Why is the string group not a Lie group? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T00:55:32Z http://mathoverflow.net/feeds/question/39831 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/39831/why-is-the-string-group-not-a-lie-group Why is the string group not a Lie group? Konrad Waldorf 2010-09-24T07:24:33Z 2010-09-24T12:28:39Z <p>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.</p> <p>[<em>In a previous version of this question I screwed up the definition and caused some confusion, see the comments below.</em>]</p> <p>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 $\pi_3$. </p> <p>What is the correct statement about Lie group structures on the string group, and how does one prove it?</p> http://mathoverflow.net/questions/39831/why-is-the-string-group-not-a-lie-group/39833#39833 Answer by David Roberts for Why is the string group not a Lie group? David Roberts 2010-09-24T07:47:19Z 2010-09-24T10:35:34Z <p>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 $SO(n)$ which is all that matters. For that matter start with the 1-connected $Spin(n)$.</p> <p>[OK, a short train ride later, now I'm home from work. To continue...]</p> <p>The fibre of the 3-connected cover is a 2-type, and in the case of $Spin(n)$ this is a $K(\mathbb{Z},2)$, so at the very least, $String(n)$ can't be finite-dimensional. If one could construct a primitive[1] $PU(\mathcal{H})$-bundle on $Spin(n)$ whose Dixmier-Douady classs was the generator $\langle -,[-,]\rangle \in H^3(Spin(n))$, then you would have an infinite-dimensional Lie group model for $String(G)$ (here $\mathcal{H}$ is a infinite-dimensional separable Hilbert space, $PU(\mathcal{H})$ is then a smooth model for $K(\mathbb{Z},2)$).</p> <p>([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.)</p> <p>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 $String$ only being a topological group).</p> http://mathoverflow.net/questions/39831/why-is-the-string-group-not-a-lie-group/39838#39838 Answer by Daniel Pomerleano for Why is the string group not a Lie group? Daniel Pomerleano 2010-09-24T08:55:32Z 2010-09-24T08:55:32Z <p>As David Roberts is saying it's conceivable the string group could be represented by an infinite dimension manifold. I'm totally agnostic on that, but as I interpret the question it's asking why it's not equivalent(as an H-space?) to a non-compact finite dimensional Lie group(David Robert also explains that for a compact simply connected Lie group we always have $\pi_3$ non vanishing). I think though the underlying space has cohomology in infinitely many dimensions. Let me illustrate this in the case of String(3). So we have a Serre spectral sequence for the fibration $K(Z,2)\mapsto String(3) \mapsto S^3$. Now thinking of Z[x] as the cohomology ring of K(Z,2), the differential has to be $d:x \mapsto e$, the generator for the cohomology of $S^3$. So using the Leibnitz rule, $x^2\mapsto 2x\otimes e$, $x^3 \mapsto 3x^2\otimes e$... etc. This means that $H^5(String(3)= Z/2Z$, H^7(String(3))=Z/3Z... etc</p>