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1d
comment Finding the fallacy in this wrong limit computing result
Now, look at the definition of $\lim_{k\rightarrow\infty} x_k = L$. It, by definition, means that for all $\epsilon > 0$, there is an $N >0$ with the property that if $k > N$, then $|x_k - L| < \epsilon$. Now, given the constant sequence $x_k = -1$, and given any $\epsilon > 0$, take $L = -1$ and $N = 1$. Then $|x_k - L| = |-1 -(-1)| = 0 < \epsilon$. I agree that infinity is tricky business in general, but when applied to constant sequences it's not.
1d
comment Finding the fallacy in this wrong limit computing result
Note that $\rho(x)$ is a slight modification of the classical Dirichlet fuction, and is well known to be discontinuous everywhere. In particular, $\lim_{k\rightarrow \infty} \rho(x_k) \neq \rho(\lim_{k\rightarrow\infty} x_k)$ in general. (What your comment effectively demonstrates is that $\rho$ is discontinuous at $x = e$).
1d
comment Finding the fallacy in this wrong limit computing result
I disagree with your first equality (and now see that I missed the fact that it's $\frac{1}{2k+1}$, not $2k+1$). The point is the any (real) odd root of $(-1)$ is $-1$, so the sequence you're looking at is literally $-1,-1, -1, ...$. This has limit $-1$.
1d
comment Finding the fallacy in this wrong limit computing result
Maybe I'm misunderstanding. Are you claiming that $\lim_{k\rightarrow \infty} (-1)^{2k+1} = -1$ or not? (If we are thinking of the sequential notion of a limit, that is, $k\in \mathbb{N}$, then the limit is $1$. If it's the analysis notion of a limit ($k\in \mathbb{R}$), then I agree there is no limit.)
1d
comment Finding the fallacy in this wrong limit computing result
I don't understand your first sentence. In computing the value of a limit, if the sequence is literally constant, then the limit is the same constant. It doesn't matter if "odd/even is not defined for 2k+1" when $k = \infty$.
2d
comment Relationship between cohomology and higher-homotopy
Is $M$ supposed to be a 3-manifold? Otherwise, I don't understand your parenthetical ($H^3(M) \cong \mathbb{Z}$).
Apr
18
comment Lifting cohomology-killing maps through the 3-sphere
@guy-in-seoul: I agree. If this question ever gets bumped again, someone should change it. (I don't think this minor of an edit is worth the bump.)
Apr
16
awarded  algebraic-topology
Apr
15
comment Maps from $D^n$ to $D^n$ with a single inverse set are open.
Is it easy to write down an example of such an $f$? I'm having trouble getting intuition as to why this should be true
Apr
15
comment The biggest degree of a map between fixed surfaces
@JesusRS: I suspect so, but don't immediately see how to make Mike's answer in the MSE link work in this case.
Apr
15
answered The biggest degree of a map between fixed surfaces
Apr
15
comment The biggest degree of a map between fixed surfaces
I think I agree.. writing it up now...
Apr
15
comment The biggest degree of a map between fixed surfaces
the cup product $f^\ast(v)\cup f^\ast(w) = 0$ for any $v,w\in H^1(S;\mathbb{Q})$. Since $H^2$ is generated by $H^1$ as an algebra for both $T$ and $S$, it follows that the map on the $H^2$ level is he $0$ map. Dualizing again, it's the $0$ map on the $H_2$ level, so is degree $0$.
Apr
15
comment The biggest degree of a map between fixed surfaces
However, it follows from math.stackexchange.com/questions/1230760/… that, actually, all maps $T\rightarrow S$ have degree $0$. According to that link, any such map $T\rightarrow S$ is NOT injective on the $H_1$ level, which persists even if we use $\mathbb{Q}$ coefficients. By dualizing, this implies the map on the $H^1$ level is not surjective, and since $H^1(T;\mathbb{Q}) = \mathbb{Q}^2$, this implies that the image of $H^1(S;\mathbb{Q})$ in $H^1(T;\mathbb{Q})$ is a line. In particular,
Apr
12
comment Is there a map from the torus to the genus 2 surface which is injective on homology?
Very nice proof!
Apr
12
comment Ricci curvature of the Grassmannian?
The Grassmanian is diffeomorphic to the homogeneous space $U(n)/U(n-k) \times U(k)$. I believe the isotropy representation is irredicuble, and if true, this implies the Grassmanian has a unique homogeneous metric, up to rescaling. So, the question then becomes: is the induced metric on $G(k,\mathbb{C}^n)$ embedded into $\mathbb{P}^N$ homogeneous? (I don't know enough about the Plucker embedding to answer the question).
Apr
9
comment Is a compact, simply-connected 3-manifold necessarily $S^3$ with $B^3$'s removed?
Thanks everyone! That was very informative. (Someday, I really need to learn some $3$-manifold topology).
Apr
8
comment Is a compact, simply-connected 3-manifold necessarily $S^3$ with $B^3$'s removed?
Do you have a reference for the "half lives, half dies" fact? Like Olivier, modulo that detail, I'm sold. For the homeo vs diffeo issue, see mathoverflow.net/questions/84532/…
Apr
8
comment Is a compact, simply-connected 3-manifold necessarily $S^3$ with $B^3$'s removed?
In general dimensions, gluing balls into spheres doesn't give you diffeomorphisms, only homeomorphisms (Alexander trick). But in dim 3, homeo and diffeo classifications are the same. What is this "half lives, half dies" thing? (I am not a 3-manifold topologist)
Apr
1
comment Advice on finding counterexamples
In addition to the helpful answers/comments, I just wanted to mention that your subgroup definition is incorrect. For example, for $\mathbb{N}\subseteq \mathbb{Z}$, we see that for $a,b\in \mathbb{N}$, $ab\in \mathbb{N}$. The fix: in addition to requiring $ab\in H$, you also have to require that $a^{-1}\in H$.