Recent MathOverflow Questions

Are localization functors always essentially surjective?

Math Overflow Recent Questions - Wed, 06/07/2017 - 05:28

Let $\mathcal{C}$ be a category and $\mathcal{W} \subseteq \text{Arr}(\mathcal{C})$ a set (or class) of arrows.

There are (at least) two notions of localization of $\mathcal{C}$ with respect to $\mathcal{W}$, a strict and a weak one. In the strict one, a functor $\lambda: \mathcal{C} \rightarrow \mathcal{D}$ is a localization iff $\lambda$ inverts all arrows in $\mathcal{W}$ and if for every functor $G: \mathcal{C} \rightarrow \mathcal{E}$ with the same property there exists a unique functor $G':\mathcal{D} \rightarrow \mathcal{E}$ with $$ G = G'\circ{}\lambda $$ It is not hard to see, that such a strict localization functor $\lambda$ must actually be surjective on objects. In the weak version of localization, one merely asks for the existence of a functor $G'$ together with an isomorphism $$ \eta: G \stackrel{\sim}{\rightarrow} G'\circ{}\lambda$$ such that for every other such pair $(G'',\eta')$ there exists a unique isomorphism $\kappa:G' \stackrel{\sim}{\rightarrow} G''$ with $$ (\kappa \circ{} \lambda) \eta = \eta' $$ Now, if such a weak localization functor would always be essentially surjective, then I (think I) can prove that the notions of strict and weak localization are essentially equivalent, in the sense that the existence of one implies that of the other.

So, are all weak localization functors essentially surjective?

local property of split exact sequence [on hold]

Math Overflow Recent Questions - Wed, 06/07/2017 - 04:00

in module category of ring A,Is a short exact sequence is split if and only if the localization of this sequence is split for every prime ideals?

Thanks!

Identify the sphere bundle of a complex line bundle $BD_{2n}\to BU(1)$

Math Overflow Recent Questions - Tue, 06/06/2017 - 15:21

I'd like to know whether it is possible to identify the sphere bundle arising as follow:

Let $\xi \colon BD_{2n}\to BU(1)$ the complex line bundle corresponding to the element $y^2 \in H^2(D_{2n};\Bbb Z) \cong \Bbb Z_2\langle x^2,y^2\rangle$ (we assume $n=0 \pmod{4}$ and $D_{2n}$ is the dihedral group of order $2n$).

Let $q\colon S(\xi) \to BD_{2n}$ be the sphere bundle of $\xi$. It's easy to see that it's a Eilenberg MacLane space $K(G,1)$ where $G$ sits in the following s.e.s. $$0 \to \Bbb Z \to G \to D_{2n} \to 0$$ which arises from the following piece of l.e.s. of a fibration $$0 \to \pi_2(\Bbb CP^{\infty}) \to \pi_1(S(\xi)) \xrightarrow{\pi_1(q)} \pi_1(BD_{2n}) \to 0$$ I'd like to identify such $G$. The Serre exact sequence in homology (integer coefficient) gives us the following exact sequence:

Since it's know that $H_2(BD_{2n})\cong \Bbb Z_2$, we have that $H_2(BU(1))\hookrightarrow H_1(S(\xi))$ which means that the exact sequence gives us $$0 \to \Bbb Z \to H_1(S(\xi)) \to \Bbb Z_2 \oplus \Bbb Z_2 \to 0$$

which implies that $H_1(S(\xi))$ cannot be a finite abelian group. In particular we already have that that $G \neq D_{\infty}$ since $H_1(D_{\infty})=ab(D_{\infty})=\Bbb Z_2\oplus \Bbb Z_2$.

From the exact sequence retrieved by the dual Blakers Massey Theorem (see here, in our case the base space is $1$-connected and the map is $1$-connected)

I was able to retrive this informations: $H^1(S(\xi);\Bbb Z)\cong \Bbb Z$, the boundary map $\delta \colon H^1(S(\xi);\Bbb Z) \to H^2(\Bbb C P^{\infty})$ is multiplication by $2$ and $H^2(S(\xi);\Bbb Z)\cong \Bbb Z_2$ since I know explicitly the map $H^2(BU(1))\to H^2(BD_{2n})$.

Is there a way to identify such $G$ completely?

Does this function belong to $L^2(\mathbb{D})$?

Math Overflow Recent Questions - Tue, 06/06/2017 - 14:20

Edit: After the answer of Prof. Eremenko to the previous version, I realized that a weaker assumption works for the main motivation of this post. so I revise the question.

The unit disk in the plane is denoted by $\mathbb{D}=\{(x,y) \in \mathbb{R}^2\mid x^2+y^2 <1\}$

Assume that $f:\mathbb{D} \to \mathbb{R}$ is a smooth (or real analytic function) such that for every analytic curve $\alpha \subset \mathbb{R}^2$ , $f$ is uniformly continuous on $\alpha \cap \mathbb{D}$.

Does this imply that $f\in L^2(\mathbb{D})$, or at least $f\in L^p(\mathbb{D})$for some $p >1$?

Motivations:

I encountered such type of functions in the "proof" of the proposition of the following note:

https://arxiv.org/abs/math/0408037

In reality, the proof is not correct, because the above assumption does not imply that $f$ has a (unique) continuous extension to the boundary. The mistake of the note was that it assumed such continuous extension.

More precisely we have a vector field $X=P\partial_x+Q\partial_y$ on the plane. We define a linear operator on the space of smooth functions on the plane with $L_X(f)=X.f=Pf_x+Qf_y$. We wish to solve the differential equation $$X.f=g$$.(This equation is called "Homological equation" in the following talk: "https://cms.math.ca/Events/Toulouse2004/abs/ss7.html#lt")

If $\gamma$ is a hyperbolic limit cycle of $X$ with period $T$ and surrounds a hyperbolic singularity, we can solve the above equation in the interior of $\gamma$ provided $\int_0^T g(\gamma(t))dt=0$. However we wish to pass from the interior of $\gamma$ to its exterior.

The true fact is that, in the interior of $\gamma$, the solution $f$ of $X.f=g$ satisfies the above Assumption. (In this post , for simplicity, we replace an arbitrary $\gamma$ by the unit circle) So the question in this post would be a motivation to consider a Sobolov space for action of $L_X$.

On a number theoretic problem

Math Overflow Recent Questions - Mon, 06/05/2017 - 07:03

Given coprime $a,b$ which are similar in size is there $T_1,T_2,|U|,|V|$, $N$ such that
$$b<T_1\equiv b \bmod N,\quad a<T_2\equiv a \bmod N$$ $$N=UT_1-VT_2 \mbox{ with }\mathsf{gcd}(a, U)=1,\mathsf{gcd}(b, V)=1$$ $$|U|,|V|\approx N^{3/4+\epsilon_1}$$ $$a,b\approx N^{1/4+\epsilon_2}$$ $$2VT_1^{-1}\equiv2UT_2^{-1}\equiv(T_1T_2)^{-1}(UT_1+ VT_2)\equiv(Ua^{-1}+ Vb^{-1})\bmod N \in\Big[-\frac N{ab},\frac N{ab}\Big]$$ holds at some small $\epsilon>0$?

There seems to be just enough room if $N^{\epsilon_1+\epsilon_2}>\zeta(2)$ holds at least on assumption $a,b$ are similar in size on a very very crude reasoning.

  1. Given $a,b$ there are roughly $a^{4-4\epsilon_2}$ choices for $T_1$ and $b^{4-4\epsilon_2}$ choices for $T_2$.

  2. In $N=UT_1-VT_2$ for each choice of $T_1,T_2$ there are $a^{3+\epsilon_1-3\epsilon_2}b^{3\epsilon_1-3\epsilon_2}$ choices for $U,V$ in $UT_1-VT_2$.

  3. Since $N$ is roughly $a^{4-4\epsilon_2}$ or $b^{4-4\epsilon_2}$ we have on average $\frac{a^{3+\epsilon_1-3\epsilon_2}b^{3\epsilon_1-3\epsilon_2}}{\max({a^{4- 4 \epsilon_2},b^{4-4\epsilon_2}})}$ which is (assuming $a$ and $b$ are similar size) $a^{2+2\epsilon_1-2\epsilon_2}$ different choices of $U,V$ for a given $N$.

  4. Since $\frac N{ab}$ is roughly $\frac{a^{4-4\epsilon_2}}{a^2}=a^{2-4\epsilon_2}$ we have that we can expect on average $a^{2\epsilon_1+2\epsilon_2}$ distinct $U,V$ to have $(T_1T_2)^{-1}(UT_1+ VT_2)\in\Big[-\frac N{ab},\frac N{ab}\Big]$.

  5. Since $a,b$ are fixed we can expect to choose $U,V$ with $\mathsf{gcd}(a, U)=1,\mathsf{gcd}(b, V)$ provided $N^{2\epsilon_1+2\epsilon_2}>\zeta(2)^2$ or $N^{\epsilon_1+\epsilon_2}>\zeta(2)$.

Thus we can expect quintuples $(T_1,T_2,|U|,|V|,N)\in\Bbb N\times\Bbb N\times\Bbb N\times\Bbb N\times\Bbb N$ that are valid.

If this reasoning is valid is there a way to make this argument rigorous?

Where can I find a copy of Bérard-Bergery's lecture notes on quaternionic manifolds?

Math Overflow Recent Questions - Mon, 06/05/2017 - 06:10

In the 1970's, Bérard-Bergery proved certain results on quaternionic Kähler manifolds, some of which are explained in the book Einstein Manifolds by Besse. Several times, Besse's book references a set of unpublished lecture notes by Bérard-Bergery for these results ([BéBer 7] in the book):

L. Bérard Bergery, Variétés quaternionniennes, Notes d'une conference a la table ronde "Varietes d'Einstein", Espalion (1979). (Unpublished).

I am looking for a copy of these lecture notes but was unfortunately unable to find any. Would anyone here have any suggestions on where to look?

abc conjecture modulo surjective polynomial and varieties

Math Overflow Recent Questions - Mon, 06/05/2017 - 06:08

In short the question is: is known that abc can't fail via elementary arguments.

A simple argument in favour of the abc conjecture is it is theorem for polynomials and abc can't fail only with polynomial identity.

Let $a,b,c,f,g \in \mathbb{Z}[x_1,x_2,\ldots,x_n]$ and $a+b=c$ and $a,b$ are coprime.

Case 1: Assume $f \mid rad(abc)$ where $f$ is surjective or the radical of $f$ at integers is sufficiently small, possibly bounded infinitely often. Then the degree argument fails and abc implies either $a,b$ are not coprime or the size argument fails. There is identity with the radical of $f$ sufficiently small and the size argument failing.

Case 2: Assume $a+b=c + f g$ where $f=0$ has infinitely many integer solutions. Also assume the degree argument fails. Then abc implies the same as in Case 1. Example of such identity is here

Is it known that abc can't fail with case 1 or case 2?

A topology for which symplectic forms are dense in skew forms

Math Overflow Recent Questions - Mon, 06/05/2017 - 05:40

Let $V$ be a vector space over an algebraically closed field. Let $S$ denote the vector space of skew-symmetric bilinear forms on $V$. When $V$ is finite dimensional the subset of $S$ consisting of forms with kernel of minimal possible dimension (either $1$ or $0$) is an open subset of $S$ with respect to the Zariski topology. What about when $V$ has countably infinite dimension? Is there a topology on $S$ such that the non-degenerate forms constitute an open dense subset? Note that if $V$ has a countable basis then it is clear that a non-degenerate form exists.

I'm especially interested in the case where the field has positive characteristic.

Should this hyphothesis exist? [on hold]

Math Overflow Recent Questions - Mon, 06/05/2017 - 04:30

First i bein with question. What if there are more than just plus and minus sign such that this sign squared give us negatine.If we represent multiplying - to - in turns, and think of munis as turn in oposite direction, we'll get this Than if we imagine third sign as half turn, we'll get our imaginary number i that I call s moreover if we cube it we'll get ns (negative s). So should it exist? Or this is just ravings of a madman who don't know math?

Question on the proof of Lemma5.1 of "A-Valued Semicircular Systems"

Math Overflow Recent Questions - Mon, 06/05/2017 - 04:21

I read the paper "A-Valued Semicircular system" recently. I am confused about the following equation in the proof of Lemma 5.1(c)(page 20): $$a \cdot j(h) = aJX(h)\Omega = aJX(h)J\Omega=JX(h)Ja\Omega =J(X(h) \cdot a^*)\Omega = j(h\cdot a^*) $$ I do not understand why $JX(h)Ja\Omega = J(X(h) \cdot a^*)\Omega$. I guess may be the right action $h \cdot a^*$ is given by $h \cdot a^* = JaJ h\Omega$ and $h$ is treated as a vector in the Hilbert space. If this interpretation is correct, then we indeed have $$a\cdot j(h) = j(h \cdot a^*), \quad j(h) \cdot a^* = j(a \cdot h)$$

However, if I adopt the interpretation of the right action above, I can not understand the following equation, since the right action here should be understood by treating $g$ as the element in Hilbert C$^*$ bimodule: $$JX(h)JX(g)a \Omega = JX(h)J(g \cdot a)$$

Would anyone let me know how should I understand the equations here? Thank you in advance.

De Morgan (Boolean algebra) algebra question about association? [on hold]

Math Overflow Recent Questions - Mon, 06/05/2017 - 03:17

This might be the completely wrong field (Not native English speaker) so let me apologize in advance. The math problem I need help with

How did he from the 4th to the 5th transform the 2 relation into 3. The only way to do it is distribution and that only applies when the relations are different not the same as in here ?

Approximation of quasi-periodic function by trigonometric polynomials

Math Overflow Recent Questions - Mon, 06/05/2017 - 03:01

The elements of the closure of $\{ \sum_{j=1}^n a_j e^{i\nu_j x}: a_j\in \mathbb{C}, \nu_j\in \mathbb{R} \}$ in the supremum-norm are called almost periodic functions. An almost periodic function $f$ is called quasi-periodic iff the frequency module

$$ \mathcal{M}(f)=\left\langle \left\{ \nu\in \mathbb{R}: \lim_{t\rightarrow \infty} \frac{1}{t} \int_0^t f(x)e^{-i\nu x}dx\neq 0 \right\} \right\rangle_{\mathbb{Z}} $$

is finitely generated as $\mathbb{Z}$-module. By definition we can approximate $f$ in supremum-norm by functions of the form $\sum_{j=1}^n a_j e^{i\nu_j x}$. I'd like to know whether one can preserve the frequencies.

Question: Is it always possible to approximate $f$ in the supremum-norm by functions $\sum_{j=1}^n a_j e^{i\nu_j x}$ with $\nu_j\in \mathcal{M}(f)$?

Maxwell's equations, geometric formulation

Math Overflow Recent Questions - Mon, 06/05/2017 - 02:35

So I've read all over online that Maxwell's equations for the electromagnetic field are$$dF = d^*F = 0,$$where $F$ takes values in the abelian Lie algebra of $\text{U}(1)$. Could somebody explain as to why this is the case?

I have followed up with some of the texts on the following MO thread:

Maxwells equations and differential forms

But I would still be interested in an explanation nonetheless from someone here.

Why is the radical of a reductive group equal to the connected component of the center?

Math Overflow Recent Questions - Mon, 06/05/2017 - 02:29

If $G$ is a connected reductive group over a perfect field $k$ (The definition given in Milne's "Algebraic Groups": $G$ is a connected group variety containing no non-trivial connected unipotent normal subgroup variety, even over the algebraic closure of $k$), why is the radical of $G$ (that is, the maximal connected solvable normal subgroup) equal to the connected component of the center? This is stated on page 7 of Shahidi's "Eisenstein series and automorphic L-functions." On page 357 of Milne's Algebraic Groups it says that the center of $G$ is of multiplicative type, and its largest subtorus is equal to the radical of $G$.

Borel's book (Linear Algebraic Groups) has a proof if $k$ is algebraically closed -- one can characterize the center of $G$ as the intersection of the Borels. But I don't know if that generalizes in some way over an arbitary field.

Convergence of modified forward-backward algorithm

Math Overflow Recent Questions - Mon, 06/05/2017 - 01:54

I've defined something with looks "almost" like a proximal operator except that is sorrounded by linear operators that "almost" cancel-out. I can't just plug it into a proximal algorithm since it might not converge in theory. I'm wondering whether something can be done to remedy the situation. Viz,

Let $g: \mathbb R^q \rightarrow (-\infty,+\infty]$ be convex lower semi-continuous function, and $D \in \mathbb R^{p \times q}$ and $E \in \mathbb R^{q \times p}$ be matrices such that

Asumptions:

  • $EE^T$ is an orthogonal projection.
  • $ED$ is an oblique projection
  • $D$ is a left-inverse of $E$ i.e $DE = I_{p \times p}$.
  • $DD^T = \text{diag}(\alpha_1,\ldots,\alpha_p)$ with $\alpha_i > 0$ for all $i$.

Consider the mapping

$$\text{prox}_{g,E,D} = D \circ \text{prox}_g \circ E : \mathbb R^p \rightarrow \mathbb R^p,\; x \mapsto D(\text{prox}_g(Ex)), $$

where $\text{prox}_g(z) := \text{argmin}_{u \in \mathbb R^q}\frac{1}{2}\|u-z\|_2^2 + g(u)$ defines the proximal operator of $g$.

Question: Under what further assumptions on $E$ and $D$

  • Is $\text{prox}_{g,E,D}$ a proximal operator w.r.t some Bregman divergence ?

  • Are the iterates $$x^{(k+1)} \leftarrow \text{prox}_{\gamma_k g,E,D}(x^{(k)} - \gamma_k f(x^{(k)})) $$ convergent ?

N.B.: Here $f$ is a smooth function with Lipschitz gradient and the $\gamma_k > 0$ are appropriately chosen scalars.

Generalized notion of divisor function?

Math Overflow Recent Questions - Sun, 06/04/2017 - 23:43

Divisor function $d(n,m)$ counts the number of $q\in\Bbb N$ with $b<q<m$ such that $n\bmod q\equiv0$.

Given $b>0$ what is the correct asymptotic, probabilistic and average case behavior of the function $f(n,m,b)$ that counts the number of $q\in\Bbb N$ with $b<q<m$ such that $n\bmod q\in[-b,b]$?

Questions about cluster $\mathcal{X}$-varieties and amalgamation

Math Overflow Recent Questions - Sun, 06/04/2017 - 20:56

I am trying to learn the amalgamation of two cluster seeds and I am reading the paper https://arxiv.org/pdf/math/0508408.pdf written by V.V. Fock and A. B. Goncharov. I am at a loss for the Lemma 2.2. on page 9.

My question is (a) I don't find a function $c$ on $K$.

(b) I don't understand the proof of Lemma 2.2, who can explain it clearly? Or offer me some relative papers. Any help will be appreciated.

Show the Cartan 3-form transgresses to the Killing form in the Weil algebra

Math Overflow Recent Questions - Sun, 06/04/2017 - 20:09

Let $G$ be a connected, reductive Lie group, and $W\mathfrak g = (S[\mathfrak g^\vee] \otimes \Lambda[\mathfrak g^\vee],\delta)$ the associated Weil algebra. This is a CDGA equipped with an action of $\mathfrak g$ by interior multiplication $\iota$ and a resulting action $\mathcal L$ of $\mathfrak g$ by Lie derivatives, $\mathcal L$-equivariantly DGA-isomorphic to the standard Koszul complex $(S[\mathfrak g^\vee] \otimes \Lambda[\mathfrak g^\vee],k)$. In particular, both it and the subalgebra $(W\mathfrak g)^G$ annihilated by the $\mathcal L$-action are contractible.

The basic subalgebra of $W\mathfrak g$, that annihilated by both the $\mathcal L$- and $\iota$-actions, is easily seen to be just the symmetric invariants $S[\mathfrak g^\vee]^G$, on which the Weil differential $\delta$ vanishes identically. By acyclicity of the big algebra, each symmetric invariant $x$ is $\delta y$ for some $y \in (W\mathfrak g)^G$, and we can consider its projection $z$ to the subalgebra $\Lambda[\mathfrak g^\vee]^G$, sending the symmetric generators to zero. The resulting element $z$ is independent of the choice of $y$, and the $x \mapsto z$ corresponds to the cohomology suspension $H^*BG \to H^* G$ in the universal bundle $G \to EG \to BG$ and induces a linear bijection between generators of $\Lambda[\mathfrak g^\vee]^G$ and $S[\mathfrak g^\vee]^G$.

There are natural elements of $S^2[\mathfrak g^\vee]^G$ and $\Lambda^3[\mathfrak g^\vee]^G$ which pull back along homomorphisms of Lie groups and are nonzero when $G$ is nonabelian, namely the Killing form $B$ and the Cartan form $(X,Y,Z) \mapsto B\big([X,Y],Z\big)$. From naturality and lack of other options (also, through hearsay), the Cartan form must be the suspension of the Killing form, but I've never seen anyone show this. But it must be possible to do very explicitly; very explicit formulae for the Weil differential and the two forms in question are possible.

Pick corresponding $B$-orthonormal bases $\theta^i$ and $u^i$ of $\Lambda^1[\mathfrak g^\vee]$ and $S^1[\mathfrak g^\vee]^G$. Let $\iota_i$ and $\mathcal L_i$ be the action of the dual basis of $\mathcal g$ and $c^i_{jk}$ the structure contents. If $x$ is the Killing form and $z$ is the Cartan form, find a simple expression in these symbols for $y$.

Fredholm property about $L^p$-extension $(p\neq 2)$ of differential operators

Math Overflow Recent Questions - Sun, 06/04/2017 - 19:36

The following is a well-known result for elliptic operators.

Theorem. Let $P: \Gamma(E)\to \Gamma(F)$ be an elliptic operator of order $m$ between vector bundles $E$ and $F$ over a compact manifold $X$. Then $P$ extends to a Fredholm map $P: W^{k,2}(E) \to W^{k-m,2}(F)$ whose index is independent of $k$ (cf. p193. Lawson-Michelsohn's book 'Spin Geometry')

This kind of extension involves Hilbert spaces and thus is relatively not hard to work with. However, concerning Floer theory (or pseudoholomorphic curvers theory), the linearized maps we concern have to be between non-Hilbert Banach spaces, such as $W^{1,p}(\mathbb R \times S^1, u^*TM)$ or $L^p(S^2, \Lambda^{0,1} \otimes_J u^*TM)$, where $p$ is always assumed to be greater than 2.

In the case $p\neq 2$, it seems much harder to show any Fredholm properties, and the proofs are usually not natural but very technical. You may agree to me if you refer to Audin-Damian's book(section 8.7) or McDuff-Salamon's book(Proposition 3.1.11 which requires Theorem C.2.3) respectively.

So, I would like to ask does there exist any general theorem concerning Fredholm properties, which is similar to the above one? If exist, what could be a general principle to show the Fredholm properties?

PS: In the above theorem, can we relax the condition that the base manifold $X$ is compact?

Integer matrices whose determinant equals their norm

Math Overflow Recent Questions - Sun, 06/04/2017 - 17:26

Let $M$ be an $2 \times 2$ matrix, with all entries in $\mathbb{N}$: $$ M= \begin{bmatrix} a & b \\ c & d \end{bmatrix} \;. $$ So $$ \mathrm{det}(M) = a d - b c \; . $$ The Euclidean norm (a.k.a. Frobenius Norm) of $M$ is the square root of the sum of the squares of its entries: $$ |M| = \sqrt{ a^2 + b^2 +c^2 + d^2 } \;. $$ I am trying to understand when $\mathrm{det}(M) = |M| = r$, $r \in \mathbb{N}$. For example, when $$ M= \begin{bmatrix} 15 & 9 \\ 3 & 3 \end{bmatrix} \;, $$ we have $$\mathrm{det}(M) = 15 \cdot 3 - 9 \cdot 3 = 18$$ and $$|M| = \sqrt{225 + 81 + 9 + 9} = \sqrt{324} = 18 \;.$$ Another solution is $$ M= \begin{bmatrix} 5 & 11 \\ 49 & 137 \end{bmatrix}\;, \;\mathrm{det}(M) = |M| = 146 \;. $$

Q. What are the solutions for $M$ a $2 \times 2$ matrix and $\mathrm{det}(M) = |M|$ a natural number $r$?

The positive-orthant point $(a,b,c,d)$ lies on the origin-centered sphere of radius $r$ in $\mathbb{R}^4$, but with the added constraint that $a d - b c = r$.

My goal was to understand when $\mathrm{det}(M) = |M|$ is a natural number $r$ for $M$ an $n \times n$ matrix of entries in $\mathbb{N}$, but already for $n=2$ it seems not entirely straightforward.

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