New answers tagged manifolds
8
votes
Homotopy equivalent but non-homeomorphic high-dimensional manifolds
There is a nice algebro-geometric example in dimension 6 in Libgober-Wood, Topology 21, no. 4, pp. 469-482 (1982). They consider smooth complete intersections $X\subset \mathbb{P}^8$ of degrees (2,3, ...
- 37.3k
23
votes
Homotopy equivalent but non-homeomorphic high-dimensional manifolds
I will answer the first question on the existence of closed simply-connected manifolds that are homotopy equivalent but not homeomorphic.
No such example exists in dimension 5. Closed simply-connected ...
- 28.5k
3
votes
Accepted
Extending $\mathbb{R}$ to a higher dimensional manifold
Already answered (twice) but let me add the comment that one can easily construct an example by taking a product of the real line with a space which has the desired properties but is not a manifold.
2
votes
Extending $\mathbb{R}$ to a higher dimensional manifold
The existing answer already settles the question as stated, but let me add that the answer is negative even among finite dimensional spaces, the Menger curve is a one-dimensional and homogeneous Peano ...
- 2,904
2
votes
Extending $\mathbb{R}$ to a higher dimensional manifold
If by ‘manifold’ you mean locally Euclidean then the answer is no. The power $\mathbb{R}^\mathbb{N}$ of the real line, the Hilbert cube, every $\ell_p$-space, …, there are very many infinite-...
- 9,910
5
votes
Naturality of Lie bracket — alternate proof
The Lie bracket can somehow be computed from flows, and the flows intertwine. In more detail, if $X$ maps to $\bar{X}$, i.e. $F'(p)X(p)=\bar{X}(F(p))$ then clearly the flows of $X$ and $\bar{X}$ ...
- 25.6k
15
votes
Accepted
Naturality of Lie bracket — alternate proof
My favorite proof of this takes a step back.
If $M$ is a manifold, let $TM$ be its tangent bundle. The total space is itself a manifold, so you can form the tangent bundle of $TM$, i.e.
$$T(TM) = T^2 ...
- 43.1k
3
votes
Naturality of Lie bracket — alternate proof
Here I present a proof of the naturality of Lie derivative, only using the abstract definition, and without referencing the naturality of Lie bracket.
Let $\phi(t,p)=\phi^t(p)$ be the flow generated ...
- 583
Top 50 recent answers are included