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I wondered how to create a symbol that looks like "#" (for connected sum) and "ff" for "fiber". The size and angles should be like the "#", but the vertical bars should look like two f`s. Another way to describe it would be a fortissimo symbol with two bars.

http://upload.wikimedia.org/wikipedia/commons/7/71/Music-fortissimo.png

enter image description here

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2 Answers 2

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Here, I introduce \varhash and \varfhash to be used in tandem (or for sans font, \sfvarhash and \sfvarfhash). I started with a \sffamily f to build things from there. In the last line, I overlay the two glyphs to see if they compare.

Since I use the current font for the f, this answer will depend on the font chosen for the document. Bu the answer can be tailored to suit.

EDITED to provide both sans and roman versions.

\documentclass{article}
\usepackage{stackengine,graphicx}
\newsavebox{\foobox}
\newcommand{\slantbox}[2][.5]{\mbox{%
        \sbox{\foobox}{#2}%
        \hskip\wd\foobox
        \pdfsave
        \pdfsetmatrix{1 0 #1 1}%
        \llap{\usebox{\foobox}}%
        \pdfrestore
}}
%%%%%% SF ITSHAPE VERSION
\def\sfvarfhash{%
  \stackinset{c}{}{b}{1.5pt}{\rule{1.45ex}{\sfrlthk}\kern1pt}{%
  \stackinset{c}{}{b}{3.7pt}{\rule{1.45ex}{\sfrlthk}\kern1pt}{%
    \textsf{\itshape ff}%
  }}%
}
\def\sfvarhash{%
  \kern1pt%
  \stackinset{c}{}{b}{1.5pt}{\rule{1.45ex}{\sfrlthk}\kern1pt}{%
  \stackinset{c}{}{b}{3.7pt}{\rule{1.45ex}{\sfrlthk}\kern1pt}{%
    \slantbox[.2]{\mysfrule\kern2pt\mysfrule\kern2.3pt}%
  }}%
  \kern1pt%
}
\def\sfrlthk{.17ex}
\def\mysfrule{\rule{\sfrlthk}{1.4ex}}
%%%%%% RM ITSHAPE VERSION
\def\varfhash{%
  \kern -1.5pt%
  \stackinset{c}{}{c}{-1.7pt}{\kern1pt\rule{1.7ex}{\rlthk}}{%
  \stackinset{c}{}{c}{1.7pt}{\kern1pt\rule{1.7ex}{\rlthk}}{%
    \kern2pt\itshape ff\kern2pt%
  }}%
}
\def\varhash{%
  \kern-.5pt%
  \stackinset{c}{}{c}{-1.7pt}{\kern1pt\rule{1.7ex}{\rlthk}}{%
  \stackinset{c}{}{c}{1.7pt}{\kern1pt\rule{1.7ex}{\rlthk}}{%
    \kern2pt\kern.2pt\slantbox[.2]{\myrule\kern2.4pt\myrule}\kern.2pt\kern2pt%
  }}%
  \kern1pt%
}
\def\rlthk{.13ex}
\def\myrule{\rule[-.33ex]{\rlthk}{1.8ex}}
\begin{document}
\sfvarfhash\sfvarhash\par
\ooalign{z\sfvarhash x\cr z\sfvarfhash x} overlaid\par
\varfhash\varhash\par
\ooalign{z\varhash x\cr z\varfhash x} overlaid\par
\par \# for comparison
\end{document}

enter image description here

And here is a version with \int instead of f, which allows more flexibility with the placement of the horizontal bars:

\documentclass{article}
\usepackage{stackengine,graphicx}
\newsavebox{\foobox}
\newcommand{\slantbox}[2][.5]{\mbox{%
        \sbox{\foobox}{#2}%
        \hskip\wd\foobox
        \pdfsave
        \pdfsetmatrix{1 0 #1 1}%
        \llap{\usebox{\foobox}}%
        \pdfrestore
}}
%%%%%% RM ITSHAPE VERSION
\def\varfhash{%
  \kern-4.1pt%
  \stackinset{c}{}{c}{-1.4pt}{\kern1pt\rule{1.6ex}{\rlthk}}{%
  \stackinset{c}{}{c}{1.4pt}{\kern1pt\rule{1.6ex}{\rlthk}}{%
    \kern3.1pt${\int}\kern-4pt{\int}$\kern2.1pt%
  }}%
  \kern-2.2pt%
}
\def\varhash{%
  \kern-.7pt%
  \stackinset{c}{}{c}{-1.4pt}{\kern.5pt\rule{1.6ex}{\rlthk}}{%
  \stackinset{c}{}{c}{1.4pt}{\kern.5pt\rule{1.6ex}{\rlthk}}{%
    \kern2.4pt\slantbox[.09]{\myrule\kern2.1pt\myrule}\kern2.4pt%
  }}%
  \kern.6pt%
}
\def\rlthk{.16ex}
\def\myrule{\rule[-.23ex]{\rlthk}{1.6ex}}
\begin{document}
$\varfhash\varhash$\par
$x\varfhash y\varhash z$\par
\ooalign{z\varhash x\cr z\varfhash x} overlaid\par
\par \# for comparison
\end{document}

enter image description here

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5
  • I think it is pretty good. I just had in mind, that it is like a "f" at the upper and lower end (as in the wikimedia picture). And maybe a little less bold, but that would be just an extra.
    – w_w
    Mar 26, 2015 at 13:32
  • @Zonan Please see revision.
    – Steven B. Segletes
    Mar 26, 2015 at 13:34
  • Ah, now I understand why you put the vertical bars so far apart: In order to cover the f' s bar. Is there a way to avoid the f' s bar? I guess the integral solution below. Otherwise I´ll just lift the lower bar of the "fiber connected sum"
    – w_w
    Mar 26, 2015 at 13:42
  • @Zonan If I use the f then one is stuck with the vertical placement of the bar. I suppose one could craft the answer in terms of \int, using a similar approach, with flexibility in the offset values of the \stackinset terms. Of course, if one uses the \int, the vertical slant will change.
    – Steven B. Segletes
    Mar 26, 2015 at 13:46
  • @Zonan I have added a version with the \int, though I still think I prefer the italic f version.
    – Steven B. Segletes
    Mar 26, 2015 at 14:10
1

One way to achieve this would be to use the double integral math symbol and strike it out.

\usepackage{ulem,amsmath,xspace}
\newcommand{\fiber}{\sout{$\iint$}\xspace}

And then use \fiber where needed. It's far from ideal, but it can do the trick.

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