10

I'm looking for the symbol that's similar to \cdot in math mode, but instead of a small circle, it's a square. Like this:

enter image description here

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15

There are several ways for doing this. One idea could be \centerdot from amssymb, but this has the defect of sitting on the baseline, despite its name.

One can fix this by observing that the apparent height of the glyph is three times the real height. So smashing the height to one third and using \vcenter will raise the symbol up to the formula axis.

\documentclass{article}
\usepackage{amssymb}

\makeatletter
\DeclareRobustCommand{\sqcdot}{\mathbin{\mathpalette\morphic@sqcdot\relax}}
\newcommand{\morphic@sqcdot}[2]{%
  \sbox\z@{$\m@th#1\centerdot$}%
  \ht\z@=.33333\ht\z@
  \vcenter{\box\z@}%
}
\makeatother

\begin{document}

$\bar{f} \sqcdot f\simeq c_{a\sqcdot b}$

\end{document}

enter image description here

Second way: import \sqbullet from the mathabx fonts.

\documentclass{article}

\DeclareFontFamily{U}{mathb}{}
\DeclareFontShape{U}{mathb}{m}{n}{
  <-5.5> mathb5
  <5.5-6.5> mathb6
  <6.5-7.5> mathb7
  <7.5-8.5> mathb8
  <8.5-9.5> mathb9
  <9.5-11.5> mathb10
  <11.5-> mathb12
}{}
\DeclareSymbolFont{mathb}{U}{mathb}{m}{n}
\DeclareMathSymbol{\sqcdot}{\mathbin}{mathb}{"0D}% name to be checked

\begin{document}

$\bar{f} \sqcdot f\simeq c_{a\sqcdot b}$

\end{document}

One could avoid wasting a symbol font with the help of \text:

\documentclass{article}
\usepackage{amsmath}

\DeclareFontFamily{U}{mathb}{}
\DeclareFontShape{U}{mathb}{m}{n}{
  <-5.5> mathb5
  <5.5-6.5> mathb6
  <6.5-7.5> mathb7
  <7.5-8.5> mathb8
  <8.5-9.5> mathb9
  <9.5-11.5> mathb10
  <11.5-> mathbb12
}{}
\DeclareRobustCommand{\sqcdot}{%
  \mathbin{\text{\usefont{U}{mathb}{m}{n}\symbol{"0D}}}%
}

\begin{document}

$\bar{f} \sqcdot f\simeq c_{a\sqcdot b}$

\end{document}

Both examples typeset the same

enter image description here

For using the glyph in the MnSymbol font, it is similar:

\documentclass{article}
\usepackage{amsmath}

\DeclareFontFamily{U}{MnSymbolC}{}
\DeclareFontShape{U}{MnSymbolC}{m}{n}{
  <-5.5> MnSymbolC5
  <5.5-6.5> MnSymbolC6
  <6.5-7.5> MnSymbolC7
  <7.5-8.5> MnSymbolC8
  <8.5-9.5> MnSymbolC9
  <9.5-11.5> MnSymbolC10
  <11.5-> MnSymbolCb12
}{}
\DeclareRobustCommand{\sqcdot}{%
  \mathbin{\text{\usefont{U}{MnSymbolC}{m}{n}\symbol{"69}}}%
}

\begin{document}

$\bar{f} \sqcdot f\simeq c_{a\sqcdot b}$

\end{document}

enter image description here

Take your pick.

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  • Great answer, thanks. You may want to check the name mathbb12... I eventually ran into compile errors, which I resolved by changing it to mathb12 (only one b). – 808sound May 3 '17 at 2:57
  • 1
    @808sound Thanks for spotting the typo – egreg May 3 '17 at 6:05
  • 2
    @egreg +1. Many compliments for your jobs. The screenshots are fantastics. – Sebastiano May 3 '17 at 7:21
  • @Sebastiano The screenshots should be credited to the Retina display. ;-) – egreg May 3 '17 at 7:23
  • Great answer! Is there a simple way to modify it for an unfilled square rather than a filled square? Simply using \square is not good enough, since I need the square to be the same size as \bullet and \textopenbullet, whereas \square is too big. – Evan Aad Nov 12 '18 at 10:38
14

Handcraft it.

\documentclass{scrartcl}

\newcommand*\sq{\mathbin{\vcenter{\hbox{\rule{.3ex}{.3ex}}}}}

\begin{document}

$f$ by $\bar f$ gives $\bar f \sq f \simeq c$ for $c$ the constant\dots

\end{document}

enter image description here

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  • 1
    Is it the case that this does not automatically handle different math styles (e.g., \scriptstyle, etc.)? – Steven B. Segletes Oct 15 '15 at 2:32
  • 1
    It looks good for me in different styles, but you could, as always, develop some \mathpalette\sqaux\relax then \if\scriptstyle#1.2ex\else\if\scriptscriptstyle#1.15ex\else.3ex\fi\fi (or whatever dimensions you choose) in place of the dimensions of \rule. – Manuel Oct 15 '15 at 7:24
9

Here, I use a \scalerel* to make a strutted rule scale to the size of \cdot. scalerel package macros work in all math styles automatically (unless disabled).

\documentclass{article}
\usepackage{scalerel}
\def\sq{\mathbin{\scalerel*{\strut\rule{1ex}{1ex}}{\cdot}}}
\begin{document}
$f$ by $\bar f$ gives $\bar f \sq f \simeq c$ for $c$ the constant\dots\par
$\scriptscriptstyle\bar f \sq f \simeq c$
\end{document}

enter image description here

Changing the size/vertical position of the \rule allows any size/position of box to be made. Here, I double the size of the square dot by using \rule[-.5ex]{2ex}{2ex}.

\documentclass{article}
\usepackage{scalerel}
\def\sq{\mathbin{\scalerel*{\strut\rule[-.5ex]{2ex}{2ex}}{\cdot}}}
\begin{document}
$f$ by $\bar f$ gives $\bar f \sq f \simeq c$ for $c$ the constant\dots\par
$\scriptscriptstyle\bar f \sq f \simeq c$
\end{document}

enter image description here

| improve this answer | |
  • Is there a simple way to modify your answer to obtain an unfilled square rather than a filled square? – Evan Aad Nov 12 '18 at 10:40
  • 1
    @EvanAad You could adapt this answer, tex.stackexchange.com/questions/459167/…, except changing the scale from that of \Delta to instead the scale of \cdot. – Steven B. Segletes Nov 12 '18 at 13:13
4

\sqbullet from mathabx perhaps?

\documentclass{article}
\usepackage{mathabx}

\begin{document}

 $f \sqbullet f$
\end{document}
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  • 2
    With the usual caveat that loading mathabx changes all symbols. – egreg Oct 14 '15 at 20:37
  • 1
    @egreg: Is \filledsquare from MnSymbol better? – user31729 Oct 14 '15 at 20:38
  • 1
    No: while mathabx symbols somewhat match CM ones, those in MnSymbol are very distinct from them. There's a canonical answer about “importing a single symbol”, though. – egreg Oct 14 '15 at 20:39
4

Use \amssymb package and \centerdot.

\documentclass{article}
\usepackage{amssymb}

\begin{document}
$\bar{f} \centerdot f\simeq c$
\end{document}
| improve this answer | |
  • Unfortunately it sits on the baseline, rather than be vertically centered to the formula axis. – egreg Oct 14 '15 at 20:48
  • @egreg, \mathord{\raisebox{1pt}{$\centerdot$}} is useful? – ferahfeza Oct 14 '15 at 20:59
  • 2
    No, it would not work in different font sizes. – egreg Oct 14 '15 at 20:59
2

You can use direct \vrule with \mathchoice TeX primitive.

\newcount\stylenum  \newdimen\styledim 
\def\varstyle#1{\mathchoice{\stylenum=0 #1}{\stylenum=1 #1}{\stylenum=2 #1}{\stylenum=3 #1}}
\def\mathaxis{\fontdimen22\ifcase\stylenum 
   \textfont\or\textfont\or\scriptfont\or\scriptscriptfont\fi2 }
\def\setstyledim{\styledim=\ifcase\stylenum .1em\or.1em\or.07em\or.05em\fi\relax}

\def\sqdot{\mathbin{\varstyle{\raise\mathaxis\hbox{\setstyledim
       \kern\styledim 
       \vrule width1.2\styledim height.6\styledim depth.6\styledim
       \kern\styledim}}}}

$f\sqdot g, f\cdot g, \scriptstyle f\sqdot g, f\cdot g$

\bye
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