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The rank of SO(3) is 1, the rank of SO(4) is 2. I'm trying to understand the definition of rank of a group with those two examples.

The rank of a group is the cardinality of the smallest generating set. The definition from Wikipedia is given in the first sentence. (Link to wikipedia: https://en.wikipedia.org/wiki/Rank_of_a_group)

Definition of generating set: "a generating set of a group is a subset such that every element of the group can be expressed as a combination (under the group operation) of finitely many elements of the subset and their inverses. "

In the case of SO(3), the group operation would be (matrix-)multiplication and there is no way one could express all the uncountably many rotations in the xy-plane with a finite product of matrices.

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    $\begingroup$ It is the rank of its Lie algebra, i.e., the dimension of a Cartan subalgebra - see this post. $\endgroup$
    – Dietrich Burde
    Commented Mar 27, 2020 at 12:52
  • $\begingroup$ Thank you very much. For my use case (finding the number of Casimir operators) it would be a detour to go over to lie groups anyways. $\endgroup$
    – Hans-Ulrich Rudel
    Commented Mar 27, 2020 at 14:21
  • $\begingroup$ Every time anyone defines a function called $f$ which can be applied to both an $X$ and a $Y$, and every $X$ is a $Y$, but the values of $f$ don't agree, Bourbaki kills a kitten. Here, rank is the culprit, but there are many others. On behalf of all of mathematics, I apologize. $\endgroup$
    – Adam Chalcraft
    Commented Mar 27, 2020 at 21:07

2 Answers 2

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Since you're asking about the Lie groups $SO(3)$ and $SO(4)$, you're looking at the wrong definition of rank.

You don't want the rank of a group meaning the minimal number of generators; for an uncountable group, that rank is uncountable, as you suspected.

Instead you want the rank of a Lie group, and I quote from that link: "For connected compact Lie groups... the rank of the Lie group is the dimension of any one of its maximal tori."

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The rank of a Lie-group is the dimension of a maximal torus. In $SO(3)$ a maximal torus is given by the rotations around just one axis, for example $SO(2)\times \{1\} < SO(3)$, and this is diffeomorphic to $S^1$, the circle. Therefore the rank of $SO(3)$ is 1 (the dimension of $S^1$).

In $SO(4)$ you have maximal tori of the form $SO(2)\times SO(2)$, so here the maximal torus really is a torus $S^1 \times S^1$, which is 2-dimensional. So the rank of $SO(4)$ is 2.

Edit: I changed maximal abelian groups to maximal tori. (Thanks for the comment)

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    $\begingroup$ "Torus" is more restrictive than "abelian subgroup": torus is an abelian subgroup consisting of semisimple (diagonalizable) elements. There might be abelian subgroups of bigger dimension than maximal tori, for example $SL(2n)$ whose maximal tori have dimension $2n-1$ has abelian subgroups of dimension $n^2$ (block-triangular matrices $\begin{pmatrix}I_n&A\\0_n&I_n\end{pmatrix}$) $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Mar 27, 2020 at 21:27

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