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We know that

  • framing structure means the trivialization of tangent bundle of manifold $M$.

  • string structure means the trivialization of Stiefel-Whitney class $w_1$, $w_2$ and half of the first Pontryagin class $p_1/2$.

question

  1. What are the relations between framing structure and string structure?

  2. When we have frame bordism $\Omega_3^{fr}=Z/{24}$ and string bordism $\Omega_3^{string}=Z/{24}$, what do we really impose on the relations between framing structure and string structure? Are there one-to-one correspondence between the frame manifold and string manifold through bordism relations? Do they all have the same generator of $Z/{24}$ by the 3-sphere $S^3$ with Lie group framing?

  3. In dimension 3, if trivial tangent bundle is $w_1 =0$. But we are considering the cobordism, so we need both dimensions 3d and bounded in 4d, but does it grantee that,

trivial tangent bundle $\Leftrightarrow$ $w_1 =0, w_2 =0$, and $p_1/2=$0?

Can we show that is true for 3d-4d cobordant relation? How about other dimensions?

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    $\begingroup$ There are various intermediate steps between an arbitrary vector bundle and a trivial vector bundle. Trivial vector bundles admit framings. The process of crawling up the Whitehead tower of the orthogonal group is what you are interested in here. The first step of the Whitehead tower is to find a map to $O_n$ that is the fiber of a fibration such that the fiber is connected. That's $SO_n \to O_n$. In the language of Whitehead towers we have killed $\pi_0$. The next step would be to kill $\pi_1$. That's what spin structures are about. The next step is to kill $\pi_2$. . . $\endgroup$
    – Ryan Budney
    Commented Dec 14, 2023 at 1:08
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    $\begingroup$ I think if you read Milnor's paper on spin structures you might get the gist of this general process. Perhaps also look at Whitehead towers in Hatcher's book. $\endgroup$
    – Ryan Budney
    Commented Dec 14, 2023 at 1:09
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    $\begingroup$ I failed to emphasize it, but the point of Whitehead towers is that you kill low-dimensional homotopy groups of a space without affecting the higher homotopy groups. As far as I know these spaces don't all have names, but the common one for the orthogonal group is $\cdots \to spin_n \to SO_n \to O_n$. These maps kill $\pi_1$ and $\pi_0$ respectively. $\endgroup$
    – Ryan Budney
    Commented Dec 14, 2023 at 1:25
  • $\begingroup$ >> "There are various intermediate steps between an arbitrary vector bundle and a trivial vector bundle. Trivial vector bundles admit framings." Reply: yes thanks, I already know what you wrote. But could we say that Trivial vector bundles admit Framings which also include the String structure. But not the other way around, the String structure does Not imply Framing? $\endgroup$
    – zeta
    Commented Dec 14, 2023 at 12:32
  • $\begingroup$ could we say that Trivial vector bundles admit Framings which also include the String structure. But not the other way around, the String structure does Not imply Framing? However, in 3d, is that Framing if and only if String structure? They imply each other? How about when the dimensions d < 7 when all the Frame bordism group = String bordism group? Does that mean d < 7, Framing if and only if String structure? Could you clarify this and maybe you can write an answer? $\endgroup$
    – zeta
    Commented Dec 14, 2023 at 12:35

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Answer:

  1. Any framed manifold is string manifold.

  2. Frame and String Bordism groups are coincide up to n=6.

To see these results, we use the filtration on BO (as Ryan suggested). First, we remark that a manifold $M$ come with a map $M \to BO$, which encode the transition maps between different charts. We can ask "can we use only determinant 1 transition maps"?, which is like asking to factor/lift $M \to BO$ as $M \to BSO \to BO$. We can repeat this procedure (try to lift) along the tower $$* \to... \to BString \to BSpin \to BSO \to BO$$ ,when any space in the tower is obtain by "killing" the lower Homotopy group of the previous space. Framed manifold is a null homotopy of the map $M \to BO$ which is like factor it throw a $*=BFrame$. So having Frame structure implies any other structure.

Let $M$ be a 3 dimensional string manifold, then we have factorization: $$M \to BString \to BO$$ ,but $Bstring$ is 7 connected so the map $M \to BString$ is actually null homotopic (as map from 3 dimensional manifold) and M have Frame structure. The same hold for any dimension $<7$ (we used the fact that map from $d$ dimension manifold to $n$ connected space, where $d<n$, is null homotopic)

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  • $\begingroup$ Thanks! voted. :) Do we know any string manifold that is not framed manifold in n=3, or other dimensions? $\endgroup$
    – zeta
    Commented Dec 14, 2023 at 15:20
  • $\begingroup$ @zeta: in dimension three having a trivial tangent bundle is equivalent to being orientable. To get a gap between trivializable and any of these higher structures (spin, string, etc) you have to go to higher dimensions. $\endgroup$
    – Ryan Budney
    Commented Dec 14, 2023 at 16:18
  • $\begingroup$ in dimension 3, if trivial tangent bundle is w1 =0. But we are considering the cobordism, so we need both dimensions 3d and bounded in 4d, but how comes it grantees that trivial tangent bundle <=> w1 =0, w2 =0, and perhaps p_1/2=0? I think that is true for 3d-4d cobordant relation? That is the main point of my question, if you read carefully. $\endgroup$
    – zeta
    Commented Dec 14, 2023 at 17:44
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    $\begingroup$ @zeta: If that's the point of your question you should ask more clearly worded questions, as it's not clear to me. You are asking quite a few rather similar questions, so it isn't clear to me you are getting at what you are really interested in. $\endgroup$
    – Ryan Budney
    Commented Dec 14, 2023 at 19:18
  • $\begingroup$ Yes, I edited: trivial tangent bundle $\Leftrightarrow$ $w_1 =0, w_2 =0$, and $p_1/2=$0? in which dimensions? $\endgroup$
    – zeta
    Commented Dec 15, 2023 at 5:31

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