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visits | member for | 4 years, 1 month |
seen | Apr 16 at 21:43 | |
stats | profile views | 2,081 |
Apr 15 |
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A neat monodromy group of a family of Kaehler manifolds
I don't think this works. |
Apr 9 |
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Reference for Kronecker-Weyl theorem in full generality
Sounds right to me. I don't understand how Peter and Greg think their comments are responsive. |
Apr 6 |
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What does it mean to speak of a homotopy fibration sequence?
An example of where things go bad in your last paragraph is where $Y$ is a point. Then the composite is already constant, but the trivial nullhomotopy is the wrong one. |
Apr 6 |
awarded | at.algebraic-topology |
Apr 5 |
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rationalization of classifying spaces
Commutativity shows that the answer to the second question in the first bullet point is "No," but does not address the first question in that bullet point. I think your second argument actually shows that finite dimensional rational $H$-spaces are commutative, so it really is the same reason. An infinite dimensional non-commutative example is given by $\Omega S^{2n}$. Its classifying space $S^{2n}$ is not an $H$-space, let alone a product of EM spaces, not even after rationalization. |
Apr 5 |
answered | What does it mean to speak of a homotopy fibration sequence? |
Apr 2 |
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Has this chain complex associated with a simplicial complex been studied before?
There is an obvious contravariant equivalence between sheaves and cosheaves on a fixed stratification with values in perfect complexes. Verdier duality is best thought of as a weird covariant equivalence, $F\mapsto\{U\mapsto H_c^*(U;F)\}$. How did you get $I$ (or $1-I$) without derived functors? Maybe you should have had them. |
Apr 2 |
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Can we reconstruct positive weight invariants in algebraic topology using algebraic geometry?
The text of Deligne's 1974 ICM address is available |
Apr 2 |
awarded | Nice Answer |
Mar 31 |
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Has this chain complex associated with a simplicial complex been studied before?
Your $I$ is $K$ of Verdier duality. Thus, the homology of $1-I$ is about self-dual sheaves. Cross-referencing $L$-theory and 2-periodic 2-torsion suggests the Rothenberg exact sequence for change of decoration. |
Mar 14 |
awarded | Yearling |
Feb 28 |
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Groups which are only defined up to conjugation
I think "groupoid" is a pretty good answer. Do you have any examples that are not groupoids, where the indeterminacy is for a reason other than choosing which of many isomorphic objects to take the automorphism group? (the vna example is a little more involved, but the key indeterminacy is exactly this.) |
Feb 2 |
answered | origin of spectral sequences in algebraic topology |
Dec 20 |
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Are $(\infty,1)$-categories $A_\infty$ categories?
There are too many uses of $A_\infty$. Theo's question is nonlinear, not even stable. Yes, $A_\infty$ should always be equivalent to associative. If you specialize Theo's question to $S$ = chain complexes, you get the situation where $A_\infty$ chain categories are equivalent to DG categories, which is the same as $A_\infty$-algebras are equivalent to DGAs. The scenario in which things are not equivalent is when you don't balance enrichments, because enriching over $\mathbb Z$-modules is extra structure, unless they are rational. |
Dec 19 |
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Modern Mathematical Achievements Accessible to Undergraduates
At least Apéry's irrationality of zeta(3) is accessible, though not quite in the 30 year window. Or at least Beukers's reformulation. |
Dec 17 |
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Does base extension reflect the property of being isomorphic?
oh, ok......... |
Dec 17 |
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Does base extension reflect the property of being isomorphic?
In the infinite field case, you're using the separable hypothesis, right? |
Nov 18 |
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Is every connected space equivalent to some B(Aut(X))?
Maybe it's a matter of taste, but this seems like an awkward phrasing to me. I'd rather first say "Is every topological group equivalent to the automorphism group of a space?" which sounds like a natural question; and only then clarify with the technical details of what I mean by "topological group" and "automorphism group of a space." |
Nov 18 |
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A simple minded Poincare duality for orbifolds?
@Geordie an example of what can go wrong is the mapping torus of a degree 2 map of a sphere. That is, glue the ends of $S\times I$ by a degree 2 map. The local cohomology is at every point the cohomology of a sphere. But the sheaf of local cohomology is not the constant sheaf $\mathbb Z$. The right thing to do is to check whether the restriction from a small open set to a point is an isomorphism, but it's actually multiplication by 2. Maybe these groups associated with points aren't actually stalks of everything. I think the relevant sheaf is loc sys $\mathbb Z[1/2]$ with monodromy 2. |
Jun 28 |
awarded | Necromancer |