In the scope of gaining space, I would like my superscript minuses to be shortened from:

$ J^{-1} $


$ J^{\text -1} $

using a rule.

  • What do you mean by "using a rule"? – user156344 Mar 27 at 9:37
  • @JouleV I believe a hyphen is meant.@Dash: What's the question? – campa Mar 27 at 9:38
  • Sorry but I don’t quite understand. You wanna shorten a short sequence of characters to a long sequence? – L. F. Mar 27 at 9:39
  • What I mean by a rule is that I don't want a solution that consists in putting "\text" in every brackets like I did. Instead, I would like a a way to define it globally. In the spirit of defining a "newcommand" perhaps. If you run my code, you'll see that the second proposition has a shorter superscript "-1". It therefore uses less space which is my final goal. I am open to other suggestions. – Dahs Mar 27 at 9:52
  • @Dash DON'T DO THAT!!!. Your readers will be grateful. – user156344 Mar 27 at 9:59

I'm not sure that “saving space” at the expense of readability should be pursued.

In the picture, top is the standard, bottom is the “space saving” version.



\AtBeginDocument{\mathcode`-="8000 }


$-J^{\csname std@minus\endcsname1}$



enter image description here

  • I think I will not pursue this idea after all. But thanks for the code anyway, it helps me understanding the "makeatletter/makeatother" stuff. I am surprised you think it looks less readable. Of course the symbol is smaller, but it looks more like the way I (and possibly others) write on paper. The minus symbol is still visible and the reader should think about an "inverse". And because I insisted on the definition of $J^{-1}$ at the beginning of the text. I don't think this should be read from 100 feet using binoculars and I can read it with myopia. Can you develop your comment ? – Dahs Mar 27 at 10:32
  • @Dash If you want that the inverse is denoted differently, it's better to define a command for it, and not monkeying with the minus sign generally. – egreg Mar 27 at 10:52
  • Why are general modifications of such sign a bad idea ? – Dahs Mar 27 at 10:56
  • @Dash Try $a^{m-n}$. – egreg Mar 27 at 10:57

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