0

I am going through some notes and it makes reference to something called the residual action.

For an example, it says if $L$ is a Lie algebra and $ K \subseteq L $ is an Abelian ideal, then the construction

$$(v +K, w ) \mapsto [v, w] $$ doesn’t depend on the choice $v$ of coset representative and hence defined a residual action.

Nowhere is it properly defined what a residual action is. Then it says if $\{I_i \}_{i=0}^{n} $ is a flag for $L$ then the following are equivalent

  1. $[L,I_i ] \subseteq I_{i-1} $ for all $i$ .

  2. $I_i $ is an ideal of $L$ and ${I_i}_i $ is an Abelian ideal chain for which the residual actions $$\rho _i : L/I_i \rightarrow \mathfrak{gl} (I_i/I_{i-1}) $$ are trivial representations for all $i$.

How is $\rho _i $ being defined? The only thing I could think of to make work was defining $$\rho_i (x+I_i )=[x+I_{i-1}, -] $$

Anyone have any ideas?

Anonmath101
  • 1,818
  • 2
  • 15
  • 30
  • I don't think the first "construction" is well-defined for general $v, w \in L$. It would make sense for $w\in K$, i.e. the quotient $L/K$ acts on $K$ for an abelian ideal $K$. – Torsten Schoeneberg Jul 10 '22 at 14:58
  • More generally, the quotient $L/I$ acts on $I/[I,I]$ for a general ideal $I$, which makes me assume that part of the definition of a "flag" in this context is that $[I_i,I_i] \subseteq I_{i-1}$, for otherwise I would not see how that residual action would be defined. If that is the case, then the stated equivalence seems easy to prove. – Torsten Schoeneberg Jul 10 '22 at 15:04
  • Is there a general notion of what it means for a residual action in this context of Lie algebras? I don’t know what the maps $\rho _i $ actually are – Anonmath101 Jul 10 '22 at 15:04
  • You write "Abelian ideal chain" though which puzzles me a bit. Because if that means all the $I_i$ are abelian, then on the one hand of course $[I_i, I_i] =0\subseteq I_{i-1}$ trivially; on the other hand, the whole exercise seems a bit overkill. – Torsten Schoeneberg Jul 10 '22 at 15:09
  • 1
    The maps $\rho_i$ are what you think they are. I would write a bit more formally that $\rho_i$ sends $x\in L/I$ to the map $$ad(x)i: y + I{i-1} \mapsto [x,y] + I_{i-1}$$ which is a linear map on $I_i/I_{i-1}$. – Torsten Schoeneberg Jul 10 '22 at 15:13
  • 1
    Here abelian ideal chain means that $I_i/ I_{i-1} $ is abelian for all $i$ – Anonmath101 Jul 10 '22 at 15:57

0 Answers0