How can I find a Chevalley basis of $B_2$?

Question

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~ :)

Answer 1

It might be useful to write out all entries of those matrices; while your notation generalises nicely to higher dimensions, it's at first confusing that the $b_i, c_i$ etc. stand for various matrices of various different shapes. Written entirely out, a general element of that split form of type $B_2$, often called $\mathfrak{so}(3,2)$, looks like this:

$$\pmatrix{0&e&f&g&h\\ -g&a&c&0&i\\ -h&d&b&-i&0\\ -e&0&j&-a&-d\\ -f&-j&0&-c&-b}$$ where all ten variables are from the ground field.

The diagonal matrices contained in this form a Cartan subalgebra, and let's say we call $\alpha$ the root sending $$\alpha: \pmatrix{0&0&0&0&0\\ 0&a&0&0&0\\ 0&0&b&0&0\\ 0&0&0&a&0\\ 0&0&0&0&b} \mapsto a-b ;$$

and $\beta$ the one sending the above element to $-a$; the other roots are $\alpha+\beta$, $\alpha+2\beta$, and all their negatives. E.g. the root space to $\beta$ is the one where everything except the letter $e$ is zero; the root space to $\alpha+\beta$ is the one where everything except $h$ is zero; the root space to $\alpha+2\beta$ consists of the matrices where everything but $j$ is zero, etc.

Let's believe the relations in Dietrich Burde's answer. How to find an explicit set of corresponding matrices?

Well, the $h$'s in any Cartan-Weyl basis are already determined by $\rho(h_\rho)=2$ and living in $[\mathfrak g_\alpha, \mathfrak g_{-\alpha}]$, giving us here

$$h_\alpha= \pmatrix{0&0&0&0&0\\ 0&1&0&0&0\\ 0&0&-1&0&0\\ 0&0&0&-1&0\\ 0&0&0&0&1}, h_\beta= \pmatrix{0&0&0&0&0\\ 0&-2&0&0&0\\ 0&0&0&0&0\\ 0&0&0&2&0\\ 0&0&0&0&0}.$$

Further, $x_\beta$ is some $$x_\beta= \pmatrix{0&x&0&0&0\\ 0&0&0&0&0\\ 0&0&0&0&0\\ -x&0&0&0&0\\ 0&0&0&0&0}$$ with $x \neq 0$ in the ground field; because we want $[x_{-\beta}, x_\beta]=h_\beta$, we need

$$x_{-\beta}= \pmatrix{0&0&0&2/x&0\\ -2/x&0&0&0&0\\ 0&0&0&0&0\\ 0&0&0&0&0\\ 0&0&0&0&0}.$$

Likewise but easier we have $$x_\alpha= \pmatrix{0&0&0&0&0\\ 0&0&y&0&0\\ 0&0&0&0&0\\ 0&0&0&0&0\\ 0&0&0&-y&0}, x_{-\alpha}= \pmatrix{0&0&0&0&0\\ 0&0&0&0&0\\ 0&-1/y&0&0&0\\ 0&0&0&0&1/y\\ 0&0&0&0&0}$$

for some $y\neq 0$. Now $x_{\alpha+\beta}$ can be determined as

$$x_{\alpha+\beta} = [x_\beta, x_\alpha]= \pmatrix{0&0&xy&0&0\\ 0&0&0&0&0\\ 0&0&0&0&0\\ 0&0&0&0&0\\ -xy&0&0&0&0}$$

and $x_{\alpha+2\beta}$ as

$$x_{\alpha+2\beta} = \frac12 [x_\beta, x_{\alpha+\beta}]= \pmatrix{0&0&0&0&0\\ 0&0&0&0&0\\ 0&0&0&0&0\\ 0&0&-\frac{1}{2}x^2y&0&0\\ 0&\frac{1}{2}x^2y&0&0&0}.$$

To find a good $x_{-\alpha-\beta}$, first set a general candidate for it as $$\pmatrix{0&0&0&0&z\\ 0&0&0&0&0\\ -z&0&0&0&0\\ 0&0&0&0&0\\ 0&0&0&0&0}.$$

But we need $[x_\beta, x_{-\alpha-\beta}]=-2x_{-\alpha}$ and/or $[x_\alpha, x_{-\alpha-\beta}]=x_{-\beta}$, both forcing $z=\frac{2}{xy}$.

I stop here because I'm confident we could continue and find the last few $x_\rho$'s; we end up with two arbitrary parameters $x,y$, which of course we could ocnveniently set to $1$.

Answer 2

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*}