Friday 1 June 2012

Why are Lie Algebras/ Lie Groups so much like crossed modules, and not?

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 $gin G$ and $vin g$. 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.

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