# Does the existence of a string structure depend on a spin structure ? (and a generalization)

## 1 Question

Let $E\to X$$\newcommand{\Z}{\mathbb{Z}} \newcommand{\R}{\mathbb{R}} \newcommand{\C}{\mathbb{C}} \newcommand{\N}{\mathbb{N}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\F}{\mathbb{F}} \newcommand{\bZ}{\mathbb{Z}} \newcommand{\bR}{\mathbb{R}} \newcommand{\bC}{\mathbb{C}} \newcommand{\bH}{\mathbb{H}} \newcommand{\bQ}{\mathbb{Q}} \newcommand{\bF}{\mathbb{F}} \newcommand{\bN}{\mathbb{N}} \DeclareMathOperator\id{id} % identity map \DeclareMathOperator\Sq{Sq} % Steenrod squares \DeclareMathOperator\Homeo{Homeo} % group of homeomorphisms of a topoloical space \DeclareMathOperator\Diff{Diff} % group of diffeomorphisms of a smooth manifold \DeclareMathOperator\SDiff{SDiff} % diffeomorphism under some constraint \DeclareMathOperator\Hom{Hom} % homomrphism group \DeclareMathOperator\End{End} % endomorphism group \DeclareMathOperator\Aut{Aut} % automorphism group \DeclareMathOperator\Inn{Inn} % inner automorphisms \DeclareMathOperator\Out{Out} % outer automorphism group \DeclareMathOperator\vol{vol} % volume \newcommand{\GL}{\text{GL}} % general linear group \newcommand{\SL}{\text{SL}} % special linear group \newcommand{\SO}{\text{SO}} % special orthogonal group \newcommand{\O}{\text{O}} % orthogonal group \newcommand{\SU}{\text{SU}} % special unitary group \newcommand{\Spin}{\text{Spin}} % Spin group \newcommand{\RP}{\Rr\mathrm P} % real projective space \newcommand{\CP}{\Cc\mathrm P} % complex projective space \newcommand{\HP}{\Hh\mathrm P} % quaternionic projective space \newcommand{\Top}{\mathrm{Top}} % topological category \newcommand{\PL}{\mathrm{PL}} % piecewise linear category \newcommand{\Cat}{\mathrm{Cat}} % any category \newcommand{\KS}{\text{KS}} % Kirby-Siebenmann class \newcommand{\Hud}{\text{Hud}} % Hudson torus \newcommand{\Ker}{\text{Ker}} % Kernel \newcommand{\underbar}{\underline} %Classifying Spaces for Families of Subgroups \newcommand{\textup}{\text} \newcommand{\sp}{^}E\to X$ be a (stable) vector bundle. This has a classifying map $X\to BO$$X\to BO$.

A $BO\langle n\rangle$$BO\langle n\rangle$-structure on $E$$E$ is (the vertical homotopy class of) a lift of the classifying map to a map $X\to BO\langle n \rangle$$X\to BO\langle n \rangle$. (For $n=2,4,8$$n=2,4,8$ this is an orientation, a spin structure, a string structure respectively.)

Since for $n>m$$n>m$, the map $BO\langle n \rangle \to BO$$BO\langle n \rangle \to BO$ factors through $BO\langle m \rangle \to BO$$BO\langle m \rangle \to BO$, a $BO\langle n\rangle$$BO\langle n\rangle$-structure induces a $BO\langle m\rangle$$BO\langle m\rangle$-structure, or, vice versa, this specific $BO\langle m\rangle$$BO\langle m\rangle$-structure can be lifted to a $BO\langle n\rangle$$BO\langle n\rangle$-structure.

Question 1.1. Given a vector bundle $X\to BO$$X\to BO$ and two $BO\langle n\rangle$$BO\langle n\rangle$-structures on it, is it possible that one of the $BO\langle n\rangle$$BO\langle n\rangle$-structures can be lifted to a $BO\langle n+1\rangle$$BO\langle n+1\rangle$-structure and the other can't?

This is not possible for $n=2$$n=2$ and $n=4$$n=4$, i.e. the question whether an oriented vector bundle admits a spin structure does not depend on the orientation, and the question whether a spin vector bundle admits a string structure does not depend on the spin structure. The reason in the first case is the obstruction to admitting a spin structure is the second Stiefel-Whitney class which is a homotopy invariant. For the second, more subtle point, of why the spin characteristic class $\frac{p_1}{2}$$\frac{p_1}{2}$ does not depend upon the choice of spin structure see [Čadek&Crabb&Vanvzura2008, Defintion p.170].

The answer is however yes for all larger $n$$n$ with $BO\langle n\rangle\ne BO\langle n+1\rangle$$BO\langle n\rangle\ne BO\langle n+1\rangle$. For example the existence of a $BO\langle 9\rangle$$BO\langle 9\rangle$-structure on a string vector bundle can depend on the string structure.

## 3 Further discussion

The map $BO\langle n+1 \rangle \to BO\langle n \rangle$$BO\langle n+1 \rangle \to BO\langle n \rangle$ has homotopy fiber $K(\pi_{n}BO,n-1)$$K(\pi_{n}BO,n-1)$ and is the pullback of the path-loop fibration

$\displaystyle \Omega K(\pi_{n}BO,n) \to PK(\pi_{n}BO,n) \to K(\pi_{n}BO,n)$
via a map $BO\langle n\rangle \to K(\pi_{n}BO,n)$$BO\langle n\rangle \to K(\pi_{n}BO,n)$, corresponding to a class $k\in H^{n}(BO\langle n\rangle;\pi_{n}BO)$$k\in H^{n}(BO\langle n\rangle;\pi_{n}BO)$. Note that this is the generator of $H^{n}(BO\langle n\rangle;\pi_{n}BO)\cong Hom(\pi_{n}BO;\pi_{n}BO)$$H^{n}(BO\langle n\rangle;\pi_{n}BO)\cong Hom(\pi_{n}BO;\pi_{n}BO)$.

In the case $n=4i$$n=4i$, it follows from this result that the Pontryagin class $p_i\in H^{4i}(BO;\Zz)$$p_i\in H^{4i}(BO;\Zz)$ maps to a certain multiple $b_i\cdot k\in H^{4i}(BO\langle 4i\rangle;\Zz)$$b_i\cdot k\in H^{4i}(BO\langle 4i\rangle;\Zz)$ (see here for the value of $b_i$$b_i$). In these cases, the generator $k$$k$ is usally denoted by $\frac{p_i}{b_i}$$\frac{p_i}{b_i}$, although $p_i\in H^{4i}(BO;\Zz)$$p_i\in H^{4i}(BO;\Zz)$ itself is indivisible.

In the cases $n=1,2$$n=1,2$ the class $k$$k$ equals $w_1\in H^1(BO;\Zz_2)$$w_1\in H^1(BO;\Zz_2)$ respectively $w_2\in H^2(BSO;\Zz_2)$$w_2\in H^2(BSO;\Zz_2)$.

By obstruction theory, it follows that a map $f:X\to BO\langle n \rangle$$f:X\to BO\langle n \rangle$ lifts to $BO\langle n+1\rangle$$BO\langle n+1\rangle$ if and only if the "characteristic" class

$\displaystyle f^*k\in H^{n}(X;\pi_{n}BO)$
vanishes.

In particular, since the Stiefel-Whitney classes of a vector bundle are independent of an orientation, this answers the question for $n=2$$n=2$.

For every $BO\langle n\rangle$$BO\langle n\rangle$-structure on a vector bundle $X\to BO$$X\to BO$ there exists an opposite $BO\langle n\rangle$$BO\langle n\rangle$-structure (inducing the opposite orientation) defined as follows: The connective cover construction is functorial, thus the non-trivial deck transformation $\tau$$\tau$ of the $2$$2$-fold cover $BSO\to BO$$BSO\to BO$ induces a self-map $\tau$$\tau$ of $BO\langle n\rangle$$BO\langle n\rangle$. Composing the $BO\langle n\rangle$$BO\langle n\rangle$-structure with this self-map of $BO\langle n\rangle$$BO\langle n\rangle$ gives the opposite structure. We have $\tau^*(k)=k$$\tau^*(k)=k$ since $\tau$$\tau$ is a self-equivalence which for $n=4k$$n=4k$ is the identity on the Pontryagin classes. Thus for a $BO\langle n\rangle$$BO\langle n\rangle$-structure and its opposite, either both lift to a $BO\langle n+1\rangle$$BO\langle n+1\rangle$-structure or both do not.

In general, we are given two maps $f,g:X\to BO\langle n \rangle$$f,g:X\to BO\langle n \rangle$ (i.e. two stable vector bundles over $X$$X$ with $BO\langle n \rangle$$BO\langle n \rangle$-structure) for which we assume that the compositions with $BO\langle n \rangle \to BO\langle m \rangle$$BO\langle n \rangle \to BO\langle m \rangle$ are homotopic (i.e. the bundles are isomorphic as bundles with $BO\langle m \rangle$$BO\langle m \rangle$-structure). We have to investigate whether it is possible that $f^*k\ne g^*k=0$$f^*k\ne g^*k=0$. Here let us assume that $BO\langle n\rangle$$BO\langle n\rangle$ and $BO\langle m\rangle$$BO\langle m\rangle$ are consecutive connective covers in the sense that

$\displaystyle BO\langle n\rangle = BO\langle n-1\rangle = \dots = BO\langle m+1\rangle \ne BO\langle m\rangle.$

Since the compositions of $f$$f$ and $g$$g$ with $BO\langle n \rangle \to BO\langle m \rangle$$BO\langle n \rangle \to BO\langle m \rangle$ are homotopic, it follows that $f$$f$ and $g$$g$ differ by a map from $X$$X$ to the homotopy fiber $K(\pi_{m}BO,m-1)$$K(\pi_{m}BO,m-1)$ of $BO\langle n \rangle \to BO\langle m \rangle$$BO\langle n \rangle \to BO\langle m \rangle$. More precisely, the map $BO\langle n \rangle \to BO\langle m \rangle$$BO\langle n \rangle \to BO\langle m \rangle$ is a map of H-spaces, and given $f,g$$f,g$ as above, there exists a map $h:X\to K(\pi_{m}BO,m-1)$$h:X\to K(\pi_{m}BO,m-1)$ such that $f$$f$ is homotopic to the composition

$\displaystyle X \stackrel{g,h}{\longrightarrow} BO\langle n \rangle\times K(\pi_{m}BO,m-1) \stackrel{id\times i}\longrightarrow BO\langle n \rangle\times BO\langle n \rangle\to BO\langle n \rangle$

where the last map is the $H$$H$-space multiplication.

Under the $H$$H$-space multiplication $k$$k$ pulls back to

$\displaystyle k\otimes 1 + 1\otimes k \in H^*(BO\langle n\rangle)\otimes H^*(BO\langle n\rangle)\subseteq H^*(BO\langle n\rangle \times BO\langle n\rangle).$

Now it follows that $f^*k=(g,ih)^*(k\otimes 1 + 1\otimes k)=g^*k+h^*i^*k$$f^*k=(g,ih)^*(k\otimes 1 + 1\otimes k)=g^*k+h^*i^*k$.

Now we choose $X=K(\pi_{m}BO,m-1)$$X=K(\pi_{m}BO,m-1)$ and $h=id$$h=id$ as the "universal" example; thus we have to know whether $i^*(k)=0$$i^*(k)=0$.

For $n=4,m=2$$n=4,m=2$ we need to know the pullback of $k=\frac{p_1}{2}$$k=\frac{p_1}{2}$ under $i: K(\Zz_2,1)\to BSpin$$i: K(\Zz_2,1)\to BSpin$. This is zero, since the reduction modulo $2$$2$ is trivial, as the reduction of $k$$k$ modulo $2$$2$ is the image of $w_4\in H^4(BO;\Zz_2)$$w_4\in H^4(BO;\Zz_2)$. See [Stong1963, p.539].

For higher $n,m$$n,m$ we show that $i^*(k)\ne 0$$i^*(k)\ne 0$ by considering reductions modulo $p$$p$, which have been computed for $p=2$$p=2$ by [Stong1963] and for $p$$p$ odd by [Giambalvo1969].

For $n=8,m=4$$n=8,m=4$ we need to know the pullback of $k=\frac{p_2}{6}$$k=\frac{p_2}{6}$ under $i: K(\Zz,3)\to BString$$i: K(\Zz,3)\to BString$. This is a non-zero class: it suffices to show that its reduction modulo $3$$3$ is nontrivial. This follows from [Giambalvo1969, Theorem 1']. (The reduction modulo $2$$2$ of $k$$k$ is $w_8$$w_8$, so that the reduction modulo $2$$2$ of $i^*k$$i^*k$ is zero.)

Thus for example the trivial stable vector bundle on $K(\Zz,3)$$K(\Zz,3)$ admits a string structure which does not lift to a $BO\langle 9\rangle$$BO\langle 9\rangle$-structure.

In all higher dimensions $n$$n$, the (reductions modulo $2$$2$ of) $i^*k$$i^*k$ are non-zero by [Stong1963, p.539].