bio | website | math.cornell.edu/~hatcher |
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location | Cornell University | |
age | ||
visits | member for | 4 years, 10 months |
seen | 3 hours ago | |
stats | profile views | 1,962 |
Sep 21 |
comment |
Contractibility of space of embeddings of a disc
@IgorBelegradek: The formula seems to work equally well for embeddings $D^2\to {\mathbb R}^2$ and diffeomorphisms of ${\mathbb R}^2$. For embeddings the term $f(tx)$ is in effect restricting embeddings to smaller and smaller concentric disks, so one still gets embeddings, and then the $1/t$ factor re-expands to keep the derivative the identity at the origin. The limit exists, just as for diffeomorphisms, and is the identity. |
Sep 21 |
comment |
Contractibility of space of embeddings of a disc
For question 2 the space is not contractible. There is a map from the space of topological embeddings to ${\mathbb R}^2 - \{0\}$ given by $f\mapsto f(1,0)$ and this map splits (i.e., has a section) by considering only embeddings which are orientation-preserving Euclidean similarities. So the space is not simply-connected. An extra condition analogous to the condition on the derivative in question 1 is needed, though it is perhaps not obvious what the best way to state this condition is. |
Sep 21 |
comment |
Contractibility of space of embeddings of a disc
Please clarify what you mean by "proper". Usually a proper embedding is one which takes boundary to boundary, which in this case would force the embeddings to be diffeomorphisms or homeomorphisms, respectively, in the two cases. |
Sep 18 |
awarded | Enlightened |
Sep 9 |
comment |
Existence of $n$-connected topological groups with $m$-dimensional action extending that of $GL(m)$ on $\mathbb{R}^m$
A simple observation: $F$ must be non-orientable, since if it is orientable, then $G$, being path-connected, must act by orientation-preserving homeomorphisms, contrary to the fact that the action of $GL(m)$ on ${\mathbb R}^m$ includes orientation-reversing homeomorphisms. |
Sep 6 |
comment |
“Economic” CW-structure for Eilenberg-MacLane spaces?
JUst to clarify: The usual construction of a homology decomposition, as in my algebraic topology book for example, shows that for each simply-connected space $X$ with finitely generated homology groups there is a CW complex $Y$ and a weak homotopy equivalence $Y\to X$, where $Y$ has the minimum number of cells in each dimension consistent with the structure of $H_*(X;{\mathbb Z})$, namely, 1 cell for each infinite cyclic summand of the homology and 2 cells for each finite cyclic summand. This applies in particular for $K(\pi,n)$'s with $n>1$ and $\pi$ finitely generated. |
Sep 5 |
comment |
Inverse cohomological isomorphisms
@Lennart Meier: Yes, thanks. Corrected now. |
Sep 5 |
revised |
Inverse cohomological isomorphisms
edited body |
Sep 4 |
answered | Inverse cohomological isomorphisms |
Sep 4 |
comment |
Inverse cohomological isomorphisms
Correction: The universal covering map $S^3 \to SO(3)/I$ has degree 120, the order of the fundamental group of $SO(3)/I$. |
Aug 24 |
awarded | Good Answer |
Aug 21 |
answered | In a fibration, can a deformation retraction of the base be lifted to the total space? |
Aug 18 |
answered | A query about Hatcher flow on arc complex |
Aug 17 |
awarded | Good Answer |
Aug 14 |
awarded | Nice Answer |
May 17 |
revised |
Intuition behind Thom class
corrected a typo in a formula |
Feb 19 |
awarded | Enlightened |
Feb 19 |
comment |
Seifert fiberable manifolds with several Seifert fiberings
@Werner Thumann: That is correct, each prism manifold has exactly two Seifert fiberings. This is an interesting contrast to lens spaces, $S^3$, and $S^1\times S^2$, each of which has infinitely many different Seifert fiberings. |
Feb 18 |
awarded | Nice Answer |
Feb 16 |
revised |
Seifert fiberable manifolds with several Seifert fiberings
added 45 characters in body |