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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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  • $\begingroup$ Konrad -- unless $n\leq2$ there is no group at all satisfying the conditions of the posting. $\endgroup$
    – algori
    Commented Sep 24, 2010 at 7:29
  • $\begingroup$ @algori could you give a reference for this (or a sketch of an argument)? $\endgroup$
    – user47274
    Commented Sep 24, 2010 at 7:32
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    $\begingroup$ 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. $\endgroup$
    – algori
    Commented Sep 24, 2010 at 7:37
  • $\begingroup$ ... or $\pi_3\neq 0$. Argh! $\endgroup$
    – algori
    Commented Sep 24, 2010 at 7:45
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    $\begingroup$ I think your definition needs to be revised to make the string group uniquely defined. In particular, String(n) should be the universal 3-connected term in the Moore-Postnikov tower of $O(n)$, and the definition given above does not exclude higher terms. Some of the commenters may be confused because it is not a covering group in the usual sense of covering spaces. $\endgroup$
    – S. Carnahan ♦
    Commented Sep 24, 2010 at 8:03

3 Answers 3

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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).

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  • $\begingroup$ David -- re $pi_3(G)=\mathbf{Z}$: what about tori? $\endgroup$
    – algori
    Commented Sep 24, 2010 at 8:10
  • $\begingroup$ I'm only thinking about the definition of String proper. $\endgroup$
    – David Roberts ♦
    Commented Sep 24, 2010 at 8:14
  • $\begingroup$ oh yes, it says "simple". Need to get some sleep. $\endgroup$
    – algori
    Commented Sep 24, 2010 at 8:18
  • $\begingroup$ David, can you complete your argument by explaining why a 2-type can't be finite-dimensional? $\endgroup$
    – Konrad Waldorf
    Commented Sep 24, 2010 at 9:02
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    $\begingroup$ Ok, it can't be finite-dimensional because it has cohomology in infintely many degrees, see Daniel's answer below. $\endgroup$
    – Konrad Waldorf
    Commented Sep 24, 2010 at 12:50
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To follow up, there is now an infinite-dimensional Lie group model of String:

  • Thomas Nikolaus, Christoph Sachse, Christoph Wockel, A Smooth Model for the String Group, Int. Math. Res. Not. 16 (2013) 3678-3721, doi:10.1093/imrn/rns154, arXiv:1104.4288.
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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 Roberts 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 $\mathrm{String}(3)$. So we have a Serre spectral sequence for the fibration $K(\mathbb{Z},2)\to \mathrm{String}(3) \to S^3$. Now thinking of $\mathbb{Z}[x]$ as the cohomology ring of $K(\mathbb{Z},2)$, 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 $H^5(\mathrm{String}(3))= \mathbb{Z}/2\mathbb{Z}$, $H^7(\mathrm{String}(3))=\mathbb{Z}/3\mathbb{Z}$... etc

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  • $\begingroup$ 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 $n$! $\endgroup$
    – Konrad Waldorf
    Commented Sep 24, 2010 at 12:39
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    $\begingroup$ $K(\mathbb{Z},2)$ is $\mathbb{CP}^\infty$, and its cohomology is a polynomial algebra on a generator in degree 2; most algebraic topology texts should cover this. $\endgroup$
    – Tyler Lawson
    Commented Sep 24, 2010 at 12:46
  • $\begingroup$ 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. $\endgroup$
    – Daniel Pomerleano
    Commented Sep 24, 2010 at 17:06

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