# Automorphism group of Lie algebra $\mathfrak{g\oplus g}$

Let $\mathfrak g$ be the Lie algebra of strictly upper triangular 3x3 matrices. How can I determine the group $\operatorname{Aut}(\mathfrak{g\oplus g},\Delta\mathfrak g)$, where $\Delta\colon\mathfrak g\to\mathfrak {g\oplus g}$ is the diagonal embedding?

By $\operatorname{Aut}(\mathfrak{g\oplus g},\Delta\mathfrak g)$ I mean the automorphisms $\phi\in\operatorname{Aut}(\mathfrak{g\oplus g})$ such that $\phi(\Delta\mathfrak g)\subset\Delta\mathfrak g$.

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Do you mean the automorphisms of $\def\g{\mathfrak g}\g\oplus\g$ that centralise the diagonal subalgebra $\Delta\g$? Or rather which normalise? Or something else? I am not acquainted with a two-argument version of 'Aut'. – Marc van Leeuwen Apr 18 '13 at 5:30
Yes, I mean $\operatorname{Aut}(\mathfrak{g\oplus g)/\operatorname{Aut}(\Delta g})$. – Earthliŋ Apr 18 '13 at 5:32
And what is $g\oplus g/\Delta g$? – Mariano Suárez-Alvarez Apr 18 '13 at 5:33
Ok, you edited that away, but now about the new version of your comment:... In general, $Aut(\Delta g)$ is not normal in $Aut(g\oplus g)$: for example, if $g$ is abelian of dimension $2$. How do you know it is in this case? – Mariano Suárez-Alvarez Apr 18 '13 at 5:39
What do you mean by that intersection? – Mariano Suárez-Alvarez Apr 18 '13 at 5:58

I really don't know what you mean by how can I find this group: you can simply compute it!

The algebra $\def\g{\mathfrak g}\g$ has a basis $\{x,y,z\}$ such that $z$ is central and $[x,y]=z$. It follows that $\g\oplus g$ has a basis $\{x_1,y_1,z_1,x_2,y_2,z_2\}$ with $[x_i,y_i]=z_i$ for $i\in\{1,2\}$ and all other brackets between elements of the basis equal to zero.

Let $f:\g\oplus\g\to\g\oplus\g$ be an automorphism.

The derived algebra $\def\h{\mathfrak h}\h=[\g,\g]$ is spanned by $\{z_1,z_2\}$ and coincides with the center of $\g$. Since this is obviously preserved by every automorphism, there are scalars $a_i$, $b_1$ such that $f(z_i)=a_iz_1+b_iz_2$. Since the restriction of $f$ to a map $\h\to\h$ must be invertible, we must have $\det\begin{pmatrix}a_1&a_2\\b_1&b_2\end{pmatrix}\neq0$.

There are scalars $p_i$, $q_i$, $r_i$, $s_i$, $t_i$, $u_i$, $v_i$, $w_i$ and elements $\xi_i$, $\zeta_i\in\h$ such that $f(x_i)=p_ix_1+q_iy_1+r_ix_2+s_iy_2+\xi_i$ and $f(y_i)=t_ix_1+u_iy_1+v_ix_2+w_iy_2+\zeta_i$. It follows that $[f(x_i),f(y_i)]=(p_iu_i-q_it_i)z_1+(r_iw_i-s_iv_i)z_2$, and this must be equal to $f([x_i,y_i])=f(z_i)=a_iz_1+a_iz_2$. It follows that we must have $a_i=p_iu_i-q_it_i$ and $b_i=r_iw_i-s_iv_i$. Since $[x_1,x_2]=[x_1,y_2]=[x_2,y_1]=[y_1,y_2]=0$, we must have \begin{align} &[f(x_1),f(x_2)]=(p_1q_2-q_1p_2)z_1+(r_1s_2-s_1r_2)z_2=0,\\ &[f(x_1),f(y_2)]=(p_1u_2-q_1t_2)z_1+(r_1w_2-s_1v_2)z_2=0,\\ &[f(x_2),f(y_1)]=(p_2u_1-q_2t_1)z_1+(r_2w_1-s_2v_1)z_2=0,\\ &[f(y_1),f(y_2)]=(t_1u_2-u_1t_2)z_1+(v_1w_2-w_1v_2)z_2=0, \end{align} and this means that \begin{align} p_1q_2-q_1p_2=r_1s_2-s_1r_2=0,\\ p_1u_2-q_1t_2=r_1w_2-s_1v_2=0,\\ p_2u_1-q_2t_1=r_2w_1-s_2v_1=0,\\ t_1u_2-u_1t_2=v_1w_2-w_1v_2=0, \end{align} So far we have expressed all the conditions that express the fact that $f$ is an isomorphism. If you want $f$ to preserve $\Delta f$, then this gives you more conditions. Write them out. Then look at what you got.

I won't write the details, because this is very boring and you are, after all, who is interested in the result! :-)

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"Just compute it" isn't what I hoped for when I asked the question, but maybe the only sensible answer. Thank you for the lead, I'm giving it a try. – Earthliŋ Apr 19 '13 at 7:22
If you compute it and see a pattern, you may then try to synthetically get a description of the group which you can present with pride to your friends. But if you do not know even how to get started, a computation (and this one is, really, a very simple one! I did the part I did directly in TeX :-) )is a great plan. – Mariano Suárez-Alvarez Apr 19 '13 at 7:24
I did the calculation. As you said it was very boring—it turns out to be a 10-dimensional subgroup of $GL(6,\mathbb R)$ times $\mathbb Z_2$ (semi-direct). – Earthliŋ Apr 28 '13 at 1:36