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

Revision as of 17:01, 21 January 2011

1 Question

Let $<!-- COMMENT: To achieve a unified layout, along with using the template below, please OBSERVE the following: besides, $...$ and $$...$$, you should use two environments: - For statements like Theorem, Lemma, Definition etc., use e.g. {{beginthm|Theorem 1|(Milnor)}} ... ... ... {{endthm}}. - For references, use e.g. {{cite|Milnor1958b}}. END OF COMMENT --> == Question == <wikitex>; Let $E\to X$ be a (stable) vector bundle. This has a classifying map $X\to BO$. A $BO\langle n\rangle$-structure on $E$ is (the vertical homotopy class of) a lift of the classifying map to a map $X\to BO\langle n \rangle$. (For $n=2,4,8$ this is an orientation, a spin structure, a string structure respectively.) Since for $n>m$,the map $BO\langle n \rangle \to BO$ factors through $BO\langle m \rangle \to BO$, a $BO\langle n\rangle$-structure induces a $BO\langle m\rangle$-structure, or, vice versa, this specific $BO\langle m\rangle$-structure can be lifted to a $BO\langle n\rangle$-structure. {{beginthm|Question}} Given a vector bundle $X\to BO$ and two $BO\langle n\rangle$-structures on it, is it possible that one of the $BO\langle n\rangle$-structures can be lifted to a $BO\langle n+1\rangle$-structure and the other can't? {{endthm}} </wikitex> == Answers == <wikitex>; This is not possible for $n=2$ and $n=4$, 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 answer is however yes for larger $n$. For example the existence of a $BO\langle 9\rangle$-structure on a string vector bundle can depend on the string structure. </wikitex> == Further discussion == <wikitex>; The map $BO\langle n+1 \rangle \to BO\langle n \rangle$ has homotopy fiber $K(\pi_{n}BO,n-1)$ and is the pullback of the path-loop fibration $$\Omega K(\pi_{n}BO,n) \to PK(\pi_{n}BO,n) \to K(\pi_{n}BO,n)$$ via a map $BO\langle n\rangle \to K(\pi_{n}BO,n)$, corresponding to a class $k\in H^{n}(BO\langle n\rangle;\pi_{n}BO)$. Note that this is the generator of $H^{n}(BO\langle n\rangle;\pi_{n}BO)\cong Hom(\pi_{n}BO;\pi_{n}BO)$. By obstruction theory, it follows that a map $f:X\to BO\langle n \rangle$ lifts to $BO\langle n+1\rangle$ if and only if the "characteristic" class $$f^*k\in H^{n}(X;\pi_{n}BO)$$ vanishes. This is for $n=1,2$ the first and second Stiefel-Whitney class of the corresponding vector bundle over $X$. In particular, since the Stiefel-Whitney classes of a vector bundle are independent of an orientation, this answers the question for $n=2$. In general, we are given two maps $f,g:X\to BO\langle n \rangle$ (i.e. two stable vector bundles over $X$ with $BO\langle n \rangle$-structure) for which we assume that the compositions with $BO\langle n \rangle \to BO\langle m \rangle$ are homotopic (i.e. the bundles are isomorphic as bundles with $BO\langle m \rangle$-structure). We have to investigate whether it is possible that $f^*k\ne g^*k=0$. Here let us assume that $BO\langle n\rangle$ and $BO\langle m\rangle$ are consecutive connective covers in the sense that $$BO\langle n\rangle = BO\langle n-1\rangle = \dots = BO\langle m+1\rangle \ne BO\langle m\rangle.$$ Since the compositions of $f$ and $g$ with $BO\langle n \rangle \to BO\langle m \rangle$ are homotopic, it follows that $f$ and $g$ differ by a map from $X$ to the homotopy fiber $K(\pi_{m}BO,m-1)$ of $BO\langle n \rangle \to BO\langle m \rangle$. More precisely, the map $BO\langle n \rangle \to BO\langle m \rangle$ is a map of H-spaces, and given $f,g$ as above, there exists a map $h:X\to K(\pi_{m}BO,m-1)$ such that $f$ is homotopic to the composition $$ X \stackrel{g,h}{\longrightarrow} BO\langle n \rangle\times K(\pi_{m}BO,m-1) \stackrel{id\times i}\longrightarrow BO\langle n \rangle\times BO\langle n \rangle\to BO\langle n \rangle$$ where the last map is the $H$-space multiplication. Under the $H$-space multiplication $k$ pulls back to $$k\otimes 1 + 1\otimes k \in H^*(BO\langle n\rangle)\otimes H^*(BO\langle n\rangle)\subseteq H^*(BO\langle n\rangle \times BO\langle n\rangle).$$ Now it follows that $f^*k=(g,ih)^*(k\otimes 1 + 1\otimes k)=g^*k+h^*i^*k$. Now we choose $X=K(\pi_{m}BO,m-1)$ and $h=id$ as the "universal" example; thus we have to know whether $i^*(k)=0$. For $n=4,m=2$ we need to know the pullback of $k=\frac{p_1}{2}$ under $i: K(\Zz_2,1)\to BSpin$. This is zero. For $n=8,m=4$ we need to know the pullback of $k=\frac{p_2}{6}$ under $i: K(\Zz,3)\to BString$. This is a non-zero class: it suffices to show that its reduction modulo 3 is nontrivial. This follows from {{cite|Giambalvo1969|Theorem 1'}}. Thus for example the trivial stable vector bundle on $K(\Zz,3)$ admits a string structure which does not lift to a $BO\langle 9\rangle$-structure. The situation is similar in higher dimensions $n$ divisible by 4. </wikitex> == References == {{#RefList:}} [[Category:Questions]] [[Category:Research questions]]E\to X$ be a (stable) vector bundle. This has a classifying map $X\to BO$.

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

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

2 Answers

This is not possible for $n=2$ and $n=4$, 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 answer is however yes for all larger $n$ with $BO\langle n\rangle\ne BO\langle n+1\rangle$. For example the existence of a $BO\langle 9\rangle$-structure on a string vector bundle can depend on the string structure.

3 Further discussion

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

$$\displaystyle \Omega K(\pi_{n}BO,n) \to PK(\pi_{n}BO,n) \to K(\pi_{n}BO,n)$$

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

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

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

This is for $n=1,2$ the first and second Stiefel-Whitney class of the corresponding vector bundle over $X$. In particular, since the Stiefel-Whitney classes of a vector bundle are independent of an orientation, this answers the question for $n=2$.

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

In general, we are given two maps $f,g:X\to BO\langle n \rangle$ (i.e. two stable vector bundles over $X$ with $BO\langle n \rangle$-structure) for which we assume that the compositions with $BO\langle n \rangle \to BO\langle m \rangle$ are homotopic (i.e. the bundles are isomorphic as bundles with $BO\langle m \rangle$-structure). We have to investigate whether it is possible that $f^*k\ne g^*k=0$. Here let us assume that $BO\langle n\rangle$ and $BO\langle m\rangle$ 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 $f$ and $g$ with $BO\langle n \rangle \to BO\langle m \rangle$ are homotopic, it follows that $f$ and $g$ differ by a map from $X$ to the homotopy fiber $K(\pi_{m}BO,m-1)$ of $BO\langle n \rangle \to BO\langle m \rangle$. More precisely, the map $BO\langle n \rangle \to BO\langle m \rangle$ is a map of H-spaces, and given $f,g$ as above, there exists a map $h:X\to K(\pi_{m}BO,m-1)$ such that $f$ 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 $H$-space multiplication.

Under the $H$-space multiplication $k$ 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 $f^*k=(g,ih)^*(k\otimes 1 + 1\otimes k)=g^*k+h^*i^*k$.

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

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

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

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

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

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

4 References

• [Giambalvo1969] V. Giambalvo, The $mod$ $p$ cohomology of $BO\langle 4k\rangle$, Proc. Amer. Math. Soc. 20 (1969), 593–597. MR0236913 (38 #5206) Zbl 0176.52601
• [Stong1963] R. E. Stong, Determination of $H^\ast ( BO(k,\cdots,\infty ),Z_2)$ and $H^\ast ( BU(k,\cdots,\infty ),Z_2)$, Trans. Amer. Math. Soc. 107 (1963), 526–544. MR0151963 (27 #1944) Zbl 0116.14702