A crossed module consists of a pair of groups $G$ and $H$ with a group homomorphism, $t:H \rightarrow G$, and $\alpha: G \times H \rightarrow H$ that defines an action of $G$ on $H$, $\tilde{\alpha}$: via $\alpha(g,h)=\tilde{\alpha}(g) (h)$. These maps satisfy, $t(\alpha(g,h))= g t(h) g^{-1}$, and $\alpha(t(h), h')= h h' h^{-1}$.
According to Baez and Lauda HDA 5 example 48, page 64, a Lie group and a Lie algebra can be used to construct a crossed module. The construction is to let $\alpha$ be defined via the adjoint map, and to let $t$ be defined as the trivial map $t(v)=1$ for all $v \in g$.
To simplify my question, suppose that $g = su(2)$, and $G=SU(2)$. Now consider $\alpha(g,v) = g v g^{-1}$ for $g\in G$ and $v\in g$. Define $t(v) = \exp{v}$. We can compute that $t(\alpha(g,h))= g t(h) g^{-1}$. However,(1)
- The Baker-Campbell-Hausdorff formula precludes that $t$ is a homomorphism, and(2)
- $\tilde{\alpha}(\exp{h}) (h')$ is rotation of $h'$ about the vector $h$ through an angle $ 2 |h|$. In other words, $\tilde{\alpha}(\exp{h}) (h') = (\exp{h}) h' (\exp{(-h)}).$
So the lack of homomorphism and the wrong Peiffer identity puts a damper on things. Is there a ``crossed module" like name for such a structure? For example, is it related to a $2$-group in any sense?
This question may be related to Theo's about Bernoulli numbers.
