Is there a proof of the non-existence of semi-simple Lie algebras whose dimensions are, for example, $4$ or $5$?
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1The smallest simple Lie algebras have dimensions $3$ and $6$. – Angina Seng Jun 08 '20 at 14:31
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I assume characteristic is zero, probably the argument works in characteristic $\neq 2,3,5$ without change.
Several answers using classification were already given.
Without classification it comes quickly. First we can suppose that the field is algebraically closed (extend scalars otherwise).
If the rank is $r$, the roots span the dual of the Cartan subalgebra and are stable under $\alpha\mapsto -\alpha$, so the number of roots is even, say $2n$ and $n\ge r$; moreover $\dim(\mathfrak{g})=r+2n$.
Hence $\dim\le 5$ implies $r\neq 1$, and $r=0$ implies $n=0$ i.e., dimension $0$. So the rank is 1, which implies the dimension is 3 or 5: this already excludes dimension 4. Finally dimension 5 is excluded because roots cannot be collinear (unless equal or opposite), that is, rank 1 implies dimension 3.
This also excludes dimension 7 since then the only possible rank is 2, which implies the dimension $r+2n$ is even.
YCor
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1how do you define rank r, and if dim<=5 how you exclude other possible rs – User not found Nov 22 '22 at 14:49