$\mathrm{String}/\mathbb{CP}^{\infty}=\mathrm{Spin}$ or a correction to this quotient group relation

Question

We know that there is a fiber sequence: $$ \dotsb \to B^3 \mathbb Z \to B \mathrm{String} \to B \mathrm{Spin} \to B^4 \mathbb Z \to \dotsb. $$

Is this fiber sequence induced from a short exact sequence?

• If so, is that $$ 1 \to B^2 \mathbb Z = B S^1= \mathbb{CP}^{\infty} \to \mathrm{String} \to \mathrm{Spin} \to 1? $$

If so, does the String group contain a normal subgroup $B^2 \mathbb Z = B S^1= \mathbb{CP}^{\infty}$?

• Is the classifying space $B S^1$ of the abelian group $S^1$ also a group? Is $\mathbb{CP}^{\infty}$ an abelian group or nonabelian group?

• So $\mathbb{CP}^{\infty}$ is a normal subgroup of $\mathrm{String}$, so $\mathrm{String}/\mathbb{CP}^{\infty}=\mathrm{Spin}$ where $\mathrm{Spin}$ is a quotient group of $\mathrm{String}$ group?

Please kindly correct me if I said anything wrong or stupid! Many thanks(giving)!

Answer 1

The topological groups $String$ and $BS^1$ are - a priori - only defined up to homotopy equivalence. In that setting, it makes sense to talk about fibre sequences, but the question for a group extension, as well as the question of whether or not a group is abelian, depend on models.

In Stephan Stolz's original work about the string group [1], reviewed in [2], he realizes String as an extension of topological groups $$ Gau(P) \to String \to Spin. $$ The group $Gau(P)$ is the group of gauge transformations in a principal $PU(H)$-bundle $P$ over $Spin$, topologized in some appropriate way. It is normal in $String$, but it is not abelian. It also is not isomorphic as topological groups to $PU(H)$, $\mathbb{CP}^\infty$, or $BS^1$, but all these groups are weakly homotopy equivalent.

When working with (infinte-dimensional, strict) 2-groups and strictly exact sequences, the work of Baez et al. [3] gives a concrete model for a $String$-2-group, fitting into an exact sequence of 2-groups $$ \mathcal{L}Spin \to String \to Spin $$ (where $Spin$ is regarded as a 2-group with only identity morphisms). The kernel 2-group $\mathcal{L}Spin$ has objects $\Omega Spin$, the based loop group, and morphisms a semi-direct product of $\Omega Spin$ and its universal central extension. By geometric realization, one obtains another group extension of topological groups, $$ |\mathcal{L}Spin| \to String \to Spin $$ with some other group $\mathcal{L}Spin$ whose homotopy type is again a $K(\mathbb Z,2)$.

Allowing non-strict (homotopy-)exact sequences of strict 2-groups, any model of the string 2-group is a central extension $$ BS^1 \to String \to Spin $$ by the abelian 2-group $BS^1$. This is telling us (at least me) that this is a good setting. A very simple such model was described lately in [4]. Note that geometric realization will not result into an exact sequence of topological groups.

[1] Stolz, Stephan, A conjecture concerning positive Ricci curvature and the Witten genus, Math. Ann. 304, No. 4, 785-800 (1996). ZBL0856.53033.

[2] Nikolaus, Thomas; Sachse, Christoph; Wockel, Christoph, A smooth model for the string group, Int. Math. Res. Not. 2013, No. 16, 3678-3721 (2013). ZBL1339.22009.

[3] Baez, John C.; Stevenson, Danny; Crans, Alissa S.; Schreiber, Urs, From loop groups to 2-groups, Homology Homotopy Appl. 9, No. 2, 101-135 (2007). ZBL1122.22003.

[4] Ludewig, Matthias; Waldorf, Konrad, Lie 2-groups from loop group extensions

Answer 2

These are sequences not of groups, but of ∞-groups, which can be modeled as simplicial groups or topological groups, equipped with a class of weak equivalences induced from simplicial sets or topological spaces via the obvious forgetful functor.

In particular, the homotopy fiber of the morphism of ∞-groups $\def\String{{\sf String}}\def\Spin{{\sf Spin}}\def\B{{\sf B}} \String→\Spin$ is the ∞-group $\def\Z{{\bf Z}}\B^2\Z$.

Normal ∞-subgroups can be defined as monomorphisms of ∞-groups that are also homotopy fibers, so the answer to the stated question about normal subgroups is positive, although the notion of a normal ∞-subgroup is rarely used in such a context.

The ∞-group $\B^2\Z$ is indeed an ∞-group. In fact, much more generally, every connective chain complex of abelian groups corresponds to a simplicial abelian group (and therefore to an abelian ∞-group), using the Dold–Kan correspondence. The ∞-group $\B^2\Z$ is obtained from the chain complex $\Z[2]$.

Finally, in the most obvious model for ∞-groups $\String$ and $\Spin$ (as simplicial groups, say), the canonical map $\String→\Spin$ is surjective in every simplicial degree, so its kernel $K$ can be computed degreewise. If we take $K$ as our model for $\B^2\Z$, then indeed $\Spin$ is isomorphic to the quotient (i.e., degreewise cokernel) $\String/K$, which in this case computes the homotopy cofiber of the map $K→\String$.