答案1
三条建议:
首先,第 4 行和第 6 行的箭头太长了。这是因为第 7 行第 3 列的条目太宽,tikz-cd
不会与列重叠。您可以使用\llap
该条目中的命令覆盖此问题。
请注意,您只能\llap
最多该条目,而不是全部,以便上面的箭头仍然指向该条目的一部分。
\begin{tikzcd}
AAAAAAAAAAAAA\arrow[rrr,equal]\arrow[dd] &&& AAAAAAAAAAA\arrow[d,"s"]\\
&&& AAAAAAAAAAAAAAAA\arrow[d,"P"]\\
AAAAAAAAAAAAAAAAAA\arrow[d] &&& AAAAAAAAAAAAAAAAAAAA\arrow[d]\\
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAA\arrow[r]\arrow[d] & I_1\arrow[r] & I_2\arrow[r,"\simeq"]\arrow[dd]& AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA\arrow[dd,equal]\\
AAAAAAAAAAAAAAA\arrow[d]\\
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA\arrow[rr] && I_3\arrow[d] & AAAAAAAAAAAAAAAAAAAAAAAAAAAAAA\arrow[d]\\
&& \llap{$BBBBBBBBBBBBBBBBBB$}CCCC\arrow[r,equal] & AAAAAAAAAAAAAAAAAAAAA\arrow[d,"f"]\\
&&& AAAAAAAAAAAAAAA
\end{tikzcd}
其次,如果您不介意稍微重新组织一下,您可以通过删除一列并将地图变为垂直来节省更多I_1
空间I_2
。
\begin{tikzcd}
AAAAAAAAAAAAA\arrow[rr,equal]\arrow[d] && AAAAAAAAAAA\arrow[d,"s"]\\
AAAAAAAAAAAAAAAAAA\arrow[d] && AAAAAAAAAAAAAAAA\arrow[d,"P"]\\
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAA\arrow[r]\arrow[d] & I_1\arrow[d] & AAAAAAAAAAAAAAAAAAAA\arrow[d]\\
AAAAAAAAAAAAAAA\arrow[d] & I_2\arrow[r,"\simeq"]\arrow[d]& AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA\arrow[d,equal]\\
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA\arrow[r] & I_3\arrow[d] & AAAAAAAAAAAAAAAAAAAAAAAAAAAAAA\arrow[d]\\
& \llap{$BBBBBBBBBBBBBBBBBB$}CCCC\arrow[r,equal] & AAAAAAAAAAAAAAAAAAAAA\arrow[d,"f"]\\
&& AAAAAAAAAAAAAAA
\end{tikzcd}
第三,如果您仍然需要更多空间,则\mu hom(G_1,G_2)
在许多条目中都有重复的表达式。您可以将该表达式分配给单个字符。例如,然后在您的图表中Let $\mathcal{M}=\mu hom(G_1,G_2)$.
替换。\mathcal{M}
答案2
aligned
您可以使用弯曲箭头将最大的对象分成两条线,以避免重复另一个大对象,从而节省大量空间。
其中aligned
之一需要\amp
技巧,因为tikzcd
符号里面&
有另一层含义。
请参阅手册以了解如何填写带圈的字母。
\documentclass{article}
\usepackage{amsmath,mathrsfs,amssymb}
\usepackage{tikz-cd}
\usepackage{rotating}
\DeclareMathOperator{\mhom}{\mu hom}
\newcommand{\uR}{\mathrm{R}}
\newcommand{\CC}{\mathbb{C}}
\begin{document}
\begin{sidewaysfigure}
\centering\footnotesize
\let\amp=& % for the inner aligned
\begin{tikzcd}
\uR p^{}_{1_!} p^{-1}_{2^a} \mhom(G_1,G_2) \arrow[rrr,equals] \arrow[dd]
%&&[-11em]&[-3em]
&&&
\uR p^{}_{1_!} p^{-1}_{2^a} \mhom(G_1,G_2) \arrow[d,"s"]
\\
&&&
\uR p^{}_{1_!}(K_W\otimes p^{-1}_{2^a} \mhom(G_1,G_2)) \arrow[d,"P"]
\\
\uR p^{}_{1_!}(K_W\otimes p^{-1}_{2^a} \mhom(G_1,G_2)) \arrow[d] &&&
\uR p^{}_{1_!}(p^{-1}_1 E_U\otimes K_W\otimes p^{-1}_{2^a} \mhom(G_1,G_2)) \arrow[d]
\\
\uR p^{}_{1_!}(\mhom(L,\Omega_{X\times Y/X}\otimes p^{-1}_{2^a}\mhom(G_1,G_2))
\arrow[r] \arrow[d] &
\mathscr{I}_1 \arrow[r] &
\mathscr{I}_2 \arrow[r,"\simeq"] \arrow[dd] &
\begin{aligned}[t]
\amp\uR p^{}_{1_!}(\uR p^{}_{12_!}(p^{-1}_{12}\mhom(\CC_{\Delta_X},\Omega_{X\times X/X})[d_X] \\
\amp\qquad \otimes p^{-1}_{2^a3}\mhom(L,\Omega_{X\times Y/X})
\otimes p^{-1}_{3^a}\mhom(G_1,G_2))) \arrow[dd,equals]
\end{aligned}
\\
\mhom(L\circ G_1,\Omega_{X\times Y/X}\circ G_2) \arrow[d]
\\
\uR\tilde{p}^{}_{1_!}(\delta^a_* E_U
\otimes \tilde{p}^{-1}_{2^a}\mhom(L\circ G_1,\Omega_{X\times Y/X}\circ G_2))
\arrow[rr] &&
\mathscr{I}_3 \arrow[dr,to path=|-(\tikztotarget.real west)] &
\mhom(\CC_{\Delta_X},\Omega_{X\times X/X})[d_X]\circ
\mhom(L,\Omega_{X\times Y/X})\circ\mhom(G_1,G_2) \arrow[d] \\
&&&
\mhom(\CC_{\Delta_X}\circ L\circ G_1,\Omega_{X\times X/X}\circ\Omega_{X\times Y/X}\circ G_2)[d_X]
\arrow[d,"\int"]
\\
&&&
\mhom(L\circ G_1,\Omega_{X\times Y/X}\circ G_2)
\end{tikzcd}
\caption{A very big diagram}
\end{sidewaysfigure}
\end{document}