Does non-commuting $\mathfrak{g}$ imply non-abelian $G$?

Question

Question 8.1 in Kristopher Tapp's introductory text on matrix groups asks to show that $SO(n)$ is non-abelian ($n>2$) by finding two elements of $so(n)$ that do not commute.

Why is this method valid?

At least in the text, he only proves that $[A,B]=0$ implies $e^A e^B = e^{A+B} = e^B e^A$, but does not mention whether the converse is true. The latter being true is the only way I could rationalize his statement.

Thanks!

Answer 1

Suppose $G$ is an abelian matrix group. We show that $\mathfrak{g}$ is abelian; that is, $[X, Y] = 0$, for all $X, Y \in \mathfrak{g}$.

For any $g \in G$, define $\mathrm{C}_g \colon G \to G$ by $h \mapsto ghg^{-1}$. Then, $\mathrm{Ad}_g\, \colon \mathfrak{g} \to \mathfrak{g}$ is defined as $\mathrm{D} \left( \mathrm{C}_g \right)_\mathrm{I}$. Since $G$ is abelian, $\mathrm{C}_g$ is the identity map, and hence $\mathrm{Ad}_g = \mathrm{Id}$. Now, $\mathrm{exp}(tX)$ is a path in $G$ beginning at $\mathrm{I}$ with initial velocity $X$. Therefore, by definition, \begin{equation} [X, Y] = \frac{\mathrm{d}}{\mathrm{dt}} \bigg|_{t=0} \mathrm{Ad}_{\mathrm{exp} \left( tX \right)} Y = \frac{\mathrm{d}}{\mathrm{dt}} \bigg|_{t=0} Y = 0. \end{equation}

Answer 2

If $A,B$ do not commute then $e^Ae^B = e^{A+B+ \frac{1}{2}[A,B]+ \cdots }$ and the higher terms are nested commutators. This is the Baker Campbell Hausdorff relation. Since elements of the Lie group close to the identity can be written as exponentials of the algebra this shows the noncommutativity. Intuitively, the higher order terms are much smaller for elements near the identity. This is just a sketch. There is probably a clever argument which avoids the BCH identity.

Answer 3

You need connectivity.

Pick any element $g\in G$ and any element $X\in\mathfrak{g}$. Consider that since $\exp(tX)\in G$, we have by the abelianess of $G$ that $g\exp(tX)g^{-1}=\exp(tX)$. So, if we differentiate we get, fortuitously, $\text{Ad}(g)X=X$.

Now, take $X$ and $Y$ in $\mathfrak{g}$. Apply the above argument with $g=\exp(tX)$ to see that $\text{Ad}(\exp(tX))Y=Y$. Differentiate to get the desired result.

Answer 4

Suppose $G$ is abelian. Then $\mathfrak{g}$ is trivially abelian (you can express $[A,B]$ for elements of the Lie algebra in terms of commutators of elements of the group). Therefore the contrapositive holds: $\mathfrak{g}$ not abelian implies $G$ is not abelian.

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As I say in the comments, I meant it was conceptually trivial that the local $\mathfrak{g}$ inherits the global abelian property.

Since people obviously thought this was worth spelling out, here is the way I have seen this done, taking the definition to be matrix commutators from a matrix Lie group: $$ g(t) = \exp(Xt),\;h(t) = \exp(Yt) \\ \text{Abelian}\implies \boxed{I = g h g^{-1} h^{-1} = I + [X,Y]t^2 + O(t^3)} $$ and hence $[X,Y]=0$.

Answer 5

The Lie algebra structure on $\def\lie#1{\mathfrak{#1}}\lie{g}$ is defined by the following two-step procedure:

• For a Lie group $G$, each element $g \in G$ acts by conjugation on $G$, $\def\on{\operatorname}\on{Ad}_g \colon h \mapsto ghg^{-1}$. This map $G \to G$ has a differential map on the tangent space at the identity, $d\on{Ad}_g \colon TG_1 \to TG_1$. By definition, $\lie{g} = TG_1$ and typically $d\on{Ad}_g$ is also written $\on{Ad}_g$, called the adjoint action of $g$ on $\lie{g}$.

• The adjoint action is a Lie group action of $G$ on $\lie{g}$; mainly, the aggregate map $\on{Ad} \colon G \times \lie{g} \to \lie{g}$ is smooth. It, too, has a differential $d\on{Ad} \colon T(G \times \lie{g})_{(1,0)} \to T\lie{g}_0$. By general principles of differential topology, we have $$T(G \times \lie{g})_{(1,0)} \cong TG_1 \times T\lie{g}_0 \cong \lie{g} \times \lie{g}$$ where we have used the fact that $\lie{g}$ is a vector space and thus isomorphic to its own tangent space at any point. Thus, $d\on{Ad}$ is some kind of bivariate map from $\lie{g}$ to itself; in fact, it's the Lie bracket.

It follows from the above definition that if $\on{Ad}_g \colon G \to G$ is the constant function $1$ for all $g \in G$, then ultimately, the Lie bracket is zero. The former is the case whenever $G$ is abelian. Taking the contrapositive, if $\lie{g}$ is nonabelian, then $G$ is necessarily as well.