Semisimple complex Lie algebra of type $A_3$ contains a Lie subalgebra of type $B_2$ as fixed points of an automorphism. I'm trying to prove the next theorem:

Let $\mathfrak{g}$ be a complex semisimple Lie algebra of type $A_3$, $\mathfrak{h}$ a Cartan subalgebra, $R$ a root system associated to the pair $(\mathfrak{g},\mathfrak{h})$ and $\Delta=\{\alpha_1,\alpha_2,\alpha_3\}$ a basis of simple roots. Let $(x_i,h_i,y_i)$ be the $\mathfrak{sl}_2$-triple associated to each  simple root $\alpha_i$, $i=1,2,3$, and $\sigma$ be the automorphism of $\mathfrak{g}$ induced by the automorphism of $R$ which interchanges $\alpha_1$ and $\alpha_3$ and fixes $\alpha_2$. Then $x_1+x_3$, $x_2$, $y_1+y_3$ and $y_2$ generate the subalgebra $\mathfrak{g}^\sigma$ of fixed points, and it is a semisimple Lie algebra of Type $B_2$. 

I know that by the semisimplicity of $\mathfrak{g}$, we have a deocmposition $\mathfrak{g}=\mathfrak{h}\oplus\bigoplus\limits_{\alpha\in R^*}\mathfrak{g}_{[\alpha]}$ and $x_i\in\mathfrak{g}_{[\alpha_i]}$, $y_i\in\mathfrak{g}_{[-\alpha_i]}$ for every $i=1,2,3$, so if $\sigma$ interchanges $\alpha_1$ and $\alpha_3$ and fixes $\alpha_2$, then it interchanges $x_1$ with $x_3$, $y_1$ with $y_3$ and fixes $x_2$ and $y_2$, so $x_1+x_3$, $x_2$, $y_1+y_3$ and $y_2$ are all in $\mathfrak{g}^\sigma$. But I don't know how to conclude that they generate that subalgebra, or that it is of type $B_2$. I would appreciate any hints or comments about this. 
 A: Looking at you question, it seems that you are looking for a reply in terms of roots which needs a fair amount of verification, but there also is a simple description in terms of linear algebra: $A_3=\mathfrak{sl}(4,\mathbb C)$ and of course, this contains $\mathfrak{sp}(4,\mathbb C)$, which has type $C_2\cong B_2$ as a subalgebra. It is also easy to see that this subalgebra coincides with the fixed points of an involutive automorphism: If $\omega$ is a non-degenerate, skew symmetric bilinear form on $\mathbb C^4$, then for $A\in\mathfrak{sl}(4,\mathbb C)$ you define $A^\dagger$ via $\omega(Az,w)=\omega(z,A^\dagger w)$ for all $z,w\mathbb C^4$. It is easy to see that $A\mapsto -A^\dagger$ is an automorphism, whose fixed points are the elements of $\mathfrak{sp}(4,\mathbb C)$ by definition. 
A: "The automorphism of $\mathfrak{g}$ induced by the automorphism of $R$" is actually a bit ambiguous, but it's clear from context that they mean the automorphism that switches $x_1 \leftrightarrow x_3$, $y_1 \leftrightarrow y_3$ and fixes $x_2$ and $y_2$. It's probably a good exercise to check that this indeed defines an automorphism and determines it completely. 
In fact, let's switch to Chevalley basis notation and call 


*

*$e_{\alpha_i} := x_i$, $e_{-\alpha_i} := y_i$ for $i=1,2,3$

*$e_{\alpha_1+\alpha_2} := [x_1, x_2]$, $e_{\alpha_2+\alpha_3} := [x_2, x_3]$, $e_{-\alpha_1-\alpha_2} := [y_1, y_2]$, $e_{-\alpha_2-\alpha_3} := [y_2, y_3]$

*$e_{\alpha_1+\alpha_2+\alpha_3} := [x_1, [x_2, x_3]] = [[x_1, x_2], x_3]$, $e_{-\alpha_1-\alpha_2-\alpha_3} := [y_1, [y_2, y_3]] = [[y_1, y_2], y_3]$


where according to what they call an $\mathfrak{sl}_2$-triple in your source you might have to throw in some minus signs in the second line.
In any case, it's a good exercise to recall that these twelve $e$'s (each of them being a basis of the root space to the root in its index), together with $h_1, h_2$ and $h_3$, are a vector space basis of $\mathfrak{g}$, and the one we want to work with: Because it makes the automorphism completey explicit. E.g, it also switches $h_1 \leftrightarrow h_3$, $e_{\alpha_1+\alpha_2} = [x_1,x_2] \leftrightarrow -[x_2,x_3] = - e_{\alpha_2+\alpha_3}$, and fixes $h_2$ as well as $e_{\alpha_1+\alpha_2+\alpha_3}$.
It's linear algebra to see that a vector space basis of the fixed space $\mathfrak{g}^\sigma$ is:


*

*$h_1+h_3$, $h_2$ (basis of the new, two-dimensional CSA $\mathfrak{h}^\sigma$)

*$e_{\alpha_1}+e_{\alpha_3}$, $e_{\alpha_2}, e_{\alpha_1+\alpha_2}-e_{\alpha_2+\alpha_3}, e_{\alpha_1+\alpha_2+\alpha_3}$ (four positive root vectors)

*$e_{-\alpha_1}+e_{-\alpha_3}, e_{-\alpha_2}$, $e_{-\alpha_1-\alpha_2}-e_{-\alpha_2-\alpha_3}, e_{-\alpha_1-\alpha_2-\alpha_3}$ (four negative root vectors)


Setting $e_{\beta_1} := e_{\alpha_1}+e_{\alpha_3}$ and $e_{\beta_2} := e_{\alpha_2}$, compute how nicely
$[e_{\beta_1}, e_{\beta_2}] = e_{\alpha_1+\alpha_2}-e_{\alpha_2+\alpha_3} =: e_{\beta_1+\beta_2}$
and
$[e_{\beta_1}, e_{\beta_1 +\beta_2}] = -2e_{\alpha_1+\alpha_2+\alpha_3} =: e_{2\beta_1 + \beta_2}$ etc.
so that $\beta_1$ and $\beta_2$ are a a basis of a root system of type $B_2$, with $\beta_1$ being the short root. Check further that $\mathfrak{h}^\sigma$ indeed acts on those new root spaces, with $h_1+h_3$  of course being the coroot of $\beta_1$, and $h_2$ the coroot of $\beta_2$. So $\mathfrak{g}^\sigma$ decomposes as
$\mathfrak{h}^\sigma \oplus $ root spaces, and hence is semisimple of type $B_2$.
It is another exercise to see how this, written out in matrices, modulo some choices (scalings and signs I would guess), is the same as what happens in Andreas Cap's answer, just from a very different viewpoint.
