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They have already mentioned and approached to the problem e.g. here There are no semisimple Lie algebras of dimension $4$, $5$, or $7$

We only know, root space decomposition without introducing root system or its classification.

Given complex Lie semi-simple algebra $L$, we have

$$L=H\oplus \bigoplus_{\alpha \in \Phi}L_\alpha$$ $$L_\alpha=\{x\in L:(\forall h\in H)\ [h,x]=\alpha(h)x\}$$ $H$ is Cartan Subalgebra. $\Phi =\{\alpha\neq 0\in H^* \;|\; L_\alpha\neq 0\}$ where $H^*$ is dual space of the Cartan Subalgebra $H$

By above decomposition we can show $$\dim(L)=\dim(H)+\vert\Phi\vert$$

We can also know that $\pm \alpha\in \Phi$ if $\mp \alpha\in \Phi$

That shows $|\Phi|=$ even.

We also know, every semisimple complex Lie Algebra $L$ has a simple subalgebra isomorphic to $\mathfrak {sl}(2,\mathbb C)$

$\bullet$For $dimL=4$:

I can show that $|\Phi|$ can't be $0$ and $4$. Trivially since $L$ is neither abelian, nor $H$ is trivial.

For $|\Phi|=2$, $dimH=2$. Can I say that $L$ would have $\mathfrak {sl}(2,\mathbb C)$ as a subalgebra but it is dimension $3$?

So as you can see I need to show brutely without invoking classifications via $A_n,B_n,G_n$ and root systems. Using basic informations and dimension formula how we can show these dimensions cannot have semisimple complex Lie structre?

User not found
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  • In my answer to the linked question I try explicitly to not use the classification, just elementary facts about roots. I would not know how to avoid those facts except for invoking something like https://math.stackexchange.com/q/4099260/96384. – Torsten Schoeneberg Nov 22 '22 at 17:02
  • I am sorry but your answer is full of possibilities come from the classifications. And I do not understand why, when $rank\Phi=1$ means $dimH=2$ in the case $L$ is dimension $4$, cannot happen? – User not found Nov 22 '22 at 17:44
  • For example in the case $dimL=5$, by dimension considerations I could only say it "may" be possible $dimH=1$ and $|\Phi|=4$ is possible, how can we get contradiction? – User not found Nov 22 '22 at 18:20

1 Answers1

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Following up discussion in comments:

I think the bare minimum one needs to know is that

a) roots come in $\pm$ pairs, so their number $\lvert \Phi \rvert$ is even;

b) the number of such pairs is $\ge (rank(\Phi)) = dim (H)$.

With this, we can show that if $rank(\Phi) \ge 3$, the Lie algebra has dimension $\ge 9$; and that for rank $2$, the lowest possible dimensions are $2+4=6$ and $2+6=8$.

Finally, one might need to convince oneself that if $rank(\Phi)=1$, there cannot be more than one pair of roots. I.e. in this case, the dimension of the Lie algebra must be $3$ (in fact, it must be $\mathfrak{sl}_2(\mathbb C)$).

For this, one might need to know that in the case at hand,

c) if $\alpha$ is a root, the only scalar multiple $c \cdot \alpha$ that also is a root is $-\alpha$.

I think without a)-c), or something equivalent to it, it would be nearly impossible to show what you want. But these three facts are more elementary than showing the roots form a "root system" in the usual sense, let alone the full classification of those!

Torsten Schoeneberg
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  • I actually thought c) was part of the definition of a root system. Or is that one of those things that differ between treatments of root systems? – Tobias Kildetoft Nov 22 '22 at 19:10
  • @TobiasKildetoft: Sure it usually is, although depending on context one might follow Bourbaki and call that a "reduced" root system (allowing also factors of type BC). But be that as it may, OP said in post and comments that they are not supposed to use the fact that "the roots (presumably: weights of a CSA in an LA) form a root system (in the sense of usual abstract definitions)". So I thought we cannot use any of those properties. I just think we really cannot do without this one here. – Torsten Schoeneberg Nov 22 '22 at 19:14
  • thank you for your answer, the only point missing now is convincing when rank is 1 there can't be more than one pair of roots. Which I also asked after our discussion in the comments: https://math.stackexchange.com/questions/4582826/complex-semi-simple-lie-algebra-dimension-5 I am religiously believe that the case but cannot convince with some enough information to back it up. – User not found Nov 22 '22 at 19:39
  • A proof for part c) that does not use any properties of root systems could possibly be built on the idea that if $\alpha$ and $c\alpha$, $c\neq\pm1$, are both roots, then the usual commutator rules tell us that $[L_{\alpha},L_{c\alpha}]\subseteq L_{(c+1)\alpha}$ et cetera. Meaning that either we can keep adding more roots, ladder operator style, or we find two roots such that the root spaces commute. The latter case should then lead to a non-trivial ideal, I think. A number of details needs to be filled in. – Jyrki Lahtonen Nov 22 '22 at 19:42