1
$\begingroup$

So I want to understand does the following addition give stricter condition for equivalence of two stably-complex manifolds?


Assume the following relation: let two stably-complex $n$-manifolds $M^n,N^n$ be complex' cobordant if

  1. They are a boundary of stably-complex $n+1$-manifold $W^{n+1}$ such that stably-complex structure is induced from $W$;
  2. Any automorphism of $M$ that preserves stably-complex structure on M could be extended to $W$ (same condition for $N$). In more detail it means that for any such automorphism $f$ pullback of a bundle equivalent to tangent bundle $f^*(TM\oplus \epsilon^k)\cong TM\oplus \epsilon^k$ is isomorphic to itself as complex bundles. (@MichaelAlbanese thank you for clarification)

My question: is this relation something new as compared to standard complex cobordism definition (i.e. condition 1)?

$\endgroup$
2
  • 2
    $\begingroup$ In $2$, what do you mean by an automorphism of $M$ that preserves the stably-complex structure on $M$? Do you mean a diffeomorphism $f : M \to M$ such that $f^*(TM\oplus\varepsilon^k) \cong TM\oplus\varepsilon^k$ as complex bundles? $\endgroup$ Dec 31, 2019 at 22:05
  • $\begingroup$ @MichaelAlbanese Yes, I think so. $\endgroup$ Dec 31, 2019 at 23:04

1 Answer 1

4
$\begingroup$

This is not the same as complex cobordism. For example, $S^6$ with the stably-complex structure given by $TS^6\oplus\varepsilon^2_{\mathbb{R}} \cong \varepsilon^4_{\mathbb{C}}$ is complex nullcobordant as it is the boundary of the closed ball $B^7$ with the stably-complex structure $TB^7\oplus\varepsilon^1_{\mathbb{R}} \cong \varepsilon^4_{\mathbb{C}}$. For any diffeomorphism $f : S^6\to S^6$, $f^*\varepsilon^4_{\mathbb{C}} \cong \varepsilon^4_{\mathbb{C}}$ and hence is an automorphism in the sense you suggested. However, not every diffeomorphism of $S^6$ extends to a diffeomorphism $B^7 \to B^7$. This is related to the existence of exotic spheres in dimension $7$.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .