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How can I find a Chevalley basis of a type $B_2$ when the related lie algebra is defined as a linear Lie algebra of elements of the form $x= \begin{pmatrix} 0 & b_1 & b_2 \\ c_1 & m & n \\ c_2 & p & q \end{pmatrix}$, where $c_1=-b_2^t$, $c_2=-b_1^t$, $q=-m^t$, $n^t=-n$, $p^t=-p$?

When trying to find such a base, the constraints, especially $[x_{\alpha}x_{\beta}]=c_{\alpha,\beta}x_{\alpha+\beta}$ for $\alpha,\beta$ independent, and $\alpha+\beta$ being a root, turn out to be very hard to follow.

Thank you~ :)

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1 Answer 1

The simple Lie algebra of type $B_2$ has positive roots $\alpha,\beta,α+β,2β+α$. A Chevalley basis is given by $$\lbrace h_{\alpha}, h_{\beta}, x_{\alpha},x_{\beta},x_{\alpha+\beta},x_{2\beta+\alpha}, x_{-\alpha},x_{-\beta},x_{-\alpha-\beta},x_{-2\beta-\alpha}\rbrace $$ with brackets \begin{align*} [h_β,x_β] & = 2x_{\beta}\\ [h_β,x_{\alpha}] & = -2x_{\alpha}\\ [h_β,x_{\alpha+β}] & = 0\\ [h_β,x_{\alpha+2β}] & = 2x_{\alpha+2β}\\ [h_α,x_β] & = -x_{\beta}\\ [h_α, x_α] & = 2x_{\alpha} \\ [h_α , x_{α+β}] & = x_{α+β}\\ [h_α , x_{α+2β}] & = 0\\ [x_β,x_{\alpha}] & = x_{α+β}\\ [x_β,x_{\alpha+\beta}] & = 2x_{α+2β}\\ [x_β,x_{-\alpha-\beta}] & = -2x_{-α}\\ [x_β,x_{-\alpha-2\beta}] & = -x_{-α-\beta}\\ [x_{\alpha},x_{-\alpha-\beta}] & = x_{-\beta}\\ [x_{\alpha+\beta},x_{-\alpha-2\beta}] & = x_{-\beta}\\ \cdots & = \cdots \end{align*}

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