\bar below symbol

Is there a command like \bar{...} but such that the bar is placed below the symbol?

Of course there is \underline{...} but then the line is adapted to the width of the symbol and I don't want that (compare with the difference between \overline{...} and \bar{...}).

I need this for math mode.

• Do you need it in text mode or in math mode? Have a look at the accents package. – egreg Jul 25 '13 at 13:02
• Maybe this from amsmath: \underset{\bar{}}{A} – Sigur Jul 25 '13 at 13:03
• @egreg That's good. Could you turn the comment into an answer please. – lpdbw Jul 25 '13 at 13:12

The accents package has the feature you're looking for:

\usepackage{accents}
\newcommand{\ubar}[1]{\underaccent{\bar}{#1}}


Why the \newcommand? Because you might change your mind or find a new way to underline a symbol, so you can just modify the definition instead of hunting through the document for occurrences of \underaccent{\bar}.

• Interesting. With this we don't need to write \bar{} like mine suggestion. Also, the space between the symbol and the bar is different. – Sigur Jul 25 '13 at 13:23
• @Sigur The accents package ensures that the correct vertical dimension is assigned to the underaccent; with \underset{\bar{}}{x} you get a full vertical sized bar; and, moreover, it will be in script style. – egreg Jul 25 '13 at 13:30

My solution is \newcommand{\ubar}[1]{\text{\b{$#1$}}} since I get a conflict of accents and amsmath packages concerning \vec command.

• Can someone please explain why this has been downvoted? I am unsure if I should use the accents package in case there is indeed an assue with the amsmath package, as @SergeyPolyakovskiy suggests, however it is unclear. I have tried searching it in other questions and it seems that such a \vec issue does exist, but some clarification would be helpful. Thanks! – Flint72 Oct 5 '14 at 21:25
• This. This answer saved my hide. The solution that is accepted here -- and referenced seemingly everywhere when it comes to the problem of making a suitable underbar expression relies on a really outdated package that, as far as I can figure, was last updated in 2006. It is not compatible with amsmath and a host of other packages. This solution works tremendously well. – Tom Feb 15 '16 at 3:51

You can play with the three parameters (1.2pt underset gap, .8ex rule length, .075ex rule width

\documentclass{article}
\usepackage{stackengine}
\begin{document}
\newcommand\barbelow[1]{\stackunder[1.2pt]{$#1$}{\rule{.8ex}{.075ex}}}
$$\overline{A} \bar{A} \barbelow{A}$$
\end{document}


• I needed to use a subscript with the \barbelow{f}, and this solution works the best - thanks! I used a slightly different set of parameters (1pt, .8ex, 0.075ex) though. – user25216 Apr 20 '20 at 21:25

Have you tried using \underbar{symbol} in a math environment?

• This only works in text mode. – egreg Feb 20 '18 at 0:24
• Just used it in math environment! It does work! – chadoulis Jan 4 '20 at 7:18
• For me using \underbar gives a bunch of weird errors having to do with closing math mode. – ifly6 Jan 15 '20 at 19:41

Use the \textrm{.1.} for text mode IN the math mode. In the brackets you can then use the text mode command \underline{.2.} to underline your text input in the {.2.} brackets.

in math mode...

\textrm{\underline{this text input will be underlined}}


... worked for me

This post was also very helpful for me as I've had a similar problem. I am using \bar and this solution

\usepackage{stackengine}
\newcommand\barbelow[1]{\stackunder[1.2pt]{$#1$}{\rule{.8ex}{.075ex}}}


to underline in math mode. Unfortunately it seems like the \bar commads provides a slightly thicker bar which is why I am wondering if there is a possibility to make \barbelow a little bit thicker as well?

• Hi, welcome. Please note that this site isn't like a normal forum. All answer-posts should contain solutions to the problem posed in the question-post. Follow-up questions, if not handled in comments should be asked via the Ask Question-link in the top right. That said, change .075ex to some higher value, the thickness of the bar is determined by that length. – Torbjørn T. Jun 20 '20 at 12:38