0

I am reading: https://i.stack.imgur.com/QPS4m.png and am not understanding their decomposition $$\mathcal{G} = \text{Ker}(ad_{a}) \oplus \text{Im}(ad_{a})$$ where $\mathcal{G}$ is a Lie algebra and $ad_{a} := [a, \cdot]$. Since a Lie algebra is a vector space plus some conditions and bracket...then isn't this always true by the classic Linear Algebra rank-nullity theorem? If so, why say this?

Sebastiano
  • 7,649
KZ-Spectra
  • 361

1 Answers1

1

If $a$ is a nilpotent element, i.e., say one where the adjoint map squares to $0$, then the image must be contained in the kernel. The dimensions adding does not mean that one is a complement to the other.

For example, the map $f:\mathbb{R}^2\to \mathbb{R}^2$ given by $(x,y)\mapsto (0,x)$ does not satisfy your statement. The image is generated by $(0,1)$, and so is the kernel.

Edit: As Torsten Schoeneberg mentioned, the map $f$ is $\mathrm{ad}_x$ for the $2$-dimensional Lie algebra with basis $x,y$, and Lie bracket $[x,y]=x$.

David A. Craven
  • 11,051
  • 1
    To make this answer perfect, add that this map $f$ "is" $ad_a$ in the two-dimensional Lie algebra with basis $a,b$ and bracket $[a,b]=a$. – Torsten Schoeneberg Aug 02 '20 at 16:14
  • @TorstenSchoeneberg I was actually just giving it as an example of a nilpotent map, I didn't think about whether it is actually an adjoint map for a Lie algebra! – David A. Craven Aug 02 '20 at 17:25