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https://mathoverflow.net/questions/400708
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6
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In mathematics and physics, especially gauge theory, there are many different but related notions of wedge products when discussing vector space- and vector bundle-valued differential forms. For example, if $U,V,W$ are finite-dimensional $\mathbb{R}$-vector spaces and $\mu:U\times V\to W$ is a bilinear map, then we always can define a wedge product of the type
$$\wedge\_{\mu}:\Omega^{k}(\mathcal{M},U)\times\Omega^{l}(\mathcal{M},V)\to\Omega^{k+l}(\mathcal{M},W)$$
via the usual detinition, i.e.
$$\alpha\wedge\_{\mu}\beta:=\sum\_{i=1}^{\mathrm{dim}\_{\mathbb{R}}(U)}\sum\_{j=1}^{\mathrm{dim}\_{\mathbb{R}}(V)}(\alpha^{i}\wedge\beta^{j})\mu(e\_{i},f\_{j})$$
where $(e\_{i})\_{i=1}^{\mathrm{dim}\_{\mathbb{R}}(V)}$ is a basis of $V$ and $(f\_{j})\_{j=1}^{\mathrm{dim}\_{\mathbb{R}}(U)}$ is a basis of $U$ and where $\alpha^{i}\in\Omega^{k}(\mathcal{M})$ and $\beta^{j}\in\Omega^{l}(\mathcal{M})$ are the coordinate forms with respect to this basis. There are many different examples where this appears in the literature: Choosing the vector space scalar product $\mu:\mathbb{R}\times V\to V$ leads to a wedge product between real-valued and $V$-values forms, choosing an $\mathrm{Ad}$-invariant inner product on a Lie algebra $\mathfrak{g}$ leads to the "trace" of $\mathfrak{g}$-valued forms, which is for example used in the definition of Chern-Simons forms, and when taking a Lie-bracket, we recover the well-known wedge product $[\cdot\wedge\cdot]$.
Now, in many applications, one also has to deal with bundle-valued forms and in this context, there are also many different type of wedge products. I would imagine that every product in this context can be defined in the following unified framework:
Let $E,F,G$ be three smooth $\mathbb{R}$-vector bundles. Furthermore, let $\mu\in\Gamma^{\infty}(E^{\ast}\otimes F^{\ast}\otimes G)$, which we can view as a smooth assignment $\mathcal{M}\ni p\mapsto\mu\_{p}$ such that $\mu\_{p}:E\_{p}\times F\_{p}\to G\_{p}$ is a bilinear map. Then we define a product of the type
$$\wedge\_{\mu}:\Omega^{k}(\mathcal{M},E)\times\Omega^{l}(\mathcal{M},F)\to\Omega^{k+l}(\mathcal{M},G).$$
For this, we take a local frame $\{e\_{a}\}\_{a=1}^{\mathrm{rank}(E)}\subset\Gamma^{\infty}(U,E)$ of $E$ and a local frame $\{f\_{b}\}\_{b=1}^{\mathrm{rank}(F)}\subset\Gamma^{\infty}(V,F)$ of $F$ on some open subsets $U,V\in\mathcal{M}$. Then we can write any $\alpha\in\Omega^{k}(\mathcal{M},E)$ and every $\beta\in\Omega^{l}(\mathcal{M},F)$ as
$$\alpha\vert\_{U}=\sum\_{a=1}^{\mathrm{rank}(E)}\alpha^{a}e\_{a}\hspace{1cm}\text{and}\hspace{1cm}\beta\vert\_{V}=\sum\_{b=1}^{\mathrm{rank}(F)}\beta^{b}f\_{b}$$
for some local coordinate forms $\alpha^{a}\in\Omega^{k}(U)$ and $\beta^{b}\in\Omega^{l}(V)$. We then define the wedge-product $\wedge\_{\mu}$ locally as
$$(\alpha\wedge\_{\mu}\beta)\vert\_{U\cap V}:=\sum\_{a=1}^{\mathrm{rank(E)}}\sum\_{b=1}^{\mathrm{rank(F)}}(\alpha^{a}\wedge\beta^{b})\mu(e\_{a},f\_{b}),$$
where $\mu(e\_{a},f\_{b})\in \Gamma^{\infty}(U\cap V,G)$ has to be understood as $\mu(e\_{a},f\_{b})(p):=\mu\_{p}(e\_{a}(p),f\_{b}(p))$ for all $p\in U\cap V$. This is well-defined and independent of the choice of local frames.
A particular example of this would be a bundle metric $\langle\cdot,\cdot\rangle\in\Gamma^{\infty}(E^{\ast}\otimes E^{\ast})$. When choosing the adjoint bundle $\mathrm{Ad}(P):=P\times\_{\mathrm{Ad}}\mathfrak{g}$ of some principal bundle $P$, we recover for example the wedge-product used in the definition of the Yang-Mills action, when using the bundle metric on $\mathrm{Ad}(P)$ induced by an $\mathrm{Ad}$-invariant inner product on $\mathfrak{g}$.
>
> Are such general type of wedge-products discussed in the literature and does anyone have some references for them?
>
>
>
Context: I am preparing some text on mathematical gauge theory and for this, I include some discussion of preliminaries including vector bundle valued differential forms and I would really like to treat them in a very general framework.
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https://mathoverflow.net/users/259525
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General wedge-product for vector bundle valued forms
|
The most general definition I know is the following. Every fiberwise bilinear form $\eta: V\_1 \times V\_2 \to W$ of vector bundles $V\_1, V\_2, W$ over $M$ gives rise to the wedge product of vector-bundle-valued differential forms by
$$(\alpha \wedge\_\eta \beta)\_m (X\_1, \dots X\_{p+q}) = \frac{1}{p!q!} \sum\_{\sigma \in S(p+q)} \mathrm{sign}(\sigma) \eta\_m\bigl(\alpha\_m(X\_\sigma(1), \dots, X\_{\sigma(p)}), \beta\_m(X\_{\sigma(p+1)}, \dots, X\_{\sigma(p+q)})\bigr)$$
where $\alpha \in \Omega^p(M; V\_1), \beta \in \Omega^q(M, V\_2), m \in M$ and $X\_i \in T\_m M$. No choice of basis or local frame needed.
A discussion of wedge products is this sense can be found e.g. in the [Monastir Lecture Notes by Neeb](https://www.math.uni-hamburg.de/home/wockel/data/monastir.pdf).
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6
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https://mathoverflow.net/users/17047
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400842
| 164,576 |
https://mathoverflow.net/questions/400813
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20
|
Say that a Diophantine equation is *almost-satisfiable* iff for each $n\in\mathbb{N}$ it has a solution mod $n$. Trivially genuine satisfiability over $\mathbb{N}$ implies almost-satisfiability, but the converse fails - see the discussion [here](https://mathoverflow.net/questions/47442/diophantine-equation-with-no-integer-solutions-but-with-solutions-modulo-every), or for a fun "nuke" note that almost-satisfiability is $\Pi^0\_1$ and so cannot coincide with satisfiability as the latter is [properly $\Sigma^0\_1$](https://en.wikipedia.org/wiki/Hilbert%27s_tenth_problem) *(and as far as I can tell that's actually non-circular! :P)*.
My question is the following: is almost-satisfiability known to be decidable?
It's plausible to me that one could whip up a Diophantine equation $\mathcal{D}\_T$ such that the behavior of a given Turing machine $T$ over the first $s$ steps is connected to the behavior of $\mathcal{D}\_T$ over something like $\mathbb{Z}/s\mathbb{Z}$ *(sort of a "Diophantine [Trakhtenbrot](https://en.wikipedia.org/wiki/Trakhtenbrot%27s_theorem)* theorem"), but I don't actually see how to do that. Certainly I don't see how to lift any of the MRDP analysis to almost-satisfiability in a useful way. On the other hand, I also don't see how to get a $\Sigma^0\_1$ definition of almost-satisfiability. Work of [Berend/Bilu](https://www.ams.org/journals/proc/1996-124-06/S0002-9939-96-03210-8/S0002-9939-96-03210-8.pdf) shows that almost-satisfiability of single-variable Diophantine equations is decidable, which is nontrivial *(in contrast to genuine solvability for single-variable equations which is a trivial application of the rational roots theorem)*, but at a glance I don't see how to generalize their arguments to multiple variables.
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https://mathoverflow.net/users/8133
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Is "almost-solvability" of Diophantine equations decidable?
|
A Diophantine equation is almost-satisfiable iff it is satisfiable over the ring $\widehat{\mathbb Z}$, the profinite completion of $\mathbb Z$ (also called by some the Prüfer ring), by a standard compactness argument (the solutions over each $\mathbb Z/n\mathbb Z$ form an inverse system of finite sets whose inverse limit is the set of solutions over $\widehat{\mathbb Z}$, and hence if the set of solutions is non-empty for each $n$, then the inverse limit is non-empty). From the direct product decomposition $\widehat{\mathbb Z}=\prod\_p \mathbb Z\_p$, this in turn reduces things to solvability over the $p$-adic integers $\mathbb Z\_p$ for all $p$. This question is solved in the paper J. Ax, Solving diophantine problems modulo every prime. Ann. Math. 85, 161–183 (1967) on pages 170,171. So the problem is decidable.
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27
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https://mathoverflow.net/users/15934
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400844
| 164,577 |
https://mathoverflow.net/questions/400840
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3
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I have seen that some papers talk of computational cost of the network and they measure it in MACs. I didn't find any clear explanation of what it is.
Could anyone explain in clear words the meaning of computational cost and why it should be taken into consideration in a network?
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https://mathoverflow.net/users/178519
|
What is the computational cost in a neural network?
|
Computational cost is simply a measure of the amount of resources the neural network uses in training or inference, which is important so you can know how much time or computing power you'll need to train or use an NN. It can measured in a variety of ways, but common ones are time and number of computations, expressed either as number of floating point operations (FLOPs) or as number of multiply-and-accumulate operations (MACs or MACCs). Since a lot of what a neural network is doing will be multiplying inputs by weights and adding them together, this is obviously a useful measure.
For a survey of the topic I'd recommend [Efficient Processing of Deep Neural Networks: A Tutorial and Survey](https://arxiv.org/abs/1703.09039).
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6
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https://mathoverflow.net/users/235087
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400846
| 164,578 |
https://mathoverflow.net/questions/400815
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3
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Let $\mathcal C, \mathcal D\subseteq 2^\omega$.
Let
$$
\DeclareMathOperator{\Either}{Either}
\Either(\mathcal C,\mathcal D)=\{A\oplus B: \text{either }A\in \mathcal C, B\in\mathcal D\text{, or }B\in \mathcal C, A\in\mathcal D\}
$$
Has this operation been named and studied in the context of Medvedev degrees (i.e., strong reducibility of mass problems)?
Its interest comes from the fact that from an element of $\Either(\mathcal C,\mathcal D)$ we cannot necessarily compute an element of $\mathcal C$ (or $\mathcal D$) uniformly.
|
https://mathoverflow.net/users/4600
|
Join-like operation and Medvedev reducibility
|
Kojiro Higuchi and Takayuki Kihara have studied operations of this flavour in their papers "Inside the Muchnik degrees" I+II ([doi Part 1](https://doi.org/10.1016/j.apal.2014.01.003),[doi Part 2](https://doi.org/10.1016/j.apal.2014.03.001)). It has been a few years since I read those, and I do not remember whether this particular operation plays a role. However, the idea of combining sets in ways that foil uniform reductions features very heavily in this work.
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3
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https://mathoverflow.net/users/15002
|
400847
| 164,579 |
https://mathoverflow.net/questions/400839
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6
|
Let $E$ be a vector bundle on some smooth algebraic variety and $E^\*$ its dual. Suppose $A$ is a sheaf (constructible or a $D$-module) on $E$. Given a linear function $f$ on $E$, we may compute the stalk at $f$ of the Fourier transform of $A$.
Now I’ve heard a slogan along the lines of “the stalk of the Fourier transform of $A$ is the vanishing cycles of $A$ along $f$.” Obviously this doesn’t really make sense; the stalk is a sheaf on the base and the vanishing cycles a sheaf on $f^{-1}(0)$. Nevertheless, my guess is that somehow this idea can be made precise.
So, is there a precise statement relating nearby and vanishing cycles to Fourier transforms of sheaves?
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https://mathoverflow.net/users/101861
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What do nearby/vanishing cycles have to do with Fourier transforms?
|
You probably want to say $A$ is a constructible sheaf / $D$-module with regular singularities. $D$-modules with irregular singularities (for example, those created by the Fourier transform) will behave very differently, even if $E$ is a vector bundle of rank $1$ on a point.
---
I'm going to pretend you asked the question for $\ell$-adic sheaves and describe the relevant phenomena there. I think the situation for $D$-modules will be analogous.
Let's start with the Fourier transform on $\mathbb A^1$. For an $\ell$-adic sheaf $A$ with tame ramification (analogue of regular singularities), the correct statement is:
The nearby cycles of the Fourier transform of $A$ at $\infty$ is isomorphic to a sum over points of $\mathbb A^1$ of the vanishing cycles of $A$ at that point. Furthermore, the Fourier transform is locally constant away from $0$, so its stalk at any point is non-canonically isomorphic to the nearby cycles of the Fourier transform at $\infty$ and thus non-canonically isomorphic to this sum.
Now let's consider the Fourier transform on $\mathbb A^n$, keeping the base still a point. The stalk of the Fourier transform at $\lambda f$ for a scalar $\lambda$ is going to be the stalk of the Fourier transform of $R f\_\* A$ at $\lambda$. Thus it will be non-canonically isomorphic to the sum of the stalks of the vanishing cycles of $R f\_\* A$ at each point of $\mathbb A^1$.
If $f$ were proper, it would then by isomorphic to the sum over $a$ of the cohomology of $f^{-1}(a)$ with coefficients in the vanishing cycles complex of $A$ along $f$. Maybe we can say that this is morally true in general. So we need to take $f^{-1}(a)$ for all $a$, not just $f^{-1}(0)$, and take the cohomology, to get the desired isomorphism. If the vanishing cycles are supported at finitely many points, this would just be the sum of the vanishing cycles at those points. Maybe that's what people who told you this were thinking of?
However, while this may be morally true, it will not be literally true in every case. The simplest counterexample I can think of is on $\mathbb A^n$, take $A$ to be the constant sheaf on the curve $xy=1$ and $f$ to be $y$. Then $A$ has no vanishing cycles on any point of $\mathbb A^2$ (none on $f^{-1}(0)$ since $A$ vanishes in a neighborhood of $f^{-1}(0)$, and none elsewhere because $A$ in each other fiber looks identical to the generic fiber) but its Fourier transform has a stalk of rank one at $y$.
---
Over a nontrivial base, I guess you want to take the vanishing cycles sheaf of $A$ along $f$ on $f^{-1}(a)$, push forward to the base, and then sum over $a$. This will be even more morally and less factually correct, but maybe there is some genericity condition under which it is literally true.
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4
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https://mathoverflow.net/users/18060
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400848
| 164,580 |
https://mathoverflow.net/questions/400720
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2
|
Let $A$ be a $C^\*$-algebra. Let $M(A)$ be its multiplier $C^\*$-algebras. The strict topology on $M(A)$ is given by
$$x\_\lambda \to x \iff \forall a\in A: (\|x\_\lambda a-xa\| + \|ax\_\lambda - ax\| \to 0 ).$$
We can identify $M(A) \cong \mathcal{L}(A)$ where $\mathcal{L}(A)$ are the adjoinable operators of the right Hilbert $A$-module $A$ with respect to the inner product $\langle a,b \rangle:= a^\*b.$ The strict topology on $\mathcal{L}(A)$ is given as follows:
$$t\_\lambda \to t \iff \forall a \in A: (\|t\_i(a)-t(a)\| + \|t\_i^\*(a) -t^\*(a)\| \to 0).$$
When we identify $A$ as a subset of $\mathcal{L}(A)$ via the mapping $ab^\* \mapsto \theta\_{a,b}$ where $\theta\_{a,b}(x) = a\langle b,x\rangle$, we obtain two notions of strict topologies on $M(A)$. I guess on bounded sets these topologies agree, but are these topologies in general equal?
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https://mathoverflow.net/users/216007
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Strict topology on the multiplier algebra
|
Yes, these topologies agree basically by definition once you understand the isomorphism $M(A)\cong \mathcal L(A)$ (since $tθ\_{a,b}=θ\_{t(a),b}$ the canonical isomorphism $A≅ \mathcal K(A)$ takes this element to $t(a)b^\*=t(ab^\*)$). The isomorphism $A \to \mathcal K(A)$ is given by $a \mapsto \theta\_{a\_1,a\_2}$ where $a=a\_1a\_2^\ast$ is any way of writing $a$ as a product of two elements (any element
in a $C^\ast$-algebra can be written as a product of two elements, e.g. if a $a= u|a|$ is the polar decomposition in the enveloping von Neumann algebra $A^{\ast\ast}$, then $a\_1 := u|a|^{1/2} \in A$ (this can be seen by approximating $t\mapsto t^{1/2}$ by polynomials) and $a = a\_1 |a|^{1/2}$).
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1
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https://mathoverflow.net/users/126109
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400856
| 164,582 |
https://mathoverflow.net/questions/400818
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4
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Let $f: Y \to X$ be an etale morphism of schemes.
>
> If $X$ has pseudo-rational singularities then does $Y$ also have pseudo-rational singularities?
>
>
>
For the definition of pseudo-rational see, for example, Definition 9.4 [here](https://arxiv.org/pdf/1703.02269.pdf).
Note that if $X$ has a resolution of singularities, then it follows from Lemma 9.3 of the paper of Kovacs linked above (and flat base change) that $Y$ has pseudo-rational singularities if $X$ does, but I would like to know whether there is an unconditional proof.
The question above is related to the following (to which I would be happy to get an answer as well):
>
> Let $R$ be an excellent noetherian local ring. If $R$ is pseudo-rational then is its completion also pseudo-rational?
>
>
>
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https://mathoverflow.net/users/519
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Is pseudo-rationality preserved by etale morphisms?
|
I believe the answer is yes, although I do not know of a reference.
We will use the original definition of pseudo-rationality due to Lipman and Teissier [[Lipman–Teissier 1981](https://doi.org/10.1307/mmj/1029002461), p. 102], and the following characterization:
**Lemma** [[Lipman–Teissier 1981](https://doi.org/10.1307/mmj/1029002461), Remark (a) on p. 102]**.** *Let $(R,\mathfrak{m})$ be an $n$-dimensional normal Cohen–Macaulay local ring. Then, $R$ is pseudo-rational if and only if for every projective birational morphism $f\colon W \to \operatorname{Spec}(R)$ where $W$ is integral, there exists a proper birational morphism $g\colon W' \to W$ such that $W'$ is normal and integral and such that*
$$\delta\_{fg}\colon H^n\_{\mathfrak{m}}(R) \longrightarrow H^n\_{(fg)^{-1}(\{\mathfrak{m}\})}(\mathcal{O}\_{W'})$$
*is injective.*
**Proposition.** *Let $(R,\mathfrak{m})$ be an $n$-dimensional noetherian local ring that is pseudo-rational, and let $\varphi\colon R \to S$ be an étale map. Then, for every prime ideal $\mathfrak{n} \subseteq S$ lying over $\mathfrak{m}$, the ring $S\_{\mathfrak{n}}$ is pseudo-rational.*
*Proof.* Since $\varphi$ is étale, $S$ is normal and Cohen–Macaulay [[Matsumura 1989](https://doi.org/10.1017/CBO9781139171762), Corollary to Theorem 23.9 and Corollary to Theorem 23.3], and $S\_\mathfrak{n}$ is of dimension $n$ [Stacks, Tag [04N4](https://stacks.math.columbia.edu/tag/04N4)].
Now let $f\_\mathfrak{n}\colon W\_\mathfrak{n} \to \operatorname{Spec}(S\_\mathfrak{n})$ be a projective birational morphism, where $W\_\mathfrak{n}$ is integral. Since $S\_{\mathfrak{n}}$ is integral and noetherian, there exists an ideal $I\_{\mathfrak{n}} \subseteq S\_{\mathfrak{n}}$ such that $W\_\mathfrak{n}$ is the blowup along $I\_{\mathfrak{n}}$ [[EGAIII$\_1$](http://www.numdam.org/item?id=PMIHES_1961__11__5_0), Corollaire 2.3.6]. Clearing denominators, there exists an ideal $I \subseteq S$ localizing to $I\_{\mathfrak{n}}$ such that the blowup $f\colon W \to \operatorname{Spec}(S)$ along $I$ localizes to $f\_\mathfrak{n}$.
By [Stacks, Tag [087B](https://stacks.math.columbia.edu/tag/087B)] or [[Rydh](https://people.kth.se/%7Edary/tamecompactification20110517.pdf), Proposition 4.14], there exists an ideal $J \subseteq R$ such that we have the commutative diagram
$$\require{AMScd}\begin{CD}
W' @>g>> W @>f>> \operatorname{Spec}(S)\\
@VVV @. @VVV\\
\operatorname{Bl}\_J\bigl(\operatorname{Spec}(R)\bigr) @= \operatorname{Bl}\_J\bigl(\operatorname{Spec}(R)\bigr) @>h>> \operatorname{Spec}(R)
\end{CD}$$
where $W' \to \operatorname{Spec}(S)$ is the blowup along $JS$, and the outer rectangle is cartesian by flat base change [Stacks, Tag [0805](https://stacks.math.columbia.edu/tag/0805)]. Replacing $\operatorname{Bl}\_J(\operatorname{Spec}(R))$ by its normalization, we may assume that $\operatorname{Bl}\_J(\operatorname{Spec}(R))$ is normal.
Now by flat base change for local cohomology [[Hashimoto–Ohtani 2008](https://doi.org/10.1307/mmj/1220879415), Theorem 6.10], the homomorphism
$$\delta\_{f\_\mathfrak{n}g\_\mathfrak{n}}\colon H^n\_{\mathfrak{n}}(S\_{\mathfrak{n}}) \longrightarrow H^n\_{(f\_\mathfrak{n}g\_\mathfrak{n})^{-1}(\{\mathfrak{n}\})}(\mathcal{O}\_{W'\_{\mathfrak{n}}})$$
is obtained from $\delta\_h$ by tensoring with $S\_{\mathfrak{n}}$, and hence is injective. Finally, since $fg$ is a blowup and $W'$ is normal by [[Matsumura 1989](https://doi.org/10.1017/CBO9781139171762), Corollary to Theorem 23.9], we see that $S\_\mathfrak{n}$ is pseudo-rational by the Lemma above. $\blacksquare$
It should be possible to prove that completions of pseudo-rational G-rings are pseudo-rational, using Néron–Popescu desingularization [[Popescu 1986](https://doi.org/10.1017/S0027763000022698), Theorem 2.4; [Popescu 1990](https://doi.org/10.1017/S0027763000002981), p. 45; [Swan 1998](https://mathscinet.ams.org/mathscinet-getitem?mr=1697953), Theorem 1.1] to reduce to showing that pseudo-rationality is preserved under étale extensions (shown above) and polynomial extensions.
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2
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https://mathoverflow.net/users/33088
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400859
| 164,583 |
https://mathoverflow.net/questions/400868
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2
|
This is related to one of my previous questions [here](https://mathoverflow.net/questions/398449/k-textth-maxima-of-n-i-i-d-chi-square-random-variables).
Let $(Z\_1, Z\_2, \ldots, Z\_n)\sim N(0, \Omega)$, where $\Omega = (1-\mu) I\_{n\times n} + \mu \boldsymbol{1}\_n\boldsymbol{1}\_n^\top $. Here $\boldsymbol{1}\_n$ denotes the vector of 1's of length $n$. Let us define $X\_i = Z\_i^2$ and I am trying to investigate the asymptotic of $k^{th}$ order statistic $X\_{(k:n)}$ where $k/n \to 1$ as $n \to \infty$.
As a special case let $k=n$. When $\mu =0$, it is known that $T\_n:=X\_{(n:n)}/\log n \overset{p}{\to} 2$.
What can we say about $T\_n$ when $0\leq \mu<1$? I have some conjectures about $T\_n$. If I am not wrong then one can use Slepian's lemma to conclude that $P(X\_{(n:n)}>t)\leq P(Y\_{(n:n)}>t)$ for $t\in \mathbb{R}$, where $\{Y\_i\}\_{i=1}^n$ are i.i.d $\chi^2\_1$. Though I am not completely sure of this fact. Here is the idea I am thinking about. Let $(W\_1, \cdots,W\_n)\sim N(0, I\_{n\times n})$. By Slepians' lemma we know $P(Z\_{(n:n)}> t)\leq P(W\_{(n:n)}>t)$. Now I am trying to take square of $Z\_{(n:n)}$ and $W\_{(n:n)}$. But squaring is not monotone and I am stuck. But I believe in asymptotic regime both of them are positive and hence my claim should work.
If this is true then I believe that $P(T\_n>2)\to 0$, i.e. heuristically speaking $T\_n \leq 2$ in limiting sense.
Any help will be appreciated. Also is any result for general $\Omega$ known? Thanks.
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https://mathoverflow.net/users/151115
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Limiting behavior of $k^{th}$ order statistics of n non-i.i.d chi square random variables
|
*(To long for a comment.)*
Let $b\_n = \sqrt{\frac{\mu}{n} + \frac{1-\mu}{n^2}} - \frac{\sqrt{1-\mu}}{n}$. Then
$$ \bigl( \sqrt{1-\mu} \, I\_{n\times n} + b\_n \mathbf{1}\_n \mathbf{1}\_n^{\top} \bigr)^2 = (1-\mu) I\_{n\times n} + \mu \mathbf{1}\_n \mathbf{1}\_n^{\top}. $$
In light of this, we can realize $(Z\_i)\_{1\leq i \leq n}$, using $(W\_i)\_{1\leq i\leq n} \sim \mathcal{N}(0,I\_{n\times n})$, by
$$ Z\_i = \sqrt{1-\mu} \, W\_i + b\_n S, $$
where $S=\sum\_{i=1}^{n} W\_i$. Then the law of iterated logarithm tells that $b\_n S = \mathcal{O}(\sqrt{\log\log n})$ almost surely, whereas the near-extreme values of $W\_i$'s will be of order $\sqrt{\log n}$.
For me, this seem to suggest that the near-maxima behavior of $(Z\_i^2)\_{1\leq i\leq n}$ will be very similar to that of the rescaled i.i.d. variables $((1-\mu)W\_i^2)\_{1\leq i\leq n}$.
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1
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https://mathoverflow.net/users/15602
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400870
| 164,585 |
https://mathoverflow.net/questions/400706
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0
|
In *Multiplicative Number Theory - Vol. I* by Montgomery and Vaughan the following result is proved.
**Theorem 7.20** Let $A(x,r)$ denote the number of $n\leq x$ such that $\Omega(n)\leq r \log \log x,$ and let $B(x,r)$ denote the number of $n\leq x$ for which $\Omega(n)\geq r \log \log x.$ If $0<r\leq 1$ and $x\geq 2$ then
$$
A(x,r)\ll x(\log x)^{r-1-r \log r}.
$$
If $1\leq r \leq R<2$ and $x\geq 2$ then
$$
B(x,r)\ll\_{R} x(\log x)^{r-1-r \log r}.
$$
How does the $R$-dependent proportionality constant vary as $R$ ranges over $[1,2)$?
Also, the indicated proof techniques are the same for both $A(x,r)$ and $B(x,r),$ but $r<R$ is required for $B(x,r)$. What happens if $r\geq 2$? Surely then even fewer $n$ satisfy the inequality $\Omega(n)\geq r \log \log n$ so the bound on $B(x,r)$ still holds. So why the restriction $r<R$?
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https://mathoverflow.net/users/17773
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Proportionality constant in Montgomery-Vaughan Theorem 7.20
|
**Edit:** *My only goal is to mark this as answered.*
As explained by Greg Martin in a comment:
Certainly $(,)$
is a decreasing function of $$,
but that doesn't imply that the bound on $(,)$
continues to hold for $\geq 2$.
The reason that $<2$
is required in the given proof is that it proceeds via upper bounds for
$$
\sum\_{n \leq x} \Omega(n),
$$
and this sum genuinely changes character when $>2$, since the integers $$
that are powers of 2 (or nearly so) are then increased by the map $n \mapsto \Omega(n)$
; for example, the largest power of $2$ less than $$ already gives a contribution of
$$x^{(\log r)/\log 2}$$
to the sum.
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0
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https://mathoverflow.net/users/17773
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400881
| 164,587 |
https://mathoverflow.net/questions/393515
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3
|
I'm currently studying Mumford's *Geometric Invariant Theory*. Unfortunately, I'm stuck understanding a detail in Theorem 1.1.
(Partial) Claim of Theorem 1.1
------------------------------
Let $X = \operatorname{Spec} R$ be an affine scheme over a characteristic-zero field $k$ and consider a reductive group action $G \curvearrowright X$. Then there is a categorical quotient of $X$ by $G$, induced by the inclusion $\phi\colon R\_0 \hookrightarrow R$ ($R\_0$ being the ring of invariants).
Part where I'm stuck
--------------------
Now, he wants to use Remark 6 on page 8 to derive the claim.
### Remark 6 on page 8
Let $G$ be a group scheme acting on a scheme $X$ (via $\sigma\colon G\times X \to X$). Furthermore, let $\phi\colon X \to Y$ be a morphism of schemes. The following conditions together imply that $(Y, \phi)$ is a categorical quotient of $G\curvearrowright X$:
1. $\phi\circ\sigma = \phi\circ p\_2$ (where $p\_2$ is the projection to the second component of the cartesian product)
2. $\phi$ induces an injective morphism of schemes $\mathcal{O}\_y \hookrightarrow \phi\_\*\mathcal{O}\_X$ that has the subsheaf of invariants as its image.
3. **If $W$ is an invariant closed subset of $X$, then $\phi(W)$ is closed in $Y$**; if $(W\_i)$ is a family of invariant closed subsets of $X$, then $\phi\left(\bigcap\_i W\_i\right) = \bigcap\_i \phi(W\_i)$.
### Mumford's proof sketch for condition (3)
I understand how Mumford proves the first two conditions, but I don't get the third one.
Mumford shows that for closed invariant sets $(W\_i)$,
$$ \overline{\phi\left(\bigcap\_i W\_i\right)} = \bigcap\_i \overline{\phi(W\_i)}. $$
Now, it suffices to prove that $\phi(W\_1)$ is closed.
Mumford then prompts me to apply the above equation to the case where $W\_1$ is arbitrary and $W\_2$ is the preimage of a closed point of $Y$, and claims that this implies that $\phi(W\_1)$ is closed.
This last claim is the one I don't understand.
My thoughts
-----------
* In the situation where $W\_2$ is the preimage of a closed point $y \in Y$, both sides of the equation we already have are subsets of $\phi(\phi^{-1}(y))$. If this set is empty, the equation just states $\varnothing = \varnothing$, not very helpful.
* Since the left hand side is either $\overline{\{y\}}$ or $\overline{\varnothing}$, the closure on the left-hand side is irrelevant in the situation of the previous bullet point.
* Let's assume that $\phi$ is surjective. Then the right-hand side is $\{y\}$ if and only if $y\in \overline{\phi(W\_1)}$ and $\varnothing$ otherwise. So if $y\in \overline{\phi(W\_1)}$, the equation implies $y\in \bigcap\_i \phi(W\_i) \subseteq \phi(W\_1)$. $\phi(W\_1)$ contains all closed points of its closure. However, what does this help, and is $\phi$ really surjective?
Hints for how to expand Mumford's proof sketch are highly appreciated.
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https://mathoverflow.net/users/237033
|
Why is the image of closed invariant subsets closed? Mumford, GIT, Theorem 1.1
|
The idea is actually rather simple. Let $W$ be a closed $G$-invariant subset of $X$, and $y$ a closed point that is not in $\phi(W)$. Note that $\phi^{-1}(y)$ is also closed and $G$-invariant. We already know that
$$
\overline{\phi(W\cap\phi^{-1}(y))}=\overline{\phi(W)}\cap\{y\}.
$$
But the LHS is empty which means that $y\notin\overline{\phi(W)}$. This tells us that $\phi(W)$ must be closed. Otherwise, we may find a closed point $y\in\overline{\phi(W)}\setminus\phi(W)$ that is not contained in $\overline{\phi(W)}$ by the argument above which is absurd. See the proof of Theorem 6.1 in Dolgachev's book "Lectures on Invariant Theory" for more details.
FYI, the morphism $\phi$ is submersive which means that it is surjective and
the induced topology on its image is the quotient topology. One could show that any categorical quotient satisfying the 3 conditions in remark 6 is submersive.
|
1
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https://mathoverflow.net/users/333428
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400903
| 164,593 |
https://mathoverflow.net/questions/400832
|
0
|
I try to find an upper bound for the mixing time of a random walk $S$ on a connected graph $L=(V,E)$ which has $k<\min\_{v\in V}d(v)$ loops at every vertex. The transition probabilities of this random walk are given by
$$p\_{v,w}=\dfrac{1}{d(v)+k};\qquad p\_{v,v}=\dfrac{k}{d(v)+k}$$
where $d(v)$ is the degree of the vertex $v$. I can easily calculate the stationary distribution $\pi(v) = \dfrac{d(v)+k}{\sum\_{v}d(v)+k|V|}$ but now I am interested in the mixing time of $S$, $t\_{mix}:=\inf\{t\geq 0|\inf\_{\mu}||\mu P^t -\pi||\_{TV}\leq 4^{-1}\}$ (see Markov Chains and Mixing Times; David A. Levin, Yuval Peres, Elizabeth L. Wilmer). For a random walk on a simple graph, without loops, I know that the mixing time is bounded from above by $C\log\left(\min\_{v\in V}\dfrac{1}{\pi(v)}\right)\Phi(L)^{-1}$ where $C$ is some constant and $\Phi(L)$ is the cheeger constant of $L$. In all approaches I have found, which use spectral graph theory, the fact that $\mathrm{trace}(A)=0$ is explicitly used for the adjacency matrix $A$ of $L$. But how does it work for graphs with loops?
Thank you for your help!
(This question is a copy from [here](https://math.stackexchange.com/questions/4214270/mixing-time-for-random-walk-on-graph-with-k-loops-on-each-vertex))
|
https://mathoverflow.net/users/333230
|
Mixing time for random walk on graph with $k$ loops on each vertex
|
The inequality you are citing should have a power 2 on the Cheeger constant (a.k.a the bottleneck ratio), so the inequality should read:
$$t\_{\rm mix} \le C\log\left(\min\_{v\in V}\dfrac{1}{\pi(v)}\right)\Phi(L)^{-2} \,.$$
This need not hold on a simple graph without loops; e.g. it fails if the graph is bipartite, where the mixing time is infinite. The inequality holds for lazy simple random walk, where the number of loops added at each vertex equals its degree. In the most general case you need to also bound the most negative eigenvalue of the chain. For the simplest case of the inequality combine inequality (12.10) and Theorem 13.10 (Due to Jerrum-Sinclair and Lawler-Sokal) in [1].
[1] Markov Chains and Mixing Times; 2nd edition, <https://pages.uoregon.edu/dlevin/MARKOV/mcmt2e.pdf>
|
1
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https://mathoverflow.net/users/7691
|
400904
| 164,594 |
https://mathoverflow.net/questions/400892
|
2
|
If $G$ is a graph with edge set $E$, let $W$ be the $\mathbb{Z}/2$-vector space generated by the elements of $E$. If $A = \{a\_1, \dots, a\_n\} \subset E$, let $\bar{A} = a\_1 + \dots + a\_n \in V$; then $\bar{A}\_1 + \bar{A}\_2 = \overline{A\_1 \Delta A\_2}$, where $\Delta$ indicates symmetric difference.
I'll define the *cycle space* of $G$ to be the subspace of $W$ generated by simple cycles of $G$. More precisely, the *cycle space* of $G$ is the subspace of $W$ generated by the set $\{\bar{C} \mid C \text{ is a simple cycle of } G\}$. We could also view the cycle space as the first simplicial homology group of $G$ over $\mathbb{Z}/2$. It is not difficult to show that the dimension of the cycle space of $G$ is the corank of the cycle matroid of $G$.
Given any matroid $M$ with ground set $E$, we could define the *circuit space* of $M$ in a completely analogous way, just using the word "circuit" instead of "simple cycle." My question is: is it always true that the dimension of the circuit space of $M$ is the corank of $M$? If not, for what types of matroids is this true? Finally, can anyone recommend good resources that deal with this sort of thing?
Thanks!
|
https://mathoverflow.net/users/202668
|
Dimension of circuit space of a matroid
|
The dimension of the circuit space of a matroid $M$ is the corank of $M$ if and only if $M$ is binary. Here is a proof. Given a basis $B$ and $e \notin B$, we let $C(e,B)$ be the unique circuit contained in $B \cup \{e\}$. We will use the following well-known characterization of binary matroids (Theorem 9.1.2 in Oxley's *Matroid Theory* text).
**Theorem.** A matroid $M$ is binary if and only if, for all bases $B$ of $M$ and all circuits $C$ of $M$, $C= \triangle\_{e \in C-B} C(e,B)$.
Here, $\triangle$ means symmetric difference, or equivalently $\sum$, if we view sets as vectors over $\mathbb{F}\_2$.
Let $\dim(M)$ denote the dimension of the circuit space of $M$ and $r^\*(M)$ denote the corank of $M$. We now prove that $\dim(M)=r^\*(M)$ if and only if $M$ is binary.
*Proof.* For each basis $B$ of $M$, let $\mathcal{C}\_B:=\{C(e,B) : e \notin B\}$. The circuits in $\mathcal{C}\_B$ are linearly independent since for each $e \notin B$, there is a unique circuit in $\mathcal{C}\_B$ containing $e$. Thus, $\dim(M) \geq r^\*(M)$, for every matroid $M$.
If $M$ is binary, then by the above theorem, $C= \triangle\_{e \in C-B} C(e,B)$, for every circuit $C$. Thus, $\dim(M)=r^\*(M)$.
If $M$ is not binary, then by the above theorem, there exists a basis $B'$ and a circuit $C'$ such that $C' \neq \triangle\_{e \in C-B'} C(e,B')$. In particular, $C'$ is not a sum of the vectors in $\mathcal{C}\_{B'}$. Thus, $\dim(M)>r^\*(M)$.
|
4
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https://mathoverflow.net/users/2233
|
400906
| 164,596 |
https://mathoverflow.net/questions/400899
|
0
|
I was searching for some results for the log-concavity of the modified Bessel function of a second type, but I failed. Has there been any known work on this? I am not even sure if it is the modified Bessel function of a second kind is indeed log-concave or not.
|
https://mathoverflow.net/users/333425
|
Log-concavity of the modified Bessel function of a second kind
|
Theorem 2(b) in [1] is equivalent to log-convexity of $K\_\nu$ for every $\nu$. This is said to be "well-known", and three references are given.
[1] Árpád Baricz, Saminathan Ponnusamy, Matti Vuorinen, *Functional inequalities for modified Bessel functions*, [DOI:10.1016/j.exmath.2011.07.001](https://doi.org/10.1016/j.exmath.2011.07.001), [arXiv:1009.4814](https://arxiv.org/abs/1009.4814)
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2
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https://mathoverflow.net/users/108637
|
400908
| 164,597 |
https://mathoverflow.net/questions/400922
|
1
|
Let $(X,d)$ be a metric space with finite Assouad dimension $0<C\_X$. It seems intuitive to me that if $\emptyset \subset Y\subseteq X$ then $Y$ is also doubling and its Assouad dimension, denoted here by $C\_Y$, should satisfy $C\_Y\leq c C\_X$ (where $c$ is some absolute constant independent of $X$ and of $Y$).
Is this true, and if so where can I find this fact?
|
https://mathoverflow.net/users/176409
|
Monotonicity of doubling dimension
|
This is Lemma 9.6(i) in J. C. Robinson, *Dimensions, embeddings, and attractors.* Cambridge Tracts in Mathematics, 186. Cambridge University Press, Cambridge, 2011.
In the proof the author says "it is obvious". I am no longer sure if it is obvious (perhaps it is) since I see a potential issue: if $Y\subset X$, then every ball in $Y$ is a restriction of a ball from $X$, but if $B$ is a ball in $X$ not centered at $Y$, then $B\cap Y$ is not a ball in $Y$ so there are less balls in $Y$ to cover. Is it an issue? I haven't checked (I still did not have my morning coffee).
|
2
|
https://mathoverflow.net/users/121665
|
400924
| 164,602 |
https://mathoverflow.net/questions/400912
|
10
|
$\DeclareMathOperator{\Spec}{Spec}$ Let $X$ be an algebraic stack.
Is there is a well-defined notion of the *residue field* of a point $x \in |X|$?
Attempts:
1. [Recall](https://stacks.math.columbia.edu/tag/04XE) that a point on a stack is an equivalence class of morphisms $\Spec k \to X$ from fields $k$. The issue is that it is not clear that there is a minimal choice of $k$ to warrant being called the residue field.
2. There is also the notion of a [residual gerbe](https://stacks.math.columbia.edu/tag/06ML) on a stack; but again it is not clear whether this comes with some kind of canonical field of definition which is compatible with 1.
3. If $X$ has a coarse moduli space $X^c$, then one could define the residue field of $x$ to be the residue field of the image of $x$ in $X^c$. This is well-defined, but seems to lose some of the subtle properties of stacks and again it's not clear whether it is compatible with 1. and 2.
I'm happy to assume my stack is sufficiently nice (e.g. smooth, DM,..)
|
https://mathoverflow.net/users/5101
|
Residue field of point on an algebraic stack
|
By definition, a residue field is an equivalence class of morphisms $\operatorname{Spec} k \to X$, i.e. of pairs of a field $k$ and an object in $X(k)$
We can upgrade that equivalence class into a category: Given fields $k$, $L$ and objects $a \in X(k) , b\in X(L)$, a morphism is a map $s \colon k \to L $ together with an isomorphism $s^\* a \to b$.
The key property that the residue field $F$ should have is that for every $k$-point of $X$ we obtain a map $F \to k$.
I claim we should define the residue field as the universal object with this property.
In other words, an element of the residue field is an assignment to each pair $k, a \in X(k)$ an element $\alpha\_a \in k$, compatible in the obvious way with morphisms: For $s \colon (k, a) \to (L,b)$ a morphisms, we have $s(\alpha\_a) = \alpha\_b$.
Elements of the residue field form a ring as there is an obvious notion of addition and multiplication. To check that they form a field, we need to check that every element is either invertible on every $k$-point or zero on every $k$-point, but this follows from the fact that we are working with a single equivalence class.
So we indeed have a universal notion of the residue field.
|
5
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https://mathoverflow.net/users/18060
|
400927
| 164,603 |
https://mathoverflow.net/questions/398359
|
8
|
Let $f$ be a smooth function defined on the unit disc $D \subset \mathbf{R}^2$ with
\begin{equation}
f \geq 0 \text{ in $D$ and } f(0) = 0.
\end{equation}
This is allowed to have a degenerate minimum at the origin, namely it is allowed that $D^2 f(0) = 0.$
**Question.** When is there $\rho \in (0,1)$ and $u \in C^1(D\_\rho)$ so that $\lvert D u \rvert^2 = f$? I would be more than happy with an answer specialised to the case where $f$ is the polynomial $(xy)^{2N}$—say with $N \geq 2$—if a general discussion is too onerous.
* As far as I understand, the equation was initially considered with a strictly positive right-hand side. One classical example is where $\lvert \nabla u \rvert^2\_g = 1$, with respect to the some Riemannian metric $g$ on $D$. One may attempt to rescale the Euclidean metric to $g = f^2 g\_e$, in order to get $\lvert \nabla u \rvert\_g^2 = f^{-2} \lvert \nabla u \rvert\_{g\_e} = 1$. However $g$ is unfortunately degenerate where $f = 0$.
* When the zero of $f$ at the origin is *non*-degenerate, then one can construct a solution to the eikonal equation via a sort of dynamic argument, as is explained in this [answer](https://mathoverflow.net/questions/82227/solutions-to-the-eikonal-equation).
|
https://mathoverflow.net/users/103792
|
When does the eikonal equation $\lvert Du \rvert^2 = f$ admit a local solution?
|
For $n=1$ or $2$, there is no $u\in C^1(D\_\rho)$ for any $\rho>0$ that satisfies $|\nabla u|^2 = (xy)^{2n}$. (Note that $f=(xy)^2$ has a *degenerate* minimum at $(0,0)$, so $n=1$ should be allowed in this discussion.) Meanwhile, for $n\ge 3$, there do exist $u\in C^1(\mathbb{R}^2)$ that satisfy $|\nabla u|^2 = (xy)^{2n}$.
The above non-existence result is special to the case $f = (xy)^{2n}$. For example, in the case $f=(xy)^{2n}(x^2{+}y^2)$, which has an even more degenerate minimum at $(0,0)$, there is a real-analytic, global solution $u(x,y) = (xy)^{n+1}/(n{+}1)$.
A few preliminaries before sketching the main argument are in order.
First, note that the disk radius $\rho>0$ actually plays no role in the problem. If $u\in C^1(D\_\rho)$ were to satisfy $|\nabla u|^2 = (xy)^{2n}$, then for any $r>0$ the scaled function $\tilde u(x,y) = r^{-(2n+1)}\,u(rx,ry)$ would satisfy $\tilde u\in C^1(D\_{\rho/r})$ and $|\nabla \tilde u|^2 = (xy)^{2n}$. Thus, one can assume that $\rho$ be arbitrarily large.
Second, one can assume, by adding a constant to $u$, that $u(0,0)=0$, so I will assume this normalization made henceforth. Then, the obvious integral inequality arising from $|\nabla u| = |xy|^n$ and the Cauchy-Schwartz inequality would imply that
$$
|u(x,y)| \le \frac{|xy|^n\sqrt{x^2+y^2}}{(2n{+}1)}.
$$
In particular, $u$ would vanish to order $2n{+}1$ at $(0,0)$ and would satisfy $u(x,0) = u(0,y) = 0$. It follows from this that $u$ could not be of differentiability class $C^{2n+2}$, since, if it were, the limit function
$$
p(x,y) = \lim\_{r\to0} \frac{u(rx,ry)}{r^{2n+1}}
$$
would exist and be a polynomial homogeneous of degree $2n{+}1$ that satisfied $|\nabla p|^2 = (xy)^{2n}$, and it is easy to show that there is no such polynomial. However, as will be seen, when $n\ge3$, there exists a $u\in C^{n-1}(\mathbb{R}^2)$ satisfying $|\nabla u|^2 = (xy)^{2n}$ and the homogeneity condition $u(rx,ry) = |r|^{2n+1}\,u(x,y)$ for all $r$. This $u$ is real-analytic away from the lines $x\pm y = 0$ but fails to be $C^n$ on these two lines.
From now on, I will assume that $u\in C^1(D\_\rho)$ (with $\rho>>0$ as large as necessary for the argument) satisfies $u(0,0)=0$
and $|\nabla u|^2 = (xy)^{2n}$. In particular, as the OP points out, $u$ satisfies the eikonal equation $|\nabla^g u|^2 = 1$ for the singular 'metric' $g = (xy)^{2n}(\mathrm{d}x^2+\mathrm{d}y^2)$, so that the gradient flow lines of $\nabla^g u$ are $g$-geodesics in the four quadrants of the $xy$-plane where $xy\not=0$.
Now, the metric $g$ has some interesting properties: First, it is homogeneous of degree $2n{+}2$, so that its family of geodesics is preserved under the scaling homothety, and, moreover, it is invariant under the discrete symmetries $(x,y)\to(-x,y)$, $(x,y)\to(x,-y)$, and $(x,y)\to(y,x)$. Consequently, it suffices to study the behavior of its geodesics in the 'first' quadrant, where $x>0$ and $y>0$, and it is immediate that the lines $y\pm x = 0$ are geodesics in the quadrants. Let the ray $\{(x,x)\ |\ x>0\}$ in the first quadrant be known as the *fundamental geodesic*. Of course, $g$ is not complete in the first quadrant, as the two boundary rays can be reached from, say, $(1,1)$, by curves of finite $g$-length.
Now, computation shows that the Gauss curvature of $g$ is $K = n(x^2{+}y^2)/(xy)^{2n+2}>0$, which suggests that nearly all of the geodesics of $g$ will avoid going to the singular boundary where $xy=0$, and, indeed, this turns out to be the case (see below). To parametrize the geodesics, it turns out to be convenient to use a parameter $t$ other than arc length. A curve $\bigl(x(t),y(t)\bigr)$ in the first quadrant parametrizes a $g$-geodesic when there is a function $\phi(t)$ satisfying the ODE system
$$
\dot x = xy\cos\phi,
\quad \dot y = xy\sin\phi,
\quad \dot\phi = n\,(x\cos\phi-y\sin\phi),\tag1
$$
and every $g$-geodesic in the first quadrant has such a parametrization, unique up to replacing $t$ by $t+t\_0$ for some constant $t\_0$.
In this case, arclength $s(t)$ along the geodesic satisfies $\dot s = (xy)^{n+1}$. [The advantage of writing the geodesic equations this way is that they extend smoothly across the singular locus $xy=0$.] Note that these equations are invariant under the homothetical scaling $(t,x,y,\phi)\to(t/r,rx,ry,\phi)$. Because of the scaling symmetry of the equations, one can extract a 2D phase portrait that makes clear the behavior of the geodesics as follows: Let $x+iy = \mathrm{e}^{u+iv}$. Then the above equations become (after a change of independent variable)
$$
u' = \cos(\phi{-}v)\,\cos v\sin v,\qquad
v' = \sin(\phi{-}v)\,\cos v\sin v,\qquad
\phi' = n\,\cos(\phi{+}v).\tag2
$$
One can now draw the $v\phi$-phase portrait, concentrating on the strip $0\le v\le \pi/2$, which represents the first quadrant in the $xy$-plane, and bearing in mind that these equations are invariant under the involution $(v,\phi)\to(\tfrac12\pi{-}v,\tfrac12\pi{-}\phi)$ and reverse under $(v,\phi)\to (v,\phi{+}\pi)$.
There are a sink at $S\_- = (v,\phi)=(\pi/4,\pi/4)$, a source at $S\_+ = (v,\phi)=(\pi/4,-3\pi/4)$, and saddles at $S\_1 = (v,\phi)=(0,\pi/2)$, $S\_2 = (v,\phi)=(\pi/2,0)$, $S\_3=(v,\phi)=(0,-\pi/2)$, and $S\_4=(\pi/2,-\pi)$. In addition to the 'trivial' separatrices that make up the boundary lines $v=0$ and $v=\tfrac12\pi$, there are four 'non-trivial' separatrices: $L\_1$ leaving $S\_1$ and going to $S\_-$, $L\_2$ leaving $S\_2$ and going to $S\_-$, $L\_3$ leaving $S\_+$ and going to $S\_3$, and $L\_4$ leaving $S\_+$ and going to $S\_4$.
Here is where the difference between the cases $n=1,2$ and the cases $n\ge 3$ becomes evident. When $n\le 2$, the fixed points $S\_\pm$ are *spiral*, i.e., the linearization of the flow at these two points has eigenvalues that are non-real (and complex conjuate), while, when $n\ge 3$, the linearizations of the flow at these two points have distinct real eigenvalues whose ratio is a real number strictly between $n{-}2$ and $n{-}1$.
One then finds that, when $n\ge 3$, the union of the two separatrices $L\_1$ and $L\_2$, together with their endpoints $S\_1$, $S\_2$, and $S\_-$ is the graph $\phi = f\_n(v)$ of a function $f\_n:[0,\tfrac12\pi]\to[0,\tfrac12\pi]$ that is real-analytic except at the midpoint $v=\tfrac14\pi$, where it is $C^{n-2}$. One then shows that the corresponding $g$-geodesics in the first quadrant define a $C^{n-2}$ foliation by geodesics that meet the boundary rays $x=0$ and $y=0$ orthogonally. Moreover, by reflecting this foliation across the lines $x=0$ and $y=0$, one can construct a foliation $\mathcal{F}$ of $\mathbb{R}^2$ minus the origin by $g$-geodesics that is real-analytic except along the lines $x\pm y=0$, where it is $C^{n-2}$. It then follows easily that there is a unique $C^{n-1}$ function $u$ that vanishes on the axes $x=0$ and $y=0$, satisfies $|\nabla u|^2=(xy)^{2n}$ and, away from the axes, the gradient lines of $u$ are the leaves of the foliation $\mathcal{F}$.
Meanwhile, when $n\le 2$, the spiral nature of the two fixed points $S\_\pm$ leads to an analysis that shows that there is no foliation of the 'first quadrant' quarter of a disk $D\_\rho$ by $g$-geodesics, which leads to the conclusion that there is no $C^1$ solution $u$ on any $D\_\rho$ to the equation $|\nabla u|^2=(xy)^{2n}$.
If there is interest, I can supply details of these arguments when I get the time.
|
8
|
https://mathoverflow.net/users/13972
|
400928
| 164,604 |
https://mathoverflow.net/questions/400929
|
0
|
Let $A$ be a subset of $\mathbb{R}^n$, and denote by $C(A)$ the space of complex-valued continuous functions defined on $A$. We know that if $A$ is compact then we can define a norm on $C(A)$ so that it is a Banach space. If $A$ is open in $\mathbb{R}^n$ then there exists a topology on $C(A)$ that makes it a Frechet space but not normable (see, for example, page 33-34 of Rudin's Functional Analysis).
Now suppose $A$ is open in $\mathbb{R}^n$. My question is if there exists a topology on $C(A)$ so that the topological vector space $C(A)$ is normable? I don't have any preference for the answer, so either yes or no is good to me.
Similarly I would like to ask the same question for $C^k(A)$, where $k\in\mathbb{N}$, for $C^\infty(A)$, and for $C^\infty(M)$ where $M$ is a smooth manifold (you may put any assumption on $M$ as you like, compact, closed, or has a boundary, etc).
P.S. I know the [Kolmogorov's normability criterion](https://en.wikipedia.org/wiki/Kolmogorov%27s_normability_criterion), but I guess there exist some topologies on $C(A)$ other than the one above, assuming $A$ is open in $\mathbb{R}^n$. So I am not sure if there is some topology on $C(A)$ that would make it normable.
|
https://mathoverflow.net/users/41686
|
About the normability of the space of continuous functions
|
Every real or complex vector space can be equipped with a norm (at least under the axiom of choice): Take a Hamel basis $B$ with coefficient functionals $\varphi\_b$ and define $\|x\|=\sum\limits\_{b\in B} |\varphi\_b(x)|$.
This norm however, is hardly ever of any use. For all your examples, natural additional assumptions would be completeness of the norm and the continuity of all evaluations $\delta\_x: f\mapsto f(x)$. The closed graph theorem then implies that the norm topology would be equal to the natural Frechet topology -- which cannot be since in the examples you have proper (non Banach) Frechet spaces.
|
4
|
https://mathoverflow.net/users/21051
|
400934
| 164,607 |
https://mathoverflow.net/questions/400926
|
3
|
Given an elementary abelian $p$-group $A$ of order $p^n$ for $n\geq 2$ and choose a subgroup $H$ of order $p$ from $Aut(A)\cong GL(n,p)$. We can use semi-direct product $A\rtimes B$ to construct a nonabelian $p$-group containing an elementary abelian maximal subgroup. Is there any other example and a complete characterization of $p$-groups satisfying this property?
|
https://mathoverflow.net/users/134942
|
nonabelian $p$-group contains an elementary abelian maximal subgroup
|
Not every such group has this form, but they can be classified.
Let $G$ be a $p$-group containing an elementary abelian $p$-group $A$ as a maximal subgroup.
Being maximal, it is an index $p$ normal subgroup. Let $g$ be a generator of the quotient $G/A$. Then $g$ acts by conjugation on $A$ as a matrix $\sigma \in GL\_n(\mathbb F\_p)$
The unique eigenvalue of $\sigma$ is $1$ and thus it can be put in Jordan normal form, writing it as a block-diagonal matrix with $m\_k$ unipotent blocks of size $k$ for some sequence $m\_k$ satisfying $\sum\_k k m\_k =n$. Furthermore, since a unipotent block of size $k$ has order $>p$ if $k>p$, we must have $m\_k =0 $ for $k>p$.
Let $\alpha \in A = g^p$. We can express any element of $G$ as $a g^r$ with $a\in A$ and $r \in \{0,\dots, p-1\}$. The multiplication table of $G$ is determined by the relations $ ga g^{-1} = \sigma(a)$ and $g^p = \alpha$. We obtain a group with $A$ as an index $p$ subgroup this way if and only if $\alpha$ is $\sigma$-invariant.
The space of $\sigma$-invariants in unipotent block of size $k$ for $\sigma$ is one-dimensional. So it might seem that the space of choices of $\alpha$ is $\mathbb F\_p^{\sum\_{k=1}^p m\_k}$. However, up to automorphisms of $\mathbb F\_p^n$ commuting with $\sigma$, if $\alpha$ has a nontrivial component in any block of size $k$ then we can cancel the component in every other block of size $k$ and every block of greater size. Thus, the only information that matters is the least $k$ such that $\alpha$ has a component in a block of size $k$, or equivalently, the least $k$ such that $\alpha$ is not in the image of $(\sigma-1)^k$. Call this $k\_\alpha$, and if $\alpha=0$, set $k\_\alpha=\infty$.
The groups with relation $g^p=\alpha\_1$ and $g^p=\alpha\_2$ are equivalent if we can write $ g^p =\alpha\_1, (ag)^p=\alpha\_2$, or in other words if $a + \sigma(a) + \sigma^2(a) + \dots + \sigma^{p-1}(a) = \alpha\_2- \alpha\_1 $. Restricting to any block of size $<p-1$, $a + \sigma(a) + \sigma^2(a) + \dots + \sigma^{p-1}(a)=0$, so $\alpha\_1,\alpha\_2$ become equivalent under this if and only if their difference is supported in blocks of size $p$. In other words, the cases $k\_{\alpha}=p$ and $k\_\alpha=\infty$ are equivalent.
The last source of equivalence of the pair $(G,A)$ is changing $g$ for a power of $g$, but this just has the effect of multiplying $a$ by a constant and thus is already accounted for.
So the classification of pairs $(G,A)$ where $G$ is a $p$-group and $A$ is a maximal elementary abelian subgroup is given by tuples $(m\_1,\dots, m\_p,k\_{\alpha})$ of nonnegative integers where $\sum\_{k=1}^p k m\_k = n$, $1\leq k\_{\alpha} \leq p-1$ or $k\_{\alpha}=\infty$, and $m\_{k\_\alpha}>0$ unless $k\_\alpha=\infty$.
$G$ is nonabelian if and only if $m\_1 \neq n$.
Semidirect products only occur in the case $k\_\alpha=\infty$, so there are lots of nonabelian groups that aren't semidirect products.
|
6
|
https://mathoverflow.net/users/18060
|
400942
| 164,610 |
https://mathoverflow.net/questions/400872
|
0
|
Let $M: \mathbb{C}[x,y] \to \mathbb{C}[x,y]$,
$(x,y) \mapsto (p,q)$, with $p,q \in \mathbb{C}[x,y]$
satisfying $\operatorname{Jac}(p,q):=p\_xq\_y-p\_yq\_x \in \mathbb{C}-\{0\}$.
Such a polynomial map is called a Keller map, and the two-dimensional Jacobian Conjecture
says that such a map is injective and surjective.
I know that there are many papers trying to prove injectivity or surjectivity of Keller maps.
My question is quite basic, and I think it is dealt with in one paper or another or perhaps in algebraic geometry books, but I am not able to find the relevant references now.
Further assume that $p(0,0)=q(0,0)=0$.
>
> **Question:** If $M$ is not injective, then is it true that there exist $c,d \in \mathbb{C}-\{0\}$ such that $p(c,0)=q(c,0)=0$ and $p(0,d)=q(0,d)=0$?
>
>
>
I ask, in other words: Given a map $\tilde{M}: \mathbb{C}^2 \to \mathbb{C}^2$, defined by $(a,b) \mapsto (p(a,b),q(a,b))$ (where $p,q \in \mathbb{C}[x,y]$ have invertible Jacobian),
if we assume that $\tilde{M}$ is not injective and $(0,0) \mapsto (0,0)$, is it true that there exist $c,d \in \mathbb{C}-\{0\}$, with $(c,0) \mapsto (0,0)$
and $(0,d) \mapsto (0,0)$.
(What I remember vaguely is something like: 'If $M$ is not injective, then it is not injective at zero'; I am not sure about my specific question. I guess it was meant that instead of $M$ we can change variables= compose it with an automorphism, and then get what I asked?).
**What I have tried:**
Write:
$p=p\_ny^n+\cdots+p\_1y+p\_0$,
where $p\_i \in k[x]$
and
$q=q\_my^m+\cdots+q\_1y+q\_0$,
where $q\_j \in k[x]$.
The assumption $p(0,0)=q(0,0)=0$ implies that $p\_0(0)=0$
and $q\_0(0)=0$.
We have:
$p(x,0)=p\_0$
and
$q(x,0)=q\_0$.
I am asking if $p\_0$ and $q\_0$ have a common root $c \in \mathbb{C}-\{0\}$
other than $0$ ($0$ is a common root of $p\_0$ and $q\_0$).
We assumed that $p\_0=xf$ and $q\_0=xg$, for some $f,g \in k[x]$.
So I ask if $f$ and $g$ have a common root $c \in \mathbb{C}-\{0\}$.
Observation: $f$ and $g$ cannot have $0$ as a common root,
since in this case $p\_0=x^2\tilde{f}$ and $q\_0=x^2\tilde{g}$,
for some $\tilde{f},\tilde{g} \in k[x]$.
But then $\operatorname{Jac}(p,q)|(0,0)=0$
($p\_x(0,0)=0$ and $q\_x(0,0)=0$).
Therefore, $f$ and $g$ are coprime or have a common root other than $0$.
If $f$ and $g$ are coprime, then $\gcd(p(x,0),q(x,0))=\gcd(p\_0,q\_0)=x$.
(Similarly, $p(0,y)$ and $q(0,y)$ have $0$ as a common root, and if they do not have other than $0$ common roots, then $\gcd(p(0,y),q(0,y))=y$).
Then $p=p\_ny^n+\cdots+p\_1y+xf$,
$q=q\_my^m+\cdots+q\_1y+xg$,
with $\gcd(f,g)=1$.
(There exist $u,v \in k[x]$ such that $uf+vg=1$.
Then, $up+vq=Wy+x$, for some $W \in \mathbb{C}[x,y]$).
Relevant papers, for example, are [Injectivity on one line](https://arxiv.org/pdf/alg-geom/9305008.pdf) and [One-one polynomial maps](https://www.jstor.org/stable/2034426) (Lemma 1).
A relevant question is [this](https://math.stackexchange.com/questions/2803632/a-sufficient-and-necessary-condition-for-mathbbcfx-gx-mathbbcx?noredirect=1&lq=1).
Now asked also in [MSE](https://math.stackexchange.com/questions/4215275/injectivity-of-keller-maps).
Thank you very much!
|
https://mathoverflow.net/users/72288
|
Injectivity of Keller maps
|
This can be achieved after appropriate changes of coordinates of the source and target. More precisely, there are automorphisms $A, B$ of $\mathbb{C}[x,y]$ such that $A \circ M \circ B$ has the property you want.
This follows e.g. from [Orevkov's result](https://www.math.univ-toulouse.fr/~orevkov/jc86.pdf) that if $\tilde M: \mathbb{C}^2 \to \mathbb{C}^2$ is a counterexample to the Jacobian conjecture, then the topological degree $d$ of $\tilde M$ is $\geq 4$. For generic $c \in \mathbb{C}^2$, $\tilde M^{-1}(c)$ has $d$ points.
**Claim:** You can choose $c$ such that not all points of $\tilde M^{-1}(c)$ are collinear.
(You are done if the claim holds, since then by changing coordinates you can map $c$ to the origin and $3$ non collinear points on $\tilde M^{-1}(c)$ respectively to the origin, a point on the $x$-axis and a point on the $y$-axis.)
There must be an elementary way to prove the claim for more general maps, but now I can only think of the following argument which applies only to non-injective Keller maps $\tilde M$: due to the "injectivity on one line" result you mentioned, the restriction of $\tilde M$ to each "vertical line" $y = b$ is non-injective. Moreover, for almost all $b \in \mathbb{C}$, the degree of $\tilde M|\_{y = b}$ is *smaller* than the global degree $d$ (since otherwise $\tilde M|\_{x = a}$ will be injective for all $a$). Therefore we can choose points $(a\_1, b), (a\_2, b), (a', b') \in \mathbb{C}^2$ which map to the same point under $\tilde M$ and such that $a\_1 \neq a\_2$ and $b \neq b'$, as required.
|
1
|
https://mathoverflow.net/users/1508
|
400945
| 164,613 |
https://mathoverflow.net/questions/400887
|
1
|
Let $f \sim \mathcal{GP}(0, K)$ be a zero-mean Gaussian process defined on a compact set $\mathcal{D} \subset \mathbb{R}^d$, where $K \colon \mathcal{D} \times \mathcal{D} \rightarrow \mathbb{R} $ is the covariance kernel. A random function sampled from such a GP can also be regarded as a member of the RKHS $\mathcal{H}$ with kernel $K$. Thus, we can consider the random variable $\|f \|\_{ \mathcal{H}}$.
It would be interesting to see the tail behavior of such a random variable. That is, can we develop an inequality of the form
\begin{align}
\mathbb{P} \big ( \| f \|\_{ \mathcal{H}} > q(\delta) \big ) \leq \delta, \qquad \forall \delta \in (0,1).
\end{align}
It would be great if we could characterize $q(\delta)$.
The motivation of this problem is from extending finite-dimensional Gaussian random vectors to infinite dimensions. For a finite-dimensional Gaussian random vector $v \sim N(0, \Sigma)$, we can easily get a tail bound for $\| v\|\_2$, the Euclidean norm of $v$.
|
https://mathoverflow.net/users/81633
|
Tail bound on the RKHS norm of a zero-mean Gaussian process
|
In fact, if the RKHS $\mathcal{H}$ is infinite dimensional, then $\mathbb P(f\in\mathcal{H})=0$ -- see e.g. [Corollary 4.10](https://arxiv.org/abs/1807.02582). So, no inequality of the desired form exists in infinite dimensions.
|
2
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https://mathoverflow.net/users/36721
|
400950
| 164,614 |
https://mathoverflow.net/questions/400947
|
1
|
A (translational) packing of a convex compact subset (with non-empty interior) $\mathcal C$ of $\mathbb R^d$ is a union
of translated non-overlapping (but perhaps touching) copies
of $\mathcal C$.
The (translational) packing density of $\mathcal C$ is the maximal proportion of $\mathbb R^d$ occupied by a suitable packing. The best packing density is $1$ and is achieved if and and only if $\mathcal C$ tiles $\mathcal R^d$. (The proof is a compacity argument.)
Is something known about convex compact sets achieving the worst packing density for a given dimension? (The worst possible packing-density in a fixed dimension is always strictly positive: The set of all convex compact sets is compact, up to the action of the affine group.)
I suspect that the worst case for $d=2$ are triangles (probably achieving only a translational packing density of $1/2$). (I guess that the $2$-dimensional case is quite accessible and has been studied by somebody.)
**Added correction:** (based on the reply by RavenclawPrefect) Triangles are indeed the worst case in dimension $2$ but a packing density of $2/3$ can be achieved: Surround every triangle by six touching triangles such that every intersection between two triangles involves a vertex of one triangle and the midpoint of an edge of the other triangle. (End of added correction)
More generally, simplices should be fairly bad in all dimensions.
Is there an interesting lower bound for packing densities for simplices?
(The notion is affine, all simplices have thus the same
packing density and considering the simplex
defined by all points with coordinate-sum $\leq 1$ in $[0,1]^d$
shows that the packing-density of a $d$-dimensional simplex is at least $1/d!$.)
Final remark: The case of packing densities for (Euclidean)
balls is a well-studied subject. However exact values are
only known in very few cases. The lack of knowledge in high dimensions is irritating even in this simple case.
|
https://mathoverflow.net/users/4556
|
Worst convex compact set for translational packings of $\mathbb R^d$
|
I believe this is not in general known for $d>2$. [Chapter 2](http://www.csun.edu/%7Ectoth/Handbook/chap2.pdf) of the *Handbook of Discrete and Computational Geometry* provides some pretty detailed information about translational packing density, which in their notation is $\delta\_T$. In particular, we have that the translational packing density of the regular tetrahedron is between $18/49\approx0.3673$ and $0.3745$ (an unusually tight range, compared to what we know for most shapes). I am not sure if this is the smallest known, but I can't think of anything known to perform worse.
The quote on page 30
>
> One could expect that the restriction to arrangements of translates of a set means a considerable simplification. However, this apparent advantage has not been exploited so far in dimensions greater than $2$.
>
>
>
seems relevant here.
Things are a little better known for *lattice* packing densities, where we further stipulate that the centers of the bodies form a lattice. We don't actually know *any* convex bodies where lattice packings aren't optimal, but I think in general we also don't have proofs of this optimality.
However, in dimension $2$ we do have an exact answer: you are correct that triangles uniquely attain the minimum, although that minimum value is $2/3$. See Theorem 1.1 from [this paper](https://www.sciencedirect.com/science/article/pii/S0723086913000789).
|
2
|
https://mathoverflow.net/users/89672
|
400951
| 164,615 |
https://mathoverflow.net/questions/400915
|
15
|
Consider a normal first course on category theory (say up to and including the statement and proof) of the adjoint functor theorem (AFT). What are the minimal assumptions for the definition of a set one needs to make in order that everything works? As far as I understand, up to and including the AFT there is very little one needs besides that fact that sets should have elements and that we have avoided Russell's paradox. So what is a minimal set of axioms allowing this to work?
|
https://mathoverflow.net/users/153228
|
Minimal set of assumptions for set theory in order to do basic category theory
|
To complement Tom Leinster's answer, let me try to be specific:
1. To form the product category $\mathcal{C} \times \mathcal{D}$, we need ordered pairs, which we can get from the axiom of **unordered pairs**.
2. It's probably a good idea to have the **empty set** $\emptyset$, so that the initial category exists.
3. My experience from type theory leads me to believe that we want function extensionality, or else we cannot reasonably work with functors and natural transformations (which are functions). Function extensionalty is equivalent to set-theoretic **extensionalty**.
4. To form the hom-set $\mathrm{Hom}(A,B) = \{f \in \mathcal{C}\_1 \mid \mathrm{dom}(f) = A \land \mathrm{cod}(f) = B\}$ we seemingly need **bounded separation**. It's a little more difficult to see whether we need unbounded separation (my guess would be that we can work pretty nicely without it).
5. To form functor categories, we need **powersets**. Indeed, given any set $A$, its powerset may be generated as the set of objects of the functor category $2^A$, where $2$ is the discrete category on two objects.
6. There are two functors form the terminal category $\mathbf{1}$ to the arrow category $\bullet \to \bullet$. If we think their coequalizer exists (in the category of small categories) then we believe in the axiom of **infinity**, because the coequalier is the monoid of natural numbers.
I am pretty sure the axiom of choice and excluded middle are not needed for general category theory, and foundation also seems quite irrelevant. How about union and replacement?
|
17
|
https://mathoverflow.net/users/1176
|
400962
| 164,620 |
https://mathoverflow.net/questions/400959
|
2
|
We work in a countable language of finite-order arithmetic, which allows us to quantify over natural numbers, sets of natural numbers, sets of sets of natural numbers, and so on. We measure the complexity of sentences with a generalization of the arithmetical and analytical hierarchies to higher subscripts. We call $\Pi^m\_n$ and $\Sigma^m\_n$ for $m, n \in \mathbb{N}$ *classes* of the arithmetical hierarchy.
I'm interested in *special* classes of the arithmetical hierarchy that are built up as follows.
1. $\Delta^0\_0$ is special.
2. If every true sentence in $\Pi^m\_n$ (resp., $\Sigma^m\_n$) follows from (i.e., can be proved from) true sentences belonging to special classes, then $\Pi^m\_n$ (resp., $\Sigma^m\_n$) is special.
3. $\Pi^m\_n$ is special if and only if $\Sigma^m\_n$ is special.
4. No other classes are special.
As an example, it's easy to see that $\Pi^0\_n$ (equivalently, $\Sigma^0\_n$) is special for all $n$. $\Delta^0\_0$ is special, and all true $\Sigma^0\_1$ sentences follow from true sentences belonging to $\Delta^0\_0$ (because the latter contains witnesses for all true $\Sigma^0\_1$ sentences). Thus, $\Pi^0\_1$ is special. We can then similarly deduce that $\Sigma^0\_2$ is special and so on.
If I'm understanding a 1961 result of Grzegorczyk, Mostowski, and Ryll-Nardzewski [1] correctly, then all true $\Pi^1\_1$ sentences follow from true first-order arithmetic sentences, so $\Pi^1\_1$ is special too.
My question is for which $m$ and $n$ is $\Pi^m\_n$ special?
[1] "Definability of sets in models of axiomatic theories" (Thanks to Ali Enayat for bringing this paper to my attention in another question of mine.)
|
https://mathoverflow.net/users/163672
|
Special classes of the arithmetical hierarchy of sentences of finite-order arithmetic
|
Per the comments, we're looking at deduction in some system based on the $\omega$-rule as opposed to standard first-order deduction (or Henkin semantics or etc.). There's a technical issue here - in my experiene the $\omega$-rule is usually formulated for *first-order arithmetic* sentences, so I'm not sure what it means to deduce a $\Pi^m\_n$ or $\Sigma^m\_n$ sentence using the $\omega$-rule for $m>0$ (this may be in the linked paper I don't have access to) - but in fact there's a coarse calculation which will apply to any reasonable interpretation I can think of:
Every version of the $\omega$-rule I can think of is $\Pi^1\_1$ - roughly, "the $\omega$-consequences of $T$" will always be $\Pi^1\_1$ relative to $T$. If (for example) we start with the true $\Pi^1\_1$ theory of arithmetic, we won't even get all the true $\Sigma^1\_1$ sentences since the true $\Sigma^1\_1$ theory of arithmetic isn't itself $\Pi^1\_1$ (more broadly, the projective hierarchy doesn't collapse).
Now your notion of specialness gives us only two tools for "climbing up" the syntactic hierarchy: deduction and complementation. In terms of Turing degree this means that we're not going to escape the $\omega$th hyperjump of $\emptyset$, which is a tiny subclass of $\Delta^1\_2$.
*(This is contra a silly claim I made originally - the point is that "$\Pi^1\_1$ in $\Sigma^1\_1$" is much *weaker* than "$\Pi^1\_2$," or more concretely that $\mathcal{O}^\mathcal{O}$ is not $\Pi^1\_2$ complete. The sense in which applying the $\omega$-rule "adds a $\Pi^1\_1$" seems to follow the former pattern if set up in a natural way. That said, I see no sense at all in which we can get outside the analytical hierarchy.)*
|
2
|
https://mathoverflow.net/users/8133
|
400963
| 164,621 |
https://mathoverflow.net/questions/400961
|
0
|
What is a toric lattice? and how can I construct one in `Macaulay2` and compute its basis? is there any alternative method to make one? Since I went through the whole documentation of the M2 but could not find anything. For example, we know that a toric lattice of $\dim=3$ is an identity matrix of size 3.
|
https://mathoverflow.net/users/333602
|
What is a toric lattice?
|
From the [documentation](https://faculty.math.illinois.edu/Macaulay2/doc/Macaulay2-1.15/share/doc/Macaulay2/NormalToricVarieties/html/_normal__Toric__Variety_lp__List_cm__List_rp.html) of the `normalToricVariety(*,*)` routine (emphasis mine):
>
> This is the general method for constructing a normal toric variety.
>
>
> A normal toric variety corresponds to a strongly convex rational
> polyhedral fan in affine space. **In this package, the fan associated to
> a normal $d$-dimensional toric variety lies in the rational vector space
> $\mathbb{Q}^d$ with underlying lattice $N = \mathbb{Z}^d$**. The fan is encoded by the minimal
> nonzero lattice points on its rays and the set of rays defining the
> maximal cones (meaning cones that are not proper subsets of another
> cone in the fan). More precisely, `rayList` lists the minimal nonzero
> lattice points on each ray (a.k.a. one-dimensional cone) in the fan.
> Each lattice point is a list of integers. The rays are ordered and
> indexed by nonnegative integers: $0,1,\dotsc,n$. Using this indexing, a
> maximal cone in the fan corresponds to a sublist of $\{0,1,\dotsc,n\}$. All
> maximal cones are listed in `coneList`.
>
>
>
I'd imagine the other routines either take the same approach or explain it in their documentation.
|
2
|
https://mathoverflow.net/users/103164
|
400970
| 164,624 |
https://mathoverflow.net/questions/400946
|
1
|
Let $\mathbb{R}$ be the set of real numbers. Given a subset $S$ of $\mathbb{R}$, let $\mathcal{T}\_S$ be the translation-invariant topology generated by $S$. That is, $\mathcal{T}\_S$ is the topology with a subbasis consisting of all translates of $S$. Suppose $A$ is a subset of $\mathbb{R}$ such that for every nonempty subset $S$ of $A$, we have that $\mathcal{T}\_S$ is disconnected. Does it follow that $\mathbb{R}$ cannot be written as a finite union of translates of $A$?
Some preliminary thoughts on this question: If we replace $\mathbb{R}$ by $\mathbb{Z}$, then the answer is no. For example, we can take $A$ to be the set of even integers, and one can verify that it has the above property. But $\mathbb{Z}$ equals the union of $A$ and $A+1$. So far, for $\mathbb{R}$ the only sets $A$ I’ve been able to come up with that have the required property are subsets of finite unions of cosets of a proper subgroup of $\mathbb{R}$. But the real numbers, unlike the integers, cannot be written as a finite union of cosets of a proper subgroup. So one approach to this problem might be to show that those are the only such $A$. However, I have no idea whether that’s true.
|
https://mathoverflow.net/users/132459
|
Subsets of $\mathbb{R}$, every nonempty subset of which generates a disconnected translation-invariant topology
|
This example is inspired by an example given op page 13 in [J. van Mill, Homogeneous subsets of the real line, Compositio Mathematica, 46 (1982) no. 1, pp. 3-13](http://www.numdam.org/item/?id=CM_1982__46_1_3_0).
Let $H$ be a Hamel base for $\mathbb{R}$ over $\mathbb{Q}$ such that
$1\in H$.
For $x\in\mathbb{R}$ let $q(x)$ denote the coefficient of~$1$ in its expression
as a linear combination of members of $H$.
Let $G=\{x:q(x)=0\}$ and let
$I=\mathbb{Q}\cap\bigcup\{(\pi+2n,\pi+2n+1):n\in\mathbb{Z}\}$.
We set $A=G+I$, so $A$ is the union of the cosets $G+q$ where $q\in I$, or $A=\{x:q(x)\in I\}$.
Notice that $A+\pi=\mathbb{R}\setminus A$, so in the translation-invariant
topology generated by $A$ the set $A$ itself is clopen, and $\mathbb{R}$
is the union of two translates of $A$.
We show that if $S$ is a nonempty subset of $A$ then $A$ is open in
the translation-invariant topology $\tau\_S$ generated by $S$, so that
$\tau\_S$ is not connected.
To begin note that $I+2\mathbb{Z}=I$,
so that $G+S+2\mathbb{Z}\subseteq A$ and the set $G+S+2\mathbb{Z}$
can be written as $G+T+2\mathbb{Z}$ for some $T\subseteq Q\cap(\pi,\pi+1)$.
Now consider the translation-invariant topology $\tau\_T$ on $\mathbb{Q}$ generated by $T$.
This topology contains nonempty sets of arbitrarily small diameter.
Let $\varepsilon>0$ and take $p,q\in T$ such that $p < \inf T+\varepsilon/2$
and $q>\sup T-\varepsilon/2$.
Then $q\in T\cap(T+(q-p))\subseteq(q-\varepsilon/2,q+\varepsilon/2)$.
It follows that $(\pi,\pi+1)\cap\mathbb{Q}$ belongs to $\tau\_T$.
But then
$$
A=G+\bigl((\pi,\pi+1)\cap\mathbb{Q}\bigr)+2\mathbb{Z}
$$
belongs to the translation-invariant topology generated by
$G+T+2\mathbb{Z}$, and hence to~$\tau\_S$.
|
3
|
https://mathoverflow.net/users/5903
|
400981
| 164,628 |
https://mathoverflow.net/questions/400978
|
5
|
Let $S$ be the (say, left) shift operator on $\ell^2(\mathbb{Z})$. For a non-zero vector $x \in \ell^2(\mathbb{Z})$, consider the set
$$X = \{ S^n v \mid n \in \mathbb{Z} \}.$$
Is this always a total set, i.e., is its span dense in $\ell^2(\mathbb{Z})$?
|
https://mathoverflow.net/users/16702
|
Do powers of the shift operator applied to a non-zero vector always yield a total set?
|
Such sets are not always total. The shift operator $S$ is unitarily equivalent to multiplication by $z$ on $L^2(S^1)$. From this perspective you can see vectors for which the set you write is not total, for example the characteristic function of an interval.
|
9
|
https://mathoverflow.net/users/2085
|
400987
| 164,629 |
https://mathoverflow.net/questions/400993
|
0
|
Let $G$ be a cyclic group of order $n$ and $K\leq AutG$ be a subgroup of the automorphism group of $G$. We denote the orbits of the natural action of $K$ on $G$ by $O\_1,\cdots, O\_s$. Let $\underline{X}\_i=\sum\_{x\in O\_i}x$ be the sum of elements in each orbit in the integral group ring $\mathbb{Z}G$. Then the $\mathbb{Z}$-span of $\underline{X}\_i$'s is a subring $\mathcal{A}$ of $\mathbb{Z}G$. Is $\mathcal{A}$ always generated by some $\underline{X}\_i$ as a ring over $\mathbb{Z}$?
|
https://mathoverflow.net/users/134942
|
generator of a subring of integral group ring
|
Not when $n=4$ and $G$ is the full automorphism group of $\mathbb Z/4$.
Then $\mathcal A$ is spanned by $\underline{X}\_1,\underline{X}\_2, \underline{X}\_4$ where $\underline{X}\_i$ is the sum of all elements of order $i$.
Then $\underline{X}\_1=1$ is the identity, $\underline{X}\_2^2 = 1$, $\underline{X}\_2 \underline{X}\_4 = \underline{X}\_4$, and $\underline{X}\_4^2 = 2 \underline{X}\_2 + 2$ so none can generate since $\underline{X}\_1=1$ simply generates $\mathbb Z$, $\underline{X}\_2$ generates the subspace spanned by $1$ and $\underline{X}\_2$, and $\underline{X}\_4$ generates the subspace spanned by $1, \underline{X}\_4, 2\underline{X}\_2$.
|
5
|
https://mathoverflow.net/users/18060
|
401001
| 164,635 |
https://mathoverflow.net/questions/400995
|
6
|
I want to build a finite CW complex such that $\pi\_1$ is non-abelian and $H\_i$ are zero for $i\geq 2.$
From Hatcher for a given group G, one can create an example of a 2-complex $X\_G$ with $\pi\_1(X\_G)=G.$ I also checked from Mayer-Vietoris that if $G$ is cyclic such complex won't have any higher homology for $i\geq 2.$ I tried to take $G=S\_3,$ the symmetric group of order 6 and from Mayer-Vietoris I get $H\_2$ is $Z.$ I believe this was a correct calculation. Or is there a way to get the groups $G$, with $G$ non-abelian, such that we can get $\pi\_1 =G$ and $H\_i=0$ for $i\geq 2.$
Any reference or idea to create such an example? Or is there a way to claim such a finite complex can't exist?
|
https://mathoverflow.net/users/333818
|
Finite CW complex with finite non-abelian fundamental group and higher homologies zero
|
**Theorem.** Let $G$ be a group. There exists a finite 3-complex $X\_G$ with $\pi\_1 X\_G = G$ and $H\_i X\_G = 0$ for $i > 1$ if, and only if, $G$ is finitely presentable and has second group homology $H\_2(G) = 0$.
The more interesting question to me is whether this is possible for a *finite 2-complex*, and Jens Reinhold's answer gives us the first step there. This is possible for the group $2I$ mentioned in my comment above.
---
**Lemma:** Let $X$ be any CW complex. Then the cokernel of the Hurewicz map $\pi\_2 X \to H\_2 X$ is isomorphic to $H\_2(\pi\_1 X)$. (Proof: attach cells to make $X$ into a $K(\pi\_1 X, 1)$; doing so kills off precisely $\pi\_2 X$ inside of $H\_2 X$.) This is exercise 23 in Hatcher section 4.2.
**Proof of theorem.** Pick any finite presentation $P$ of $G$ and construct the presentation complex $X\_P$. This is a finite complex with $\pi\_1 X\_P = G$, and with $H\_2(X\_P)$ a free abelian group $\Bbb Z^k$ on a finite number of generators. Because the cokernel of $\pi\_2(X\_P) \to H\_2(X\_P)$ is precisely $H\_2(\pi\_1 X\_P) = H\_2(G) = 0$, it follows that $\pi\_2(X\_P) \to H\_2(X\_P) = \Bbb Z^k$ is surjective. Now choose $k$ maps $\rho\_i: S^2 \to X\_P$ so that these give a basis of the second homology group, and set $$X\_G = X\_P \cup\_{i=1}^k D^3,$$ attaching these 3-cells along the $\rho\_i$. The resulting complex has $H\_i X\_G = 0$ for $i > 1$ by a Mayer-Vietoris argument, while $\pi\_1 X\_G = \pi\_1 X\_P \cong G$ by the van Kampen theorem.
|
7
|
https://mathoverflow.net/users/40804
|
401008
| 164,638 |
https://mathoverflow.net/questions/401033
|
1
|
Let $A,B$ are two $p\times p$ positive definite matrices such that $0<\delta\_0\leq \min\{\lambda\_{\min}(A), \lambda\_{\min}(B)\}\leq \max\{\lambda\_{\max}(A), \lambda\_{\max}(B)\}\leq \delta\_1$. Also assume that $\Vert A-B\Vert\_{op}\leq \varepsilon$. Can we upper bound $\Vert A^{-1} - B^{-1} \Vert\_{op}$ in terms of $(\delta\_0, \delta\_1, \varepsilon)$. I tried several things with the definition
$$
\Vert A^{-1} - B^{-1} \Vert\_{op} = \max\_{u,v\in \mathbb{S}^{p-1}} u^\top (A^{-1}-B^{-1})v.
$$
One easy observation is that $\Vert A^{-1}-B^{-1}\Vert\_{op}\leq \vert 1/\lambda\_{\min}(A) - 1/\lambda\_{\max}(B)\vert$. But I think this does not help me in any way. Any help will be appreciated.
**Edit:**
It could be possible that one can not get such bound. For a related question see [here](https://math.stackexchange.com/questions/2800167/matrix-inverse-is-a-uniformly-continuous-function-for-uniformly-positive-definit).
|
https://mathoverflow.net/users/151115
|
If $\Vert A-B\Vert_{op}\leq \varepsilon$ then $A^{-1}$ and $B^{-1}$ are uniformly close
|
$A^{-1}-B^{-1}=B^{-1}(B-A)A^{-1}$ so
$\|A^{-1}-B^{-1}\| \le \delta\_0^{-2} \epsilon $. This is the best possible bound in terms of the given parameters, as you can see by considering 2 by 2 diagonal matrices: Consider
$A=$diag$(\delta\_0,\delta\_1)$ and $B=$diag$(\delta\_0+\epsilon,\delta\_1+\epsilon)$ where $\delta\_1$ is large and $\epsilon \to 0$.
|
5
|
https://mathoverflow.net/users/7691
|
401035
| 164,643 |
https://mathoverflow.net/questions/401043
|
-1
|
I am trying to evaluate a fairly simple summation:
$\sum\_{k=1}^n ka^kb^{n-k}$
Which is related to the common identity for $\sum\_{k=1}^n ka^k$ available on Wikipedia.
I've previously seen lengthy lists of obscure summation formulas in the comments but could not find any this time via the search function.
I found Henry Goulds list at <https://dokumen.tips/documents/combinatorial-identities-a-standardized-set-of-tables-listing-500-binomial.html>, but it only has formulas involving binomial coefficients.
Is there a list of obscure summation identities? And in particular is there a published identity for this sum?
I got to this sum after some operations on a difference of powers factorization.
|
https://mathoverflow.net/users/334014
|
List of obscure summation identities
|
As suggested by the OP, I post my comment as an answer:
Try out Wolfram Alpha. The code for your example is <https://www.wolframalpha.com/input/?i=Sum%5B+k+a+%5Ek+b%5E%28n-k%29+%2C+%7B+k%2C+1%2C+n+%7D+%5D>
|
0
|
https://mathoverflow.net/users/37436
|
401044
| 164,644 |
https://mathoverflow.net/questions/401057
|
3
|
In a paper I was reading, it was mentioned that if $M$ is a closed Riemannian manifold, then by fixing a basis for $L^2(M)$ consisting of eigenfunctions of the Laplacian, the space of smoothing operators on $L^2(M)$ can be identified with the algebra of matrices $a\_{ij}$ such that
$$\sup\_{i,j}i^k j^l |a\_{ij}| <\infty$$
for all $k,l\in\mathbb N$.
**Question 1:** Could someone elaborate on how this identification can be done, or point me to a reference?
**Question 2:** Can this also be done if $M$ is non-compact?
|
https://mathoverflow.net/users/78729
|
Identification of smooth operators with rapidly decreasing matrices
|
This is, of course, a long story incorporating many strands but I will try to give a quick overview. Firstly, it is, as so often, convenient to skip to a more general framework. In your case, this would be that of an unbounded self-adjoint operator $T$ on Hilbert space (here that would be the Laplacian--more later).
One can associate with it a Frechet space $H^\infty(T)$ (the intersection of the domains of definitions of its powers) and a $DF$-space $H^{-\infty}(T)$ which are in duality. In the case of classical differential operators (the most frequent examples occur with the Laplacian and Schrödinger operators), the former is a space of test functions, the latter of distributions. This is a fairly direct consequence of the spectral theorem (in the form that any such operator can be represented as one of multiplication by a measurable function on an $L^2$-space).
If the spectrum of $T$ is discrete and consists of a sequence $(\lambda\_n)$ of eigenvalues which are such that $|\lambda\_n|$ is asymptotically like $n^\alpha$ for some positive $\alpha$, then the situation is particularly transparent. $H^\infty$ and $H^{-\infty}$ are a nuclear Frechet space and Silva space respectively. The eigenfunctions of $T$ form a basis for both spaces and they are identifiable with the sequence spaces $s$ and $s´$ of rapidly decreasing resp. slowly increasing sequences via coefficients.
The smoothing operators, i.e., continuous linear operators from $H^{-\infty}$ into
$H^\infty$ are then identified with the two variable version of $s$ in a standard way.
It is classical that the Laplacian satisfies these condition in many cases, e.g.,on a closed manifold, a compact manifold (with suitable boundary conditions--Dirichlet or Neumann), as does the Schrödinger operator for suitable potential functions. The requisite estimates on the eigenvalues go under the generic name of Weyl inequalities.
You cannot, however, expect such results in the non-compact case. Here the Schrödinger operator is, perhaps, more appropriate.
|
2
|
https://mathoverflow.net/users/317800
|
401061
| 164,646 |
https://mathoverflow.net/questions/401059
|
1
|
*EDIT (August 9, 2021):* I would like to ask a more general question. The original question that was fully answered is below the line.
For a positive real number $x$, denote the fractional part $x-[x]$ of $x$ by $\langle x \rangle$.
Let $\ell>0$ be an integer. Is
$$\Phi\_{\ell} := \liminf\_{n>0 \text{ not a } {\ell}^{\text{th}} \text{ power}} \ \sum\_{k = 1}^{\ell^2} \left\langle n^{\frac{k}{\ell}}\right\rangle$$
equal to zero?
---
Fix a prime $p$. Is it known whether or not
$$\Theta\_p := \liminf\_{n>0 \text{ not a } p^{\text{th}} \text{ power}} \ \sum\_{k = 1}^{p-1} \left\langle n^{\frac{k}{p}}\right\rangle$$
is zero?
|
https://mathoverflow.net/users/14233
|
Bounding the fractional parts of the $p^{\text{th}}$ roots of $n,n^2,...,n^{p-1}$
|
Let $n=m^p+1$ for some large enough $m$. For $0<k<p$ we then have $m^{kp}<n^k<m^{kp}+O(m^{(k-1)p})=m^{kp}(1+O(m^{-p}))$, so $$m^k<n^{k/p}<m^k(1+m^{-p})^{k/p}\leq m^k(1+O(m^{-p}))=m^k+o(1).$$ Hence $\langle n^{k/p}\rangle=o(1)$ as $m\to 0$ for each $0<k<p$, and in particular the $\liminf$ in your question is zero.
The $\liminf$ in the edited question also tends to zero, though to see this now we have to consider more terms of the Taylor expansion. Specifically, let us again pick $n=m^\ell+1$, where now $m$ to be divisible by a suitable number, to be specified later. We may assume $\ell\nmid k$ and pick $d$ such that $d\ell<k<(d+1)\ell$. Then we have
$$n^{k/\ell}=m^k(1+m^{-\ell})^{k/\ell}=m^k(1+a\_1m^{-\ell}+a\_2m^{-2\ell}+\dots+a\_d m^{-d\ell}+O(m^{-(d+1)\ell}))\\
=m^k+a\_1m^{k-\ell}+a\_2m^{k-2\ell}+\dots+a\_d m^{k-d\ell}+O(m^{k-(d+1)\ell}),$$
where $a\_i$ are some rational numbers. If we pick $m$ divisible by all of their denominators, then the terms up to $a\_d m^{k-d\ell}$ will be integers, and the $O(m^{k-(d+1)\ell})$ term will tend to zero since the exponent is negative. This at least shows that $n^{k/\ell}$ can be arbitrarily close to an integer. So see that the fractional part is small we also have to see that the error term is positive. This follows from the fact that the coefficient of the next term in the Taylor expansion, $a\_{d+1}m^{k-(d+1)\ell}$, has coefficient
$$a\_{d+1}=\binom{k/\ell}{d+1}=\frac{k/\ell(k/\ell-1)\dots(k/\ell-d)}{(d+1)!}$$
which is positive by our choice of $d$.
The same argument should work if you replace $\ell^2$ in the sum by any constant (possibly dependent on $\ell$, but not $n$).
|
4
|
https://mathoverflow.net/users/30186
|
401062
| 164,647 |
https://mathoverflow.net/questions/401020
|
7
|
Does one need the axiom of replacement in the [small object argument](https://ncatlab.org/nlab/show/small+object+argument) and in the [transfinite construction of free algebras](https://ncatlab.org/nlab/show/transfinite+construction+of+free+algebras)?
My motivation for the question is that I heard that the axiom of replacement is never needed in practice.
If the answer is "yes", how can one avoid the axiom of replacement in practical applications of these two theorems?
Note that one can't use ordinals if one doesn't allow the axiom of replacement (I think).
|
https://mathoverflow.net/users/333306
|
Does the small object argument need replacement?
|
The way it is usually presented, certainly yes. As you point out it usually refers to possibly uncountable regular ordinals, which would usually mean von Neumann ordinals. Once you have the regular ordinal, say $\kappa$, then regardless of whether $\kappa$ is a von Neumann ordinal or just a well ordered set, you need to define a sequence of objects $K\_\alpha$ for each $\alpha < \kappa$, and take colimits at each limit stage, and overall at the end. That kind of argument does seem to require replacement, as far as I can see. Certainly it is possible to define a sequence of sets, even of length $\omega$ whose colimit does not provably exist in the category of sets under $\mathbf{ZFC}$ minus replacement (since $V\_{\omega + \omega}$ is a model of that theory, and we can define the sequence $K\_n := V\_{\omega + n}$).
On the other hand, there are two alternative versions of the small object argument in Swan, [*W-types with reductions and the small object argument*](https://arxiv.org/abs/1802.07588) that don't use replacement (and also don't use definitions by recursion into universes of small types, which is, roughly speaking, the type theoretic version of replacement). The first can be carried out in a topos with natural number object and satisfying the choice principle $\mathbf{WISC}$, which I believe holds for any Grothendieck topos under the assumptions of $\mathbf{ZFC}$ minus replacement. The second is specific to the case of monic generating cofibrations in presheaf toposes, but does not require $\mathbf{WISC}$, so would work in a metatheory of $\mathbf{ZF}$ minus replacement. Both arguments are for a slightly non standard definition of cofibrantly generated, and they have a different format to the usual proof using ordinals. Instead of defining a sequence of objects along an ordinal, there are just two steps, first define a $W$-type, and then either quotient it (for the first argument) or carve out a subobject (for the second).
|
7
|
https://mathoverflow.net/users/30790
|
401065
| 164,649 |
https://mathoverflow.net/questions/401064
|
15
|
[A recent algorithm](https://www.maths.ox.ac.uk/node/38304) unknots in quasipolynomial time. But I want to know what happens to the crossing number. Assuming your unknot has $n$ crossings, if I remember correctly it might be necessary to increase $n$ temporarily. But to what? $n+C$? $C\*n$? Even worse? (I don't even know if the mentioned algorithm can be translated to actual Reidemeister moves - it might not be necessary if the goal is just *recognizing* the unknot.)
|
https://mathoverflow.net/users/11504
|
Unknot recognition - how tangled does it get?
|
Joel Hass and Jeff Lagarias proved that one can transform any unknot diagram with $n$ crossings into the standard unknot diagram using not more than $2^{cn}$ Reidemster moves. They were able to obtain the explicit value for $c$ of $c=10^{11}$. See [here](https://www.ams.org/journals/jams/2001-14-02/S0894-0347-01-00358-7/S0894-0347-01-00358-7.pdf). This was subsequently improved by Marc Lackenby to a polynomial bound in the paper in the Annals "A polynomial upper bound on Reidemeister moves" In that paper, an unknot with at most $n$ crossings is shown to have a transformation into the standard unknot diagram with at most $(236n)^{11}$ Reidemeister moves; at no stage does the diagram have more than $(7n)^2$ crossings.
|
23
|
https://mathoverflow.net/users/127690
|
401066
| 164,650 |
https://mathoverflow.net/questions/400965
|
9
|
In [Infinitesimal analysis without the Axiom of Choice](https://doi.org/10.1016/j.apal.2021.102959), Hrbacek and Katz have shown that it is possible to formulate an axiomatic theory which provides a formalisation of calculus procedures which make use of infinitesimals (known as SPOT, an acronym of its axioms).
Elsewhere in another somewhat related article by Katz I have read that SPOT is conservative over traditional Zermelo–Fraenkel set theory and so does not depend on the axiom of choice or on the existence of ultra-filters. Can someone explain in terms more suitable for a non-expert what it means for SPOT to be “conservative” over ZF, and also why this implies no dependence on the axiom of choice?
|
https://mathoverflow.net/users/119114
|
SPOT as a conservative extension of Zermelo–Fraenkel
|
In plain terms, the conservativity of SPOT over ZF means that if a particular statement S in the language of ZF is provable in SPOT, then ZF can already prove S (with a possibly different proof). Note that ZF does not include the axiom of choice.
More formally, the conservativity of SPOT over ZF is a statement about *formal proofs;* it asserts that for every proof $\pi$ of a statement S from the axioms of SPOT, there is a proof $\pi'$ of S from the axioms of ZF.
A much older conservativity proof was noted by Georg Kreisel in 1956. Kreisel observed (on the basis of Gödel's work on the [constructible universe](https://en.wikipedia.org/wiki/Constructible_universe)) that if S is an *arithmetical statement* (i.e., a first order sentence formulated in the usual language of Peano Arithmetic), and S is provable in ZFC + GCH (where ZFC is ZF plus the axiom of choice, and GCH is the general form of the continuum hypothesis) then S is already provable in ZF alone.
So, by Kreisel's observation, if one manages to prove an arithmetical statement (e.g., Goldbach's conjecture) using a "fancy" proof that uses the axiom of choice and/or the continuum hypothesis, there is another "spartan" proof in ZF alone of the same statement. This [FOM post](https://cs.nyu.edu/pipermail/fom/2015-May/018735.html) of mine provides more detail (and further links).
A vast generalization of Kreisel's observation was proved by Shoenfield in 1961, it is known as the Shoenfield [absoluteness](https://en.wikipedia.org/wiki/Absoluteness) theorem, and it is one of the cornerstones of modern set theory.
|
18
|
https://mathoverflow.net/users/9269
|
401076
| 164,653 |
https://mathoverflow.net/questions/401048
|
2
|
Let $\ell^n: [0,\infty)\to [0,1]$ be right-continuous and increasing functions s.t. $\ell^n(0)=0$. Given $x>0$ and Brownian motion $(B\_t)\_{t\ge 0}$, can we prove
$$\limsup\_{n\to\infty}\mathbb P[\exists s\in [0,t]:~ x+B\_s\le \ell^n(s)]\le \mathbb P[\exists s\in [0,t]:~ x+B\_s\le \limsup\_{n\to\infty}\ell^n(s)],\quad \forall t>0?$$
PS : It appears that the pathwise inequality
$$\limsup\_{n\to\infty} {\bf 1}\_{\{\exists s\in [0,t]: x+B\_s\le \ell^n(s)\}}\le {\bf 1}\_{\{\exists s\in [0,t]: x+B\_s\le \limsup\_{n\to\infty}\ell^n(s)\}}$$
does not hold.
|
https://mathoverflow.net/users/nan
|
Question concerning an inequality on probabilities of hitting times in a paper
|
$\newcommand{\ep}{\varepsilon}\newcommand{\de}{\delta}$This is not quite obvious, and it has hardly anything to do with the reverse Fatou lemma.
Indeed, for all $s\in[0,t]$, let
\begin{equation\*}
l\_n(s):=\sup\_{m\colon m\ge n}\ell^m(s),
\end{equation\*}
so that
\begin{equation\*}
\ell^n(s)\le l\_n(s)\downarrow l(s):=\limsup\_n\ell^n(s). \tag{0}
\end{equation\*}
So,
\begin{align\*}
&\limsup\_n P(\exists s\in[0,t]\ x+B\_s\le\ell^n(s)) \\
\le &\limsup\_n P(\exists s\in[0,t]\ x+B\_s\le l\_n(s)) \\
=&\lim\_n P(\exists s\in[0,t]\ x+B\_s\le l\_n(s))=P(A),
\end{align\*}
where
\begin{align\*}
A&:=\{\forall n\ \exists s\in[0,t]\ x+B\_s\le l\_n(s)\} \\
& =\{\forall n\ge m\ \exists s\in[0,t]\ x+B\_s\le l\_n(s)\};
\end{align\*}
here and in what follows, $m$ is any natural number.
So, it is enough to show that
\begin{equation\*}
P(A)\overset{\text{(?)}}\le P(C), \tag{1}
\end{equation\*}
where
\begin{equation\*}
C:=\{\exists s\in[0,t]\ x+B\_s\le l(s)\}.
\end{equation\*}
We shall actually show that
\begin{equation\*}
P(A\setminus C)\overset{\text{(?)}}=0, \tag{2}
\end{equation\*}
which will of course imply (1).
Suppose that event $A$ occurs. For all $n$, let
\begin{equation\*}
s\_n:=\inf\{s\in[0,t]\colon x+B\_s\le l\_n(s)\};
\end{equation\*}
of course, $s\_n$ is a random variable (r.v., with values in $[0,t]$ on $A$), depending on the random path of the Brownian motion $(B\_t)$.
Moreover, since $l\_n(s)\downarrow l(s)$ for all $s\in[0,t]$, we have
\begin{equation\*}
s\_n\uparrow s\_\*
\end{equation\*}
for some r.v. $s\_\*$, with values in $[0,t]$ on $A$.
Consider first the case when $A$ occurs and $s\_n<s\_\*$ for all $n$. Then for all $n$ there is some $t\_n\in[s\_n,s\_\*)$ such that $x+B\_{t\_n}\le l\_n(t\_n)$.
Also, $l\_n(s)$ is nondecreasing in $s\in[0,t]$. So, for all $n$, we have $x+B\_{t\_n}\le l\_n(s\_\*)$.
So,
\begin{equation\*}
x+B\_{s\_\*}=\lim\_n(x+B\_{t\_n})\le\lim\_n l\_n(s\_\*)=l(s\_\*).
\end{equation\*}
Thus,
\begin{equation\*}
A\cap\{\forall n\ s\_n<s\_\*\}\subseteq C. \tag{2.5}
\end{equation\*}
If $A$ occurs and $s\_n=s\_\*=t$ for some $n$ (and hence for all large enough $n$), then for such $n$ we have $x+B\_t\le l\_n(t)$ and hence $x+B\_t\le l(t)$. Thus,
\begin{equation\*}
A\cap\{\exists n\ s\_n=s\_\*=t\}\subseteq C. \tag{2.75}
\end{equation\*}
If $A$ occurs and if $s\_n=s\_\*<t$ for some $n$ (and hence for all large enough $n$) and if $s\_\*$ is a point of discontinuity of the function $l$, then for large enough $n$ we have $x+B\_{s\_\*}=x+B\_{s\_n}\le l\_n(s\_\*+)$, so that $x+B\_{s\_\*}\le l^+(s\_\*)$, where $l^+(s):=\lim\_n l\_n(s+)$.
So,
\begin{equation\*}
x+B\_d\le l^+(d) \tag{3}
\end{equation\*}
at some point $d\in D$, where $D$ is the set of all points of discontinuity of the nondecreasing function $l$; there are at most countably many such points.
Note that for all $u\in[0,t)$ and all $s\in(u,t]$ we have $l(s)=\lim\_n l\_n(s)\ge\lim\_n l\_n(u+)=l^+(u)$.
So, $l(s)\ge l^+(u)$ for all $u\in[0,t)$ and all $s\in(u,t]$.
Now suppose that $C$ does not occur, so that $x+B\_s>l(s)$ for all $s\in[0,t]$ and hence $x+B\_s> l^+(d)$ for each $d\in D$ and all $s\in(d,t]$. In view of, say, the (local) law of the iterated logarithm for the Brownian motion, for each $d\in[0,t)$ the event
$\{x+B\_d\le l^+(d),\ x+B\_s>l^+(d)\ \forall s\in(d,t]\}$ has the zero probability.
In view of (3) and because the set $D$ is at most countable, we conclude that
\begin{equation\*}
P(A\cap\{\exists n\ s\_n=s\_\*<t,s\_\*\in D\}\setminus C)=0. \tag{4}
\end{equation\*}
Suppose finally that $A$ occurs and $s\_n=s\_\*<t$ for some $n$ (and hence for all large enough $n$) and $s\_\*$ is a point of continuity of the function $l$. Take now any real $\ep>0$. Then there is some real $\de>0$ such that $l(s\_\*+\de)\le l(s\_\*)+\ep$. So, for all large enough $n$,
\begin{equation\*}
x+B\_{s\_\*}=x+B\_{s\_n}\le l\_n(s\_n+\de)=l\_n(s\_\*+\de)\to l(s\_\*+\de)\le l(s\_\*)+\ep.
\end{equation\*}
Letting now $\ep\downarrow0$, we get $x+B\_{s\_\*}\le l(s\_\*)$. So,
\begin{equation\*}
A\cap\{\exists n\ s\_n=s\_\*<t,s\_\*\notin D\}\subseteq C. \tag{5}
\end{equation\*}
Collecting (2.5), (2.75), (4), and (5), we confirm (2), as desired.
|
2
|
https://mathoverflow.net/users/36721
|
401084
| 164,657 |
https://mathoverflow.net/questions/401068
|
2
|
I'm studying the book *Etale cohomology and the Weil conjecture* by Freitag, Kiehl and I have some questions on the subchapter introducing the machinery associating to an étale presheaf
a sheaf (that is the "sheafification" procedure in étale world): see pages 11-13.
Let $A$ be a commutative, unital, Noetherian ring and $\operatorname{Et}(A)$ the category
of étale extensions of $A$ (étale means that the extensions $A \to B$ are flat, unramified
$A$-algebras of finite type).
We follow the notation from the book
and call as an étale presheaf (of set, abelian groups,...; for sake of simplicity say sets) a covariant functor
$$ \mathcal{F}: \operatorname{Et}(A) \to (Set), (Ab),... $$
Now we want to associate to $\mathcal{F}$ another étale presheaf $\widetilde{\mathcal{F}}$ as follows:
Let $B \in \operatorname{Et}(A)$ and
$\mathcal{B}:= (B \to B\_i, i \in I) $ an *étale cover of $B$*, that's a finite family
$(B \to B\_i, i \in I) $ of étale maps , when the images of $\operatorname{Spec} B\_i$
cover all elements of $\operatorname{Spec} B$.
(These form naturaly a category; a morphism $\mathcal{B} \to \mathcal{B}'$ between
$$ \mathcal{B}= (B \to B\_i, i \in I), \mathcal{B}'= (B \to B\_j', j \in J) $$
is given by a map $\sigma:J \to I$ inducing $B\_{\sigma(j)} \to B\_j'$ (datails can be worked out
easily).
Now we associate to an etale cover $\mathcal{B}= (B \to B\_i)$ the object $\mathcal{F}(\mathcal{B})$
defined as set of all $I$-tuples $(s\_i)\_{i \in I}$ with $s\_i \in \mathcal{F}(B\_i)$ such that
all $s\_i$ fulfil the **compatibility condition**,
that is for every $i,j$ should hold
$$f^l\_i(s\_i)= f^r\_j(s\_j),$$
where the involved maps are
$f^l\_i: \mathcal{F}(B\_i) \to \mathcal{F}(B\_i \otimes\_B B\_j)$
(respectively $f^r\_j: \mathcal{F}(B\_j) \to \mathcal{F}(B\_i \otimes\_B B\_j)$) are
naturally induced by family of canonical maps $ l\_i: B\_i \to B\_i \otimes\_B B\_j, b\_i \mapsto b\_i \otimes 1$,
(resp $ r\_j: B\_i \to B\_i \otimes\_B B\_j, b\_i \mapsto b\_i \otimes 1$ )
We define the presheaf $\widetilde{\mathcal{F}}$ as direct limit over all coverings of $B$
$$ \tilde{\mathcal{F}}(B):= \varinjlim\_{\mathcal{B}} \mathcal{F}(\mathcal{B})$$
(details like checking thats this is a functor etc. I omit here)
Now we come to my **Questions**. The book states some very important facts without
giving a reference. And I'm looking for proofs of several statements
there.
**Question 1a:** It is stated (p. 12) that for every $B \in \operatorname{Et}(A)$
and etale cover $\mathcal{B}= (B \to B\_i)$ the induced map
$\widetilde{\mathcal{F}}(B) \to \widetilde{\mathcal{F}}(\mathcal{B})$
is injective (note that $\widetilde{\mathcal{F}}$ is in general still
a presheaf). How can it be proved?
**Question 1b:** Futhermore it is claimed that if for all $B$ and
$\mathcal{B}= (B \to B\_i)$ the maps
$$ \mathcal{F}(B) \to \mathcal{F}(\mathcal{B}) $$
are *injective*, then $\widetilde{\mathcal{F}}$ is already an étale sheaf.
Why that's true? (Rmk.: combined with statement of Question 1a this
implies that the application of double tilde
$\widetilde{\widetilde{\mathcal{F}}}$ gives a sheaf)
**Question 2:** Why is the double tilde functor (= sheafification as
we saw above modulo the two questions) $\widetilde{\widetilde{\mathcal{F}}}$ is
exact? That is if we apply if to short exact sequence $0 \to \mathcal{A} \to \mathcal{B} \to \mathcal{C} \to 0$ of étale presheaves we obtain a ses of sheaves?
(recall, that by definition a sequence $0 \to \mathcal{A} \to \mathcal{B} \to \mathcal{C} \to 0$ of presheaves on site $\operatorname{Et}(A)$ is ses iff $0 \to \mathcal{A}(B) \to \mathcal{B}(B) \to \mathcal{C}(B) \to 0$ is ses for *every* $B \in \operatorname{Et}(A)$.)
Is $\widetilde{\mathcal{F}}$ exact as functor between etale presheaves?
|
https://mathoverflow.net/users/108274
|
Some facts about sheafification functor on étale site
|
A good trick for answering questions of this type is that there are multiple foundational resources in étale cohomology. If a detail isn't explained in one you're reading, you can quickly check another one - the notation should hopefully be similar enough that you can transfer the proof over. In particular, the Stacks project tends to give a lot of details, and I found [an answer to question 1](https://stacks.math.columbia.edu/tag/00WB) and [a partial answer to question 2](https://stacks.math.columbia.edu/tag/00WJ) there. (You should probably try to solve them as an exercise before doing this.)
The way I would say it.
Question 1a: Suppose we are given two elements of $\widetilde{\mathcal F}(B)$ whose image in $\widetilde{\mathcal F}(\mathcal B)$ is equal. We must show they are equal. By assumption, the two elements arise from covers $\mathcal A, \mathcal C$ of $\mathcal B$ and families of sections on these covers satisfying a compatibility condition. Their pullbacks to $\widetilde{\mathcal F}(\mathcal B)$ are given by the pullbacks of those families of sections to $\mathcal B \times\_{B} \mathcal A$ and $\mathcal B\times\_B \mathcal C$ respectively. If those are equal, then by the definition of direct limit, there is a common refinement where they are equal. But this common refinement is also a common refinement of $\mathcal A$ and $\mathcal C$ where our two families of sections are equal, so again by the definition of directly limit, the two starting elements of $\mathcal F(B)$ are equal.
Question 1b: Well, one must show that given a cover $U\_1,\dots, U\_n$ of $B$, the map from sections of $\widetilde{\mathcal F}$ on $B$ to compatible sections on $U\_1,\dots, U\_n$ is an isomorphism. So let's construct an inverse map. Given a compatible section of $\widetilde{\mathcal F}$ on a cover $U\_1,\dots, U\_n$, we obtain for each $i$ a cover $\mathcal U\_i$ of $U\_i$ and a compatible family of sections of $\mathcal F$ on that cover. We would like to make $\bigcup\_i \mathcal U\_i$ a cover of $B$ and this family of sections a compatible family. The only issue is that when we pass to an intersection $\mathcal U\_i \times \mathcal U\_j$, the pullbacks of the sections from $U\_i$ and $U\_j$ are only necessarily equal in $\widetilde{\mathcal F}$, not in $\mathcal F$. But given that $\mathcal F \to \widetilde{\mathcal F}$ is injective, we get the desired compatibility also in $\mathcal F$.
Question 2: No, $\tilde{F}$ is not exact as a functor between etale presheaves. To make a counterexample, take any exact sequence $0 \to \mathcal A \to \mathcal B \to \mathcal C \to 0$ of sheaves whose sequence of global sections $0 \to H^0( X, \mathcal A) \to H^0(X,\mathcal B)\to H^0(X,\mathcal C)$ is not right exact. Let $\mathcal C'$ be the quotient of $\mathcal B$ by $\mathcal A$ as presheaves, i.e. $\mathcal C'(U) = \mathcal B(U)/\mathcal A(U)$.
Then $0 \to \mathcal A \to \mathcal B \to \mathcal C' \to 0$ is a short exact sequence of presheaves. Its sheafification is simply $ \mathcal A \to \mathcal B \to \mathcal C$, which is a short exact sequence of sheaves, but not a short exact sequence of presheaves, since its global sections are not short exact.
Instead, the proof that sheafification is exact makes use of the fact that $\widetilde{\widetilde{\mathcal F}}$ is a left adjoint and therefore right exact, without proving right exactness for $\widetilde{\mathcal F}$.
|
2
|
https://mathoverflow.net/users/18060
|
401093
| 164,661 |
https://mathoverflow.net/questions/400769
|
5
|
Say $A\_0$ is an ordinary abelian variety over ${\mathbf{F}}\_q$. Call $\mathcal{A}$ the canonical lift of $A\_0$ over $R := W({\mathbf{F}}\_q)$. It carries a lift of the $q$-th power map on $A\_0$. We call $\phi : \mathcal{A}\to\mathcal{A}$ this lift. It exists by functoriality of the canonical lift.
Call $K = \text{Frac}(R)$ and choose a field embedding $K\subset\mathbf{C}$.
Call $A$ the complex torus $(\mathcal{A}\times\_K\mathbf{C})(\mathbf{C})$ and $F$ the endomorphism of $A$ induced by the Frobenius lift on $\mathcal{A}$, i.e.
$$F = (\phi \times\_{\text{Spec}(R)}\text{id}\_{\mathbf{\text{Spec}(C)}})^{\rm an}:A\to A.$$
$F$ acts on the Betti cohomology $H^\*(A,\mathbf{C})$. By the Weil conjectures (or by an argument of Serre, in this case), its eigenvalues are of the form $q^{\*/2}\zeta$ for algebraic numbers $\zeta$ of complex absolute value $1$.
>
> Suppose $v \in H^{2m}(A,\mathbf{C})$ is an eigenvector of $F$ whose eigenvalue is of the form $q^m\zeta$ for $\zeta$ a root of unity. Is $v$ a class of type $(m,m)$?
>
>
>
|
https://mathoverflow.net/users/nan
|
Ordinary abelian varieties and Frobenius eigenvalues
|
Let's examine how $\phi$ acts on the algebraic Dolbeaut cohomology $$H^1(\mathcal A\_K , \mathcal O\_{\mathcal A})+ H^0 ( \mathcal A\_K, \Omega^1\_{\mathcal A}).$$
I claim its eigenvalues on $H^1(\mathcal A\_K , \mathcal O\_{\mathcal A})$ are units and its eigenvalues on $H^0 ( \mathcal A\_K, \Omega^1\_{\mathcal A})$ are $q$ times units.
For the first claim, we can use the Artin-Schreier exact sequence $$H^1 ( A\_{0, \overline{\mathbb F\_q}}, \mathbb Z/p) \to H^1 ( A\_{0, \overline{\mathbb F\_q}} , \mathcal O\_{ A\_0} )\to H^1 ( A\_{0, \overline{\mathbb F\_q}} , \mathcal O\_{ A\_0} ) ,$$ the image of whose first arrow is a basis for $H^1 ( A\_{0, \overline{\mathbb F\_q}} , \mathcal O\_{ A\_0} )$ on which Frobenius acts invertibly, showing that Frobenius acts invertibly on $H^1 ( A\_0, \mathcal O\_{A\_0})$ and thus invertibly on $H^1 (\mathcal A, \mathcal O\_{\mathcal A})$.
For the second claim, we can use the fact that the pullback of a polarization along $\phi$ is $q$ times that polarization, so Frobenius is a symplectic similitude with similitude character $q$ for the form on $H^1(\mathcal A\_K)$ induced by that polarization, hence for each eigenvalue $\lambda$ that is a $p$-adic unit, $q/\lambda$ must also be an eigenvalue.
Using that claim and the isomorphism $$\wedge ^a H^1(\mathcal A\_K , \mathcal O\_{\mathcal A})\otimes \wedge^b H^0 ( \mathcal A\_K, \Omega^1\_{\mathcal A}) \to H^a ( \mathcal A\_K, \Omega^b\_{\mathcal A})$$ we conclude that all eigenvalues of $\Phi$ on $$H^a ( \mathcal A\_K, \Omega^b\_{\mathcal A})$$ are $q^b$ times a $p$-adic unit.
So any eigenvalues in degree $2m$ of the form $q^m$ times a unit must occur in $H^{m,m}$, as desired.
|
4
|
https://mathoverflow.net/users/18060
|
401096
| 164,662 |
https://mathoverflow.net/questions/401095
|
-1
|
This is inspired by a recent [math.SE question](https://math.stackexchange.com/questions/4215964/does-there-exist-n-m-in-mathbbn-such-that-lvert-left-frac32-ri).
Given that mathematicians like to come up with theoretical constructs which do not necessarily always have any practical purpose (but sometimes provide lots of fun anyway): to generalize the above question, let's define a "distance function" on reals greater than $1$ by $$d(x,y):=\inf\_\limits{m,n\in\mathbb N}|x^m-y^n|.$$ As $0\not\in\mathbb N$, this function is supposedly rather pathological. I think that certain values of $d(x,y)>0$ can at least be found as minima, e.g. if $x=\frac{\sqrt{5}+1}2$, then $x^n$ is close to a [Lucas number](https://en.wikipedia.org/wiki/Lucas_number), so if $y$ then is an integer (or a $k$th root of an integer), the question of finding $d(x,y)$ essentially boils down to the question whether [Lucas numbers can be infinitely often odd powers](https://mathoverflow.net/questions/342483/squares-in-lucas-sequences) (which is very probably not the case).
Maybe there are any other fun facts that can be said about the function $d$, starting with the question:
>
> Is it possible that $d$ is (against intuition) continuous in each variable?
>
>
>
Note that one could also ask:
>
> Does it satisfy the triangle inequality?
>
>
>
Now for that, we would obviously need to have $d(x,y)=d(x^k,y)=d(x,y^k)$ for all $x,y>1$ and all $k\in\mathbb N$, but as soon as there is a $d(x,y)>0$ given by a minimum (as probably for the case above), this fails.
Any other ideas?
|
https://mathoverflow.net/users/29783
|
A pathological (?) function involving powers
|
The answer to both of your questions is no.
Notice that $d(2, 3) = 1$. However, for any $\varepsilon > 0$, there exists a number $2 - \varepsilon < x < 2$ such that $d(x, 3) = 0$ and thus it is not a continuous function. To see this, take $n$ to be some large positive integer and $m$ the unique integer satisfying
$$2^{m - 1} < 3^n < 2^m$$
or equivalently
$$\left( 2^{1 - \frac{1}{m}} \right)^m < 3^n < 2^m$$
therefore there is a number $2^{1 - \frac{1}{m}} < x < 2^m$ such that $x^m = 3^n$ and therefore $d(x, 3) = 0$. Clearly, taking $n \to \infty$ we can let $x$ be arbitrarily close to $2$. In fact, this argument shows that for any $y > 1$, the set of $x > 1$ such that $d(x, y) = 0$ is dense in the set of numbers that are larger than $1$.
As for the triangle inequality, it is easy to see that $d(2, 3) = 1$ and $d(2, 16) = 0$. Furthermore, $d(3, 16) > 1$ because the only solutions to the equation
$$2^m - 3^n = \pm 1$$
in positive integers $m, n$ are $(m, n) = (1, 1), (2, 1), (3, 2)$, which shows that there are no solutions to
$$16^m - 3^n = \pm 1$$
and, thus, $d(3, 16) > d(3, 2) + d(2, 16)$.
|
1
|
https://mathoverflow.net/users/88679
|
401097
| 164,663 |
https://mathoverflow.net/questions/327495
|
7
|
I am interested in two related constructions which give us either the cohomology or the $T \times \mathbb{C}^\*$-equivariant $K$-theory of flag varieties.
Let $G$ be a semisimple, simply connected algebraic group, with $T \subset B \subset G$ a chosen maximal torus and Borel subgroup. In order to gain geometric information about the flag variety $G/B$, we make use of a collection of $\mathbb{P}^1$ bundles. To be a little more explicit:
In what I would (perhaps erroneously) call the Bernstein-Gelfand-Gelfand approach, to find the cohomology of $G/B$, we would use minimal parabolic subgroups $P\_i$ and maps $G/B \rightarrow G/P\_i$. This gives $G/B$ the structure of a $\mathbb{P}^1$-bundle over $G/P\_i$. In this situation we can use the Leray-Serre spectral sequence to obtain $H^\*(G/B)$ in terms of $H^\*(\mathbb{P}^1)$ and $H^\*(G/P\_i)$, and in particular we can get our hands on classes $[\overline{X}\_{s\_i}]$ where $X\_{s\_i}$ is the Schubert cell associated to a simple reflection $s\_i$.
In the construction of $K^{T \times \mathbb{C}^\*}(G/B)$, following Ginzburg (and possibly originally due to Kazhdan and Lusztig?), we construct a different $\mathbb{P}^1$-bundle. Namely, the $G$-diagonal orbits in $G/B \times G/B$ are parametrized by $w \in W$. Let $Y\_{s\_i}$ denote the orbit associated to $(B/B, s\_iB/B)$, and $\overline{Y\_{s\_i}}$ its orbit closure. Then via projection onto the first factor, $\pi\_1 :\overline{Y\_{s\_i}} \rightarrow G/B$ is a $\mathbb{P}^1$-bundle. Ginzburg goes on to construct all sorts of sheaves in this setup, and constructs the affine Hecke algebra geometrically.
The two constructions should be related in the following way: "The $T\_{s\_i}$ action on $K^{T \times \mathbb{C}^\*}(T^\*G/B)$ is given by $e^{\lambda} \mapsto \frac{e^{\lambda}-e^{s\_{i} (\lambda)}}{e^{\alpha\_i}-1}-q \frac{e^{\lambda}-e^{s\_{i}(\lambda)+\alpha\_i}}{e^{\alpha\_i}-1}.$ The fraction on the left is the Demazure operator associated to $s\_i$, which is used to find the $K$-theory of $G/B$" (here I am paraphrasing from Chriss-Ginzburg Thm 7.2.16).
I know I am mixing cohomology and all sorts of $K$ theory here, but it seems that there should be a more accessible topological relationship.
My question is: on the most simple (purely topological) level, how do these two distinct $\mathbb{P}^1$-bundles give us similar information about the cohomology (or $K$-theory) of $G/B$?
I feel like I've simultaneously included too many details and left out too many details, and I'd be happy to edit for clarification.
|
https://mathoverflow.net/users/119460
|
Geometric interpretations of nil-Hecke ring and affine Hecke algebra
|
The subvariety $\overline{Y}\_{s\_i}\subset G/B \times G/B$ is the fiber product $G/B\times\_{G/P\_i}G/B$. The set $\overline{Y}\_{s\_i}$ is the saturation for the diagonal $G$-action of $\{B/B\}\times P\_i/B$, by definition, and of course, that also lies in the fiber product; since they are smooth irreducible varieties of the same dimension, inclusion shows they are equal.
|
4
|
https://mathoverflow.net/users/66
|
401106
| 164,666 |
https://mathoverflow.net/questions/401108
|
6
|
I am interested to know examples of topological groups $G$ for which the intersection $\bigcap\{H\leq G\mid H\text{ open}\}$ of all open subgroups of $G$ is the trivial subgroup but for which the intersection $\bigcap\{N\trianglelefteq G\mid N\text{ open}\}$ of all open *normal* subgroups is not the trivial subgroup.
Clearly (1) must be totally disconnected (2) cannot inject into a pro-discrete group by a continuous homomorphism and (3) it can't contain a topological subgroup isomorphic to $\mathbb{Q}$. I imagine that topological groups fitting this description exist and perhaps some are even important in some area I am not familiar with.
Does such a topological group exist? If so, is there an abundance of "standard" examples?
|
https://mathoverflow.net/users/5801
|
Intersection of all open subgroups vs. the intersection of all open normal subgroups
|
$S\_\infty$, the group of all permutations of $\mathbb{N}$, has a neighborhood base of the identity of open subgroups. (In fact a Polish group with that property is isomorphic to a closed subgroup of $S\_\infty$).
But without thinking about exactly which ones are open, $S\_\infty$ has a very limited supply of normal subgroups outlined here: <https://math.stackexchange.com/a/166472/29633>
|
6
|
https://mathoverflow.net/users/6342
|
401109
| 164,667 |
https://mathoverflow.net/questions/401036
|
2
|
Let $V$ be a Ternary rings of operators(TRO) i.e. closed subspace of $B(H,K)$ such that $xy^\*z \in V$ for all $x,y,z \in V$. A subspace $I$ of $V$ is called a left (right)TRO ideal provided $VV^\*I \subset I$$(IV^\*V \subset V)$.
>
> Sum of closed left and right ideals is closed provided one of the ideal has bounded approximate identity.
>
>
>
I was informed that results on sum of closed TRO ideals is known from some old paper(~1974) of Kirchberg but unfortunately I could not find it. Can someone please help me to trace that paper? Thank you so much.
|
https://mathoverflow.net/users/129638
|
Looking for an old paper of Kirchberg
|
I guess you don't need approximate identity assumption to prove it (in any case what does approximate identity mean for a TRO?). I am not sure which paper of Kirchberg you are referring to, but indeed Kirchberg has proved that the sum of a closed left ideal and a closed right ideal is closed (for $C^\*$-algebras) and this fact generalizes to TROs ($C^\*$-spaces in Kirchberg's terminology) via the linking algebra construction. For the proof, see Section 4 of Kirchberg's paper "On restricted perturbations..." JFA 1995 (<https://mathscinet.ams.org/mathscinet-getitem?mr=1322640>).
Added: Associated with a closed left TRO ideal $I$ is a closed left ideal
$$L:=\left[\begin{matrix} [VI^\*] & I \\ I^\* & [V^\*I]\end{matrix}\right]$$
of the linking $C^\*$-algebra
$$A:=\left[\begin{matrix} [VV^\*] & V \\ V^\* & [V^\*V]\end{matrix}\right].$$
Likewise for a right TRO ideal.
|
1
|
https://mathoverflow.net/users/7591
|
401113
| 164,668 |
https://mathoverflow.net/questions/401070
|
-3
|
This question is a follow-up to [Are there infinitely many L-rigs?](https://mathoverflow.net/questions/372349/are-there-infinitely-many-l-rigs) which is already pretty convoluted.
Define the $\varphi$-evaluation morphism at a complex number $s$ as $\epsilon\_{\varphi,s}:F\mapsto \varphi(F)(s)$ where $F$ is a map from $\mathbb{C}$ to itself. Consider a "general covariance condition" to naturally constrain the elements $\varphi$ of the automorphism group $G\_{\mathcal{L}}$ of some L-rig $\mathcal{L}$ by requiring this group to be invariant under the map $\iota$ that maps $\varphi$ to $\varphi^{-1}$ as well as under any element $g$ of $\operatorname{Aut}(G\_{\mathcal{L}})$, so we require the equality $\epsilon\_{\varphi,s}=\epsilon\_{g(\varphi),s}$ to hold for all $(g,\varphi,s)$. That way we'd get in particular $\varphi(F)(s)=\varphi^{-1}(F)(s)$, so that every automorphism of a given L-rig $\mathcal{L}$ would be of order at most $2$. Can we prove that $G\_{\mathcal{L}}$ is itself of order at most $2$?
|
https://mathoverflow.net/users/13625
|
Structure of the automorphism group of an L-rig
|
This an answer following an argument from Wojowu: as we require the equality $\epsilon\_{g(\varphi),s}=\epsilon\_{\varphi,s}$ to hold for all $(g,\varphi,s)$, and thus for all $s$, this means that $g(\varphi)=\varphi$ for all $\varphi$, so that $g$ is the identity. So the automorphism group of $G\_{\mathcal{L}}$ is trivial and $G\_{\mathcal{L}}$ is of order at most $2$.
|
0
|
https://mathoverflow.net/users/13625
|
401118
| 164,669 |
https://mathoverflow.net/questions/401125
|
4
|
In his paper "Paul Levy's Isoperimetric Inequality" (published as appendix C in *Metric Structures for Riemannian and Non-riemannian Spaces*), Gromov claims that if $H$ is a minimal $n$-dimensional hypersurface dividing a Riemannian into two pieces of fixed volume, $v$ is any point and $h \in H$ satisfies $dist(H,v)=dist(h,v)$, then $h$ is a non-singular point of $H$. This is because there is a sphere, centred at the midpoint of the geodesic from $h$ to $v$, which is tangent to $H$ at $h$, and hence the tangent cone to $H$ at $h$ is contained in a half-space.
Gromov states that hence, the point $h$ is non-singular. As justification, Gromov cites Almgren's 1976 book *Existence and regularity almost everywhere of solutions to elliptic variational problems with constraints*. I tried to follow this reference, but I found Almgren's book very hard to follow.
How does Almgren prove that an isoperimetric surface is nonsingular at any point where the tangent space is contained in a half-space? Are there any more accessible books or papers which contain this result? In particular, would it apply to arbitrary $(\Lambda,r\_0)$ surfaces, or only to isoperimetric surfaces?
|
https://mathoverflow.net/users/106263
|
Nearest point is always regular for isoperimetric hypersurfaces
|
I think you're inadvertently opening a big can of worms. The question can be answered by a combination of two facts: the absence of branch points in (almost-)minimising hypersurfaces and Allard's regularity theorem.
Specifically, the tangent cones to $H$ at $h$ must be multiples of an $n$-dimensional hyperplane $P$ say, with some multiplicity $Q \in \mathbf{Z}\_{>0}$. The tangent cones cannot be more complicated minimal cones, as for example a Frankel-type argument demonstrates. Let $$ \mathbf{C} \in \mathrm{VarTan}(H,h)$$ be a tangent cone to $H$ at $h$: this is a (singular) minimal surface. Knowing that $\mathbf{C}$ is a stationary varifold is enough for now. By construction the cone is supported in a closed half-space, for example $$\mathrm{spt} \, \mathbf{C} \subset \{ X \in \mathbf{R}^{n+1} \mid X^{n+1} \geq 0 \}.$$
The intersection of $\mathbf{C}$ with the unit sphere $\partial B$ defines a stationary varifold contained in a hemisphere. Now on the one hand, as $\partial B$ has positive Ricci curvature, $\mathrm{spt} \, \mathbf{C}$ and $\partial B \cap \{ X^{n+1} = 0 \}$ must intersect: this is Frankel's theorem. On the other hand, this intersection must be tangential, and the maximum principle forces them to coincide: $$\mathrm{spt} \, \mathbf{C} = \{ X^{n+1} = 0 \}.$$
Therefore letting $P = \{ X^{n+1} = 0 \}$ one has $$\mathbf{C} = Q \lvert P \rvert.$$
When $H$ is minimising (or almost-minimising), then it cannot have branch point singularities, and necessarily $Q = 1$.
Therefore the tangent cones are multiplicity one tangent planes, and by Allard regularity $H$ must be smooth in a neighbourhood of the point $h$.
|
2
|
https://mathoverflow.net/users/103792
|
401127
| 164,672 |
https://mathoverflow.net/questions/397619
|
15
|
I have been wrestling with the following problem for some time now. If possible, I would prefer a hint rather than a full solution, because I would like to "solve" it myself.
Let $f(z)$ be a power-series and $[z^n]\{-\}$ denote the $n$'th coefficient. Show that the following holds, whenever $[z^0]{f(z)}=1$:
$$
\exp\left[\sum\_{n,m>0}\sum\_{j>0}j[z^{n+j}]\{f(z)^n\}[z^{m-j}]\{f(z)^m\} \frac{q^{n+m}}{(nm)}\right] = \sum\_{n\geq 0}[z^n]\{f(z)^{n-1}\}q^n\,.
$$
**Remark:** This identity appears in comparing two generating series of the same geometric invariants computed using different methods.
**Edit 1:** Alternatively, if one wishes to get rid of the exponential, one can take a logarithm of the equation and derivative with respect to $q$. Then we are left to solve the following:
$$
\left[\sum\_{n,m>0}\sum\_{j>0}j[z^{n+j}]\{f(z)^n\}[z^{m-j}]\{f(z)^m\} \frac{(n+m)}{(nm)}q^{n+m}\right]\sum\_{l\geq 0}[z^{l}]\{f(z)^{l-1}\}q^l = \sum\_{n\geq 0}n[z^{n}]\{f(z)^{n-1}\}q^n\,.
$$
**Edit 2:** To avoid it being pointed out again (also see Timothy's answer). It is easy to rewrite the identity using Lagrange inversion (this is where this identity comes from in the first place). The issue appears when one is only allowed to sum over $n\in \mathbb{Z}\_{>0}$. Timothy used for this the notation $[q^{>0}]\{w(q)^{-j}\}$. I have been unable to do anything with this expression, hence the more complicated form above.
|
https://mathoverflow.net/users/109370
|
Comparing two power-series
|
I decided to summarize the two main proofs that I liked. The first one was motivated by the answer by esg. The second one can be found in the answer of Alex Gavrilov and is made more explicit. I am very grateful for their help.
**1. proof:**
Using Gessel [(2.4.4)](https://arxiv.org/abs/1609.05988) and the unique solutions $H(q)=qf(q)$, we can write
\begin{align\*}
&\sum\_{j>0}\sum\_{n,m>0}[z^{n+j}]\{f(z)^n\}[z^{m-j}]\{f(z)^m\}\frac{p^n}{n}\frac{q^m}{m}= -\sum\_{j>0}\frac{1}{j}H^j(p)H^{-j}(q) \\
&-\sum\_{j>0}\sum\_{-j\leq n<0}j[z^{n+j}]\{f(z)^n\}[z^{m-j}]\{f(z)^m\}\frac{p^n}{n}\frac{q^m}{m}-\sum\_{j>0}[t^{-1}]\big\{t^{-j-1}\log\Big(R(t)\Big)\big\}H^j \\
&=\log\Big(1-\frac{H(p)}{H(q)}\Big)+ \sum\_{n>0}\sum\_{j\geq n}[z^{j-n}]\big\{R^{-n}(z)\big\}H^jp^{-n}+\log\Big(R(H)\Big)
\end{align\*}
Therefore we obtain after taking exponential and limit $p\to q$:
$$
\lim\_{p\to q}\frac{H(p)-H(q)}{p-q}\frac{1}{R\big(H(q)\big)}\frac{p}{H(p)} =\Big(\frac{dq}{dH}\Big)^{-1}\Big(\frac{q}{H(q)}\Big)^2=\frac{1}{f\big(H(q)\big)}\Big(1-qf'(H)\Big)\,.
$$
The last term is clearly the right hand side by the Lagrange inversion formula.
**2. proof**
Without loss of generality assume that $f(q)$ is a polynomial in $q$. We will use the following identity
\begin{align\*}
[q^{>0}]H^{-j}(q)&=\sum\_{n>0}[z^n]\big\{H^{-j}(z)\big\}q^n = \frac{1}{2\pi i} \oint\_{|z|=R}\sum\_{n>0}\Big(\frac{q}{z}\Big)^n\frac{1}{z}H^{-j}(z)dz\\
&= \frac{1}{2\pi i} \oint\_{|z|=R}\Big(\frac{1}{z-q}-\frac{1}{z}\Big)H^{-j}(z)dz\,.
\end{align\*}
Then we get
$$
\sum\_{j>0}\frac{1}{j}H^j(q)[q^{>0}]H^{-j}(q)=\frac{1}{2\pi i}\oint\_{|z|=R}\Big(\frac{1}{z-q}-\frac{1}{z}\Big)\log\Big(\frac{H(z)}{H(z)-H(q)}\Big)dz
$$
Note that
\begin{align\*}
&\frac{1}{2\pi i}\oint\_{|z|=R}\Big(\frac{1}{z-q}-\frac{1}{z}\Big)\log\Big(\frac{z}{z-q}\Big)dz\\
&= -\int\_{0}^{2\pi}\Big(\frac{\frac{r}{R}e^{2\pi i(\theta-\tau)}}{1-\frac{r}{R}e^{2\pi i(\theta-\tau)}}\Big)\log\Big(1-\frac{r}{R}e^{2\pi i(\theta-\tau)}\Big)d\tau=0\,,
\end{align\*}
where the integral vanishes, because it is proportional to the integral of a total derivative of a $2\pi$ periodic function.
Therefore, we may work instead with
$$
\frac{1}{2\pi i}\oint\_{|z|=R}\Big(\frac{1}{z-q}-\frac{1}{z}\Big)\log\Big(\frac{H(z)\big(z-q\big)}{z\big(H(z)-H(q)\big)}\Big)dz\,.
$$
Taking the residues of this integrand at $z=q$ and $z=0$, we obtain again
$$
-2\log\Big(\frac{H(q)}{q}\Big)-\log\big(H'(q)\big)\,.
$$
|
1
|
https://mathoverflow.net/users/109370
|
401129
| 164,673 |
https://mathoverflow.net/questions/401130
|
4
|
Let $\pi \colon E \rightarrow \mathbb{CP}^1$ be a complex vector bundle. It is a well-known fact that a Dolbeault operator on $\pi\colon E \rightarrow \mathbb{CP}^1$ gives a holomorphic structure on $E$.
My questions are derived from this fact:
1. Let $\{D\_t\}\_{t\in [0,1]}$ be a smooth family of Dolbeault operators on $\pi \colon E \rightarrow \mathbb{CP}^1$. This family induces a family of holomorphic vector bundles whose underlying complex vector bundle is the given one. Are these holomorphic vector bundles isomorphic as holomorphic vector bundles?
2. Suppose that the answer to the first question is no in general. What condition on $\{D\_t\}\_{t\in [0,1]}$ except for a constant family gives us a family of isomorphic holomorphic vector bundles?
Any comments and references are welcome. Thank you in advance.
|
https://mathoverflow.net/users/41200
|
Family of Dolbeault operators on complex vector bundles over $\mathbb{CP}^1$
|
The answer to the first question is in fact no: consider the family of Dolbeaut operators on the complex vectorbundle of degree 0 and rank 2 underlying $$V=\mathcal O(-1)\oplus\mathcal O(1).$$ Consider the family of operators $$\bar\partial^t=\begin{pmatrix}\bar\partial^{\mathcal O(-1)} & t\, \gamma \\0& \bar\partial^{\mathcal O(1)} \end{pmatrix}$$ parametrised by $t\in\mathbb C$, where $\gamma\in\Gamma(CP^1,\bar K K)=\Omega^2(CP^1,\mathbb C)$ is a 2-form with non-vanishing integral. By Serre-duality the bundle $\mathcal O(1)$ is a holomorphic subbundle w.r.t. the holomorphic bundle defined by the Dolbeaut operator $\bar\partial^t$ if and only if $t=0$.
Concerning 2: You should compute the dimensions $D(d)$ of $H^0(CP^1, (V,\bar\partial^t)\otimes \mathcal O(d))$ for some $d\in\mathbb Z.$ If the dimension is constant for all $d\in\mathbb Z$ the bundles are isomorphic by Grothendieck splitting. But it is enough to restrict to some $d$'s, depending on the actual situation. For example, if the holomorphic bundle is trivial at one point in the (connected) family, then all bundles are isomorphic if the dimension $D(-1)=0$ stays trivial.
|
2
|
https://mathoverflow.net/users/4572
|
401134
| 164,674 |
https://mathoverflow.net/questions/400967
|
15
|
I have a vague memory of an infinite game due to Ernst Specker with
the following properties:
(1) It is a two-person perfect information game, where the players
move alternately.
(2) The possible moves depend only on the current position.
(3) There is no winning strategy where each move is based only on
the current position
(4) There is a winning strategy (for one of the players) if the
strategy uses the current position and the history of previous moves.
Can someone provide a reference or explanation for this game?
|
https://mathoverflow.net/users/2807
|
An infinite game possibly due to Ernst Specker
|
I don't know about the game attributed to Specker, but here is a
simple game with your desired features.
Let us call it the **Chocolatier's game**. There are two players,
the Chocolatier and the Glutton. To begin play, the Chocolatier
serves up finitely many unique and exquisite chocolate creations on
a platter, and then the Glutton chooses one of them to eat. Play
continues — at each stage the Chocolatier adds finitely many
additional chocolates to the platter, and the Glutton consumes one
of those available. The excess uneaten chocolates accumulate on the platter as play progresses.
After infinite play, the Glutton wins if every single chocolate
that was served was eventually consumed. Otherwise, the Chocolatier
wins.
The Glutton can easily win simply by paying attention to the order
in which the chocolates were added, and consuming them in that
order. If only finitely many chocolates are added at each stage, the Glutton should simply make sure to consume them before moving on to the
chocolates that were added at later stages.
Indeed, the Glutton can win even if the Chocolatier places countably many chocolates on each turn. At turn $n=2^k(2m+1)$, let the Glutton eat the $k$th
chocolate added at stage $m$, if any, and otherwise eat
arbitrarily.
But what about strategies that depend only on the current position,
that is, the assortment of chocolates on the platter?
In one sense, it is easy to see that there can be no such winning
strategy for the Glutton. If the Glutton will choose a particular
chocolate from a given assortment, then let the Chocolatier simply
replace it with an identical chocolate type on the next move, and again on all subsequent moves. If
the strategy does not know the history, then the Glutton will
choose it again every time, and the other chocolates will never be
eaten.
Perhaps it makes a more interesting game, however, to say that the Chocolatier loses if chocolate types are ever repeated. And for this version of the game, there are some interesting things to say.
If there are only countably many chocolate types available in all,
then again the Glutton has a winning strategy that depends only on
the current assortment on offer. Namely, the Glutton should fix an
enumeration of the possible chocolate types that might appear, and
at each stage select the chocolate that appears earliest in this
order — it is the tastiest-looking chocolate as defined by
that priority. With this strategy he will succeed in eating all the
chocolates, because at the limit, if any chocolate was left, there
would have to be a tastiest-looking one (earliest in the
enumeration), and this would have been eaten once the finitely many
tastier chocolates had been consumed.
If the Chocolatier were uncountably creative, however, and able to
serve up uncountably many different chocolate types, then this
argument breaks down. Indeed, in this case I claim there is no
winning strategy for the Glutton that depends only on the chocolate
assortment on offer. And indeed, there is no such strategy that
works even if we should insist that the Chocolatier present an
assortment only of size two at each stage.
To see this, suppose the Glutton will follow a fixed strategy that
selects a particular chocolate to eat from amongst any two
chocolates.
For any given chocolate $c$, if there are infinitely many others
$d\_n$ that would be preferred to it by the strategy, if presented
as a pair $\{c,d\_n\}$, then the Chocolatier can present these pairs
$\{c,d\_0\}$, $\{c,d\_1\}$, $\{c,d\_2\}$, in turn. At each stage,
$d\_n$ would be consumed and the Chocolatier can present
$\{c,d\_{n+1}\}$. In the end, the inferior chocolate $c$ would never
have been selected and so the Chocolatier will win.
So if this is a winning strategy for the Glutton, then we may
assume that every chocolate is preferred to all but finitely many
of the others. But this is simply impossible with an uncountable
set. To see this, take any countably infinite set of chocolates and
close under the finite witnesses of strictly preferred chocolates.
One thereby constructs a countably infinite set $D$ of chocolates
so that any chocolate in $D$ is preferred to any chocolate not in
$D$. Since there were uncountably many chocolate types, there is
some $c\notin D$. This contradicts our assumption that there are
only finitely many chocolates preferred to a given chocolate.
Let me remark that in the version of the game where we do not allow
the Chocolatier to repeat chocolate types (as opposed to saying that this is allowed, but causes the Chocolatier to lose), then we should really
include the list of already-consumed chocolates as part of the
position, and this complicates the arguments above. (I asked a followup question about this at [The Chocolatier's game](https://mathoverflow.net/q/401151/1946).) I suspect but
do not yet know that the Glutton can have no winning strategy that
depends only on these more general kinds of positions in the uncountable case.
|
23
|
https://mathoverflow.net/users/1946
|
401136
| 164,675 |
https://mathoverflow.net/questions/401140
|
2
|
Let $\epsilon\_1,...,\epsilon\_n$ be i.i.d. random signs, $\mathbf{u}\_1,...,\mathbf{u}\_n$ i.i.d. uniform random vectors on the unit sphere $\mathbb{S}^{d-1}$, assuming $d$ even, and $\mathbf{v}\_1,...,\mathbf{v}\_n$ be their half-truncations, that is $\mathbf{v}\_i[j] = \mathbf{u}\_i[j]$ for all $j \in \left \{1,...,d/2\right\}$, else $\mathbf{v}\_i[j] = 0$, where $\mathbf{v}\_i[j]$ denotes the $j$-th entry of $\mathbf{v}\_i$. We assume that the $\epsilon\_i$'s are independent from the $\mathbf{u}\_i$'s.
I am looking for nontrivial lower bounds on the following quantity.
$$\mathbb{E}\left |\sum\_{i = 1}^n \epsilon\_i \| \mathbf{v}\_i\|^2\_2 \right |$$
Any help would be greatly appreciated!
|
https://mathoverflow.net/users/334327
|
Lower bounds on random process
|
$\newcommand\ep\epsilon\newcommand\v{\mathbf v}$Conditioning on the $\ep\_i$'s and using Jensen's and [Szarek's](http://pldml.icm.edu.pl/pldml/element/bwmeta1.element.bwnjournal-article-smv58i1p13bwm) inequalities, we have
$$E\Big|\sum\_1^n\ep\_i\|\v\_i\|\_2^2\Big|
\ge E\Big|\sum\_1^n\ep\_ic\_i\Big|\ge\frac1{\sqrt2}\sqrt{\sum\_1^nc\_i^2},$$
where
$$c\_i:=E\|\v\_i\|\_2^2=1/2,$$
by symmetry. So,
$$E\Big|\sum\_1^n\ep\_i\|\v\_i\|\_2^2\Big|
\ge \frac{\sqrt n}2.$$
---
This lower bound is of course sharp, up a universal constant factor, because
$$E\Big|\sum\_1^n\ep\_i\|\v\_i\|\_2^2\Big|
\le\sqrt{E\Big(\sum\_1^n\ep\_i\|\v\_i\|\_2^2\Big)^2}
=\sqrt{\sum\_1^n E\|\v\_i\|\_2^4}
= \sqrt{n E\|\v\_1\|\_2^4}
\le\sqrt n.$$
|
3
|
https://mathoverflow.net/users/36721
|
401142
| 164,676 |
https://mathoverflow.net/questions/401149
|
3
|
Given a group $G$, suppose $G$ admits a non-elementary acylindrical action
on a Gromov hyperbolic space $S$.
I heard that stabilizer of a pair of points on $\partial S$ in the acylindrically hyperbolic group is either finite or virtually cyclic but couldn't find a reference. I wonder if someone knows where it is and could tell me.
|
https://mathoverflow.net/users/104837
|
The stabilizer of a pair of points in the acylindrically hyperbolic group is either finite or virtually cyclic
|
I do not know a reference where this statement exactly is proved, but Theorem 1.1 from Osin's article *Acylindrically hyperbolic groups* does most of the work.
It implies that, if the stabiliser $H$ of a pair of points at infinity $\alpha,\omega \in \partial S$ is not virtually infinite cyclic, then it has bounded orbits. But $H$ already quasi-preserves a (quasi-)geodesic $\gamma$ between $\alpha$ and $\omega$. Some basic hyperbolic geometry then implies that $H$ actually quasi-fixes $\gamma$, and it follows from the acylindricity condition that $H$ must be finite.
|
2
|
https://mathoverflow.net/users/122026
|
401155
| 164,679 |
https://mathoverflow.net/questions/401147
|
-1
|
Consider the projective plane $\mathbb{P}^2\_{\overline{\mathbb{C}(t)}}$ over the algebraic closure of the function field $\mathbb{C}(t)$.
Take the point $p\_0 = [0:1:0]\in \mathbb{P}^2\_{\overline{\mathbb{C}(t)}}$ and eight more general points $p\_1,\dots,p\_8\in \mathbb{P}^2\_{\overline{\mathbb{C}(t)}}$ which are not defined over $\mathbb{C}(t)$. Assume that there is a cubic curve $C \subset \mathbb{P}^2\_{\overline{\mathbb{C}(t)}}$, defined over $\mathbb{C}(t)$, passing through $p\_0,p\_1,\dots,p\_8$. We can write $C$ as
$$C = \{ \alpha\_0(t) x^3 + \alpha\_1(t)x^2y + \alpha\_2(t)x^2z + \alpha\_3(t)xy^2 + \alpha\_4(t)xyz + \alpha\_5(t)xz^2 + \alpha\_6(t)y^3 + \alpha\_7(t)y^2z + \alpha\_8(t)yz^2 + \alpha\_9(t)z^3 = 0 \}$$
where $\alpha\_0,\dots,\alpha\_9$ are polynomials in $t$, and extend $C$ to a fibration on $\mathbb{P}^1$ as
$$C' = \{A\_0(s,t) x^3 + A\_1(s,t)x^2y + A\_2(s,t)x^2z + A\_3(s,t)xy^2 + A\_4(s,t)xyz + A\_5(s,t)xz^2 + A\_6(s,t)y^3 + A\_7(s,t)y^2z + A\_8(s,t)yz^2 + A\_9(s,t)z^3 = 0 \}$$
where now $A\_0,\dots,A\_9$ are homogeneous polynomials on $\mathbb{P}^1$.
Can we find such a curve $C'$ such that the polynomials $A\_0,\dots,A\_9$ appearing in the expression of $C'$ are of degree one or at least $A\_9$ is of degree one?
|
https://mathoverflow.net/users/14514
|
Coefficients of elliptic curves over function fields
|
Are you asking if, for all tuples $p\_1,\dots, p\_8$, there exists such a $C'$ with $A\_9$ of degree one? This is false, assuming $A\_9=0$ does not count as degree one.
We can for example choose one of the $p\_i$ to equal $[0: 0 : 1]$ for $t=1$ and and one to equal $[0,0,1]$ for $t=2$. Then regardless of which $C'$ we take, $A\_9$ will equal $0$ for $t=1$ and $t=2$ and thus cannot have degree one.
|
1
|
https://mathoverflow.net/users/18060
|
401164
| 164,685 |
https://mathoverflow.net/questions/401157
|
7
|
Consider the category $\mathbf{Top}$ of topological spaces, the category $\mathbf{Topos}$ of toposes and geometric morphisms, and the category $\mathbf{Loc}$ of locales. Let
$$\mathrm{Sh}\colon\mathbf{Top}\to\mathbf{Topos}$$
be the functor sending a space $X$ to the topos of sheaves on $X$. Does this functor have a left or a right adjoint?
Of course, $\mathrm{Sh}$ factorizes as
$$\mathbf{Top}\to\mathbf{Loc}\to\mathbf{Topos},$$
since the sheaves on $X$ are the sheaves on the locale of open subsets of $X$.
It is well-known that $\mathbf{Top}\to\mathbf{Loc}$ has a right adjoint and $\mathbf{Loc}\to\mathbf{Topos}$ has a left adjoint ($\mathbf{Loc}$ is reflective in $\mathbf{Topos}$). So the question whether $\mathrm{Sh}$ has a left adjoint reduces to the question whether $\mathbf{Top}\to\mathbf{Loc}$ has a left adjoint and the question whether $\mathrm{Sh}$ has a right adjoint reduces to the question whether $\mathbf{Loc}\to\mathbf{Topos}$ has a right adjoint.
*Remark.* I already asked this on math.SE but I didn't get an answer.
|
https://mathoverflow.net/users/333306
|
Does the functor $\mathrm{Sh}\colon\mathbf{Top}\to\mathbf{Topos}$ have an adjoint?
|
In this answer, Topos is interpreted as a 2-category.
(As a side remark, the 1-category of toposes does not make sense
until one picks a specific model for toposes and geometric morphisms, and different models
need not be equivalent as 1-categories.
For the 1-categorical framework to make sense, at the very least
one needs to organize toposes into a relative category,
so that different models can be shown to be Dwyer–Kan equivalent as relative categories.)
>
> whether Top→Loc has a left adjoint
>
>
>
The functor Top→Loc does not have a left adjoint because it does not preserve finite products.
For example, the product of rational numbers with themselves
as locales and as topological spaces produces nonisomorphic locales.
In particular, rational numbers form a topological group, but not a localic group.
>
> whether Loc→Topos has a right adjoint
>
>
>
Loc→Topos does not have a right adjoint because it does not preserve homotopy colimits.
For example, suppose G is a discrete group acting on a point.
Then the colimit of this action in locales is again a point.
But the homotopy colimit of this action in toposes
is the delooping of G, which is not equivalent to a point.
Thus, both Loc→Topos and Top→Topos do not have a right adjoint functor.
|
9
|
https://mathoverflow.net/users/402
|
401166
| 164,687 |
https://mathoverflow.net/questions/401103
|
2
|
Let $v(x, t) = \mathbb E [f(x + W\_t)]$ with a Brownian motion $W$. Then, Malliavin calculus leads to the sensitivity in $x$:
$$\partial\_x v(x, t) = \frac{1}{t} \mathbb E [ f(x + W\_t) W\_t ].$$
I am interested in $u'(x)$ with $u$ defined by
$$u(x) = \mathbb E \int\_0^T f(x + W\_t) dt$$
for some function $f$. First we can write
$$u'(x) = \int\_0^T \partial\_x v(x, t) dt = \int\_0^T \frac{1}{t} \mathbb E[f(x+W\_t) W\_t] dt$$
whenever the last integral exists.
[Question] What is the sufficient conditions on $f$ to have the $u'(x)$? In particular, when $f$ is not differentiable.
[Discussion] For instance, if $f$ is a constant, then $u'(x) = 0$; if $f$ is an identity, $u'(x) = T$. More generally, if $f'$ exists with polynomial growth, then $u'$ can be computed using the following alternative formula:
$$u'(x) = \mathbb E[ \int\_0^T f'(x+W\_t) dt].$$
|
https://mathoverflow.net/users/5656
|
Existence of the derivative of functionals of Brownian motion
|
$\newcommand{\ep}{\varepsilon}\newcommand{\R}{\mathbb R}\newcommand{\vpi}{\varphi}\newcommand{\De}{\Delta}\newcommand{\Om}{\Omega}$The derivative $u'(x)$ exists and the equality
\begin{equation\*}
u'(x) =\int\_0^T \frac{dt}t\, Ef(x+W\_t)W\_t \tag{1}
\end{equation\*}
holds under very mild restrictions on $f$, just a bit stronger than the restrictions on $f$ needed for $u(x)$ itself to exist.
Indeed, suppose that the function $f\colon\R\to\R$ is Borel-measurable and such that
\begin{equation\*}
\int\_0^T dt\, E|f(x+W\_t)|<\infty; \tag{2}
\end{equation\*}
for (2) to hold, it is enough that
\begin{equation\*}
|f(u)|\le Ce^{au^2}\quad\text{for some real $C$}, \\
\text{some real $a<\frac1{2T}$, and all real $u$. } \tag{3}
\end{equation\*}
Then, by Fubini's theorem,
\begin{equation\*}
\begin{aligned}
u(x)&=\int\_0^T dt\,Ef(x + W\_t) \\
&=\int\_0^T dt\int\_\R dz\,f(x+z\sqrt t)\vpi(z) \\
&=\int\_0^T \frac{dt}{\sqrt t}\int\_\R du\,f(u)\vpi\Big(\frac{x-u}{\sqrt t}\Big),
\end{aligned}
\tag{4}
\end{equation\*}
where $\vpi$ is the standard normal density.
Let us show that **condition (3) is sufficient for (1)** as well:
For all real $s$, let
\begin{equation\*}
\psi(s):=|\vpi'(s)|=|s|\vpi(s).
\end{equation\*}
Take any real $x$, any $\ep\in(0,T/2]$, and any
\begin{equation\*}
b\in\Big(a,\frac1{2T}\Big). \tag{4.5}
\end{equation\*}
Then
\begin{equation\*}
\begin{aligned}
\max\Big\{\psi\Big(\frac{y-u}{\sqrt t}\Big)\colon \ep\le t\le T,|y-x|\le1\Big\}
&\le C\_{\ep,x,b,T}e^{-bu^2},
\end{aligned}
\tag{5}
\end{equation\*}
where $C\_{\ep,x,b,T}$ is a real number depending only on $\ep,x,b,T$, but not on $(t,u)$. Also,
\begin{equation\*}
\begin{aligned}
&\int\_\ep^T \frac{dt}{\sqrt t}\int\_\R du\,
\Big|\partial\_x\Big(f(u) \vpi\Big(\frac{x-u}{\sqrt t}\Big)\Big| \\
&=\int\_\ep^T \frac{dt}t\int\_\R du\,|f(u)|
\psi\Big(\frac{x-u}{\sqrt t}\Big)
\end{aligned}
\tag{6}
\end{equation\*}
and
\begin{align\*}
\int\_\ep^T \frac{dt}t\int\_\R du\,|f(u)|e^{-bu^2}<\infty, \tag{7}
\end{align\*}
by (3) and (4.5).
Moreover, for $y\in[x-1,x+1]$, in view of (3) and the condition $\ep\in(0,T/2]$,
\begin{equation\*}
\begin{aligned}
&\int\_0^\ep \frac{dt}{\sqrt t}\int\_\R du\,
\Big|\partial\_y\Big(f(u) \vpi\Big(\frac{y-u}{\sqrt t}\Big)\Big| \\
&=\int\_0^\ep \frac{dt}{\sqrt t}\,\int\_\R dz\,\big|f(y+z\sqrt t)\big|\,|z|\,\vpi(z) \\
&\le CC\_{x,T}\int\_0^\ep \frac{dt}{\sqrt t}\,\int\_\R dz\,e^{z^2/3}\vpi(z) \to0
\end{aligned}
\tag{8}
\end{equation\*}
as $\ep\downarrow0$, where $C\_{x,T}$ is a real number depending only on $x,T$, but not on $\ep$.
Now we can write
\begin{equation\*}
\begin{aligned}
& \int\_0^T \frac{dt}t\, Ef(x+W\_t)W\_t \\
=&\int\_0^T \frac{dt}t\,\int\_\R dz\,f(x+z\sqrt t)z\sqrt t\,\vpi(z) \\
=&\int\_0^T \frac{dt}t\int\_\R du\,f(u)
\frac{u-x}{\sqrt t}\vpi\Big(\frac{x-u}{\sqrt t}\Big) \\
=&\int\_0^T \frac{dt}{\sqrt t}\int\_\R du\,
\partial\_x\Big(f(u)
\vpi\Big(\frac{x-u}{\sqrt t}\Big)\Big) \\
=& \partial\_x\int\_0^T dt\int\_\R du\,f(u)
\frac1{\sqrt t}\vpi\Big(\frac{x-u}{\sqrt t}\Big) \\
=& \partial\_x\int\_0^T dt\, Ef(x+W\_t)=u'(x);
\end{aligned}
\tag{9}
\end{equation\*}
the interchange of the differentiation and integration (the fourth equality in display (9)) is possible by [Lemma 2.3](https://projecteuclid.org/journals/annals-of-probability/volume-43/issue-5/Exact-Rosenthal-type-bounds/10.1214/14-AOP942.full), in view of (5), (6), (7), and (8). The penultimate equality in (9) holds by (4).
Thus, it is proved that (3) is sufficient for (1).
|
2
|
https://mathoverflow.net/users/36721
|
401167
| 164,688 |
https://mathoverflow.net/questions/401034
|
14
|
I am trying to bound a function that includes $\sum\limits\_{\substack{d < n^{1/3} \\ d \mid n}} 1$.
Is there an upper bound known for this sum, either in general or in terms of $\sum\limits\_{\substack{d \mid n}} 1$? Or in general is there a bound for $\sum\limits\_{\substack{d < n^{1/k} \\ d \mid n}} 1$? Any help is appreciated.
Edit: I am realizing that a lower bound for this sum in terms of $\sum\limits\_{\substack{d \mid n}} 1$ would also be useful if anyone can help with that.
|
https://mathoverflow.net/users/333969
|
How many divisors of $n$ are below $n^{1/3}$?
|
One thing you asked for is a lower bound.
Following FusRoDah, I will let $d\_k(n)$ be the number of divisors of $n$ of size less than $n^{1/k}$, and $d(n)$ be the number of divisors of $n$.
Then I claim $$ d\_1(n) \leq d\_3(n) (d\_3(n)+5),$$ giving an explicit lower bound of size roughly $d\_1(n)^{1/2}$.
Proof: First note that $2 d\_2(n) =d\_1(n)$ since, for $d$ a divisor, $n/d$ is also a divisor, so at least half the divisors have size at most $\sqrt{n}$.
Then $d\_2(n)$ is at most the number of divisors of $n$ of size $\leq \sqrt{n}$ that can be written as a product of two divisors of size $< n^{1/3}$ plus the number that cannot be written as a product. We bound both separately.
The number that can be written as a product of two is certainly at most $\frac{d\_3(n) (d\_3(n)+1)}{2}$.
If $d$ cannot be written as a product, then writing it as a product of a sequence of primes and taking the products of initial segments of the sequence, we must skip straight from $\leq d / n^{1/3}$ to $\geq n^{1/3}$, so one of the prime divisors must be at least $n^{2/3}/d \geq n^{1/6}$. Then the product of the remaining prime divisors is $\leq d/ n^{1/6} \leq n^{1/3}$, so is $<n^{1/3}$ unless $d$ has three prime divisors of size exactly $n^{1/6}$ (but there can be exactly one such $d$ and it is easily absorbed). Thus, this one large prime divisor must have size $n^{1/3}$.
The number of such $d$ is then at most $d\_3(n)$ times the number of prime divisors of $n$ of size $\geq n^{1/3}$, which is at most $2$ (since the bound is trivially true when $n$ is the perfect cube of a prime). This gives
$$d\_2(n) \leq \frac{d\_3(n) (d\_3(n)+1)}{2} + 2 d\_3(n) $$ and multiplying by $2$ we get the stated bound.
This lower bound is of roughly the correct shape, since if $n$ is the product of $r$ primes of size about $n^{1/r}$, for $r$ large, then $d\_1(n)=2^r$ while $$d\_3(n) \approx \left( \frac{1}{ (1/3)^{1/3} (2/3)^{2/3} } \right)^r = \left( \frac {27}{4} \right)^{r/3} = (2^r)^{.918 \dots } . $$
So the true optimal bound is indeed lower than $d\_1(n)$ by a power, although possibly a smaller one.
Similar arguments should get polynomial lower bounds for $d\_k(n)$ for all $k$.
|
11
|
https://mathoverflow.net/users/18060
|
401172
| 164,691 |
https://mathoverflow.net/questions/401154
|
1
|
Let $(W\_t)\_{t\ge 0}$ be a standard Brownian motion. For each $t\in [0,1]$, it is known that, e.g. from Burkholder-Davis-Gundy's inequality
$$\mathbb E\big[\sup\_{s\in [t,t+\Delta t]}|W\_s-W\_t|^p\big]=O(\Delta t^{p/2}),\quad \forall p\ge 1,$$
where $O$ refers to "of order". Do we have an estimate of
$$\mathbb E\big[\sup\_{s,t\in[0,1],~ |s-t|\le\Delta t}|W\_s-W\_t|^p\big]?$$
A student asked me this question when I taught Euler's scheme applied to SDEs and its convergence, but I cannot find any reference. Any answers, comments or references are appreciated.
|
https://mathoverflow.net/users/261243
|
On the "uniform continuity" of Brownian motion under expectation
|
For $n\in\mathbb{Z}\_{\geq 0}$ and $0\leq i< 2^n$, denote
$$
X\_{n,i}=\sup\_{t\in[i2^{-n}, (i+1)2^{-n}]}{|W\_t-W\_{i2^n}|}.
$$
Let $n$ be such that $2^{-{n}}<|\Delta t|\leq 2^{-{n+1}}$. Then,
$$\sup\_{s,t\in[0,1],~ |s-t|\le\Delta t}|W\_s-W\_t|^p\leq 4\sup\_{i}|X\_{i,n}|^p,$$ because such $s$ and $t$ must belong to a union of three consecutive diadic intervals of length $2^{-n}$. For a fixed $n$, $X\_{n,i}$ are i. i. d. random variables; moreover, by Brownian scaling,
$$
X\_{n,i}\stackrel{\mathcal{D}}{=}2^{-n/2}X\_{0,0}.
$$
If $F(t)$ denotes the cumulative disctribution function of $X\_{0,0}$, then we have, with $N=2^n,$
$$\mathbb{P}(\sup\_{i}|X\_{i,n}|\leq t)=F(N^{1/2} t)^{N}.$$
It is well known that $\mathbb{P}(\sup\_{[0,1]} W\_t>a)=2\mathbb{P}(W\_t>a)\leq 2e^{-a^2/2}$ for $a$ large enough, so that
$F(t)\geq 1-4e^{-t^2/2}$, and
$$
\mathbb{P}(\sup\_{i}|X\_{i,n}|>t)\leq 1-(1-4e^{-Nt^2/2})^{N}.
$$
Let $T\_n:=\sqrt{2\log 2}n^\frac12 N^{-\frac12}$, then, plugging into the above,
$$
\mathbb{P}(\sup\_{i}|X\_{i,n}|>kT\_n)\leq 1-e^{N\log(1-4\cdot 2^{-k^2n})}\leq 10\cdot 2^{(1-k^2)n}.
$$
That is to say, the median of $\sup\_{i}|X\_{i,n}|$ is of order at most $T\_n$ and its tail decays very sharply after that, implying
$$\mathbb{E}\sup\_{i}|X\_{i,n}|^p=O(T\_n^p)$$ for all $p$, or
$$\mathbb{E}\left(\sup\_{s,t\in[0,1],~ |s-t|\le\Delta t}|W\_s-W\_t|^p\right)=O\left((|\Delta t\log (\Delta t)|^\frac{p}{2}\right).$$
These estimates don't lose much and they are of course very similar to Lévy's theorem on the modulus of continuity for Brownian motion.
|
1
|
https://mathoverflow.net/users/56624
|
401174
| 164,692 |
https://mathoverflow.net/questions/400976
|
6
|
Voisin uses the fact "If $X$ is a K3 surface with an ample line bundle $\mathcal L$ such that $\mathcal L$ generates $\mathop{\mathrm{Pic}}(X)$ and $(\mathcal L^2) = 4t - 2$, then every smooth curve $C \in \lvert\mathcal L\rvert$ satisfies $K\_{t, 1}(C, K\_C) = 0$." to prove the Green conjecture holds for generic curves of even genus. My question is why given an even integer $g$, there always exists a K3 surface $X$ and a smooth curve $C \subseteq X$ of genus $g$ satisfying the conditions above.
|
https://mathoverflow.net/users/129738
|
Existence of curves of arbitrary genus on some K3 surface
|
This should be a consequence of the surjectivity of the period map for K3 surfaces. I believe with this in mind the reasoning is somewhat standard, but it's useful to try and make it explicit. The underlying strategy is as follows: 1) identify a non-empty locus $\mathcal{W}$ in the period domain to which a K3 surface $X$ with the desired property could be mapped into via the period map, then 2) use surjectivity of the period map to assert existence of such $X$.
To this end
* Let's make the following definitions of, respectively, the K3
lattice, the moduli space of marked K3 surfaces[1], and
the period domain: $$\Lambda := E\_8(-1)^{\oplus 2}\oplus U^{\oplus
3}$$ $$N := \{(X,\varphi)\}/\sim$$ $$ D := \{[v] \in
\mathbb{P}\Lambda\_{\mathbb{C}} : (v)^2 = 0\text{ and }
(v,\overline{v}) > 0\}. $$
Then the period map $\mathcal{P} : N \rightarrow D \subset
\mathbb{P}\Lambda\_{\mathbb{C}}$ sending $(X,\varphi)$ to
$[\varphi\_{\mathbb{C}}(H^{2,0}(X))]$ is surjective (see e.g. Theorem
4.1 in "Lectures on K3 surfaces" by D. Huybrechts).
* Now suppose $V \subset \Lambda$ is any sublattice. Then $V^{\perp}$
(in particular) determines a closed slice
$$D\_V :=
D\cap\mathbb{P}(V^{\perp})\_{\mathbb{C}}$$ of the period domain. If
$\mathcal{P}(X,\varphi) \in D\_V$ this then means
$\varphi\_{\mathbb{C}}(H^{2,0}(X)) \subset (V^{\perp})\_{\mathbb{C}}$
which[2] implies $\varphi\_{\mathbb{C}}(H^{1,1}(X)) \supset
V\_{\mathbb{C}}$ and thus[3] $$ \varphi(NS(X)) \supset V.
$$
* Finally, let $$ D\_V^{\circ} := D\_V \setminus
\bigcup\_{V'\not\subset V}D\_{V'}. $$ Then clearly if
$\mathcal{P}(X,\varphi) \in D\_V^{\circ}$ we have $$ \varphi(NS(X))
\cong V. $$ Since $D\_V^{\circ}$ is the complement of a countable
union of proper closed subsets, it is non-empty.
* So for the desired
result, it now suffices to take $\mathcal{W} = D\_V^{\circ}$ where $V = \langle \lambda \rangle$ for
$\lambda \in \Lambda$ such that $(\lambda)^2 = 4t - 2$. The
existence of such $\lambda$ can be verified for any $t \geq 1$ using our explicit
knowledge of the lattice $\Lambda$ (as indicated above) - e.g.
$\lambda = (2t-1,1) \in U$ has the desired property, thinking of $U
\cong \mathbb{Z}^2$ with intersection form $\left[\begin{array}{cc}0
& 1\\1 & 0\end{array}\right]$.
---
[1] if $X$ is a K3 surface with marking $\varphi : H^2(X;\mathbb{Z}) \xrightarrow \cong \Lambda$ (an isometry) then $(X,\varphi) \sim (X',\varphi')$ if and only if $\varphi' = \varphi\circ f^\*$ for some isomorphism $f : X \rightarrow X'$.
[2]using the fact that $\varphi$ is an isometry and is defined over $\mathbb{Z}$ (and thus commutes with conjugation on $\Lambda\_{\mathbb{C}}$), one sees this as follows:
$$
\begin{array}{rcl}
\varphi\_{\mathbb{C}}(H^{2,0}(X)) & \subset & (V^{\perp})\_{\mathbb{C}}\\
\implies \varphi\_{\mathbb{C}}(H^{0,2}(X)) = \varphi\_{\mathbb{C}}(\overline{H^{2,0}(X)}) & \subset &(V^{\perp})\_{\mathbb{C}} = \overline{(V^{\perp})\_{\mathbb{C}}}\\
\implies V\_{\mathbb{C}} = (V^{\perp})\_{\mathbb{C}}^{\perp} & \subset & \varphi\_{\mathbb{C}}(H^{2,0}(X)\oplus H^{0,2}(X))^{\perp}\\
& = & \varphi\_{\mathbb{C}}((H^{2,0}(X)\oplus H^{0,2}(X))^{\perp})\\
& = & \varphi\_{\mathbb{C}}(H^{1,1}(X))
\end{array}
$$
[3]using the same facts as above, one sees this as follows:
$$
\begin{array}{rcl}
\varphi(NS(X)) & = & \varphi\_{\mathbb{C}}(H^{1,1}(X)\cap H^2(X;\mathbb{Z}))\\
& = & \varphi\_{\mathbb{C}}(H^{1,1}(X))\cap \varphi\_{\mathbb{C}}(H^2(X;\mathbb{Z}))\\
& = & \varphi\_{\mathbb{C}}(H^{1,1}(X))\cap \Lambda\\
& \supset & V\_{\mathbb{C}}\cap \Lambda\\
& = & V.
\end{array}
$$
|
4
|
https://mathoverflow.net/users/76148
|
401185
| 164,696 |
https://mathoverflow.net/questions/365820
|
7
|
Given a (fibrant) simplicially enriched category $\mathcal{C}$, I'm interested in the possibility of replacing it with a weakly equivalent one (in Bergner model structure) such that all the mapping spaces are minimal Kan complexes, that I read about for example in $\textit{Higher Topos Theory}$, section 2.3.3.
The naive idea, that is, individually replacing every mapping space with a minimal model, is bound not to work, because the construction of minimal models for a Kan complex is overtly not functorial, since it relies on an inductive choice of representatives for homotopy classes of simplices in each dimension.
Is there a way to obviate this issue and somehow choose such representatives in a way that is compatible with compositions in $\mathcal{C}$? I expect the construction won't give a functor $s \mathcal{Cat} \to s \mathcal{Cat}$ even in case of a positive answer, but I would be quite content with performing this on a single given simplicially enriched category.
In case of a negative answer, is there any other way that I can obtain a simplicially enriched category with the desired property? Feel free to add any conditions you like on $\mathcal{C}$ if it helps. Thanks!
|
https://mathoverflow.net/users/134438
|
Locally minimal simplicial categories
|
It's not possible in general to ensure that all the hom-spaces in a simplicial category are minimally fibrant. Here's a counterexample [inspired by Isbell](https://mathoverflow.net/a/128629/2362).
Consider $Set$ with its cartesian monoidal structure (the same approach, *mutatis mutandis*, will work with the cocartesian monoidal structure). Let $C \subset Set$ be the subcategory with two objects, $\{\emptyset\}$, $\mathbb N$, and with morphisms the bijections. Then after making a choice of bijection $\mathbb N \cong \mathbb N \times \mathbb N$, the groupoid $C$ inherits a monoidal structure equivalent to the restriction of the cartesian monoidal structure on $Set$. As Zhen Lin shows at the linked argument of Isbell, there is no strict monoidal structure on $C$ equivalent to this monoidal structure.
(Of course, in line with the strictification theorem, the monoidal structure can be strictified after passing to an equivalent groupoid which is not skeletal; the point is that the flexibility gained by having extra isomorphic copies of each object is essential; this monoidal category cannot be *simultaneously* skeletized and strictified.)
Let $BC$ be the 1-object bicategory (with groupoidal homs) associated to $C$, and let $bC$ be any simplicial category corresponding to $BC$. Suppose that $D$ were a simplicial category with minimally-fibrant hom-spaces equivalent to $bC$. By passing to a skeleton, we may assume that $D$ has one object.
Now, we know the homotopy type of the hom-space of $D$ -- it is the classifying space of $C$ -- which is a minimal Kan complex! By uniquness of minimal models, this implies that the hom-space of $D$ is in fact *isomorphic* to the classifying space of $C$. Therefore the composition on $D$ -- which is strictly associative and unital -- restricts to a strict monoidal structure on $D$ equivalent to the one we've chosen. This contradicts Isbell's theorem that no such strictification exists.
**EDIT:** I think Isbell's argument uses some non-invertible morphisms, so doesn't quite apply to $C$ as above. Here is an alternative argument to Isbell's in the cocartesian variation. That is, let $C' \subset Set$ be the subcategory with two objects $\emptyset, \mathbb N$, and morphisms the bijections. The cocartesian monoidal structure on $Set$ restricts to a monoidal structure on $C'$ after we choose a bijection $\mathbb N \cong \mathbb N \amalg \mathbb N$. Without loss of generality, this bijection corresponds to the partition of $\mathbb N$ into the even numbers and odd numbers. Such a monoidal structure cannot be strictified. To see this, observe that if $f: \mathbb N \to \mathbb N$ is any nontrivial bijection, then $(1 \otimes f) \otimes 1$ moves some number which is congruent to 2 mod 4, and fixes any number which is not 2 mod 4, whereas $1 \otimes (f \otimes 1)$ moves some number which is congruent to 1 mod 4, and fixes any number which is not 1 mod 4. Therefore these two bijections are distinct, contradicting the strictness of the monoidal structure.
|
4
|
https://mathoverflow.net/users/2362
|
401186
| 164,697 |
https://mathoverflow.net/questions/159554
|
5
|
I'm looking for a reference for the statement that almost every partial order on $n$ elements has trivial automorphism group. I've been told that this is a folklore result. Does anyone know of a good reference?
|
https://mathoverflow.net/users/22055
|
Reference for statement that almost every $n$-element partial order has trivial automorphism group
|
Prömel (1987) proves a more general statement of rigidity for many classes of structures. In particular he has:
Corollary 2.3. Let $P^u(n)$ denote the number of unlabeled partial orders
on an $n$-element set. Then there exists a constant $s$ such that for all $n$
$$
P^u(n) \le \frac{P(n)}{n!} \left(1 + \frac{s}{2^{n/4}} \right),
$$
and as a consequence,
Corollary 2.3.a. Almost all partial orders are rigid, i.e., have no non-trivial automorphism.
*Prömel, Hans Jürgen*, [**Counting unlabeled structures**](http://dx.doi.org/10.1016/0097-3165(87)90061-6), J. Comb. Theory, Ser. A 44, 83-93 (1987). [ZBL0618.05029](https://zbmath.org/?q=an:0618.05029).
|
3
|
https://mathoverflow.net/users/171662
|
401199
| 164,701 |
https://mathoverflow.net/questions/401025
|
15
|
The *Cantor bijection* given by
$$(x,y)\longmapsto {x+y\choose 2}-{x\choose 1}+1$$
is a bijection from $\{1,2,3,\dotsc\}^2$ onto $\{1,2,3,\dotsc\}$.
It can be generalized to bijections $\varphi\_d:\{1,2,3,\dotsc\}^d
\longrightarrow \{1,2,3,\dotsc\}$ given by
$$(x\_1,\dotsc,x\_d)\longmapsto (d+1\bmod 2)+(-1)^d\sum\_{k=1}^d(-1)^k{x\_1+\dotsb+x\_k\choose k}$$
where $(d+1\bmod 2)$ equals $1$ if $d$ is even and $0$ otherwise. (The proof is a sort of double induction on $d$ and on the sum $x\_1+x\_2+\dotsb+x\_d$.)
It is of course possible to consider compositions of the above formulæ in order to get additional, more complicated polynomial bijections.
A straightforward counting argument shows that we obtain in this way $d! s\_d$ different polynomial bijections between $\{1,2,\dotsc\}^d$ and $\{1,2,\dotsc\}$ where $s\_1,s\_2,\dotsc$
are the little Schroeder numbers with generating series
$$\sum\_{n=1}^\infty s\_nq^n=\frac{1+q-\sqrt{1-6q+q^2}}{4}\ .$$
Are there other "exotic" polynomial bijections (between $\{1,2,\dotsc\}^d\longrightarrow \{1,2,\dotsc\}$)?
(The answer is obviously "no" for $d=1$ and unknown for $d=2$.
I do not know if an "exotic" bijection is known for $d=3$.)
**Added for completeness:** Sketch of a proof that $\varphi\_d$ is a bijection:
We set
$$A\_d(n)=\{(x\_1,\ldots,x\_d)\in\{1,2,\ldots\}^d\ \vert\ x\_1+x\_2+\ldots+x\_d=n\}$$
and $A\_d(\leq n)=A\_d(d)\cup A\_d(d+1)\cup
\ldots\cup A\_d(n)$.
It is enough to prove that $\varphi\_d$
induces a bijection
between $A\_d(\leq n)$ and $\{1,\ldots,{n\choose d}\}$.
This is clearly true for $d=1$ (and arbitrary $n$) and for $n=d$ with arbitrary $d$.
Since $(x\_1,\ldots,x\_d)\longmapsto (x\_1,\ldots,x\_{d-1})$ is a bijection between $A\_d(n)$ and $A\_{d-1}(\leq n-1)$
we have (using slightly abusing notations for sets)
\begin{align\*}\varphi\_d(A\_d(n))&=
{n\choose d}-\left(\varphi\_{d-1}(A\_{d-1}(\leq n-1))-1\right)\\
&=\left\{{n\choose d}-{n-1\choose d-1}+1,\ldots,{n\choose d}\right\}\\
&=\left\{{n-1\choose d}+1,\ldots,{n\choose d}\right\}
\end{align\*}
which ends the proof of the induction step.
|
https://mathoverflow.net/users/4556
|
Are there exotic polynomial bijections from $\mathbb N^d$ onto $\mathbb N$?
|
[Wikipedia says](https://en.wikipedia.org/wiki/Fueter%E2%80%93P%C3%B3lya_theorem) "The generalization of the Cantor polynomial in higher dimensions" is $$(x\_1,\ldots,x\_n) \mapsto x\_1+\binom{x\_1+x\_2+1}{2}+\cdots+\binom{x\_1+\cdots +x\_n+n-1}{n}$$ Note that this is not equivalent to your generalisation $$(x\_1,\ldots,x\_d)\mapsto (d+1\bmod 2)+ \sum\_{k=1}^d(-1)^{k+d}{x\_1+\dotsb+x\_k\choose k}$$ so the word "*The*" is misleading. Lew, Morales, and Sánchez proved (*Diagonal polynomials for small dimension*, Math. Sys. Th. 29 (1996) 305–310) that these two families give the only "diagonal polynomials" up to permutation of variables for dimension 3, where a "diagonal polynomial" is a bijection $\mathbb{N}^d \to \mathbb{N}$ which orders all tuples satisfying $x\_1 + \cdots + x\_d = k$ before any tuple satisfying $x\_1 + \cdots + x\_d = k + 1$. In the same edition Morales and Lew gave $2^{d-2}$ inequivalent (i.e. not related by permutation of variables) diagonal polynomials for higher $d$ (*An enlarged family of packing polynomials on multidimensional lattices*, Math. Sys. Th. 29 (1996) 293-303).
In follow-up papers, Morales (*Diagonal polynomials and diagonal orders on multidimensional lattices*, Th. Comput. Sys. 30 (1997) 367–382) showed that this was an undercount, and in particular listed 6 inequivalent diagonal polynomials for $d=4$; and Sánchez described *A family of $(n-1)!$ diagonal polynomial orders of $\mathbb{N}^n$*, Order 12 (1997).
Fetter, Arredondo, and Morales proved in *The diagonal polynomials of dimension four*, Adv. Applied Math. 34 (2005) 316-334 that the six polynomials which Morales and Sánchez describe are all the diagonal polynomials of dimension 4 (up to permutation of variables). As far as I know it's an open question whether there are diagonal polynomials outside Sánchez's family for $d > 4$.
(Disclosure: I have not been able to read all of these papers, as I don't currently have access to a library with journal subscriptions, so I'm relying on the reports of the two I've found online).
I'm not sure to what extent these diagonal polynomials count as "exotic", but it is at least interesting to note that there's more than one per dimension, and this is far too long for a comment anyway.
|
14
|
https://mathoverflow.net/users/46140
|
401201
| 164,702 |
https://mathoverflow.net/questions/401198
|
3
|
Let $V^n$ be a Stein space(or Stein manifold) in $\mathbb{C}^N$. I want to construct a Stein space(or Stein manifold) $W^{n+1}$ such that $H\_i(V;\mathbb{Z})=H\_{i+1}(W; \mathbb{Z}).$
If we take the suspension of $V,$ is it a Stein space? Or can I get a Stein space $W^{n+1}$ which has the same homotopy as the suspension of $V$?
|
https://mathoverflow.net/users/333818
|
How to get a Stein space which has homotopy type of suspension of another Stein space
|
I'm not sure what you mean by "suspension" of $V$ here. The notion of suspension I have in mind (doubling the cone of $V$ over its base) doesn't yield a manifold, and even if it did it would give an odd-dimensional manifold, so the answer to your question would be no.
About the homotopy type, the answer is yes. Since an Stein $n$-manifold has the homotopy type of an $n$-dimensional cell complex and suspension only increases dimension by 1, it's enough to know that every cell complex of dimension at most $n$ there is a (complex) $n$-dimensional Stein manifold with the same homotopy type. This was proved by Eliashberg: Stein manifolds (up to deformation) can be described by (Legendrian) handlebody diagrams, and you can get infinitely many of those thickening a cell complex to a handle decomposition. (This can be done in an essentially arbitrary way if $n>2$, and with a little bit of care when $n=2$.)
The general case is contained in Eliashberg's paper (*Topological characterization of Stein manifolds of dimension > 2*, Int. Math. J. **1**, 1990), and is discussed in Cieliebak and Eliashberg's *From Stein to Weinstein and back*.
The case of $n = 2$ is mentioned in the two previous references, and is studied in depth in Gompf's *Handlebody construction of Stein surfaces* (Ann. Math. (2) **148**, no. 2, 1998); you can also look at Ozbagci and Stipsicz's book *Surgery on contact 3-manifolds and Stein surfaces*.
|
6
|
https://mathoverflow.net/users/13119
|
401205
| 164,705 |
https://mathoverflow.net/questions/401210
|
2
|
For a sufficient large field $k$ with characteristic 2, $S\_3$ and $D\_{10}$ both do have the property that the trivial module over their group algebra has nontrivial self-extension. Puig's work on nilpotent blocks shows that being p-nilpotent for the considered characteristic is sufficient to guarantee the trivial module has nontrivial self-extension and $S\_3$ and $D\_{10}$ are 2-nilpotent. But if we change the characteristic of $k$ from 2 to 3,5 respectively, direct computation shows that the trivial module over the group algebras $kS\_3$ and $kD\_{10}$ only have trivial self-extension respectively. However we know that they are all solvable, thus being solvable is not a sufficient condition on $G$ for the trivial module over $kG$ has nontrivial self-extension. Fixed a algebraically closed field $k$ of characteristic $p$, is there any other criterion for the trivial module over the group algebra $kG$ has nontrivial self-extension except for that $G$ is $p$-nilpotent?
|
https://mathoverflow.net/users/134942
|
Nontrivial self-extension of trivial $kG$-module
|
$\mathrm{Ext}^1\_{kG}(k,k)$ classifies these extensions, and it's not too hard to show that $\mathrm{Ext}^1\_{kG}(k,k) \cong H^1(G;k) \cong \hom(G, k)\cong \bigoplus\_{i\in I}\hom(G^{ab},\mathbb F\_p)$ where $|I| = \dim\_{\mathbb F\_p}(k)$.
In particular, this is nonzero if and only if $\hom(G^{ab},\mathbb F\_p)\neq 0$, if and only if $G^{ab}$ has a nontrivial $p$-group quotient. This is equivalent to $G$ having a nontrivial $p$-group quotient, which is weaker than being $p$-nilpotent.
|
8
|
https://mathoverflow.net/users/102343
|
401212
| 164,706 |
https://mathoverflow.net/questions/359594
|
6
|
Write $n$ iid draws of $(x,y)$ as $(x^n, y^n)$. Fix $R\in (0,H(x))$. What is the min of $n^{-1}H(y^n|f(x^n))$ over maps $f$ with range $\lbrace 1,\dots,\exp nR\}$, taking $n\to \infty$?
|
https://mathoverflow.net/users/10668
|
Can information be extracted more precisely using more random trials?
|
The characterization is given in terms of a so-called auxiliary random variable. It is as explicit of an answer as you'll get, unless you consider very special cases (like jointly Gaussian $X,Y$, or binary-valued $X,Y$). Namely, you have
$$
\lim\_{n\to\infty}\min\_{f: x^n \mapsto f(x^n)\in \{1,\dots,2^{nR}\}} \frac{1}{n}H(Y^n|f(X^n)) = \inf\_{U : Y-X-U, I(X;U)\leq R} H(Y|U).
$$
In the expression on the right, we look over all random variables $U$, jointly distributed with $X,Y$, such that $Y$ and $U$ are conditionally independent given $X$, and $I(X;U)\leq R$, where $I$ denotes the usual mutual information.
This is a special case of the indirect source coding problem (or, more generally CEO problem) under logarithmic loss. The information bottleneck problem corresponds to the unconstrained optimization problem associated to that on the right (i.e., the IB functional is equivalent to the Lagrangian of the thing on the right).
Here is a quick sketch of the proof.
Claim (Lower Bound): For any $f: x^n \mapsto f(x^n) \in \{1,\dots, 2^{nR}\}$, there is a random variable $U$ satisfying $Y-X-U$, $I(X;U)\leq R$ and $\frac{1}{n}H(Y^n|f(X^n)) \geq H(Y|U)$.
Proof: By the chain rule, write
$$
\frac{1}{n}H(Y^n|f(X^n)) = \frac{1}{n}\sum\_{i=1}^n H(Y\_i|Y^{i-1},f(X^n))= \frac{1}{n}\sum\_{i=1}^n H(Y\_i|V\_i),
$$
where we define $V\_i := (Y^{i-1},f(X^n))$. Observe that, since $(X^n,Y^n)$ are iid draws of $(X,Y)$, we have $Y\_i - X\_i - V\_i$ for each $i=1,\dots, n$. Now, on a common probability space with $(X,Y)$ construct a random variable $U$ as follows. Given $X=x$, draw $Q$ uniformly from $\{1,\dots, n\}$, and take $U = (V\_Q,Q)$, where $V\_Q|\{Q=i, X=x\} \sim p\_{V\_i|X\_i}(\cdot | x)$. Evidently, we have $Y - X - U$, and by definition of $U$,
$$
\frac{1}{n}H(Y^n|f(X^n)) = \frac{1}{n}\sum\_{i=1}^n H(Y\_i|V\_i) = H(Y|U),
$$
since the $Y\_i$'s are iid. Now, for each $i=1,\dots, n$, we have $X\_i - (X^{i-1},f(X^n)) - (Y^{i-1},f(X^n))$ (again, using the fact that $(X^n,Y^n)$ are iid draws of $X,Y$ together with the fact that $f$ is a function of the $x\_i$'s), so properties of entropy and the data processing inequality give
$$
R \geq \frac{1}{n}I(X^n ; f(X^n)) = \frac{1}{n}\sum\_{i=1}^n I(X\_i ; X^{i-1},f(X^n)) \geq \frac{1}{n}\sum\_{i=1}^n I(X\_i ; V\_i) = I(X;U).
$$
Claim (Upper Bound): If $Y-X-U$ satisfy $I(X;U)\leq R$, then for $n$ sufficiently large, there is a function $f: x^n \to f(x^n) \in \{1,\dots, 2^{nR}\}$ with $\frac{1}{n}(H(Y^n| f(X^n))+o(n)) = H(Y|U)$
Proof: Fix any random variable $U$ on a common probability space with $(X,Y)$ such that $Y-X-U$ and $I(X;U) < R$. Generate $2^{nR}$ sequences $U^n(1), \dots, U^n(2^{nR})$ independently, according to $p\_U^{\otimes n}$. By standard typicality arguments, if we observe $X^n$, then with high probability there is an index $f(X^n)\in \{1,\dots, 2^{nR}\}$ such that $U^n(f(X^n))$ is jointly typical with $X^n$ (and therefore also $Y^n$ by the Markov Lemma). By standard properties of (conditional) typical sets, we have
$$
\frac{1}{n}H(Y^n| f(X^n)) \sim H(Y|U).
$$
|
1
|
https://mathoverflow.net/users/99418
|
401219
| 164,708 |
https://mathoverflow.net/questions/401151
|
26
|
I have a question about the Chocolatier's game, which I had
introduced in [my recent answer to a question of Richard
Stanley](https://mathoverflow.net/a/401136/1946).
To recap the game quickly, the Chocolatier offers up at each stage
a finite assortment of chocolates, and the Glutton chooses one to
eat. At each stage of play, the Chocolatier can extend finitely the chocolate assortment on offer, and the Glutton chooses from those currently available. After infinite play, the Glutton
wins if every single chocolate that was offered is eventually consumed.
In this post, I am interested in the version of the game where the
Chocolatier is not allowed to repeat chocolate types — each
new chocolate on offer is a uniquely exquisite new creation.
As I explain at the other the post, the Glutton clearly has a winning
strategy, simply by keeping track of when new chocolates are added
and making sure to organize the consumption so that every chocolate
is eventually eaten. (And this idea works even when the Chocolatier
is allowed to extend the offers countably infinitely, and not necessarily just finitely.)
Furthermore, in the case where there are only countably many possible chocolate
types, then the Glutton has a winning strategy that
depends only on the chocolates currently on offer, not requiring
any knowledge of the game history. The strategy is simply to fix an
enumeration of all the possible chocolate types and then eat the
chocolate available that appears earliest in that enumeration. None
could be left at the limit, since it would have been chosen once
the earlier ones had been consumed.
Meanwhile, in the case where there are uncountably many chocolate
types available, I had proved that the Glutton has no such strategy
that depends only on the chocolates currently on offer.
My question is whether we can extend this argument also to allow
the strategy to depend on the set of chocolates already eaten.
**Question.** Does the Glutton have a winning strategy in the
Chocolatier's game which depends only on the set of chocolates
currently on offer and the set of chocolates already eaten?
I conjecture a negative answer on sufficiently large uncountable
sets and perhaps on all uncountable sets.
Here is an alternative equivalent formulation of the game, the catch-up
covering game on a set $X$. The first player plays an increasing
chain of finite subsets
$$A\_0\subset A\_1\subset A\_2\subset\cdots\subset X$$
and the second player chooses elements $a\_n\in A\_n$. After infinite
play, the second player wins if $\bigcup\_n
A\_n=\{a\_0,a\_1,a\_2,\ldots\}$. Of course, the second player can win, by looking at the history of how elements were added to the sets,
but the question is whether there is a winning strategy that at
move $n$ depends only on the current set $A\_n$ and the set of
already-chosen elements $\{a\_k\mid k<n\}$. The argument on the previous post shows that on an uncountable set $X$ there can be no such winning strategy that depends only on the difference set $A\_n\setminus\{a\_k\mid k<n\}$, which is the set of elements currently available for choosing anew.
|
https://mathoverflow.net/users/1946
|
The Chocolatier's game: can the Glutton win with a restricted form of strategy?
|
*(Not an answer; promoted from a comment on another answer)*
If we modify the game so that the glutton can remember (only) the last chocolate they ate, they have a winning strategy as follows:
Well-order $X$. At each step, let $c\in X$ be the last chocolate we ate. If, among the chocolates offered to us, there is any less than $c$, eat the greatest offered chocolate which is less than $c$. If not, eat the greatest chocolate offered overall.
This strategy means that "most" of the time, we eat chocolate in decreasing sequences, only jumping upwards in the order when we have nowhere further to decrease to. This works because each decreasing sequence includes every chocolate that was available at the start of the sequence and each sequence is finite because of well-ordering.
---
This strategy might extend to the original problem - if one could come up with a choice function $f:P\_{>0}(X) \rightarrow X$ such that, for every sequence of distinct terms $x\_0,x\_1,x\_2,\ldots$ from $X$ and every $n\in\mathbb N$, there were infinitely many $m\in \mathbb N$ such that
$$f(\{x\_0,x\_1,\ldots,x\_m\})=x\_n.$$
Using such a function, one could replace "last chocolate we ate" in the strategy with "$f$ applied to the set of chocolates we've eaten" - and we would then find similar decreasing sequences by, whenever we eat a chocolate, looking for when we next choose that chocolate as $c$ in the strategy, and looking for which chocolate we eat at that step (if smaller) and so on. I'm not sure whether finding such a function is easier than solving the original problem, however (or even possible).
|
13
|
https://mathoverflow.net/users/64294
|
401221
| 164,709 |
https://mathoverflow.net/questions/401218
|
5
|
Say $f \in L^p[a,b]$, with $p \in \mathbb{N}, p > 1 $. Does its Fourier Series converge in the metric space $L^p[a,b]$? Does the series converge pointwise? And at which conditions?
Say now $p = 1$, Does its Fourier Series converge in the metric space $L^1[a,b]$? Does the series converge pointwise? And under which conditions?
|
https://mathoverflow.net/users/334508
|
Convergence of Fourier series
|
1. **Convergence in $L^p$, $p>1$.**
True, by [M. Riesz's Theorem](https://www.jstor.org/stable/1993749). This is a standard topic in every harmonic analysis course, with several readable proofs.
2. **Convergence pointwise almost everywhere, $p>1$.**
True, by the [Carleson-Hunt Theorem](https://terrytao.wordpress.com/2020/05/14/247b-notes-4-almost-everywhere-convergence-of-fourier-series/#more-11764). This result is over 50 years old, but famously difficult. Though the techniques used are now standard, there aren't really any easy proofs, even for continuous functions.
3. **Convergence in $L^1$.**
False, and not very difficult to prove. Suppose that Fourier series converged in $L^1$. Let $S\_N$ be the operator that takes a (Schwartz, say) function and returns the $N$th partial Fourier sum. I claim that these operators are not bounded uniformly in the $L^1$ norm, which is enough to give a contradiction by the uniform boundedness principle. For indeed applying it to the Fejer kernel $K\_M$ gives $$\|S\_NK\_M\|\_{L^1}\to\|D\_N\|\_{L^1}$$ as $M\to\infty$. Where $D\_N$ is the Dirichlet kernel. But $\|D\_N\|\_{L^1}$ is unbounded, so $S\_N$ cannot be uniformly bounded.
4. **Convergence pointwise almost everywhere, $p=1$.**
False, due to [Kolmogorov](https://iopscience.iop.org/article/10.1070/RM1983v038n04ABEH004205) almost 100 years ago. In fact, it is possible to find an $L^1$ function whose Fourier series diverges *everywhere*. This is not an easy counterexample, but it is presented in some courses in harmonic analysis.
|
25
|
https://mathoverflow.net/users/142740
|
401224
| 164,710 |
https://mathoverflow.net/questions/401196
|
2
|
Let $\Omega \subset \mathbb{R}^n$ be an open open subset. Let $u,v\colon \Omega\to \mathbb{R}$ be two functions such that at least one of them is compactly supported. Assume each of $u$ and $v$ can be presented as a difference of two bounded subharmonic functions in $\Omega$. Thus in particular the distributional Laplacians $\Delta u,\Delta v$ are well defined as signed measures on $\Omega$.
**Question.** Is it true that
$$\int\_\Omega u(x)\Delta v(x) dx=\int\_\Omega v(x)\Delta u(x) dx?$$
**Remark.** (1) The expressions under the both integrals are well defined as signed measures with compact support. Thus both sides make sense.
(2) The simplest unknown to me case is $n=2$.
|
https://mathoverflow.net/users/16183
|
Comparing integrals of bounded subharmonic functions
|
Without loss of generality, $u$ has compact support $K\subset\Omega$. Therefore the (signed) measure $\Delta u$ is supported in $K$ as well. Let $(\phi\_k)$ be a sequence of smooth (radial) mollifiers such that $\phi\_k\*u$ is supported in $K^\delta$ (the closed $\delta$ neighborhood of $K$, with $\delta>0$ so small that $K^\delta\subset\Omega$). Additionally, suppose $0\le\phi\_k$, $\int\_{\Bbb R^n}\phi\_k(x)\phantom{!}dx=1$, and $\phi\_k$ is supported in the ball $B(0,\delta/k)$, for each $k\ge 1$.
Then $\lim\_k \phi\_k\*u=u$ pointwise and boundedly, because (for example) $u$ is finely continuous. Likewise $\lim\_k\phi\_k\*v=v$. Then
$$
\eqalign{
\int\_\Omega u(x)\cdot\Delta v(dx)
&=\lim\_k\int\_\Omega (\phi\_k\*u)(x)\cdot\Delta v(dx)\cr
&=\lim\_k\int\_\Omega \Delta(\phi\_k\*u)(x)\cdot v(x) \phantom{b}dx\cr
&=\lim\_k\int\_\Omega (\phi\_k\*\Delta u)(x)\cdot v(x) \phantom{b}dx\cr
&=\lim\_k\int\_\Omega (\phi\_k\*v )(x)\phantom{b}\Delta u(dx)\cr
&=\int\_\Omega v(x)\cdot\Delta u(dx).\cr
}
$$
Here the second equality is just the definition of $\Delta v$ as a distribution; the third likewise; the fourth is Fubini.
(Edited per suggestion of makt, to fix the vacuity noticed by Mateusz Kwaśnicki.)
|
0
|
https://mathoverflow.net/users/42851
|
401231
| 164,714 |
https://mathoverflow.net/questions/401237
|
0
|
Prove that this sum holds for all positive integers $k$. I'm quite sure this is right but I can't see immediately how to go about proving it. This will help resolve a problem regarding sums of binomial coefficients that I'm working on. Any ideas?
|
https://mathoverflow.net/users/265714
|
Prove for all $k \in \mathbb{N}$, that $\sum_{j=0}^{2k+1} {n+j-1\choose j} + \sum_{j=0}^{2k+1}(-1)^j{n+2k+2\choose j} = 0$
|
It's known that $\frac1{(1-x)^n}=\sum\_{i\geq0}\binom{n+j-1}jx^j$. Combined with the relation $\frac1{1-x}\frac1{(1-x)^n}=\frac1{(1-x)^{n+1}}$, one finds that
$\sum\_{k\geq0}x^k\sum\_{j=0}^k\binom{n+j-1}j=\sum\_{k\geq0}x^k\binom{n+k}k$. In particular, one gathers that
$$\sum\_{j=0}^{2k+1}\binom{n+j-1}j=\binom{n+2k+1}{2k+1}. \tag1$$
On the other hand, the binomial relation $\binom{a}i+\binom{a}{i-1}=\binom{a+1}i$ and telescoping sums imply that
\begin{align\*}
\sum\_{j=0}^{2k+1}(-1)^j\binom{n+2k+2}j
&=\sum\_{j=0}^{2k+1}(-1)^j\left[\binom{n+2k+1}j+\binom{n+2k+1}{j-1}\right] \\
&=\sum\_{j=0}^{2k+1}(-1)^j\binom{n+2k+1}j-\sum\_{j=0}^{2k+1}
(-1)^{j-1}\binom{n+2k+1}{j-1} \\
&=\sum\_{j=0}^{2k+1}(-1)^j\binom{n+2k+1}j-\sum\_{j=0}^{2k}
(-1)^j\binom{n+2k+1}j\\
&=-\binom{n+2k+1}{2k+1}. \tag2
\end{align\*}
The OP's claim follows from adding (1) and (2).
|
2
|
https://mathoverflow.net/users/66131
|
401243
| 164,719 |
https://mathoverflow.net/questions/401181
|
10
|
In the $\infty$-world, [connective spectra play the role of abelian groups](https://ncatlab.org/nlab/show/abelian+infinity-group), while [$\mathbb{E}\_\infty$-spaces play that of commutative monoids](https://ncatlab.org/nlab/show/E-infinity+space). This may be rephrased by saying that we may identify the $\infty$-categories of spectra and $\mathbb{E}\_\infty$-spaces with the $\infty$-categories $\mathsf{Mon}\_{\mathbb{E}\_\infty}(\mathcal{S})$ and $\mathsf{Grp}\_{\mathbb{E}\_{\infty}}(\mathcal{S})$ of $\mathbb{E}\_\infty$-monoids/groups in the $\infty$-category of anima $\mathcal{S}$.
Now, the $1$-categories $\mathsf{Ab}$ and $\mathsf{CMon}$ come equipped with tensor products $\otimes\_{\mathbb{Z}}$ and $\otimes\_{\mathbb{N}}$. These correspond in homotopy theory to the tensor products $\otimes\_{\mathbb{S}}$ and $\otimes\_{\mathbb{F}}$ of connective spectra and $\mathbb{E}\_\infty$-spaces.
While the tensor product of connective spectra is widely discussed in the literature, I'm finding it a bit difficult to find references for that of $\mathbb{E}\_\infty$-spaces. So far I've found the following:
* Gepner–Groth–Nikolaus, *Universality of multiplicative infinite loop space machines*, [arXiv:1305.4550](https://arxiv.org/abs/1305.4550), which establishes in Theorem 5.1 a universal property for the tensor product $\otimes\_{\mathbb{F}}$ as the unique functor making the free $\mathbb{E}\_\infty$-monoid functor
$$
\mathcal{S}\_\*\to\mathsf{Mon}\_{\mathbb{E}\_{\infty}}(\mathcal{S})
$$
into a symmetric monoidal functor.
* Blumberg–Cohen–Schlichtkrull, *Topological Hochschild homology of Thom spectra and the free loop space*, [arXiv:0811.0553](https://arxiv.org/abs/0811.0553), which establishes a point-set model for $\mathbb{E}\_{\infty}$-spaces, called *$\*$-modules*, rectifying $\mathbb{E}\_\infty$-spaces to strict monoids in $\*$-modules. See also [MO 92866](https://mathoverflow.net/questions/92866/).
* Sagave–Schlichtkrull, *Diagram spaces and symmetric spectra*, [arXiv:1103.2764](https://arxiv.org/abs/1103.2764), which establishes another point-set model for $\mathbb{E}\_{\infty}$-spaces, called *$\mathcal{I}$-spaces*, similarly rectifying $\mathbb{E}\_\infty$-spaces to strict monoids in $\mathcal{I}$-spaces. See also [arXiv:1111.6413](https://arxiv.org/abs/1111.6413).
* Lind, *Diagram spaces, diagram spectra, and spectra of units*, [arXiv:0908.1092](https://arxiv.org/abs/0908.1092), which proves that $\mathcal{I}$-spaces and $\*$-modules define equivalent homotopy theories.
**Questions:**
* What are some other references discussing the tensor product of $\mathbb{E}\_\infty$-spaces?
* What is the unit of this tensor product?
* Finally, what are some concrete examples of it?
|
https://mathoverflow.net/users/130058
|
Tensor products of $\mathbb{E}_\infty$-spaces
|
The article by Gepner-Groth-Nikolaus is the canonical reference for the tensor product of $E\_\infty$-spaces. In the end it is quite a formal construction so there is not that much to say. A useful point of view that does not appear in loc. cit. is that this tensor product comes from the Lawvere theory of commutative monoids. To explain this, consider the $(2,1)$-category $\mathrm{Span}(\mathrm{Fin})$ whose objects are finite sets and whose morphisms are spans $I\leftarrow K\rightarrow J$. It has the following universal property: for any $\infty$-category $\mathcal C$ with finite products, there is an equivalence
$$
\mathrm{CMon}(\mathcal C) = \mathrm{Fun}^\times(\mathrm{Span}(\mathrm{Fin}),\mathcal C),
$$
where $\mathrm{Fun}^\times$ is the $\infty$-category of functors that preserve finite products. Since $\mathrm{Span}(\mathrm{Fin})$ is self-dual, this means that $E\_\infty$-spaces are finite-product-preserving presheaves on $\mathrm{Span}(\mathrm{Fin})$:
$$
\mathrm{CMon}(\mathcal S) = \mathcal P\_\Sigma(\mathrm{Span}(\mathrm{Fin})).
$$
This was first studied in the [thesis of J. Cranch](https://arxiv.org/pdf/1011.3243.pdf). From this perspective, the direct sum and tensor product are the Day convolutions of $\sqcup$ and $\times$ on $\mathrm{Span}(\mathrm{Fin})$ (here $\times$ means the usual product of finite sets, which is not the categorical product in $\mathrm{Span}(\mathrm{Fin})$; the latter is the same as the categorical coproduct, i.e., the disjoint union $\sqcup$). For example, $E\_\infty$-semirings can be described as right-lax symmetric monoidal functors $(\mathrm{Span}(\mathrm{Fin}),\times)\to(\mathcal S,\times)$ that preserve finite products.
**The unit.** As Rune already explained, the unit for the tensor product of $E\_\infty$-spaces is the free $E\_\infty$-space on a point, that is the groupoid $\mathrm{Fin}^\simeq$ of finite sets with the $E\_\infty$-structure given by disjoint union. This is equivalently the presheaf on $\mathrm{Span}(\mathrm{Fin})$ represented by the point, which is the unit for $\times$ on $\mathrm{Span}(\mathrm{Fin})$.
Here are a few examples I could think of. Let $E\in \mathrm{CMon}(\mathcal S)$.
**Tensoring with a free $E\_\infty$-space.** Let $X\in\mathcal S$. Then
$$
\left(\coprod\_{n\geq 0} (X^n)\_{h\Sigma\_n}\right) \otimes E = \operatorname{colim}\_X E,
$$
where the colimit is taken in $\mathrm{CMon}(\mathcal S)$. This follows from the case $X=\*$ using that $\otimes$ preserves colimits in each variable.
**Tensoring with $\mathbb S$.** Tensoring with the sphere spectrum $\mathbb S$ is the same as group-completing:
$$
\mathbb S\otimes E = E^\mathrm{gp}.
$$
For example, for a ring $R$,
$$
\mathbb S\otimes \mathrm{Proj}(R) = K(R).
$$
where $\mathrm{Proj}(R)$ is the groupoid of finitely generated projective $R$-modules, and $K(R)$ is the K-theory space.
**Tensoring with $\mathrm{Fin}^\simeq[n^{-1}]$.** Another localization of $\mathrm{CMon}(\mathcal S)$ is obtained by inverting integers (or rather, finite sets). The inclusion of the full subcategory of $E\_\infty$-spaces on which multiplication by $n$ is invertible has a left adjoint $E\mapsto E[n^{-1}]$, which is equivalent to tensoring with $\mathrm{Fin}^\simeq[n^{-1}]$. But unlike in the cases of either abelian monoids or spectra, $\mathrm{Fin}^\simeq[n^{-1}]$ is not just the sequential colimit of multiplication by $n$ maps; it is obtained from the latter by killing suitable perfect subgroups of its fundamental groups, in the sense of Quillen's plus construction, to ensure that $n$ acts invertibly.
**Tensoring with $\mathbb N$.** Let $\mathrm{FFree}\_{\mathbb N}$ be the 1-category of finite free $\mathbb N$-modules. There is a functor
$$
\mathrm{Span}(\mathrm{Fin}) \to \mathrm{FFree}\_{\mathbb N}
$$
sending a finite set $I$ to $\mathbb N^I$, inducing an adjunction
$$
\mathrm{CMon}(\mathcal S) = \mathcal P\_\Sigma(\mathrm{Span}(\mathrm{Fin})) \stackrel{\mathrm{str}}\rightleftarrows \mathcal P\_\Sigma(\mathrm{FFree}\_{\mathbb N}).
$$
Objects in the right-hand side are sometimes called *strictly commutative monoids* (the group-complete ones are connective $H\mathbb Z$-module spectra). Tensoring with $\mathbb N$ amounts to strictifying a commutative monoid in this sense:
$$
\mathbb N\otimes E = E^\mathrm{str}.
$$
Unlike $\mathbb S$, $\mathbb N$ is not an idempotent semiring, that is, strictifying is not a localization. Indeed, $\mathbb N\otimes\mathbb N$ is an $E\_\infty$-space whose group completion is the "integral dual Steenrod algebra".
**Tensoring with $\mathrm{Vect}\_\mathbb{C}^\simeq$.** Let $\mathrm{Vect}\_\mathbb{C}^\simeq=\coprod\_{n\geq 0} BU(n)$, where $U(n)$ is regarded as an $\infty$-group (despite the notation, this is not really the core of an $\infty$-category of vector spaces). This is an $E\_\infty$-space whose group completion is $\mathrm{ku}$. There is a related $\infty$-category $2\mathrm{Vect}\_{\mathbb C}$ whose objects are finite sets and whose morphisms are matrices of complex vector spaces. As in the previous example we get an adjunction
$$
\mathrm{CMon}(\mathcal S) = \mathcal P\_\Sigma(\mathrm{Span}(\mathrm{Fin})) \rightleftarrows \mathcal P\_\Sigma(2\mathrm{Vect}\_{\mathbb C}).
$$
An object in the right-hand side is roughly speaking a commutative monoid such that $U(n)$ acts on the multiplication by $n$ map in a coherent way. Tensoring with $\mathrm{Vect}\_\mathbb{C}^\simeq$ gives the free commutative monoid with such structure.
|
10
|
https://mathoverflow.net/users/20233
|
401245
| 164,720 |
https://mathoverflow.net/questions/401244
|
10
|
$\newcommand{\K}{\mathrm{K}}$The abelian group completion functor $\K\_0\colon\mathsf{CMon}\to\mathsf{Ab}$ satisfies
$$
\K\_0(A)
\cong
\mathbb{Z}\otimes\_{\mathbb{N}}A,
$$
naturally in $A\in\mathrm{Obj}(\mathsf{CMon})$, where
* $\mathbb{Z}$ is the additive monoid of integers (i.e. $\K\_0(\mathbb{N})$, the group completion of $\mathbb{N}$);
* $\otimes\_\mathbb{N}$ is the tensor product of commutative monoids.
---
**Question.** Does the $\mathbb{E}\_{\infty}$-group completion functor $\K\_0\colon\mathsf{Mon}\_{\mathbb{E}\_\infty}(\mathcal{S})\to\mathsf{Grp}\_{\mathbb{E}\_\infty}(\mathcal{S})$ similarly satisfies
$$\K\_0(X)\cong QS^{0}\otimes\_\mathbb{F}X,$$
where now
* $QS^0$, the stabilization of $S^0$, is the $\mathbb{E}\_{\infty}$-group completion of $\mathbb{F}=\coprod\_{n\in\mathbb{N}}\mathbf{B}\Sigma\_{n}$, the groupoid of finite sets and permutations;
* $\otimes\_{\mathbb{F}}$ is the [tensor product of $\mathbb{E}\_{\infty}$-spaces](https://mathoverflow.net/questions/401181)?
|
https://mathoverflow.net/users/130058
|
Group completion of $\mathbb{E}_{\infty}$-monoids via tensor products
|
Yes, for the same reason. Let me sketch a proof.
1- $QS^0\otimes X$ is group-complete. Indeed, its $\pi\_0$ is $\mathbb Z\otimes \pi\_0(X)$, and that's a group for the usual reasons. Another way to prove it is to prove that the shear map for $X\otimes Y$ is (the shear map of $X)\otimes Y$, which can be seen by noting that $\otimes$ commutes with coproducts and hence finite products in each variable.
2- There is a natural transformation $X\to QS^0\otimes X$ given by tensoring $\mathbb F\to QS^0$ by $X$, and this induces a natural transformation $X^{gp}\to QS^0\otimes X$.
3- Both sides commute with colimits (a colimit of grouplike $E\_\infty$-spaces is grouplike so I don't have to worry about whether I'm talking about colimits in monoids or grouplike monoids), therefore to check that this map is an equivalence, it suffices to do so for $X= \mathbb F$, and for that one it is a tautology.
Another way to phrase this is to use the following sequence of natural equivalences (and using point 1- for the last one):
$X^{gp} = QS^0\otimes\_{QS^0} X^{gp} = (QS^0\otimes\_\mathbb F X)^{gp}= QS^0\otimes X$
The second natural equivalence comes from the fact that group completion is symmetric monoidal, and $(QS^0)^{gp}\simeq QS^0$.
|
9
|
https://mathoverflow.net/users/102343
|
401247
| 164,722 |
https://mathoverflow.net/questions/397778
|
16
|
I've been looking into Apéry's irrationality proof of $\zeta (3)$, and one of the first questions I instantly had, was how did he derive the following continued fraction?
$$\begin{equation\*} \zeta (3)=\dfrac{6}{5+\overset{\infty }{\underset{n=1}{\mathbb{K}}}\dfrac{-n^{6}}{34n^{3}+51n^{2}+27n+5}}\end{equation\*}$$
Furthermore, is it possible to get a similar continued fraction for $\zeta(5)$, $\zeta(7)$ or $G$?
A rapidly converging central binomial series was [recently found](https://arxiv.org/abs/1207.3139) for Catalan's constant:
$$G = \frac{1}{2} \sum\_{n=0}^{\infty} (-1)^n \frac{(3n+2) 8^n}{(2n+1)^3 \binom{2n}{n}^3}$$
which is in similar spirit to
$$\zeta(3)=\frac{5}{2}\sum\_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^3 \binom{2n}{n}}$$
which Apéry utilised. If I understand correctly, this gives us the first stage of an analogue of Apéry's proof for $G$. The next stage in his proof is to use a fast recursion formula that approaches $\zeta(3)$:
$$n^3 u\_n + (n-1)^3 u\_{n-2} = (34n^3 - 51n^2 + 27n - 5) u\_{n-1},\, n \geq 2$$
A similar recursive formula that approaches $G$ was [found:](https://arxiv.org/abs/math/0201024)
$$(2n+1)^2 (2n+2)^2 p(n) u\_{n+1}-q(n)u\_n = (2n-1)^2 (2n)^2 p(n+1) u\_{n-1}$$
where
$$p(n) = 20n^2 - 8n + 1$$
$$q(n) = 3520n^6 + 5632n^5 +2064n^4-384n^3-156n^2+16n+7$$
Now as Zudilin mentions in his paper, 'the
analogy is far from proving the desired irrationality of $G$', but why exactly is this recursive formula not good enough to prove the irrationality of $G$?
A major part of Apéry's proof is to define the double sequence:
$$c\_{n,k} := \sum\_{m=1}^{n} \frac{1}{m^3} + \sum\_{m=1}^{k} \frac{(-1)^{m-1}}{2m^3 \binom{n}{m} \binom{n+m}{m}}$$
Where does this come from, and how could one create a similar sequence instead that follows from the above binomial series for $G$?
|
https://mathoverflow.net/users/174578
|
Extending Apéry's proof to Catalan's constant?
|
**Summary**:
* The continued fraction, the recurrence and the explicit form of the sequence are interchangeable and for the Apéry numbers, we don't know what come first. This extend to other constructions for other constants.
* The approximation for the Catalan's constant $G$ fails because it doesn't converge too fast. i.e the growth ratio between the inclusion function and the sequence diverges (or when raising it to the power of $\frac{1}{n}$ and taking limit to $\infty$, is $> 1$)
This is in the spirit of [Fischler's expository article](http://www.numdam.org/item/SB_2002-2003__45__27_0/), [Zudilin's paper](https://arxiv.org/abs/math/0201024) and [Zeilberger-Zudilin paper](https://arxiv.org/pdf/1912.10381.pdf) on automatic discovery of proofs of irrationality. I will try to be rigorous to the extent possible for my knowledge.
First, we will modify a bit the criteria for irrationality. How the criteria for irrationality that appears in Poorten's paper implies the criteria that appear in these papers? Let's recall it. Let $\beta$ a real number. If $\frac{p\_{n}}{q\_{n}}$ is a rational sequence with $\frac{p\_{n}}{q\_{n}} \neq \beta$ and
$$\left|\beta-\frac{p\_{n}}{q\_{n}}\right| < \frac{1}{q\_{n}^{1+\delta}}$$
then $\beta$ is irrational.
If you multiply by $q\_{n}$
$$\left|q\_{n}\beta-p\_{n}\right| < \frac{1}{q\_{n}^\delta}$$
with $0<\delta<1$ that exists. If we make the same procedure (power to $\frac{1}{n}$ and then taking limit), for these kind of sequences, the right side simply yields a constant (e.g $\mu$) to the power of $\delta$.
$$\lim\_{n \rightarrow \infty} \frac{1}{q\_{n}^{\frac{1}{n}\delta}}=\frac{1}{\mu^{\delta}}<1$$
The last part is just because we need only a $\delta$ to exist, and it could a very small one, close to 0, but no 0.
In summary, the criteria get translated into
$$\lim\_{n \rightarrow \infty}\left|q\_{n}\beta-p\_{n}\right|^{\frac{1}{n}}=L<1$$
Second, it is necessary to mention that the explicit form of a sequence, its recurrence relation, its continued fraction, integral representations, etc. are all *complementary* descriptions of the sequence and each of them help us to prove the ingredients of a Apéry-like proof, namely:
1. The existence of a linear form over two sequences satisfying the previous irrationality criteria
2. The existence of an "inclusion" for the sequences, i.e, multiplying the sequences by a function give integer sequences
3. Our approximation is different to 0 for infinite $n$.
Point 1) requires to know asymptotics estimates of the sequences and for this purpose the recurrence description is useful. Also we need to estimate the error between the number and our approximation. Again you can use the recurrence or an integral representation (like in Beukers's [proof](https://londmathsoc.onlinelibrary.wiley.com/doi/abs/10.1112/blms/11.3.268)). Point 2) is generally proved using the explicit form of the sequences. And for Point 3) you have many options. As you can see, it's flexible and depends of the nature of the sequence.
The technical considerations are a) The asymptotics of a sequence given by a holonomic recurrence can be obtained using the algorithm in [Wimp and Zeilberger paper](https://www.sciencedirect.com/science/article/pii/0022247X85902094) or the method of Flajolet and Sedgewick presented in the book "Analytic Combinatorics". *(Some cautious here, and indeed I would like to consider this in another post. The Wimp-Zeilberger algorithm is an explanation of the [Birkhoff-Trjitzinsky method](https://projecteuclid.org/journals/acta-mathematica/volume-60/issue-none/Analytic-theory-of-singular-difference-equations/10.1007/BF02398269.full), which hasn't been verified completely. Also, since calculations are complicated, we require of a algorithmic package to find the terms, like Kauer's Mathematica package. It doesn't mean that some previously given asymptotics are wrong, but we need to check further details)*. These are the reasons why we will avoid calculating an full asymptotics (even if we can) and we will deal with expressions of the kind $\lim\_{n \rightarrow \infty}a\_{n}^{\frac{1}{n}}$ , that can be tackle using the Poincaré-Perron theorem plus some assumption of [minimality](https://link.springer.com/chapter/10.1007/0-387-27602-5_8) of the solutions plus this [paper of Perron](https://link.springer.com/article/10.1007%2FBF01458689), also knows as Perron's second theorem. Modern approaches to the proof use this setup to avoid more technicalities. For a summary of the full asymptotic of the Apéry numbers see [this](https://link.springer.com/chapter/10.1007/0-387-27602-5_8), page 379.
b) it's a bit more tricky and it's required to calculate the $p$-adic valuation of sub-terms in the explicit form, which generally are combination of powers and binomial coefficients. Here also we need the inclusion function to be smaller in comparison to the sequence's asymptotics, to reach point 1).
Let's try some examples with $\zeta (3)$ and the Catalan's constant to visualize this.
For $\zeta(3)$, the recurrence is given by
$$n^3 u\_n -(34n^3 - 51n^2 + 27n - 5) u\_{n-1}+ (n-1)^3 u\_{n-2} = 0\quad n \geq 2$$
To use the Poincaré-Perron (PP) theorem, we need to transform it into a form $u\_{n}+a(n)u\_{n-1}+b(n)u\_{n-2}=0$ with
$$a=\lim\_{n \rightarrow \infty} a(n) \quad b=\lim\_{n \rightarrow \infty} b(n)$$
Dividing by $n^3$, we obtain
$$u\_n -(34 - 51n^{-1} + 27n^{-2} - 5n^{-3}) u\_{n-1}+ (1
+3n^{-1}+3n^{-2}+n^{-3}) u\_{n-2} = 0\quad n \geq 2$$
Then, we calculate the roots of a characteristic polynomial $x^2+ax^2+b=0$, i.e, $x^2-34x^2+1=0$, and the conclusion of the theorem is that if $x\_{n}$ and $y\_{n}$ are individual independent solutions of the recurrences, then
$$\lim\_{n \rightarrow \infty}\frac{x\_{n+1}}{x\_{n}}=\lambda\_{1} \quad \lim\_{n \rightarrow \infty}\frac{y\_{n+1}}{y\_{n}}=\lambda\_{2}$$
where $\lambda\_{1}$ and $\lambda\_{2}$ and the roots and $|\lambda\_{1}|\leq|\lambda\_{2}|$. This alone doesn't give asymptotic estimates of the general solutions, so we will use some improvements of the theorem. If $a(n)\sim n^{C\_{1}}$ , $b(n)\sim n^{C\_{2}}$ and $C\_{1}=C\_{2}$, then the solution given by $y\_{n}$ is [minimal](https://link.springer.com/chapter/10.1007/0-387-27602-5_8) (page 370), in the sense that for any other independent solution of the recurrence, we can call it $h\_{n}$ (i.e $h\_{n}$ is not a multiple of $y\_{n}$ for some initial conditions), we have:
$$\lim\_{n \rightarrow \infty}\frac{y\_{n}}{h\_{n}}=0$$
In this case, we can use the following [estim](https://dalspace.library.dal.ca/handle/10222/13298)[ates](https://link.springer.com/article/10.1007%2FBF01458689),
$$\lim\_{n \rightarrow \infty} a\_{n}^{\frac{1}{n}} = \lim\_{n \rightarrow \infty} b\_{n}^{\frac{1}{n}}=17+12\sqrt{2}=\alpha$$
In the case of $G$, the recurrence is
$$(2n+1)^2 (2n+2)^2 p(n) u\_{n+1}- q(n)u\_n - (2n-1)^2 (2n)^2 p(n+1) u\_{n-1}=0$$
We have a similar limit, if we divide by $(2n+1)^{2}(2n+2)^2 p(n)$ and then taking limits on $a(n)$ and $b(n)$, we arrive to the characteristic polynomial $x^2-11x-1=0$ and therefore we obtain
$$\lim\_{n \rightarrow \infty} a\_{n}^{\frac{1}{n}} = \lim\_{n \rightarrow \infty} b\_{n}^{\frac{1}{n}}=\frac{11}{2} + \frac{5 \sqrt{5}}{2}=\delta$$
From the recurrence, you can also estimate the error between the number and our approximation. In the case of $\zeta(3)$, using the recurrence, we have:
$$a\_{n}b\_{n-1}-a\_{n-1}b\_{n}=\frac{6}{n^{3}}$$
$$\left|\zeta (3)-\frac{a\_{n}}{b\_{n}}\right| = \sum\_{k=n+1}^{\infty}\frac{6}{n^{3}b\_{k}b\_{k-1}}$$
For $G$ we will find a similar expression since:
$$(2n+1)^2 (2n+2)^2 p(n)(a\_{n+1}b\_{n}-a\_{n}b\_{n+1})=-(2(n-1)+1)^2 (2(n-1)+2)^2 p(n+1)(a\_{n}b\_{n-1}-a\_{n-1}b\_{n})$$
$$=+(2(n-2)+1)^2 (2(n-2)+2)^2 p(n+2)(a\_{n-1}b\_{n-2}-a\_{n-2}b\_{n-1})$$
$$=\textrm{.....}$$
$$\therefore a\_{n+1}b\_{n}-a\_{n}b\_{n+1}=\frac{13}{32}(-1)^{n}\frac{p(2n)}{(2n+1)^2 (2n+2)^2 p(n)}$$
and a similar bound holds.
Now you have to prove the inclusions, for $\zeta (3)$ the inclusion is given by $2D\_{n}^{3}a\_{n} \in \mathbb{Z}$, and for $G$ Zudilin proved the inclusions
$$2^{4n+3} D\_{n} b\_{n} \in \mathbb{Z} \quad 2^{4n+3}D\_{2n-1}^3 a\_{n} \in \mathbb{Z}$$
where $D\_{n}=\textrm{lcm}(1,2,....,n)=e^{\psi(n)}$ ($\psi$(n) is the Chebyshev $\psi$ function). So, for $G$, if we multiply $a\_{n}$ and $b\_{n}$ by $2^{4n+3}D\_{2n-1}^3$, they become integers. Note: Trivially we have that $D\_{n}|D\_{2n-1}^3$. We will define the inclusion of $a(n)$ as $\textrm{Incl}\_{a,n}$ (or by analogy, $\textrm{Incl}\_{b,n}$ for $b\_{n}$) to be the function that we need to multiply to become $a\_{n}$ (or $b\_{n}$) an integer. I won't give too much detail on this step, but the proof of both of them relies on the technical argument explained before (the explicit form sometimes yields this easily by means of hypergeometric manipulations)
Finally, we have that
$$\left|q\_{n}\beta-p\_{n}\right|<\frac{Cq\_{n}}{b\_{n}^{2}}$$
Since for these sequences, we have $q\_{n}=b\_{n}\textrm{Incl}\_{a,n}$
$$\left|q\_{n}\beta-p\_{n}\right|<\frac{C\textrm{Incl}\_{a,n}}{b\_{n}}$$
Let's define $\phi$
$$\phi=\lim\_{n \rightarrow \infty}\textrm{Incl}\_{a,n}^{\frac{1}{n}}$$
For $\zeta(3)$
$$\phi=\lim\_{n \rightarrow \infty}e^{3\frac{\psi(n)}{n}}=e^{3}$$
by the prime number theorem $\lim\_{n \rightarrow \infty}\frac{\psi(n)}{n}=1$
Thus, if you raise the approximation to the power of $\frac{1}{n}$ and them apply the limit to $\infty$, the behaviour of the right side if governed by $\frac{\phi}{\alpha}$ (or $\delta$ in denominator, in the case of $G$). If this ratio is bigger that one, the right bound is not useful. If it less that 1, we can prove irrationality.
For $\zeta(3)$, $\frac{e^3}{\alpha} \approx 0.59$ and for $G$, $\frac{e^6 2^4}{\gamma}\approx 582$. The huge difference in the ratios is what Zudilin indicated. Since for $G$ the value is bigger that one, therefore the sequence is not useful to prove irrationality. Even if we use the conjectured inclusions in the conclusion remarks, we obtain $\frac{e^4 2^4}{\gamma}\approx 79$. This illustrates very well the interplay between the approximation, the asymptotics of the inclusion function and the asymptotics of our sequence.
In the literature, you can find many approaches to construct these sequences: first you can work with an explicit form of the sequences (many of them are obtained from hypergeometric identities), and then you guess a recurrence for them (using the Gosper-Zeilberger algorithm; see [this](https://www2.math.upenn.edu/%7Ewilf/AeqB.html)). If you have the recurrence, the continued fraction description is immediate. Or viceversa, you try to find recurrences that gives origin to integer or "quasi-integer" sequence, by searching the space of coefficients in the recurrence and then prove a closed form for the sequences using combinatorial techniques. Therefore, we still don't have a general way to come up with sequences like the Apéry numbers, i.e, we're looking for ["well-poised"](https://www.sciencedirect.com/science/article/pii/S0377042704003966) sequences in the sense of they almost satisfy the three ingredients, but the seed or the intuition to know where to look comes from the researcher. So the answer to your questions are: where does the continued fraction comes? From the recurrence. Where does the recurrence comes? From the explicit form with the $c\_{n,k}$ of Poorten's paper. Where does the explicit form with the $c\_{n,k}$ comes? From combinatorial identities and a bit of luck. You can start wherever you want, for example see [this](http://www.numdam.org/item/STNG_1977-1978__6__A6_0/), specially Remarque 2.
The interesting question is of course, what is common in the sequences that allow us to prove the irrationality of $\zeta(2)$ and $\zeta(3)$? This question is deep, and goes back to [Zagier's paper](https://people.mpim-bonn.mpg.de/zagier/files/tex/AperylikeRecEqs/APERY.ps.gz): the generating functions of the sequences that are used in the proofs admitted a modular parametrization, a fact that was proved rigorously by [Beukers](http://www.numdam.org/item/AST_1987__147-148__271_0/) and is related also to the differential equation satisfied by the generating function (Picard-Fuchs equations and variants). Some sequences that are ["well-poised"](https://www.sciencedirect.com/science/article/pii/S0377042704003966) in the sense of they almost satisfy the three ingredients don't possess this modular behaviour. In contrast, some family of sequences that arise in the massive search of recurrences exhibit this phenomenon. Surprisingly, the latter give sequences that satisfy the point 2) almost immediately with small inclusions, which is not trivial, and there's still work in progress to understand why. Also some of them satisfy the condition 1) qualitatively, depending of which constant are approximating, of course. But they fail in the final calculation of the irrationality criteria. For more details, see [this](https://lfant.math.u-bordeaux.fr/seminar/slides/2019-11-05T10:00--Henri_Cohen.pdf) and this [post](https://mathoverflow.net/questions/311469/has-ap%C3%A9rys-proof-of-the-irrationality-of-zeta3-ever-been-used-to-prove-the).
I'm in the combinatorial side, so an expert in the work of Zagier and Beukers can complement this part better.
Comment: In many parts, I'm using the positivity, rationality and monoticity of the sequences to simplify the argument. Point 3) is in general easy to prove.
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16
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https://mathoverflow.net/users/302667
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401250
| 164,723 |
https://mathoverflow.net/questions/401233
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2
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I'm looking for a simple example of the following (so that I can get better intuition for it). If possible, I'd like a 2 dimensional rational example.
$X$ is an irreducible projective variety, with an open subset $U$ that is (isomorphic to) the total space of a line bundle over an irreducible variety - in other words such that there exists an irreducible $Y$ where $\pi: U \rightarrow Y$ is a line bundle. We can look at $Z=\bigcup\_{y\in Y} (\overline{\pi^{-1}(y)}\setminus \pi^{-1}(y))$ (where the closure is taken in $X$) - the union of the ''limits'' of the fibers of $\pi$. I want an example where $Z$ is not connected.
The intuition is that the ''limit'' of a fiber over a special point $y$ in $Y$ might actually be ''far away'' from the ''limits'' of the fibers of all the points ''close to'' $y$.
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https://mathoverflow.net/users/3077
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Example sought: disconnected closures of fibers of line bundles
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Take the blowup $X = \text{Bl}\_p(\mathbb{P}^1\times \mathbb{P}^1)$ with exceptional $E$ and $l,l' \subset X$ the proper transforms of the rulings through $p \in \mathbb{P}^1\times \mathbb{P}^1$. Then $U := X\setminus (E\cup l)$ is the trivial line bundle over $\mathbb{P}^1$ but $Z$ will be the disconnected set $l\cup (l'\cap E) \setminus (l\cap E)$.
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4
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https://mathoverflow.net/users/76148
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401253
| 164,724 |
https://mathoverflow.net/questions/401252
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10
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I paraphrase part of the wikipedia article on the Weyl character formula: [Weyl character formula](https://en.wikipedia.org/wiki/Weyl_character_formula).
If $\pi$ is an irreducible finite-dimensional representation of a complex semisimple Lie algebra $\mathfrak{g}$ and $\mathfrak{h}$ is a choice of Cartan subalgebra of $\mathfrak{g}$, then the Weyl character formula states that the character $\operatorname{ch}\_\pi$ of $\pi$ is given by
$$ \operatorname{ch}\_\pi(H) = \frac{\sum\_{w \in W} \epsilon(w) e^{w(\lambda+\rho)(H)}}{\prod\_{\alpha \in \Delta^+}(e^{\alpha(H)/2} - e^{-\alpha(H)/2})},$$
where $W$ is the Weyl group, $H \in \mathfrak{h}$, $\epsilon(w)$ is the determinant of the action of $w \in W$ on the Cartan subalgebra $\mathfrak{h}$, $\Delta^+$ denotes the set of positive roots of $(\mathfrak{g}, \mathfrak{h})$, $\lambda$ denotes the highest weight of $\pi$ and $\rho$ is half the sum of all positive roots of $(\mathfrak{g}, \mathfrak{h})$ (i.e. half the sum of all the elements of $\Delta^+$).
If $\mathfrak{g} = \mathfrak{sl}(n)$, it is known that the denominator of the RHS of the Weyl character formula can be written as a determinant (actually a Vandermonde determinant). While it does seem counterproductive, since the numerator looks simple enough (well, to some extent), yet I am interested whether or not the numerator can also be written as a determinant, at least for $\mathfrak{g} = \mathfrak{sl}(n)$, though I am also interested in the other cases too. After all, the Weyl group in this special case is $S\_n$, the symmetric group on $n$ elements, and an $n \times n$ determinant can be expanded as an alternating sum over $S\_n$. So this does seem promising. I apologize if it turns out to be a trivial question perhaps (I currently have limited access to online journals etc.).
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https://mathoverflow.net/users/81645
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Can the numerator in Weyl's character formula be written as a determinant?
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The classical definition of the Schur polynomials, which considerably predates the Weyl character formula, is as a ratio of two determinants (a so-called "bialternant"): see, e.g., [https://en.wikipedia.org/wiki/Schur\_polynomial#Definition\_(Jacobi's\_bialternant\_formula)](https://en.wikipedia.org/wiki/Schur_polynomial#Definition_(Jacobi%27s_bialternant_formula)).
Of course the Schur polynomials are $\mathfrak{sl}\_n$ characters. There are similar things in other types too, if you are interested in that (for example, see Proposition 1.1 in Okada's "Applications of Minor Summation Formulas to Rectangular-Shaped Representations of Classical Groups", <https://doi.org/10.1006/jabr.1997.7408> ).
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14
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https://mathoverflow.net/users/25028
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401254
| 164,725 |
https://mathoverflow.net/questions/401268
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2
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**Problem:**
>
> let $P\_1(x), P\_2(x), Q\_1(y), Q\_2(y)$ be some polynomials of degree $d$ in $\mathbb{F}\_p$. Let
> \begin{equation}
> A := \{ (x, y) \in \mathbb{F}\_p^2 : P\_1(x) = Q\_1(y) \},\\
> B := \{ (x, y) \in \mathbb{F}\_p^2 : P\_2(x) = Q\_2(y) \}.
> \end{equation}
> Let us also assume that equations $P\_1(x) = Q\_1(y)$ and $P\_2(x) = Q\_2(y)$ are independent in certain sense: for example, polynomials $P\_1(x) - Q\_1(y)$ and $P\_2(x) - Q\_2(y)$ do not have common divisors.
> Can one imply that
> $$|A \cap B| \ll \max{(|A|, |B|)} ?$$
>
>
>
This is clear that each of the equations $P(x) = Q(y)$ may have at most $dp$ solutions, and this is nearly attained when, for example, $P(x) = x^d + 3, Q(y) = y^d + 3$, where $d | p - 1$.
There is the following variation of Bombieri's result by Chalk and Smith (see Theorem 2 in [here](http://matwbn.icm.edu.pl/ksiazki/aa/aa18/aa18121.pdf), page 202):
**Theorem.** Let $(b\_1, b\_2) \in \mathbb{F}\_p^2$ be nonzero and $f(x, y) ∈ \mathbb{F}\_p[x, y]$ be a polynomial of degree $d \geqslant 1$ with the following property: there is no $c \in \mathbb{F}\_p$ for which the polynomial $f(x, y)$ is divisible by $b\_1x + b\_2y + c$. Then
$$
\bigg|
\sum\_{(x, y):f(x,y)=0}
e^{2\pi i(b\_1x+b\_2y)/p}
\bigg| \leqslant 2d^2p^{1/2}.
$$
This is roughly saying, that as long as $P(x) - Q(y)$ does not have any linear divisors, then sets $A, B$ are equidistributed over $\mathbb{F\_p} \times \mathbb{F\_p}$.
This gives an intuition, that sets $A$ and $B$ should be somehow independent, and probably even an inequality $|A \cap B| \ll |A||B|/p^2 \ll d^2$ could have place, upon a certain condition of independence of polynomials $P\_1, P\_2, Q\_1, Q\_2$.
Are there any known results of this type?
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https://mathoverflow.net/users/334675
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Common roots to "independent" equations $P_1(x) = Q_1(y)$ and $P_2(x) = Q_2(y)$ in $\mathbb{F}_p \times \mathbb{F}_p$
|
The bound $d^2$ is, indeed, correct. It follows from Bézout's theorem, see [here](https://en.wikipedia.org/wiki/B%C3%A9zout%27s_theorem#Plane_curves). In your case, affine plane curves are exactly $A$ and $B$, the condition on common divisor is the lack of common component. You can take $F=\mathbb F\_p$ and $E=\overline{\mathbb F\_p}$.
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2
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https://mathoverflow.net/users/101078
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401270
| 164,730 |
https://mathoverflow.net/questions/398544
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26
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Here is a concrete, if seemingly unmotivated, aspect of the question I am interested in:
>
> **Question 1.** Let $a$ and $b$ be two elements of a (noncommutative) semiring $R$ such that $1+a^3$ and $1+b^3$ and $\left(1+b\right)\left(1+a\right)$ are invertible. Does it follow that $1+a$ and $1+b$ are invertible as well?
>
>
>
The answer to this question is
* "yes" if $ab=ba$ (because in this case, $1+a$ is a left and right divisor of the invertible element $\left(1+b\right)\left(1+a\right)$, and thus must itself be invertible; likewise for $1+b$).
* "yes" if $R$ is a ring (because in this case, $1+a$ is a left and right divisor of the invertible element $1+a^3 = \left(1+a\right)\left(1-a+a^2\right) = \left(1-a+a^2\right)\left(1+a\right)$, and thus must itself be invertible; likewise for $1+b$).
* "yes" if $1+a$ is right-cancellable (because in this case, we can cancel $1+a$ from $\left(1+a\right) \left(\left(1+b\right)\left(1+a\right)\right)^{-1} \left(\left(1+b\right)\left(1+a\right)\right) = 1+a$ to obtain $\left(1+a\right) \left(\left(1+b\right)\left(1+a\right)\right)^{-1} \left(1+b\right) = 1$, which shows that $1+a$ is invertible), and likewise if $1+b$ is left-cancellable.
I am struggling to find semirings that are sufficiently perverse to satisfy none of these cases and yet have $\left(1+b\right)\left(1+a\right)$ invertible. (It is easy to find cases where $1+a^3$ is invertible but $1+a$ is not; e.g., take $a = \begin{pmatrix} 0&1\\ 0&0 \end{pmatrix}$ in the matrix semiring $\mathbb{N}^{2\times 2}$.)
The *real* question I'm trying to answer is the following (some hopefully reasonably clear handwaving included):
>
> **Question 2.** Assume we are given an identity that involves only positive integers, addition, multiplication and taking reciprocals. For example, the identity can be $\left(a^{-1} + b^{-1}\right)^{-1} = a \left(a+b\right)^{-1} b$ or the positive [Woodbury identity](https://en.wikipedia.org/wiki/Woodbury_matrix_identity) $\left(a+ucv\right)^{-1} + a^{-1}u \left(c^{-1} + va^{-1}u\right)^{-1} va^{-1} = a^{-1}$. Assume that this identity always holds when the variables are specialized to arbitrary elements of an arbitrary **ring**, assuming that all reciprocals appearing in it are well-defined. Is it then true that this identity also holds when the variables are specialized to arbitrary elements of an arbitrary **semiring**, assuming that all reciprocals appearing in it are well-defined?
>
>
>
There is a natural case for "yes": After all, the same claim holds for **commutative** semirings, because in this case, it is possible to get rid of all reciprocals in the identity by bringing all fractions to a common denominator and then cross-multiplying with these denominators. However, this strategy doesn't work for noncommutative semirings (and even simple-looking equalities of the form $ab^{-1} = cd^{-1}$ cannot be brought to a reciprocal-free form, if I am not mistaken). Question 1 is the instance of Question 2 for the identity
\begin{align}
\left(1+a^3\right)^{-1} \left(1+b^3\right)^{-1} \left(1+a\right) \left(\left(1+b\right)\left(1+a\right)\right)^{-1} \left(1+b\right) = \left(1+a^3\right)^{-1} \left(1+b^3\right)^{-1}
\end{align}
(where, of course, the only purpose of the $\left(1+a^3\right)^{-1} \left(1+b^3\right)^{-1}$ factors is to require the invertibility of $1+a^3$ and $1+b^3$). Indeed, if $\alpha$ and $\beta$ are two elements of a monoid such that $\beta\alpha$ is invertible, then we have the chain of equivalences
\begin{align}
\left(\alpha\text{ is invertible} \right)
\iff
\left(\beta\text{ is invertible} \right)
\iff
\left( \alpha \left(\beta\alpha\right)^{-1} \beta = 1 \right)
\end{align}
(easy exercise).
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https://mathoverflow.net/users/2530
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Subtraction-free identities that hold for rings but not for semirings?
|
Tim Campion's [idea](https://mathoverflow.net/a/401257) works, though his example needs a little fixing. As in Tim's answer, we will find a rig with two elements $X$ and $Y$ such that $X+Y=1$ but $XY \neq YX$.
Let $(M,+,0)$ be any commutative monoid. Let $R$ be the set of endomorphisms of $M$ obeying $\phi(x+y)=\phi(x)+\phi(y)$ and $\phi(0)=0$. Then $R$ is a rig, with $(\alpha+\beta)(x) = \alpha(x) + \beta(x)$ and $(\alpha \beta)(x) = \alpha(\beta(x))$.
Let $M$ be $\{ 0,1,2 \}$ with $x+y \mathrel{:=} \max(x,y)$. Define
\begin{gather\*}
\alpha(0) = 0,\ \alpha(1) = 0,\ \alpha(2) = 2 \\
\beta(0) = 0,\ \beta(1) = 1,\ \beta(2) = 1.
\end{gather\*}
Then $\alpha+\beta=\mathrm{Id}$ but $\alpha \beta \neq \beta \alpha$.
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11
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https://mathoverflow.net/users/297
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401273
| 164,732 |
https://mathoverflow.net/questions/401278
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1
|
If I start with a, say, 3-CW complex $X$ which can be embedded in $\mathbb{R}^5$, I can get a neighbourhood $U$ of $X$ which has the same homotopy type of $X$. Then $U$ is a $5-$ dimensional open manifold. Can I get a close manifold (compact without a boundary) $M$, of dimension $6$ (or some higher dimension) such that $M$ and $X$ have the same homotopy type?
|
https://mathoverflow.net/users/333818
|
How can I construct a closed manifold from a finite CW complex?
|
Take $X=S^3$. Then no closed manifold of dimension at least 6 has the same homotopy type.
|
9
|
https://mathoverflow.net/users/334338
|
401280
| 164,736 |
https://mathoverflow.net/questions/401283
|
1
|
We consider the sequence $n\longmapsto {n\choose k}+1$
for $k\geq 1$ a fixed integer. For $k\geq 3$ odd,
this sequence seems to contain surprisingly few prime numbers
while there are many primes (perhaps roughly the expected
amount) among the first terms of this sequence for even $k\geq 2$.
Is there an explanation for this observation (or is it an artefact)?
The following table gives the number of values $n\leq 10^5$
such ${n\choose k}+1$ is prime for $k=1,..,15$:
$$\begin{array}{cc}
k&\sharp(\mathcal P\cap \{{n\choose k}+1\ \vert\ n=k,\dots,10^5\})\\
1&9592\\2&9863\\3&3\\4&6765\\5&5\\6&6203\\
7&7\\8&3092\\9&8\\10&4364\\11&21\\12&2817\\13&19\\
14&2968\\15&16
\end{array}$$
(with $\mathcal P=\{2,3,5,7,11,\ldots\}$ denoting the set of primes).
The small subsets $\mathcal S\_k=\{n\_1,\ldots,n\_{i\_k}\}\_k$ of $\{1,\ldots,10^5\}$ corresponding to prime-numbers
${n\choose k}+1$ for $n\in \mathcal S\_k$ and $k\geq 3$ odd are given by
$$\{3,4,5\}\_3,\ \{5,6,9,11,14\}\_5,\ \{7,9,11,14,20,29,104\}\_7,$$
$$\{9,10,14,35,39,71,119,839\}\_9,$$
$$\{11,12,13,17,19,29,32,34,44,65,69,76,83,109,153,197,279,791,1385,6929,13859\}\_{11}$$
$$\{13,17,20,27,34,44,51,55,69,87,104,119,142,209,251,263,359,857,923\}\_{13},$$
$$\{15,16,17,19,38,83,89,90,131,714,1091,1286,2001,2309,4003,6551\}\_{15}.$$
For $k=3$, I checked that there are no additional primes (of the form ${n\choose 3}+1$) for $n$ up to $10^6$.
**Added after accepting the answer of user334725:** The analogous phenomenon exists for ${n\choose k}-1$ and $k$ even.
(This question is by the way an illustration of the fact that experimental maths cannot be done simultaneously with clear thinking: Playing with the computer shuts generally my brain down!)
|
https://mathoverflow.net/users/4556
|
There seem to be only few primes of the form ${n\choose k}+1$ if $k\geq 3$ is odd
|
When $k$ is odd, writing $f\_k(n)={n\choose k}+1$ you have $f\_k(-1)=0$ as a polynomial evaluation, and removing the denominator gives you the extra $k!$ in integers.
I.e., $5!f\_5(n)=n(n-1)(n-2)(n-3)(n-4)+120$, so $$5!f\_5(-1)=(-1)(-2)(-3)(-4)(-5)+120=-120+120=0.$$ As $-1$ is a root of $f\_k(n)$, then $n+1$ divides it (again consider denominators in the application).
Note also that all members of the "small sets" have $n+1$ dividing $k!$, showing that such sets are finite.
|
7
|
https://mathoverflow.net/users/334725
|
401284
| 164,737 |
https://mathoverflow.net/questions/401282
|
5
|
Let $Q=(Q\_0,Q\_1)$ be the following quiver, $Q\_0$ consist of 2 vertices, denoted by 1,2. $Q\_1$ consist a loop at 1 called $\gamma$, an arrow $\alpha$ from 1 to 2 and an arrow $\beta$ from 2 to 1. The relation $\rho$ is $\{\beta\alpha, \beta\gamma, \gamma\alpha, \gamma^m\}$ for some integer $m\geq 2$. Given a field $k$, is the quiver algebra $kQ$ always of finite representation type?
|
https://mathoverflow.net/users/134942
|
Is this quiver with relations of finite representation type
|
An easy way to see that the algebra is of infinite representation type (for any $m \geq 2$) is to observe that it is a [string algebra](http://www.math.uni-bonn.de/people/schroer/fd-atlas-files/FD-BiserialAlgebras.pdf), and that you have strings of arbitrary length, each corresponding to an indecomposable module.
For instance the strings $(\alpha \beta \gamma^{-1})^t$, $t \geq 1$, correspond to indecomposable modules of vector space dimension $3t+1$.
|
8
|
https://mathoverflow.net/users/18756
|
401294
| 164,741 |
https://mathoverflow.net/questions/401289
|
29
|
I came across this problem while doing some simplifications.
So, I like to ask
>
> **QUESTION.** Is there a closed formula for the evaluation of this series?
> $$\sum\_{(a,b)=1}\frac{\cos\left(\frac{a}b\right)}{a^2b^2}$$
> where the sum runs over all pairs of positive integers that are relatively prime.
>
>
>
|
https://mathoverflow.net/users/66131
|
Closed formula for a certain infinite series
|
Apply Möbius summation, the formula for $\sum\_{n>=1}\cos(2\pi n x)/n^2$ to obtain:
$$11/4-45\zeta(3)/\pi^3=1.00543...\;$$
|
49
|
https://mathoverflow.net/users/81776
|
401304
| 164,743 |
https://mathoverflow.net/questions/401302
|
3
|
Let $R$ be a dvr and $U\to \text{Spec}(R)$ an affine smooth $R$-scheme with non-empty special fiber $U\_0$.
Let $Z\subset U$ be a closed subset. Assume the intersection of $Z$ with $U\_0$ is empty.
>
> Is $Z$ empty?
>
>
>
If $U\to \text{Spec}(R)$ was proper then the answer would be yes because the image of $Z$ could only be the closed point in $\text{Spec}(R)$.
A closed point of $Z$ could map to the generic point of $\text{Spec}(R)$ without properness of $U\to \text{Spec}(R)$, so $Z$ could be contained in the generic fiber, so the answer should be no.
I'm just having a bit of trouble picturing the situation visually.
|
https://mathoverflow.net/users/nan
|
Non-empty closed subsets with empty special fiber
|
$R=\mathbb{Z}\_p, U=\mathrm{Spec}\:\mathbb{Z}\_p\times \mathbb{Q}\_p, Z=\mathrm{Spec}\:\mathbb{Q}\_p\neq \emptyset$
|
3
|
https://mathoverflow.net/users/334839
|
401305
| 164,744 |
https://mathoverflow.net/questions/401241
|
3
|
Let $G=(V,E)$ be a simple, undirected graph, finite or infinite, with $V \neq \emptyset$. Is the following statement true?
>
> There is a cardinal $\kappa \leq |V|$ and an injective map $\psi : V \to {\cal P}(V)$ such that for $v\neq w\in V$ we have: $$\{v,w\} \in E \; \text{ if and only if }\; \big|\big(\psi(v) \setminus \psi(w)\big)\cup\big(\psi(w)\setminus \psi(v)\big)\big| < \kappa.$$
>
>
>
|
https://mathoverflow.net/users/8628
|
Representing graphs by sets of small symmetric difference
|
Statement fails for $G = K\_{2, 3}$. Proof is either with computer search, or by case analysis (an attempt follows).
Let the parts of $G$ be $v\_0, v\_1$ and $u\_0, u\_1, u\_2$ respectively. Consider $\Delta = |\psi(u\_0) \triangle \psi(u\_1)| + |\psi(u\_0) \triangle \psi(u\_2)| + |\psi(u\_1) \triangle \psi(u\_2)|$. On one hand, $\Delta \geq 3|\kappa|$, and on the other hand, considering contribution of each bit (element of the underlying set of the image of $\psi$), $\Delta \leq 2 \cdot 5$. This implies $|\kappa| \leq 3$.
$|\kappa| = 1$ is trivially impossible. $|\kappa| = 2$, together with injectiveness of $\psi$, implies (WLOG) $\psi(v\_0) = \varnothing$, $\psi(u\_i) = \{i\}$. The only $\psi(v\_1)$ at distance at most $1$ from all $\psi(u\_i)$ is $\varnothing$, which clearly fails.
If $|\kappa| = 3$, then pairwise distances between $u\_i$ are (in some order) $3, 3, 4$. WLOG $\psi(u\_0) = \varnothing$, $\psi(u\_1) = \{0, 1, 2\}$, $\psi(u\_2) = \{0, 3, 4\}$. $\psi(v\_i)$ must be at distance at most $2$ from each of them. The only suitable set is $\{0\}$, thus choosing $\psi(v\_i)$ is impossible.
|
2
|
https://mathoverflow.net/users/106512
|
401317
| 164,750 |
https://mathoverflow.net/questions/401291
|
3
|
Given a $C^\*$-algebra $A$, we write $\Omega(A)$ for its space of characters, i.e. its non-zero algebra homomorphisms $A \to \mathbb{C}$. If $X$ is a compact Hausdorff space, it is well-known that
$$X \to \Omega(C(X)): x \mapsto \operatorname{ev}\_x$$
is a homeomorphism of topological spaces.
---
Let $X,Y$ be compact Hausdorff spaces.
Let a $\*$-homomorphism $\pi: C(X) \to C(Y)$ be given. Then we can associate a continuous map $Y \to X$ by the composition
$$Y \cong \Omega(C(Y)) \to \Omega(C(X)) \cong X$$
where the middle map sends $\chi \mapsto \chi\circ \pi.$
---
Question: Is it possible to explicitly write down a formula for this map $Y \to X$? I guess this can be answered if we can write down an explicit inverse for the homeomorphism
$$X \to \Omega(C(X)).$$
|
https://mathoverflow.net/users/216007
|
A $*$-homomorphism $C(X) \to C(Y)$ gives a continuous map $Y \to X$
|
Given a maximal ideal $\mathfrak{m} \subseteq C(X)$, the corresponding point $x \in X$ or rather its singleton $\{x\}$ is the intersection of all zero sets $Z(f)$ of all functions $f \in \mathfrak{m}$.
So the map $\varphi : Y \to X$ associated to $\pi : C(X) \to C(Y)$ is defined by
$$\{\varphi(y)\} = \bigcap\_{f \in C(X),\, \pi(f)(y)=0} Z(f).$$
But as Yemon Choi points out, $\pi(f)=f \circ \varphi$ can also be seen as the definition of $\varphi$!
|
5
|
https://mathoverflow.net/users/2841
|
401319
| 164,751 |
https://mathoverflow.net/questions/401321
|
7
|
Say that structures $\mathfrak{A},\mathfrak{B}$ with the same underlying set are **parametrically equivalent** iff every primitive relation/function in one is definable (with parameters) in the other. For example, every group $\mathfrak{G}=(G;\*,{}^{-1})$ is parametrically equivalent to its "torsor reduct" $\mathfrak{T}\_\mathfrak{G}=(G; (x,y,z)\mapsto x\*y^{-1}\*z)$.
Parametrically equivalent structures can still have very different combinatorial properties. In analogy with the notion of vertex transitive graphs, say that a structure $\mathfrak{A}$ is **point-transitive** iff the natural action of its automorphism group is $1$-transitive - that is, if for every $a,b\in\mathfrak{A}$ there is some $f\in Aut(\mathfrak{A})$ with $f(a)=b$. If $\mathfrak{G}$ is a nontrivial group then $\mathfrak{G}$ is not point-transitive *(no automorphism can move the identity)* but $\mathfrak{T\_G}$ is point-transitive *(for each $g\in G$ the map $a\mapsto g\*a$ is in $Aut(\mathfrak{T\_G}$))*.
I'm broadly interested in understanding the information captured by the family of automorphism groups of parametric equivalents of a given structure (see also [here](https://math.stackexchange.com/questions/4213906/must-the-poset-of-automorphism-group-variants-be-upwards-directed)). Motivated by the group/torsor example, one question which seems like it should be easy to answer is: which structures have a parametric equivalent whose automorphism group acts $1$-transitively? But even for simple structures things aren't very clear to me. So I'd like to start with a simple example:
>
> Is the field of rationals $\mathfrak{Q}=(\mathbb{Q};+,\times)$ parametrically equivalent to a point-transitive structure?
>
>
>
Per the torsor example above, the answer is affirmative for $(\mathbb{Q};+)$. It's also affirmative for $(\mathbb{Q};+,<)$ and various similar expansions, by the same construction, so the existence of such a parametric equivalent isn't connected to any obvious model-theoretic tameness property. However, multiplication seems to complicate things.
|
https://mathoverflow.net/users/8133
|
Is $\mathbb{Q}$ "equivalent" to a structure with transitive automorphism group action?
|
Consider $(\mathbb{Q},R)$, where $R$ is the 6-ary relation defined by $$R(a,b,c,d,e,f) \iff (a-b)(c-d)=(e-f)$$
The automorphism group of this structure includes at least the translations $x\to x+h$. Since those translations include $x \to x+(b-a)$, there is always a translation taking $a$ to $b$, so the group of automorphisms is point-transitive.
We can define $x+y$ and $xy$ in terms of $R$ as the unique $s$ and $t$ for which $R(1,0,s,x,y,0)$ and $R(x,0,y,0,t,0)$ respectively -- i.e. using the equations
\begin{align}
(1-0)(s-x)&=(y-0)\\
(x-0)(y-0)&=(t-0)
\end{align}
Finally, we can define $R$ in terms of $+$ and $\times$ directly by $ac+bd+f=ad+bc+e$.
|
6
|
https://mathoverflow.net/users/nan
|
401326
| 164,753 |
https://mathoverflow.net/questions/401348
|
3
|
Let $X$ be a subset of $\mathbb R^2$ consisting of $n$ distinct points. Let $d\_1(X)$ be the number of pairs of points of $X$ on distance $1$ from each other. Define
$$d\_1(n)=\sup\_{X\subset \mathbb R^2|, |X|=n}d\_1(X).$$
In particular $d\_1(1)=0$, $d\_1(2)=1$, $d\_1(3)=3$, $d\_1(4)=5$, etc.
**Question.** I wonder what is known about the asymptotic behaviour of $d\_1(n)$ and about its upper bound?
It should be at least $n\log(n)$ as the following example shows:
**Example.** It is easy to see that $d\_1(2^d)\ge d2^d$, indeed, one can take the set $Y$ consisting of $d$ unit vectors and then take $2^Y$ consisting of sums of vectors over all $2^d$ subsets of $Y$. Then for generic $Y$ we have $|2^Y|=2^d$ and $d\_1(2^Y)=d2^d$. By fiddling a bit with $Y$ one can increase $d\_1(2^Y)$.
|
https://mathoverflow.net/users/13441
|
Planar subsets with many pairs of points on distance $1$
|
This is the so-called "Erdős unit distances problem"; see for instance the [related Wikipedia entry](https://en.wikipedia.org/wiki/Unit_distance_graph#Number_of_edges) or this [recent survey by Szemerédi](https://doi.org/10.1007/978-3-319-32162-2_15).
As you might expect, a good deal is known, but the problem is by no means resolved.
|
9
|
https://mathoverflow.net/users/25028
|
401351
| 164,762 |
https://mathoverflow.net/questions/401342
|
2
|
Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space, $(X\_n:\Omega\rightarrow \mathbb{R}^m)\_n$ be a sequence of i.i.d. random variables and let $L:\mathbb{R}^m\rightarrow [0,\infty)$ be Lipschitz. Let $\mu\_n:=\frac1{n} \sum\_{k=1}^n \delta\_{X\_k}$. Are there conditions under which:
$$
\mathbb{P}\left(|\mathbb{E}\_{X\sim\mu\_n}[L(X)]-\mathbb{E}\_{X\sim Law(X\_1)}[L(X)]|\geq t\right)\leq \exp\left(
I(t)
\right),
$$
where $I$ is a [good rate function](https://en.wikipedia.org/wiki/Rate_function)?
|
https://mathoverflow.net/users/298030
|
Concentration Inequality for Bounding Lipschitz Empirical Lass
|
Your inequality is trivial and useless as written. On its left-hand side we have a probability which is $\le1$ and goes to $0$ as $t\to\infty$, whereas on the right-hand side we have an expression which is $\ge1$ and goes to $\infty$ as $t\to\infty$, because for any good rate function $I$ on $[0,\infty)$ we have $I(t)\to\infty$ as $t\to\infty$. Also, the left-hand side of your inequality depends on $n$, whereas the right-hand side does not. Overall, this is not how a good rate function is used.
The corrected version of your inequality is as follows:
\begin{equation\*}
P(|T\_n|\ge t)\le e^{-nI(t)} \tag{1}
\end{equation\*}
for some good rate function $I$ and all real $t\ge0$, where
\begin{equation\*}
T\_n:=\frac1n\,\sum\_1^n Y\_k,\quad Y\_k:=L(X\_k)-EL(X\_k).
\end{equation\*}
Moreover, the conditions that the $X\_k$'s take values in $\mathbb R^m$ and that $L$ is Lipschitz are of no relevance. Instead, all we need here is that the $Y\_k$'s are iid zero-mean real-valued random variables.
Once all this preliminary cleaning is done, we can now say that for (1) to hold (for some good rate function $I$ and all real $t\ge0$), it is sufficient that
\begin{equation\*}
Ee^{h|Y\_1|}<\infty \tag{2}
\end{equation\*}
for some real $h>0$. (It is also easy to see that (2) is also necessary for (1).)
Indeed, assume (2) holds for some real $h>0$. By Markov's inequality, for any $t\ge0$,
\begin{equation\*}
P(T\_n\ge t)\le e^{-ntx+nl(x)}
\end{equation\*}
for all $x\ge0$, where
\begin{equation\*}
l(x):=\ln Ee^{xY\_1}
\end{equation\*}
and hence
\begin{equation\*}
P(T\_n\ge t)\le e^{-nI\_+(t)},
\end{equation\*}
where
\begin{equation\*}
I\_+(t):=\sup\_{x\ge0}(tx-l(x)).
\end{equation\*}
Similarly,
\begin{equation\*}
P(-T\_n\ge t)\le e^{-nI\_-(t)},
\end{equation\*}
where
\begin{equation\*}
I\_-(t):=\sup\_{x\ge0}(tx-l(-x)),
\end{equation\*}
so that
\begin{equation\*}
P(|T\_n|\ge t)\le\min(1,e^{-nI\_+(t)}+e^{-nI\_-(t)}). \tag{3}
\end{equation\*}
Note that $I\_+(t)\ge th-l(h)\to\infty$ as $t\to\infty$, and similarly $I\_-(t)\to\infty$ as $t\to\infty$. Also, the functions $I\_\pm$ are nondecreasing and lower semi-continuous, being the pointwise suprema of a family of nondecreasing continuous functions; so, the functions $I\_\pm$ are also left-continuous. Also, $I\_\pm(0)=0$ -- because $l(0)=0$, $l'(0)=EY\_1=0$, and $l$ is convex, so that $l(x)\ge0$ for all $x\ge0$.
So, there exists
\begin{equation\*}
t\_\*:=\max\{t\ge0\colon e^{-I\_+(t)}+e^{-I\_-(t)}\ge1\}\in(0,\infty),
\end{equation\*}
and then $e^{-I\_+(t)}+e^{-I\_-(t)}\ge1$ for $t\in[0,t\_\*]$ and $e^{-I\_+(t)}+e^{-I\_-(t)}<1$ for $t\in(t\_\*,\infty)$.
Defining now the function $I$ by the requirement that
\begin{equation\*}
e^{-I(t)}=e^{-I\_+(t)}+e^{-I\_-(t)}
\end{equation\*}
for $t>t\_\*$, with $I(t):=0$ for $t\in[0,t\_\*]$, we get
\begin{equation\*}
\min(1,e^{-nI\_+(t)}+e^{-nI\_-(t)})\le e^{-nI(t)}
\end{equation\*}
for all natural $n$ and all real $t\ge0$.
Thus, (3) yields (1).
|
6
|
https://mathoverflow.net/users/36721
|
401355
| 164,764 |
https://mathoverflow.net/questions/401345
|
6
|
Let $\mu$ be a probability measure with finite support on integers or the real line with the property that $\mu( 0) \le p$ for a fixed $0<p <1$. Let $S\_n$ denote the random walk starting at $0$, where each step has distribution $\mu$. Denote by $p\_n$ the probability that $S\_n=0$, so $p=p\_1$.
Is there a function $f(n)$ depending only on $p$ (and not on the entire distribution of $\mu$) such that $p\_n \le f(n)$ and $f(n)$ tends to zero as $n \to \infty$?
If $f(n)$ can also depend on $\mu$, then this is a very classical problem with precise asymptotic expressions for $p\_n$. However, all these results seem to depend on other parameters of $\mu$. I would appreciate any comments or pointers to papers and books with similar results.
|
https://mathoverflow.net/users/3635
|
Uniform upper bounds for the return probability of random walks on $ \mathbb{R}$
|
The actual (negative) answer was given in fedja's comment.
Since fedja said he would wait for somebody to come up with a reference for the relaxed statement, here we go:
Let $X$ be a random variable (r.v.) with distribution $\mu$. By (say) inequality (2.5) of Chapter III in [Petrov's book](https://link.springer.com/book/10.1007/978-3-642-65809-9), for some universal real constant $C>0$ and all real $t>0$
$$P(S\_n=0)\le C/\sqrt{n(1-Q(t))},$$
where
$$Q(t):=\sup\_{x\in\mathbb R}P(x\le X\le x+t).$$
Letting now $t\downarrow0$, we get
$$P(S\_n=0)\le f(n):=C/\sqrt{n(1-q)},$$
where
$$q:=Q(0)=\sup\_{x\in\mathbb R}P(X=x).$$
Assuming now that $q<1$, we get $f(n)\to0$ as $n\to\infty$.
|
3
|
https://mathoverflow.net/users/36721
|
401360
| 164,765 |
https://mathoverflow.net/questions/401364
|
6
|
The [1991 paper](https://www.sciencedirect.com/science/article/pii/0022404991900306#!) of Lewis, “*Is there a convenient category of spectra?*” proved that it is impossible to have a point-set model for spectra satisfying the following criteria:
1. There is a symmetric monoidal smash product $\wedge$;
2. We have an adjunction $\Sigma^\infty\dashv\Omega^\infty$;
3. The sphere spectrum $\mathbb{S}$ is the unit for $\wedge$;
4. There is either a natural transformation
$$(\Omega^\infty E)\wedge(\Omega^\infty F)\Rightarrow\Omega^\infty(E\wedge F)$$
or a natural transformation
$$\Sigma^\infty(E\wedge F)\Rightarrow(\Sigma^\infty E)\wedge(\Sigma^\infty F),$$
either of which is then required to satisfies the usual coherence conditions for monoidal functors.
5. There is a natural weak equivalence $\Omega^\infty\Sigma^\infty X\dashrightarrow\lim\_{n\in\mathbb{N}}(\Omega^n\Sigma^nX)$.
Since Lewis's paper, a number of model categories of spectra have appeared, each of which satisfies some, but not all, of the requirements (1)–(5). For instance, the category of $\mathbb{S}$-modules of Elmendorf–Kriz–Mandell–May satisfy (1)–(4), but not (5).
A modern point of view regarding spectra is that they are the $\infty$-categorical analogue of abelian groups in the sense that $\mathsf{Grp}\_{\mathbb{E}\_{\infty}}(\mathcal{S})\cong\mathsf{Sp}\_{\geq0}$. Similarly, the $\infty$-categorical analogue of commutative monoids are $\mathbb{E}\_{\infty}$-spaces, the $\mathbb{E}\_{\infty}$-monoids in the symmetric monoidal $\infty$-category of spaces $\mathcal{S}$.
Is there an analogue of Lewis's argument for $\mathbb{E}\_{\infty}$, giving a similar list of nice properties we may expect of a point-set model of $\mathbb{E}\_{\infty}$-spaces, but such that there is no such point-set model satisfying all of them?
Moreover, in this case, how do the current point-set models for $\mathbb{E}\_{\infty}$-spaces (such as [$\*$-modules](https://arxiv.org/abs/0811.0553), [$\Gamma$-spaces](https://www.sciencedirect.com/science/article/pii/0040938374900226), [$\mathcal{I}$-spaces](https://arxiv.org/abs/1103.2764)) fare in such a list?
|
https://mathoverflow.net/users/130058
|
Lewis's convenience argument for $\mathbb{E}_{\infty}$-spaces
|
For $E\_\infty$ spaces, homotopy-theoretically there is a functor $L: \mathcal{S} \to E\_\infty \mathcal{S}$ with a right adjoint $R$. The only property on this list that really needs replacing on this list is property (5): the unit
$$
X \to RL(X)
$$
should be homotopy equivalent to the natural inclusion
$$
X \to Free\_{E\_\infty}(X) \simeq \coprod\_{k \geq 0} E \Sigma\_k \times\_{\Sigma\_k} (X^k)
$$
into the free $E\_\infty$-space on $X$. (Yes, yes, possibly a version with basepoints, I know)
I *believe* that all three of the models of $E\_\infty$ spaces that you list (commutative monoids in $\*$-modules, $\Gamma$-spaces, commutative $\mathcal{I}$-space monoids) satisfy properties (1)-(4) and fail the analogue of property (5), each due to an issue about whether an input to an adjunction is cofibrant/fibrant. For $\Gamma$-spaces, for example, the map $X \to RL(X)$ only adds a disjoint basepoint. Perhaps someone with more experience with the other models would be able to fill in those stories better.
|
9
|
https://mathoverflow.net/users/360
|
401367
| 164,768 |
https://mathoverflow.net/questions/398427
|
8
|
This is a follow-up question to this [MO question](https://mathoverflow.net/questions/374180/function-of-x-1-x-2-x-3-x-4-that-factors-in-two-ways-as-phi-1-x-1-x-2/374197#374197), which was asked by Richard Stanley in a comment to my answer there.
Let $S$ be a commutative monoid and $f(x\_1, \dots, x\_n)$ be a function from $S^n$ to $S$. Given a partition $\alpha$ of $[n]$, we say that $f$ *factors with respect to $\alpha$*, if for each $A \in \alpha$ there exists a function $f\_A$ (which only depends on the variables $x\_i$ for $i \in A$) such that $f=\prod\_{A \in \alpha} f\_A$. Given two partitions $\alpha$ and $\beta$ of $[n]$, $a \wedge b$ is the partition of $[n]$ whose sets are the non-empty sets of the form $A \cap B$ for $A \in \alpha$ and $B \in \beta$.
>
> **Question.** Is it true that if $f$ factors with respect to both $\alpha$ and $\beta$, then $f$ also factors with respect to $\alpha \wedge \beta$?
>
>
>
My answer to the linked question shows that the answer is **yes** if $S$ is a group, but the proof uses the fact that inverses exist. My proof also works for non-abelian groups (as long as you are careful what factoring means), but for this question I am happy to assume that $S$ is commutative.
|
https://mathoverflow.net/users/2233
|
Functions over monoids which factor in two different ways
|
Take $S$ to be the monoid on the set $\{0,2,3,4,5,6\}$ with operation $x\oplus y=\min(x+y,6).$
Let $g:\{0,1\}^3\to S$ be the function $(x,y,z)\mapsto \min(x+y+z+4,6).$ Using any surjective $h:S\to \{0,1\}$ this can be converted to $f(x,y,z)=g(h(x),h(y),h(z)).$ I'll just work with $g.$
$g$ factors with respect to $12,3$ and $1,23$: $g(x,y,z)=(x+y+2)\oplus(z+2)=(x+2)\oplus(y+z+2).$
But $g$ does not factor with respect to $1,2,3.$ Suppose for contradiction that $g(x,y,z)=f\_1(x)\oplus f\_2(y)\oplus f\_3(z)$ for all $x,y,z.$ We must have $f\_i(1)=f\_i(0)+1$ for each $i,$ because $g(0,0,1)=g(0,1,0)=g(1,0,0)=5$ and $g(0,0,0)=4.$ This implies $f\_i(0)\geq 2.$ But then $f\_1(0)\oplus f\_2(0)\oplus f\_3(0)\geq 6> g(0,0,0).$
|
4
|
https://mathoverflow.net/users/164965
|
401397
| 164,775 |
https://mathoverflow.net/questions/401385
|
1
|
Is there a $P$ time definable sequence of **succinct polynomial sized representation of balanced bipartite graphs** whose number of perfect matchings is a primorial?
For factorial a complete bipartite graph suffices.
Motivations:
1. The speed of growth of the function defining the sequence might capture prime gap information (for instance it might be able to provide lower bound on worst possible gap).
2. Since $n\#\approx 2^{O(n)}$ holds the sequence might be a sequence of planar graphs of sizes $O(n)$ in number of vertices and edges providing a way to count primes better.
3. The graphs perhaps satisfy symmetry as these objects usually exist for a reason (for complete balanced bipartites the symmetry group is canonical and perhaps for primorial as well there is a nice symmetry group and perhaps might hint at the construction of these family of bipartite graphs).
Input is $n$ but I seek succinct representation of a fixed family and the representation may be $polylog(n)$ sized expressing an exponentially larger balanced bipartite graph with perfect matching a primorial.
---
Succinct representation of balanced complete bipartite graphs could be a circuit of $2\lceil\log\_2(n)\rceil$ input bits where the first half of the input bits represent one color and other half another color and the circuit trivially answers $1$ for all $n^2$ pairs of inputs to imply all pairs of vertices pairing the colors are connected. A boolean circuit for it is trivial as it always outputs $1$.
|
https://mathoverflow.net/users/10035
|
Succinct polynomial sized representation of balanced bipartite graphs whose perfect matching count is a primorial
|
Consider the graph $G\_k$ with vertex set $$\{u\_1, \ldots, u\_k, v\_1, \ldots, v\_k\}$$ and edges $$\{(u\_1, v\_1), \ldots, (u\_1, v\_k)\} \cup \{(u\_2, v\_1), \ldots, (u\_k, v\_{k-1})\} \cup \{(u\_2, v\_k), \ldots, (u\_k, v\_k)\}$$ It has $k$ perfect matchings, because once $u\_1$ is assigned to $v\_i$ this forces the assignments $$\{(u\_2, v\_1), \ldots, (u\_i, v\_{i-1}), (u\_{i+1}, v\_{i+1}), \ldots, (u\_k, v\_k)\}$$
Therefore a disjoint union of $G\_p$ for all prime $p \le n$ has $n\#$ perfect matchings.
As noted in my earlier comment, this answers the question but disappoints you on all points of the motivation.
---
For a compact circuit representation similar to the encoding for the complete bipartite graph given in the question, encode both $u\_i$ and $v\_i$ from $G\_p$ as $(p, i)$, so that the full input (the encoding of a $u$ vertex as $(p,i)$ and the encoding of a $v$ vertex as $(q,j)$) is $4 \lceil \lg n \rceil$ bits. Then the circuit needs to encode $$(p = q) \wedge (i = 1 \vee i = j \vee i = j+1)$$Addition of $1$ to a $\lg n$-bit number and equality testing of two $\lg n$-bit numbers can both be done in $O(\lg n)$ gates.
|
4
|
https://mathoverflow.net/users/46140
|
401399
| 164,776 |
https://mathoverflow.net/questions/401398
|
2
|
At 1st we consider some weak statement of Chevalley–Warning theorem for any finite field: If $f$ is a homogeneous polynomial of degree $d$ with $n$ independent variables over a finite field $F$. Then if $ n > d $ then there is a non trivial solution of this homogeneous polynomial in $ F^{n} / \{0,0,...,0\} $.
Now if we consider $f$ is a homogeneous polynomial of degree $d$ with $n$ independent variables over the rational field $\mathbb{Q}$ and if $ n > d $ does there exist always a nontrivial solution in $ \mathbb{Q}^{n} / \{0,0,...,0\} $? The answer is no when $r$ is even as for an example $ x\_{1}^{d} + x\_{2}^{d}+.....+ x\_{n}^{d} = 0 $ have only one solution $\{0,0,...,0\} $ in $ \mathbb{Q}^{n}$.
My question is when $ d $ is odd does there exist always a nontrivial solution in $ \mathbb{Q}^{n} / \{0,0,...,0\} $? Describe when $ d = 3 $ and $ n= 5 $ especially.
|
https://mathoverflow.net/users/215016
|
Chevalley–Warning theorem for rational field $\mathbb{Q} $
|
The condition that $d$ is odd just implies that there is no real obstruction to the existence of rational points, but there could still be $p$-adic obstructions for some prime $p$.
As an example, let $p$ be a prime and $a$ an integer which is coprime to $p$ for which $a \bmod p$ is not a cube (this necessarily implies that $p \equiv 1 \bmod 3$).
Consider the cubic surface:
$$X: \quad x\_1^3 - a x\_2^3 + p(x\_3^3 - a x\_4^3) = 0.$$
It is a fun exercise to show that this has no non-trivial rational solution; the proof actually shows that there is no non-trivial $p$-adic solution. Note that there is obviously a non-trivial solution mod $p$, e.g. $(0,0,1,1) \bmod p$, as predicted by Chevalley-Warning.
A more interesting question is whether having a real point and a $p$-adic point is sufficient to guarantee a rational point; again this is not true in general, with smooth cubic surfaces giving counter-examples (this is called the Hasse principle in the literature). A famous example, due to Cassels and Guy, is:
$$5x\_1^3 + 12x\_2^3 + 9x\_3^3 + 10x\_4^3 = 0.$$
Proving that there is no non-trivial rational point in this case is not so elementary (it requires cubic reciprocity). From a modern perspective there is a Brauer--Manin obstruction to the Hasse principle in this case.
I suspect from the question you may have a specific example in mind you are studying. If you allow many variables (roughly $d2^d$) then the circle method can actually show that the Hasse principle holds, which may be sufficient for your application.
**Addendum:** This is a summary of the alternative construction given in the comments, written up for completeness.
Let $d \in \mathbb{N}$. Let $D$ be a division algebra over $\mathbb{Q}$ of dimension $d^2$. This can be shown to exist using the fundamental exact sequence for the Brauer group of $\mathbb{Q}$ from class field theory, combined with the fact that the period = index for central simple algebras over number fields.
Consider the [reduced norm](https://en.wikipedia.org/wiki/Central_simple_algebra) on $D$. After choosing a $\mathbb{Q}$-basis for $D$, this defines a homogeneous polynomial $N$ of degree $d$ in $d^2$ variables. We claim that the equation $N(\mathbf{x}) = 0$ has no non-trivial solution.
To see this, we use the fact that an element of a central simple algebra has reduced norm $0$ if and only if it is not invertible. But by construction $D$ is a division algebra, hence every non-zero element is invertible, hence only $\mathbf{0}$ has $0$ norm, as claimed.
Note: Division algebras satisfy the Hasse principle, hence this failure of existence of rational solutions is explained by a failure of existence of $p$-adic solutions for some $p$.
It is possible, though painful, to make these norm form equations explicit. The first issue is writing down explicit division algebras, which Will Sawin did in the comments.
As an example, consider the quaternion algebra $(a,b)$. The reduced norm here is just the [Pfister form](https://en.wikipedia.org/wiki/Pfister_form):
$$x\_1^2 - ax\_2^2 - bx\_3^2 + abx\_4^2.$$
Taking $b = p $ to be an odd prime and $a<0$ such that $a \bmod p$ is a quadratic non-residue, one obtains an example which does not non-trivially represent zero $p$-adically, but has real solutions. (The above cubic surface example is a variant of this equation).
|
4
|
https://mathoverflow.net/users/5101
|
401403
| 164,777 |
https://mathoverflow.net/questions/401415
|
0
|
Let $\Omega$ be a convex body$^{\boldsymbol{1}}$ in $\mathbb{R}^n$ where $n$ is a positive integer. Fix a positive integer $k$ and some $0<\alpha\leq 1$. Let $k\_1> k\_2>0$. Does there necessarily exist a diffeomorphism $\phi^{k,\alpha}\in C(\Omega,\Omega)$ satisfying:
$$
\lVert\phi-1\_{\Omega}\rVert\_{k,\alpha}= k\_1 \text{ and } \lVert\phi-1\_{\Omega}\rVert\_{\infty}\leq k\_2,
$$
where $\lVert\cdot\rVert\_{k,\alpha}$ is the [usual norm on the Hölder space](https://en.wikipedia.org/wiki/H%C3%B6lder_condition#H%C3%B6lder_spaces) $C^{k,\alpha}(\Omega,\mathbb{R}^n)$ and $\lVert\cdot\rVert\_{\infty}$ is the familiar [sup-norm](https://en.wikipedia.org/wiki/Uniform_norm) on $C(\Omega,\mathbb{R}^n)$.
Intuitively, I imagine this can be constructed by starting with some “small homeomorphism” $\tilde{\phi}:\Omega\rightarrow \Omega$ and then smoothing it out/mollifying it. But I don't know how to formalize this idea, or if it is even true.
**Edit$^{\boldsymbol{1}}$:**
Following *@Pietro Majer*'s point; I should mention that I also assume that $\Omega$ is a convex body in $\mathbb{R}^n$ (so non-empty interior) and that $n\in \mathbb{Z}^+$ *(so $\Omega$ cannot be a point).*
|
https://mathoverflow.net/users/176409
|
Existence of a Hölder homeomorphism satisfying prescribed norm constraints
|
As it is the answer is no, by the following counter-example
$$.$$
|
1
|
https://mathoverflow.net/users/6101
|
401420
| 164,780 |
https://mathoverflow.net/questions/401417
|
3
|
The Wiener measure is (in the classical sense) a Gaussian measure on the Banach space $C[0,1]:=\{f:[0,1] \to \mathbb{R} \mid f\text{ is continuous and } f(0)=1\}$.
The Wiener process is a stochastic process whose definition can be found in any textbook. In any text, the stochastic integrals or stochastic differential equations use the notation $dW$ frequently, which should denote the Wiener measure, I suppose.
However, I cannot figure out the exact relation between the Wiener 'process' and 'measure'. Wikipedia says that the Wiener process induces the Wiener measure but what exactly does that mean?
I am afraid this question might not belong to MO but I ask here. Could anyone please clarify and help me understand?
|
https://mathoverflow.net/users/56524
|
What exactly is the relation between the Wiener process and Wiener measure?
|
The Wiener measure $w$ is the distribution of the Wiener process/random function $W$ on $C[0,1]$; that is,
$$P(W\in A)=w(A)$$
for all Borel sets $A\subseteq C[0,1]$. Here "Borel sets" can be replaced by "open sets" or "closed sets".
Equivalently,
$$Ef(W)=\int\_{C[0,1]}f\,dw$$
for all (say) nonnegative Borel-measurable functions $f\colon C[0,1]\to\mathbb R$. Here "nonnegative Borel-measurable" can be replaced by (say) "bounded continuous".
|
4
|
https://mathoverflow.net/users/36721
|
401422
| 164,781 |
https://mathoverflow.net/questions/401414
|
3
|
I am trying to upper bound the variance of a centered tree and I would like to get an upper bound which would look like : $$\sum\limits\_{\substack{ (l\_1, ..., l\_d) \neq (k\_1, ..., k\_d), \\ \sum\_{j=1}^d l\_{j} = \sum\_{j=1}^d k\_{j} = k } } \frac{k!}{k\_{1}! ... k\_{d} !} \frac{k!}{l\_{1}!... l\_{d} !} \left(\frac{1}{2} \right)^{\sum\_{j=1}^d \mathbb{1}\_{l\_j \neq k\_j}} \leq C\_d d^{2k} \varepsilon(k)$$
where $C\_d$ is any constant depending only on $d$ and $\varepsilon(k)$ tends to $0$ as $k$ tends to infinity. We are summing over both $l\_j$s and $k\_j$s. We know that a single sum of multinomial coefficients equals $d^k$. Therefore, writing $k =qd+r$, I tried to distinguish two parts of the sum, a part $A$ where $\left(1/2\right)^{\sum\_{j=1}^d \mathbb{1}\_{l\_j \neq k\_j}} \leq (1/2)^q$ and we have $$A \leq \left(\frac{1}{2}\right)^q d^{2k}, $$ and another part $B$ which contains much less terms (only the terms where $\sum\_{j=1}^d \mathbb{1}\_{l\_j \neq k\_j} \leq q-1$) with something polynomial in $k$ (i.e $k^d+...$) but I've been unsuccessful so far.
The overall sum comes from, with $x \in [0,1]^d$ a given point and $X$ a uniform random variable on $[0,1]^d$, $$\mathbb{E} \left[ \left( \sum\limits\_{\substack{(k\_1, ..., k\_d), \\ \sum\_{j=1}^d k\_{j} = k } } \frac{k!}{k\_{1}! ... k\_{d} !} \left(\frac{1}{d}\right)^k \prod\_{j=1}^d \mathbb{1}\_{\lceil 2^{k\_j} x\_j\rceil = \lceil 2^{k\_j} X\_j\rceil} \right)^2\right] $$
where $\lceil y \rceil$ corresponds to the first integer $m$ such that $m \geq y$.
I welcome any suggestions or ideas and would be happy to discuss this problem with you !
|
https://mathoverflow.net/users/335858
|
Upper bound for the crossed-terms of a sum of multinomial coefficients
|
I think that you cannot hope for $\epsilon(k)$ going to zero. Using the identity
$$(x\_1+\ldots+x\_d)^k=\sum\_{0\leq k\_1,k\_2,\ldots,k\_d\leq k\_1+\ldots+k\_d=k}\frac{k!x\_1^{k\_1}\cdots x\_d^{k\_d}}{k\_1!\cdots k\_d!}$$
we get the lower bound
$$2^{-d}d^{2k}-\sum\_{0\leq k\_1,\ldots,k\_d\leq k\_1+\ldots+k\_d=d}\left(\frac{k!}{k\_1!\cdots k\_d!}\right)^2$$ (take the square of the above identity and set $x\_1=\ldots=x\_d=1$) on your quantity.
We have yet to show that the
correction
$$C=\sum\_{0\leq k\_1,\ldots,k\_d\leq k\_1+\ldots+k\_d=d}\left(\frac{k!}{k\_1!\cdots k\_d!}\right)^2$$
is $o(d^{2k})$.
Since contributions to
$$A=\sum\_{0\leq k\_1,\ldots,k\_d\leq k\_1+\ldots+k\_d=d}\frac{k!}{k\_1!\cdots k\_d!}$$
behave roughly like a multiple of a Gaussian centered at $\frac{1}{d}(k,k,\ldots,k)$ with variance of order $\sqrt{k}$, the size of $C$ is roughly (up to a constant asymptotically) given by
$$\frac{A^2}{\sqrt{k}^d}=d^{2k}\sqrt{k}^{-d}$$ which ends the proof.
|
1
|
https://mathoverflow.net/users/4556
|
401430
| 164,783 |
https://mathoverflow.net/questions/401234
|
3
|
I want to prove the following: (Here, $W^{2,2}$ is a Sobolev space as defined in [Evans](https://bookstore.ams.org/gsm-19-r), chapter 5; $S$ is a Schwartz space; and if $A$ is a distribution and $a$ a function, then $\langle A, a\rangle$ means $A(a)$).
>
> **Theorem.** Let $\newcommand{\C}{\mathbb C}\newcommand{\R}{\mathbb R}C\in]0,\infty[$. For every $f\in L^2 (\mathbb R;\C)$ there exists a $g\in (L^2 \cap W^{2,2}\_{\text{loc}})(\R;\C)$ such that (in the weak sense)
> \begin{equation}
> -g'' + C^2 g = f
> \end{equation}
> and an explicit solution is given by
> \begin{equation}
> g(t) = \frac{1}{2C} \int\_{\R} e^{-C|t-\tau|} f (\tau)\,\mathrm d\tau.
> \end{equation}
>
>
>
---
**My attempt.** (Skip to the bottom for my question)
Consider the tempered distribution
\begin{equation\*}\begin{split}
\mathscr Z: S(\R;\C)&\to\C, \\
\phi&\mapsto\int\_\R e^{-{C\lvert t\rvert}} \phi(t)\,\mathrm dt = \int\_0^\infty e^{-Ct}\phi(t)\,\mathrm dt+\int\_{-\infty}^0 e^{Ct} \phi(t)\,\mathrm dt.
\end{split}\end{equation\*}
Then we have, for all $\phi\in S(\R;\C)$,
\begin{equation\*}
\langle\mathscr Z',\phi\rangle=-\langle\mathscr Z,\phi'\rangle=\int\_0^\infty e^{-Ct}\phi'(t)\,\mathrm dt+\int\_{-\infty}^0 e^{Ct} \phi'(t)\,\mathrm dt.
\end{equation\*}
Integrating both terms by parts, where the exponential term gets differentiated and $\phi'$ gets integrated, we get
\begin{equation\*}\begin{split}
-\langle\mathscr Z',\phi\rangle &= \left[e^{-Ct}\phi(t)\right]^\infty\_0+C\int\_0^\infty e^{-Ct}\phi(t)\,\mathrm dt+\left[e^{Ct}\phi(t)\right]^0\_{-\infty}-C\int\_{-\infty}^0 e^{Ct}\phi(t)\,\mathrm dt \\
&= C\int\_0^\infty e^{-Ct}\phi(t)\,\mathrm dt-C\int\_{-\infty}^0 e^{Ct}\phi(t)\,\mathrm dt.
\end{split}\end{equation\*}
Integrating both terms by parts in the same way, we get (where $\delta\_0$ is the Dirac distribution at $0$)
\begin{equation\*}\begin{split}
\langle\mathscr Z'',\phi\rangle &= \langle-\mathscr Z',\phi'\rangle \\
&= C\left[e^{-Ct}\phi(t)\right]^\infty\_0+C^2\int\_0^\infty e^{-Ct}\phi(t)\,\mathrm dt - \left(C \left[e^{Ct}\phi(t)\right]\_{-\infty}^0-C^2\int\_{-\infty}^0 e^{Ct}\phi(t)\,\mathrm dt\right) \\
&= C^2 \langle\mathscr Z,\phi\rangle-2C\phi(0) = C^2 \langle\mathscr Z,\phi\rangle-2 C\langle{\delta\_0,\phi}\rangle.
\end{split}\end{equation\*}
---
Now I would like to finish by writing
$$g = \frac{\mathscr Z\* f}{2C}$$ and therefore
$$-g''+C^2 g = \frac{-(\mathscr Z'' \* f)+C^2 (\mathscr Z\* f)}{2C} = \frac{-((C^2\mathscr Z-2C\delta\_0)\*f)+C^2 (\mathscr Z\* f)}{2C}=\delta\_0\*f = f.$$
**My question:** However, to do this, I formally use
$$ (A\*a)'=(A'\*a)$$
when $A$ is a distribution. Is there some result that justifies this? And furthermore, does this result also imply that $g$, defined as the convolution of $f$ with $\mathscr Z$, can be written as a $W^{2,2}\_{\text{loc}}$ function?
|
https://mathoverflow.net/users/129831
|
How to rigorously differentiate the convolution of a distribution and a $L^2$ function?
|
The last step is formally justified by **15.8, differentiation property** of José Sebastião e Silva's "Integrals and orders of growth of distributions." (The paper is currently available [here](http://jss100.campus.ciencias.ulisboa.pt/Publicacoes/Artigos-de-Investigacao/Inv-JSS/Integrals%20and%20orders%20of%20growth%20of%20distributions.pdf).)
More precisely: $\mathscr K$ is defined by a continuous function. Also, by Lemma 8.2 of Haim Brezis' *Functional Analysis, Sobolev Spaces and Partial Differential Equations* (2010), we have, in the weak sense, since $f\in L^2\subset L^1\_{\text{loc}}$, $F'=f$ where $F(x):=\int\_0^x f$ is continuous.
Therefore, one can indeed apply 15.8 mentioned above.
|
1
|
https://mathoverflow.net/users/129831
|
401437
| 164,787 |
https://mathoverflow.net/questions/401439
|
0
|
Let $x=(x\_1,\ldots,x\_d)$ be uniformly distributed on the unit-sphere in $\mathbb R^d$.
>
> **Question.**
> What is a good upper-bound for Wasserstein distance between $N(0,1/d)$ and the marginal distribution of $x\_1$ ?
>
>
>
|
https://mathoverflow.net/users/78539
|
Wasserstein distance between $N(0,1/d)$ and the marginal distribution of $x_1$ when $x=(x_1,\ldots,x_d)$ is uniform on the unit-sphere in $R^d$
|
Let $g \sim N(0, (1/d)I\_d)$ be independent of $x$. Then $g\_1 \overset{\mathcal L}{=} \|g\| x\_1$, so $(x\_1, \|g\|x\_1)$ is a coupling between the marginal distribution of $x\_1$ and $N(0, 1/d)$.
The norm $\|g\|$ is sharply concentrated around $1$, with fluctuations of order $1/\sqrt{d}$, so the Wasserstein distance is bounded above by
$$\mathbb E|x\_1 - \|g\|x\_1| = (\mathbb E|1 - \|g\||) \mathbb E|x\_1| = O(d^{-1}),
$$
where we have used the fact that $\mathbb E|x\_1|^2 = (1/d)\mathbb E\|x\|^2 = 1/d$ by isotropy of the distribution of $x$, so that $\mathbb E|x\_1| \le \sqrt{\mathbb E|x\_1|^2} = 1/\sqrt{d}$ thanks to Jensen's inequality.
|
4
|
https://mathoverflow.net/users/37014
|
401444
| 164,789 |
https://mathoverflow.net/questions/401441
|
30
|
### Summary
Someone claims $\mathbb{R}$ can be constructed as the following intriguing quotient, which is related to Gromov's bounded cohomology. I want to find out if it is true.
$$\frac{\bigl\{f:\mathbb{Z} \to \mathbb{Z} \mathrel| \mbox{ the set } \{f(m+n)-f(m)-f(n) \mathrel| m, n \in \mathbb{Z}\} \mbox{ is bounded}\bigr\}}{\{f:\mathbb{Z} \to \mathbb{Z} \mathrel| f \mbox{ is bounded}\} }.$$
**EDIT** See KConrad's [comment](https://mathoverflow.net/questions/401441/a-natural-construction-of-real-numbers#comment1026184_401507) below. A similar construction, described in [Hermans - An elementary construction of the real
numbers, the $p$-adic numbers and the
rational adele ring](https://www.universiteitleiden.nl/binaries/content/assets/science/mi/scripties/bachelor/2017-2018/hermans-bscthesis.pdf), yields $\mathbb{Q}\_p$ and the rational adele ring $\mathbb{A}$.
---
### Main text
In [A'Campo - A natural construction for the real numbers](https://arxiv.org/abs/math/0301015v1), a natural
construction of the real numbers is given as follows. (**EDIT**: In this post I only address the bijection and the ring structure. The correspondence is complete; for that please refer to the paper.)
**Definition (Bounded cochains)**
Define $C^{n} = C^{n}(\mathbb{Z})$ to be $\operatorname{Map}(\mathbb{Z}^{\times
n}, \mathbb{Z})$ and $C^n\_b = C^{n}\_b(\mathbb{Z})$ to be the
subset consisting of functions $f$ having bounded image, i.e.
$\operatorname{Card}(\operatorname{Im}(f)) < \infty$.
**Definition (Differentials)**
Define $d: C^n \to C^{n+1}$ to be such that
$$df(x\_1,\dotsc,x\_{n+1}) = f(x\_2,\dotsc,x\_{n+1}) + \sum\_{k=1}^{n}(-1)^{k} f(x\_1, \dotsc, x\_{k-1}, x\_k+x\_{k+1}, \dotsc, x\_{n+1}) + (-1)^{n+1}f(x\_1,\dotsc,x\_n).$$
Obviously, $d(C^n\_b) \subseteq C^{n+1}\_b$, so $C^n\_b \subseteq
d^{-1}(C^{n+1}\_b)$.
**Algebraic Operations**
Clearly, $C^1$ has a ring structure, where addition is given by
point-wise addition, and multiplication is given by function
composition.
**Claim. $\mathbb{R} \simeq d^{-1}(C^2\_b)/(C^1\_b)$**
This claim is made in page 1 (definition of $\mathbb{R}$) and
page 6 (that $\mathbb{R}$ is the usual $\mathbb{R}$) of the
paper. An explicit map $\Phi: d^{-1}(C^2\_b) \to \mathbb{R}$ is
given in page 4 as
$$\lambda \mapsto \left[\left(\frac{\lambda (n+1)}{n+1}\right)\_{n \in
\mathbb{N}}\right]$$
using Cauchy sequences.
**Question** Why is $\ker(\Phi) = C^1\_b$? By the
definition of the equivalence on the set of Cauchy sequences,
$\Phi(\lambda)$ represents $0 \in \mathbb{R}$ if and only if
>
> For each $\epsilon > 0$, there exists an $N \in \mathbb{N}$
> such that $\frac{|\lambda(n+1)|}{(n+1)} < \epsilon$ whenever $n > N$.
>
>
>
However, $\lambda: \mathbb{Z} \to \mathbb{Z}$ that sends $n$ to
$\lfloor\sqrt{|n|}\rfloor$ is one such element that is not in $C^1\_b$
(namely, not bounded).
**EDIT** As Anthony Quas points out [below](https://mathoverflow.net/questions/401441/a-natural-construction-of-real-numbers#comment1026025_401441), such $\lambda$ isn't in the preimage of $d$. You can see this by taking $m = n \to \infty$. Still, I'm curious about a direct proof for the kernel being $C^1\_b$. This is given in Anthony Quas's [answer](https://mathoverflow.net/a/401445).
### Related
* [Category-theoretic description of the real numbers](https://math.stackexchange.com/q/839848) (Mathematics Stack Exchange)
* Gromov's bounded cohomology, see
[Ivanov - Notes on the bounded cohomology theory](https://arxiv.org/abs/1708.05150) and the 9th page
of A'Campo's paper.
|
https://mathoverflow.net/users/124549
|
A natural construction of real numbers?
|
So here is my attempt to reconstruct the construction...
Suppose $f\colon\mathbb Z\to\mathbb Z$ satisfies $|f(m+n)-f(m)-f(n)|\le M$ as $m,n$ run over $\mathbb Z$. Then setting $m=n=2^k$, we see
$|f(2^{k+1})-2f(2^k)|\le M$, from which it follows that $f(2^k)/2^k$ is a Cauchy sequence, and so converges to some $\alpha\in\mathbb R$.
Now given $n\in (2^{k-1},2^k]$, let its binary expansion be
$n=2^{k-1}+2^{j\_1}+\ldots+2^{j\_r}$ (with $r< k\le \log\_2 n$).
Inductively, we can show $\Big|f(n)-\big[f(2^{k-1})+\ldots +f(2^{j\_r})\big]\Big|\le rM$, from which, together with the above, we can deduce that the full sequence $f(n)/n$ converges to $\alpha$.
On the other hand, given $\alpha\in\mathbb R$, if one defines $f(n)=\lfloor \alpha n\rfloor$, then it is easy to see that $f(n+m)-f(n)-f(m)$ takes values in $\{0,1\}$, so that the map is surjective.
Finally, suppose $f(n)/n\to 0$ and $|f(n+m)-f(n)-f(m)|\le M$ is bounded. We have to show that $f$ is bounded. If $|f(n\_0)|\ge 2M$ for some $n\_0$, then $|f(2n\_0)|\ge 2|f(n\_0)|-M$, from which we see inductively that $f(2^kn\_0)\ge (2^k+1)M$, contradicting the assumption that $f(n)/n\to 0$.
|
27
|
https://mathoverflow.net/users/11054
|
401445
| 164,790 |
https://mathoverflow.net/questions/401432
|
2
|
This is a reference request/nomenclature question. Let $A \subseteq \mathbb{P}^n$ be a finite set of points not contained in a hyperplane (over some field), and let $\sigma\_r(A)$ be the $r$-th secant variety to $A$. This secant variety forms a *subspace arrangement*, i.e., a finite union of linear subspaces of $\mathbb{P}^n$. Is there a specific name for subspace arrangements of this form? Surely such arrangements have been studied, and I would be very grateful for a reference.
|
https://mathoverflow.net/users/150898
|
Secant variety to a zero-dimensional projective variety
|
I believe this would be a dual arrangement of a star arrangement.
A star arrangement is a union of subspaces defined as follows. Let $H\_1,\dotsc,H\_d$ be a collection of hyperplanes and fix an integer $c$. The codimension $c$ star arrangement $X\_c$ is the union of intersections of $c$ of the $H\_i$, over all size $c$ subsets of $H\_1,\dotsc,H\_d$. Usually there is some hypothesis of linear generality so that any $c$ of the $H\_i$'s are independent. See for example <https://arxiv.org/abs/1203.5685>, <https://arxiv.org/abs/1801.04579>.
Given a subspace arrangement $\mathcal{A} = \{W\_1,\dotsc,W\_s\}$ of subspaces in $\mathbb{P}V$, the dual arrangement is $\mathcal{A}^\* = \{W\_1^\perp,\dotsc,W\_s^\perp\}$ in $\mathbb{P}V^\*$.
Well, the subspace $W = H\_{i\_1} \cap \dotsb \cap H\_{i\_c}$ is dual to $W^\perp = \operatorname{span}\{H\_{i\_1}^\perp,\dotsc,H\_{i\_c}^\perp\}$. So the secant varieties you're asking about are dual arrangements of star arrangements. I don't know a better or alternative name and I'm not aware of any work on these arrangements specifically, but the authors of the papers I linked might be able to give more information if you write to them.
By the way, these dual arrangements of star arrangements are not necessarily themselves star arrangements. Let $A$ be a set of $5$ general points in $\mathbb{P}^3$ so $\sigma\_2(A)$ is a set of $10$ lines. This arrangement of lines has $4$ through each node (point where lines meet) and $3$ in each plane spanned by two meeting lines. This arrangement of lines isn't a star arrangement. A star arrangement given by pairwise intersections of $5$ general hyperplanes would again consist of $10$ lines, but this time with $4$ in each plane spanned by two meeting lines, and $3$ lines through each node.
|
2
|
https://mathoverflow.net/users/88133
|
401462
| 164,796 |
https://mathoverflow.net/questions/401460
|
8
|
Let $R$ be a ring of global dimension $1$. Then I have seen the claim (in a paper, and in this MO post [When do chain complexes decompose as a direct sum?](https://mathoverflow.net/questions/32854/when-do-chain-complexes-decompose-as-a-direct-sum)) that any chain complex over $R$ is equivalent to its cohomology as an object of the derived category of $R$-module. How does one prove such a thing? (Either a sketch of the argument or a reference would be appreciated).
|
https://mathoverflow.net/users/59235
|
Chain complexes split in the derived category over rings of global dimension 1
|
One reference is H. Krause, "Derived categories, resolutions, and Brown representability", Contemporary Math. vol.436, AMS, 2007, p.101-139 or <https://arxiv.org/abs/math/0511047> , Section 1.6.
Another possible reference is L. Positselski, O.M. Schnürer, "Unbounded derived categories of small and big modules: Is the natural functor fully faithful?", J. Pure Appl. Algebra 225 (2021) Paper No. 106722 or <https://arxiv.org/abs/2003.11261> , Appendix A.
|
8
|
https://mathoverflow.net/users/2106
|
401464
| 164,797 |
https://mathoverflow.net/questions/401459
|
4
|
Given a positive integer $d$, does there exist an integer $n$ that depends only on $d$ (or perhaps also on the dimension of $X$), such that for any degree $d$ finite étale covering $\pi: \widetilde X \to X$ of projective varieties and very ample line bundle $\mathcal L$ on $X$, $\pi^\ast(\mathcal L)^{\otimes n}$ is very ample.
|
https://mathoverflow.net/users/129738
|
Pullback of very ample line bundles under finite étale covering
|
The answer is no. Take for $X$ a (smooth) plane curve of degree $2p+3$. There exists a line bundle $M$ on $X$ with $M^{2}=K\_X$ and $h^0(M)=0$. Then $\eta :=M(-p)$ is a line bundle of order 2 in $JX$, giving rise to a double étale covering $\pi :\tilde{X}\rightarrow X $. Put $\mathscr{L}=\mathscr{O}\_X(1)$. Then $$H^0(\tilde{X},\pi ^\*\mathscr{L}^p)=\pi^\* H^0(X,\mathscr{L}^p)\oplus \pi^\*H^0(X,\mathscr{L}^p\otimes \eta )=\pi^\*H^0(X,\mathscr{L}^p)\,.$$ This means that the map defined by $\pi ^\*\mathscr{L}^p$ factors through $\pi $, hence $\pi ^\*\mathscr{L}^p$ is not very ample.
|
8
|
https://mathoverflow.net/users/40297
|
401465
| 164,798 |
https://mathoverflow.net/questions/401476
|
4
|
Say that an algebra $\mathfrak{A}$ (in the sense of universal algebra) is **point-transitive** iff for every $a,b\in\mathfrak{A}$ there is a $\pi\in Aut(\mathfrak{A})$ with $\pi(a)=b$. While genuinely point-transitive algebras are somewhat rare, many naturally-occurring algebras yield *the same clone as* a point-transitive algebra. Specifically, given an algebra $\mathfrak{A}$ let $\mathsf{Cl}(\mathfrak{A})$ be the smallest set of functions from some finite Cartesian power of $\mathfrak{A}$ to $\mathfrak{A}$ which contains all the constant functions and projection functions, all the primitive functions of $\mathfrak{A}$ itself, and is closed under composition. **(EDIT: as Keith Kearnes states [below](https://mathoverflow.net/a/401506/8133), this is actually the *polynomial clone* of $\mathfrak{A}$.)** Then:
* Each *group* $(G;\*,{}^{-1},e)$ yields the same clone as its "torsor reduct" $(G;(a,b,c)\mapsto a\*(b^{-1}\*c))$, which is clearly point-transitive since each $x\mapsto y\*x$ is an automorphism.
* A similar trick works with *rings*: each ring $(R;0,1,+,-,\times)$ yields the same clone as $(R; (a,b,c,d,e)\mapsto (a-b)(c-d)+e)$, and the latter is again point-transitive via maps of the form $x\mapsto x+y$. (This is basically [due to Matt F.](https://mathoverflow.net/a/401326/8133))
Say that an algebra $\mathfrak{A}$ is **almost-point-transitive** iff we have $\mathsf{Cl}(\mathfrak{A})=\mathsf{Cl}(\mathfrak{B})$ for some point-transitive algebra $\mathfrak{B}$ with the same underlying set. I'd like to just ask "Which algebras are almost-point-transitive?," but I don't see how to make that precise in the right way (e.g. to avoid "the almost-point-transitive ones" as an answer). Instead, the following seems like it might be more immediately approachable:
>
> If $\mathfrak{A}$ is almost-point-transitive and $\mathfrak{B}\in\mathsf{HSP}(\mathfrak{A})$, must $\mathfrak{B}$ be almost-point-transitive as well?
>
>
>
I strongly suspect that the answer is negative but I don't see how to construct a counterexample.
|
https://mathoverflow.net/users/8133
|
Do almost-point-transitive algebras generate almost-point-transitive varieties?
|
Let me distinguish between **clone** and **polynomial clone**.
The former is the smallest composition-closed
collection of operations on $A$
containing the primitive operations of $\mathbb A$ and the projections,
while the latter is the smallest composition-closed
collection of operations on $A$
containing the primitive operations of $\mathbb A$, the projections,
and the constant operations.
In fact, let me write $\mathbb A\_A$ for the constant expansion of $\mathbb A$,
and then refer to the clone of $\mathbb A\_A$ when I want to
talk about the polynomial clone of $\mathbb A$.
The $1$-unary
algebra $\mathbb A =\langle \mathbb Z; f(x)\rangle$,
where $f^{\mathbb A}(x)=x+1$,
provides a negative answer to the question.
This algebra has transitive automorphism group,
since all powers of $f^{\mathbb A}(x)$ are automorphisms.
The terms in this language in the variable
$x$ are $x, f(x), f^2(x), \ldots$, so each proper subvariety
of the variety of all $1$-unary algebras is
axiomatizable by some set of
identities of the form $f^m(x)\approx f^n(x)$,
$m\neq n$, or of the form $f^m(x)\approx f^n(y)$.
None of these hold in $\mathbb A$, so $\mathbb A$
generates the variety of all $1$-unary algebras.
Let $\mathbb B = \langle \mathbb Z; f(x)\rangle$
where $f^{\mathbb B}(x)=|x|$.
$\mathbb B$ is a $1$-unary algebra whose basic operation
has proper range.
The algebra $\mathbb B\_B$ has the property
that its clone consists of essentially unary
operations only, it contains all the constant
operations on the domain, and it contains
a nonsurjective unary operation $f^{\mathbb B}(x)$.
Let $\mathbb C$ be any algebra defined on the same
set as $\mathbb B$ whose polynomial clone
agrees with that of $\mathbb B$.
Then $\mathbb C\_C$ has the property
that its clone consists of essentially unary
operations only, it contains all the constant
operations on the domain, and it contains
a nonsurjective unary operation $f^{\mathbb B}(x)$.
Since the clone of the reduct $\mathbb C$
contains all the nonconstant operations
of the clone of $\mathbb C\_C$,
the clone of $\mathbb C$
must contain $f^{\mathbb B}(x)$.
The range of $f^{\mathbb B}(x)$ is closed under
any automorphism of $\mathbb C$, so the automorphism
group of $\mathbb C$ is not transitive.
Altogether this shows that (i) $\mathbb A$
has transitive automorphism group (so $\mathbb A$ is point-transitive), but
(ii) the variety generated by $\mathbb A$
contains an algebra $\mathbb B$
that is not almost-point-transitive.
|
4
|
https://mathoverflow.net/users/75735
|
401506
| 164,805 |
https://mathoverflow.net/questions/401497
|
0
|
Let $(x\_n)\_{n\in\mathbb{N}}$ be a non-increasing sequence in [0,1], (i.e. $x\_n\ge x\_{n+1},n\in\mathbb{N} $), such that $x\_n\ge\frac{1}{n},n\in\mathbb{N} $.
If we fix $k\in\mathbb{N}$ is there necessarily a lower bound c>0 for the fractions $\frac{x\_{kn}}{x\_n}$ for all $n\in\mathbb{N}$?
I was thinking that if it is true then maybe $\frac{x\_{kn}}{x\_n}\ge\frac{1}{k}$.
|
https://mathoverflow.net/users/336624
|
Lower bound for $\frac{x_{kn}}{x_n}$, where $(x_n)_{n\in\mathbb{N}}$ is a non-increasing sequence in [0,1] with $x_n\ge\frac{1}{n}$
|
Such a lower bound does not exist in general.
E.g., for natural $j$ and natural $n\in((j-1)!,j!]$, let
$x\_n:=1/(j-1)!$, with $x\_1:=1$, so that $x\_n\ge1/n$ for all natural $n$. Also, for any fixed natural $k\ge2$ and all natural $j\ge k$, we have $kj!\in(j!,(j+1)!]$ and hence
$$\frac{x\_{kj!}}{x\_{j!}}=\frac{(j-1)!}{j!}\to0$$
as $j\to\infty$, so that
$$\inf\_{n\ge1}\frac{x\_{kn}}{x\_n}=0.$$
|
1
|
https://mathoverflow.net/users/36721
|
401510
| 164,808 |
https://mathoverflow.net/questions/401483
|
3
|
In addition to classic two-valued logic, there are *many* many-valued logics, including Łukasiewicz's and Kleene's three-valued logics, Gödel's many-valued logic $G\_k$, and infinite-valued fuzzy logic and probability logic.
I wonder: Was ever the case that some kind of "0-valued" and "1-valued" logics came up in a natural way in a certain context?
|
https://mathoverflow.net/users/244671
|
0-valued and 1-valued logics?
|
There is an article by C. L. Hamblin titled *One-valued Logic,* The Philosophical Quarterly, 66 (1967), 38-45, that you may find interesting.
|
2
|
https://mathoverflow.net/users/8027
|
401515
| 164,811 |
https://mathoverflow.net/questions/401498
|
1
|
I have the following problem: I'm given a linear bounded operator $P\in \mathcal{L}(L^2([a,b]))$, $a,b\in \mathbb{R}$ and I want to find a sequence of approximating linear bounded operators $(P\_n)\_{n\geq 1}$ satisfying the following conditions:
1. $P\_n \to P$ in $\mathcal{L}(L^2([a,b]))$ as $n\to \infty$ (i.e. in the norm topology);
2. $P\_n(H^1\_0([a,b])) \subset H^1\_0([a,b])$ for every $n\geq 1$;
3. $P\_n : H^1\_0([a,b]) \to H^1\_0([a,b])$ is bounded.
Here, $H^1\_0([a,b])$ denotes the Sobolev space of order $1$ with Dirichlet boundary conditions.
Is it possible to find such an approximating sequence for a general operator $P\in \mathcal{L}(L^2([a,b]))$?
Thank you very much in advance!
|
https://mathoverflow.net/users/248499
|
Approximating linear bounded operator on $L^2([a,b])$
|
Here's a more explicitly worked out version of my comment above. Consider $P(\sin nx)=\sin 2^n x$. This is an isometry, so in particular bounded in $L^2(0,\pi)$. It can not be approximated in the desired way.
Indeed, suppose we had an operator $Q$ with $\|P-Q\|<\epsilon$ that is also bounded on $H^1$. Since $\|\sin nx \|\_{H^1}\simeq n$, this would have to map to functions $g=Q(\sin nx)$ with $\|g'\|\_2\lesssim n$. However, we also have $\|g-\sin 2^n x\|\_2<\epsilon$. This is not possible because the second inequality forces $g$ to mimic the oscillations of $\sin 2^n x$, but that will make $\|g'\|\_2$ too large.
We have approximately $2^n$ intervals, of size $\simeq 2^{-n}$ each, on which $\sin 2^n x\ge 3/4$. If we had $g(x)\le 1/4$ throughout such an interval $I$, then $\|g-\sin 2^nx\|^2\_{L^2(I)}\gtrsim 2^{-n}$. Thus we can have at most $\simeq \epsilon 2^n$ such intervals.
In other words, for a fraction close to one (as $\epsilon\to 0$) of the $2^n$ intervals where $|\sin 2^n x|\ge 3/4$, the function $g$ will also take a corresponding value $|g|\ge 1/4$ on these intervals. So $g$ oscillates between $\pm 1/4$ at least $\gtrsim 2^n$ times.
However, when $g$ changes from $1/4$ to $-1/4$ on $J$, then
$$
\frac{1}{2} \le \int\_J |g'| \le \|g'\|\_{L^2(J)} |J|^{1/2} ,
$$
and thus in particular $\|g'\|^2\_{L^2(J)}\gtrsim 1$. Since we have $\simeq 2^n$ such intervals, $\|g'\|\_2$ comes out much too large.
|
2
|
https://mathoverflow.net/users/48839
|
401528
| 164,816 |
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