How to let three subscripts and three superscripts to be aligned

Question

Suppose I need to get a formula like $L^{2,3/4,5/6}_{1,2,3}$ in Latex. Then $2$--$1$, $3/4$--$2$, $5/6$--$3$ is not aligned. See Then how can I let these three to be aligned?

Answer 1

Let us make things complicated.

\documentclass{article}
\usepackage{calc}
\begin{document}
$L^{2,3/4,5/6}_{1,\makebox[\widthof{3/4}]{$\scriptstyle 2,$}\makebox[\widthof{5/6}]{$\scriptstyle 3\hphantom{,}$}}$
\end{document}

I agree with Benjamin's comments. Hence an attempt to increase the spaces.

\documentclass{article}
\usepackage{calc}
\begin{document}
$L^{2,3/4,5/6}_{1,\makebox[\widthof{3/4}]{$\scriptstyle 2,$}\makebox[\widthof{5/6}]{$\scriptstyle 3\hphantom{,}$}}$
\end{document}

One can also resort to reduce the space before and after / by using

\newcommand*{\mybar}{\kern-.2ex/\kern-.2ex}

instead of /.

Answer 2

Just for fun, with low level commands:

\documentclass{article}
\newcommand{\stacked}[2]{%
  \vcenter{\tabskip=0pt
    \gdef\separator{/}
    \halign{$\scriptstyle##$\hfil&&$\scriptstyle\separator##$\hfil\cr
      #1\cr
      \noalign{\nointerlineskip\kern.2ex\gdef\separator{\phantom{/}}}
      #2\crcr
      \noalign{\kern.275ex}
    }%
  }%
}

\begin{document}
$L^{2/}_{1,}L\stacked{2,3&4,5&6}{1,&2,&3}$

\end{document}

---

If centered alignment is preferred, here's a (rather complicated) way.

\documentclass{article}
\usepackage{xparse}
\newlength{\commawidth}
\newcommand{\dostacked}[2]{%
  \settowidth\commawidth{$\scriptstyle,\,$}
  \vcenter{
    \tabskip=0pt
    \halign{\hfil$\scriptstyle##\vphantom{/}$\hfil\tabskip\commawidth
            &&\hfil$\scriptstyle##\vphantom{/}$\hfil\cr
      #1\cr
      \noalign{\nointerlineskip\kern.2ex}
      #2\crcr
      \noalign{\kern.275ex}
    }%
  }%
  \kern-\commawidth
}
\ExplSyntaxOn
\NewDocumentCommand{\nospacecomma}{}
 {
  \rlap{$\scriptstyle,$}
 }
\cs_set_eq:NN \egreg_dostacked:nn \dostacked
\cs_generate_variant:Nn \egreg_dostacked:nn { xx }
\NewDocumentCommand{\stacked}{mm}
 {
  \seq_set_split:Nnn \l_tmpa_seq { , } { #1 }
  \seq_set_split:Nnn \l_tmpb_seq { , } { #2 }
  \egreg_dostacked:xx
   { \seq_use:Nn \l_tmpa_seq { \nospacecomma& } }
   { \seq_use:Nn \l_tmpb_seq { \nospacecomma& } }
 }
\ExplSyntaxOff

\begin{document}
$L^{2/}_{1}L\stacked{2,3/4,5/6}{1,2,3}X$

\end{document}

The first L^{2/}{1} is just for comparing the heights, while the trailing $X$ is for checking that no space is added at the end.

Not the best code. But, hey, it works!

Answer 3

Using the tabstackengine package introduced at Writing a table with equally spaced columns, based on the widest column, we make the sub/super array a stack, and only have to tweak the space between the sub/super line and the vertical placement of the subscript line. A long stackgap of 6pt (vs. 7pt) matches the default sub/super placement, but the slashes in the superscript tend to interfere, so I increased it.

The alignment of the stack is centered, though an optional argument of [l] or [r] on the \tabbedVectorStack can change that.

\documentclass{article}
\usepackage{tabstackengine}
\begin{document}
\setstackgap{L}{7pt}% SUB- SUPER- BASELINE SHIFT
\( L_1^2 \textrm{~versus~} L\raisebox{0.5pt}% SUB- BASELINE
  {\scriptsize\tabbedVectorstack{2,&3/4,&5/6\\1,&2,&3}} \)
\end{document}

Answer 4

Code

\[
L^{2,3/4,5/6}_{1, \hspace{9pt}2, \hspace{8pt}3}
\]

Second Version: where phantom is used instead. Thanks for the comments, leading to improvement.

Code:

\[
L^{2,3/4,5/6}_{1,\phantom{3/}2,\phantom{5/}3}  % This (shown here) aligns to the right.
\]

\[
L^{2,3/4,5/6}_{1,2\phantom{/4},3\phantom{/6}}  % This (not shown here) aligns to the left.
\]

\[
L^{2,3/4,5/6}_{1,\phantom{3}2\phantom{4},\phantom{5}3\phantom{6}} % This (also not shown here) aligns to the center.
\]

Conclusion: Pair up with superscripts and use phantom appropriately.

Answer 5

This is very simple and it almost works:

Almost: $ L^{2,}_{1,} {}^{3/4,}_{2,} {}^{5/6}_3 $

However, the first superscript (2,) is smaller than the others, hence slightly lower. So here's how to fix it without guesswork about the dimensions (now simplifed thanks to @sgmoye):

After: $ L^{2,\mathstrut}_{1,} {}^{3/4,}_{2,} {}^{5/6}_3 $

Note that in other examples, some exponents might be too big to fit in the space TeX reserves with a \mathstrut; you'd then need to reserve vertical space in a more general way. Here's a (somewhat clumsy) way to do that:

Version 3: $ L^{2,\llap{\phantom{\scriptsize /}}}_{1,} {}^{3/4,}_{2,} {}^{5/6}_3 $

Here, our model for vertical space is a (scriptsize) slash; but it could be anything. \phantom makes it invisible, and \llap ensures that it claims no horizontal space.

Answer 6

L_{1,\phantom{3}2\phantom{4},\phantom{5}3\phantom{6}}^{2,3/4,5/6}

ADDENDUM: The \phantom macro puts a blank space. The space size is equal to the horizontal size of the enclosed argument. For example, \phantom{3} puts a blank space with a length which is equal to the horizontal size of 3, etc... So, it mimics the effect of something in there that we can not see it ( that is why it's call "phantom" ).