Does the existence of a string structure depend on a spin structure ? (and a generalization)
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(For $n=2,4,8$ this is an orientation, a spin structure, a string structure respectively.) | (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 | + | 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, | $BO\langle m \rangle \to BO$, 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. | this specific $BO\langle m\rangle$-structure can be lifted to a $BO\langle n\rangle$-structure. | ||
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{{endthm}} | {{endthm}} | ||
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== Answers == | == Answers == | ||
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− | This is not possible for $n=2$ and $n=4$, i.e. the question whether an oriented vector bundle admits a spin structure | + | 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 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 $\frac{p_1}{2}$ does not depend upon the choice of spin structure see \cite{Čadek&Crabb&Vanvzura2008|Defintion p.170}. |
− | 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 | + | |
− | + | ||
− | + | ||
+ | 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. | ||
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$$\Omega K(\pi_{n}BO,n) \to PK(\pi_{n}BO,n) \to K(\pi_{n}BO,n)$$ 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 $H^{n}(BO\langle n\rangle;\pi_{n}BO)\cong Hom(\pi_{n}BO;\pi_{n}BO)$. | $$\Omega K(\pi_{n}BO,n) \to PK(\pi_{n}BO,n) \to K(\pi_{n}BO,n)$$ 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 $H^{n}(BO\langle n\rangle;\pi_{n}BO)\cong Hom(\pi_{n}BO;\pi_{n}BO)$. | ||
+ | In the case $n=4i$, it follows from [[Obstruction classes and Pontrjagin classes (Ex)|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 [[Obstruction_classes_and_Pontryagin_classes|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. | ||
+ | |||
+ | In the cases $n=1,2$ the class $k$ equals $w_1\in H^1(BO;\Zz_2)$ respectively $w_2\in H^2(BSO;\Zz_2)$. | ||
+ | |||
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 | 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. | $$f^*k\in H^{n}(X;\pi_{n}BO)$$ vanishes. | ||
− | + | ||
In particular, since the Stiefel-Whitney classes of a vector bundle are independent of an orientation, this answers the question for $n=2$. | 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) | 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 | 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 | + | (i.e. the bundles are isomorphic as bundles with $BO\langle m \rangle$-structure). |
We have to investigate whether it is possible that $f^*k\ne g^*k=0$. | 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 | Here let us assume that $BO\langle n\rangle$ and $BO\langle m\rangle$ are consecutive connective covers in the sense that | ||
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For $n=4,m=2$ we need to know the pullback of $k=\frac{p_1}{2}$ under $i: K(\Zz_2,1)\to BSpin$. | 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. | + | 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 {{cite|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 {{cite|Stong1963}} and for $p$ odd by {{cite|Giambalvo1969}}. | ||
For $n=8,m=4$ we need to know the pullback of $k=\frac{p_2}{6}$ under $i: K(\Zz,3)\to BString$. | 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 is a non-zero class: it suffices to show that its reduction modulo $3$ is nontrivial. |
− | This follows from {{cite|Giambalvo1969|Theorem 1'}}. | + | This follows from {{cite|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 {{cite|Stong1963|p.539}}. | ||
</wikitex> | </wikitex> | ||
== References == | == References == | ||
{{#RefList:}} | {{#RefList:}} | ||
− | + | [[Category:Questions]] | |
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[[Category:Research questions]] | [[Category:Research questions]] |
Latest revision as of 10:51, 4 September 2012
Contents |
[edit] 1 Question
Let be a (stable) vector bundle. This has a classifying map .
A -structure on is (the vertical homotopy class of) a lift of the classifying map to a map . (For this is an orientation, a spin structure, a string structure respectively.)
Since for , the map factors through , a -structure induces a -structure, or, vice versa, this specific -structure can be lifted to a -structure.
Question 1.1. Given a vector bundle and two -structures on it, is it possible that one of the -structures can be lifted to a -structure and the other can't?
[edit] 2 Answers
This is not possible for and , 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 does not depend upon the choice of spin structure see [Čadek&Crabb&Vanvzura2008, Defintion p.170].
The answer is however yes for all larger with . For example the existence of a -structure on a string vector bundle can depend on the string structure.
[edit] 3 Further discussion
The map has homotopy fiber and is the pullback of the path-loop fibration
In the case , it follows from this result that the Pontryagin class maps to a certain multiple (see here for the value of ). In these cases, the generator is usally denoted by , although itself is indivisible.
In the cases the class equals respectively .
By obstruction theory, it follows that a map lifts to if and only if the "characteristic" class
In particular, since the Stiefel-Whitney classes of a vector bundle are independent of an orientation, this answers the question for .
For every -structure on a vector bundle there exists an opposite -structure (inducing the opposite orientation) defined as follows: The connective cover construction is functorial, thus the non-trivial deck transformation of the -fold cover induces a self-map of . Composing the -structure with this self-map of gives the opposite structure. We have since is a self-equivalence which for is the identity on the Pontryagin classes. Thus for a -structure and its opposite, either both lift to a -structure or both do not.
In general, we are given two maps (i.e. two stable vector bundles over with -structure) for which we assume that the compositions with are homotopic (i.e. the bundles are isomorphic as bundles with -structure). We have to investigate whether it is possible that . Here let us assume that and are consecutive connective covers in the sense that
Since the compositions of and with are homotopic, it follows that and differ by a map from to the homotopy fiber of . More precisely, the map is a map of H-spaces, and given as above, there exists a map such that is homotopic to the composition
where the last map is the -space multiplication.
Under the -space multiplication pulls back to
Now it follows that .
Now we choose and as the "universal" example; thus we have to know whether .
For we need to know the pullback of under . This is zero, since the reduction modulo is trivial, as the reduction of modulo is the image of . See [Stong1963, p.539].
For higher we show that by considering reductions modulo , which have been computed for by [Stong1963] and for odd by [Giambalvo1969].
For we need to know the pullback of under . This is a non-zero class: it suffices to show that its reduction modulo is nontrivial. This follows from [Giambalvo1969, Theorem 1']. (The reduction modulo of is , so that the reduction modulo of is zero.)
Thus for example the trivial stable vector bundle on admits a string structure which does not lift to a -structure.
In all higher dimensions , the (reductions modulo of) are non-zero by [Stong1963, p.539].
[edit] 4 References
- [Giambalvo1969] V. Giambalvo, The cohomology of , Proc. Amer. Math. Soc. 20 (1969), 593–597. MR0236913 (38 #5206) Zbl 0176.52601
- [Stong1963] R. E. Stong, Determination of and , Trans. Amer. Math. Soc. 107 (1963), 526–544. MR0151963 (27 #1944) Zbl 0116.14702
- [Čadek&Crabb&Vanvzura2008] M. Čadek, M. Crabb and J. Vanžura, Obstruction theory on 8-manifolds, Manuscripta Math. 127 (2008), no.2, 167–186. MR2442894 (2009f:55015) Zbl 1157.55011