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.

  • $\begingroup$ I tried to fix the html: on my computer, at least, everything was displaying as the same link. (Note: the "preview" feature for writing question uses a different implementation of MarkDown than the actual website does for posting. In particular, preview runs in the browser, whereas when a question is posted the server does the work.) I still don't know why the link to Wikipedia fails. $\endgroup$ – Theo Johnson-Freyd May 30 '10 at 23:36
  • 1
    $\begingroup$ Incidentally, I don't have an answer to your questions, but I wanted to make a remark about the HDA5 example. Namely, a good way to think about the Lie algebra is as the infinitesimal neighborhood of the identity in the group. A better way is to take the group over the dual numbers $K[e]/e^2$, and the Lie algebra is precisely the points near the identity. Then the map "$t$" in the HDA5 example simply takes a point in this larger group to its "shadow", and so clearly respects all structure. Whereas in your construction, you are trying to take an "infinitesimal" quantity and scale it up. $\endgroup$ – Theo Johnson-Freyd May 30 '10 at 23:46
  • $\begingroup$ @Theo 1 and 2. Thanks for both. I am having a bad time with html and TeX on this site. A recent post was edited a few times by others. Arrgh! $\endgroup$ – Scott Carter May 31 '10 at 13:44

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