The string group $String(n)$ is by definition a 3-connected cover of $Spin(n)$. This definition determines the homotopy type of the string group.


[*In a previous version of this question I screwed up the definition and caused some confusion, see the comments below.*]


A common argument is saying that "the string group cannot be a Lie group because it has vanishing $\pi\_3$". This is obviously not a complete argument because $(\mathbb{R},+)$ is a nice Lie group with vanishing $\pi\_3$. 


What is the correct statement about Lie group structures on the string group, and how does one prove it?



Why is the string group not a Lie group?
The result is that a compact, connected simple Lie group $G$ has $\pi\_3(G) = \mathbb{Z}$. Simple covering space or subgroups arguments should get you to $\mathrm{SO}(n)$ which is all that matters. For that matter start with the 1-connected $\mathrm{Spin}(n)$.


[OK, a short train ride later, now I'm home from work. To continue...]


The fibre of the 3-connected cover is a 2-type, and in the case of $\mathrm{Spin}(n)$ this is a $K(\mathbb{Z},2)$, so at the very least, $\mathrm{String}(n)$ can't be finite-dimensional. If one could construct a primitive[1] $PU(\mathcal{H})$-bundle on $\mathrm{Spin}(n)$ whose Dixmier-Douady classs was the generator $\langle -,[-,]\rangle \in H^3(\mathrm{Spin}(n),\mathbb{Z})$, then you would have an infinite-dimensional Lie group model for $\mathrm{String}(G)$ (here $\mathcal{H}$ is a infinite-dimensional separable Hilbert space, $PU(\mathcal{H})$ is then a smooth model for $K(\mathbb{Z},2)$, if we take the norm topology, making it a Banach Lie group).


([1] Primitive in the sense that for the group operations $G\times G\to G$ and $(-)^{-1}:G\to G$ there are bundle maps covering them.)


I don't know if this is possible or not, but I'm sure this idea has occurred to someone before, and since we haven't seen it, there might be a reason (well, I haven't seen it and everyone goes on about $\mathrm{String}\_G$ only being a topological group).


