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Anton Fetisov's user avatar
Anton Fetisov's user avatar
Anton Fetisov's user avatar
Anton Fetisov
  • Member for 14 years, 1 month
  • Last seen more than a week ago
  • Moscow
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My first question - on Affine Schemes in Algebraic Geometry
@Yosemite Sam, thank you for the pointer! My answer is more or less a reformulation of @Sasha 's one. It's just a step forward: you don't need to check his equality for all l.r.s. $Y$, it suffices to check only affine ones, because all (pre)sheaves are colimits of representables. And then it becomes exactly the representability condition for $X$.
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My first question - on Affine Schemes in Algebraic Geometry
@Qiaochu Yuan, I didn't really describe what a scheme is, I just pointed the direction. There is Zariski-type Grothendieck topology on the category of affine schemes and the corresponding sheaves are schemes. This allows to associate a topological space to any scheme. The structure sheaf appears as a sheaf of morphisms from an open subset to the affine line (affine scheme represented by $\mathbb{Z}[T]$). So description of affine schemes as representable functors is meaningful, although possibly (like any representability) too difficult to check in practice without more concrete theorems.
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Doing geometry using Feynman Path Integral?
Try the book "The Feynman Integral and Feynman's Operational Calculus", by G.W. Johnson and M.L. Lapidus. Chapter 20 contains some discussion of Witten's know invaraiants, Atiyah-Singer index theorem and some more.
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A 2-category of chain complexes, chain maps, and chain homotopies?
@Qiauchu, yes, you can factor out higher homotopies, but this is inherently evil. I just wanted to underline that such approach doesn't generalize neatly to higher categories. Also the question was about homotopies per se, and they behave badly.
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A 2-category of chain complexes, chain maps, and chain homotopies?
Strictly speaking, what you get this way isn't a bicategory, it is a $(\infty,1)$-category, as they call it in nLab. Your multiplication of 2-cells is defined and associative only up to a coherent action of 3-cells. And multiplication of 3-cells is ok only up to 4-cells. This is just the same in topology: if you have a homotopy from $f$ to $g$ and from $g$ to $h$, you don't have a uniquely defined homotopy from $f$ to $h$! There are numerous ways to contract $[0;2]$-interval into a $[0;1]$-interval.
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The main theorems of category theory and their applications
@David: Yes, of course. I just didn't feel like discussing technical conditions. The existence of such internal language is, in my opinion, a prominent categorical fact by itself. It is completely non-obvious. The study of semantics for this language also takes many pages and is categorical in nature. The problems of set theory are of little importance here.
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The main theorems of category theory and their applications
Actually, you can only recover base category up to Cauchy completion. A trivial example is two different, but Morita equivalent rings. Or do you mean something more intricate?
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Is the Mendeleev table explained in quantum mechanics?
No, I don't get your sarcasm. As said before, if you're fine with the common amount of rigour in Mendeleev's table, then you should be just as fine with the common explanations in QM textbooks, so what's the point?
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Is the Mendeleev table explained in quantum mechanics?
Trivially, it's not a mathematically precise statement, so by the standards of total rigour it's vague. It's more precise (very precise) in predicting the ground state electron configurations, but less precise in predictions of chemical properties. Exceptions are rare, especially for low atomic weights, but they exits and well-known to chemists. I'm not a chemist myself, so I'll just give a few simple links: chemwiki.ucdavis.edu/Inorganic_Chemistry/… en.wikipedia.org/wiki/…
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tensor/hypermatrix analogues of $GL(n,\mathbb{C})$?
A linear representation of group G is, by definition, a homomorphism $G \to GL(n,\Bbb{K})$ for some field $\Bbb{K}$. So all linear representations have this form, without doubt. Are you asking if any other representations, besides linear, are interesting?
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Is the Mendeleev table explained in quantum mechanics?
It's strange to call "rigorous" a treatment where the charge of the nucleus is much larger then the number of electrons. In the cases considered both are around 10 - some sort of large number! Also, in chemical reactions the atoms are weakly ionized, almost neutral, so in fact $N\approx Z$. The words "use the model of non-interacting fermions" clearly hide much of the work and make a huge leap of faith from the initial equation to the answer.
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Proofs without words
A typical fake refutation. You don't need to define Lebesgue measure to do manipulations in geometry. All operations can be defined geometrically if I associate a number X with the segment of length X, and define $X \mapsto X^2$ as a function, mapping a segment to a square with such side. In fact, even many of infinite summations can be done geometrically, using the obvious topology and metric on shapes. Thanks to this formalistic tradition it took 100 years of pain to get from non-trivial Lebesgue construction to much more natural motivic integration.
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