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Suppose we want to construct a semisimple Lie algebra $L$ of dimension $n$.

We know that if we can write $L = L_1 \oplus L_2 \oplus \cdots \oplus L_s$ as a direct sum of simple ideals, then $L$ must be semisimple.

I am not quite sure of what simple ideals we could use here to get a Lie algebra for a given dimension, any suggestions? Thanks.

arnav_de
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maddiemoo
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    It is easy to see that there is a complex semisimple Lie algebra in dimension $3,6$ and any $n\ge 8$, but not in dimension $1,2,4,5,7$, see here for example. This follows easily from the dimension formulas of the complex simple Lie algebras. – Dietrich Burde May 21 '22 at 13:47
  • I don't fully understand that post unfortunately, is there a way we could construct such a semisimple Lie algebra, let's say of dimension 6 then, as you state this is possible, using a direct sum of simple ideals? – maddiemoo May 21 '22 at 13:57
  • Yes, exactly. We can do with only direct sums of $A_1$ and $A_2$ to obtain all dimensions $n\ge 13$, see my answer. This is number theory, i.e., the coin problem. And of course $\dim (A_1\oplus A_1)=6$. – Dietrich Burde May 21 '22 at 14:03

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We have $\dim(A_1)=\dim \mathfrak{sl}_2(\Bbb C)=3$ and $\dim(A_2)=\dim \mathfrak{sl}_3(\Bbb C)=8$. By the Frobenius coin problem we know that we can obtain every integer $n\ge 3\cdot 8-3-8=13$ as a linear combination $3x+8y=n$, i.e., direct sums of $A_1$ and $A_2$ yield a semisimple Lie algebra of every dimension $n\ge 13$. For dimensions less than $13$ we can check by hand that only $n=1,2,4,5,7$ cannot arise.

Dietrich Burde
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