TITLE: Why is the commutator defined differently for groups and rings?
QUESTION [36 upvotes]: The commutator of two elements in a group is defined as $[g, h] = g^{−1}h^{−1}gh.$
In a ring, the commutator of two elements is $[a, b] = ab - ba.$
I'm asking because a ring is a (abelian) group under addition, so I would have expected it to be $[g, h] = -g-h+g+h.$

REPLY [5 votes]: You can relate the commutator for a Lie group $G$ to the (ring-type) commutator of its Lie algebra $\mathfrak{g}$.
Say $g$ lies infinitesimally close to the group identity, $g = \exp(\epsilon\,X)$ for some $X \in \mathfrak{g}$.  Likewise say $h = \exp(\epsilon\,Y)$ for some $Y \in \mathfrak{g}$.  Then, $ghg^{-1}h^{-1}$ also lies infinitesimally near the identity of $G$.  We compute using the Baker-Campbell-Hausdorff formula that its generator is
$$
\log ghg^{-1}h^{-1} = \epsilon^{2}[XY] + \mathrm{O}(\epsilon^3)\,.
$$
If you want to make further contact with the ring commutator, think of the Lie bracket as "coming from" the commutator with respect to the associative product in $\mathrm{U}\mathfrak{g}$, the universal enveloping algebra of $\mathfrak{g}$, so that
$$
[XY] = XY - YX
$$
in a genuine way.