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Could someone please tell me what I am missing in the following argument. Either my understanding of the exact statement of Ado's theorem is wrong, or there is a flaw in my argument below.

For a finite dimensional Lie algebra ${\bf g}$ consider the adjoint representation $\phi:{\bf g}\rightarrow {\bf gl}\left({\bf g}\right); \phi\left(X\right)={\bf ad}_X$. We have $\ker\left(\phi\right)={\cal z}\left({\bf g}\right)$ the centre of ${\bf g}$, so we can think of ${\bf g}$ as the direct sum $\left({\bf g} / {\cal z}\left({\bf g}\right)\right) \bigoplus {\cal z}\left({\bf g}\right)$ of the matrix Lie algebra of little ad matrices and the centre ${\cal z}\left({\bf g}\right)$. But then, suppose the centre is m-dimensional; since Abelian it can be represented as the algebra of $m\times m$ diagonal matrices. So the whole Lie algebra can be realised as the algebra of block-diagonal matrices of the form:

$\left(\begin{array}{cc}{\bf ad}_X & {\bf 0} \\\\ {\bf 0} & {\bf \Lambda}\end{array}\right)$

where ${\bf \Lambda}$ is an $m\times m$ diagonal matrix.

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  • $\begingroup$ Why can we think of $\mathbf g$ as a direct sum? $\endgroup$ – darij grinberg Apr 19 '11 at 9:37
  • $\begingroup$ (NB: direct sum of WHAT structures?) $\endgroup$ – darij grinberg Apr 19 '11 at 9:37
  • $\begingroup$ Hello Darij: the centre ${\cal z}\left({\bf g}\right)$ is an ideal - the formatting should be clearer now: $\endgroup$ – Selene Routley Apr 19 '11 at 9:43
  • $\begingroup$ Grand Charles, many thanx $\endgroup$ – Selene Routley Apr 19 '11 at 9:47
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    $\begingroup$ Incidentally, a proof (in characteristic $0$) of Ado's theorem is available in the Lie Theory lecture notes on my website: math.berkeley.edu/~theojf/LieQuantumGroups.pdf . From start to finish, the proof is basically Chapter 4. The main inputs are the Zassenhaus extension lemma (which lets you build modules (with good control of what acts nilpotently) of a large Lie algebra from modules of pieces of the Lie algebra) and Levi's theorem, which breaks every Lie algebra as a semidirect product of a solvable part and a semisimple, and hence is a better version of your block decomposition. $\endgroup$ – Theo Johnson-Freyd Apr 19 '11 at 14:55
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Your argument fails because the bracket can (sometimes) take pairs of elements into the centre. Therefore the direct sum as vector spaces isn't necessarily a direct sum of Lie algebras. For nilpotent Lie algebras, such as the Heisenberg algebra, you can't make the claim about block-diagonal form.

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