# Adjusting the position of \dot

I'm writing a paper that uses the symbol \dot{\mathcal{U}} which for me (in PDFLaTeX with Palatino math fonts) looks like this:

As you can imagine, from further away, the dot is invisible. Is there some easy way of increasing the space between the character and the dot, or moving it to the right, so it's more visible?

• In this post, the goal was decrease the space of the dot from a line. I think the two answers should be applicable to your case just in reverse by increasing the distance. – dustin Aug 15 '13 at 2:08
• @dustin, I think the idea is to move the dot to the right, not vertically. (At least, this is what I would think to be most appropriate.) – Sean Allred Aug 15 '13 at 2:21
• @SeanAllred that is what I would want but I interpreted increase space between the character and the dot as meaning vertical distance. – dustin Aug 15 '13 at 2:28
• @dustin if only everyone thought in boxes :-) the world would be a much more TeX-friendly place – Sean Allred Aug 15 '13 at 2:35
• The U seems to be drawn from the Zapf Chancery font, which is completely unapt for a math symbol font. The palatino package doesn't show that symbol, as it doesn't affect math. Please, add a minimal example showing how you get that output. You should try also \usepackage{newpxtext,newpxmath} or \usepackage{mathpazo} for using math with the Palatino font and not \usepackage{palatino}. – egreg Aug 15 '13 at 9:15

You can define a new macro that provides a slightly displaced dot, say \dispdot[<disp>]{<stuff>}. Below I've defined \dispdot to give a .2ex displacement. Supplying a different <disp> will change the location of the \dot:

\documentclass{article}
\newcommand{\dispdot}[2][.2ex]{\dot{\raisebox{0pt}[\dimexpr\height+#1][\depth]{$#2$}}}% \dispdot[<displace>]{<stuff>}
\begin{document}
$d\mathcal{U}, d\dot{\mathcal{U}}, d\dispdot{\mathcal{U}}, d\dispdot[.5ex]{\mathcal{U}}$
\end{document}


If need be, it would be possible to capture the math style.

For horizontal displacement, you could define \dispdot to give a 3mu displacement (actually, due to symmetry it is half the actual displacement):

\documentclass{article}
\newcommand{\dispdot}[2][3mu]{\dot{#2\mkern#1}\mkern-#1}% \dispdot[<disp>]{<stuff>}
\begin{document}
$d\mathcal{U}, d\dot{\mathcal{U}}, d\dispdot{\mathcal{U}}, d\dispdot[5mu]{\mathcal{U}}$
\end{document}


Note that math units are very small, so adjustments will be very tiny.

This answer originally followed the logic of Vertical spacing between character and \hat, but has been revised to handle differernt math sizes, shown below in \textstyle, \scriptstyle, and \scriptscriptstyle.

Setting \dydot adjusts the vertical shift of the dot.

\documentclass{article}
\usepackage{stackengine}
\usepackage{scalerel}% IF YOU NEED IT FOR SCRIPTSTYLE MATH
\def\stacktype{L}
\stackMath
\def\dydot{.2ex}
\newcommand\caldot[1]{\ThisStyle{%
\stackon[\dydot]{\SavedStyle\mathcal{#1}}
{\SavedStyle\dot{\phantom{\mathcal{#1}}}}}}
\begin{document}
$$A \caldot U$$
$$\scriptstyle A \caldot U$$
$$\scriptscriptstyle A \caldot U$$
\end{document}


• Thanks; I tried this out, and it works pretty well, though something is wrong with the sizing in subscripts. – Ben Webster Aug 15 '13 at 4:58
• @BenWebster Revised to handle mathstyles. – Steven B. Segletes Aug 15 '13 at 9:42