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The generalized homology theory of the Thom spectrum $MO=\varinjlim\Sigma^nMTO_n$ is bordism theory:\begin{equation*}\pi_k(MO\wedge X)=\Omega^O_k(X)\end{equation*}These groups form the ring of (unoriented) $X$-cobordism classes of (unoriented) manifolds.

But what information do the Madsen-Tillmann spectra $MTO_n$ contain? Does the homotopy ring\begin{equation*}\pi_k(MTO_n\wedge X)=\text{ ?}\end{equation*}yield any useful classification of manifolds?

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  • $\begingroup$ I am not sure about the homology but the cohomology classifies fiber bundles with fiber manifolds having $O_n$-structure on the tangent bundle. Presumably the same applies to the homology. $\endgroup$
    – user43326
    Aug 23, 2014 at 8:47
  • $\begingroup$ Can you clarify what you mean by "fiber bundles with fiber manifolds having $O(n)$-structure on the tangent bundle"? I'm not sure what the "fiber manifold" is. Do you mean the base space? What are the fibers? $\endgroup$ Aug 25, 2014 at 4:47
  • $\begingroup$ $\Omega_k^{SO}(X)$ should classify oriented $k$-manifolds; that is, $k$-manifolds with $SO(k)$-structures on their tangent bundles. Or does $\pi_k(MSO_k\wedge X)$ give this classification? $\endgroup$ Aug 25, 2014 at 4:48
  • $\begingroup$ That should read $\pi_k(MTSO_k\wedge X)$. $\endgroup$ Aug 25, 2014 at 4:57
  • $\begingroup$ I meant to say "whose fibers are manifolds"... $\endgroup$
    – user43326
    Aug 25, 2014 at 7:53

1 Answer 1

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This is an exercise in understanding the Pontrjagin--Thom correspondence. The group $\pi_k(MTO(n) \wedge X_+)$ is in bijection with tuples of

  1. a $(n+k)$-manifold $M$,
  2. an $n$-dimensional vector bundle $V \to M$,
  3. a stable isomorphism $\varphi: V \oplus \epsilon^k \oplus \epsilon^\ell \cong TM \oplus \epsilon^\ell$ for $\ell \gg 0$, and
  4. a continuous map $f : M \to X$.

This data is taken up to cobordism in the obvious way. Note that the spectrum $MTO(n)$ is not connective, which corresponds to the fact that the above makes sense for negative $k$.

On the other hand, for a $d$-dimensional manifold $X$ the cohomology theory $[X,MTO(n)]$ is represented by tuples of

  1. a $(d+n)$-dimensional manifold $E$ with a proper map $\pi : E \to X$,
  2. an $n$-dimensional vector bundle $V \to E$,
  3. a stable isomorphism $\varphi : TE \oplus \epsilon^\ell \cong V \oplus \pi^*(TX) \oplus \epsilon^\ell$,

again taken up to cobordism in the obvious way.

The point that user43326 is referring to is that if $\pi : E^{d+n} \to X^d$ is a smooth fibre bundle with compact $n$-dimensional fibres, then we may define $V := T_\pi E = \mathrm{Ker}(D\pi : TE \to TX)$ to be the vertical tangent bundle and choose a splitting of the short exact sequence $$0 \to T_\pi E \to TE \to \pi^*(TX) \to 0$$ of vector bundles on $E$. This gives an isomorphism $\varphi : TE \cong V \oplus \pi^*(TX)$, and so the tuple $(\pi: E \to X, V, \varphi)$ represents a class in $[X, MTO(n)]$. (However, despite what user43326 said, it is not true that all classes in this cohomology theory arise in this way.)

The reason that $$\mathrm{hocolim}_{n \to \infty} \Sigma^n MTO(n) \simeq MO$$ can be easily seen from the first description above. Concretely, a class in $\pi_k$ of the homotopy colimit is represented by a tuple

  1. a $k$-manifold $M$,
  2. an $n$-dimensional vector bundle $V \to M$ for some $n \gg0$,
  3. a stable isomorphism $\varphi : V \oplus \epsilon^\ell \cong TM \oplus \epsilon^{n-k} \oplus \epsilon^\ell$, for $\ell \gg0$,

taken up to cobordism. By taking $n$ large enough, and destabilising the isomorphism, we get $\varphi: V \cong TM \oplus \epsilon^{n-k}$, and so the last two pieces of data cancel out: we are left with just $k$-manifolds up to cobordism.

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  • $\begingroup$ Thank you for the detailed response. This is very helpful. $\endgroup$ Aug 26, 2014 at 23:36
  • $\begingroup$ Have the groups $\pi_k(MTO(n))$ and $\pi_k(MTSO(n))$ been computed? $\endgroup$ Aug 26, 2014 at 23:37
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    $\begingroup$ Very few homotopy groups of $MTO(n)$ or $MTSO(n)$ have been computed. By the description above, the map $s: MTO(n) \to \Sigma^{-n} MO$ induces an isomorphism on homotopy groups is negative degrees, so in this range the groups are known. The homotopy groups of $MTO(n)$ in small positive degrees can be analysed quite effectively by i) computing the cohomology of the homotopy fibre of $s$, ii) running the Adams spectral sequence on it. (All of this can be done for the oriented version too.) For $n=1$ or $2$ I have some charts for the Adams $E^2$ pages of $MTO(n)$ and $MTSO(n)$ on my webpage. $\endgroup$ Aug 27, 2014 at 8:17

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