[Math] What are examples when the equality of some invariants is good enough in algebraic topology

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As far as my understanding goes, most of the tools of algebraic topology (homotopy groups, homology groups, cup product, cohomology operations, Hopf invariant, signature, characteristic classes, knot invariants…) are mostly designed with "discrimination" in mind: if two spaces/maps/bundles/… do not have the same invariant, then they are not "equivalent" for the correct notion of equivalence. And of course two things having the same invariants in some arbitrary big class is not enough for the things to be equivalent (e.g. there are spaces that have the same homotopy and homology groups that are not homotopy equivalent).

But sometimes computing only a few of those invariants is enough:

  • If a knot has the same crossing number as the unknot (zero), then it's the unknot. (Maybe a silly example.)
  • If a space has the same homotopy groups as an Eilenberg–MacLane space (i.e. they all vanish except one), then it has the same homotopy type as an Eilenberg–MacLane space.
  • If a closed (connected) 3-manifold has the same fundamental group as the 3-sphere, then it's actually homeomorphic to the 3-sphere (the famous Poincaré conjecture / Perelman theorem).
  • The result that made me think of this question: under some technical conditions, if a simplicial operad has the same rational homology as the little $n$-disks operad (as Hopf $\Lambda$-operads), then it is rationally equivalent to the little $n$-disks operad (Fresse–Willwacher).

The above results are the ones I could think of (and are probably very skewed towards my interests).

What are some other examples of situations where an algebro-topological invariant, that in general is not enough to characterize an object, is enough to actually completely characterize the object?

Best Answer

Classification of simply-connected 4-5 and 6-manifolds.

  • The homotopy type of a simply connected compact 4-manifold only depends on the intersection form on the second homology group. Moreover a famous theorem of M. Freedman implies that the homeomorphism type of the manifold only depends on this intersection form, and on a $\mathbb{Z}/2\mathbb{Z}$-invariant, the Kirby–Siebenmann invariant
  • Barden in 1965 has given the following classification result for simply-connected compact smooth manifolds. Let $M$ and $N$ be simply-connected, closed, smooth 5-manifolds and let $\phi:H_2(M)\cong H_2(N)$ be an isomorphism preserving the linking form and the second Stiefel-Whitney class. Then $\phi$ is realised by a diffeomorphism.
  • A system of invariants is also known for simply-connected 6-manifolds, it is described in the paper "CUBIC FORMS AND COMPLEX 3-FOLDS" by Okonek and Van de Ven (L'enseignement mathématique, 1995).
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