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If $\mathfrak{g}$ is a semisimple $\Rightarrow$ $\mathfrak{h} \subset \mathfrak{g} $ imply...

Discussion in 'Mathematics' started by Matheus Manzatto, Oct 8, 2018.

  1. Let $\mathfrak{g}$ be a semisimple Lie Algebra, $\langle \cdot,\cdot\rangle$ the Killing form and $\mathfrak{h}\subset \mathfrak{g}$ a subalgebra of $\mathfrak{g}$.

    I'm trying to find a counterexample (or prove) the following affirmation:

    Affirmation: $\mathfrak{h} \cap \mathfrak{h}^\perp = \{0\}$.

    where $\mathfrak{h}^\perp = \{X \in \mathfrak{g};\ \langle X, Y\rangle =0, \ \forall \ Y\in \mathfrak{h} \}$.

    Does anyone know how to prove this or have a nice counterexample?

    NB: In my opinion this affirmation seems false, however all spaces that I tried $\mathfrak{h} \cap \mathfrak{h}^\perp = \{0\}$ holded.

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