Subalgebra of one less dimension

[jostpuur]

Let g be a complex Lie algebra, of dimension greater than one. Does there always exist a subalgebra of dimension dim(g)-1?

If the claim is not true, is it true for some particular dimension? I can see that the claim is trivially true for dim(g)=2, but what about for example dim(g)=3?

The o(3,R) spanned by e_1,e_2,e_3, (with real coefficients) satisfying [e_1,e_2]=e_3, [e_2,e_3]=e_1, [e_3,e_1]=e_2, doesn't have a two dimensional subalgebra, but the obvious counter example attempt o(3,C) (the same basis, but complex coefficients) to the claim doesn't work, because now e_1+ie_2 and e_3 actually span a two dimensional subalgebra, [e_1+ie_2,e_3] = i(e_1 + ie_2).

 

[fresh_42]

Interesting question. A complex Lie algebra can always be written an $\mathfrak{g} = \mathfrak{T} \ltimes \mathfrak{R}$ and the radical has an ideal of codimension one, say $\mathfrak{I} \subseteq \mathfrak{R}$. Now we can build $\mathfrak{T} \oplus \mathfrak{I}$ with $[\mathfrak{T},\mathfrak{I}] \subseteq \mathfrak{R}$ but we need that $\mathfrak{I}$ is an ideal of $\mathfrak{g}$ which I think is not generally the case, because the semisimple part $\mathfrak{T}$ acts "too transitive" on $\mathfrak{R}$. But for solvable, complex Lie algebras, i.e. $\mathfrak{T}=\{\,0\,\}$, we even have an ideal of codimension one.

Dimension two is trivially true, since there is only an Abelian and one solvable Lie algebra. The question hasn't much to do with dimension, rather with the operation of $\mathfrak{T}$ on $\mathfrak{R}$ as mentioned above. Low dimensions just don't have enough structure available for $\mathfrak{T}$.

Simple algebras as the orthogonal Lie algebra you mentioned, or more generally semisimple Lie algebras always have trivial radical $\mathfrak{R}=\{\,0\,\}$. Hence we must ask if there is always a subalgebra $\mathfrak{S}$ of codimension $1$ in a simple Lie algebra. This would automatically create an example in the general case, too, as we can build $\mathfrak{S}\oplus \mathfrak{R}$.

So let's assume $\mathfrak{g} = \mathfrak{h} \oplus \sum_{\alpha \in \Delta^{-}} \mathbb{C}E_\alpha \oplus \sum_{\alpha \in \Delta^{+}} \mathbb{C}E_\alpha$. We can certainly not choose a vector from the CSA $\mathfrak{h}$ since it is generated by the rest. But if we choose a vector $E_\alpha$ as missing basis vector, then we can recover it with elements of $\mathfrak{h}$, as it is within a chain of roots, except in the case $| \Delta^{+}|=1$ where $\mathfrak{g}=\mathfrak{sl}(2,\mathbb{C})$ and its Borel subalgebra $\mathfrak{B(sl}(2,\mathbb{C}))$ is the only example of a simple Lie algebra with a subalgebra of codimension $1$.