1
$\begingroup$

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

$\endgroup$
5
  • $\begingroup$ 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

1
$\begingroup$

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 $\mathfrak{g}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.