# Why is the string group not a Lie group?

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?

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Konrad -- unless $n\leq2$ there is no group at all satisfying the conditions of the posting. – algori Sep 24 '10 at 7:29
@algori could you give a reference for this (or a sketch of an argument)? – Michele Triestino Sep 24 '10 at 7:32
Michele -- a covering group of a Lie group is a Lie group. So either the connected component of the unit is a torus of $\pi_3\neq 0$. I presume the author meant something else but I'm not sure what exactly. – algori Sep 24 '10 at 7:37
... or $\pi_3\neq 0$. Argh! – algori Sep 24 '10 at 7:45
thank you! you are right – Michele Triestino Sep 24 '10 at 7:48

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 $Spin(n)$.

[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 $Spin(n)$ 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 $K(\mathbb{Z},2)$).

([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 $String$ only being a topological group).

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David -- re $pi_3(G)=\mathbf{Z}$: what about tori? – algori Sep 24 '10 at 8:10
I'm only thinking about the definition of String proper. – David Roberts Sep 24 '10 at 8:14
oh yes, it says "simple". Need to get some sleep. – algori Sep 24 '10 at 8:18
David, can you complete your argument by explaining why a 2-type can't be finite-dimensional? – Konrad Waldorf Sep 24 '10 at 9:02
Ok, it can't be finite-dimensional because it has cohomology in infintely many degrees, see Daniel's answer below. – Konrad Waldorf Sep 24 '10 at 12:50

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

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Daniel, sorry for this stupid question, can you specify a reference for the cohomology of $K(\mathbb{Z},2)$? If it really has cohomology in infinitely many degrees, than one can use David's argument above to show that $String(n)$ cannot be finite-dimensional for all $n$! – Konrad Waldorf Sep 24 '10 at 12:39
$K(\mathbb{Z},2)$ is $\mathbb{CP}^\infty$, and its cohomology is a polynomial algebra on a generator in degree 2; most algebraic topology texts should cover this. – Tyler Lawson Sep 24 '10 at 12:46
Aha, thanks! – Konrad Waldorf Sep 24 '10 at 12:56
Well it just seemed a little subtle because the string group is only defined up to H-space equivalence so it seemed possible that you could have an infinite manifold that is homotopy equivalent to a f.d. one. – Daniel Pomerleano Sep 24 '10 at 17:06