I have gotten myself quite confused with dimensions and ranks of Lie group and Lie algebras. As far as I understand:

_The $\bf{rank}$ of a Lie algebra is its number of Casimir operators (linearly independent operators built from elements of the Lie algebra which commute with all elements of the Lie algebra)
_The $\bf{rank}$ of a Lie group is the dimension of any one of its Cartan subgroups. The rank of a Lie group is equal to the rank of the corresponding Lie Algebra
_The $\bf{dimension}$ of a Lie group is its number of continuous parameters. This is equal to the number of generators of its simply connected part: $$\rho(\alpha_1, ..., \alpha_n) = \exp(i\alpha_aT^a) $$where $\alpha^a$ are the parameters, $T^a$ are the generators and $a$ runs from 1 to dim(Lie group)

The set of all linear combinations of the generators forms a vector space, which together with the Lie bracket forms the Lie Algebra. In this way the generators form a basis for the Lie Algebra.

_The $\bf{dimension}$ of a Lie algebra is its dimension as a vector space. This is greater than or equal to the $\bf{rank}$.

Is this correct so far ?

Can I conclude that dim(Lie Group) = cardinality(Lie Algebra basis) = dim(Lie Algebra) ?
I feel I have misunderstood this last part

Any help understanding any of these terms is appreciated, I am having trouble with ranks and dimensions of Lie algebras and Lie Groups

  • $\begingroup$ Fine, sure; for hairsplitting you could try the math SE.... $\endgroup$ Mar 31, 2021 at 17:31
  • $\begingroup$ Is knowing whether dim(Lie Algebra) = dim(Lie Group) of no use to a Physicist ? Have you perhaps taken the harmless nerdy banter mathematicians = hairsplitters, physicists = sloppy too seriously ? $\endgroup$
    – Mr Lolo
    Mar 31, 2021 at 17:46
  • $\begingroup$ I am happy to take my question to Mathematics SE if it is better suited. This question came up while I was studying a course as part of my Physics degree, hence my choice of SE $\endgroup$
    – Mr Lolo
    Mar 31, 2021 at 17:48
  • 1
    $\begingroup$ It's the same dimension. One is the dimension of the Lie group, the other is the dimension of the Lie Algebra understood as a vector space. $\endgroup$ Mar 31, 2021 at 18:29
  • $\begingroup$ Great, many thanks ! $\endgroup$
    – Mr Lolo
    Mar 31, 2021 at 18:33

1 Answer 1


A Lie group $G$ is a differentiable manifold with additional structure so$-$as any other differentiable manifold$-$its dimension is given by the dimension of the Euclidean space which is locally homeomorphic to. For this reason, one can show that the dimension of the tangent (vector) space $T_p G$ at any point $p\in G$ is equal to the dimension of $G$ as a manifold. Finally, the Lie algebra $\mathfrak{g}$ of $G$ can be identified with the tangent space at its identity, that is $\mathfrak{g} \cong T_1 G$ is a Lie algebra isomorphism, which in turn implies that $$\operatorname{dim}\mathfrak{g} = \operatorname{dim} T_1 G = \operatorname{dim} G.$$

Let's turn to ranks. As well as the rank of a finite dimensional Lie group is the dimension of any of its Cartan subgroups, the rank of a finite dimensional Lie algebra is equal to the dimension of any of its Cartan subalgebras. Hence the notion of rank of a Lie algebra is analogous to the one of rank of a Lie group. In fact, if $\mathfrak{g}$ is the Lie algebra associated to $G$, then $$\operatorname{rank}\mathfrak{g} = \operatorname{rank} G.$$

Naturally, a Cartan subgroup of $G$ is again a Lie group with the same identity element as $G$, so we could expect some kind of relation between the Lie algebra of that subgroup and the Lie algebra of $G$. Actually, if $G$ is real and connected, then the Lie algebra associated with any of its Cartan subgroups is a Cartan subalgebra of $\mathfrak{g}$.


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