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A manifold is a topological space that locally resembles Euclidean space near each point. More precisely, each point of an n-dimensional manifold has a neighbourhood that is homeomorphic to the Euclidean space of dimension n.

7 votes

$\mathbb{CP}(2)$ from gluing boundary of 4-ball

There was a gap in my first answer to the first question, discussed in the comments. I hope it's clear that this doesn't constitute a full answer to OP's question, which remains interesting. First que …
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7 votes
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A question on the manifold $ \{n\otimes n-m\otimes m:n,m\in S^2,(n,m)=0\} $

I'm not sure exactly what you were claiming, but here is a correct claim. $\pi_1(N)$ has six order four elements $\{\pm i, \pm j, \pm k\}$, and there are three natural maps $p_1, p_2, p_3: N \to \Bbb{ …
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17 votes
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Are different categories of manifolds non-equivalent (as abstract categories)?

Your method of looking for automorphisms of objects works for $C^k$ manifolds as well, $0 < k \le \infty$. … This follows from a result of Filipkiewicz (which I found in Kathryn Mann's excellent survey): If $M, N$ are smooth manifolds without boundary and $\varphi: \text{Diff}^p(M) \to \text{Diff}^q(N)$ is an …
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12 votes
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Existence of normal microbundles

Not all locally flat submanifolds have a normal microbundle, but they do stably. Rourke-Sanderson prove that there is a PL embedding $S^{19} \times I \to S^{29}$ with no topological normal microbundle …
26 votes
Accepted

Critical dimensions D for "smooth manifolds iff triangulable manifolds"

In dimensions $\leq 2$, homology manifolds are in fact manifolds (Theorem 16.32 in Bredon's sheaf theory) and topological manifolds admit unique PL structures, so we see that there is a PL homeomorphism … Then $X_{5 + k} = X_5 \times T^k$ is also triangulable, being a product of triangulable manifolds. …
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18 votes
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Is there a closed non-smoothable 4-manifold with zero Euler characteristic?

To construct manifolds with nontrivial Kirby-Siebenmann invariant we should apply Freedman's theorem: simply connected topological 4-manifolds are determined by their intersection form and Kirby-Siebenmann …
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16 votes
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Obstruction of spin-c structure and the generalized Wu manifods

Define the Wu manifold $W(n) = SU(n)/SO(n)$, the inclusion $SO \to SU$ given by thinking of $\Bbb C^n = \Bbb R^n \otimes \Bbb C$ (that is, including real matrices into complex matrices). Note that $W( …
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15 votes
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Non-triangulable 4-manifold as a boundary of some 5 manifold

Hsu proves that these two invariants precisely classify 4-dimensional oriented topological manifolds up to bordism (and in particular $\text{ks}$ is an invariant of topological bordism). … There is a formula for spin 4-manifolds: $\text{ks}(X) = \sigma(X)/8 \pmod{2}$. In particular, $\text{ks}(X_{E8}) = \sigma(E8)/8 = 1$. Thus the $E8$-manifold is not null-bordant. …
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3 votes

Do $\mathbb{HP}^2\#\overline{\mathbb{HP}^2}$ and $\mathbb{OP}^2\#\overline{\mathbb{OP}^2}$ a...

Here is a way of viewing the construction in Mikhail Katz' answer. Define $\Bbb{KP}^n \# \overline{\Bbb{KP}}^n$ in the following way: you can identify $\Bbb{KP}^n$ with the disc of elements of norm a …
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12 votes
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Is the connected sum of a triangulable manifold with a non-triangulable manifold a non-trian...

The middle term is the homology cobordism group of 3-manifolds and the last map is the Rokhlin invariant of 3-manifolds. These invariants are additive under connected sum. … In dimension 4 triangulable manifolds are automatically PL and therefore smooth. …
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71 votes
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Not all manifolds can be triangulated: In which dimensions?

This is probably true more generally than closed manifolds, but I'm feeling a little paranoid.) … Their theorem 2.1 (plus Manolescu's result) implies that there are non-triangulable manifolds in every dimension $n \geq 5$. However, all orientable 5-dimensional manifolds are triangulable. …
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