frac {i+\cl{\frac {u+2v} {3}}-1-\frac {u-v} {3}-\frac {1} {3}} {2}\\
\le\frac {i+\frac {u+2v} {3}+\frac {2} {3}-1-\frac {u-v} {3}-\frac {1} {3}}
{2}=\frac {v+i-2} {2}+\frac {2} {3}.
\endmultline$$
Combined with (3.26), this implies that (3.24) is satisfied, unless
$j=(v+i-1)/2$. But if we plug this into (3.30), we obtain $i\ge
(2u+v)/3-1/3$, a contradiction to our assumption $i\le \cl{(2u+v)/3}-2$.

Collecting our results so far, we have seen that if
$u-v\equiv 1,2\pmod3$, then each $(i,j)$-entry in region~I satisfies
(3.23) or (3.24). If $u\equiv v\pmod3$, region~I also contains entries
from row $i=(2u+v)/3-1$. First let $j\le (u+2v)/3-2$. Then it is
immediate that (3.24) is satisfied. If $j=(u+2v)/3-1$, then (3.25) is
satisfied. 
This shows that if $u\equiv v\pmod3$ then an $(i,j)$-entry in region~I
satisfies (3.23), (3.24), or (3.25).

ad Region II. Here, by (3.27), we have
$$i+j\ge \cl{\frac {u-1} {2}}+\fl{\frac {u+2v-1} {2}}+1=u+v.$$
Hence, (3.22) is satisfied, except when $i=\cl{(u-1)/2}$ and
$j=\fl{(u+2v-1)/2}+1$. But in that case there holds (3.23), apart
from a few exceptional cases. For, if $u\ne0,2$ then
$$\cl{\frac {u-1} {2}}\le \frac {u+\fl{\frac {u-1} {2}}-1} {2}.$$
Since $v$ is nonnegative it follows that
$$\cl{\frac {u-1} {2}}\le \frac {u+\(\fl{\frac {u+2v-1} {2}}+1\)-2}
{2},\tag3.31$$
which is nothing but (3.23) with the current choices of $i$ and $j$.
Thus, (3.23) is satisfied except when $u=0$ or $u=2$.
But (3.31), and hence
(3.23), holds in more cases. Namely, by inspection, if $u=0$, then
(3.31) holds for $v\ge2$, and if $u=2$, then
(3.31) holds for $v\ge1$. So the only cases in which (3.31) is not
true are $u=v=0$, $u=0$ and $v=1$, $u=2$ and $v=0$.
Starting 
from the back, the case $u=2$, $v=0$ does not bother us, since in that
case region~II is empty (there is no $i$ satisfying (3.27)).
The case $u=0$, $v=1$ is one of the exceptional cases that are
treated separately. Finally, in case $u=v=0$ we have
$i=\cl{(u-1)/2}=0$ and $j=\fl{(u+2v-1)/2}+1=0$. Hence, (3.25) is
satisfied.

ad Region III. We argue as in the considerations concerning
region~II. In fact, the arguments given there can be used word by
word, with $i$ and $j$ interchanged, and with $u$ and $v$
interchanged.

ad Region~IV. By (3.29) we have
$$i+j\ge \fl{\frac {2u+v-1} {2}}+1+\fl{\frac {u+2v-1} {2}}+1\ge 
\frac {3u+3v} {2}+1>u+v.$$
Hence, (3.22) is satisfied.

\smallskip
Consequently, if we are not in one of the cases $u=0$, $v=1$, or
$u=1$, $v=0$, then the rows
$\cl{(u-1)/2},\dots,
\cl{(2u+v)/3}-2+\chi(u\equiv v\pmod3),\fl{(2u+v-1)/2}+
1,\dots,u+v$ are rows with zeros in columns
$\cl{(v-1)/2},\dots,\cl{(u+2v)/3}-1,\mathbreak
\fl{(u+2v-1)/2}+1,\dots,u+v$.
These are
$$\cl{\frac {2u+v} {3}}-1+\chi(u\equiv v\kern-5pt\pmod3)-\cl{\frac {u-1} {2}}
+u+v-\fl{\frac {2u+v-1} {2}}
\tag3.32$$
rows, containing possibly nontrivial entries in only
$$\cl{\frac {v-1} {2}}+\fl{\frac {u+2v-1} {2}}-\cl{\frac {u+2v}
{3}}+1\tag3.33$$
columns. By simple algebra, the difference between (3.32) and (3.33)
equals
$$u-v+\cl{\frac {v-u} {3}}+\cl{\frac {2v-2u}
{3}}+\chi(u\equiv v\kern-5pt\pmod3).\tag3.34$$
As is easily verified, the expression (3.34) equals 1 always. So we
have found $N+1$ rows (with $N$ the expression in (3.33)) that
actually live in $\Bbb R^N$ ($\Bbb R$ denoting the set of real numbers). 
Hence, they must be linearly dependent.
This implies that the determinant of $\Cal N$ must be 0.

\smallskip
Finally we settle the cases $u=0$, $v=1$, and $u=1$, $v=0$. 
If $u=0$ and $v=1$ then the matrix $\Cal N$ is a
$2\times 2$ matrix (cf\. Figure~1) in which row 1 vanishes. For,
$i=1$ and $j=0$ satisfy (3.20), (3.21), and (3.24), while $i=1$ and
$j=1$ satisfy (3.20), (3.21), and (3.22). Hence, $\det(\Cal N)=0$.
Similarly, if $u=1$ and $v=0$ then the matrix $\Cal N$ is a
$2\times 2$ matrix in which column 1 vanishes. For,
$i=0$ and $j=1$ satisfy (3.20), (3.21), and (3.23), while $i=1$ and
$j=1$ satisfy (3.20), (3.21), and (3.22). Hence again, $\det(\Cal N)=0$.

\smallskip
Altogether, this implies that $P(q^{-u},q^{-v};n)=0$, as we observed
earlier. And this is what we wanted to prove.\quad \quad \qed


\enddemo
As last lemma we prove a characterization theorem for the ``big
factor" on the right-hand side of (1.4) (more precisely, on the
right-hand side of (3.1)), which we wish to identify as $P(x,y;n)$.
We shall eventually show that $P(x,y;n)$ has all the
properties (1)--(4) 
that are stated below and thus be able to finish the proof of
Theorem~2.
\proclaim{Lemma 3}The polynomial 
$$Q(x,y;n)=\prod _{i=0} ^{n-1}\(q^{-2 i^2} 
(q^2;q^2)_i \)\sum _{k=0} ^{n} (-1)^k q^{n k} {n\brack k}_q {y^ k} 
(x;q)_{k} (y;q)_{n-k}
\tag3.35$$
satisfies the following four properties:
\roster
\item"(1)" $Q(x,y;n)$ is a polynomial in $x$ and $y$, of degree $n$ in
$x$, and also of degree $n$ in $y$.
\item"(2)" $Q(q^{-u},q^{-v};n)=0$ for all nonnegative integers $u$ and $v$ with
$u+v\le n-1$.
\item"(3)" $Q(x/y,y;n)$ is a polynomial in $x$ and $y$.
\item"(4)" $\dsize Q(-q^{-n}/y,y;n)=(-1)^nq^{\binom n2}(-q;q)_n
\prod _{i=0} ^{n-1}\(q^{-2 i^2} 
(q^2;q^2)_i \)y^n$, for any nonnegative integer $n$.
\endroster
Moreover, the conditions {\rm(1)--(4)} determine a polynomial in $x$
and $y$ uniquely.
\endproclaim
\demo{Proof}{\it ad\/} (1). This is obvious from the definition
(3.35).

\smallskip
{\it ad\/} (2). We have $(q^{-u};q)_k=0$ for $k>u$. Hence, if $k>u$ the
corresponding summand in the sum in (3.35) vanishes for $x=q^{-u}$ and
$y=q^{-v}$. Now let $k\le u$. Because of $u+v\le n-1$ it follows that
$k<n-v$, or equivalently, $n-k>v$. But this implies $(q^{-v};q)_{n-k}=0$.
Therefore also any summand with $k\le u$ vanishes for $x=q^{-u}$ and
$y=q^{-v}$. Thus, $Q(q^{-u},q^{-v};n)=0$, as desired.

\smallskip
{\it ad\/} (3). This is obvious from the definition
(3.35).

\smallskip
{\it ad\/} (4). Setting $x=-q^{-n}/y$ in (3.35), we get
$$\frac {Q(-q^{-n}/y,y;n)} {\prod _{i=0} ^{n-1}\(q^{-2 i^2} 
(q^2;q^2)_i \)}=\sum _{k=0} ^{n} (-1)^k q^{n k} {n\brack k}_q {y^ k} 
(-q^{-n}/y;q)_{k} (y;q)_{n-k},$$
or after little manipulation,
$$\frac {Q(-q^{-n}/y,y;n)} {\prod _{i=0} ^{n-1}\(q^{-2 i^2} 
(q^2;q^2)_i \)}=(y;q)_n\sum _{k=0} ^{n}\frac {(-q^{-n}/y;q)_k\,
(q^{-n};q)_k} {(q^{1-n}/y;q)_k\, (q;q)_k}\(-q^{n+1}\)^k.$$
The sum on the right-hand side can be summed by means of the
$q$-Vandermonde summation \cite{\GaRaAA, (1.5.2); Appendix (II.7)},
$$\sum _{k=0} ^{n}\frac {(a;q)_k\, (q^{-n};q)_k} {(c;q)_k\, (q;q)_k}
\(\frac {cq^n} {a}\)^k=
{{{(\let\over/ {c\over a};q)}_{n}}\over {{(\let\over/ c;q)}_{n}}},
$$
where $n$ is a nonnegative integer.
Thus we obtain
$$\frac {Q(-q^{-n}/y,y;n)} {\prod _{i=0} ^{n-1}\(q^{-2 i^2} 
(q^2;q^2)_i \)}=(-1)^nq^{\binom n2}(-q;q)_n\,y^n,
$$
as desired.

\smallskip
Finally we have to confirm that indeed the properties (1)--(4)
determine a polynomial in $x$ and $y$ uniquely.

Let $H(x,y)$ be a polynomial in $x$ and $y$
satisfying conditions (1)--(4). Because of (1), $H(x,y)$ can be
written in the form
$$H(x,y)={\sum _{0\le i,j\le n}
^{}}a_{ij}\,y^{n-j}\,(x;q)_i\,(y;q)_j,\tag3.36$$
with uniquely determined coefficients $a_{ij}$. Now, in (3.36) 
we set $x=1$ and $y=q^{-v}$, $0\le v\le n-1$. Because of (2), we obtain
$0=\sum _{j=0} ^{v}a_{0j}\,q^{-v(n-j)}(q^{-v};q)_j.$
From this system of equations we get $a_{0j}=0$ for $0\le j\le
n-1$. Similarly, by using (2) with $x=q^{-1},q^{-2},\dots,q^{-(n-1)}$, we get
$a_{ij}=0$ whenever $i+j\le n-1$. 

Thus, $H(x,y)$ can be written in the form
$$H(x,y)=\underset i+j\ge n\to{\sum _{i,j\ge0}
^{}}a_{ij}\,y^{n-j}\,(x;q)_i\,(y;q)_j.$$
Next, property (3) comes into effect. According to that property, we 
have that
$$H(x/y,y;n)=\underset i+j\ge n\to{\sum _{i,j\ge0} ^{}}
a_{ij}\,y^{n-i-j}\,(y-x)(y-qx)\cdots(y-q^{i-1}x)\,(y;q)_j\tag3.37$$
is a polynomial in $x$ and $y$. Let $(i,j)$ be a pair with
$a_{ij}\ne0$, where $i+j$ is maximal, and if there are several such pairs
then choose one with maximal $i$. Then there occurs a term
$y^{n-i-j}x^i$ in the corresponding summand in (3.37) which does not
cancel. If $i+j>n$ then the exponent of $y$ is negative. However, 
this would contradict (3).

Therefore, $H(x,y)$ can be written in the form
$$H(x,y;n)=\sum _{k=0} ^{n}b_k\,y^k\,(x;q)_k\,(y;q)_{n-k},$$
where we set $b_k:=a_{k,n-k}$.

Now we apply (4). We have
$$\align (-1)^nq^{\binom n2}(-q;q)_n
\prod _{i=0} ^{n-1}\(q^{-2 i^2} 
(q^2;q^2)_i \)&y^n=H(-q^{-n}/y,y;n)\\
&=\sum _{k=0} ^{n}b_k\,y^{k}\,(-q^{-n}/y;q)_k\,(y;q)_{n-k}\\
&=\sum _{k=0} ^{n}b_k\,q^{\binom
k2-nk}\,(-yq^{n-k+1};q)_k\,(y;q)_{n-k}.
\tag3.38\endalign$$
It is straight-forward to see that the polynomials
$$(-yq^{n-k+1};q)_k\,(y;q)_{n-k},\quad  k=0,1,\dots,n,$$ 
are linearly
independent. Hence, by comparison of coefficients, equation (3.38) 
determines the coefficients $b_k$, $k=0,1,\dots,n$, uniquely, which
implies that $H(x,y)$ is uniquely determined.


This completes the proof of the Lemma.\quad \quad \qed
\enddemo

Finally we are in the position to finalize the proof of Theorem~2.

\smallskip
{\it Proof of Theorem~2.} We verify that $P(x,y;n)$ has properties
(1)--(4) in Lemma~3. Once this is done, it follows from Lemma~3 that
$P(x,y;n)$ equals the polynomial $Q(x,y;n)$ in (3.35), which is
exactly what we want to show. This would complete the proof of
Theorem~2.

Now, $P(x,y;n)$ satisfies property (1) because of Lemma~1, and it
satisfies property (2) because of Lemma~2. 

To check property (3),
replace $x$ by $x/y$ in (3.2) and then multiply both sides by
$y^{\binom n2}$. Thus we obtain
$$\multline \det_{0\le i,j\le
n-1}\big(q^{-2ij}\,(xq^i;q)_{j}\,(y-xq^{2i-j+2})\cdots(y-xq^{2i+1})\,
(yq^{2j-i+2};q)_{i}\\
\hskip3cm\cdot(-xq^{i+j+2};q)_{n-j-1}\,
(1-yq^{2j-i}-yq^{2j-i+1}+xq^{i+j+1})\big)\\
=\prod _{i=0} ^{n-1}
\Big((y-x^2q^{2i+1})\cdots(y-x^2q^{3i}) \,(xyq^{2i+1};q)_i\Big)
\cdot P(x/y,y;n).
\endmultline
$$
The left-hand side and the product on the right-hand side are polynomial. 
Besides, we know that the
product on the right-hand side divides the left-hand side. So
evidently, $P(x/y,y;n)$ is a polynomial in $x$ and $y$.

Finally, regarding property~(4), set $x=-q^{-n}/y$ in (3.2). Then, in
the matrix on the left-hand side, 
the term $(q^{-n+i+j+2})_{n-j-1}$ appears. This term
vanishes whenever $i+j\le n-2$. Hence, the matrix has triangular form.
So the determinant is easily calculated. Property~(4) then follows
after simplification.\quad \quad \qed


\head\tenpoint\bf Appendix: Some basic hypergeometric identities\endhead

Here we prove the basic hypergeometric identities that are needed in
the text. We start with the summation needed in the proof of 
Theorem~1.
\proclaim{Lemma A1}Let $n$ be a positive integer. Then
$$\multline \sum_{k = 0}^{\fl{n/2}}
{{ \left( 1 - {q^{3 k - 2 n}} \right)  }\over
   {\left( 1 - {q^{-2 n}} \right) }}
{{({\let \over / -{q^{-n}}}; q) _{k} \,
 ({\let \over / b}; q) _{k} \, ({\let \over / c}; q) _{k}  \,
       ({\let \over / {{{q^{1 - 2 n}}}\over {b c}}}; q) _{k}  \,
              ({\let \over / {q^{-n}}}; {q^2}) _{k} \,
            ({\let \over / {q^{1 - n}}}; {q^2}) _{k}  }\over 
    { ({\let \over / q}; q) _{k}  \, ({\let \over / {q^{1 - n}}}; q)
_{k}\,
            ({\let \over / {{{q^{2 - 2 n}}}\over b}}; {q^2}) _{k}  \,
       ({\let \over / {{{q^{2 - 2 n}}}\over c}}; {q^2}) _{k}  \,
     ({\let \over / b c q}; {q^2}) _{k}    }}
{q^k}
\\={{({\let \over / b}; q) _{n}  \,({\let \over / c}; q) _{n}  \,
        ({\let \over / b c {q^n}}; q) _{n} \,
      ({\let \over / -q}; q) _{n-1}
     }\over {({\let \over / b}; {q^2}) _{n}  \,
      ({\let \over / c}; {q^2}) _{n}\,  ({\let \over / b c q}; {q^2}) _{n}
}}.
\endmultline\tag A.1$$
\endproclaim
\demo{Proof} We start with Rahman's transformation formula \cite{\RahmAC,
(3.12); \GaRaAA, (3.8.13)}
$$\multline 
\sum_{k = 0}^{\infty}{{\left( 1 - a {q^{3 k}} \right)  }\over
   {{1 - a}}}
          {{{\let \over / (b;q)_k\, (c;q)_k\, ({{a q}\over {b
c}};q)_k}} \over
         {{\let \over / (q;q)_k\, ({{a q}\over d};q)_k\, (d;q)_k}}} 
{{{\let \over / (a;q^2)_k\, (d;q^2)_k\, ({{a q}\over
d};q^2)_k}} \over {{\let
         \over / ({{a {q^2}}\over b};q^2)_k\, ({{a {q^2}}\over
c};q^2)\, (b c q;q^2)_k}
         }}
{q^k}
\\=    { {\let \over / (a {q^2};q^2)_\infty\, (b q;q^2)_\infty\, (c
q;q^2)_\infty\, ({{a {q^2}}\over {b c}} ;{q^2})
    _\infty} \over {\let \over / (q;q^2)_\infty\, ({{a {q^2}}\over
b};q^2)_\infty\, ({{a {q^2}}\over c};q^2)_\infty\,
    (b c q ;{q^2}) _\infty} }\\
\times  \sum _{k=0} ^{\infty}{{\let \over / (b;q^2)_k\,
(c;q^2)_k\, ({{a q}\over
    {b c}};q^2)_k }\over { \let \over / (q^2;q^2)_k \,
(d q;q^2)_k\, ({{a {q^2}}\over d};q^2)_k }}
q^{2k}
.
\endmultline\tag A.2$$
Setting $a=q^{-2n}$ in this transformation formula, 
with $n$ integral and $n\ge1$, we
obtain a summation for a {\it terminating} series,
$$\multline
\sum_{k = 0}^{n}{{\left( 1 - {q^{3 k - 2 n}} \right)  }\over
    {\left( 1 - {q^{-2 n}} \right)}}
{{({\let \over / b}; q) _{k} \, ({\let \over / c}; q) _{k}  \,
({\let \over / {{{q^{1 - 2 n}}}\over {b c}}}; q) _{k}  }\over
{        ({\let \over / q}; q) _{k} \,
({\let \over / {{{q^{1 - 2 n}}}\over d}}; q) _{k}  \,
({\let \over / d}; q) _{k}  }}\\
\cdot
{{      ({\let \over / {q^{-2 n}}}; {q^2}) _{k}\,
                ({\let \over / d}; {q^2}) _{k}  \,
                ({\let \over / {{{q^{1 - 2 n}}}\over d}}; {q^2}) _{k}  
    }\over 
      {       ({\let \over / {{{q^{2 - 2 n}}}\over b}}; {q^2}) _{k} 
\,
        ({\let \over / {{{q^{2 - 2 n}}}\over c}}; {q^2}) _{k} \,
 ({\let \over / b c q}; {q^2}) _{k}  
           }}
{q^k} = 0.
\endmultline\tag A.3$$
Now we let $d$ tend to $q^{-n}$. The effect is that all terms with
$n/2< k<n$ vanish. Hence, only the terms with $0\le k\le n/2$ and the
one with $k=n$ remain. Thus, if we simplify the resulting $(k=n)$-term
and put it on
the right-hand side, we obtain exactly (A.1).\quad \quad \qed
\enddemo

Finally we prove the summation which is needed in the proof of
Theorem~2.
\proclaim{Lemma A2}Let $n$ be a positive integer. Then
$$\multline \sum_{k = 0}^{{\fl{n/2}}}
        \left( 1 + q - {B^2} {q^{1 + 2 n + k}} - B {q^{1 + 2 k}} \right) 
\frac {(1-q^{3k-2n})} {(1-q^{-2n})}\\
\cdot\frac {         ({\let \over / -{q^{-n}}}; q) _{k}  \,
            ({\let \over / {B^2} {q^{2 n}}}; q) _{k}     \,
    ({\let \over / {{{q^{-1 - 2 n}}}\over B}}; q) _{k}  } 
     {            ({\let \over / -{{{q^{1 - 2 n}}}\over B}}; q) _{k} 
\,
            ({\let \over / {q^{1 - n}}}; q) _{k}    \,
         ({\let \over / q}; q) _{k}       }
{{ 
           ({\let \over / {q^{-n}}}; {q^2}) _{k}  \,
        ({\let \over / {q^{1 - n}}}; {q^2}) _{k}  
   }\over 
      {  ({\let \over / B {q^2}}; {q^2}) _{k}  \,
        ({\let \over / B {q^3}}; {q^2}) _{k}  }}{q^{2 k}} \\= 
q^{-\binom n2}
       \left( 1 + q - B {q^{1 + 2 n}} - {B^2} {q^{1 + 3 n}} \right)  
\frac {  ({\let \over / -q}; q) _{n-1}  } 
   {    ({\let \over / - B {q^n}  }; q) _{n}}
  {{ 
       ({\let \over / {B^2} {q^{2 n}}}; q) _{n} }\over 
     {({\let \over / B {q^2}}; q) _{n}   }}.
\endmultline\tag A.4$$
\endproclaim
\demo{Proof} 
For the left-hand side of (A.4) we have
$$\multline 
\sum_{k = 0}^{{\fl{n/2}}}
        \left( 1 + q - {B^2} {q^{1 + 2 n + k}} - B {q^{1 + 2 k}} \right) 
\frac {(1-q^{3k-2n})} {(1-q^{-2n})}\\
\cdot\frac {{         ({\let \over / -{q^{-n}}}; q) _{k}  \,
            ({\let \over / {B^2} {q^{2 n}}}; q) _{k}     \,
    ({\let \over / {{{q^{-1 - 2 n}}}\over B}}; q) _{k}  }} 
     {{            ({\let \over / -{{{q^{1 - 2 n}}}\over B}}; q) _{k} 
\,
            ({\let \over / {q^{1 - n}}}; q) _{k}    \,
         ({\let \over / q}; q) _{k}       }}
{{ 
           ({\let \over / {q^{-n}}}; {q^2}) _{k}  \,
        ({\let \over / {q^{1 - n}}}; {q^2}) _{k}  
   }\over 
      {  ({\let \over / B {q^2}}; {q^2}) _{k}  \,
        ({\let \over / B {q^3}}; {q^2}) _{k}  }}{q^{2 k}} \\= 
q\frac {{\let \over / \left( 1 + B {q^{2 n}} \right)  
       \left( 1 - {B^2} {q^{2 n}} \right)  
       \left( 1 - {B^2} {q^{1 + 2 n}} \right)  }} 
  {{\let \over / ({1 - {B^3} {q^{1 + 4 n}}})}}
\sum _{k=0} ^{\fl{n/2}}
\frac {(1-q^{3k-2n})} {(1-q^{-2n})}\hskip3cm\\
\hskip2cm
\cdot \frac {{\let \over / (-{q^{-n}};q)_k\, ({B^2} {q^{2 +
        2 n}};q)_k\, ({{{q^{-1 - 2 n}}}\over B};q)_k}} 
      {{\let \over / (-{1\over {B {q^{2 n}}}};q)_k\, 
      ({q^{1 - n}};q)_k\, (q;q)_k}}
\frac {{\let \over / ({q^{-n}};q^2)_k\, ({q^{1 - n}};q^2)_k}} 
      {{\let \over / (B {q^2};q^2)_k\, (B {q^3};q^2)_k}} q^k
\\+
\frac {\left( 1 - B q \right)  \left( 1 - B {q^{1 + 2 n}} \right)  } 
    {({1 - {B^3} {q^{1 + 4 n}}})}
\sum _{k=0} ^{\fl{n/2}}
\frac {(1-q^{3k-2n})} {(1-q^{-2n})}
\frac {{\let \over / (-{q^{-n}};q)_k\, ({B^2} {q^{2 n}};q)_k\,
        ({1\over {B {q^{2 n}}}};q)_k}} 
      {{\let
        \over / (-{{{q^{1 - 2 n}}}\over B};q)_k\, ({q^{1 - n}};q)_k\,
     (q;q)_k}}\\\hskip3cm \cdot
\frac {{\let \over / ({q^{-n}};q^2)_k\, ({q^{1 - n}};q^2)_k}} 
      {{\let \over / (B q;q^2)_k\, (B {q^2};q^2)_k}} q^k.
\endmultline$$
Each of the series on the right-hand side of this identity 
can be summed by means of Lemma~A1.
After little manipulation we arrive at (A.4).\quad \quad \qed
\enddemo


\NoBlackBoxes

\Refs

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\endRefs
\enddocument
