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According to wikipedia, every semisimple Lie algebra is a subalgebra of the special linear Lie algebra $\mathfrak{sl}_n$ for some $n\ge 1$. But I do not know any proof of this, nor am I able to imagine an intuitive reason for this remarkable fact. Can you give me a hint or an indication on how to demonstrate this?

Davius
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2 Answers2

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Let $\mathfrak g$ be a semisimple Lie algebra over a field $\Bbbk$ with characteristic $0$. According to Ado's theorem, $\mathfrak g$ is isomorphic to a subalgebra $\mathfrak s$ of $\mathfrak{gl}_n$, for some $n\in\Bbb N$. But $\mathfrak{gl_n}$ is isomorphic to a subalgebra of $\mathfrak{sl}_{n+1}$: it's the subalgebra which consits of matrices of the form $\left[\begin{smallmatrix}M&0\\0&-\operatorname{tr}(M)\end{smallmatrix}\right]$, with $M\in\mathfrak{gl}_n$.

José Carlos Santos
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    In characteristic $0$, we do not even need to use the (neat!) fact that $\mathfrak{gl}n \hookrightarrow \mathfrak{sl}{n+1}$. Rather, if $\rho: \mathfrak{s} \rightarrow \mathfrak{gl}_n(K)$ is any Lie algebra homomorphism, then $im(\rho) \subseteq \mathfrak{sl}_n(K)$. Because in char. $0$, semisimple LA's are perfect, i.e. every element can be written as sum of commutators; so since $\rho$ is a homomorphism, every element in its image can be written as a sum of commutators, but commutators of elements of $\mathfrak{gl}_n$ are in $\mathfrak{sl}_n$. – Torsten Schoeneberg Oct 17 '22 at 17:57
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Since a semisimple Lie algebra $L$ over any field $K$ has trivial center, because the solvable radical is zero, the adjoint representation is faithful. So we obtain an embedding of Lie algebras $$ L\hookrightarrow \mathfrak{gl}_n(K)\hookrightarrow\mathfrak{sl}_{n+1}(K), $$ without refering to Ado's Theorem, or Iwasawa's Theorem for prime characteristic.

Dietrich Burde
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  • I prefer this answer for using much less machinery than the other, and working in any characetristic. Unfortunately, my argument that the image of any representation is automatically contained in $\mathfrak{sl}$, see comment to other answer, would not work in positive characteristic, or would it? – Torsten Schoeneberg Oct 17 '22 at 17:59
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    @TorstenSchoeneberg Thank you! I know that a semisimple Lie algebra need not be a direct sum of ideals in characteristic $p$, and that then the Lie algebra need not be perfect. You have pointed this out here, with reference to Sakellar, Remark 2.2. on page $9$. On the other hand, the embedding using the trace seems very elementary and is a good argument, I think. – Dietrich Burde Oct 17 '22 at 18:49