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I came across the claim here which states that there are no complex semisimple Lie algebras of dimension $4$, $5$, or $7$. As the problem suggests, we can take a Cartan subalgebra $H$ and root system $\Phi$ so that $L=H\oplus(\oplus_{\alpha\in\Phi}L_\alpha)$ where $$L_\alpha=\{x\in L:(\forall h\in H)\ [h,x]=\alpha(h)x\}$$

As shown in Humphreys' Introduction to Lie Algebras, every $L_\alpha$ has dimension $1$, so $\dim(L)=\dim(H)+\vert\Phi\vert$. I know $\Phi$ spans $H^*$, so $\dim(H)\leq\vert\Phi\vert$. Furthermore, I know that for any $\alpha\in\Phi$, the only multiples of $\alpha$ in $\Phi$ are $\alpha$ and $-\alpha$. Since $\mathbb{C}$ has characteristic $0$, this implies $\vert\Phi\vert$ is even.

Finally, another fact which may be useful is that given nonzero vectors $x_\alpha\in L_\alpha$ and $y_\alpha\in L_{-\alpha}$, if we take $h_\alpha=[x_\alpha,y_\alpha]\in H$, then $x_\alpha$, $y_\alpha$, and $h_\alpha$ span a three-dimensional subspace of $L$ isomorphic to $\mathfrak{sl}(2,\mathbb{C})$. However, I am not sure how to connect these pieces to give the desired conclusion.

As a follow-up question, would the claim still be true if I replaced the field with an arbitrary field of characteristic $0$, or is it also necessary for the field to be algebraically closed?

Anonymous
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1 Answers1

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One little fact that will help resolve this is that $\dim(H)=rank(\Phi)$, the rank of the root system.

Now the rest is done by a look at the classification of root systems. If we don't have that ready, we can even get by with just the low-dimension / low-rank cases by hand:

A root system of rank $\ge 3$ must contain at least six roots (a basis and their negatives), so in this case we would already have $rank(\Phi)+\lvert \Phi\rvert \ge 9$ (usually much larger in fact, but dimension $9$ indeed occurs for $\mathfrak{sl}_2\oplus \mathfrak{sl}_2 \oplus \mathfrak{sl}_2$).

Root systems of rank $2$ are implicitly classified at the beginning of every lecture on root system when one discusses the relations which two roots can have to each other. Turns out the possibilities are $A_1 \times A_1, A_2, B_2=C_2$, and $G_2$; while the first indeed contains four roots and describes the semisimple, six-dimensional $\mathfrak{sl}_2 \oplus \mathfrak{sl}_2$, all the others contain $\ge 6$ roots and thus correspond to Lie algebras of dimension $\ge 2+6 =8$ (actually, $A_2$ corresponds to the $8$-dimensional $\mathfrak{sl}_3$, and $B_2=C_2$ to the $10$-dimensional $\mathfrak{so}_5 \simeq \mathfrak{sp}_4$; the dimension of the exceptional Lie algebra of type $G_2$ is $14=2+12$).

There's only one root system of rank $1$: $A_1$, which corresponds to the $3$-dimensional $\mathfrak{sl}_2$. So there's nothing that could make up a semisimple Lie algebra of dimension $4,5,$ or $7$ (or dimension $1$ or $2$, for that matter; I'd think the next non-occurring dimension is $11$, edit: As Jason DeVito and Dietrich Burde point out in comments (thanks!), actually all higher dimensions do occur.)

The above implicitly assumed that we work over an algebraically closed field of characteristic $0$. As YCor points out in a comment, this is enough to conclude for any base field $k$ of characteristic $0$. Namely, if $L$ is a semisimple Lie algebra over $k$ of $k$-dimension $n$, and $K\vert k$ is any field extension, then the scalar extension $L\otimes_k K$ is a semisimple Lie algebra of $K$-dimension $n$. Apply to an algebraic closure of $k$.

Torsten Schoeneberg
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