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 Yearling
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7h
comment Free basis and free group
I don't know. I would consider $\langle a,b,c: b=c, b=c^2\rangle$ to be free group despite the fact that there is a relation among the generators. (The point is that the group I've presented it isomorphic to $\mathbb{Z}$)
19h
awarded  Yearling
Jul
20
comment Tangent space manifold
I'd suggest that, instead of trying to cook up such an $f$, instead, find a matrix $A:\mathbb{R}^n\rightarrow \mathbb{R}^n$ and consider $A\circ \varphi$.
Jul
20
comment All riemannian isometries between open subsets of $\mathbb{R}^n$ are affine
This can't be true without additional assumptions. For example, consider $(0,1)\cup (2,3)\subseteq \mathbb{R}$. As a Riemannian manifold, there is an isometry which fixes $(0,1)$ and flips $(2,3)$, but there is no affine isometry of $\mathbb{R}$ which accomplishes this. Perhaps the theorem is true if your open set is connected?
Jul
20
comment Which Riemannian Manifolds are scale-free?
Not every flat non-compact manifold is scale free. The flat cylinder $S^1\times \mathbb{R}$ isn't scale free because the length of the shortest closed geodesic changes when you scale. Is there any scale free Riemannian manifold other than flat $\mathbb{R}^n$?
Jun
29
comment Principle of Transfinite Induction
Your last line should read "if the continuum hypothesis is true", rather than AC ;-);
Jun
26
comment A function between covering spaces.
It's not true without some kind of compabiliity between $f$, $p_1$, and $p_2$. For a family of counterexamples, let $X_1 = X_2 = X$ be any space which admits a surjective continuous non-covering map $f$ and let $p_1 = p_2 = Id_X$. For a concrete example, let $X = \mathbb{R}$ with $f(x) = x^3 - x$ This isn't a covering as $f^{-1}(0) = \{-1,0,1\}$, so $0$ three preimages, but $100$ only has one. If I had to guess, the assumption $p_1 = p_2\circ f$ should make it work.
Jun
12
answered Pushforward injective
Jun
4
comment If compact simply connected manifold has the same rational homotopy groups as $S^n$ or $\mathbb{C}P^n$, must it have the same cohomology ring?
together with a paper of mine in Diff. Geo. and Appl. about classification of 4 and 5-d biquotients which shows that the only rationally elliptic 4 manifolds are those with second betti number at most $2$. (I'm sure this was well known before my paper, but I don't remember a different reference off hand.)
Jun
4
comment If compact simply connected manifold has the same rational homotopy groups as $S^n$ or $\mathbb{C}P^n$, must it have the same cohomology ring?
On the other hand, I have no idea what happens when you connect sum more than 2 pieces. I do know that such manifolds $M$ are rationally hyperbolic, which implies that the function $f(n) = \dim \bigoplus_{k= 0}^n \pi_k(M)\otimes\mathbb{Q}$ grows exponentially. A reference for the fact that two summands fit into such a bundle is Totaro's paper "Cheeger manifolds and the classification of biquotients". A reference for the case of more than two summands is combining Felix-Halperin-Thomas's Rational homotopy theory book (the section on rational ellipticity)....
Jun
4
comment If compact simply connected manifold has the same rational homotopy groups as $S^n$ or $\mathbb{C}P^n$, must it have the same cohomology ring?
@MikeMiller: Both spaces $B_{\pm} = \mathbb{C}P^2\sharp \pm \mathbb{C}P^2$ are the base space of a torus bundle $T^2\rightarrow S^3\times S^3\rightarrow B_{\pm}$. It follows from the LES homotopy sequence that, other than the fact that $\pi_2(B_{\pm}) = \mathbb{Z}^2$, $\pi_k(B_{\pm}) \cong \pi_k(S^3\times S^3)$. More generally, $\mathbb{C}P^n\sharp \pm \mathbb{C}P^n$ is the base space of a $T^2$ bundle with total space $S^3\times S^{2n-1}$, so the homotopy groups don't depend on the choice of $\pm$.
Jun
1
comment Homogeneous metric on a homogeneous space $G/K$ - is this the same as a $G$ - invariant metric?
I would use the term "homogeneous metric" to mean "a metric for which the isometry group acts transitively". So, yes, $\tilde{g}$ is a homogeneous metric according to this usage.
May
30
comment Homogeneous metric on a homogeneous space $G/K$ - is this the same as a $G$ - invariant metric?
Short answer: a homogeneous Riemannian manifold is any Riemannian manifold whose isometry group acts transitively on the manifold. Every such manifold is isometric to one of the form $(G/K,\tilde{g})$.
May
30
awarded  Revival
May
20
comment $\bigvee _{n\in\mathbb{N}}\mathbb{S}^1$ and its homeomorphisms
@Dejan: Thanks ;-)
May
20
comment $\bigvee _{n\in\mathbb{N}}\mathbb{S}^1$ and its homeomorphisms
Are you sure this is true? What if $X$ is the union of $S_n$ where $S_n$ is the circle given by the equation $(x-n)^2 + y^2 = n^2$? (Think of starting with a small circle, and drawing larger and larger circles around it, all meeting at some predetermined point.)
May
20
comment Give an example of a sequence in $l^{\infty}$ which cannot be generated by the linear operator.
I'm not sure what you mean by the supremum of a series. Yes, the sequence which consists of alternating 1s and 0s should work fine.
May
20
comment Give an example of a sequence in $l^{\infty}$ which cannot be generated by the linear operator.
The (partial sums of the ) harmonic series are unbounded.
May
20
comment Give an example of a sequence in $l^{\infty}$ which cannot be generated by the linear operator.
Your logic is getting to the right idea: You want a bounded sequence without a limit.
May
19
comment Showing two spaces are homotopy equivalent
I would define the deformation retract in pieces. First, collapse $D_2$ down to $S^2\ cup I$, then collapse $I$.