# Tag Info

Accepted

### Not all manifolds can be triangulated: In which dimensions?

In dimensions up to three, every manifold is triangulable (this is classical). In dimension 4, there are simply connected non-triangulable manifolds (such as the E8 manifold); in fact, a closed 4-...
• 9,243
Accepted

### Do rings of smooth functions differ from rings of continuous functions?

No. In both the smooth function ring and the continuous function ring a maximal ideal $\frak m$ consists of the functions vanishing at some point. In the smooth case $\frak m/\frak m^2$ is the ...
• 54.1k
Accepted

### Questions on J. F. Nash's answer about his errors in the proof of embedding theorem

Igor already answered questions 1 and 2. What Nash wrote is an attempt to describe to a non-expert audience the solution scheme he had for non-compact manifolds. In the noncompact case he proceeds ...
• 36.5k

• 95.5k

### Closed manifold with non-vanishing homotopy groups and vanishing homology groups

As suggested by Lennart Meier, the connected sum $M=P\#P$ of two copies of the Poincaré homology sphere gives an example. The homotopy groups $\pi_n(M)$ are nonzero for all $n>1$ because the ...
• 19.4k
Accepted

### Manifolds as Cauchy completed objects

(Expanding on Phil Tosteson’s comment.) No: the Cauchy-completion characterisation doesn’t hold for the PL, topological, or complex-analytic cases. The key technical point is that split idempotents ...
Accepted

### A question on connected sum of compact manifolds

Let us suppose that $dim(M)\geq 3$ then we have that: $\pi_1(M \# M)\cong \pi_1(M)*\pi_1(M)$, $H_*(M;\mathbb{Z})\cong H_*(M;\mathbb{Z})\oplus H_*(M;\mathbb{Z})$ when $*< dim(M)$. As $M$ is ...
• 9,792
Accepted

### Is there a closed non-smoothable 4-manifold with zero Euler characteristic?

The Kirby-Siebenmann invariant in $H^4(M;\Bbb Z/2)$, an obstruction to smoothability, is additive under connected sum in dimension 4. In even dimensions, $\chi(M \# N) = \chi(M) + \chi(N) -2$. To ...
• 9,243
Accepted

### Transitive embedding of the projective plane $\Bbb R P^2$ into the $4$-sphere

Yes. You take each vector $v \in \mathbb{R}^3$ to the vector $v \cdot v \in \operatorname{Sym}^2\mathbb{R}^3=\mathbb{R}^6$. This takes each unit vector $v$ to the same place as $-v$. So it descends to ...
• 25.3k
Accepted

### Is there a closed manifold whose universal cover is $\mathbb{R}^n\setminus\{x_1, \dots, x_k\}$ for some $k > 1$?

If we demand that the universal cover is homeomorphic / diffeomorphic to $\mathbb{R}^n \setminus \{x_1,\ldots,x_k\}$ with $k>1$ the answer is no, there are no such closed manifolds. Each missing ...
• 1,356
Accepted