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Search options answers only not deleted created 2011-09-28 - 2012-09-28
8 votes

Calculating the Riemann Curvature tensor out of sectional curvature

There is an explicit, but complicated, formula for going from the sectional curvature back to the curvature tensor in equation (1.10) on page 16 of Cheeger and Ebin's book.
Arch Stanton's user avatar
4 votes
Accepted

Is the space of directions an inner metric space for inner metric space of curvature $\ge k$?

I'm not familiar with that reference but a standard example for this is by Stephanie Halbeisen "On tangent cones of Alexandrov spaces with curvature bounded below". The example is necessarily infinite …
Vitali Kapovitch's user avatar
9 votes
Accepted

SL(2,C) Chern-Simons theory in genus 1

Let me call your $\omega$ as $\omega_I$. The symplectic form you get from the Chern-Simons action is $k\omega_I+s\omega_K$, where $\omega_K$ is one of the Kähler forms on the Hitchin space, which, in …
Pavel Safronov's user avatar
2 votes

Subsets of $\mathbb{R}^+$ closed under addition

Look at Kist and Leestma - Additive semigroups of positive real numbers. It would seem in some sense to answer the question.
Benjamin Steinberg's user avatar
4 votes

Subsets of $\mathbb{R}^+$ closed under addition

Nobody can even catalog the set of all subsets of positive integers closed under addition (aka subsemigroups of the infinite cyclic semigroup). See Repnitskii, Vladimir - On subsemigroup lattices wit …
user avatar
9 votes

Why are free groups residually finite?

Karl Auinger and I came up with the following proof which proves residually p and even stronger properties (e.g., residually finite of square-free exponent). Let me first introduce the main construct …
Benjamin Steinberg's user avatar
14 votes
Accepted

Fubini's theorem without completeness or $\sigma$-finiteness conditions

You do not need $\sigma$-finiteness of the measure in Fubini theorem, although it is an hypothesis that can be assumed with no loss of generality, in that the support of an integrable function is, of …
Pietro Majer's user avatar
  • 60.5k
13 votes

What is the Krull dimension of the ring of holomorphic functions on a complex manifold?

Are you also looking for holomorphic manifolds with $\dim \mathcal O=\infty$? In that case, in the paper by Sasane On the Krull Dimension of Rings of Transfer Functions [Acta Applicandae Mathematicae …
Francesco Polizzi's user avatar
20 votes

Examples of theorems with proofs that have dramatically improved over time

A favorite of mine is the chirality of the trefoil knot, which can be proved easily using the Jones polynomial or some of its relatives. Louis Kauffman's paper "New invariants in the theory of knots", …
18 votes

Examples of theorems with proofs that have dramatically improved over time

Kurosh's original proof of the subgroup theorem for free products used messy Kurosh systems. This was improved by covering space proofs (or equivalently covering groupoid proofs). One might argue the …
21 votes

Examples of theorems with proofs that have dramatically improved over time

There are several examples from Tauberian theory. Around 1930, Karamata surprised people by giving much simpler proofs of Littlewood's original Tauberian theorems for power series. Wiener's Tauberia …
10 votes

Faà di Bruno's formula for inverse functions?

See Warren P. Johnson, Combinatorics of Higher Derivatives of Inverses, American Mathematical Monthly, Vol. 109, No. 3 (Mar., 2002), pp. 273-277, http://www.jstor.org/stable/2695356
Ira Gessel's user avatar
2 votes

Faà di Bruno's formula for inverse functions?

This is sometime called the Lagrange inversion formula.
Liviu Nicolaescu's user avatar
1 vote
Accepted

Faà di Bruno's formula for inverse functions?

Riordan's Combinatorial identities has a chapter on partition polynomials that may be helpful. It specifically covers the question you are asking, but is in umbral calculus.
user avatar
1 vote

Understand Witten's "QFT and Jones Polynomials" - how does he get to the twisted Dirac opera...

Well well well, it seems like Witten has played a pretty nasty joke on all of us... And all the authors who copied from him apparently fell for it as well! But I found a possible resolution to the pr …
moep's user avatar
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