\documentclass[crop,border=2mm]{standalone}

\usepackage{amsfonts, amsmath, amssymb}

\begin{document}
\begin{tabular}{l}
{\huge type B2, s=3, subset=[1]}\\ \\


$\displaystyle
\begin{array}{r|l l l}
	{\scriptstyle i+j=1}&L_{1,1}L_{1,2}^{3}L_{2,2}L_{2,3}^{3}L_{2,4}&&\\
	{\scriptstyle i+j=3}&0&\mathbb{C}L_{1,1}^{3}L_{1,2}^{6}L_{2,2}^{3}L_{2,3}^{6}L_{3,3}L_{2,4}^{3}L_{3,4}&\\
	{\scriptstyle i+j=5}&0&0&L_{1,1}L_{1,2}^{3}L_{2,2}L_{2,3}^{3}L_{2,4}\\
	\hline h^{i,j}&{\scriptstyle j-i=1}&{\scriptstyle j-i=3}&{\scriptstyle j-i=5}
\end{array}
$ \\ \\


$\displaystyle
\begin{array}{r|l l l}
	{\scriptstyle i+j=1}&189&&\\
	{\scriptstyle i+j=3}&0&544&\\
	{\scriptstyle i+j=5}&0&0&189\\
	\hline h^{i,j}&{\scriptstyle j-i=1}&{\scriptstyle j-i=3}&{\scriptstyle j-i=5}
\end{array}
$ \\ \\


$\displaystyle
\begin{array}{rll}
	\text{module}&\text{multiplicity}&\text{dimension} \\ \hline \text{all}&&922 \\
	L\left(\alpha_{1}+\alpha_{2}\right)&5&5\\
	L\left(\alpha_{1}+ 2\alpha_{2}\right)&12&10\\
	L\left( 2\alpha_{1}+ 2\alpha_{2}\right)&5&14\\
	L\left( 2\alpha_{1}+ 3\alpha_{2}\right)&12&35\\
	L\left( 2\alpha_{1}+ 4\alpha_{2}\right)&5&35\\
	\mathbb{C}&1&1\\
	L\left( 3\alpha_{1}+ 3\alpha_{2}\right)&1&30\\
	L\left( 3\alpha_{1}+ 4\alpha_{2}\right)&1&81
\end{array}
$ \\ \\

\end{tabular}
\end{document}
