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The study of differentiable manifolds and differentiable maps. One fundamental problem is that of classifying manifolds up to diffeomorphism. Differential topology is what Poincaré understood as topology or “analysis situs”.
3
votes
1
answer
419
views
Simply-connected 4-manifolds can be blown up and down to complex projective planes. How abou...
There is a nice theorem that for every simply connected, closed, smooth, oriented manifold $M$, we have for some $m \in \mathbb{N}$:
$$M \# \left(\mathop{\#}^m \left(\mathbb{C}\mathbb{P}^2 \# \overlin …
7
votes
0
answers
201
views
What are (the motivations for Dominic Joyce's definition of) smooth functions for manifolds ...
In this article, Joyce defines a new kind of smooth map of manifolds with corners.
The standard requirement would be that a smooth map is smooth in every chart. He calls such maps weakly smooth. In …
7
votes
2
answers
793
views
How can Wall's theorem be generalised to non-simply connected manifolds?
In a sense, this is a follow-up to this question.
By work of Freedman and Wall, it is known that if two simply-connected 4-manifolds $M$ and $N$ are homeomorphic, then there is $k \in \mathbb{N}$ suc …
10
votes
1
answer
274
views
Are there 4d state sum models, extended TQFTs or chain mail invariant that detect smooth str...
A state sum model is a smooth invariant defined on smooth triangulated, or PL manifolds, by summing a local partition function over labels attached to the elements of the triangulation.
Typical exampl …
7
votes
0
answers
444
views
Differential topology on arbitrary fields
What do the differential topology theories on arbitrary fields have in common?
Different differential topology theories
There is "ordinary" differential topology on real manifolds, with its rich re …
0
votes
Theoretical physics: Why not just $\mathbb{R}^4$?
If you accept that quantum gravity with matter should be a Topological Quantum Field Theory and that TQFTs probably can't distinguish simply connected homotopy equivalent 4-manifolds, you should come …
10
votes
1
answer
773
views
What are Kirby diagrams of candidate exotic 4-manifolds?
It is an open problem whether there exist smooth manifolds homeomorphic, but not diffeomorphic to the standard $S^4$. The same is true for the 4-torus and several other manifolds. Handle decomposition …
4
votes
0
answers
273
views
Cap product for (co)homology from handle decompositions/Kirby diagrams
Since handle decompositions and Morse functions are intimately related, I'm imagining that a given explicit handle decomposition allows for an explicit description of the cellular complex and thus of …
5
votes
0
answers
95
views
Pull back group cohomology onto handle decomposition
A construction encountered in the Dijkgraaf-Witten invariant uses the following ingredients:
An oriented, (assumed here to be smooth) manifold $M^n$
A finite group $G$ (and a field, chosen to be $\m …
3
votes
0
answers
100
views
What's a completely computational/syntactical model for handle decompositions of manifolds?
Simplicial sets, CW complexes
Simplicial sets can be described completely algebraically, by specifying a family of sets, and maps between them satisfying certain relations. This description can be di …
11
votes
Accepted
Construction of invariants of 4-manifolds with the Kirby calculus
Disclaimer: Shameless self-advertising.
Yes, it can be done, and it's really beautiful! You can define the Crane-Yetter invariant with Kirby calculus, and possibly other TQFTs ("dichromatic model …
4
votes
2
answers
700
views
Are there Kirby diagrams with 3-handles?
Let $M\colon \partial_- M \to \partial_+ M$ be an oriented, compact cobordism. Assume that there is a handle decomposition with at most one 0-handle, and denote the handle bodies by $M_i, i \in \{0,\d …
14
votes
1
answer
647
views
Is there a PL, or topological, bordism hypothesis?
The bordism hypothesis says that the $(\infty, n)$-category of smooth, framed $n$-bordisms, $(n-1)$-dimensional boundaries, and corners down to points, is freely generated symmetric monoidal with dual …
10
votes
2
answers
598
views
What are TQFTs that are multiplicative under connected sums? Do bordisms with connected sum ...
In general, one extracts a manifold invariant from a TQFT by interpreting the closed manifold as a bordism from the empty set to the empty set. The TQFT sends this bordism to a homomorphism of the gro …
31
votes
2
answers
2k
views
A manifold is a homotopy type and _what_ extra structure?
Motivation: Surfaces
Closed oriented 2-manifolds (surfaces) are "classified by their homotopy type". By this we mean that two closed oriented surfaces are diffeomorphic iff they're homotopy equivalen …