
答案1
\documentclass{article}
\usepackage{amsmath,stackengine}
\DeclareMathOperator*\dtimes{\times\mkern-7.95mu
\ensurestackMath{\raisebox{.1pt}{\stackanchor[-.00pt]{\times}{\times}}}
\mkern-8mu\times
}
\begin{document}
\[
\dtimes_{\lambda \in\Gamma} \mathcal{T}_\lambda
\qquad
\mathcal{T}_\lambda \dtimes \mathcal{T}_\lambda
\]
\end{document}
和這個
答案2
在這裡,有不同的方法。\varhash
從包中使用mathabx
:如果您願意,可以輪換:該描述在第二個 MWE 中採用與\rotatebox
一起使用的命令\usepackage{graphicx}
。
\documentclass[a4paper,12pt]{article}
\usepackage{amsmath,amssymb}
\usepackage{mathabx}
\usepackage{scalerel}
\DeclareMathOperator*{\slantedhash}{\scalerel*{\varhash}{\sum}}
\begin{document}
Using \verb|\varhash| as sum operator:
\[\underset{\lambda\in\mathcal{T}_{\lambda}}{\slantedhash}\mathcal{T}_{\lambda}\]
Using \verb|\varhash| as binary operator:
\[\mathcal{T}_{\lambda} \mathbin{\varhash} \mathcal{T}_{\lambda}\]
\end{document}
stix
透過旋轉命令使用字體\equalparallel
:
\documentclass[a4paper,12pt]{article}
\usepackage{amsmath,amssymb}
\usepackage{stix}
\usepackage{graphicx}
\begin{document}
\[\underset{\lambda\in\mathcal{T}_{\lambda}}{\rotatebox[origin=c]{-45}{$\equalparallel$}}\mathcal{T}_{\lambda}, \quad \mathcal{T}_{\lambda}\rotatebox[origin=c]{-45}{$\equalparallel$} \mathcal{T}_{\lambda}\]
\end{document}
包裝上boisik
有符號:\smashtimes
。或使用符號的經典旋轉\#
\documentclass[a4paper,12pt]{article}
\usepackage{amsmath,amssymb}
\usepackage{graphicx}
\begin{document}
\[\underset{\lambda\in\Gamma}{\rotatebox[origin=c]{-20}{$\#$}}\mathcal{T}_{\lambda}, \quad \mathcal{T}_{\lambda}\mathbin{\rotatebox[origin=c]{-20}{$\#$}} \mathcal{T}_{\lambda}\]
\end{document}