Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\phi(\Delta\mathfrak g)\subset\Delta\mathfrak g$.

share|improve this question
    
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
2  
What do you mean by that intersection? –  Mariano Suárez-Alvarez Apr 18 '13 at 5:58
show 3 more comments

1 Answer

up vote 1 down vote accepted

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! :-)

share|improve this answer
    
"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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.