unary vs binary operators (LaTeX)

[David Rosoff]

Hi,

I would like to know whether one can force the binary operator \ast
(asterisk character) to act as a unary operator throughout a document, or
alternatively, whether one can force the spacing to be the same either way
it is used.

(Context: I'm writing about the Hodge star operator, which is a unary
operator, but frequently it is iterated. LaTeX seems to be parsing it as a
unary operator in the context $\ast \omega$, and binary in the context
$\ast \ast \omega$ or even $\omega + \ast \omega$.)

Thanks in advance,

-dave

[Maurizio Loreti]

David Rosoff <ros...@math.washington.edu> writes:

> I would like to know whether one can force the binary operator \ast
> (asterisk character) to act as a unary operator

\mathord

--
Maurizio Loreti http://www.pd.infn.it/~loreti/mlo.html
Dept. of Physics, Univ. of Padova, Italy ROT13: ybe...@cq.vasa.vg

[Donald Arseneau]

David Rosoff <ros...@math.washington.edu> writes:

> I would like to know whether one can force the binary operator \ast
> (asterisk character) to act as a unary operator throughout a document, or

Just \ast or * also?

\DeclareMathSymbol{*}{\mathop}{symbols}{"03} % \ast
\DeclareMathSymbol{\ast}{\mathop}{symbols}{"03}

Donald Arseneau as...@triumf.ca

[Dan Luecking]

On 10 Jun 2003 09:18:38 +0200, Maurizio Loreti <m...@foobar.it> wrote:

>David Rosoff <ros...@math.washington.edu> writes:
>
>> I would like to know whether one can force the binary operator \ast
>> (asterisk character) to act as a unary operator
>
>\mathord

I believe Mr Loreti means \renewcommand{\ast}{\mathord{*}},
which will use precisely the symbol that * alone uses, but
convert it to an ordinary symbol. In some contexts
\mathop{*}\nolimits
might be preferred.

Dan

--
Dan Luecking Department of Mathematical Sciences
University of Arkansas Fayetteville, Arkansas 72701
luecking at uark dot edu

[Maurizio Loreti]

Dan Luecking <luec...@uark.edu> writes:

> On 10 Jun 2003 09:18:38 +0200, Maurizio Loreti <m...@foobar.it> wrote:
>
> >David Rosoff <ros...@math.washington.edu> writes:
> >
> >> I would like to know whether one can force the binary operator \ast
> >> (asterisk character) to act as a unary operator
> >
> >\mathord
>
> I believe Mr Loreti means \renewcommand{\ast}{\mathord{*}},

Well, no :-0

My suggestion (a little bit too short, I realize) was not to redefine
the original command (\ast may be needed as a true binary operator
elsewhere) but to define a new one or to use a straight \mathord{\ast}
in the text. BTW, I am not so sure that * and \ast are required to
generate the same character for every available font; use
\mathord{\ast} and not \mathord{*}.

[David Rosoff]

On 11 Jun 2003, Maurizio Loreti wrote:

> Dan Luecking <luec...@uark.edu> writes:
>
> > On 10 Jun 2003 09:18:38 +0200, Maurizio Loreti <m...@foobar.it> wrote:
> >
> > >David Rosoff <ros...@math.washington.edu> writes:
> > >
> > >> I would like to know whether one can force the binary operator \ast
> > >> (asterisk character) to act as a unary operator
> > >
> > >\mathord
> >
> > I believe Mr Loreti means \renewcommand{\ast}{\mathord{*}},
>
> Well, no :-0
>
> My suggestion (a little bit too short, I realize) was not to redefine
> the original command (\ast may be needed as a true binary operator
> elsewhere) but to define a new one or to use a straight \mathord{\ast}
> in the text. BTW, I am not so sure that * and \ast are required to
> generate the same character for every available font; use
> \mathord{\ast} and not \mathord{*}.

Thanks for the tips, works great. I would rather have not used this
operator at all: what a lousy choice for a unary operator!

Cheers,

-dave

[Dan Luecking]

On 11 Jun 2003 07:22:40 +0200, Maurizio Loreti <m...@foobar.it> wrote:

>Dan Luecking <luec...@uark.edu> writes:
>
>> On 10 Jun 2003 09:18:38 +0200, Maurizio Loreti <m...@foobar.it> wrote:
>>
>> >\mathord
>>
>> I believe Mr Loreti means \renewcommand{\ast}{\mathord{*}},
>
>Well, no :-0
>
>My suggestion (a little bit too short, I realize) was not to redefine
>the original command (\ast may be needed as a true binary operator
>elsewhere) but to define a new one or to use a straight \mathord{\ast}
>in the text. BTW, I am not so sure that * and \ast are required to
>generate the same character for every available font; use
>\mathord{\ast} and not \mathord{*}.

As I said in the snipped part, one is guaranteed that \mathord{*}
will produce the same character as * alone (so of course if *
is not correct, don't use it!).

Also, the original poster asked that \ast have certain behavior and
so one has to redefine \ast for this to happen.