Do we define the rank of a Lie algebra by using the geometric multiplicity of the zero eigenvalues of the adjoint representation? Chapter 11 of Pontrjagin's "Topological groups" (the later editions) tells how to obtain a root system for the given Lie algebra, $L$.  It begins by considering the ``multiplicity'' of the zero eigenvalue of the adjoint representation $${\tt ad}\, a : x \mapsto [a, x]$$ on $L$, for each $a \in L$.  Is this the algebraic multiplicity or the geometric multiplicity?
 A: It's a bit hard to answer this question without having a look at the book (which I don't have available). Also, the question in the title is not exactly the same as in the question body.
However, the sources which I know work with the algebraic multiplicity here. This is certainly the case in Bourbaki's book on Lie Groups and Algebras (ch. VII §2 no. 2) -- they define the rank of $\mathfrak g$ as the highest $r \in \mathbb N_0$ such that $T^r$ divides the characteristic polynomials of all $ad_\mathfrak{g} a$ for $a \in \mathfrak g$.
As a sanity check, this article also does not explicitly say whether it works with geometric or algebraic multiplicities to define the rank, but it does say that e.g.
$\mathfrak{g} := \pmatrix{0&\ast & \ast \\
0&0&\ast\\
0&0&0}$
has rank $3$ -- more generally, any nilpotent (fin-dim., in characteristic $0$) Lie algebra has rank equal to its dimension. But you'll notice that e.g. the geometric multiplicity of the eigenvalue $0$ to $ad_{\mathfrak g} a$ of $a = \pmatrix{0&1 & 0 \\
0&0&0\\
0&0&0}$ is $2$, not $3$. And of course any nilpotent non-abelian Lie algebra would have elements where the geometric multiplicity of the eigenvalue $0$ is strictly less than its algebraic multiplicity.
As a final note, since you mention root spaces, probably the discussion restricts to semisimple Lie algebras at some point. If it's not explicitly proved in your source, it's a worthwhile exercise that in that case, it does not matter whether you work with geometric or algebraic multiplicities to define the rank. This follows e.g. immediately from the root space decomposition, once one knows that the rank is also the dimension of any Cartan subalgebra.
