%This is an AmS-TeX file %producing an output of 23 pages \input amstex \input amsppt.sty \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: % \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: % \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 \@tempdimb\@tempdima\advance\@tempdimb -\wd\@linechar \if@negarg \hskip -\@tempdimb \else \hskip \@tempdimb \fi \multiply\@tempdima \@m \@tempcnta \@tempdima \@tempdima \wd\@linechar \divide\@tempcnta \@tempdima \@tempdima \ht\@linechar \multiply\@tempdima \@tempcnta \divide\@tempdima \@m \advance\@clnht \@tempdima \ifdim \@linelen <\wd\@linechar \hskip \wd\@linechar \else\@upordown\@clnht\copy\@linechar\fi} \def\@hline{\ifnum \@xarg <0 \hskip -\@linelen \fi \vrule height\Pfadd@cke width \@linelen depth\Pfadd@cke \ifnum \@xarg <0 \hskip -\@linelen \fi} \def\@getlinechar(#1,#2){\@tempcnta#1\relax\multiply\@tempcnta 8 \advance\@tempcnta -9 \ifnum #2>0 \advance\@tempcnta #2\relax\else \advance\@tempcnta -#2\relax\advance\@tempcnta 64 \fi \char\@tempcnta} \def\Vektor(#1,#2)#3(#4,#5){\unskip\leavevmode \xcoord#4\relax \ycoord#5\relax \raise\ycoord \Einheit\hbox to0pt{\hskip\xcoord \Einheit \Vector@(#1,#2){#3}\hss}} \def\Vector@(#1,#2)#3{\@xarg #1\relax \@yarg #2\relax \@tempcnta \ifnum\@xarg<0 -\@xarg\else\@xarg\fi \ifnum\@tempcnta<5\relax \@linelen=#3\Einheit \ifnum\@xarg =0 \@vvector \else \ifnum\@yarg =0 \@hvector \else \@svector\fi \fi \else\@badlinearg\fi} \def\@hvector{\@hline\hbox to 0pt{\@linefnt \ifnum \@xarg <0 \@getlarrow(1,0)\hss\else \hss\@getrarrow(1,0)\fi}} \def\@vvector{\ifnum \@yarg <0 \@downvector \else \@upvector \fi} \def\@svector{\@sline \@tempcnta\@yarg \ifnum\@tempcnta <0 \@tempcnta=-\@tempcnta\fi \ifnum\@tempcnta <5 \hskip -\wd\@linechar \@upordown\@clnht \hbox{\@linefnt \if@negarg \@getlarrow(\@xarg,\@yyarg) \else \@getrarrow(\@xarg,\@yyarg) \fi}% \else\@badlinearg\fi} \def\@upline{\hbox to \z@{\hskip -.5\Pfadd@cke \vrule width \Pfadd@cke height \@linelen depth \z@\hss}} \def\@downline{\hbox to \z@{\hskip -.5\Pfadd@cke \vrule width \Pfadd@cke height \z@ depth \@linelen \hss}} \def\@upvector{\@upline\setbox\@tempboxa\hbox{\@linefnt\char'66}\raise \@linelen \hbox to\z@{\lower \ht\@tempboxa\box\@tempboxa\hss}} \def\@downvector{\@downline\lower \@linelen \hbox to \z@{\@linefnt\char'77\hss}} \def\@getlarrow(#1,#2){\ifnum #2 =\z@ \@tempcnta='33\else \@tempcnta=#1\relax\multiply\@tempcnta \sixt@@n \advance\@tempcnta -9 \@tempcntb=#2\relax\multiply\@tempcntb \tw@ \ifnum \@tempcntb >0 \advance\@tempcnta \@tempcntb\relax \else\advance\@tempcnta -\@tempcntb\advance\@tempcnta 64 \fi\fi\char\@tempcnta} \def\@getrarrow(#1,#2){\@tempcntb=#2\relax \ifnum\@tempcntb < 0 \@tempcntb=-\@tempcntb\relax\fi \ifcase \@tempcntb\relax \@tempcnta='55 \or \ifnum #1<3 \@tempcnta=#1\relax\multiply\@tempcnta 24 \advance\@tempcnta -6 \else \ifnum #1=3 \@tempcnta=49 \else\@tempcnta=58 \fi\fi\or \ifnum #1<3 \@tempcnta=#1\relax\multiply\@tempcnta 24 \advance\@tempcnta -3 \else \@tempcnta=51\fi\or \@tempcnta=#1\relax\multiply\@tempcnta \sixt@@n \advance\@tempcnta -\tw@ \else \@tempcnta=#1\relax\multiply\@tempcnta \sixt@@n \advance\@tempcnta 7 \fi\ifnum #2<0 \advance\@tempcnta 64 \fi \char\@tempcnta} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \def\Diagonale(#1,#2)#3{\unskip\leavevmode \xcoord#1\relax \ycoord#2\relax \raise\ycoord \Einheit\hbox to0pt{\hskip\xcoord \Einheit \Line@(1,1){#3}\hss}} \def\AntiDiagonale(#1,#2)#3{\unskip\leavevmode \xcoord#1\relax \ycoord#2\relax %\advance\xcoord by -0.05\relax \raise\ycoord \Einheit\hbox to0pt{\hskip\xcoord \Einheit \Line@(1,-1){#3}\hss}} \def\Pfad(#1,#2),#3\endPfad{\unskip\leavevmode \xcoord#1 \ycoord#2 \thicklines\ZeichnePfad#3\endPfad\thinlines} \def\ZeichnePfad#1{\ifx#1\endPfad\let\next\relax \else\let\next\ZeichnePfad \ifnum#1=1 \raise\ycoord \Einheit\hbox to0pt{\hskip\xcoord \Einheit \vrule height\Pfadd@cke width1 \Einheit depth\Pfadd@cke\hss}% \advance\xcoord by 1 \else\ifnum#1=2 \raise\ycoord \Einheit\hbox to0pt{\hskip\xcoord \Einheit \hbox{\hskip-\PfadD@cke\vrule height1 \Einheit width\PfadD@cke depth0pt}\hss}% \advance\ycoord by 1 \else\ifnum#1=3 \raise\ycoord \Einheit\hbox to0pt{\hskip\xcoord \Einheit \Line@(1,1){1}\hss} \advance\xcoord by 1 \advance\ycoord by 1 \else\ifnum#1=4 \raise\ycoord \Einheit\hbox to0pt{\hskip\xcoord \Einheit \Line@(1,-1){1}\hss} \advance\xcoord by 1 \advance\ycoord by -1 \fi\fi\fi\fi \fi\next} \def\hSSchritt{\leavevmode\raise-.4pt\hbox to0pt{\hss.\hss}\hskip.2\Einheit \raise-.4pt\hbox to0pt{\hss.\hss}\hskip.2\Einheit \raise-.4pt\hbox to0pt{\hss.\hss}\hskip.2\Einheit \raise-.4pt\hbox to0pt{\hss.\hss}\hskip.2\Einheit \raise-.4pt\hbox to0pt{\hss.\hss}\hskip.2\Einheit} \def\vSSchritt{\vbox{\baselineskip.2\Einheit\lineskiplimit0pt \hbox{.}\hbox{.}\hbox{.}\hbox{.}\hbox{.}}} \def\DSSchritt{\leavevmode\raise-.4pt\hbox to0pt{% \hbox to0pt{\hss.\hss}\hskip.2\Einheit \raise.2\Einheit\hbox to0pt{\hss.\hss}\hskip.2\Einheit \raise.4\Einheit\hbox to0pt{\hss.\hss}\hskip.2\Einheit \raise.6\Einheit\hbox to0pt{\hss.\hss}\hskip.2\Einheit \raise.8\Einheit\hbox to0pt{\hss.\hss}\hss}} \def\dSSchritt{\leavevmode\raise-.4pt\hbox to0pt{% \hbox to0pt{\hss.\hss}\hskip.2\Einheit \raise-.2\Einheit\hbox to0pt{\hss.\hss}\hskip.2\Einheit \raise-.4\Einheit\hbox to0pt{\hss.\hss}\hskip.2\Einheit \raise-.6\Einheit\hbox to0pt{\hss.\hss}\hskip.2\Einheit \raise-.8\Einheit\hbox to0pt{\hss.\hss}\hss}} \def\SPfad(#1,#2),#3\endSPfad{\unskip\leavevmode \xcoord#1 \ycoord#2 \ZeichneSPfad#3\endSPfad} \def\ZeichneSPfad#1{\ifx#1\endSPfad\let\next\relax \else\let\next\ZeichneSPfad \ifnum#1=1 \raise\ycoord \Einheit\hbox to0pt{\hskip\xcoord \Einheit \hSSchritt\hss}% \advance\xcoord by 1 \else\ifnum#1=2 \raise\ycoord \Einheit\hbox to0pt{\hskip\xcoord \Einheit \hbox{\hskip-2pt \vSSchritt}\hss}% \advance\ycoord by 1 \else\ifnum#1=3 \raise\ycoord \Einheit\hbox to0pt{\hskip\xcoord \Einheit \DSSchritt\hss} \advance\xcoord by 1 \advance\ycoord by 1 \else\ifnum#1=4 \raise\ycoord \Einheit\hbox to0pt{\hskip\xcoord \Einheit \dSSchritt\hss} \advance\xcoord by 1 \advance\ycoord by -1 \fi\fi\fi\fi \fi\next} \def\Koordinatenachsen(#1,#2){\unskip \hbox to0pt{\hskip-.5pt\vrule height#2 \Einheit width.5pt depth1 \Einheit}% \hbox to0pt{\hskip-1 \Einheit \xcoord#1 \advance\xcoord by1 \vrule height0.25pt width\xcoord \Einheit depth0.25pt\hss}} \def\Koordinatenachsen(#1,#2)(#3,#4){\unskip \hbox to0pt{\hskip-.5pt \ycoord-#4 \advance\ycoord by1 \vrule height#2 \Einheit width.5pt depth\ycoord \Einheit}% \hbox to0pt{\hskip-1 \Einheit \hskip#3\Einheit \xcoord#1 \advance\xcoord by1 \advance\xcoord by-#3 \vrule height0.25pt width\xcoord \Einheit depth0.25pt\hss}} \def\Gitter(#1,#2){\unskip \xcoord0 \ycoord0 \leavevmode \LOOP\ifnum\ycoord<#2 \loop\ifnum\xcoord<#1 \raise\ycoord \Einheit\hbox to0pt{\hskip\xcoord \Einheit\Punkt\hss}% \advance\xcoord by1 \repeat \xcoord0 \advance\ycoord by1 \REPEAT} \def\Gitter(#1,#2)(#3,#4){\unskip \xcoord#3 \ycoord#4 \leavevmode \LOOP\ifnum\ycoord<#2 \loop\ifnum\xcoord<#1 \raise\ycoord \Einheit\hbox to0pt{\hskip\xcoord \Einheit\Punkt\hss}% \advance\xcoord by1 \repeat \xcoord#3 \advance\ycoord by1 \REPEAT} \def\Label#1#2(#3,#4){\unskip \xdim#3 \Einheit \ydim#4 \Einheit \def\lo{\advance\xdim by-.5 \Einheit \advance\ydim by.5 \Einheit}% \def\llo{\advance\xdim by-.25cm \advance\ydim by.5 \Einheit}% \def\loo{\advance\xdim by-.5 \Einheit \advance\ydim by.25cm}% \def\o{\advance\ydim by.25cm}% \def\ro{\advance\xdim by.5 \Einheit \advance\ydim by.5 \Einheit}% \def\rro{\advance\xdim by.25cm \advance\ydim by.5 \Einheit}% \def\roo{\advance\xdim by.5 \Einheit \advance\ydim by.25cm}% \def\l{\advance\xdim by-.30cm}% \def\r{\advance\xdim by.30cm}% \def\lu{\advance\xdim by-.5 \Einheit \advance\ydim by-.6 \Einheit}% \def\llu{\advance\xdim by-.25cm \advance\ydim by-.6 \Einheit}% \def\luu{\advance\xdim by-.5 \Einheit \advance\ydim by-.30cm}% \def\u{\advance\ydim by-.30cm}% \def\ru{\advance\xdim by.5 \Einheit \advance\ydim by-.6 \Einheit}% \def\rru{\advance\xdim by.25cm \advance\ydim by-.6 \Einheit}% \def\ruu{\advance\xdim by.5 \Einheit \advance\ydim by-.30cm}% #1\raise\ydim\hbox to0pt{\hskip\xdim \vbox to0pt{\vss\hbox to0pt{\hss$#2$\hss}\vss}\hss}% } \catcode`\@=13 \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$$ For convenience, let us denote the determinant on the left-hand side of (2.1) by $D_1(x,y;n)$. Our proof of (2.1) is divided into four steps. In steps~1 and 2 we show that the right-hand side of (2.1) divides $D_1(x,y;n)$ as a polynomial in $x$ and $y$. Then, in Step~3 we show that the (total) degree in $x$ and $y$ of $D_1(x,y;n)$ is $6\binom n2$, which is exactly the degree of the right-hand side, so that $D_1(x,y;n)$ is a constant multiple of the right-hand side. Finally we show in step~4 that this constant equals 1. \smallskip {\it Step 1. $\prod _{i=0} ^{n-1}(x^2yq^{2i};q)_{i}$ is a factor of $D_1(x,y;n)$.} Let us first concentrate on a typical factor $(1-x^2yq^{2i+l})$, $0\le i\le n-1$, $0\le lu+v.\tag3.22$$ 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 $kv$. 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