Achim Krause
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Is there any use for $\sin(\sin x)$?
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53 votes

The intensity of light diffracted at a slit as a function of the angle actually involves a term $\sin\left(\frac{\alpha\beta}{2}\sin(\theta)\right)$, see https://en.wikipedia.org/wiki/...

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Simply-connected rational homology spheres
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38 votes

In dimension 4, we have the following: Simply-connectedness implies that $H_1(M)=0$. The condition that $M$ be a rational homology sphere implies that $H_2(M), H_3(M)$ are finitely generated torsion ...

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If $\Omega X \simeq \Omega Y$ then is $X \simeq Y$ for $X,Y$ simply connected?
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37 votes

To expand on my comment: As written (i.e. without requiring a map $f:X\to Y$), this is false in general. For an example, write $S^2$ as homogeneous space, $S^2=SU(2)/U(1)$. This exhibits $S^2$ as the ...

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What is the group of pointed homotopy classes of maps from $S^3 \times S^3$ to $S^3$?
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32 votes

We have that $S^3 \simeq \Omega \mathbb{H}P^{\infty}$, so by adjointness we can as well consider the group of maps $[\Sigma S^3 \times S^3, \mathbb{H}P^{\infty}]$. It is well-known that $[X,\Omega Y] \...

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What should I call a "differential" which cubes, rather than squares, to zero?
28 votes

I would just call it a module over the truncated polynomial algebra $k[D]/D^3$. Your two flavors of homology appear as positive odd-degree and even-degree groups in $\operatorname{Ext}^*_{k[D]/D^3}(k,...

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Does the isomorphic of the fundamental groups imply the existence of a mapping inducing an isomorphism?
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21 votes

No and no. For an explicit counterexample to 1. (which is also a counterexample to 2.) take the map $\mathbb{R}P^2\to \mathbb{R}P^{\infty}$.

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Is $[X, \_]$ a homology theory?
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21 votes

This holds only for compact objects (i.e. finite CW spectra), since it is easy to see that additivity fails otherwise (the other axioms of homology theories are satisfied). The usual way to obtain a ...

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Is $SU(3)/SO(3)$ cobordant with a mapping torus?
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18 votes

The mapping torus $T$ of the complex-conjugation-map $\mathbb{C}P^2 \rightarrow \mathbb{C}P^2$ does the job. For example by running the Serre spectral sequence with local coefficients, you obtain ...

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When is the augmentation ideal projective as RG-module?
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16 votes

Okay, this happens precisely in the obvious case, namely if all primes dividing the order $|G|$ are invertible in $R$. To see this, note that $\operatorname{Ext}^*_{R[G]}(R,R)$ is group cohomology of $...

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Which singular homology classes can be represented by embedded manifolds?
16 votes

The question in the title differs from the question spelled out in the post: In the title, you ask for embedded manifolds, in the post you ask for just maps from manifolds. I think the version of the ...

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What is the first interesting matric Toda bracket in the stable homotopy of the sphere?
16 votes

I know this post is quite old, but in case you are still interested, or anyone else is, I thought about sharing my recent thoughts about the topic. After all, this is the second result on "matric toda ...

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Hairy ball theorem for odd-dimensional spheres
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15 votes

The Lefschetz fixed point theorem implies that any $f: S^n \to S^n$ without fixed points has degree $(-1)^{n+1}$. But an even map $S^n \to S^n$ has even degree, since it factors as $$ S^n \xrightarrow{...

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loop space of a finite CW-complex
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14 votes

This is true for finite $\pi_1$ and false for infinite $\pi_1$: Let $\widetilde{X}$ denote the universal cover of $X$, then $\Omega\widetilde{X}$ is the unit connected component of $\Omega X$, and $\...

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Homology equivalence and isomorphism on $\pi_1$ not enough for homotopy equivalence?
13 votes

This is a modification of Anders's suggestion: Take $X=M(\mathbb{Z}[\frac{1}{2}],2)$ to be the Moore space with $H_2(X) = \mathbb{Z}[\frac{1}{2}]$. We can have $\mathbb{Z}/2$ act on $X$ this by a ...

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Simplest example of non-trivial Toda bracket in spaces
13 votes

The definition you've most likely encountered is the following: For maps $W\xrightarrow{f} X \xrightarrow{g} Y \xrightarrow{h} Z$, such that adjacent maps compose to $0$, $g$ extends to a map on ...

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Motivation for the definition of complex orientable cohomology theory
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12 votes

As you wrote, complex orientability can be characterized by the cohomology of $\mathbb{C}P^\infty$: $E$ is complex orientable if $E^*(\mathbb{C}P^\infty)$ splits according to the cell structure of $\...

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RIng that is flat over a subring as a right module but not as a left module
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10 votes

Take the associative algebra over a field $k$, with generators $x$ and $y$ subject to the relation $xy=0$. This admits a basis consisting of monomials of the form $y^a x^b$. It thus contains a subring ...

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Is every map from a simplicial complex to a sphere homotopic to a simplicial map to the boundary of a k-simplex?
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9 votes

Take $X = \partial \Delta^3$. Then homotopy classes of maps from $|X|$ to $S^2$ correspond to elements of $\pi_2(S^2)\cong \mathbb{Z}$. But there are only finitely many maps of simplicial complexes $\...

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Is there a definition of reduced $E_\infty$ ring?
9 votes

As Harry points out, the usual approach (For example in Lurie's SAG) is to think of the map $R\to \pi_0(R)$ for a connective $E_\infty$ ring as some kind of quotient by a nil-ideal. This is justified ...

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Is the cohomology ring $H^*(BG,\mathbb{Z})$ generated by Euler classes?
9 votes

Here is another short argument for why this cannot hold in general: The cohomology of the alternating groups $H^*(A_n)$ stabilizes, and I think the first degree in which it stays nontrivial is 3 (But ...

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Where does this clever choice of differential come from? (calculating $\mathrm{H}^1_{\mathrm{dR}}(E/k))$
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8 votes

Yes, $\omega$ is an invariant differential, also characterized by being nowhere vanishing (including the point at infinity). At every point, $dx$ and $dy$ together span the cotangent space (which is $...

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Embedding an icosahedron
8 votes

The answer to your question is actually "yes", but maybe not in the way you wanted. Indeed, the full (oriented) symmetry group of the icosahedron is isomorphic to $A_5$. The stabilizer of a ...

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($1$-)pullbacks of Kan complexes
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8 votes

Take any simplicial set $X$ which is not a Kan complex. Let $K$ be a Kan replacement of $X$, and let $L$ be a Kan replacement of the pushout $K\amalg_X K$. Then the two maps $K\to L$ are levelwise ...

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Does a non-singular matrix have a large minor with disjoint rows and columns and full rank?
8 votes

The matrix with zeroes on the diagonal and ones everywhere else is nonsingular, but all its "good" minors of size bigger than 1 are singular, since they have all entries equal to 1.

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Higher homotopy groups of irreducible 3-manifolds
7 votes

It's wrong for finite fundamental group, as then the universal cover is closed and has nonvanishing $\pi_3$ by Hurewicz. It's true for infinite fundamental group, again by Hurewicz applied to the ...

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A very elementary question on the definition of sheaf on a site
7 votes

Yes, you have to include the case $i=j$. Just look at what happens in the case of a single $U_1$, in order for this to boil down to the concept of a single effective epimorphism (introduced in the ...

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Given $f: X \to Y$, $g: X \to Z$, when does it exists $h: Y \to Z$ such that $hf \simeq g$?
7 votes

By replacing $f: X\to Y$ by a CW inclusion you can attack this with classical obstruction theory. I think the obstructions will lie in relative cohomology groups with local coefficients, $H^{n+1}(Y,X; ...

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$(B\mathbb Z/p^{\infty})^{\wedge}_p\rightarrow (BS^1)^{\wedge}_p$ induced by inclusion is an equivalence
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7 votes

Think of the Prüfer group as $\mathbb{Q}/\mathbb{Z}$ and of $S^1$ as $\mathbb{R}/\mathbb{Z}$. Here $\mathbb{Q}$ is discrete, $\mathbb{R}$ has its usual topology and is contracible. The map $\mathbb{Q} ...

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Trying to relate the fundamental groupoid to vector bundles
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6 votes

This does give you a vector bundle, and it comes with a flat (i. e. path homotopy invariant) parallel transport. You cannot get all vector bundles, but you can get exactly the ones with such a ...

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Integral versus real (universal) characteristic classes
6 votes

This is also equivalent to the corresponding statement for homology: Since $Hom(C,\mathbb{Z}) \otimes \mathbb{R} = Hom(C,\mathbb{R})$ for $C$ a free abelian group, the cochain complex with $\mathbb{R}$...

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