答案1
\documentclass{article}
\usepackage{stackengine}
\newcommand\tritens{\mathbin{\ensurestackMath{%
\stackengine{.3pt}{\triangle}{\scriptstyle\times}
{O}{c}{F}{F}{L}}}
}
\begin{document}
$\dot F \tritens \dot H = \triangle^! (...)$
\end{document}
如果\mathbin
不需要间距,可以将其删除:
\newcommand\tritens{\ensurestackMath{%
\stackengine{.3pt}{\triangle}{\scriptstyle\times}
{O}{c}{F}{F}{L}}
}
答案2
我猜:
\documentclass{article}
\usepackage{amsmath,amssymb}
\usepackage{graphicx}
\makeatletter
\newcommand{\tritensor}{\mathbin{\mathpalette\tritensor@\relax}}
\newcommand{\tritensor@}[2]{%
\ooalign{%
$\m@th#1\triangle$\cr
\hidewidth\scalebox{0.8}{$\times$}\hidewidth\cr
}%
}
\makeatother
\makeatother
\begin{document}
$\tritensor\boxtimes$
$\dot{F}\tritensor\dot{H}=\triangle^{!}(\dot{F}\boxtimes\dot{H})$
\end{document}
另一种方法是使用宽度为 的等边三角形\boxtimes
,这样可以更好地区分特殊张量积和三角形算子。
\documentclass{article}
\usepackage{amsmath,amssymb}
\usepackage{graphicx}
\usepackage{pict2e}
\makeatletter
\newif\if@tritensor
\newcommand{\btensor}{\mathbin{\btensor@}}
\newcommand{\btensor@}{%
\text{%
\settowidth{\unitlength}{$\m@th\boxtimes$}%
\linethickness{\fontdimen8\textfont3}%
\begin{picture}(1,1)
\roundjoin
\polygon(0.075,0.03)(0.925,0.03)(0.5,0.733)
\Line(0.2952,0.4097)(0.6748,0.03)
\Line(0.7048,0.4097)(0.3252,0.03)
\end{picture}%
}%
}
\makeatother
\begin{document}
$\dot{F}\btensor\dot{H}=\triangle^{!}(\dot{F}\boxtimes\dot{H})$
$\scriptstyle\dot{F}\btensor\dot{H}=\triangle^{!}(\dot{F}\boxtimes\dot{H})$
\end{document}
答案3
我建议\triangletimes
纳入stix
包裹:
\documentclass{article}
\usepackage{amsmath}
\usepackage{stix}
\begin{document}
$\triangletimes$
\end{document}
然而,它并不是您想要的符号(十字架较小)。