lazy latex: how use '^' and '_' without mapping argument in parenthesis

Question

For e.g. take

$G^\times$

here we do not need to write the curly parenthesis, as in

$G^{\times}$

What is it that one has to add to a newly declared command, e.g.

\newcommand{\foo}{\ensuremath \times\ast 2}

so that

$G^\foo$

yields the same as

$G^{\foo}$

thx

Welcome to TeX.SX! You can have a look at our starter guide to familiarize yourself further with our format. — barbara beeton, Jul 30 '13 at 15:17
\newcommand*{\foo}{{\times\ast 2}} works like a charm (\ensuremath is not needed and even frowned upon). — Qrrbrbirlbel, Jul 30 '13 at 15:19
is equivalent to \newcommand{\foo}{\ensuremath \times\ast 2} — arolle, Jul 30 '13 at 15:20
if you're always going to use \foo as a superscript you really don't want \ensuremath. but that wasn't your question. if you just add another outer pair of braces -- \newcommand{\foo}{{\ensuremath \times\ast 2}} it should withstand the indignity of being used without braces as a superscript. — barbara beeton, Jul 30 '13 at 15:22
@Sigur It is not relevant for this question (and even more so as \foo doesn’t take any arguments). The macro is defined via \def and not \long\def, see What's the difference between \newcommand and \newcommand*? — Qrrbrbirlbel, Jul 30 '13 at 15:24
I have a feeling this has been asked before... can't find it, though... — cgnieder, Jul 30 '13 at 15:25
@barbarabeeton this does not work when setting <0 or alike as a superscript. — arolle, Jul 30 '13 at 15:39
@arolle -- it does work if you define \newcommand{\foo}{{<0}} and use the definition in the subscript. the important thing is that the compound itself must be defined in this way, with the extra braces. there's no way to redefine ^ or _ to automatically do this. but that's not what your question asked. — barbara beeton, Jul 30 '13 at 15:47
@arolle maybe worth reading: How bad for TeX is omitting braces {}, even if the result is the same? — cgnieder, Jul 30 '13 at 15:49

barbara beeton · Answer 1 · 2013-07-30T15:51:44.930

you can "protect" the string you intend to use in sub- or superscripts by doubling the outer braces; only the outermost pair of braces gets stripped off when a definition is "internalized". thus

\newcommand{\foo}{{\ensuremath \times\ast 2}}

will have the desired effect.

however:

you really don't need \ensuremath if you're always going to use this in a sub/superscript.
it's not as "informative" when viewed in the input, since it's not obvious that \foo is a compoind object. omitting the braces at the point of use gets one out of the habit of using braces, which can get one into trouble with previously existing definitions that are "unprotected" compounds, such as \neq, defined as \def\neq{\not=} in fontmath.lts (inherited from plain.tex).

edit: a comment from the op states that this doesn't work "when setting <0 or alike as a superscript". that's true, and there's no way around it, except to make a definition, e.g. \newcommand{\lszero}{{<0}}, and use that. (that is what the original question asked for.)

best not to use \ensuremath if you do use it it needs to be {{\ensuremath{\times\ast 2}}} otherwise the ensuremath just takes\times` as its argument. — David Carlisle, Jul 30 '13 at 20:26

lazy latex: how use '^' and '_' without mapping argument in parenthesis

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