# Is the dimension of a Lie group equal to the dimension of the corresponding Lie algebra?

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

• Fine, sure; for hairsplitting you could try the math SE.... Commented Mar 31, 2021 at 17:31
• 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 ? Commented Mar 31, 2021 at 17:46
• 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 Commented Mar 31, 2021 at 17:48
• 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. Commented Mar 31, 2021 at 18:29
• Great, many thanks ! Commented Mar 31, 2021 at 18:33

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 $$\mathfrak{g}$$.