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There is a theorem that any finite dimensional semisimple Lie algebra $L$ is in fact a direct sum of simple ideals $I_1, I_2, ..., I_n$, and I am now studying its proof in Humphreys' "Introduction to Lie algebras" (section 5.2). It starts with any ideal $I$ in $L$. We introduce $$I^{\bot}=\{x \in L: \kappa(x,y) = \mathrm{Trace}(ad_x ad_y) = 0, \forall y\in L\},$$ which is an ideal, and by the Cartan's Criterion and semi-simplicity of L, we have $I \cap I^{\bot} = 0$, which I understand.$\\$

The fact I need is that $I \oplus I^{\bot} = L$, or, equivalently $\mathrm{dim} \ I + \mathrm{dim} \ I^{\bot} = \mathrm{dim} \ L$. $\\$

In other source I found that it is implied by the non-degenarativity of a Killing form $\kappa(x,y) = \mathrm{Trace}(ad_x ad_y)$ for a semisimple $L$ $\left( \text{ non-degenerativity means that the set } S = \{x \in L: \kappa (x,y) = 0, \ \forall \ y \in L \} \text{ is equal to 0} \right)$, but I don't see how.

Torsten Schoeneberg
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NewOne
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  • See also e.g. page 2 of https://www.dpmms.cam.ac.uk/study/IB/LinearAlgebra/2008-2009/bilinear-08.pdf. This is kind of basic linear algebra and the only connection to Lie algebras is the non-degeneracy of the Killing form for semisimple Lie algebras. Cf. also https://math.stackexchange.com/q/1441119/96384 or https://math.stackexchange.com/q/970309/96384. – Torsten Schoeneberg Dec 06 '20 at 22:20
  • @TorstenSchoeneberg Thank you, this helps – NewOne Dec 06 '20 at 22:42

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The fact that the bilinear form $b$ defined on $L$ is non degenerated is equivalent to the fact that $d(b):L\rightarrow L^*$ defined by $d(b)(x)(y)=b(x,y)$ is an isomorphism. $I^{\perp}$ is identified with the anihilator $I^0$ of $I$ with this isomorphism and $dim(I)+dim(I^0)=dim(L)$.

Tsemo Aristide
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  • Thank you! I understand the idea and if d is an isomorphism, I see, that all is done. I understand the linearity and injectivity, but not the fact that d is surjective. How does it follow from the fact that b is non degenerate ? – NewOne Dec 06 '20 at 21:55
  • Okay, now I understand, that as d is injective, its image must have the dimension equal at least $\text{dim} \ L$, which is equal to $\text{dim} \ L*$, so it should be surjective, am I right? – NewOne Dec 06 '20 at 22:36