New answers tagged at.algebraic-topology
2
votes
Construction of the Mayer-Vietoris spectral sequence
Here is a purely algebro-topological way of obtaining the Mayer–Vietoris spectral sequence for any homology theory $h_{\bullet}$ (the cohomological version is analogous). It is obtained by combining ...
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6
votes
Accepted
Codimension zero embeddings and maps with small fibers
Isn't the following an example?
Build $M$ starting from the plane by attaching an infinite sequence of handles whose size decreases to zero.
Take for $N$ the disjoint union of $M_k$, where $M_k$ is ...
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11
votes
Reduction of structure group and classifying spaces
To begin, I should mention that the proof of this equivalence is convincingly sketched in Stephen A. Mitchell's "Notes on principal bundles and classifying spaces", see Theorem 10.1 on page ...
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0
votes
Accepted
An attempt at an alternative calculation of the rank of $\pi_n(MO)$
$\newcommand{\a}{\mathfrak a}\newcommand{\Hom}{\operatorname{Hom}}\newcommand{\Z}{\mathbb Z}$
This can be proven by only assuming that $H^*(MO)$ is a free module. Indeed, due to $H^*(MO)$ being a ...
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6
votes
The simplicial Nerve
Just for the historical record. That is the way that Jean-Marc Cordier defined the homotopy coherent nerve in his paper in 1982. His notion was, in turn, based on the earlier work by Rainer Vogt and ...
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10
votes
Accepted
Are there any tests for knowing whether a topological space admits a CW structure?
Every compact topological manifold $M$ has the homotopy type of CW-complex. So, in the compact case, there is no such invariant of the type you are looking for (or, at least, such an invariant cannot ...
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14
votes
Accepted
What does Robert Stong mean when he says $H^*(MO(k))$ is a free Steenrod algebra in dimension less than $2k$?
First, let's clarify what the statement means:
The Steenrod algebra is a graded algebra, and $H^*(X)$ for any spectrum $X$ is a graded module over it. This means that the action $\mathcal{A}_2\otimes ...
- 10.8k
3
votes
Accepted
The double of the genus two handlebody minus three tori
Your manifold $M$ is not well-defined. This is because you made a non-trivial choice - namely of the pants decomposition $P$. We could perhaps think of your question as asking about a particular &...
- 28.1k
1
vote
Question about maps on cofibers being zero
The answer is no. Here is a counterexample. Let $p$ be a prime.
Let $F = \mathbb Z / p \otimes_{\mathbb Z}^L (-) : D(\mathbb Z) \to D(\mathbb Z)$, and let $\phi = p : \mathbb Z \to \mathbb Z$. Take $Z'...
- 63.9k
3
votes
Cohomology version of Moore space
As Allen Hatcher answered, there is no space whose cohomology is a countable direct sum of $\mathbb{Z}$'s in a single degree, and the cohomology of a wedge of spheres is instead a product.
However, ...
- 10.8k
5
votes
Is this true of the frame bundle $\operatorname{Fr}(M)$?
There is a tiny confusion of language here. If you are talking about the orthonormal frame bundle you have a Riemannian manifold not just a manifold. The frame bundle has structure group $Gl_n(\...
- 3,409
8
votes
Cohomology version of Moore space
One needs to distinguish between the direct sum and the direct product of a collection of groups. For a countably infinite collection of copies of $\mathbb Z$ the direct sum of these groups is a free ...
- 20k
4
votes
Is this true of the frame bundle $\operatorname{Fr}(M)$?
A lifting of the map $Fr(M)\to EO(n)$ along the map $(B\phi^\ast)EO(n)\to EO(n)$ always exists (up to homotopy, which I think is what you mean), simply because the space $EO(n)$ is contractible and ...
- 55.9k
2
votes
Is it always possible to connect the endpoints of a smooth injective path, so the resulting path is smooth, closed and simple?
Solution of your original question: A smooth curve cannot intersect infinitely many times lines in all directions.
Let $\gamma: T\to \Gamma$ be the parametrization
of your curve, where $T$ is the unit ...
- 91.8k
14
votes
Accepted
What does $\mathrm{Conf}_n(M)^{h S_n}$ look like?
In general, A nice enough finite dimensional space with a free $\Sigma_i$-action does not admit any homotopy fixed points. This is because a homotopy fixed point provides a splitting of the ...
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6
votes
What does $\mathrm{Conf}_n(M)^{h S_n}$ look like?
[UPDATE: Connor Malin's answer is definitely better than mine, but I will leave mine here in case the approach turns out to be useful for some other purpose.]
Put $X=\operatorname{Conf}_n(M)$ and $Y=\...
- 56.9k
5
votes
Algebraic theorems with no known algebraic proofs
Okninski proved that $M$ is a monoid and $K$ is a field of characteristic $0$ such that the monoid algebra $KM$ is von Neumann regular, then $M$ is locally finite using analytic methods. No algebraic ...
14
votes
Algebraic theorems with no known algebraic proofs
The Scott-Wiegold conjecture posits that the free product of three non-trivial cyclic groups can never be normally generated by a single element.
It was eventually proven by Jim Howie, by studying the ...
5
votes
Algebraic theorems with no known algebraic proofs
Dirichlet's theorem on primes in arithmetic progression: if $a$ and $b$ are any relatively prime positive integers, then there are infinitely many prime numbers of the form $a+nb$ for positive ...
16
votes
Algebraic theorems with no known algebraic proofs
There should be many examples in algebraic number theory, as it often interacts nontrivially with analytic number theory. For example, I am pretty sure there is no algebraic proof of Bauer's theorem (...
10
votes
Do the two orientations on an orientable manifold $M$ witness the same lifts of $\tau_M: M \to B\text{O($n$)}$ to $B\text{SO($n$)}$?
Up to homotopy, there is a fibration
$$
BSO_n \to BO_n \to B\mathbb Z_2.
$$
The space of orientations of $M$ is the (homotopy) fiber of the induced map of mapping spaces
$$
\text{map}(M,BSO_n) \to \...
- 18.8k
4
votes
Algebraic theorems with no known algebraic proofs
Infinitesimal Thurston Rigidity in Holomorphic Dynamics
This concerns eigenvalues of a natural operator on cohomology.
The statement is Galois invariant. The only known proof is not.
Algebraic proofs ...
19
votes
Algebraic theorems with no known algebraic proofs
Artin's theorem on positive polynomials, which solves Hilbert's 17th problem in the affirmative, apparently still has no algebraic proof.
Theorem (Artin): If $f \in \mathbb{R}[X_1,\dots,X_n]$ is ...
7
votes
Algebraic theorems with no known algebraic proofs
The proof (by Hecke I believe) that in a number field the ideal class of
the different is always the square of a class.
39
votes
Algebraic theorems with no known algebraic proofs
Here is my favorite one (though not so elementary).
Theorem (Grothendieck). Let $X$ be a smooth projective variety over an algebraically closed field $k$. Then, the etale fundamental group $\pi^{\rm ...
6
votes
Accepted
When does a cofibrantly generated model category have this factorization property?
I've encountered that condition a few time. Here is what I know about it:
If that property is satisfied for a model category $M$ then the full subcategory $C$ of finite $I$-cell complex is itself a ...
- 42.4k
32
votes
Algebraic theorems with no known algebraic proofs
An example comes from the connection of function fields and Riemann surfaces: Let $K$ be an algebraically closed field of characteristic $0$, $t$ a variable, and $L/K(t)$ be a finite Galois extension. ...
7
votes
Algebraic theorems with no known algebraic proofs
The fundamental theorem of algebra states that every non-constant single-variable polynomial with coefficients in $\mathbb C$ has at least one complex root; i.e., the field of complex numbers is ...
6
votes
Accepted
Express fundamental group of $\mathcal H/\Gamma$ by $\Gamma$
The statement
I know if we remove the elliptic points then the $\pi_1$ is exactly $\Gamma$.
is wrong. You can see this by considering $\Gamma=PSL(2,\mathbb Z)$.
What you probably meant to say is ...
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3
votes
Homotopy between posets
No. You are right that $H_i(X,F)$ can be defined for any functor (and is an interesting object to study), but it does not typically vanish for $i > 0$ and $X$ contractible. E.g. take $X$ to a be ...
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