You could, of course, use TikZ for this:

The symbol will scale with your font size, since it uses ex
to define the path.
\documentclass{article}
\usepackage{tikz}
\begin{document}
\newcommand\shield{%
\tikz [baseline] \draw (0,1.75ex) -- (0,0.75ex) arc [radius=0.75ex, start angle=-180, end angle=0] -- (1.5ex,1.75ex) -- cycle;%
}
A shield: \shield
\end{document}
If you're feeling fancy, you could parametrise it a bit:

\documentclass[border=3mm]{standalone}
\usepackage{tikz}
\begin{document}
\newcommand\shield[1][]{%
\tikzset{
shield width/.store in=\shieldwidth,
shield width=1.5ex,
shield height/.store in=\shieldheight,
shield height=1.75ex
}%
\tikz [baseline,#1] \draw (0,\shieldheight) -- (0,\shieldwidth/2) arc [radius=\shieldwidth/2, start angle=-180, end angle=0] -- (\shieldwidth,\shieldheight) -- cycle;%
}
A shield: \shield
A wide shield: \shield[shield width=2ex]
A tall shield: \shield[shield height=3ex]
\end{document}
The following example constructs the symbol in the following way:
- The top part with the round line end of the vertical bar is used as branch line.
- The length of the branch is the same as in the plus symbol. But the length can
be changed by redefining
\UpDownYFactor
to a value different than 1
.
- The intersection point is the same as in the plus symbol at the mathematical center axis.
- The symbol adopts in size to the current math style.
- The side bearings have the same width as in the plus symbol.
- The names for the provided symbols are the same as in package
MnSymbol
: \upY
and \downY
.
- The symbols act as binary operators. It can be changed by replacing
\mathbin
.
Example file:
\documentclass{article}
\usepackage{graphicx}
\usepackage{trimclip}
\makeatletter
\providecommand*{\upY}{%
\mathbin{%
\mathpalette\@updownY{0}%
}%
}
\providecommand*{\downY}{%
\mathbin{%
\mathpalette\@updownY{1}%
}%
}
\providecommand*{\UpDownYFactor}{1}
\newcommand*{\@updownY}[2]{%
% #1: math style
% #2: 0 = up, 1 = down
\sbox0{$#1+\m@th$}%
\dimen2=.5\dimexpr\wd0-\ht0-\dp0\relax
% => \dimen2: side bearing
\sbox2{$#1\vcenter{}$}%
\dimen4=\dimexpr\ht0-\ht2\relax
% => \dimen4: branch length
\setbox0=\hbox to 0pt{%
\hss
\clipbox{%
0pt %
{\dimexpr\totalheight-\UpDownYFactor\dimen4\relax} %
0pt %
-\dimen2%
}{$#1|$}%
\hss
}%
\ht0=\dimexpr\ht0-\dimen2\relax
\kern\dimen2 %
\raise\ht2\hbox{%
\ifnum#2=0 %
{\rotatebox{120}{\copy0}}%
\copy0 %
{\rotatebox{-120}{\copy0}}%
\else
{\rotatebox{60}{\copy0}}%
{\rotatebox{180}{\copy0}}%
{\rotatebox{-60}{\copy0}}%
\fi
}%
\kern\dimen2 %
}
\makeatother
\begin{document}
\[\begin{array}{c}
a \upY b \downY c
\\
\scriptstyle
a \upY b \downY c
\\
\scriptscriptstyle
a \upY b \downY c
\\
\rlap{$\upY$}{\downY}
\end{array}\]
\end{document}

For analyzing the center:
\usepackage{pdfrender}
...
\[
\rlap{%
\sbox0{$\vcenter{}$}%
\sbox2{${\upY}{\downY}{\upY}$}%
\raise\ht0\hbox{\vrule height.01pt depth.01pt width\wd2}%
}%
\pdfrender{TextRenderingMode=Stroke, LineWidth=.01pt}
{\upY}\rlap{$\upY$}{\downY}{\downY}
\]

Best Answer