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https://mathoverflow.net/questions/413773
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4
|
Let $\overline{\mathbb{F}\_p}$ be the algebraic closure of $\mathbb{F}\_p$. Let $W({-})$ denote the functor of taking $p$-typical Witt vectors. Then the extension $\mathbb{F}\_p\rightarrow \overline{\mathbb{F}\_p}$ induces a map $f: \mathbb{Z}\_p=W(\mathbb{F}\_p)\rightarrow W(\overline{\mathbb{F}\_p})$.
Is $f$ split as a $\mathbb{Z}\_p$-module map?
As far as I understand, $W(\overline{\mathbb{F}\_p})$ is built out of $\mathbb{Z}\_p$ by adjoining prime-to-$p$-th-power roots of unity, so it seems like you could try sending all those adjoined roots to zero. But some roots live in $\mathbb{Z}\_p$, so you better not send those to zero.
|
https://mathoverflow.net/users/163893
|
Splitting the Witt vectors of $\overline{\mathbb{F}_p}$
|
$W(\overline{\mathbb F\_p})$ is the completion of the direct limit of $W(\mathbb F\_{p^{n!}})$. The ring $W(\mathbb F\_{p^{n!}})$ is a free module of rank $n$ over $W(\mathbb F\_{p^{(n-1)!}})$, so the natural map $W(\mathbb F\_{p^{(n-1)!}}) \to W(\mathbb F\_{p^{n!}})$ has a splitting.
Combining all these splittings, we get a splitting of the direct limit. Since it is a $\mathbb Z\_p$-module homomorphism, it extends to the completion, giving a splitting of the completion.
|
7
|
https://mathoverflow.net/users/18060
|
413774
| 168,785 |
https://mathoverflow.net/questions/413725
|
9
|
Compare the following two results:
>
> Thm A) Let $A$ be a commutative $C^\*$-algebra and let $X$ be its Gelfand spectrum. Gelfand duality says that there's a natural isometric $\*$-isomorphism from $A$ to $C(X)$, the $C^\*$-algebra of continuous functions on $X$.
>
>
>
>
> Thm B) Let $A$ be a commutative ring and $X=\operatorname{Spec}A$. There's a natural sheaf of rings $\mathscr{O}$ on $X$ such that $A=\Gamma(X,\mathscr{O})$.
>
>
>
Both theorems are very close. **I wonder if it's possible to obtain theorem A as a particular case of theorem B, or a variant thereof.**
|
https://mathoverflow.net/users/131975
|
Is there a relation between Gelfand duality and the spectrum of a ring (with its Zariski topology)?
|
Yes, both Theorem A and Theorem B are special cases of a more general construction.
Denote by $R$ the category of commutative unital C\*-algebras or the category of commutative rings.
Denote by $R'$ the full subcategory of $R$ given by reduced objects in $R$, meaning the only nilpotent element is zero.
(All commutative unital C\*-algebras are reduced.)
Given an object $X∈R$, we can consider its poset $\def\Spec{\mathop{\sf Spec}}\Spec X$ of quotient objects (= subobjects in the opposite category) that belong to $R'$.
One can show that $\Spec X$ is a [locale](https://ncatlab.org/nlab/show/locale), the (localic) Zariski/Gelfand spectrum of $X$.
Furthermore, assuming the axiom of choice, this locale is [spatial](https://ncatlab.org/nlab/show/spatial+locale), so it corresponds to a topological space, namely, the traditional Zariski/Gelfand spectrum of $X$.
Given an open element $U$ of $\Spec X$ corresponding to a reduced quotient $X→Q$, its kernel $I⊂X$ is a radical ideal
and we can consider the localization $X[S^{-1}]∈R$,
defined using a universal property in the category $R$,
where $S=\{a∈X\mid I⊂\sqrt{(a)}\}$.
The assignment $$U↦X[S^{-1}]$$ defines a sheaf on $\Spec X$ valued in $R$.
This is precisely the structure sheaf of $\Spec X$
for both the Gelfand spectrum and the Zariski spectrum.
Of course, this construction is applicable to many other categories $R$:
* finitely presented entire functional calculus algebras, recovering the Stein duality for globally finitely presented Stein spaces (i.e., complex geometry).
* finitely generated germ-determined C^∞-rings, recovering the Dubuc duality for C^∞-loci (i.e., differential geometry);
* other dualities for algebraic geometry, such as formal schemes, etc.
* various versions of the above for derived geometry;
* Boolean algebras, recovering the Stone duality for compact totally disconnected Hausdorff spaces;
* complete Boolean algebras, recovering the Stonean duality for compact extremally disconnected Hausdorff spaces;
* localizable Boolean algebras, recovering the [Gelfand-type duality](https://mathoverflow.net/questions/23408/reference-for-the-gelfand-duality-theorem-for-commutative-von-neumann-algebras) for [compact strictly localizable enhanced measurable spaces](https://mathoverflow.net/questions/20740/is-there-an-introduction-to-probability-theory-from-a-structuralist-categorical/20820#20820).
|
12
|
https://mathoverflow.net/users/402
|
413776
| 168,786 |
https://mathoverflow.net/questions/413744
|
2
|
Let $X$ be a semi-normal projective variety and $p: \widetilde{X} \to X$ be the normalization. Suppose that $\widetilde{X}$ is smooth and there exists two *smooth* divisors $D\_1, D\_2 \subset \widetilde{X}$ such that $D\_1 \cong D\_2 \cong X\_{\mathrm{sing}}$ and $p$ induces as isomorphism between $\widetilde{X} \backslash (D\_1 \cup D\_2)$ and $X\backslash X\_{\mathrm{sing}}$. Note that, the non-singular divisors $D\_1$ and $D\_2$ map to $X\_{\mathrm{sing}}$ under the morphism $p$. Can we conclude that $X$ has double point singularities along $X\_{\mathrm{sing}}$?
|
https://mathoverflow.net/users/38832
|
When is the singularity of a semi-normal variety a double point singularity
|
I think this is true if we assume that the isomorphisms $D\_1 \cong X\_{\mathrm{sing}} \cong D\_2$ are also induced by $p$, at least in characteristic $0$ (though I think everything below is ok away from characteristic $2$). **Edit:** We also need to assume that the $D\_i$ are disjoint in $\widetilde{X}$. If $D\_1 \cup D\_2$ is singular in $\widetilde{X}$ then we can also get worse singularities, see the edit below.
Let $\tau : D\_1 \cong D\_2$ denote the isomorphism induced by $p$ and let $\bar{X}$ be the quotient of $\widetilde{X}$ by the equivalence relation generated by $x \sim \tau(x)$. The quotient exists as a scheme in this case since $D\_i \subset X$ are closed embeddings (this is an example of a pinching or Ferrand pushout, see e.g. [this question and its answers](https://mathoverflow.net/questions/64294/gluing-along-closed-subschemes?rq=1)). Then $\bar{X}$ is semi-normal with the required nodal singularities. Moreover, the map $p$ factors through a map $q : \bar{X} \to X$ by the universal property of quotients. By assumption $q$ is a bijection on points and isomorphism on residue fields so by semi-normality of $X$, $q$ is an isomorphism.
If we don't assume that the isomorphism $D\_1 \cong X\_{\mathrm{sing}} \cong D\_2$ is induced by $p$, then this doesn't have to be true. For example, we can let $\widetilde{X}$ be two copies of $\mathbb{P}^2$ and $D\_i$ be the conic $x^2 + y^2 + z^2$ the $i^{th}$ plane. Then we can consider the equivalence relation generated by identifying the two conics in the natural way as well as identifying $(x,y,z) \in D\_i$ with $(y,x,z) \in D\_i$. Then the quotient $X$ of $\widetilde{X}$ by this equivalence relation is semi-normal and $D\_i \cong \mathbb{P}^1 \cong X\_{\mathrm{sing}}$ but its singularities are not nodal. At a general point of $X\_{\mathrm{sing}}$, the singularities will look like $\mathbb{A}^1$ times the union of the $4$ coordinate axes in $\mathbb{A}^4$ and $p|\_{D\_i}$ is $2$-to-$1$ onto $X\_{\mathrm{sing}}$.
You might also want to take a look at Sections 5.1, 9.1 and 10.2 in Kollár's book *Singularities of the minimal model program*
**Edit:** Here is an example when $D\_i$ are smooth but $D\_1 \cup D\_2$ is not. Let $D\_i$ be the coordinate axes in $\mathbb{A}^2$ and consider the equivalence relation that identifies the two axes by swapping them. The resulting semi-normal surface $X$ is $\mathrm{Spec}$ of the the subring
$$
\{f(x,y) \mid f(t,0) = f(0,t)\} \subset k[x,y].
$$
I think this is isomorphic to $u^3 - uvw + w^2 = 0$ in $\mathbb{A}^3$ which is not nodal. The singularities get worse if $D\_1 \cup D\_2$ has worse singularities, e.g. if $D\_i$ meet in a tacnode.
|
3
|
https://mathoverflow.net/users/12402
|
413780
| 168,789 |
https://mathoverflow.net/questions/413654
|
4
|
Let $A$ be a von Neumann algebra and let $H$ be a (separable) Hilbert space.
It is known (see e.g., Section IV, Thm. 5.5 of Takesaki I) that there exists a Hilbert space $K$ such that $A \subset \mathbb{B}(K)$ such that any normal $\*$-homomorphism $\varphi : A \to \mathbb{B}(H)$ can be written as
$$ \varphi(a) = v^\* a v,$$
where $v: H \to K$ is a partial isometry with $v^\*v = \mathrm{id}\_H$ and $v v^\* \in A^\prime \subseteq \mathbb{B}(K)$.
We are led to consider the sets $\mathrm{Hom}(A, \mathrm{B}(K))$ with its u-topology (defined by seminorms $\varphi \mapsto \|\omega \circ \varphi\|\_{A\_\*}$, where $\omega \in \mathrm{B}(H)\_\*$ is an element of the predual of $\mathrm{B}(H)$) and
$$V(A, H) := \{v \in \mathrm{B}(H, K) \mid v^\*v = \mathrm{id}\_H, vv^\* \in A^\prime\}$$
with the strong operator topology.
By the result stated above, the map $V(A, H) \to \mathrm{Hom}(A, \mathrm{B}(H))$ is surjective.
**Q: Is this map a Serre fibration?**
If not, what are the problems here, and can we assume something on $A$ or change the topologies somehow to ensure this?
Edit: So really, what I am interested in is the question whether the map above admits some kind local sections (this is precisely what the definition of Serre fibration is about). In other words, I would like to know if a family of representations (parametrized by some suitably nice space) can be lifted to a family of implementing partial isometries, at least locally. This would be a family version of the result of Takesaki cited above.
---
So maybe let me write something about what I tried and where I failed.
(1) Let us first prove that the map $V(A, H) \to \mathrm{Hom}(A, \mathrm{B}(H))$ is continuous.
To this end, let $\omega \in \mathrm{B}(H)\_\*$ be positive, which can be written as $\omega(A) = \mathrm{tr}(WA)$ for some positive trace-class operator $W = \sum\_{i=1}^\infty w\_i e\_i \otimes e\_i^\*$.
Let $v\_n \to v$ be a convergent sequence in $V(A, H)$ and let $\varphi\_n$, $\varphi$ be the corresponding sequences of homomorphisms $A \to \mathrm{B}(H)$.
We have to show that $\mathrm{sup}\_{\|a\| \leq 1} |\omega \circ \varphi\_n(a) - \omega \circ \varphi(a)|$ converges to zero.
If $\|a\| \leq 1$, then
$$|\omega \circ \varphi\_n(a) - \omega \circ \varphi(a)| = |\mathrm{tr}(W v\_n^\* av\_n - Wv^\*av)| \\
\leq \sum\_{i=1}^\infty w\_i |\langle e\_i, v\_n^\* a v\_n e\_i - v^\*ave\_i\rangle|\\
\leq \sum\_{i=1}^\infty w\_i \Bigl(|\langle (v\_n-v)e\_i, a v\_n e\_i\rangle| + |\langle ae\_i, (v\_n - v) e\_i\rangle|\Bigr)\\
\leq 2 \sum\_{i=1}^\infty w\_i \|(v\_n-v)e\_i\|,
$$
which converges to zero.
(2) The fibers are as follows: It is not hard to see that if two elements $v\_1, v\_2 \in V(A, H)$ induces the same homomorphism $\varphi : A \to \mathrm{B}(H)$, then $v\_1 = wv\_2$ for a partial isometry $w \in A^\prime$ with $ww^\* = v\_1 v\_1^\*$, $w^\*w = v\_2 v\_2^\*$.
Conversely, given $v$ implementing $\varphi$, then replacing it by $wv$ for a partial isometry $w \in A^\prime$ with $w^\*w = vv^\*$ gives another element of $V(A, H)$ implementing the same $\varphi$.
(3) So what I tried was the following: Fix a projection $p \in A^\prime$, and look at the subset $V\_p(A, H)$ of all $v$ such that $vv^\* = p$, and let $\mathrm{Hom}\_p(A, \mathrm{B}(H))$ be the set of those homomorphisms that are implemented by such a $v$.
Then after fixing a basepoint $v\_0 \in V\_p(A, H)$, then by (2), we have identifications
$$ V\_p(A, H) \approx \mathrm{U}(pH)$$
given by sending $u \in \mathrm{U}(pH)$ to $uv\_0$, and two elements $u\_1$, $u\_2$ correspond to the same element of $\mathrm{Hom}(A, \mathrm{B}(H)$ if and only if $u\_2 u\_1^\* \in \mathrm{U}(A^\prime)$.
Hence there is a continuous bijection
$$\mathrm{U}(pKp)/(\mathrm{U}(pKp)\cap \mathrm{U}(A^\prime)) \to \mathrm{Hom}(A, \mathrm{B}(H)).$$
However, it is now clear to me how to show that this is a homeomorphism.
Also, I am not sure how to attack this if we don't fix $p$ in advance.
|
https://mathoverflow.net/users/16702
|
Families of representations of von Neumann algebras
|
**Theorem.**
The map $V(A,H)\to\operatorname{Hom}(A,\mathbb{B}(H))$ is open.
We write $\omega\_{\xi,\eta}$ for the linear functional $x\mapsto \langle x\xi,\eta\rangle$
and $\omega\_\xi$ for $\omega\_{\xi,\xi}$.
It is an elementary fact that if $\omega\_\xi=\omega\_{\eta}$ on a von Neumann algebra $A$,
then $a\xi\mapsto a\eta$ extends to a partial isometry $u\in A'$ such that $u\xi=\eta$.
We can trim $u^\*u$ a little, if necessary, and make it satisfy $1-u^\*u \sim 1-uu^\*$ (Murray--von Neumann equivalence),
at the cost of $\|u\xi-\eta\|<\epsilon/2$, where $\epsilon>0$ is arbitrary small.
Then $u$ extends to a unitary element in $A'$, still denoted by $u$,
which satisfies $\|u\xi-\eta\|<\epsilon$.
The following perturbation lemma is well-known and follows from
the theory of standard form (see [Takesaki, Section IX]).
**Lemma.**
For any $\epsilon>0$, there is $\delta>0$ which satisfies the following.
For any von Neumann algebra $A\subset B(K)$ and any unit vectors
$\xi,\eta\in K$, if $\|(\omega\_\xi-\omega\_\eta)|\_A\|<\delta$,
then there is a unitary element $u\in A'$ such that
$\|u\xi - \eta\|<\epsilon$.
We postpone the proof of this lemma and prove the theorem.
Let $v\_0\in V(A,H)$ and an SOT neighborhood
$$G=\{ v\in V(A,H) : \forall i\ \|(v-v\_0)\xi\_i\|<\epsilon\}$$
be given.
Here $\xi\_1,\ldots,\xi\_n\in H$ are unit vectors and
$\epsilon>0$.
Take $\delta>0$ from the Lemma for $n^{-1/2}\epsilon$.
Now suppose $v\in V(A,H)$ is such that
$$\| (\omega\_{\xi\_i,\xi\_j}\circ\operatorname{Ad}\_v
- \omega\_{\xi\_i,\xi\_j}\circ\operatorname{Ad}\_{v\_0})|\_A \|<\delta/n$$
for all $i,j$.
We consider the unit vector
$\xi=n^{-1/2}\left[\begin{smallmatrix} \xi\_1 & \cdots & \xi\_n\end{smallmatrix}\right]^T\in H^n$ and view $\mathbb{B}(H^n)=\mathbb{M}\_n\otimes\mathbb{B}(H)$.
Then
$$\|(\omega\_{(1\otimes v)\xi} - \omega\_{(1\otimes v\_0)\xi})|\_{\mathbb{M}\_n\otimes A}\|<\delta.$$
Thus by Lemma, one finds a unitary element
$u\in A'\cong (\mathbb{M}\_n\otimes A)'\cap\mathbb{B}(H^n)$ such that
$\|(1\otimes u)(1\otimes v)\xi - (1\otimes v\_0)\xi\|<n^{-1/2}\epsilon$.
This implies that $uv\in G$, which finishes the proof.
**Proof of Lemma.**
We may assume $K = p(L^2A \otimes \ell\_2)$, where $L^2A$
is a standard representation of $A$ and
$p\in (A\otimes \mathbb{C}1)'\cap\mathbb{B}(L^2A\otimes\ell\_2)$.
Fix a unit vector $\delta\_0\in\ell\_2$.
There are unique vectors $|\xi|$ and $|\eta|$ in the positive cone $(L^2A)\_+$
such that
$\omega\_\xi=\omega\_{|\xi| \otimes\delta\_0}$ and
$\omega\_\eta=\omega\_{|\eta| \otimes\delta\_0}$ on $A \otimes \mathbb{C}1$
(see [Takesaki, Theorem IX.1.2.(iv)]).
There are partial isometries $v$ and $w$ in $(A\otimes \mathbb{C}1)'\cap\mathbb{B}(L^2A\otimes\ell\_2)$ such that $\xi=v|\xi|$ and $\eta=w|\eta|$.
By the generalized Powers--Stormer inequality ([Takesaki, Theorem IX.1.2]),
one has $\| |\xi| - |\eta| \|^2 \le \|(\omega\_\xi-\omega\_\eta)|\_A\|$.
Hence $t:=wv^\*\in p(A\otimes \mathbb{C}1)'p = A'\cap \mathbb{B}(K)$ satisfies
$$\| t \xi - \eta \| = \| w(v^\*\xi-w^\*\eta)\|\approx 0.$$
Let $t=u|t|$ be the polar decomposition. Since $\|t\|\le1$ and
$\|t\xi\|\approx\|\eta\|=1$, one has $|t|\xi \approx \xi$ and $u\xi\approx\eta$.
We can further replace the partial isometry $u\in A'$ with a
unitary element without affecting $u\xi$ much.
|
6
|
https://mathoverflow.net/users/7591
|
413781
| 168,790 |
https://mathoverflow.net/questions/413778
|
3
|
Let $G$ be a finite abelian group, $X$ and $Y$ be two non-empty subsets of $G$ of equal size. Suppose that for each irreducible character $\chi$ of $G$ we have $\sum\_{x\in X}\chi(x)=\sum\_{y\in Y}\chi(y)$. Is it true that $X=Y$ in general?
|
https://mathoverflow.net/users/36341
|
Equality of subsets of abelian groups
|
$\DeclareMathOperator\Irr{Irr}$You can see this from the fact that for abelian groups, irreducible characters form a $\mathbb{C}$-basis of the space of functions from $G$ to $\mathbb{C}$. These functions correspond bijectively to linear transformations from the group algebra $\mathbb{C}G$ to $\mathbb{C}$, and $\Irr(G)$ is also a basis for the space of such linear transformations. Therefore, $\chi(\sum\_{x\in X}x)=\sum\_{x\in X}\chi(x)=\sum\_{y\in Y}\chi(y)=\chi(\sum\_{y\in Y}y)$ for each $\chi\in \Irr(G)$ implies that $f(\sum\_{x\in X}x)=f(\sum\_{y\in X}y)$ for every linear transformation $f:\mathbb{C}G\rightarrow \mathbb{C}$. This happens if and only if $\sum\_{x\in X}x=\sum\_{y\in Y}y$.
There is another way to see this. In general, for split finite dimensional algebra $A$ over a field $k$, the common zero set of irreducible characters $\Irr\_k(A)$ is $J(A)+[A,A]$, where $J(A)$ is the Jacobson radical of $A$ and $[A,A]$ is the commutator subalgebra of $A$ generated by elements of the form $[a,b]=ab-ba$.
If $G$ is a finite abelian group, then the group algebra $\mathbb{C}G$ is finite dimensional, split, semisimple and commutative. Hence, both $J(A)$ and $[A,A]$ are trivial, and the above theorem says that the common zero set of irreducible complex characters is trivial. In particular, if two elements $a$, $b$ of the group algebra satisfy $\chi(a)=\chi(b)$ for each $\chi\in \Irr(G)$, then $a-b$ is in the common zero set, which is $\{0\}$.
|
6
|
https://mathoverflow.net/users/91692
|
413782
| 168,791 |
https://mathoverflow.net/questions/413788
|
1
|
Given an increasing function $f:[0,\infty)\to[0,\infty)$, we can define $$F(x)=\int\_x^{x+1} f(t)dt,$$
which is continuous, increasing function satisfying $$f(x)\leq F(x)\leq f(x+1).$$
Question) For such $f$ can we construct a continuous increasing function $g:[0,\infty)\to[0,\infty)$ such that for all $x\in [0,\infty),$
$$c\_1 f(x)\leq g(x) \leq c\_2 f(x),$$
for some positive constant $c\_1,c\_2?$
|
https://mathoverflow.net/users/184109
|
Given an increasing function, need to construct a continuous increasing function equivalent to given function
|
Using the integer-part function, let
$$f(x) = x ^ {\lfloor x \rfloor +1}$$
Then there is no $g$ satisfying the given condition for this $f$.
For any such $g$, let $b = \lfloor c\_2/c\_1 \rfloor +2$. Then
\begin{align}
b-1<x<b \implies & g(x) \le c\_2 x^b\\
b<x<b+1 \implies & g(x) \ge c\_1x^{b+1}
\end{align}
By the continuity of $g$, $g(b)$ must satisfy both of those inequalities, so
$$c\_2 b^b \ge g(b)
\ge c\_1 b^{b+1}$$
$$\frac{c\_2}{c\_1}\ge b$$
which is impossible.
|
3
|
https://mathoverflow.net/users/nan
|
413792
| 168,794 |
https://mathoverflow.net/questions/413736
|
14
|
The standard formulation of the univalence axiom for a universe type $U$ is that, for all $X : U$ and $Y : U$, the canonical map $(X =\_U Y) \to (X \simeq Y)$ is an equivalence.
As we (usually) cannot form the type of universe types, the usual univalence axiom is actually an axiom scheme, consisting of an instance of the univalence axiom for each universe type.
We can generalise (the form of) the univalence axiom as follows.
Given a type $E (b)$ depending on $b : B$, we might say that $E (b)$ is univalent over $b : B$ if the canonical map $(b\_0 =\_B b\_1) \to (E (b\_0) \simeq E (b\_1))$ is an equivalence.
Of course, not all dependent types are univalent, but the univalence axiom for $U$ is precisely the condition that $X$ is univalent over $X : U$.
**Question.**
Without mentioning universes (or otherwise internalising the condition of being a universe type), could we formulate an axiom scheme or inference rule that is equivalent to the usual univalence axiom scheme in the presence of universe types?
One consequence of the usual univalence axiom scheme is that, for every type $E (b)$ depending on $b : B$, there is a type $E' (b')$ univalent over $b' : B'$ and a map $\chi : B \to B'$ such that $E (b) \equiv E' (\chi (b))$.
Indeed, given a univalent universe $U$ such that, (for every $b : B$) $E (b) : U$, we may take $B' \equiv U$, $E' (b') \equiv b'$, and $\chi \equiv (\lambda b : B . E (b))$.
Anyway, we can take the image factorisation of $\chi$ to replace $B'$ with something smaller, but I'm not sure if this can be done respecting the judgemental equality $E (b) \equiv E' (\chi (b))$.
If it could, it seems to me we would have a higher inductive type characterisation of $B'$ and $E'$.
Applying this to a possibly non-univalent universe type would yield a univalent universe type, so by replacing "universe" with "univalent universe" in the appropriate places we would be able to interpret univalent type theory.
Does this work?
|
https://mathoverflow.net/users/11640
|
How to formulate the univalence axiom without universes?
|
One possibility along these lines is large eliminations for higher inductive types. For instance, here is a large elimination rule for the higher inductive interval type $\mathsf{I}$ with $0,1:\mathsf{I}$ and $\mathsf{seg}:\mathsf{Id}\_{\mathsf{I}}(0,1)$:
$$
\frac{\vdash A \,\mathsf{type} \quad \vdash B \,\mathsf{type} \quad \vdash e : \mathsf{Equiv}(A,B)}{x:\mathsf{I}\vdash E(x)\,\mathsf{type} \quad E(0)\equiv A \quad E(1) \equiv B \quad \mathsf{trans}^E\_{\mathsf{seg}}= e}
$$
(Everything in some ambient context $\Gamma$, of course.) Note that this can be stated without any universe, using only the "is a type" judgment that we always have in dependent type theory. (I'm assuming function extensionality here, so that it doesn't matter what kind of equality or homotopy we have in the final law $\mathsf{trans}^E\_{\mathsf{seg}}= e$.)
Now suppose we have an interval type with this rule, and a universe type $\mathsf{U}$. Suppose moreover that the large elimination rule relativizes to $\mathsf{U}$, i.e. if $A$ and $B$ belong to $\mathsf{U}$ then so does $E$; this seems like a reasonable thing to include when "adding a universe type" to a theory that already had large eliminations. We will prove that $\mathsf{U}$ is univalent, using the encode-decode method.
Given $A,B:\mathsf{U}$, let $\mathsf{encode} : \mathsf{Id}\_{\mathsf{U}}(A,B) \to \mathsf{Equiv}(A,B)$ be defined by transport or path induction; this is the map we want to show to be an equivalence. To define $\mathsf{decode}$ in the other direction, suppose given $e:\mathsf{Equiv}(A,B)$. Then by large elimination we have $E:\mathsf{I} \to \mathsf{U}$ with $E(0)\equiv A$ and $E(1)\equiv B$ and $\mathsf{trans}^E\_{\mathsf{seg}}= e$. Now we can define $\mathsf{decode}(e) \equiv \mathsf{ap}\_{E}(\mathsf{seg}) : \mathsf{Id}\_{\mathsf{U}}(A,B)$. Then $\mathsf{encode}(\mathsf{decode}(e)) = \mathsf{trans}^E\_{\mathsf{seg}}= e$. On the other hand, by path induction it's easy to prove $\mathsf{decode}(\mathsf{encode}(p)) = p$ for any $p:\mathsf{Id}\_{\mathsf{U}}(A,B)$. Thus, $\mathsf{decode}$ is an equivalence, as desired.
To my knowledge, this was first observed in [this post](https://groups.google.com/g/homotopytypetheory/c/HfCB_b-PNEU/m/blVhWikFFFMJ) to the homotopy type theory mailing list in 2014, and I don't know of another reference. That discussion also considered "equivalence induction" (Corollary 5.8.5 in the HoTT Book) as a form of universe-free univalence, although [as Peter noted](https://golem.ph.utexas.edu/category/2011/04/homotopy_type_theory_iv.html#c037370) in 2011 that can't be phrased without a universe type unless you add some other kind of polymorphism.
|
8
|
https://mathoverflow.net/users/49
|
413795
| 168,796 |
https://mathoverflow.net/questions/413770
|
12
|
Please excuse that the following will be a somewhat soft question.
Let $(M,d)$ be a metric space and $X(\omega)$ a random variable on $M$ with distribution $\mu$.
Assume now that $M = \overline{B\_1^n(0)}$ the $n$-dimensional closed ball of radius $1$ around $0$ and that $\mu$ has a density funciton $f\_\mu$. Assume further that $f\_\mu >> 1$ near $\partial M$ and that $f\_\mu<<1$ near $0$.
Can we formally make sense of the idea that $X$ "most likely sees" $\partial M$ as a nonzero homology class, but that $X$ might also "see" this homology class might also be $0$ (because there are chains $\partial M$ is the boundary of, but all of them cover a region of very small probability)? Can we define a homology or homotopy groups for $(M,d,\mu)$?
|
https://mathoverflow.net/users/127781
|
Is there a homotopy/homology-theory for probability spaces?
|
One possible approach is via persistent homology. It is outlined in the paper "[Persistent homology for metric measure spaces, and robust statistics for hypothesis testing and confidence intervals](https://web.ma.utexas.edu/users/blumberg/metricmeasure1.pdf)" by Blumberg, Gal, Mandell and Pancia
In a nutshell, the idea is the following. Choose a finite sample $S$ of points in $M$, with probabilities determined by the density function $f\_\mu$. You get a finite metric subspace of $M$, whose points cluster in areas of high probability. To this ``cloud of points'' $S$ you can associate a diagram of simplicial complexes, indexed by a parameter $\epsilon$. Roughly speaking, you form the complex by connecting the points in $S$ that are close enough to each other, and also filling in the simplices that are small enough. Where "close enough" and "small enough" depends on the parameter $\epsilon$. The persistent homology of this diagram records the features of the homology that persist over a large range of values for $\epsilon$. Persistent homology provides a possible answer to your question.
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10
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https://mathoverflow.net/users/6668
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413798
| 168,797 |
https://mathoverflow.net/questions/413804
|
10
|
The following question is about if it is compatible to add to $\sf ZF$ an axiom asserting the existence of a countable transitive model of $\sf ZF$ such that for every strictly increasing function $f$ on the ordinals, we have a transitive countable model of $\sf ZF $ satisfying: $$\forall\alpha>0:\beth\_\alpha = \aleph\_{f(\alpha)}$$
Formally:
$ \exists M: M \equiv \operatorname {CTM}(\mathsf {ZF}) \land \forall f \subseteq M \ \big{(}\\f: \operatorname {Ord}^M \to \operatorname {Ord}^M \land \forall \alpha \forall \beta \, ( \beta > \alpha \to f(\beta) > f(\alpha) ) \\ \implies \\ \exists N : N \equiv \operatorname {CTM}(\mathsf {ZF}) \land \operatorname {Ord}^N =\operatorname {Ord}^M \land (N \models \forall \alpha > 0 : \beth\_\alpha=\aleph\_{f(\alpha)})\big{)} $
Where "$\equiv\operatorname {CTM}(\mathsf {ZF})$" means "*is a countable transitive model of ZF*"
So this is to say that the generalized continuum hypothesis can fail everywhere and in everyway.
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https://mathoverflow.net/users/95347
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Can GCH fail everywhere every way?
|
No. An early nontrivial constraint on the $\beth$ function comes from Kőnig's Theorem, that for all infinite $\kappa$, $\mathrm{cf}(2^\kappa)>\kappa$. This implies that we cannot have $\beth\_\alpha = \aleph\_{f(\alpha)}$ for all $\alpha$, when $f(1) = \omega$, nor when $f(\omega+1)$ is a cardinal below $\aleph\_\omega$.
Another constraint is Silver's Theorem, that if GCH holds below a singular of uncountable cofinality, then it holds at that singular as well.
Other constraints come from Shelah's PCF theory. Shelah showed that if $\aleph\_\omega$ is a strong limit, then $2^{\aleph\_\omega} < \aleph\_{\omega\_4}$.
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21
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https://mathoverflow.net/users/11145
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413807
| 168,800 |
https://mathoverflow.net/questions/413766
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9
|
$\newcommand\Sq{\mathit{Sq}}$Recall that a (graded) module $V^\ast$ over the Steenrod algebra $\mathcal A^\ast$ is said to be *unstable* if $\Sq^i v = 0$ for $i > |v|$. The motivating example, of course, is that if $V^\ast = H^\ast(X)$ for a space $X$ with its natural $\mathcal A^\ast$ structure, then $V^\ast$ is unstable.
The category of $\mathcal A^\ast$-modules which are finite-dimensional over $\mathbb F\_p$ is dual to the category of $\mathcal A\_\ast$-comodules which are finite-dimensional over $\mathbb F\_p$, where $\mathcal A\_\ast$ is the dual Steenrod algebra. So the instability condition should be expressible from this dual perspective.
**Question 1:** Let $V\_\ast$ be a finite-dimensional graded $\mathbb F\_p$-vector space equipped with the structure of an $\mathcal A\_\ast$-comodule. Under what conditions is the dual vector space $V^\ast$ unstable (with its natural $\mathcal A^\ast$-module structure)?
Ideally, the condition would be expressed in terms of Milnor's $\{\xi\_m, \tau\_n\}$ generators. In principle it should be straightforward to do the translation, but I am a bit intimidated by the Adem relations. I am also stuck already because even before dualizing, I don't know how to express the instability condition in terms of the generating set $\{\Sq^{2^k} \Sq^{2^{k-1}}\dotsm \Sq^1\}$, which I suppose leads to a subsidiary question:
**Question 2:** Let $V^\ast$ be a finite-dimensional $\mathcal A^\ast$-module. In terms of the generating set $\{\Sq^{2^k} \Sq^{2^{k-1}} \dotsm \Sq^1\}$, when is $V^\ast$ unstable?
I'd be happy to know the answer for $p=2$, $p>2$, or both.
|
https://mathoverflow.net/users/2362
|
What is an unstable dual-Steenrod comodule?
|
Normally people think about Steenrod comodules as graded $\mathbb{Z}/2$-modules equipped with a graded coaction $\psi\colon M\_\*\to M\_\*[\xi\_1,\xi\_2,\dotsc]$. However, it is equivalent to consider ungraded modules with coaction $\psi\colon M\to M[\xi\_0^{\pm 1},\xi\_1,\xi\_2,\dotsc]$; the grading is recovered by the formula
$$ M\_d = \{m : \psi(m)=\xi\_0^dm \pmod{\xi\_k:k>0}\}, $$
and the original coaction is recovered by setting $\xi\_0=1$.
Now the action is unstable iff $\psi(M)\leq M[\xi\_0,\xi\_1,\dotsc]$ (so no negative powers of $\xi\_0$ are involved).
If $M$ has a compatible product then we can reformulate things in terms of the scheme $X=\text{spec}(M)$. A Steenrod action is equivalent to an action of the group scheme $\text{Aut}(G\_a)$ on $X$ (where $G\_a$ is the additive formal group). The main instability condition (that certain Steenrod operations should vanish) is equivalent to saying that the action extends over the monoid $\text{End}(G\_a)$. (It is easy to see that there is at most one extension.) In the multiplicative context one also wants to impose the condition $\text{Sq}^k(x)=x^2$ when $|x|=k$. This is equivalent to saying that the Frobenius endomorphism $F\_{G\_a}\in\text{End}(G\_a)$ should act on $X$ as the Frobenius endomorphism $F\_X$.
|
15
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https://mathoverflow.net/users/10366
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413808
| 168,801 |
https://mathoverflow.net/questions/413786
|
3
|
It's known that if a compact Lie group $G$ acts freely on a compact manifold $M$, then the orbit space $M/G$ is a manifold. If we only assume that $G$ acts almost freely (i.e. $G\_x$ is finite for any $x\in M$ and there are only finitely many $x$ such that $G\_x$ is not trivial)，then can we deduce that $M/G$ is a orbifold or even a good orbifold (i.e. the universal covering is a manifold)?
And is there any reference for such kind of actions?
|
https://mathoverflow.net/users/356774
|
Almost free Lie group action
|
I think the answer to the first question is yes and the answer to the second one is no:
Yes, the quotient is an orbifold. The action of the finite group $G\_x$ in a neighbourhood of $x$ can be linearized (at least if the action is by diffeomorphisms, I don't know about $C^0$ regularity), and the quotient $M/G$ is locally modelled on $G\_x \backslash T\_xM / T\_x (G\cdot x)$.
No this orbifold is not good in general. For instance, you can glue a solid torus with a trivial circle fibration to a solid torus with a Seifert fibration with one singular fiber in the center and get a closed Seifert 3-manifold with one singular fiber. The fibration is given by the orbits of an action of S^1 and the quotient orbifold is a sphere with a single orbifold point, the simplest example of an orbifold not covered by a manifold.
More generally, I think every orbifold $M$ if dimension $n$ is the quotient of a manifold $P$ by an almost free action of the orthogonal group $O\_n$ ($P$ is the principal $O\_n$-bundle associated with the orbifold tangent bundle equipped with an orbifold Riemannian metric).
|
4
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https://mathoverflow.net/users/173096
|
413810
| 168,802 |
https://mathoverflow.net/questions/413829
|
0
|
Let $\nu : G \rightarrow H$ be a surjective group homomomorphism with kernel $N$, $H$ abelian, and $G$ finitely generated.
The rational abelianization of $N$, $H\_1(N)$ is a $\mathbb{C}[H]$-module, and we say that it is nilpotent if some power of the augmentation ideal $I\_H \subset \mathbb{C}[H] $ is in the annihilator of the module.
In some paper I read that, as an easy consequence of the Hilbert Nullstellensatz, $H\_1(N)$ is nilpotent if and only if $$\operatorname{Specm}(\mathbb{C} H) \cap \operatorname{supp}\_{\mathbb{C} H}\left(H\_{1} N\right) \subseteq\{1\}, $$ where the support means the subset of prime ideals containing the annihilator, and "$1$" is the ideal $I\_H$.
Can someone explain what version of the Nullstellensatz is applied here? I understand only the direct implication….
The result in question is [Dimca, Hain, and Papadima - The abelianization of the Johnson kernel](https://arxiv.org/abs/1101.1392) Lemma 3.3.
|
https://mathoverflow.net/users/170754
|
Nullstellensatz and nilpotence of a module
|
If we call the annihilator by $J$, then the statement reads
>
> $J$ contains some power of the maximal ideal $I\_H$ if and only if the intersection of the set of prime ideals containing $J$ with the set of maximal ideals is contained in $\{I\_H\}$
>
>
>
which simplifies to
>
> $J$ contains some power of the maximal ideal $I\_H$ if and only if the set of maximal ideals containing $J$ is contained in $\{I\_H\}$
>
>
>
The Nullstellensatz says:
>
> The intersection of all maximal ideals containing $J$ is equal to the radical of $J$
>
>
>
which, if the set of maximal ideals containing $J$ is contained in $\{I\_H\}$, implies the radical of $J$ is either the unit ideal or $I\_H$. Since each generator of $I\_H$ is contained in the radical of $J$, some power of each generator is contained in $J$, and since there are finitely many generators, some power of $I\_H$ is contained in $J$.
|
2
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https://mathoverflow.net/users/18060
|
413837
| 168,813 |
https://mathoverflow.net/questions/413772
|
3
|
I have a triangulation of a surface without boundary in $\mathbb{R}^3$.
The triangulation gives a unit normal pointing outwards for each triangle. I need to find some point in the interior of the surface.
I will outline my general strategy. I let $V$ denote the set of vertices of the triangulation and compute a number that I will call the fineness
$$
\delta = \min\_{x, y \in V \mid x\neq y} \lVert x-y\rVert.
$$
I pick some vertex $v$ and consider the point
$$
p = v -\frac{\delta}{4}u,
$$
where $u$ is some unit vector. It is possible to show that $p$ is closer to $v$ than any other vertex in the triangulation, so it suffices to choose $u$ such that it lies below all of the triangles containing $v$.
If $i$ indexes the triangles containing $v$ as a vertex and $n\_i$ is the outer normal of the triangle $i$, choosing $u$ such that $\langle u, n\_i\rangle >0$ for all $i$ will put $p$ below the surface.
We can map this onto a linear algebra problem by writing $u$ as a weighted combination of the outer normals: $u = \sum\_i w\_i n\_i$. If all of the inner products of the outer normals were non-negative, we could solve this by taking the $w\_i$ to be the components of the Perron–Frobenius eigenvector of the matrix $A\_{i,j} = \langle n\_i, n\_j\rangle$. However, the surface can have regions of high curvature, so $A$ is not always non-negative.
A natural notion of the normal at $v$ is the angle weighted averaged of the normals. This lead me to try taking $w\_i = \theta\_i$, where $\theta\_i$ is angle of the triangle $i$ at $v$. However, this alone does not guarantee $\langle u, n\_i \rangle > 0$ for all $i$. This seems like some iterative approach is needed, where the weights corresponding to positive inner products should be decreased and those corresponding to non-positive ones should increase. I think one strategy should be to add to the weights of non-positive terms and then normalize all of the weights so that $\sum\_i w\_i = 1$. If all inner products are positive, we terminate the iterative procedure. Of course, we would have to make $u$ a unit vector at the end of this.
My questions are: Am I on the right track? Is there a simpler approach? If I am on the right track, what update rule for the weights will guarantee convergence?
Edit: The shapes that I am working with are definitely not convex. The data points are from intestinal organoids, which can roughly be shaped like octopi holding broccoli.
|
https://mathoverflow.net/users/475087
|
Finding a point inside a surface
|
Let me suggest another approach, conceptually simple but maybe not as easy
to implement as Matt F.'s algorithm.
Let $P$ be the polyhedron.
Let $F$ be any face of $P$ with centroid $p$, and $\vec{n}$ the outward normal vector to $F$.
Shoot a ray $r$ from outside of $P$ in direction $-\vec{n}$ through $p$.
Ignore all intersections until the ray reaches $p$.
The ray $r$ must now penetrate to the interior of $P$
on the immediate other side of $F$. Track $r$ until it hits another face of $P$.
(It might hit a vertex or an edge of the face.)
Let this first hit-point be $q$.
Then $(p+q)/2$ is necessarily strictly interior to $P$.
This method requires careful ray-triangle intersection
(assuming all faces are triangles).
I implemented this as part of code to answer point-in-polyhedron queries
in Section 7.5 of
[Computational Geometry in C](http://cs.smith.edu/%7Ejorourke/books/compgeom.html).
|
2
|
https://mathoverflow.net/users/6094
|
413838
| 168,814 |
https://mathoverflow.net/questions/412879
|
5
|
A Silver forcing "below $2^n$" is defined e.g. in Definition 7.4.11 of [Tomek Bartoszyński and Haim Judah, Set Theory: on the structure of the real line, A. K. Peters, Wellesley, 1995.]. It is called *Infinitely equal forcing **EE*** there. In the same book, in Lemma 7.4.15 the authors show that *EE preserves p-points*. However, the proof of this lemma seems to be not complete/correct: the choice of conditions $p^{n+1}$ is not clear, as needed extensions of $p^n$ for various $r\_n^j$ may "contradict" each other. (This would not be a problem if conditions were $2^n$-splitting trees, i.e., if the forcing were more like the Sacks rather than Silver.)
Do you know another source/reference for the proof that EE preserves p-points? Or perhaps I am missing something and the proof in the book is actually complete?
|
https://mathoverflow.net/users/115702
|
Silver-like forcing preserves p-points (reference request)
|
David Chodounský and Osvaldo Guzmán showed in arXiv:1703.02082, [There are no P-points in Silver extensions](https://arxiv.org/abs/1703.02082) that *There are no P-points in Silver extensions*. They prove that
>
> after adding a Silver real no ultrafilter from the ground model can be extended to a P-point, and this remains to be the case in any further extension which has the Sacks property.
>
>
>
In particular, any free ultrafilter from the ground model will not be a p-point, and hence cannot be an ultrafilter, as Silver forcing is proper.
Silver is a complete subforcing of the forcing **EE** (right? In each entry of the generic, use the first bit — this will be a Silver generic) so also **EE** destroys all p-points.
(Edit 2022: Published 2019 in the Israel Journal of Mathematics, vol 232 (2) 759–773, <https://doi.org/10.1007/s11856-019-1886-2>)
|
6
|
https://mathoverflow.net/users/14915
|
413843
| 168,816 |
https://mathoverflow.net/questions/408632
|
3
|
Let $p, q \in \mathbb{Z}$.
Let $\operatorname{wt}(n)$ is [A000120](https://oeis.org/A000120), number of $1$'s in binary expansion of $n$ (or the binary weight of $n$) and
$$n=2^{t\_1}(1+2^{t\_2+1}(1+\dots(1+2^{t\_{wt(n)}+1}))\dots)$$
Then we have an integer sequence given by
\begin{align}
a(0, m)& = 1\\
a(n, m)& = \sum\limits\_{j=0}^{2^{\operatorname{wt}(n)}-1}m^{\operatorname{wt}(n)-\operatorname{wt}(j)}\prod\limits\_{k=0}^{\operatorname{wt}(n)-1}(1+\operatorname{wt}(\left\lfloor\frac{j}{2^k}\right\rfloor))^{t\_{k+1}+1}
\end{align}
I conjecture that
$$a(n, -1) = \sum\limits\_{j=0}^{2^{\operatorname{wt}(n)}-1}(-1)^{\operatorname{wt}(n)-\operatorname{wt}(j)}a(f(n,j),0)$$
and
$$a(n, 0) = \sum\limits\_{j=0}^{2^{\operatorname{wt}(n)}-1}a(f(n,j),-1)$$
where $a(n,-1)$ is [A329369](https://oeis.org/A329369), number of permutations of ${1,2,...,m}$ with excedance set constructed by taking $m-i$ ($0 < i < m$) if $b(i-1) = 1$ where $b(k)b(k-1)...b(1)b(0)$ ($0 \leqslant k < m-1$) is the binary expansion of $n$. The excedance set of a permutation $p$ of ${1,2,...,m}$ is the set of indices $i$ such that $p(i) > i$; it is a subset of ${1,2,...,m-1}$.
and $a(n,0)$ is [A284005](https://oeis.org/A284005),
\begin{align}
a(0, 0)& = 1\\
a(n,0)& = (1+\operatorname{wt}(n))a(\left\lfloor\frac{n}{2}\right\rfloor,0)
\end{align}
and finally $f(n,k)$ is [A295989](http://oeis.org/A295989), irregular triangle $T(n, k)$, read by rows, $n \geqslant 0$ and $0 \leqslant k <$ [A001316](http://oeis.org/A001316)$(n)$: $T(n, k)$ is the $(k+1)$-th nonnegative number $m$ such that $n \operatorname{AND} m = m$ (where $\operatorname{AND}$ denotes the bitwise $\operatorname{AND}$ operator).
\begin{align}
T(n, 0)& = 0\\
T(2n, k)& = 2T(n,k)\\
T(2n+1, 2k)& = 2T(n,k)\\
T(2n+1, 2k+1)& = 2T(n,k) + 1
\end{align}
In other words
$$a(n, -1) = \sum\limits\_{j=0}^{n}(-1)^{\operatorname{wt}(n)-\operatorname{wt}(j)}(\binom{n}{j}\operatorname{mod} 2)a(j,0)$$
and
$$a(n, 0) = \sum\limits\_{j=0}^{n}(\binom{n}{j}\operatorname{mod} 2)a(j,-1)$$
Is there a way to prove it?
|
https://mathoverflow.net/users/231922
|
Sum with products turned into subsequences
|
Actually, these two conjectures are actually equivalent.
We need a lemma: Lucas' Theorem: $\binom{a}{b}$ is odd if and only if $a\&b=b$ where $\&$ is bitwise AND operation. Let $S\_j$ be the set of integers $i$ such that $i\&j=i$. Thus, if the first relation
$$a(n, -1) = \sum\limits\_{j=0}^{n}(-1)^{\operatorname{wt}(n)-\operatorname{wt}(j)}(\binom{n}{j}\operatorname{mod} 2)a(j,0)$$
holds, we have
$$a(n, -1) = \sum\limits\_{j\in S\_n}(-1)^{\operatorname{wt}(n)-\operatorname{wt}(j)}a(j,0)$$
And therefore,
$$\sum\_{j=0}^{n}(\binom{n}{j}\operatorname{mod} 2)a(j,-1)=\sum\_{j\in S\_n}a(j,-1)=\sum\_{j\in S\_n}\sum\_{k\in S\_j}(-1)^{\text{wt}(j)-\text{wt}(k)}a(k,-1)$$
Swapping the sums, we have the right hand side equals to
$$\sum\_{k\in S\_n}a(k,-1)\sum\_{j\text{ such that }k\in S\_j,j\in S\_n}(-1)^{\text{wt}(j)-\text{wt}(k)}$$
All the other $k\in S\_n$ vanishs the sum $\sum\_{j\text{ such that }k\in S\_j,j\in S\_n}(-1)^{\text{wt}(j)-\text{wt}(k)}$ because for every bit that is $1$ in $n$ and $0$ in $k$, it contribute a factor $1$ if that bit in $j$ is $0$ and $-1$ if that bit in $j$ is $1$. Therefore, the sum $$\sum\_{k\in S\_n}a(k,-1)\sum\_{j\text{ such that }k\in S\_j,j\in S\_n}(-1)^{\text{wt}(j)-\text{wt}(k)}$$ is actually $a(n,-1)$, which implies the conjecture $a(n, 0) = \sum\limits\_{j=0}^{n}(\binom{n}{j}\operatorname{mod} 2)a(j,-1)$. Similarly, we can derive the first conjecture using the second.
So, we only need to proof $a(n, 0) = \sum\limits\_{j=0}^{n}(\binom{n}{j}\operatorname{mod} 2)a(j,-1)=\sum\_{j\in S\_n}a(j,-1)$.
Notice that the sum $\sum\_{j\in S\_n}a(j,-1)$, if we take the combinatorial argument as granted, it counts the following permutations: for all the $k${th} bit of $n$ that is $0$, $\sigma(k)\le k$. This is because all the other bits when summed, cancelled out the restrictions. (If the $k$th bit of $n$ is $1$, that all the $S\_k$ with $k$th bit $1$ with have $\sigma(k)> k$, and all the $S\_k$ with $k$th bit $0$ with have $\sigma(k)\le k$. So These two cancelled out if we sum the elements in $S\_n$.) So we claim the number of such permutations is $a(n,0)$.
Actually, we can count the permutations as follows: for the $0$ bits of $n$ (call it $i\_1,i\_2,\dots,i\_k$), we select the permutation one by one. We select $\sigma(i\_l)$ to be one of $1,2,\dots,i\_l$ except $\sigma(i\_1),\dots,\sigma(i\_{l-1})$. So this is $i\_l-(\text{number of zeroes before $i\_l$})$, that is, the number of ones before $i\_l$. The rest of them we choose it freely, so it is $(\text{number of ones})!$. That is, we can assign $t$ to the $t$th one in the binary representation of $n$, or, the number of $1$ not after that bit, and take the product. Let $n=b\_rb\_{r-1}\dots b\_0$ be its binary representation. So, the overall number of permutation is $\prod\_{u=1}^r (\text{the number of ones not later than digit }b\_u)$, and thus equals $a(n,0)$ (because the combinatorial meaning of $a(n,0)$ is such.)
|
3
|
https://mathoverflow.net/users/170895
|
413850
| 168,818 |
https://mathoverflow.net/questions/409285
|
0
|
Let $f(n)$ be [A153733](https://oeis.org/A153733), remove all trailing ones in binary representation of $n$. Here
\begin{align}
f(2n)& = 2n\\
f(2n+1)& = f(n)\\
\end{align}
Then we have an integer sequence given by
\begin{align}
a(0)& = 1\\
a(2n+1)& = a(n)\\
a(4n+2)& = 2a(n)\\
a(4n)& = 2a(2n)-a(n)
\end{align}
Here $a(n)$ is [A124758](https://oeis.org/A124758), product of the parts of the compositions in standard order.
Let
$$b(n) = \sum\limits\_{k=0}^{n}(\binom{n}{k}\operatorname{mod} 2)a(k)$$
Then I conjecture that
\begin{align}
b(0)& = 1\\
b(2n+1)& = b(n) + b(2n)\\
b(2n)& = b(n) + b(f(n-1))\\
\end{align}
Is there a way to prove it?
|
https://mathoverflow.net/users/231922
|
Modulo $2$ binomial transform of A124758
|
The first relation $b(0)=0$ is trivial.
By the condition, $a(n)$ is the product of number of consecutive zeroes plus $1$, if we write $n$ in binary. For instance, if $n=1220$, we have $n=100 1100 0100$ in binary, thus $a(n)=3\times 4\times 3=36$. Notice that if $n=2^u\times v$ where $v$ is an odd number, we have $a(n)=(u+1)a(v)$
We need a lemma: Lucas' Theorem: $\binom{a}{b}$ is odd if and only if $a\&b=b$ where $\&$ is bitwise AND operation. Let $S\_n$ be the set of $i$ if $i\&n=i$.
Now the second relation is clear. Notice that if $k\in S\_{2n+1}$ if there exist $k'\in S\_n$ such that $k=2k'$ or $k=2k'+1$. The even ones consists of exactly the elements in $S\_{2n}$. So we can divide $S\_{2n+1}$ into odd part and even part, We have
$$b(2n+1)=\sum\_{k\in S\_{2n+1}}a(k)=\sum\_{k\in S\_{2n+1},k\equiv 0\mod 2}a(k)+\sum\_{k\in S\_{2n+1},k\equiv 1\mod 2}a(k)$$
We know that
$$\sum\_{k\in S\_{2n+1},k\equiv 0\mod 2}a(k)=\sum\_{k\in S\_{2n}}a(k)=b(2n)$$
And since the odd numbers, when deleting the ending $1$ the $a(\cdot)$ value won't change. Thus,
$$\sum\_{k\in S\_{2n+1},k\equiv 1\mod 2}a(k)=\sum\_{\in S\_{2n}}a(k)=b(n)$$
Therefore we have $b(2n+1)=b(2n)+b(n)$
To prove the third one, we state some of the properties of
$S\_{n}$: Let $n=a\_ma\_{m-1}\dots a\_{k+1}10\dots 0$ ($k$ zeros in the end). Then, $f(n-1)=a\_ma\_{m-1}\dots a\_{k+1}0$. Notice that there are exactly one $1$ less from $f(n-1)$ to $n$, so we do a ``bijection''. Notice that $S(2n)$ can be divided into $b\_mb\_{m-1}\dots b\_{k+1}10\dots 0$ ($k$ zeros in the end) and $b\_mb\_{m-1}\dots b\_{k+1}00\dots 0$ ($k=1$ zeros in the end), differed by the $k+1$th from the last digit. Lett a certain pair of them be $u=b\_mb\_{m-1}\dots b\_{k+1}10\dots 0$, $v=b\_mb\_{m-1}\dots b\_{k+1}00\dots 0$ and $w=b\_mb\_{m-1}\dots b\_{k+1}0$. What we only need to prove is $(a(u)+a(v))=a((u/2)+a(v/2))+a(w)$. Summing over these pairs we have those $a(u)+a(v)$ sums to $b(2n)$, $a(u/2)+a(v/2)$ sums to $b(n)$ and $a(w)$ sums to $b(f(n-1))$.
Now we finish the proof. Let $b\_mb\_{m-1}\dots b\_{k+1}$ end in $q$ zeroes, that is, $b\_{k+q+1}=1$, $b\_{k+q}=\dots=b\_{k+1}=0$. So
$$a(u)-a(u/2)=a(b\_mb\_{m-1}\dots b\_{k+1}10\dots 0)\text{($k$ zeroes)}-a(b\_mb\_{m-1}\dots b\_{k+1}10\dots 0)\text{($k-1$ zeroes)}=(k+1)a(b\_mb\_{m-1}\dots b\_{k+1}1)-ka(b\_mb\_{m-1}\dots b\_{k+1}1)=a(b\_mb\_{m-1}\dots b\_{k+1}1)=a(b\_mb\_{m-1}\dots b\_{k+1})=qa(b\_mb\_{m-1}\dots b\_{k+q+1})$$
$$a(v)-a(v/2)=a(b\_mb\_{m-1}\dots b\_{k+1}00\dots 0)\text{($k+1$ zeroes, $k+q+1$ zeroes in the end totally)}-a(b\_mb\_{m-1}\dots b\_{k+1}00\dots 0)\text{($k$ zeroes, $k+q$ zeroes in the end totally)}=(k+q+1)a(b\_mb\_{m-1}\dots b\_{k+q+1})-(k+q)a(b\_mb\_{m-1}\dots b\_{k+q+1})=a(b\_mb\_{m-1}\dots b\_{k+q+1})$$
$$a(w)=a(b\_mb\_{m-1}\dots b\_{k+1}0)\text{($q+1$ zeroes in the end totally)}=(q+1)a(b\_mb\_{m-1}\dots b\_{k+q+1})$$
So we have $a(u)-a(u/2)+a(v)-a(v/2)=a(w)$, which yields $(a(u)+a(v))=a((u/2)+a(v/2))+a(w)$.
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2
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https://mathoverflow.net/users/170895
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413853
| 168,820 |
https://mathoverflow.net/questions/413667
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11
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By definition, a *1-motive* over an algebraically closed field $k$ is the data
$$
M = [X\stackrel{u}{\to}G]
$$
where $X$ is a free abelian group of finite type, $G$ is a semi-abelian variety over $k$, and $u:X \to G(k)$ is a group morphism.
As far as I understand, this was one of the earlier attempts of coming up with a workable notion of a mixed motive. As evidence, Deligne shows that 1-motives over $\mathbb{C}$ are equivalent to mixed Hodge structures (without torsion) of type $\{(0,0),(0,-1),(-1,0),(-1,-1)\}$.
The problem is that I don't really understand the intuition behind this definition. I assume that Deligne had concrete examples in mind when he came up with this formulation. Pehaps some extensions of $H^1(A)$, where $A$ is an abelian variety?
How should I interpret the data in the definition of a 1-motive? My naïve view on motives is that they should be compatible systems of realizations with "geometric origin", and extensions thereof. How to reconcile these two points of view? Perhaps what I lack is simply a bag of toy examples to help me think about it.
|
https://mathoverflow.net/users/nan
|
How should I think about 1-motives?
|
I don't fully understand the hostility to thinking of motives in terms of compatible systems. It's true that a motive is not just a compatible system of realizations (or else we would call it a "compatible system" and not a motive) but it's also true that studying compatible systems has always been a good way to get intuition for many aspects of the theory of motives. Maybe a better thing to say is that a motive should be a *geometric recipe* which produces a realization in any existing or future cohomology theory.
You asked how to reconcile this perspective on 1-motives with the definition of 1-motives. I guess this means how to derive a compatible system of realizations from a map $X \to G$.
The realization of the 1-motive $X \to G$ in any cohomology theory should be the extension of $H^1(G)$ by $H^1(X^\vee)$. (More properly I guess this is the dual of the realization).
You ask if this is an extension of $H^1(A)$. The group $G$ is always an extension of an abelian variety $A$ by a torus $T$, and $H^1(G)$ is an extension of $H^1(T)$ by $H^1(A)$. So we see that we are potentially extending $H^1(A)$ in both directions.
Which extension? The idea is we should morally get the $H^1$ of the quotient space $G/X$. Since this space is not very well-behaved, we can simulate this, in an arbitrary cohomology theory, as follows:
We observe that any class in $H^1(G)$ (ideally, completely represented by a torsor) is multiplicative in the sense that its pullback to $G \times G$ under the multiplication map is equal to the sum of its pullbacks under the two projections. For torsors, we can fix an isomorphism of torsors realizing this.
An element of $H^1( X \to G)$ is a class in $H^1(G)$ together with a trivialization of its pullback to $H^1(X)$ which is multiplicative in the sense that the pullback of this trivialization to $X \times X$ along the multiplication is equal to the sum of its pullbacks under the two projections (using the above isomorphism to define this sum).
You can follow this recipe in any theory and get a concrete description of the extension.
Furthermore, in every reasonable cohomology theory you will get a clue to the idea that this construction produces *all* the extensions with "geometric origin". For example, in $\ell$-adic cohomology theories, you will see that extensions of $H^1(G)$ by $X^\vee$ correspond to a Selmer group, and extensions arising from this construction are the image of the map from the group of rational points to the Selmer group. So as long as the Tate-Shafarevich group is finite, there will not be room to fit any more extensions.
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7
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https://mathoverflow.net/users/18060
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413864
| 168,824 |
https://mathoverflow.net/questions/413858
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0
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Let $u: (0,\infty) \times \mathbb R \to \mathbb R$. Suppose that $\int\_{\mathbb R} u(t,x) dx \ge 0$ (but not necessarily $u >0$). Let $A:(0,\infty) \to \mathbb R$ with $A \ge 0$. Let $\alpha \ge 0$.
Suppose that we know
>
> $$\frac{d}{dt} \alpha\int\_{\mathbb R} u(t,x) dx + A(t) \le \int\_{\mathbb R}u^2(t,x) dx.$$
>
>
>
Can we prove a **Gronwall-type inequality** that implies a bound like
$$\alpha\int\_{\mathbb R} u(t,x) dx + \int\_0^t A(t) dt \le e^t\alpha\int\_{\mathbb R}u(0,x) dx$$
(or with something else related to $u(0,x)$ on the right-hand side).
This would be true if there was no square on the right-hand side by the classic Gronwall inequality, but I wonder if we can still say something in this case.
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https://mathoverflow.net/users/nan
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Gronwall-type inequality with nonlinearity
|
The stated inequality cannot hold: Let $\alpha=1$, take $u(t,x)= U$ if $x\in[0,1]$ and $0$ otherwise, and $A(t) = U^2$. Then all the $d/dt$ are 0, but the desired inequality,
$$U + tU^2 \le e^{t}U,$$
fails for $t=\log U$ and $U\to +\infty$.
|
0
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https://mathoverflow.net/users/141760
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413870
| 168,825 |
https://mathoverflow.net/questions/412127
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1
|
If I had to partition the unit square $[0,1]\times[0,1]$ into $k^2$ rectangles such that the sum of their diagonals is minimum possible, I would simply choose the $k \times k$ grid of squares. Now suppose we also have a collection of $nk^2$ points in general position inside the unit square and impose the additional requirement that each rectangle in the partition should contain exactly $n$ points.
1. Is there an efficient algorithm to determine an optimal/near-optimal partition?
2. What if we slightly relax the assumption that each rectangle should contain an equal number of points?(as is necessary when the total number of points is not divisible by $k^2$)
This question is related to a procedure called data discretization in the field of data mining and statistics. See below.
[[1](https://ai.stanford.edu/%7Eronnyk/disc.pdf)] James Dougherty, Ron Kohavi, Mehran Sahami, Supervised and Unsupervised Discretization of Continuous Features, Machine Learning Proceedings 1995
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https://mathoverflow.net/users/8028
|
Partitioning unit square with equal frequency rectangles
|
Here is an answer inspired by redistricting and the [shortest split line](https://en.wikipedia.org/wiki/Gerrymandering#Shortest_splitline_algorithm) algorithm.
For any rectangle with $mn$ points, consider the $2m-2$ ways of dividing it horizontally or vertically into two rectangles with an integral multiple of $n$ points in each. Among these possible divisions of the rectangle into $R$ and $R’$, we can choose the one for which $E[R]+E[R’]$ is minimal, where $E$ is an appropriate scoring function.
Now apply this technique to divide the unit square into two rectangles, and apply it recursively to each of the rectangles that result.
For the scoring function of a rectangle $I\times J$ with $mn$ points, we can take $E[I\times J]=\sqrt{A^2 m^2/i^2+B^2 i^2}$ where $A=|I|$, $B=|J|$, $i=\max(1,\min(m,\sqrt{mA/B}))$. This gives an optimal sum of diagonal lengths if the distribution of points is uniform and $i$ is integral. Then the scoring is quick, and the total algorithm requires a number of steps which is $O(k^4)$.
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1
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https://mathoverflow.net/users/nan
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413874
| 168,828 |
https://mathoverflow.net/questions/413849
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4
|
Let $X$ be an irreducible projective variety over $\mathbb{C}$ (note that I do not assume $X$ smooth) and let $ p : X \longrightarrow S$ be a projective surjective morphism. For any open $U \subset S$, I consider the natural map:
$$ \pi : \mathrm{Hilb}^2\_{U}(p^{-1}(U)) \longrightarrow S^2(p^{-1}(U)/U)),$$
where $\mathrm{Hilb}^2\_U(p^{-1}(U))$ is the punctual Hilbert scheme parametrizing relative subschemes of $p^{-1}(U)$ of length $2$ and $S^2(p^{-1}(U)/U))$ is the relative symmetric square of $p^{-1}(U)$ over $U$. Let me finally denote by $\mathcal{H}\_U$ the closure in $\mathrm{Hilb}^2\_U(p^{-1}(U))$ of :
$$ \pi^{-1}\left(S^{2}(p^{-1}(U)/U) \backslash \Delta\_{p^{-1}(U)/U} \right), $$
where $\Delta\_{p^{-1}(U)/U)}$ is the relative diagonal.
**Question** : I would like to know if there exists a non-empty $U$ such that $\mathcal{H}\_U$ is irreducible?
The comment below [this question](https://mathoverflow.net/questions/212571/hilbert-scheme-of-relative-subschemes-of-lenght-2) seems to suggest (by generic smoothness) that it could be true if $X$ is smooth. However I am interested in the general case. I would also be interested in a reference (or a short proof) if the answer to the question happens to be positive.
**Edit** : In a former (naive) version of the question, I asked if the whole relative Hilbert scheme could be irreducible over some non-empty open $U$. The answer is trivially "no", as observed by Jason Starr in the comment below.
|
https://mathoverflow.net/users/37214
|
irreducibility punctual Hilbert scheme of relative subschemes of length $2$
|
If I understand the question correctly, the subset $\mathcal{H}\_U$ is the closure of the locus parametrizing two distinct points. Over this locus, the Hilbert-Chow morphism $\pi$ is an isomorphism so irreducibility of $\mathcal{H}\_U$ is equivalent to irreducibility of $S^2(p^{-1}(U)/U \setminus \Delta\_{p^{-1}(U)/U})$ which it turn would follow from irreducibility of $S^2(p^{-1}(U)/U)$.
---
**Lemma:** Suppose $p : X \to S$ is a flat and proper surjective morphism of varieties such that $S$ is irreducible and the generic fiber of $p$ is geometrically irreducible. Then the symmetric powers $S^n(X/S)$ are irreducible.
**Proof:** Since flatness is stable under base change and composition, the $n$-fold fiber powers $X^n\_S := X \times\_S \ldots \times\_S X$ are flat over $S$. Moreover, the generic fiber of $X^n\_S \to S$ is the $n$-fold product of the generic fiber of $p$ and thus geometrically irreducible. By [Tag 0559](https://stacks.math.columbia.edu/tag/0559), there exists a non-empty open subvariety $U \subset S$ such that the pullback $X^n\_U \to U$ has irreducible fibers. By flatness, $X^n\_S$ is the closure of $X^n\_U$ and by [this answer](https://math.stackexchange.com/a/584919/111541), $X^n\_U$ is irreducible. Thus $X^n\_S$ is irreducible and we conclude that $S^n(X/S) = X^n\_S/\mathfrak{S}\_n$ is irreducible.
**Corollary:** Suppose $p : X \to S$ is a proper surjection of varieties with geometrically irreducible generic fiber. Then there exists a nonempty open subset $U \subset S$ such that $S^n(p^{-1}(U)/U)$ is irreducible.
**Proof** By generic flatness, there exists a nonempty open subset $U \subset S$ over which $p$ is flat and this open set does the job by the Lemma.
---
If we don't assume that the generic fiber of $p$ is geometrically irreducible, then the result is false. For example we can take $p : X \to S$ to be the map $\mathbb{G}\_m \to \mathbb{G}\_m$ given by $z \mapsto z^5$. Then I believe that $S^2(X/S)$ has 3 irreducible components which all map $5$-to-$1$ onto $S$. One of them is the diagonal which we remove but then whats left is still not irreducible and doesn't become irreducible if we shrink $S$.
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3
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https://mathoverflow.net/users/12402
|
413879
| 168,831 |
https://mathoverflow.net/questions/413739
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8
|
Let $X$ be a (smooth) manifold. [It's well known](https://math.stackexchange.com/questions/451223/the-stone-%C4%8Cech-compactification-of-a-space-by-the-maximal-ideals-of-the-ring-of/451318#451318) that its Stone-Cech compactification $\beta X$ is homeomorphic to $\operatorname{Specm}(C(X))$, with its Zariski topology.
**Is $\beta X$ also homeomorphic to $\operatorname{Specm}(C^\infty(X))$?**
|
https://mathoverflow.net/users/131975
|
Let $X$ be a manifold. Is it true that $\beta X\cong \operatorname{Specm}(C^\infty(X))$?
|
Indeed, $\operatorname{Specm}(C^\infty(X))$ is homeomorphic to $\beta X$ (through an explicit homeomorphism that will be described below). Essentially all the theory described in Gillman & Jerison's classic book *Rings of Continuous Functions* (1960) applies: for completeness of MathOverflow, let me recall how this works.
If $f$ is a continuous real-valued function on $X$, its **zero-set** is the set $Z(f) := \{x\in X : f(x)=0\}$. The zero-set of a continuous function is closed, and conversely, as $X$ is a metric space, every closed set is the zero-set of a continuous function. But in fact, even more is true: *every closed set of $X$ is the zero-set of a smooth function* (as follows, e.g., from [here](https://math.stackexchange.com/questions/2279872/find-a-smooth-function-f-in-c-infty-such-that-a-f-10-and-b-f-11)). Because of this, the notions of “z-filter” and “z-ultrafilter” below do not depend on whether we are talking about zero-sets of continuous functions or of smooth functions (or, indeed, closed sets).
Now let us make the following definitions:
* A set $\mathscr{F}$ of zero sets is said to be a **z-filter** iff it contains $X$, and is closed under enlargement (i.e., if $Z \subseteq Z'$ are zero-sets and $Z\in \mathscr{F}$ then $Z'\in \mathscr{F}$) and finite intersection.
* A z-filter is said to be a **z-ultrafilter** iff it is proper (i.e., does not contain $\varnothing$), and maximal for inclusion among proper z-filters.
Now let $R$ be either the ring $C(X)$ of continuous real-valued functions on $X$ or the ring $C^\infty(X)$ of smooth real-valued functions on $X$. Define two maps $\mathscr{Z}\colon \{\text{ideals of $R$}\} \to \{\text{z-filters}\}$ and $\mathscr{I}\colon \{\text{z-filters}\} \to \{\text{ideals of $R$}\}$ as follows:
* If $I$ is an ideal of $R$, then $\mathscr{Z}(I) := \{Z(f) : f\in I\}$. This is indeed a z-filter, as is easy to check (for enlargement, note that if $Z(f) \subseteq Z(f')$ with $f\in I$ then $Z(f') = Z(ff')$; and for finite intersection, note that $Z(f\_1) \cap Z(f\_2) = Z(f\_1^2 + f\_2^2)$).
* If $\mathscr{F}$ is a z-filter, then $\mathscr{I}(\mathscr{F}) := \{f\in R : Z(f)\in \mathscr{F}\}$. This is indeed an ideal as is easy to check (note that $Z(g\_1 f\_1+\cdots+g\_r f\_r) \supseteq Z(f\_1)\cap \cdots \cap Z(f\_r)$ if $f\_1,\ldots,f\_r$ are in $\mathscr{I}(\mathscr{F})$).
These functions are not bijections, but still, they preserve inclusions and we trivially have $\mathscr{Z}(I) \subseteq \mathscr{F}$ iff $I \subseteq \mathscr{I}(\mathscr{F})$ (when $I$ is an ideal of $R$ and $\mathscr{F}$ a z-filter); from this follows $\mathscr{Z}(\mathscr{I}(\mathscr{F})) \subseteq \mathscr{F}$ for any z-filter $\mathscr{F}$, and $\mathscr{I}(\mathscr{Z}(I)) \supseteq I$ for any ideal $I$ of $R$, but in fact it is clear that the former is easily seen to be an equality: $\mathscr{Z}(\mathscr{I}(\mathscr{F})) = \mathscr{F}$ (the latter inclusion, $\mathscr{I}(\mathscr{Z}(I)) \supseteq I$ can be proper in general as evidenced by the ideal of functions vanishing to order $2$ at a point). Also note that an ideal $I$ is the unit ideal $R$ iff $\mathscr{Z}(I)$ is the improper z-filter $\mathscr{Z}(R)$ (the one consisting of every zero-set) because $Z(f)$ is empty iff $f$ is invertible in $R$.
At this point, it is easy to check that: if $M$ is a maximal ideal of $R$ then $\mathscr{Z}(M)$ is a z-ultrafilter, and if $\mathscr{U}$ is a z-ultrafilter then $\mathscr{I}(\mathscr{U})$ is a maximal ideal. So $\mathscr{Z}$ and $\mathscr{I}$ restrict to bijections between maximal ideals of $R$ and z-ultrafilters.
But because the above worked just as well for $R$ being the ring $C(X)$ of continuous real-valued or the ring $C^\infty(X)$ of smooth real-valued functions, we can conclude that their maximal ideals are in bijection by the compositions of the corresponding $\mathscr{Z}$ and $\mathscr{I}$ functions, and in fact since $\mathscr{I}\_{C^\infty(X)}(\mathscr{F}) = \mathscr{I}\_{C(X)}(\mathscr{F}) \cap C^\infty(X)$ the bijection from maximal ideals of $C^\infty(X)$ to those of $C(X)$ can more simply be described as $M \mapsto M \cap C^\infty(X)$. Furthermore, the Zariski topologies correspond under this bijection, because (since every z-filter is of the form $\mathscr{Z}(I)$) they both correspond to the Zariski topology on the set of z-ultrafilters whose closed sets are given by the sets of ultrafilters containing a given z-filter.
(It might be worth while examining exactly which properties are required for a ring $R$ of “functions” on $X$ to make the above reasoning ensure that $\operatorname{Specm}(R) = \beta X$.)
**PS:** It might also be interesting to consider what happens for *bounded* functions. The maximal ideals of the ring $C^\*(X)$ of bounded continuous functions on $X$ are also naturally in bijection with the Stone-Čech compactification, but the bijection between ideals of $C(X)$ and $C^\*(X)$ is *not* the naïve $M \mapsto M \cap C^\*(X)$. So I'm not sure what happens to bounded smooth functions (I didn't give it thought).
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8
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https://mathoverflow.net/users/17064
|
413891
| 168,834 |
https://mathoverflow.net/questions/413888
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1
|
In an earlier [positing](https://mathoverflow.net/questions/413804/can-gch-fail-everywhere-every-way) to $\mathcal MO$, it appears that the answer to if the $\sf GCH$ can fail everywhere in *every way* is to the negative, this is the case in $\sf ZFC$, however it also appears that matters are more free in absence of $\sf AC$. So, here the quenstion is if the opposite direction in provable in $\sf ZF$.
The following question is about if it is true in $\sf ZF$ that for every stage $V\_\alpha$ of the cumulative hierarchy there is a cardinal $\kappa > |V\_\alpha|$ such that the cardinality of $V\_{\alpha+1} \neq \kappa$ in all models of $\sf ZF$?
Formally, this is:
$ \forall M: (M \models \mathsf {ZF}) \implies \\\forall \alpha \in M \, \exists \kappa \in M \big{(} (M \models |V\_\alpha| < \kappa) \land \forall N: N \approx M \to N \models|V\_{\alpha+1}| \neq \kappa \big{)}$
Where "||" refers to cardinality defined after Scott; $N \approx M$ means that $N$ is a model of $\sf ZF$ that shares the same ordinals and cardinals with $M$, i.e., $\operatorname {Ord}^M = \operatorname {Ord}^N$ and $\operatorname {Card}^M = \operatorname {Card}^N$
In the case of $\sf ZFC$, this seems to be the case, well at least for regular $\alpha$, and the singlular seems to be even more restricted. But, is this the case in $\sf ZF$?
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https://mathoverflow.net/users/95347
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Is existence of a cardinal that witness non-failure of GCH everywhere everyway, a theorem of ZF?
|
Here's a somewhat trivial answer.
Note that $V\_\alpha$, for an infinite $\alpha$, have a particularly nice set of properties which follow from the fact that $|V\_\alpha\times V\_\alpha|=|V\_\alpha|$.
Now, easily, if $\sf AC$ fails, we can find arbitrarily high such cardinals. Simply take $|V\_\alpha|+\aleph(V\_\alpha)$, where $\aleph(x)$ is the Hartogs number of $x$. By Tarski's lemma, this cardinal is not idemmultiple.
If $\sf AC$ holds, then refer to the previous question you've asked.
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3
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https://mathoverflow.net/users/7206
|
413899
| 168,836 |
https://mathoverflow.net/questions/413901
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4
|
Suppose $L'$ is a fixed cyclic galois extension over $\mathbb {Q} $ of degree $4$.Now we know that there exists also a degree $k$ extension $L$ over $L'$ but the extension $(L/\mathbb{Q})$ may not be the galois extension. So my question is, can we always find such cyclic extension $(L/L')$ of degree $k$ such that $(L/\mathbb Q) $ is also a cyclic galois extension of degree $4k$?
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https://mathoverflow.net/users/215016
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Common Galois extension over $\mathbb Q $
|
If $k$ is odd, then yes. If $L'/\mathbb{Q}$ is a cyclic extension of degree $4$, choose an extension $M/\mathbb{Q}$ that is cyclic of degree $k$. Then the compositum $L'M/\mathbb{Q}$ will have ${\rm Gal}(L'M/\mathbb{Q}) \cong {\rm Gal}(L'/\mathbb{Q}) \times {\rm Gal}(M/\mathbb{Q})$ which will be cyclic of order $4k$.
If $k$ is even, the answer is no. In particular, if $L'/\mathbb{Q}$ is a cyclic extension that is not totally real (like $L' = \mathbb{Q}(\zeta\_{5})$), then it is not possible to find a degree $8$ extension $L/\mathbb{Q}$ with ${\rm Gal}(L/\mathbb{Q}) \cong \mathbb{Z}/8\mathbb{Z}$. If such an extension were to exist and $c \in {\rm Gal}(L/\mathbb{Q})$ is complex conjugation, then $c^{2} = 1$ and this implies that $\langle c \rangle = {\rm Gal}(L/L')$, which forces $c|\_{L'} = 1$ and hence $L'$ must be totally real.
It should be possible to give necessary and sufficient conditions on when it is possible to find such an extension $L$. It is well-known that a quadratic extension $K/\mathbb{Q}$ can be embedded in a $\mathbb{Z}/4\mathbb{Z}$ extension if and only if $K$ is totally real and the odd primes that ramify in $K$ are all $\equiv 1 \pmod{4}$. There is an extensive literature on Galois embedding problems. For a short intro, see [Section 3.4 of Klüners and Malle's paper "A database for field extensions of the rationals"](https://www.mathematik.uni-kl.de/%7Emalle/download/datbase.pdf). For much more detail, read Chapter 4 of Malle and Matzat's "Inverse Galois Theory".
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7
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https://mathoverflow.net/users/48142
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413903
| 168,837 |
https://mathoverflow.net/questions/413893
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0
|
The Weyl group of $\frak{g}\_2$ is the dihedral $D\_6$. Let us denote its longest element by $w\_0$. How many reduced decompositions does $w\_0$ have?
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https://mathoverflow.net/users/378228
|
Number of reduced decompositions of the dihedral group $D_6$
|
The weak order of a dihedral group looks like a polygon (see e.g. Figure 3.1 in the book "Combinatorics of Coxeter groups" by Björner and Brenti). Hence there are 2 reduced decompositions of $w\_0$ (= maximal chains from bottom to top) in these cases.
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2
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https://mathoverflow.net/users/25028
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413907
| 168,838 |
https://mathoverflow.net/questions/413892
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3
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Suppose that $V$ is a finite-dimensional real vector space and that $W\subseteq \operatorname{GL}(V)$
is a subgroup generated by reflections (elements $s$ of order $2$ whose locus of fixed points $H\_s$ is a hyperplane.)
Assume that $W$ contains only finitely many reflections. So the $H\_s$ divide $V$ into finitely many regions (chambers).
(1) Is there a reference for the fact (which is a slight variant of Th. 1 on p. 93 of Bourbaki, Groupes et algèbres de Lie, 4–6) that $W$ acts simply transitively on the set of chambers (and so is finite)? I believe this to be well known exactly as it is stated but cannot locate it.
(2) Is it also true (and is there a reference?) that if $S$ is the set of reflections
defined by the walls of a single chamber then $(W,S)$ is a Coxeter system and the given action of $W$ on $V$ is the geometric representation of this Coxeter system?
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https://mathoverflow.net/users/8726
|
Finiteness of a reflection group
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There are two differences between what this question is asking and standard results that are easy to find in the literature: (a) We don't assume ahead of time that $W$ is finite, but only that it has finitely many reflections, and (b) We don't assume ahead of time that the reflections all fix some Euclidean metric. We can get these assumptions back without much trouble:
(a) Here is a direct argument that "finitely many reflections" implies "finite". *See the edit history of this question for a more complicated argument with a gap. (Thanks to the OP for pointing out the gap.) The nature of the gap leads to the key point in this new argument.*
Assume that $W$ contains finitely many reflections and consider the hyperplane arrangement consisting of its reflecting hyperplanes. The *regions* of the arrangement are the closures of connected components of the complement.
The action of $W$ takes regions to regions, because the facets of regions are defined by hyperplanes and the action of $W$ takes reflecting hyperplanes to reflecting hyperplanes. (Given a hyperplane $H\_t$ and an element $w\in W$, the hyperplane $wH\_t$ is $H\_{wtw^{-1}}$.) If $W$ is infinite, then each region $R$ has an infinite stabilizer $W^R$, and since $R$ has finitely many extreme rays and elements of $W^R$ permute the extreme rays, there are infinitely many elements of $W$ that fix each extreme ray of $R$ setwise.
In particular, let $w$ be a nontrivial element of $W$ that fixes each extreme ray of $R$ pointwise. If $R'$ is an adjacent region to $R$, then since $w$ fixes their common facet and sends regions to regions, $w$ fixes $R'$ as a set as well, and thus fixes every ray of $R'$ as a set. We conclude that $w$ fixes every ray of every region.
Said another way, every ray consists of eigenvectors of $w$ (with positive, real eigenvalues). Furthermore, taking nonzero vectors from different rays of $R$, we obtain a basis of eigenvectors. Thus the ambient space decomposes as a direct sum of eigenspaces. If there is only one eigenspace, then $w$ scales the whole space by the eigenvalue, which is not $1$, since $w$ is not the identity. This is a contradiction, because the determinant if $w$ is not $\pm1$, but $W$ is generated by elements with determinant $-1$.
If there are multiple eigenspaces, then since the rays are contained in the eigenspaces and the reflecting hyperplanes are spanned by rays, each reflecting hyperplane contains all but one of the eigenspaces and intersects the other in codimension $1$. We see in this case that $W$ is a direct product of factors, each of which is isomorphic to a group generated by reflections on one of the eigenspaces. By induction on dimension, each of these is finite.
(b) There is a standard argument that for any finite group of linear transformations, there is a Euclidean symmetric bilinear form preserved by the group.
In case you have trouble finding the exact statements you want:
I have a chapter *[Finite Coxeter Groups and the Weak Order](https://nreadin.math.ncsu.edu/papers/regions.pdf)* in the book Lattice Theory: Special Topics and Applications, Volume 2, (ed. George Grätzer and Friedrich Wehrung), Springer 2016 that does these things. When I wrote this, I wrote down some results like this that surely were known but that I didn't know references for (and also many results that I did know references for). It's safe to assume that the results in that chapter *about the basic Coxeter group setup* are not new, but at least it's a place to cite things to if you can't find older results. References below are to that chapter.
The standard argument (b) is given as Proposition 10-2.8 (see also Proposition 10-2.7), but surely this argument is in many references. Once you have run that argument, so that you know you have a Euclidean reflection arrangement, the answer to your question (1) is probably in Bourbaki, or see Theorem 10-2.5. The answer to your question (2) is Theorem 10-2.9. The notes on page 557 cite Theorem 10-2.9 to Coxeter in the 1930's, but I don't recall whether that result is easily quotable from Coxeter or whether it's just implicit in the arguments there.
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2
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https://mathoverflow.net/users/5519
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413913
| 168,840 |
https://mathoverflow.net/questions/357924
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-1
|
The calculation of the area of the $\mathbb{R}^2$ plane depends on filtering used. I think, the most natural filtering is along the radius in polar coordinates:
$$S\_{\mathbb{R}^2}=\int\_0^\infty 2\pi r dr=2\pi\left(\frac{\tau^2}2+\frac1{24}\right)=\pi\tau^2+\frac\pi{12}$$
where $\tau=\int\_0^\infty dx$.
The regularized value of this area is $0$.
On the other hand, the area of a disk with radius $\tau$ (equal to the length of the real semi-axis) is $S\_c=\pi\tau^2$, and its regularized value is $-\frac\pi{12}$.
Thus, $S\_{\mathbb{R}^2}-S\_c=\frac\pi{12}$. I wonder, where this area difference comes from? Does it originate from the fact that the plane should not be considered a disk of infinite radius? Or it is some glitch of integration technique?
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https://mathoverflow.net/users/10059
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Are the shapes of the $\mathbb{R}^2$ plane and a disk of infinite radius different? Or otherwise, why their areas differ by $\frac\pi{12}$?
|
Apparently, the discrepancy comes from my use of non-natural definition of multiplication of divergent integrals. When using a more natural and intuitive [Levi-Civita field kind of construction](https://mathoverflow.net/q/413873/10059), the multiplication gives
$\int\_0^\infty dx\cdot \int\_0^\infty dx=\omega^2=2\int\_0^\infty x dx$.
So, the areas in both cases are $\pi\omega^2$, where $\omega=\int\_0^\infty dx$.
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1
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https://mathoverflow.net/users/10059
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413914
| 168,841 |
https://mathoverflow.net/questions/413908
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1
|
I am new to complex analysis and polynomials and I am looking for tutorials/books/articles for Mahler measure$$M(p((z))= \left|a\_0\right| \prod\limits\_{i=1}^d\max\{1,|\alpha\_i|\}
$$ of univariate polynomials and special polynomials like cyclotomic polynomials, polynomials over finite fields, the bounds of Mahler measure and its various relations to concepts like entropy, efficient numerical estimation of Mahler measure and the like. Could someone kindly refer me to the related literature? I would be highly obliged for the same.
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https://mathoverflow.net/users/158175
|
Mahler measure literature
|
In addition to the references at Wikipedia, see James McKee and Chris Smyth, Around the Unit Circle – Mahler Measure, Integer Matrices and Roots of Unity, <https://link.springer.com/book/10.1007/978-3-030-80031-4> and Brunault, F., & Zudilin, W. (2020), Many Variations of Mahler Measures: A Lasting Symphony (Australian Mathematical Society Lecture Series), Cambridge University Press, doi:10.1017/9781108885553.
Here's a description of the second book, taken from the publisher's website:
The Mahler measure is a fascinating notion and an exciting topic in contemporary mathematics, interconnecting with subjects as diverse as number theory, analysis, arithmetic geometry, special functions and random walks. This friendly and concise introduction to the Mahler measure is a valuable resource for both graduate courses and self-study. It provides the reader with the necessary background material, before presenting the recent achievements and the remaining challenges in the field. The first part introduces the univariate Mahler measure and addresses Lehmer's question, and then discusses techniques of reducing multivariate measures to hypergeometric functions. The second part touches on the novelties of the subject, especially the relation with elliptic curves, modular forms and special values of L-functions. Finally, the Appendix presents the modern definition of motivic cohomology and regulator maps, as well as Deligne–Beilinson cohomology. The text includes many exercises to test comprehension and challenge readers of all abilities.
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3
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https://mathoverflow.net/users/3684
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413921
| 168,844 |
https://mathoverflow.net/questions/413597
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4
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([This is an old MSE question from me](https://math.stackexchange.com/questions/3223033/a-question-on-a-possible-cyclic-sieving-phenomenon), which did not get any answer, and when looking back seems interesting to post it here:)
Let $G$ be a finite group. Consider the set $X\_G:=\cup\_{H\le G} G/H$, where the disjoint union is taken over all left cosets of the subgroups $H \le G$. Let $S \subset G$ be a generating set for $G$ and $|g| :=$ word length with respect to $S$. For a left coset $gH$ define $|gH| := \min\_{ h\in H} |gh|$.
Define $\pi(G,t) = \sum\_{x \in X\_G} t^{|x|}$. Then $\pi(G,1) = |X\_G|$.
For $G =\mathbb{Z}/(n) \equiv C\_n$ the cyclic group and $S=\{+1\}$, it seems that
$\pi(G,-1) = $ number of odd divisors of $n$ (OEIS: A001227). (1)
For $G =\mathbb{Z}/(n) \equiv C\_n$ the cyclic group and $S=\{\pm1\}$, it seems that
$\pi(G,-1) = $ #(of divisors of $n$ of the form $4m+1$)$-$#(of divisors of $n$ of the form $4m+3$), (OEIS: A002654).
So this $\pi(G,t)$ is a polynomial which when inserting the $2$-th roots of unity, counts something. My conjecture is, that $(X\_G, C\_2, \pi(G,t))$ exhibits the cyclic sieving phenomenon, but for this I need a group action from the cyclic group $C\_2$ to $X\_G$ such that (1) is fullfilled:
$\pi(C\_n,-1) = |X\_{C\_n}^{-1}| = $ number of odd divisors of $n$.
However, I tried $c\*gH = g^{-1}H$ for $C\_2 = \{1,c\}$ but this action, although natural, does not give the desired property, and I do not have a polynomial for this action to get a cyclic sieving phenomenon.
So my question is:
1. For the cyclic group $G:=C\_n$, how does one define an action from $C\_2$ to $X\_{C\_n}$ such that (1) is fullfilled? You do not have to prove that (1) is fullfilled, numerical coincidences should be ok for the first.
2. If there is such an action in 1), is it possible to define this for an arbitrary finite group $G$, hence $C\_2$ acts on $X\_G$?
Thanks for your help!
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https://mathoverflow.net/users/165920
|
A question on a possible cyclic sieving phenomenon?
|
Here is a sketch for the case of the cyclic group and $S$ containing only its generator.
Let $m$ be odd and $n=2^\ell m$. Then
$
(-1)\ast g^k H = \{ g^m e^{-1} | e \in g^k H \}
$
defines an action of $C\_2=\{+1, -1\}$ on the set of cosets of subgroups of the cyclic group $\langle g\rangle$ of order $n$:
$$
(-1)\ast\{ g^m e^{-1} | e \in g^k H \}
= \{ g^m (g^m e^{-1})^{-1} | e \in g^k H \}.
$$
The fixed points of this action are the cosets of the form $\langle g^k\rangle$ with $m = j k$:
$$
(-1)\ast\{g^{i k} | 0\leq i < n/k \}
= \{g^{j k - i k} | 0\leq i < n/k \}
= \{g^{(j-i) k} | 0\leq i < n/k \}.
$$
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2
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https://mathoverflow.net/users/3032
|
413931
| 168,847 |
https://mathoverflow.net/questions/412667
|
6
|
We know that there are non-standard models of arithmetic, and in such models there are non-standard proofs of standardly unprovable sentences. Now, we can imagine a version of representability relative to some non-standard model of some arithmetical theory such that the said function would be what I would call a "non-standard computation". At the same time we can imagine the said computation otherwise correct, as in it would be the same as adding some real/oracle to a TM. Is there a nice way to connect these concepts? Is it a trivial connection? Can it give us insights about the base theory X or the sentence S such that X + "sentence S" $\vdash$ TM(oracle) halts?
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https://mathoverflow.net/users/467143
|
Relationship between non-standard computation and TM(oracle)?
|
**Edit:** I think the answers given so far do not completely address the original question nor Gro-Tsen's follow up questions. I believe I can address some of these points so I have greatly expanded my original answer. In brief, no reasonable notion of "computation according to a nonstandard model of arithmetic" seems to be equivalent to "standard computation with an oracle for some real." Additionally there is a pretty good characterization of which sets of Turing degrees can constitute the sets computable according to some nonstandard model of arithmetic.
**Some terminology and standard results**
Suppose $\mathcal{M}$ is a nonstandard model of arithmetic (meaning either PA or some weak fragment of PA containing a sufficient amount of induction to make all arguments below work).
First, some standard terminology. Pick some reasonable way of using natural numbers to code finite sequences and let $a(i)$ denote the $i^\text{th}$ element in the sequence coded by $a$. Define the *standard system* of $\mathcal{M}$, written $SSy(\mathcal{M})$, to be the set $$\{x \subseteq \mathbb{N} \mid \exists a\in M\, (\forall i \in \mathbb{N}\, (i \in x \iff a(i) = 1))\}.$$ In other words a subset of $\mathbb{N}$ is in the standard system of $\mathcal{M}$ if it is an initial segment of some $\mathcal{M}$-finite sequence coded by an element of $\mathcal{M}$.
If it is known that the standard system of any nonstandard model of PA has some special properties. First, it is closed under relative computability—i.e. if $x$ is in the standard system and $y \le\_T x$ then $y$ is also in the standard system. Second, it is closed under computable joins—if $x$ and $y$ are both in the standard system then so is $x \oplus y$. Third, if $T$ is an infinite binary tree, some representation of which is in the standard system, then the standard system also contains some infinite path through $T$.
I believe these results are originally due to Dana Scott. A careful exposition of them can be found in Kaye's book on models of arithmetic. Unfortunately I don't have a copy of that book on hand so I can't provide a reference so instead I'll link to [these notes](https://cpb-us-w2.wpmucdn.com/blog.nus.edu.sg/dist/4/10956/files/2019/01/3_wkl-2elsf04.pdf) by Tin Lok Wong (see Theorem 3.5 in particular).
Scott actually proved something more, which will be relevant below. Any set of reals which satisfies the three properties above is called a *Scott set.* The results mentioned above show that the standard system of any nonstandard model of PA is a Scott set. A partial converse is also true: Scott showed that any countable Scott set is the standard system of some model of PA and Knight and Nadel later extended this to Scott sets of size $\aleph\_1$ (which finishes the story if we assume CH).
Lastly, let's note that no Scott set is equal to the set of reals computable from some oracle. To see why, let $A$ be a Scott set and suppose $A$ is equal to the reals computable from $x$. Then $x \in A$. But if we let $T$ be an infinite binary tree computable from $x$ with no infinite path computable from $x$ then we can see that $A$ must contain an infinite path through $T$ and thus not every real in $A$ is computable from $x$.
**Computation According to a Nonstandard Model**
Let $\mathcal{M}$ be a nonstandard model of arithmetic and define "computation according to $\mathcal{M}$ as follows: say that a set $x \subseteq \mathbb{N}$ is "computable according to $\mathcal{M}$ if there is some (possibly nonstandard) number $e \in M$ such that $\mathcal{M} \models \forall n\, (\varphi\_e(n) \downarrow)$ and $x = \{n \in \mathbb{N} \mid \mathcal{M} \models \varphi\_e(n) = 1\}$—in other words, $\mathcal{M}$ believes that the program coded by $e$ converges on every input and $x$ consists of those natural numbers on which $\mathcal{M}$ *believes* that $e$ converges and is equal to $1$.
It is easy to see that every set in $SSy(\mathcal{M})$ is also "computable according to $\mathcal{M}$." The converse is also true. Indeed, it is provable in PA that for any index for a program $e$ and any number $b$, there is a number $a$ coding a sequence which records those numbers less than $b$ on which the program coded by $e$ converges and outputs $1$ (i.e. $a$ codes a sequence of length $b$ and $a(i) = 1$ if and only if $i < b$ and $\varphi\_e(i) = 1$). Now suppose $e$ is a (possibly nonstandard) index for a program. Let $b$ be any nonstandard number and let $a$ be a nonstandard number coding the sequence we have just mentioned. Then the set $\{i \in \mathbb{N} \mid a(i) = 1\}$ is exactly the subset of $\mathbb{N}$ "computed by $e$ according to $\mathcal{M}$" and it is clearly in the standard system of $\mathcal{M}$.
Since the set of reals "computable according to $\mathcal{M}$" is exactly $SSy(\mathcal{M})$ and $\mathcal{M}$ is a Scott set, the set of reals "computable according to $\mathcal{M}$" cannot be equal to the set of reals computable from some oracle.
**What if the program only needs to converge on every natural number?**
Gro-Tsen proposed a number of modifications of the above definition of "computation according to $\mathcal{M}$." The first modification is to relax the requirement on what it means for a program to converge on all inputs. Above we required that $\mathcal{M}$ believes that the program coded by $e$ converges on all inputs in $\mathcal{M}$. What if we only require that $\mathcal{M}$ believes it converges on all inputs in $\mathbb{N}$? In other words instead of $\mathcal{M} \models \forall n\, (\varphi\_e(n) \downarrow)$ we only require $\forall n \in \mathbb{N}\,(\mathcal{M} \models \varphi\_e(n)\downarrow)$?
The answer is that this doesn't change anything. In particular, suppose $e$ is an element of $\mathcal{M}$ which codes a program which $\mathcal{M}$ believes converges on every input in $\mathbb{N}$. By [overspill](https://en.wikipedia.org/wiki/Peano_axioms#Overspill) there must be some nonstandard number $b$ such $\mathcal{M}$ believes that the program coded by $e$ converges on all inputs less than $b$. Let $e'$ be an index for a program which does the same thing as $e$ on all inputs less than $b$ and immediately halts and outputs $0$ on any input greater than $b$. Then the program coded by $e'$ converges on every input in $\mathcal{M}$ and corresponds to the same subset of $\mathbb{N}$ as $e$.
**What if the program index has to be a (standard) natural number?**
The next modification proposed by Gro-Tsen is more serious. What if we require that the index $e$ be a standard natural number rather than an arbitrary element of $\mathcal{M}$? The answer is that with this new requirement, the sets "computable according to $\mathcal{M}$ still cannot coincide with the sets computable from some oracle, except in one very special case.
To explain things more clearly, let me make another definition. Let's define the *computable standard system* of $\mathcal{M}$, written $CSSy(\mathcal{M})$, to be the set $$\{x \subseteq \mathbb{N} \mid \exists e \in \mathbb{N} \, \forall n \in \mathbb{N} \, (\mathcal{M} \models \varphi\_e(n)\downarrow \text{ and } (n \in x \iff \mathcal{M} \models \varphi\_e(n) = 1)\}$$ (note that this is not standard terminology). As Joel David Hamkins pointed out in his answer, $CSSy(\mathcal{M})$ can never be equal to $SSy(\mathcal{M})$ for any nonstandard model. The question is then: when is $CSSy(\mathcal{M})$ equal to the set of reals computable from some oracle?
The answer is that this can only occur if $CSSy(\mathcal{M})$ is exactly the set of computable reals. However, unlike the situation above, this can occur even if $\mathcal{M}$ is nonstandard.
For example, suppose $\mathcal{M}$ is a model of the true first order theory of the natural numbers. Then for any $e\in \mathbb{N}$, the program coded by $e$ will have the same behavior in $\mathcal{M}$ as it does in the "real world" so $CSSy(\mathcal{M})$ will not contain any noncomputable reals. And by compactness, it is possible for $\mathcal{M}$ to be nonstandard.
However, if $CSSy(\mathcal{M})$ contains any noncomputable real then it must actually be a Scott set and thus cannot be equal to the set of reals computable from some oracle. The only difficult part of this is to show that $CSSy(\mathcal{M})$ is closed under finding infinite paths through infinite binary trees so here's a proof of that part.
First, suppose $e \in \mathbb{N}$ is an index for a program and $n \in \mathbb{N}$ is an input such that $\varphi\_e(n)$ diverges but $\mathcal{M}$ believes it converges (such an $e$ and $n$ must exist if $CSSy(\mathcal{M})$ contains any noncomputable sets). Let $T$ be an infinite binary tree which is in $CSSy(\mathcal{M})$. Let $a$ be the program which does the following on input $i$: first, it waits for $\varphi\_e(n)$ to converge. If it converges after $m$ steps then $a$ looks for the largest $k \leq m$ such that $T$ contains a path of length $k$. If $i \leq k$ then it prints the $i^\text{th}$ element of the leftmost path of length $k$ in $T$. Otherwise it prints $0$. By overspill, $\mathcal{M}$ believes $T$ contains a path of nonstandard length so in $\mathcal{M}$ the program coded by $a$ will print a valid path through ($\mathcal{M}$'s version of) $T$ and when restricted to $\mathbb{N}$ that is an honest infinite path through $T$. However, since the description we gave above of $a$ only depended on the standard numbers $e$ and $n$, $a$ is itself standard and thus this infinite path is in $CSSy(\mathcal{M})$.
**What if we require the program to converge on all inputs in $\mathcal{M}$?**
In the definition of $CSSy(\mathcal{M})$ above, I only required the program $e$ to converge on all inputs in $\mathbb{N}$, not all inputs in $\mathcal{M}$. What if we change this? It turns out this doesn't really change anything about the above proof (basically because we were able to ensure that $a$ converges on all inputs) and so in this new version, $CSSy(\mathcal{M})$ is still either the set of computable reals or a Scott set (though perhaps a different Scott set than under the old definition).
**What if we don't intersect with $\mathbb{N}$ in the definitions above?**
It might seem a bit funny that when we looked at sets "computed according to $\mathcal{M}$ we were always taking an intersection with $\mathbb{N}$—i.e. we looked at $\{n \in \mathbb{N} \mid \varphi\_e(n) = 1\}$ rather than $\{n \in M \mid \varphi\_e(n) = 1\}$. This is for good reason. As mentioned by Joel David Hamkins, overspill implies that any set of the latter form is either finite or contains nonstandard numbers. So if we are ultimately interested in sets of natural numbers then this version of the question is very boring (the only sets of natural numbers "computable according to $\mathcal{M}$" are finite sets).
**What are the sets of Turing degrees in each of these scenarios?**
Gro-Tsen also asked about the possible sets of Turing degrees corresponding to each definition of "computable according to $\mathcal{M}$." Here, there is a pretty good partial answer. For each non-trivial definition above, the reals "computable according to $\mathcal{M}$" must always comprise a Scott set (though the converse is not clear). And quite a lot is known about sets of Turing degrees which form a Scott set so quite a lot is known about sets of Turing degrees which are those "computable according to $\mathcal{M}$" for some nonstandard $\mathcal{M}$.
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4
|
https://mathoverflow.net/users/147530
|
413942
| 168,849 |
https://mathoverflow.net/questions/413592
|
4
|
To explain what we are looking for, let's have a quick review on some points in Fourier transform on periodic functions in both continuous and discrete cases. We emphasize that our attention is just concerned with the abelian groups $\mathbb{R}$, $\mathbb{Z}$, $\mathbb{Z\_n}$ and the unit circle $\textbf{T}$.
On $\textbf{T}$, Fourier transform is a $\*$-homomorphism from $L^1(\textbf{T})$ to $c\_0(\mathbb{Z})$. One of the main question in this context is characterization of some type of functions for which the inversion formula to be hold (point-wisely and uniformly) i.e.,
$$f(x)=\sum \hat{f}(n)e^{inx}.$$
Finding the largest space satisfying the inversion formula is still an open problem. On the other hand, outstanding efforts of Kolomogrov, Kahane, Katznelson and Carleson provide sharp information concerning the points for which the identity $f(x)=\sum \hat{f}(n)e^{inx}$ fails. These two approaches have a long history and are well-organized on some books like *Trigonometric series* written by Zigmond or *Fourier Analysis* by Duoandikoetxea.
Q. On the abelian groups $\mathbb{R}$, $\mathbb{Z}$, $\mathbb{Z\_n}$, What are the main topics of Fourier transform (references ?) for anyone who wants to follow.
|
https://mathoverflow.net/users/84390
|
The main topics (issues, problems) of the Fourier transform
|
Saccone considered the space of all functions on $\mathbb{T}$ with *uniformly convergent* Fourier series. Call this space $U$. There is a natural norm given by $$\|f\|\_U = \sup\_{n\in\mathbb{N}} \|S\_nf\|\_{L^{\infty}}$$ where $S\_nf$ is the $n$th partial sum of the Fourier series of $f$. $U$ is complete with this norm.
It was shown in [[Saccone2000](https://www.ams.org/journals/proc/2000-128-06/S0002-9939-99-05361-7/)] that $U$ is not closed under pointwise multiplication of functions, so it's not a uniform algebra. Also, as a Banach space, $U$ has Pelczynski property (V) and Dunford-Pettis property.
---
Edit/Correction: $(\exists f,g\in U\hspace{4mm} fg\notin U)$ was not shown in [[Saccone2000](https://www.ams.org/journals/proc/2000-128-06/S0002-9939-99-05361-7/)]. Saccone refers to an [article](https://mathscinet.ams.org/mathscinet-getitem?mr=188695) by Kahane & Katznelson in 1965.
|
3
|
https://mathoverflow.net/users/164350
|
413943
| 168,850 |
https://mathoverflow.net/questions/413875
|
1
|
I am working on proving the following: Let $\rho(x)= \frac{2}{2+x^2}$, $\theta >1$ (assumed integer here) and $B \subset H^1\_{ul}$,(uniformly local Sobolev space), be any subset which is bounded in $H\_{ul}^\theta$. Then $B$ is pre-compact in
$$H^1\_\rho = \{ u: \mathbb R \to \mathbb R : u, \partial\_x u \in L^2\_{loc} (\mathbb R),\quad \| u\|\_{H^1\_\rho}=\| u\rho\|\_{H^1} < \infty\}.$$
I need to show that $B$ admits the finite covering in $H^1\_\rho$ by balls of radius less than $\epsilon $. Taking $\chi\_\beta$ as a smooth cut off function and decomposing $u \in B$ into two parts $u= v+u$ where $v= \chi\_\beta u$ and $w= u \chi\_\beta$, and
$$\chi\_\beta(X) =
\begin{cases}
1,& |x| \ge \beta +1\\
0, & |x| \le \beta
\end{cases}.$$
Taking the following norm
$$\|v\|^2\_{H^1\_\rho}=\|u\chi\_\beta\rho\|^2\_{H^1} =\|u\chi\_\beta\rho(x)\|^2\_{L^2} + \| \partial\_x (u\chi\_\beta\rho)\|^2\_{L^2}.$$
Here I can choose $\beta$ sufficiently large to get the first $L^2$ small as I want but in the second norm after using the product rule I obtain the derivative of the cut-off function which I do not know what it is and how to deal with it? Could enlighten me, please.
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https://mathoverflow.net/users/471464
|
First derivative of cut off function
|
I constructed a function that satisfies what I want. Let $f(x) = e^{-1/x},\,\, x>0$ and vanishes everywhere. Now we define $\varphi\_\beta(x) = f(\beta +1 - |x|)$ and $\psi\_\beta (x)=f(|x|- \beta)$. Therefore the cut-off function is
$$\chi\_\beta(x) = \frac{\psi\_\beta (x)}{\psi\_\beta (x) + \varphi\_\beta (x)}.$$
|
1
|
https://mathoverflow.net/users/471464
|
413946
| 168,851 |
https://mathoverflow.net/questions/413922
|
3
|
Consider the compact group $ G=\operatorname{SO}\_3(\mathbb{R}) $. The closed subgroups of $ G $ (other than the trivial group 1 and the whole group $ G $) are exactly $ O\_2$, $\operatorname{SO}\_2 $ and the finite groups $ C\_n$, $D\_{2n}$, $T \cong A\_4$, $O \cong S\_4$, $I \cong A\_5 $ (cyclic groups with $ n $ elements, dihedral groups with $ 2n $ elements and the three symmetry groups of the platonic solids). The normalizers of these groups are as follows:
\begin{align\*}
G&=N\_G(G)=N\_G(1) \\
O\_2&=N\_G(O\_2)=N\_G(\operatorname{SO}\_2)=N\_G(C\_n) \\
I&= N\_G(I) \\
O&=N\_G(O)=N\_G(T)=N\_G(D\_4) \\
D\_{4n} &= N\_G(D\_{2n})
\end{align\*}
where in the last equation $ n \geq 3 $. We say a (closed) subgroup is maximal if it is maximal among all proper closed subgroups of $ G $.
Observe that in the example above the maximal subgroups exactly coincide with the self-normalizing subgroups. Namely,
$$
O\_2, I,O.
$$
That the maximal subgroups are all self-normalizing is not too surprising. The normalizer of a closed subgroup is always closed. Thus a maximal subgroup is always either normal or self-normalizing. Since $ G $ is simple, adjoint (i.e. center-free), and connected that means the maximal subgroups must be self-normalizing. However I am a bit surprised that the reverse holds. That is, that every self-normalizing subgroup of $ G $ is maximal. That inspires my question:
**For a closed subgroup of a compact Lie group does self-normalizing imply maximal?**
This is true for the compact Lie group $ \operatorname{SO}\_3(\mathbb{R}) $ and thus also true for $ \operatorname{SU}\_2 $. What about the generic case? I am especially interested in $ \operatorname{SU}\_3 $.
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https://mathoverflow.net/users/387190
|
Self-normalizing implies maximal for subgroup of compact Lie group
|
$\DeclareMathOperator\SO{SO}\DeclareMathOperator\U{U}\DeclareMathOperator\O{O}$Denote by $\U'(n)$ the normalizer of $\U(n)$ in $\mathrm{GL}\_{2n}(\mathbf{R})$. It is not hard to see that $\U(n)$ has index 2 in $\U'(n)$, which is generated by $\U(n)$ and by the coordinate-wise complex conjugation. Moreover, $\U'(n)$ is maximal in $\O(2n)$ (if $n\ge 2$).
In $G=\SO(5)$, consider the subgroup $H=\SO(5)\cap (\U'(2)\times \O(1))$ (which contains $\U(2)$ with index 2). I claim that $H$ is self-normalized, but not maximal, in $G=\SO(5)$.
The subgroup $H$ is not maximal in $G$ because it is properly contained in $L=SO(5)\cap (\O(4)\times \O(1))$.
It is self-normalized. Indeed, its action on $\mathbf{R}^5$ has the irreducible decomposition $4\oplus 1$, which is preserved by the normalizer. Hence, the normalizer of $H$ in $G$ equals the normalizer of $H$ inside $\SO(5)\cap (\O(4)\times \O(1))=L$. Using that $\U'(2)$ is maximal in $\O(4)$ one can deduce easily that $H$ is maximal in $L$. Since $H$ is not normal in $L$, it is self-normalized in $L$. Hence $H$ is self-normalized in $G$.
|
3
|
https://mathoverflow.net/users/14094
|
413949
| 168,853 |
https://mathoverflow.net/questions/413933
|
3
|
I would appreciate a reference to support this statement that
appears under the **Geodesic** entry of the
[CRC Encyclopedia of Mathematics](https://www.routledge.com/The-CRC-Encyclopedia-of-Mathematics-Third-Edition---3-Volume-Set/):
>
> "no matter how badly a sphere is distorted,
> there exists an infinite number of closed geodesics on it.
> This general result, demonstrated in the early 1990's,
> extended earlier work of Birkhoff, ..."
>
>
>
|
https://mathoverflow.net/users/6094
|
Infinite number of closed geodesics on distorted sphere
|
Will Jagy answered my question:
>
> Bangert, Victor. "On the existence of closed geodesics on two-spheres." *International Journal of Mathematics* 4, no. 01 (1993): 1-10.
> [doi](https://doi.org/10.1142/S0129167X93000029).
>
>
>
"...one obtains the existence of infinitely many closed geodesics for every Riemannian metric on $\mathbb{S}^2$."
|
3
|
https://mathoverflow.net/users/6094
|
413956
| 168,854 |
https://mathoverflow.net/questions/413958
|
0
|
Let $G$ be a simply connected Lie group. Is it true that any finite dimensional representation of its Lie algebra is the differential of a representation of $G$?
A reference would be helpful.
Sorry if the question is too basic.
|
https://mathoverflow.net/users/16183
|
Representations of simply connected Lie groups
|
A finite-dimensional representation of a Lie algebra is in particular a homomorphism of finite-dimensional Lie algebras. Hence your question is answered in the affirmative e.g. by Th. 3.27 in F. Warner's book "Foundations of differential geometry and Lie groups".
|
3
|
https://mathoverflow.net/users/468624
|
413960
| 168,855 |
https://mathoverflow.net/questions/413924
|
12
|
This question is certainly somewhat opinion-based, but hopefully not hopelessly so.
The granddaddy of all applications for an efficient period finding or factoring capability (e.g. Shor's algorithm) is obviously breaking the public-key cryptography systems that currently encrypt Internet traffic.
But if we could eventually implement Shor's algorithm on a large-scale quantum computer, would it have any *other* useful applications, whether for pure math research or out in the "real world"?
This question is inspired by an [essay by Boaz Barak](https://www.boazbarak.org/Papers/skeptics.pdf), in which he mentions that he doesn't know of any reason why efficient factoring is inherently interesting other than for cryptography.
Two scope clarifications:
1. I know that the quantum Fourier transform at the heart of Shor's algorithm has other potential practical applications, e.g. for solving large linear systems via the HHL algorithm, but I'm specifically wondering about period finding and factoring, and not general ultra-fast Fourier transforms.
2. I know that the *existence* of an efficient quantum factoring algorithm has huge implications for computational complexity theory: it provides perhaps the strongest evidence we have (a) that **BPP** != **BQP**, (b) against the extended Church-Turing thesis, and (c) depending on your philosophical beliefs, perhaps against the real-world feasibility of building a fault-tolerant quantum computer. But I'm wondering about actually *executing* such an algorithm, not about its existence or properties.
|
https://mathoverflow.net/users/95043
|
Would efficient factoring have any *other* useful applications?
|
I'm not sure about "real world", but studies around [multiplicative functions](https://en.wikipedia.org/wiki/Multiplicative_function) (e.g., [aliquot sequences](https://en.wikipedia.org/wiki/Aliquot_sequence)) will definitely benefit from the availability of a fast factorization method. At very least it will allow to verify, refine, or refute existing conjectures with extensive computational evidence.
Also, there is a [list of OEIS sequences "needing factors"](https://oeis.org/wiki/OEIS_sequences_needing_factors), which gives other examples of topics unrelated to cryptography but relying on integer factorization.
|
9
|
https://mathoverflow.net/users/7076
|
413962
| 168,857 |
https://mathoverflow.net/questions/413939
|
2
|
Let $X$ be a continuous time stochastic process, and denote by $\mathcal F\_t$ its natural filtration. We define $\mathcal F\_z = \mathcal F\_0$ for all $z \leq 0$.
$X$ is said to be *strongly predictable* if there exists some $r > 0$, and an $\mathcal F\_{t-r}$ adapted process $Y$ such that
$$\lim\_{T \to \infty} \mathbb E \left [\int\_{T}^\infty |X\_t - Y\_t| \, dt \right ] = 0.$$
**Question:** Suppose $X$ is a strongly predictable martingale. Is it true that $X\_t$ converges almost surely as $t \to \infty$?
|
https://mathoverflow.net/users/173490
|
A martingale convergence theorem
|
$\newcommand{\F}{\mathcal{F}}$The answer is yes.
Indeed, by time rescaling, without loss of generality (wlog) $r=1$.
Take any random variable (r.v.) $Z$ with $EZ=0$, and let $Z'$ be an independent copy of $Z$. Then, by Jensen's inequality, for any real $z$ we have $E|Z|\le E|Z-Z'|=E|(Z-z)-(Z'-z)|\le2E|Z-z|$, so that
\begin{equation\*}
E|Z|\le2E|Z-z|. \tag{1}
\end{equation\*}
Letting $E\_t:=E(\cdot|\F\_t)$ and using (1) with $Z=X\_t-X\_{t-1}$ and real $t\ge1$, we get
\begin{equation\*}
\begin{aligned}
E\_{t-1}|X\_t-X\_{t-1}|&\le2E\_{t-1}|(X\_t-X\_{t-1})-(Y\_t-X\_{t-1})| \\
&=2E\_{t-1}|X\_t-Y\_t|,
\end{aligned}
\end{equation\*}
since $Y\_t-X\_{t-1}$ is $\F\_{t-1}$-measurable.
So, $E|X\_t-X\_{t-1}|\le2|X\_t-Y\_t|$. So, for any real $u\ge0$ and any natural $T$,
\begin{equation\*}
S\_u:=\sum\_{n=T}^\infty E|X\_{n+u}-X\_{n-1+u}|\le2\sum\_{n=T}^\infty E|X\_{n+u}-Y\_{n+u}|
\end{equation\*}
and hence
\begin{equation\*}
\begin{aligned}
\int\_0^1 du\, S\_u &\le2\sum\_{n=T}^\infty \int\_0^1 du\, E|X\_{n+u}-Y\_{n+u}| \\
& =2\int\_T^\infty dt\, E|X\_t-Y\_t|<\infty
\end{aligned}
\end{equation\*}
if $T$ is large enough, by your displayed condition. So, $S\_u<\infty$ for some $u\in[0,1]$. By time shift, wlog $u=1$. So,
\begin{equation\*}
\sum\_{n=T}^\infty E|X\_{n+1}-X\_n|=S\_1<\infty. \tag{2}
\end{equation\*}
So, $(X\_n)$ converges in $L^1$ and hence $(E|X\_n|)$ is bounded. So, by [Doob's martingale convergence theorem](https://en.wikipedia.org/wiki/Doob%27s_martingale_convergence_theorems#Statement_for_discrete-time_martingales),
\begin{equation\*}
X\_n\to Y \tag{3}
\end{equation\*}
as $n\to\infty$ almost surely (a.s.) for some (real-valued) r.v. $Y$.
Next, by [Doob's martingale inequality](https://en.wikipedia.org/wiki/Doob%27s_martingale_inequality),
\begin{equation\*}
P(\max\_{t\in[n,n+1]}|X\_t-X\_n|>h)\le\frac{E|X\_{n+1}-X\_n|}h
\end{equation\*}
for any real $h>0$.
So, by the [Borel–Cantelli lemma](https://en.wikipedia.org/wiki/Borel%E2%80%93Cantelli_lemma) and (2), for each real $h>0$ a.s. there will be only finitely many natural $n$ such that $\max\_{t\in[n,n+1]}|X\_t-X\_n|>h$. That is, $\max\_{t\in[n,n+1]}|X\_t-X\_n|\to0$ a.s. as $n\to\infty$.
Thus, by (3), $X\_t\to Y$ a.s. as $t\to\infty$, as claimed.
|
3
|
https://mathoverflow.net/users/36721
|
413966
| 168,859 |
https://mathoverflow.net/questions/410576
|
1
|
Given a simply connected locally compact group $G$, is it true that $G$ admits enough finite dimensional representations (over any field and not necessarily continuous) to separate points in $G$, what about over $\mathbb{C}$ and we require the representations to be continuous?
Again, this question is a follow-up of [this one](https://mathoverflow.net/questions/410186/examples-of-locally-compact-groups-that-do-not-admit-enough-finite-dimensional-r), and it seems better to ask it separately here.
|
https://mathoverflow.net/users/128540
|
Does every locally compact, simply connected group admit enough finite dimensional representations?
|
The connected Lie groups whose points are separated by the finite-dimensional complex representations are exactly the linear Lie groups, for instance by Th. 5.3 in [Beltiţă and Neeb - Finite-dimensional Lie subalgebras of algebras with continuous inversion](http://dx.doi.org/10.4064/sm185-3-3).
|
4
|
https://mathoverflow.net/users/468624
|
413972
| 168,861 |
https://mathoverflow.net/questions/413969
|
4
|
In the " A Primer on Mapping Class
Groups
Benson Farb and Dan Margalit"
We have :
**Proposition** 1.10 Let $\alpha$ and $\beta$ be two essential simple closed curves in a surface $S$. Then $\alpha$ is isotopic to $\beta$ if and only if $\alpha$ is homotopic to $\beta$.
Proof. One direction is vacuous since an isotopy is a homotopy. So suppose that $\alpha$ is homotopic to $\beta$. We immediately have that $i(\alpha, \beta)=0$. By performing an isotopy of $\alpha$, we may assume that $\alpha$ is transverse to $\beta$. If $\alpha$ and $\beta$ are not disjoint, then by the bigon criterion they form a bigon. A bigon prescribes an isotopy that reduces intersection. Thus we may remove bigons one by one by isotopy until $\alpha$ and $\beta$ are disjoint.
In the remainder of the proof, we assume $\chi(S)<0$; the case $\chi(S)=0$ is similar, and the case $\chi(S)>0$ is easy. Choose lifts $\widetilde{\alpha}$ and $\widetilde{\beta}$ of $\alpha$ and $\beta$ that have the same endpoints in $\partial \mathbb{H}^{2}$. There is a hyperbolic isometry $\phi$ that leaves $\widetilde{\alpha}$ and $\widetilde{\beta}$ invariant and acts by translation on these lifts. As $\widetilde{\alpha}$ and $\widetilde{\beta}$ are disjoint, we may consider the region $R$ between them. The quotient $R^{\prime}=$ $R /\langle\phi\rangle$ is an annulus; indeed, it is a surface with two boundary components with an infinite cyclic fundamental group. A priori, the image $R^{\prime \prime}$ of $R$ in $S$ is a further quotient of $R^{\prime}$. However, since the covering map $R^{\prime} \rightarrow R^{\prime \prime}$ is single-sheeted on the boundary, it follows that $R^{\prime} \approx R^{\prime \prime}$. The annulus $R^{\prime \prime}$ between $\alpha$ and $\beta$ gives the desired isotopy.
>
> how we can prove the case $\chi(S)=0$ and the case $\chi(S)>0$ ? why The annulus $R^{\prime \prime}$ between $\alpha$ and $\beta$ gives the desired isotopy ? How we can prove $R^{\prime \prime}$ desired isotopy ?
>
>
>
I think if $\chi(S)=0$ then $2-2g-(b+n)=0$ so we have two case $g=0,1$ then we have a surface with $b+n=2,0$ then $\alpha$ and $\beta$ are isotopic.
|
https://mathoverflow.net/users/155706
|
About isotopy and homotopy
|
Once you have found an annulus $R \subset S$ whose two boundary components are $\alpha$ and $\beta$, by definition of "annulus" there exists a homeomorphism $H : S^1 \times [0,1] \to R$. The composition
$$S^1 \times [0,1] \xrightarrow{H} R \hookrightarrow S
$$
then defines an isotopy in $S$ from $\alpha$ to $\beta$.
If $\chi(S)=0$ the surface $S$ is a torus or annulus, and we can then obtain the desired annulus $R$ using a Euclidean structure on $S$ in a manner similar to how the hyperbolic structure is used when $\chi(S)<0$.
In the case $\chi(S)>0$ the surface $S$ is a sphere or disc, in which case by the Schönflies theorem no essential simple closed curves exist and so the proposition is vacuously true.
|
3
|
https://mathoverflow.net/users/20787
|
413980
| 168,864 |
https://mathoverflow.net/questions/413992
|
8
|
Let $X$ be a topological space (feel free to add some simplifying assumptions here, like “completely regular” provided at least the case of finite-dimensional manifolds is covered). Let $f,g \in C^\*(X)$ where $C^\*(X)$ denotes the ring of bounded continuous real-valued functions on $X$. Denote $f^\beta, g^\beta \colon \beta X\to \mathbb{R}$ the continuous functions corresponding to $f,g\colon X\to \mathbb{R}$ under the canonical isomorphism $C(\beta X) \cong C^\*(X)$ where $\beta X$ is the Stone-Čech compactification of $X$, and let $Z(f^\beta),Z(g^\beta)$ be their zero-sets, i.e., $Z(f^\beta) = \{p\in\beta X : f^\beta(p)=0\}$.
**Question:** how can we tell whether $Z(f^\beta) \subseteq Z(g^\beta)$ merely by looking at $f,g$ (without looking at the Stone-Čech compactification)?
To give an idea of the flavor of criteria I'm looking for, let me observe that:
**Fact:** $Z(f^\beta) = \varnothing$ iff $f$ is bounded away from $0$ on $X$ (i.e. $\exists \varepsilon>0. |f|\geq\varepsilon$). (Proof: if $|f|\geq\varepsilon$ then clearly $|f^\beta|\geq \varepsilon$ so $Z(f^\beta)=\varnothing$. But conversely, if $Z(f^\beta)=\varnothing$, since $\beta X$ is compact, the image of $|f^\beta|$ is a compact subset of $\mathbb{R}\_{>0}$, so it is lower bounded by some $\varepsilon>0$.) Equivalently, this means that $f$ is invertible in $C^\*(X) \cong C(\beta X)$.
Based on the above fact, I first thought that $Z(f^\beta) \subseteq Z(g^\beta)$ meant $|g|\leq C|f|$ on $X$ for some constant $C$, but this is not correct (take $X=\mathbb{R}$ and $f\colon x\mapsto \min(1,x^2)$ and $g\colon x\mapsto \min(1,|x|)$: then $f^\beta$ and $g^\beta$ both vanish only at $0$ but we don't have $|g|\leq C|f|$). Maybe something like “$Z(f)\subseteq Z(g)$ and there exists $K\subseteq X$ compact and $C$ such that $|g|\leq C|f|$ outside $K$”?
**Motivation:** thinking about the PS in [this answer](https://mathoverflow.net/questions/413739/let-x-be-a-manifold-is-it-true-that-beta-x-cong-operatornamespecmc-inf/413891#413891) made me ask whether, for $X$ a manifold and $f$ bounded continuous on $X$, we can find $g$ bounded and *smooth* on $X$ such that $Z(f^\beta) = Z(g^\beta)$ (and to find a criterion for the latter, looking at $Z(f^\beta) \subseteq Z(g^\beta)$ first seems natural).
|
https://mathoverflow.net/users/17064
|
When are the zero sets of two continuous functions in the Stone-Čech compactification included in one another?
|
Lemma: Suppose that $X$ is a compact Hausdorff space. Let $f,g:X\rightarrow[0,\infty)$ be continuous functions. Then the following are equivalent:
1. $Z(f)\subseteq Z(g)$.
2. For all $\epsilon>0$, there exists a $\delta>0$ where $f^{-1}[0,\delta]\subseteq g^{-1}[0,\epsilon]$.
3. There exists a function $u:[0,\infty)\rightarrow[0,\infty)$ that is continuous at the point $0$ with $u(0)=0$ and where $g\leq u\circ f$.
4. There exists a continuous function $v:[0,\infty)\rightarrow[0,\infty)$ with $v(0)=0$ and where $g\leq v\circ f$.
5. There exists a continuous bijection $w:[0,\infty)\rightarrow[0,\infty)$ with $w(0)=0$ and where $g\leq w\circ f$ (the mapping $w$ is necessarily a homeomorphism and increasing).
Proof: $5\rightarrow 4,4\rightarrow 3.$ These are trivial.
$4\rightarrow 5.$ Suppose that $v:[0,\infty)\rightarrow[0,\infty)$ is a bijection with $v(0)=0$ and $g\leq v\circ f$. Then let $w:[0,\infty)\rightarrow[0,\infty)$ be the function defined by letting $w(x)=v(x)+x$. Then $g\leq w\circ f$ and $w(0)=0$.
$3\rightarrow 4.$ There is some continuous $v:[0,\infty)\rightarrow[0,\infty)$ where $v(0)=0$ and where either
$\min(\max(g),u(y))\leq v(y)$ for all $y$. Therefore, for all $x$, we have
$g(x)\leq u(f(x))$, so $g(x)\leq\min(\max(g),u(f(x)))\leq v(f(x)).$
$5\rightarrow 2$. Suppose that $\epsilon>0$. Then set $\delta=w^{-1}(\epsilon)$. If $x\in f^{-1}[0,\delta]$, then $f(x)\leq\delta$, so
$g(x)\leq w(f(x))\leq w(\delta)=\epsilon.$ Therefore $x\in g^{-1}[0,\epsilon]$.
$2\rightarrow 1$. If $\epsilon>0$, then there is a $\delta>0$ where
$Z(f)\subseteq f^{-1}[0,\delta]\subseteq g^{-1}[0,\epsilon]$. Therefore,
$Z(f)\subseteq\bigcap\_{n=1}^{\infty}g^{-1}[0,1/n]=Z(g)$.
$1\rightarrow 3.$ Suppose that $Z(f)\subseteq Z(g)$. If $Z(f)=\emptyset$, then $\min(f)>0$, so just select a suitable function $u$ with
$u(y)\geq\max(g)$ whenever $y\geq\min(f)$. Now, assume $Z(f)\neq\emptyset$.
Let $u:[0,\infty)\rightarrow[0,\infty)$ be the mapping such that
$u(y)=\max\{g(x)\mid f(x)\leq y\}=\max g[f^{-1}(-\infty,y]].$ Then $u(0)=\max\{g(x)\mid f(x)\leq 0\}=0$. Furthermore, if $x\_{0}\in X$, then
$u(f(x\_{0}))=\max\{g(x)\mid f(x)\leq f(x\_{0})\}$. Therefore,
$g(x\_{0})\leq u(f(x\_{0}))$. I claim that $u$ is upper semicontinuous (and therefore continuous at $0$).
Suppose that $y\in u^{-1}(-\infty,c)$. Then $\max g[f^{-1}(-\infty,y]]=\max\{g(x)\mid f(x)\leq y\}=u(y)<c$, so
$g[f^{-1}(-\infty,y]]\subseteq(-\infty,c)$. Therefore,
$f^{-1}(-\infty,y]\subseteq g^{-1}(-\infty,c)$. By compactness, there is some $z>y$ where $f^{-1}(-\infty,z]\subseteq g^{-1}(-\infty,c)$. However, if $s<z$, then $f^{-1}(-\infty,s]\subseteq g^{-1}(-\infty,c)$, so
$g[f^{-1}(-\infty,s]]\subseteq(-\infty,c)$. Therefore,
$u(s)=\max(g[f^{-1}(-\infty,s]])<c$ as well. Therefore, since there is a neighborhood $U$ of $y$ where $u(s)<c$ whenever $s\in U$, the function $u$ is upper-semicontinuous.
Q.E.D.
Theorem: Let $X$ be a completely regular space with compactification $C$. Suppose that $f,g:X\rightarrow[0,\infty)$ are continuous functions that extend to continuous maps $\overline{f},\overline{g}:C\rightarrow[0,\infty)$. Then the following are equivalent:
1. $Z(\overline{f})\subseteq Z(\overline{g})$.
2. For all $\epsilon>0$, there exists a $\delta>0$ where $f^{-1}[0,\delta]\subseteq g^{-1}[0,\epsilon]$.
3. There exists a function $u:[0,\infty)\rightarrow[0,\infty]$ that is continuous at the point $0$ with $u(0)=0$ and where $g\leq u\circ f$.
4. There exists a continuous function $v:[0,\infty)\rightarrow[0,\infty)$ with $v(0)=0$ and where $g\leq v\circ f.$
5. There exists a continuous bijection $w:[0,\infty)\rightarrow[0,\infty)$ with $w(0)=0$ and where $g\leq w\circ f$.
Now suppose that $X$ is a locally compact regular space with compactification $C$, and suppose that $f,g:X\rightarrow[0,\infty)$ and $f,g$ extend to mappings $\overline{f},\overline{g}:C\rightarrow[0,\infty)$. The following result characterizes when $Z(\overline{f}|\_{C\setminus X})\subseteq Z(\overline{g}|\_{C\setminus X})$, so one can use the following result to produce more characterizations of when $Z(\overline{f})\subseteq Z(\overline{g})$ when $X$ is locally compact.
Theorem: Suppose that $X$ is a non-compact locally compact regular space with compactification $C$. Let $f,g:X\rightarrow[0,\infty)$ be bounded continuous functions, and let $\overline{f},\overline{g}:C\rightarrow[0,\infty)$ be the continuous extensions of $f,g$ to the domain $C$. Then the following are equivalent.
1. $Z(\overline{f}|\_{C\setminus X})\subseteq Z(\overline{g}|\_{C\setminus X})$.
2. For each $\epsilon>0$, there exists a $\delta>0$ and a compact $K\subseteq X$ such that if $x\in X\setminus K$, then $f(x)<\delta\rightarrow g(x)<\epsilon.$
3. There exists a continuous bijection $u:[0,\infty)\rightarrow[0,\infty)$ with $u(0)=0$ and a function $A:X\rightarrow[0,\infty)$ where
$A^{-1}[\epsilon,\infty)$ is compact for each $\epsilon>0$
and where $g\leq A+(u\circ f)$.
Proof: $2\rightarrow 1$. Suppose that $c\_{0}\in Z(f|\_{C\setminus X})$. Therefore, let $(x\_{d})\_{d\in D}$ be a net that converges to $c\_{0}$. Then
for each $\epsilon>0$, there is some $\delta>0$ and compact set $K\subseteq X$ where if $x\in X\setminus K$ and $f(x)\leq\delta$, then $g(x)\leq\epsilon$. Then there is some $d\_{0}\in D$ where if $d\leq d\_{0}$, then $x\_{d}\not\in K$ and where $f(x\_{d})\leq\delta$. In this case, we have
$g(x\_{d})\leq\epsilon$. Therefore, we conclude that $g(x\_{d})\_{d\in D}\rightarrow 0$, so since $g$ is continuous, we know that $g(c\_{0})=0$, and therefore $c\_{0}\in Z(g|\_{C\setminus X})$. Thus,
$Z(f|\_{C\setminus X})\subseteq Z(g|\_{C\setminus X})$.
$1\rightarrow 3$. By the above results, we know that there is a continuous mapping $u:[0,\infty)\rightarrow[0,\infty)$ such that
$\overline{g}|\_{C\setminus X}\leq u\circ\overline{f}|\_{C\setminus X}$. Therefore, let $A:X\rightarrow[0,\infty)$ be the mapping defined by
$A=\max(0,g-(u\circ f))$. Then
$g=g-(u\circ f)+(u\circ f)\leq A+(u\circ f),$ and $\overline{A}(c)=0$ whenever $c\in C\setminus X$, so $A^{-1}[\epsilon,\infty)$ is a compact subset of $X$ whenever $\epsilon>0$.
$3\rightarrow 2$. Suppose that $g\leq A+(u\circ f)$. Then for all $\epsilon>0$, there is a $\delta>0$ with $u(\delta)<\epsilon$. Therefore, suppose that $c\in C\setminus X$, and $\overline{f}(c)\leq\delta$. Then
$$\overline{g}(c)\leq\overline{A}(c)+(u\circ\overline{f})(c)=u(\overline{f}(c))\leq u(\delta)<\epsilon.$$ In particular, there is no $c\in C\setminus X$ with $\overline{f}(c)\leq\delta$ and $\overline{g}(c)\geq\epsilon$. Therefore, if we set $K=\overline{f}^{-1}[0,\delta]\cap\overline{g}[\epsilon,\infty)$, then $K$ is a closed subset of $C$, so $K$ is compact, but $K$ is also a subset of $X$. Therefore, if $x\in X\setminus K$, then $f(x)\leq\delta\rightarrow g(x)<\epsilon$.
Q.E.D.
|
7
|
https://mathoverflow.net/users/22277
|
413997
| 168,868 |
https://mathoverflow.net/questions/413985
|
2
|
I have a doubt that assails me.
The technique of gluing along edges between manifolds is generally considered in the topological context.
I don't know if there are other gluing techniques.
I was wondering if there were theorems regarding the possibility of obtaining smooth gluing between isometric manifolds to each other.
That is, if there is an isometry between two smooth manifolds (even local, not necessarily global), is there always at least one way to glue them so that the resulting manifold is still smooth?
Could you give me some information about it and possibly point out some references?
|
https://mathoverflow.net/users/90594
|
Isometry and gluing between smooth manifolds - some references
|
Let (Mi)i=1,2 be smooth manifolds with boundary, Ni ⊆ ∂Mi unions of connected components of the boundaries of M1 and M2, respectively, and φ : N1 → N2 a diffeomorphism. Then there exists a smooth structure on the space M1 ∪φ M2 that arises by gluing M1 to M2 along N1 ≃ N2. This structure is unique up to a diffeomorphism that leaves all points from the original boundaries of M1 and M2, including N1 ≃ N2, fixed.
This statement is copied from <http://awgd.org/talks/morse/notes2.pdf> , the reference given there is Munkres “Elementary Differential Topology”.
|
1
|
https://mathoverflow.net/users/39082
|
414003
| 168,869 |
https://mathoverflow.net/questions/412251
|
5
|
Robert Bryant's answer to [Isometric embedding of SO(3) into an euclidean space](https://mathoverflow.net/questions/161295/isometric-embedding-of-so3-into-an-euclidean-space) mentions that there is an isometric embedding of the round tetrahedral space $ SO\_3/A\_4 $ into the round sphere $ S^6 $.
I like that fact. Dreaming along those lines, here are some more facts: (and I am interested in any similar examples that MO knows about)
-$S^4$ contains isometrically embedded copies of the the round projective plane $ \mathbb{R}P^2 $ and the round prism manifold $ S^3/Q\_8 $ (this happens to be the manifold of complete flags in $ \mathbb{R}^3 $) where $ Q\_8 $ is the eight element quaternion group
-This doesn't quite fit the theme of roundness but $ S^5 $ contains a weird distorted (not round but still Riemannian homogeneous, in particular with isometry group $ SU\_2 $) copy of $ \mathbb{R}P^3 $ . This can be seen as the unit tangent bundle of the sphere $ S^2 $ inside $ T(\mathbb{R}^3)=\mathbb{R}^6 $. This weird $ \mathbb{R}P^3 $ is isometrically double covered by the Berger sphere $ S^3 $.
-$ S^6 $ contains an isometrically embedded round tetrahedral space $ SO\_3/A\_4 $ as previously mentioned
-$ S^8 $ contains a round $ \mathbb{R}P^3 $
-$ S^N $ contains a round projective space $ \mathbb{R}P^n $ where $ N=\frac{(n+1)(n+2)}{2}-2 $, moreover $ N $ is the minimal such dimension.
Moreover all these embeddings are equivariant with respect to the action of the Isometry group (which is transitive-- these manifolds are all Riemannian homogeneous in addition to being round).
I am aware that for sufficiently large $ N $ every Riemannian manifold isometrically embeds in round $ S^N $. What I am really looking for are **(equivariant) isometric embeddings of round (homogeneous) Riemannian manifolds into low dimensional round spheres.** The properties in parentheses are especially interesting.
Disclaimer: ~everything is smooth I know Nash-Kuiper does very low dimensions for $ C^1 $~
|
https://mathoverflow.net/users/387190
|
Embedding round manifolds into low dimensional spheres
|
Here are a couple more examples I whipped up with the help of MSE/MO. I think they are enough to constitute an answer.
As I mentioned in my question Robert Bryant inspired me with his example of
* The tetrahedral space $ SO\_3/A\_4 $ as the orbit of $ xyz $ with respect to the representation of $ SO\_3 $ on the seven dimensional space of harmonic homogeneous polynomials of degree 3 in three real variables $ x,y,z $. Since the representation is orthogonal this gives an isometric embedding into the round sphere $ S^6 $ inside of this seven dimensional real representation.
So the other stuff I cooked up is all the other Riemannian homogeneous spherical 3 manifolds
* The homogenous lens spaces $ L\_{n,1} $ as the stabilizer of $ u^n $ in representation of $ SU\_2 $ on the space of homogeneous polynomials of degree $ n $ in two complex variables $ u,v $. This representation is complex dimension $ n+1 $ so this isometrically embeds the lens space into the round sphere $ S^{2n+1} $
* for the even lens spaces with $ n \geq 3 $ $ L\_{2n,1} \cong SU\_2/C\_{2n} \cong SO\_3(\mathbb{R})/C\_n $ we can do much better by realizing them as the orbit with respect to the representation of $ SO\_3 $ on the $ 2n+1 $ dimensional real vector space of degree $ n $ homogeneous harmonic polynomials in three real variables $ x,y,z $ of the polynomial
$$
Re[(x+iy)^n]+2Im[(x+iy)^n]
$$
here $ 2 $ can be replaced by any real number other than $ 0 $ or $ \pm 1 $.
* The homogeneous prism manifolds as the orbit with respect to the representation of $ SO\_3 $ on the $ 2n+1 $ dimensional real vector space of degree $ n $ homogeneous harmonic polynomials in three real variables $ x,y,z $ of the polynomial
$$
Re[(x+iy)^n]= \sum\_{k=0}^D (-1)^k {n \choose 2k}
x^{n-2k}y^{2k}
$$
where $ D $ is the floor of $ n/2 $ (so $ n/2 $ for even case and $ (n-1)/2 $ for odd case. Note that this polynomial is harmonic because it is the real part of the holomorphic function $ w^n=(x+iy)^n $. Since this is a $ 2n+1 $ real dimensional orthogonal representation this yields an isometric embedding of round prism manifold $ SO\_3/D\_{2n} $ into the round sphere $ S^{2n} $
* The round octahedral space $ SO\_3/S\_4 $ embeds isometrically in $ S^8 $ as the orbit of the harmonic polynomial
$$
x^4+y^4+z^4-3y^2z^2-3x^2z^2-3x^2y^2
$$
* round icosahedral space/ Poincare homology sphere embeds isometrically in $ S^{12} $ as the orbit of the harmonic polynomial
$$
21(2-\sqrt{5})(x^2-\phi^2y^2)(y^2-\phi^2z^2)(z^2-\phi^2x^2)+(x^2+y^2+z^2)^3
$$
where $ \phi $ is the golden ratio.
Indeed I claim these dimensions
* $ d=2n+1 $ for the lens spaces $ L\_{2n,1}\cong SO\_3/C\_n $ ( $n \geq 3 $)
* $ d=2n+1 $ for the prism spaces $ SO\_3/D\_{2n} $ ($ n \geq 2 $)
* $ d=7 $ for the tetrahedral space $ SO\_3/A\_4 $
* $ d=9 $ for the octahedral space $ SO\_3/S\_4 $
* $ d=13 $ for the icosahedral space $ SO\_3/A\_5 $ (note this this almost saturates the bound from Gunther's version of the Nash Embedding Theorem of $ d=14 $ for isometrically embedding a generic compact 3 manifold )
are minimal for isometrically equivariantly embedding these (round) spherical 3 manifolds in Euclidean space $ \mathbb{R}^d $. And moreover that the final three dimensions are minimal even among all isometric ( not necessarily equivariant) embeddings.
|
0
|
https://mathoverflow.net/users/387190
|
414018
| 168,877 |
https://mathoverflow.net/questions/412777
|
2
|
*Previously [asked and bountied at MSE](https://math.stackexchange.com/questions/4338046/what-are-the-symmetry-groups-of-exponentiation-only-terms):*
---
Let $\mathfrak{E}=(\mathbb{N};\mathit{exp})$ be the algebra in the sense of universal algebra consisting of the natural numbers with just exponentiation. To each term $t(x\_1,...,x\_n)$ in which each variable $x\_i$ ($1\le i\le n$) actually appears$^1$ we can assign the group $$E\_t=\{\sigma\in S\_n:\forall a\_1,...,a\_n\in\mathbb{N}[t(a\_1,...,a\_n)=t(a\_{\sigma(1)},...,a\_{\sigma(n)})]\}.$$ For example, allowing standard notational conveniences we have $E\_{x^y}$ is trivial but $E\_{(x^y)^z}\cong S\_2$.
I'm curious which groups arise, up to isomorphism, as $E\_t$s *(in the language of [this earlier question of mine](https://mathoverflow.net/questions/412563/what-term-symmetry-groups-can-an-algebra-admit), I'm asking for a description of $\mathbb{G}(\mathfrak{E})$)*. The above trick is the only useful thing I can think of, and [in a sense](https://math.stackexchange.com/a/2080920/28111) is in fact all there is, but it already gives rise to some complexity: for example, at a glance the term $$[[((a^b)^c)^d]^{[((p^q)^r)^s]}]^{[((w^x)^y)^z]}$$ yields a semidirect product of $(S\_3)^3$ and $S\_2$, but we can then "carve out" some of that group by reusing the same variable multiple times. Intuitively I suspect that each $E\_t$ can be built up from full permutation groups via semidirect products + [something else rather simple], but it seems potentially messy. There are many specific groups which seem (to me) to be plausible counterexample candidates, including the $A\_n$s and $C\_n$s for "large enough" values of $n$, but I haven't had any luck figuring out the situation with even such fairly ~~simple~~ low-complexity groups.
---
$^1$The answer to this specific question would not change if we allowed terms in which some variables don't appear; however, for general structures $\mathfrak{A}$ this restriction can be impactful *(e.g. if we take $\mathfrak{A}$ to be an algebra consisting of a single bijection from the square of the underlying set to itself)*, so I've included it here for consistency.
|
https://mathoverflow.net/users/8133
|
Possible symmetry groups of power terms
|
Based on [an observation by MSE user Pilcrow](https://math.stackexchange.com/questions/4358618/do-cyclic-groups-appear-as-symmetry-groups-of-exponentiation-only-terms#comment9100944_4358618), it seems I've been overcomplicating this:
For simplicity, let "$[x\_1,x\_2,...,x\_k]$" be shorthand for the right-associating exponent term $$x\_1^{(x\_2^{(...^{x\_k})})}.$$ Then for each $k\in\mathbb{N}$ and each subgroup $G$ of $S\_k$, we can consider the term $$t\_G:=w^{\prod\_{\sigma\in G}[x\_{\sigma(1)},x\_{\sigma(2)}...,x\_{\sigma(k)}]}$$ (allowing the obvious abuse of notation for brevity), with $w,x\_1,x\_2,...,x\_k$ distinct variables.
Since $w$ obviously can't be swapped with any of the $x\_i$s there is a canonical embedding $i:E\_{t\_G}\rightarrow S\_k$, and it's not hard (if a bit tedious) to show that in fact we have $i[E\_{t\_G}]=G$. So every finite group occurs as the symmetry group of some exponentiation-only term. For example, $C\_4$ is represented by $w^{[x\_1,x\_2,x\_3,x\_4]\cdot[x\_2,x\_3,x\_4,x\_1]\cdot[x\_3,x\_4,x\_1,x\_2]\cdot[x\_4,x\_1,x\_2,x\_3]}$, or a bit less abbreviatedly by $$w^{[x\_1^{(x\_2^{(x\_3^{x\_4})})}]\cdot[x\_2^{(x\_3^{(x\_4^{x\_1})})}]\cdot[x\_3^{(x\_4^{(x\_1^{x\_2})})}]\cdot[x\_4^{(x\_1^{(x\_2^{x\_3})})}]}.$$
|
1
|
https://mathoverflow.net/users/8133
|
414021
| 168,878 |
https://mathoverflow.net/questions/414017
|
0
|
The following is a cross-post of [this](https://math.stackexchange.com/questions/4354605/automating-proofs-via-indicator-functions) question on math.SE, which did not attract any comment and may therefore be too research-oriented for math.SE.
---
It is a common technique in measure theory to prove something for indicator functions / elementary functions, generalize it to positive-valued functions and to measurable functions via $X=X^+ − X^−$. It is used e. g. to prove multiple integral properties or "independent random variables are uncorrelated".
This technique is a bit boring to execute, if one has shown a question already for the indicator functions. Is it somehow possible to describe the set of all statements which are true for all measurable functions if the result is proven for only indicator functions / elementary functions? I'm in search of something like the [Transfer Principle](https://en.wikipedia.org/wiki/Transfer_principle#Transfer_principle_for_the_hyperreals) in nonstandard analysis.
|
https://mathoverflow.net/users/136236
|
Automating proofs via indicator functions?
|
I think, you are looking for a "monotone class argument", see the wikipedia entry for the [monotone class theorem](https://en.wikipedia.org/wiki/Monotone_class_theorem). In a nutshell, prove that the property holds for indicators and is preserved under finite sums, scalar multiplication and under monotonically increasing limits and conclude that it holds for the full class.
|
2
|
https://mathoverflow.net/users/9652
|
414030
| 168,879 |
https://mathoverflow.net/questions/414041
|
1
|
Let $n\geq1$ be an integer. Take the matrix $M(n)$, with entries, $M\_{i,j}(n)=\sin\left(\frac{(i+j)\pi}2\right)$ if $i\neq j$ and $M\_{i,i}(n)=x\_i$.
I wish to ask (this question has been modified from my previous post):
>
> **QUESTION.** Is there any interpretation/meaning to the matrix $M(n)$ and its determinant?
> $$\det M(n)=x\_1x\_2\cdots x\_n-\sum\_{\substack{1\leq i\_1<\cdots<i\_{n-2}\leq n\\
> \binom{n+1}2-(i\_1+\cdots+i\_{n-2})\,\equiv\, 1\,\, \text{mod}\, 2}}x\_{i\_1}\cdots x\_{i\_{n-2}}.$$
>
>
>
**Example.** If $n=4$ then (updated using Fedor's rewrite)
\begin{align\*}
\det\begin{pmatrix} x\_1&-1&0&1 \\ -1&x\_2&1&0 \\ 0&1&x\_3&-1 \\ 1&0&-1&x\_4 \end{pmatrix}
&=x\_1x\_2x\_3x\_4-x\_1x\_2-x\_1x\_4-x\_2x\_3-x\_3x\_4 \\
&=x\_1x\_2x\_3x\_4\left(1-(x\_1^{-1}+x\_3^{-1})(x\_2^{-1}+x\_4^{-1})\right).
\end{align\*}
|
https://mathoverflow.net/users/66131
|
Interpret this matrix and its determinant
|
Well,
\begin{align\*}
\sum\_{\substack{1\leq i\_1<\cdots<i\_{n-2}\leq n\\
\binom{n+1}2-(i\_1+\cdots+i\_{n-2})\,\equiv\, 1\,\, \text{mod}\, 2}}x\_{i\_1}\cdots x\_{i\_{n-2}}
&=x\_1\cdots x\_n\cdot \left(\sum\_{i+j\,\text{is odd}}(x\_ix\_j)^{-1}\right)\\
&=x\_1\cdots x\_n(x\_1^{-1}+x\_3^{-1}+\ldots)(x\_2^{-1}+x\_4^{-1}+\ldots).
\end{align\*}
|
3
|
https://mathoverflow.net/users/4312
|
414052
| 168,884 |
https://mathoverflow.net/questions/414047
|
0
|
A well known result in Ramsey theory is: If the set of positive integers is partitioned into a finite number of sets, then at least one of these sets will contain a solution to $x+y=z$
By "property" I mean arithmetic statements like "will contain a solution to $x+y=z$" and by "invariant under partition" I mean that whenever we partition the integers by a finite number of sets, at least one of the sets will have such property.
Is there any area of Ramsey theory dedicated to studying such types of problems? Where can I find references on more problems like that?
|
https://mathoverflow.net/users/103391
|
References for properties which are invariant under partition of $\mathbb{Z}$ by a finite number of sets
|
What you are looking for is called [partition regularity](https://en.wikipedia.org/wiki/Partition_regularity), see the linked Wikipedia article for many examples of situations where this naturally occurs. The underlying set can be anything, it needn't be the set of integers $\mathbb Z$. Moreover, I would not consider this concept to be part of Ramsey theory; it is more widely used in combinatorics and combinatorial number theory.
|
4
|
https://mathoverflow.net/users/14233
|
414055
| 168,885 |
https://mathoverflow.net/questions/414048
|
0
|
Probably $\beta \mathbb N$ is not an absolute retract (is there an easy argument for this?), but I'd be interested to know what happens in the class of extremally disconnected (compact) spaces. Is it an absolute retract therein?
|
https://mathoverflow.net/users/15129
|
Is the Čech–Stone compactification of the integers always a retract of an extremally disconnected space?
|
$\beta\mathbb{N}$ is not a retract of a Tychonoff cube because it is not connected; it also not a retract of a Cantor cube, not even a continuous image, see problem 3.12.12 in Engelking's book.
It is an absolute retract for ED spaces: if it is embedded in the compact ED space $X$ then $\mathbb{N}$ is relatively discrete subspace of $X$ and we have pairwise disjoint open sets $O\_n$ in $X$ with $O\_n\cap\beta\mathbb{N}=\{n\}$.
Then the closure $K$ of the union $O=\bigcup\_nO\_n$ is clopen and by extremal disconnectedness $K=\beta O$. Now extend the retraction $r$ from $O$ onto $\mathbb{N}$, defined by $r(x)=n$ if $x\in O\_N$, to $\beta r:K\to\beta\mathbb{N}$. As $r$ is the identity on $\mathbb{N}$ the extension $\beta r$ is the identity on $\beta\mathbb{N}$.
|
3
|
https://mathoverflow.net/users/5903
|
414056
| 168,886 |
https://mathoverflow.net/questions/414032
|
0
|
Let $X$ and $Y$ be topological spaces. By a simple function $\phi: X\to Y$ we mean a finite range Borel measurable function.
Q. Is the point-wise limit of a sequence of simple functions a Borel measurable function?
|
https://mathoverflow.net/users/84390
|
Is the point-wise limit of simple functions a measurable function?
|
The answer to your question is Yes provided the topology of $Y$ is such that for each non-empty open set $O\subset Y$ there is a strictly increasing sequence $(O\_k)$ of open sets:
$$
\overline O\_k\subset O\_{k+1}\subset O,\quad k=1,2,\ldots,
$$
with $\bigcup\_kO\_k=O$. For suppose $(f\_n)$ is a sequence of Borel functions from $X$ to $Y$ with pointwise limit $f$. With $O$ and the $O\_k$ as above,
$$
f^{-1}(O)=\bigcup\_{k=1}^\infty\bigcup\_{n=1}^\infty\bigcap\_{m=n}^\infty f\_m^{-1}(O\_k).
$$
This shows that $f^{-1}(O)$ is a measurable subset of $X$. Because the Borel $\sigma$-field on $Y$ is generated by the open sets, it follows that $f$ is Borel measurable.
In particular, the condition above is true if $Y$ is metrizable.
|
4
|
https://mathoverflow.net/users/42851
|
414058
| 168,887 |
https://mathoverflow.net/questions/414054
|
4
|
Sheafification is needed in limits and colimits of condensed abelian groups? If I have a functor $T: i \mapsto T\_i$ from an index category to condensed abelian groups the limit and colimit of this functor are just $S \mapsto \lim\_i T\_i (S)$ and $S \mapsto \text{colim}\_i T\_i (S)$ or sheafification is needed for it to be a condensed abelian group?
In particular if I have a map $\phi: T \rightarrow Q$ of condensed sets the kernel and the cokernel are just $S \mapsto \ker \phi\_S$ and $S \mapsto \text{coker} \phi\_S$?
Thank you!
|
https://mathoverflow.net/users/130868
|
Limits and colimits in the category of condensed abelian groups
|
I'll ignore the set-theoretic issues since I don't understand them well enough to say anything about them.
As with any site, the limits (and in particular, the kernel) of sheaves may be computed pointwise. On the other hand, the colomits are usually not computed pointwise (cokernels included). For example, an exact sequence of condensed Abelian groups $0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0 $
gives a long exact sequence of cohomology
$$
0 \rightarrow A(S) \rightarrow B(S) \rightarrow C(S) \rightarrow H^1(S,A) \rightarrow \dotsc
$$
This means that the cokernels should be given by the pointwise quotient if the map $H^1(S,A) \rightarrow H^1(S,B)$ is injective (this happens, for example, when $H^1(S,A)=0$).
|
7
|
https://mathoverflow.net/users/146933
|
414062
| 168,888 |
https://mathoverflow.net/questions/414011
|
2
|
I want a reference to the definition of the Lie derivative of a smooth function $f:G \to \mathbb R$ on a Lie group $G$ in the direction of an element $\theta$ of the Lie algebra $\mathfrak G$.
I can see in many places the definition of the Lie derivative with respect to a vector field, and I also understand that an element of a Lie algebra $\theta$ can be considered as a left (or right) invariant vector field whose value at the identity is $\theta$.
But I would really like a sweet, short reference that combines both of these into one definition, or at least states this explicitly that this can be done.
What I really want is something accessible to engineers with a weak math background. As an example, think of $SO(n)$ as embedded in $\mathbb R^{n^2}$, and suppose $f:SO(n) \to \mathbb R$. Then the 'Lie derivative' of $f$ in the direction of a an anti-symmetric matrix $A$ at $Q$ is
$$ \mathcal L\_A(Q) = \frac d{dt} f\big(Q(I+tA)\big) \Bigg|\_{t=0} .$$
|
https://mathoverflow.net/users/42291
|
Lie derivative on Lie group in the direction of an element of Lie algebra
|
What about the formula $(D\_X\varphi)(g)=\frac{\rm d}{{\rm d}t}\Bigl\vert\_{t=0}\varphi(g\exp\_G(tX))$ for $\varphi\in C^\infty(G)$, $g\in G$, and $X\in \mathfrak g$, where $G$ is a Lie group with its Lie algebra $\mathfrak g$? See for instance Eq. (5) in Ch. II of the book by S.Helgason, "Differential geometry, Lie groups, and symmetric spaces".
|
1
|
https://mathoverflow.net/users/468624
|
414065
| 168,889 |
https://mathoverflow.net/questions/414067
|
1
|
Recall that an ideal of a commutative ring is said to be a nil ideal if each of its elements is nilpotent. I am looking for a non-zero nil ideal of a commutative ring that is idempotent.
|
https://mathoverflow.net/users/338309
|
An example of a commutative ring with a non-zero nil ideal that is idempotent
|
Consider the ring $A:=k[X^{\mathbb{Q}\_{\geq 0}}] / X$ of polynomials with non-negative rational exponents with the relation $X^1=0$. Then the ideal $I:=\operatorname{span}\_k\{X^a \mid 0<a<1\}$ is nil, because every $X^a$ is nil. But it is also idempotent because $X^a=(X^{a/2})^2$.
|
7
|
https://mathoverflow.net/users/3041
|
414071
| 168,891 |
https://mathoverflow.net/questions/412064
|
5
|
This is in some sense a generalization of the [question](https://mathoverflow.net/questions/410387/decomposition-of-manifolds-with-toroidal-boundary) I asked some time ago. I am very sorry if this question is too basic for MathOverflow, but I just started learning about some more detailed things of the topology of 3-manifolds quite recently for some project and hence, I am still missing some concepts.
**My general question is the following:**
>
> I am interested in the class of all compact, orientable and connected
> 3-dimensional topological manifolds $\mathcal{M}$ with boundary
> $\partial\mathcal{M}\cong\_{\mathrm{homeo.}}\Sigma\_{g}$ where
> $g\in\mathbb{N}\_{0}$ and $\Sigma\_{g}$ denotes the orientable genus-g
> surface. Is there a way to "divide" this class into different types?
>
>
>
Obviously, I am not asking for a classification in the strict sense, but just a way to distinguish such manifolds by their construction. As an example, in Moishe Kohan's great answer to the question I linked above, he explained that all 3-manifolds (with the properties as above) with $\partial\mathcal{M}=T^{2}$, where $T^{2}=S^{1}\times S^{1}$ denotes the 2-torus, is of one (and only one) of the following types:
1. $\mathcal{M}$ is homeomorphic to the solid torus $\overline{T}^{2}=D^{2}\times S^{1}$.
2. $\mathcal{M}$ is homeomorphic to $\overline{T}^{2}\#\mathcal{N}$, where $\mathcal{N}$ is some closed, connected and orientable 3-manifold, which is not $S^{3}$ and where $\#$ denotes the (internal, oriented) connected sum.
3. $\mathcal{M}$ has incompressible boundary.
Is a similar trichotomy also true for the more general case with $\partial\mathcal{M}\cong\_{\mathrm{homeo.}}\Sigma\_{g}$?
|
https://mathoverflow.net/users/259525
|
"Classification" of (orientable) 3-manifolds with genus-g-surface as their boundary
|
You can find the discussion of the characteristic compression body in Section 3.3 of [Bonahon’s survey.](https://www.math.csi.cuny.edu/abhijit/gt3m/bonahon-gs3m.pdf) See Theorem 3.7 for the irreducible case, which together with the uniqueness of connect sum in Theorem 3.1 I think gives the sort of generalization of the trichotomy you are seeking. Either the manifold has incompressible boundary, or it has compressible boundary and it is a handlebody, or it has compressible boundary and is obtained from a compression body (with genus g compresible boundary) attached to some manifolds with incompressible boundary. I think this gives a reference for John Pardon’s comment.
|
5
|
https://mathoverflow.net/users/1345
|
414077
| 168,893 |
https://mathoverflow.net/questions/414080
|
0
|
We face many places to find the collision probability of two sets (or more) in my case the cryptographic hash functions. We can formalize as;
Given two sets of random variables $\mathbf{A}$ and $\mathbf{B}$ compromised $m$ and $n$ elements, respectively. Consider $t$ discrete values where each variable $\{A\_1, A\_2,\ldots,A\_m\}$ and $\{B\_1, B\_2,\ldots,B\_n\}$ can assume any of the $t$ values with equally likely probability. Clearly, a collision is at least one element of $\mathbf{A}$ is also in $\mathbf{B}$.
The problem investigated in
* 2003, [Collision probability between sets of random variables](https://www.sciencedirect.com/science/article/abs/pii/S0167715203001688) by Michael C. Wendl.
**The probability of no collision** $P\_0(m,n,t)$ is given in the below theorem;
* **Theorem 1** (Probability of success). The probability of no collisions between two sets of random variables $\{A\_1, A\_2,\ldots,A\_m\}$ and $\{B\_1, B\_2,\ldots,B\_n\}$, the elements of which can assume any of $t$ discrete values with equally likely probability is $$ P\_0(m,n,t) = \frac{1}{t^{m+n}}\sum\_{i=1}^{m}\sum\_{j=1}^{n} S\_2(m,i)S\_2(n,j) \prod\_{k=0}^{i+j-1} t-k$$
where $S\_2$ represents [Stirling numbers of the second kind](https://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind).
When $t$ becomes very large like $2^{64}$ we cannot use this theorem to calculate the probability. Therefore we need a good approximation as in the birthday problem.
Is there any good approximation for this theorem/problem that can be used easily to calculate large values assuming that $A\_1,\dots,A\_m,B\_1,\dots,B\_n$ are **independent and identically distributed random variables each uniformly distributed** over the set $[t]:=\{1,\dots,t\}$.
|
https://mathoverflow.net/users/91106
|
A good approximation for collision probability between (two) sets of random variables
|
We are assuming that $A\_1,\dots,A\_m,B\_1,\dots,B\_n$ are iid random variables each uniformly distributed over the set $[t]:=\{1,\dots,t\}$.
By the [Bonferroni inequalities](https://en.wikipedia.org/wiki/Boole%27s_inequality#Bonferroni_inequalities), for the probability of at least one collision
\begin{equation}
p:=P\Big(\bigcup\_{(i,j)\in[m]\times[n]} \{A\_i=B\_j\}\Big),
\end{equation}
we have
\begin{equation}
p\_1-p\_2\le p\le p\_1,
\end{equation}
where
\begin{equation}
p\_1:=\sum\_{(i,j)\in[m]\times[n]} P(A\_i=B\_j)=\frac{mn}t
\end{equation}
and
\begin{equation}
p\_2:=\sum\_{
\substack{(i\_1,j\_1)\in[m]\times[n], \\
(i\_2,j\_2)\in[m]\times[n]\setminus\{(i\_1,j\_1)\}}
} P(A\_{i\_1}=B\_{j\_1},A\_{i\_2}=B\_{j\_2})=\frac{mn(mn-1)}{t^2}
=o\Big(\frac{mn}t\Big)
\end{equation}
if $mn=o(t)$.
So,
\begin{equation}
p\sim\frac{mn}t
\end{equation}
if $mn=o(t)$.
Thus, the probability of no collisions is
\begin{equation}
1-p=1-\frac{mn}t\,(1+o(1))
\end{equation}
for very large $t$, that is, for $t$ much greater than $mn$.
|
3
|
https://mathoverflow.net/users/36721
|
414086
| 168,895 |
https://mathoverflow.net/questions/414051
|
7
|
Let $k=\mathbb{R}$ or $\mathbb{C}$ and let $A$ be a finite-dimensional $k$-algebra. If $A$ is simple, then the Skolem-Noether theorem says that any two algebra homomorphisms $f, g: A \to M\_n(k)$ are conjugate in $M\_n(k)$, i.e., there exists a matrix $X \in M\_n(k)$ such that for all $a \in A$, $f(a) = X g(a) X^{-1}$.
It we do not assume that $A$ is simple, then this result is generally false (see e.g. [the example given by Denis Serre](https://mathoverflow.net/a/84600) at [Conjugation between commutative subalgebras of a matrix algebra?](https://mathoverflow.net/questions/84591/conjugation-between-commutative-subalgebras-of-a-matrix-algebra)).
**My question is:** Is this result still true if $f$ and $g$ are "close"? That is, if $f$ and $g$ are close, then they are conjugate?
Here by "close" I mean that for some small $\varepsilon>0$, we have $\lVert f(a) - g(a)\rVert \leq \varepsilon \lVert a\rVert$, for some submultiplicative norm on $A$.
|
https://mathoverflow.net/users/16702
|
Are nearby subalgebras of matrix algebras conjugate?
|
I'll describe a 1-parameter family of nonisomorphic 4-dimensional subalgebras of $M\_4(K)$. Consider, for $t\in K^\*$, the matrices
$$X=\begin{pmatrix} 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix},\;Y\_t=\begin{pmatrix} 0 & 1 & 1 & 0\\ 0 & 0 & 0 & t\\ 0 & 0 & 0 & -t \\ 0 & 0 & 0 & 0 \end{pmatrix},\;Z=\begin{pmatrix} 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}.$$
Then $XY\_t=tZ$, $Y\_tX=Z$, and all other pairwise products (including squares) between these matrices are zero. Hence they form the basis of a 3-dimensional (non-unital) subalgebra $N\_t$. These algebras are pairwise non-isomorphic, except $N\_t\simeq N\_{t^{-1}}$ (indeed, in $N\_t$, the system of equations $$\{xy=\lambda yx;\quad x^2=y^2=0\}$$ has a solution $(x,y)$ with $xy\neq 0$ iff $\lambda\in\{1,t,1/t\}$).
Let now $A\_t$ be the 4-dimensional unital subalgebra with basis $(I,X,Y\_t,Z)$. These are also pairwise nonisomorphic (except $t\leftrightarrow t^{-1}$), since $N\_t$ consists of the set of nilpotent elements in $A\_t$.
Since they are not isomorphic, they are not conjugate.
Now let $A$ be 5-dimensional, with basis $(I,X,Y,Z,Z')$ with $I$ identity, $XY=Z'$, $YX=Z$, and all other products (not involving $I$) being zero. Let $f\_t:A\to A\_t\subset M\_4(K)$ map $I\mapsto I$, $X\mapsto X$, $Y\mapsto Y\_t$, $Z\mapsto Z$, $Z'\mapsto tZ$. Then $f\_t$ is a surjective homomorphism. However they have non-isomorphic images (up to $t\leftrightarrow t^{-1}$). While for $K$ being real or complex numbers, for $t'$ close enough to $t$, the homomorphisms $f\_t$ and $f\_{t'}$ are close.
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11
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https://mathoverflow.net/users/14094
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414089
| 168,896 |
https://mathoverflow.net/questions/414074
|
2
|
By 1 step breach of the GCH I mean the following: $$ 2^{\aleph\_{\alpha}} = \aleph\_{\alpha+2}$$
Now, it is known that there are more constrains on the cardinality of power sets at singlular cardinals that their cardinality is influenced by the behaviour of the continuum below them! The main theorem supporting that is Silver's Theorem which states that: if GCH holds below a singular of uncountable cofinality, then it holds at that singular as well.
My question is about if a similar result to Silver's theorem is true in $\sf ZFC$, that is if the GCH is breached in the same manner before a singular of uncountable cofinality, then that breach holds at that singular as well! Formally, that is: $$ \operatorname {singular}(\aleph\_\lambda) \land \\ \operatorname {cf}(\aleph\_\lambda) > \omega \land \\ \forall \alpha < \lambda \, (2^{\aleph\_\alpha}= \aleph\_{\alpha+2}) \\\to \\ 2^{\aleph\_\lambda}= \aleph\_{\lambda+2}$$
In English: if there is 1 step breach of the GCH below a singular of uncountable cofinality, then there is a 1 step breach of GCH at that singular itself.
Of course the more general theorem would be about $n$-step breach of GCH for a finite $n$, or it might be possible to even extend it to some infinite values?
|
https://mathoverflow.net/users/95347
|
If GCH is breached the same way before a singular of uncountable cofinality, would that breach extend to that singular?
|
One can collapse $2^\lambda$ to have cardinality $\lambda^+$ without adding $\lambda$-sequences even if $\lambda$ is singular. Therefore one could start with $2^{\aleph\_\alpha} = \aleph\_{\alpha+2}$ for all $\alpha < \omega\_1$ by Foreman-Woodin and then force $2^{\aleph\_{\omega\_1}} = \aleph\_{\omega\_1+1}$ without changing $P(\aleph\_{\omega\_1})$ and hence preserving $2^{\aleph\_\alpha} = \aleph\_{\alpha+2}$ for all $\alpha < \omega\_1$. This shows that the answer to your question is no.
|
5
|
https://mathoverflow.net/users/102684
|
414091
| 168,898 |
https://mathoverflow.net/questions/414069
|
3
|
Given a set of distinct real numbers $(x\_j)$ and non-zero complex $(c\_j)$, then the large sieve says that
$$\limsup\_{N\to\infty}\frac{1}{N}\sum\_{n=1}^{N}\left|\sum\_{j}c\_j e^{2\pi i n x\_j}\right|^2\leq \sum\_{j}|c\_j|^2.$$
Are there lower bounds saying that
$$\liminf\_{N\to\infty}\frac{1}{N}\sum\_{n=1}^{N}\left|\sum\_{j}c\_j e^{2\pi i n x\_j}\right|^2\geq \varepsilon \,\,?$$
I am aware of the paper titled [Lower bounds for expressions of large sieve type](https://arxiv.org/pdf/1105.1307.pdf), but this deals with a different formulation than the one I am interested in. Namely, it only refers to the dual equality and it includes more terms than I do. Any help would be greatly appreciated.
|
https://mathoverflow.net/users/159298
|
Are there general large sieve lower bounds?
|
If the $x\_j$'s are distinct modulo $1$ (which is the natural assumption), then
$$\lim\_{N\to\infty}\frac{1}{N}\sum\_{n=1}^{N}\left|\sum\_{j}c\_j e^{2\pi i n x\_j}\right|^2=\sum\_{j}|c\_j|^2.$$
Indeed, let us assume (without loss of generality) that the $x\_j$'s lie in $[0,1]$. Then
$$\sum\_{n=1}^{N}\left|\sum\_{j}c\_j e^{2\pi i n x\_j}\right|^2=\sum\_{j,k}c\_j\overline{c\_k}\sum\_{n=1}^N e^{2\pi in(x\_j-x\_k)},$$
where the inner sum equals $N$ for $j=k$, and has absolute value not exceeding $\csc(\pi(x\_j-x\_k))$ for $j\neq k$. The result follows.
**Remark.** The above argument coupled with Corollary 1 in [Montgomery: The analytic principle of the large sieve](https://projecteuclid.org/euclid.bams/1183540922) gives that
$$\sum\_{n=1}^{N}\left|\sum\_{j}c\_j e^{2\pi i n x\_j}\right|^2=(N+\Delta)\sum\_j|c\_j|^2,$$
where $|\Delta|\leq\max\_{j\neq k}\|x\_j-x\_k\|^{-1}.$
|
5
|
https://mathoverflow.net/users/11919
|
414093
| 168,900 |
https://mathoverflow.net/questions/316326
|
3
|
Let $$S(n) = \sum\_{p \le n} b(n-p),$$ where
$b(a)=1$ is $a$ is a sum of two squares of positive integers and $b(a)=0$ otherwise.
Trivially by PNT we have
$$S(n) \le \sum\_{p \le n}1 \ll \frac{n}{\log n}.$$
Could we do better or the above estimate is the best possible?
|
https://mathoverflow.net/users/95838
|
Sieve bound for the sum of two squares
|
One can do a bit better. For simpler presentation assume that we instead consider the function $b'$ that is the indicator function of integers all of whose prime divisors are $1 \mod 4$. We have $b'\leq b$ but $b'$ and $b$ agree on square-free odd integers and any proof for $b'$ can be adapted to $b$. Furthermore, we have $\sum\_{p\leq n} b'(n-p)\ll \sum\_{d\mid P'(z) \atop (d,n)=1 }\lambda\_d^+ \#\{p\leq n: p\equiv n \mod d \} $ by the upper bound of Brun. Here $P'(z)$ is the product of all primes $p<z$ with $p\equiv 1 \mod 4$. Using Bombieri--Vinogradov we obtain $$\ll \frac{n}{\log n } \sum\_{d\mid P'(z) \atop (d,n)=1 }\lambda\_d^+\phi(d)^{-1}\ll \frac{n}{\log n } \prod\_{p<z \atop p\equiv 1 \mod 4} (1-1/p) \ll \frac{n}{(\log n)^{3/2} } $$ since one can take $z$ so that $\log z\gg \log n$. Hence, one can save $\sqrt{\log n}$ from the trivial estimate.
|
4
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https://mathoverflow.net/users/9232
|
414097
| 168,902 |
https://mathoverflow.net/questions/414098
|
0
|
The article [Quantifiers and Quantification](https://plato.stanford.edu/entries/quantification/#QuaProModLog) in Stanford Encyclopedia of Philosophy gives the reference below, but the article is not in the Journal of Symbolic Logic, 35.
May someone help me with finding Kaplan's article?
Kaplan, D., 1970, “S5 with Quantifiable Propositional Variables”, Journal of Symbolic Logic, 35: 355.
|
https://mathoverflow.net/users/37385
|
Where is D. Kaplan's “S5 with Quantifiable Propositional Variables” published?
|
I find another more precise citation elsewhere online as:
David Kaplan. S5 with quantifiable propositional variables. *The Journal of Symbolic Logic*, 35(2):355, 1970.
It seems possible it occurs within these pages: <https://doi.org/10.2307/2270571> ("Meeting of the Association for Symbolic Logic").
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3
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https://mathoverflow.net/users/25028
|
414100
| 168,903 |
https://mathoverflow.net/questions/414102
|
1
|
Let $(\mathcal{X},\Sigma,P)$ be a Polish probability measure space, and $(\mathcal{X}^n,\Sigma^{\otimes n},P^n)$ be the product of its $n$ copies. Let $t: x^n \in \mathcal{X}^n \mapsto L\_{x^n} \in \mathcal{P}(\mathcal{X})$ be the empirical measure function, where $L\_{x^n}$ is the empirical measure of the $n$-length sequence $x^n$, and $\mathcal{P}(\mathcal{X})$ is the set of probability measures on $(\mathcal{X},\Sigma)$. We consider two kinds of subsets of $\mathcal{P}(\mathcal{X})$:
1. A subset $A$ of $\mathcal{P}(\mathcal{X})$ such that $t^{-1} (A)$ is $\Sigma^{\otimes n}$-measurable.
2. A subset $B$ of $\mathcal{P}(\mathcal{X})$ such that $B$ is measurable with respect to the Borel $\sigma$-algebra induced by the weak topology.
Obviously, $B$ is a special case of $A$. My question is: For any $A$ above, can we find a sequence $B\_k$ as the above such that $P\circ t^{-1} (B\_k)\to P\circ t^{-1} (A)$ as $k\to \infty$? Are there references for this question? Thanks.
|
https://mathoverflow.net/users/126001
|
How fine is the Borel $\sigma$-algebra induced by the weak topology?
|
Yes, and you can even do the approximation by a single set. All the structure you need is that $t$ is a measurable function between Polish spaces.
The function $t$ is continuous and hence measurable, and the space $\mathcal{P}(\mathcal{X})$ is again Polish. Let $A$ satisfy the condition in 1. Then $t^{-1}(A)$ is a Borel set in $\mathcal{X}^n$ and, therefore, $t\big(t^{-1}(A)\big)$ an [analytic](https://en.wikipedia.org/wiki/Analytic_set) subset of $\mathcal{P}(\mathcal{X})$. Analytic sets are [universally measurable](https://en.wikipedia.org/wiki/Universally_measurable_set), so $P\circ t^{-1}\Big(t\big(t^{-1}(A)\big)\Big)$ is well defined. Since every measurable set in the completion is the union of a Borel set and a subset of a null set, there exists a Borel set $B\subseteq t\big(t^{-1}(A)\big)$ such that $$P\circ t^{-1}(B)=P\circ t^{-1}\Big(t\big(t^{-1}(A)\big)\Big)=P\circ t^{-1}(A).$$
|
1
|
https://mathoverflow.net/users/35357
|
414114
| 168,908 |
https://mathoverflow.net/questions/414113
|
3
|
Let $G$ be a compact Lie group with algebra $\mathfrak{g}$. Let $\beta $ be an element in the dual of the Lie algebra $\mathfrak{g}$. We denote by $G\_\beta$ the stabilizer subgroup of $\beta$ by the cooadjoint action. Let $V$ be symplectic vector space on which $G\_\beta$ acts.
So, we have $V$ is symplectic, $G/G\_\beta$ is symplectic (since it can be identified with a cooadjoint orbit which is symplectic), How can we show that $M:= G \times\_{G\_\beta} V$ is also symplectic ?
Let $[g,v] \in M$, we have $T\_{[g,v]} M = T\_gG \times T\_vV$
How can we define a symplectic form $\omega\_{[g,v]}: T\_{[g,v]} M \times T\_{[g,v]} M \rightarrow \mathbb{C}$ ?
|
https://mathoverflow.net/users/172459
|
Define a symplectic structure on $G \times_{G_\beta} V$, where $V$ is symplectic
|
Your tangent space is too big. It has rank $\dim G + \dim V$ but it should have rank $\dim M = \dim G - \dim G\_\beta + \dim V$.
I think that the tangent space at $[g, v]$ is actually $T\_{g} (G/G\_\beta) \times T\_vV$. On the first factor you have a symplectic form $\omega\_\beta$ and on the second factor the symplectic form $\omega\_v$. On the whole tangent space you can then take block-diagonal sum $\omega\_\beta \oplus \omega\_v$.
|
1
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https://mathoverflow.net/users/6818
|
414123
| 168,911 |
https://mathoverflow.net/questions/414124
|
11
|
I am reading the following paper [1998(H.Hudzik)](https://www.jstor.org/stable/44152979) P.574
It reads using L'Hopital rule$$\liminf\_{u\to\infty} \frac{1/\varphi(1/u)}{\psi(u)}=\liminf\_{u\to\infty}\frac{\varphi'(u)}{\psi'(u)u^2[\varphi(1/u)]^2}.$$
That means we can apply L'Hopital for lower limits i.e. $$\liminf\_{u\to\infty} \frac{f(u)}{g(u)}=\liminf\_{u\to\infty}\frac{f'(u)}{g'(u)}?$$
But I only know the classical one. Is there someone can give me some reference to check this formula? Or if possible someone can give a proof?
|
https://mathoverflow.net/users/147009
|
L'Hopital rule for upper and lower limit?
|
The full L'Hopital rule says that
$$\liminf \frac{f'}{g'}\leq\liminf\frac{f}{g}\leq\limsup\frac{f}{g}\leq\limsup\frac{f'}{g'}.$$
So in the special case when the limit of $f'/g'$, exists, the limit of $f/g$
also exists and is equal to the limit of $f'/g'$.
This general rule is proved by integration.
|
23
|
https://mathoverflow.net/users/25510
|
414138
| 168,918 |
https://mathoverflow.net/questions/414139
|
22
|
There are [many models](https://arxiv.org/pdf/math/0610239.pdf) for $(\infty,1)$-categories: simplicial categories, Segal categories, complete Segal spaces, and quasi-categories.
Doubtlessly the model most used to do higher category theory in is the model of quasi-categories, due to the work of Lurie ([Higher Topos Theory](https://www.math.ias.edu/~lurie/papers/HTT.pdf), [Kerodon](https://kerodon.net/)), who calls them $\infty$-categories.
Just browsing through these books, I noticed that however, simplicial categories are used too, for instance to *construct* examples of quasi-categories, as in [the construction of the quasi-category of spaces](https://kerodon.net/tag/00TZ): there is a simplicial category of Kan complexes, and to get the quasi-category of Kan complexes we take the homotopy coherent nerve (also called simplicial nerve in HTT I think) of that.
**Question:** If simplicial categories are a more practical model for actually constructing examples, why are quasi-categories used for most of the theory? That is, what advantages do quasi-categories have over simplicial categories that outweigh the complications of constructing quasi-categories directly?
|
https://mathoverflow.net/users/475383
|
Why are quasi-categories better than simplicial categories?
|
As a preface, I think that this question should be viewed as analogous to "what are the advantages of ZFC over type theory" or vice versa. We're talking about foundations -- in principle, it doesn't matter what foundations you use; you end up with an equivalent model-independent theory of $\infty$-categories.
The paradigm is "use simplicial categories for examples; use quasicategories for general theorems". There are a lot of constructions which are simpler in quasicategories. I tend to think that a lot of the difference is visible at the model category level: the Joyal model structure on $sSet$ is just much nicer to work with than the Bergner model structure on $sCat$ (of course, the latter is still theoretically very important, at the very least for the purposes of importing examples which start life as simplicial categories). Some of these differences are:
0. The Joyal model structure is defined on a presheaf category.
1. The Joyal model structure is cartesian, making it much easier to talk about functor categories.
2. In the Joyal model structure, every object is cofibrant.
There's a synergy between (1) and (2) -- if $X$ is a quasicategory and $A$ is any simplicial set, then the mapping simplicial set $Map(A,X)$ gives a correct model for the functor category from $A$ to $X$ -- you don't need to do any kind of cofibrant replacement of $A$. This is nicely explained in [Justin Hilburn's answer](https://mathoverflow.net/a/414140/2362).
(2) is quite convenient. For example, in general frameworks like Riehl and Verity's [$\infty$-cosmoi](http://nlab-pages.s3.us-east-2.amazonaws.com/nlab/show/infinity-cosmos), a lot of headaches are avoided by assuming something like (2).
Here are a few examples of some things which are easier in quasicategories -- I'd be curious to hear other examples folks might mention!
* The join functor is very nice.
* Consequently (in combination with the nice mapping spaces), limits and colimits can be defined pretty cleanly.
An example of a theorem proven in HTT using quasicategories which I imagine would be hard to prove (maybe even to formulate) directly in simplicial categories is the theorem that an $\infty$-category with products and pullbacks has all limits. The proof uses the fact that the nerve of the poset $\omega$ is equivalent to a 1-skeletal (non-fibrant) simplicial set, and relies on knowing how to compute co/limits indexed by non-fibrant simplicial sets like this.
* The theory of cofinality is very nice, arising from the (left adodyne, left fibration) weak factorization system on the underlying category -- I imagine it would be quite complicated with simplicial categories.
Roughly at this point in the theory, though, one starts to have enough categorical infrastructure available that it becomes more possible to think "model-independently", and the differences start to matter less.
Here's a few more:
* When you take the maximal sub-$\infty$-groupoid of a quasicategory, it is literally a Kan complex, ready and waiting for you to do simpicial homotopy with. This is especially nice when you take the maximal sub-$\infty$-groupoid of a mapping object -- which doesn't quite make the model structure simplicial, but it's kind of "close".
* The theory of fibrations is pretty nice in quasicategories -- just like in ordinary categories, left, right, cartesian, and cocartesian fibrations are "slightly-too-strict" notions which are very useful and have nice properties like literally being stable under pullback. I don't know what the theory of such fibrations looks like in simplicial categories.
* The fact that $sSet$ is locally cartesian closed is sneakily useful. Even though $Cat\_\infty$ is not locally cartesian closed, there's a [pretty good supply](http://nlab-pages.s3.us-east-2.amazonaws.com/nlab/show/Conduch%C3%A9+functor) of exponentiable functors, and it's not uncommon to define various quasicategories using the right adjoint to pullback of simplicial sets. (Rule of thumb: in HTT, when Lurie starts describing a simplicial set by describing its maps in from simplices over a base, 90% of the time he's secretly describing the local internal hom of simplicial sets.)
|
31
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https://mathoverflow.net/users/2362
|
414141
| 168,920 |
https://mathoverflow.net/questions/414135
|
6
|
Let us consider $C(\mathbb{R})$, the space of continuous functions on the reals.
Q. Does there exist a sequence $\{f\_n\}$ in $C(\mathbb{R})$ such that for every $f\in C(\mathbb{R})$ one may find a subsequence $\{f\_{n\_k}\}$ of $\{f\_n\}$ pointwise converging to $f$, i.e., $f(x)=\lim f\_{n\_k}(x)$ for all $x$?
|
https://mathoverflow.net/users/84390
|
Some special sequence in $C(\mathbb{R})$
|
Yes. Take the countable set $P$ of all polynomials with rational coefficients and enumerate it somehow so that $P=\{p\_1,p\_2,\dots\}$.
Given any $f\in C(\mathbb R)$, for each natural $k$ there exists some natural $n\_k$ such that $\sup\_{x\in[-k,k]}|p\_{n\_k}(x)-f(x)|<1/k$. Here without loss of generality we may assume that $n\_1<n\_2<\cdots$. Then $p\_{n\_k}\to f$ pointwise on $\mathbb R$. (Moreover, this convergence is locally uniform.)
|
11
|
https://mathoverflow.net/users/36721
|
414142
| 168,921 |
https://mathoverflow.net/questions/292936
|
5
|
I am curious about the Hölder exponent obtained by the De Giorgi-Nash-Moser theory, as a function of the ellipticity.
More precisely: suppose $u$ satisfies weakly
$$
D\_i(a^{ij}D\_ju)=f
$$
on the $d$-dimensional ball of radius $R$, with $0$ Dirichlet boundary conditions, with the matrix $(a^{ij})\_{i,j=1..d}$ bounded from below and above by $\lambda I$ and $\Lambda I$, respectively, $\lambda>0$. For the right-hand side, let's just take $f\in L\_\infty$. Then we know that $u\in C^{\alpha}(B\_R)$, with some $\alpha=\alpha(d, \Lambda/\lambda)$, my question is whether something about the dependence on $\Lambda/\lambda$ is known. The analogous question for interior regularity (which might be easier) could also be of interest. Thanks!
|
https://mathoverflow.net/users/100941
|
Dependence of the Hölder exponent in De Giorgi-Nash-Moser
|
(By scaling we can take $\lambda=1$ for simplicity. I will also take $f=0$, and discuss just the interior estimates.)
The precise exponent is known in two dimensions (see Piccinini & Spagnolo 1972). It is $\Lambda^{-1/2}$. You can find the "worst" coefficient field in the obvious way: the diffusion matrix $a(x)$ has eigenvalue $\Lambda$ in the $x$ direction, and eigenvalue $\lambda=1$ in directions orthogonal to $x$. You can prove this is the worst as Piccinini and Spagnolo do, with a beautiful monotonicity formula type computation.
In higher dimensions, the precise exponent is not known. It is however conjectured based on the same radial example, extended in the obvious way in $d>2$. The exact exponent should be (according to my personal notes, hopefully I have not made a mistake)
\begin{equation\*}
\alpha(d,\Lambda) = \frac12 \left( -(d-2) + \sqrt{(d-2)^2 + \frac{4(d-1)}{\Lambda} } \right) .
\end{equation\*}
Okay, forget about the exact formula, the point is that this scales like $c(d) \Lambda^{-1}$ for $\Lambda \gg1$ in all dimensions $d>2$.
Ok, what is the best exponent that can be proved? The best exponent which has been proved is this:
\begin{equation\*}
\alpha(d,\Lambda) = \exp \left( - C \Lambda^{1/2} \right).
\end{equation\*}
This is the best exponent one could ever hope to have based on current methods! Indeed, all known methods of proof use a decay of oscillation iteration to get the $C^{\alpha}$ estimate (except in $d=2$, more on that below). The best decay of oscillation estimate is related to the best constant in the Harnack inequality, and this is known to be $\exp( C \Lambda^{1/2})$. There are very easy examples you can make to confirm you can do no better. The Harnack inequality with this constant was proved in an amazing paper by Giusti and Bombieri (Inventiones, also 1972).
So, to summarize: in $d>2$, we cannot prove the estimate with the conjectured optimal exponent because we do not possess a proof powerful enough to do so. All known proofs cannot get there. The bound we have is hopeless pathetic next to what we think is true, so the situation is bad, very bad! Anything better than the Giusti-Bombieri exponent would be, in my opinion, a major breakthrough.
But then what is going on in $d=2$? There is a proof of the conjectured bound, so do we have a better proof? Why yes-- indeed we do. In $d=2$, the De Giorgi-Nash $C^{\alpha}$ estimate has an easy proof. It follows from the hole-filling argument, which is just the Caccioppoli inequality (i.e., the usual $L^2$ energy estimate), plus the Sobolev inequality. This is basically because the Sobolev inequality says that in $d=2$ the space $H^1$ almost embeds into $L^\infty$. Since you can actually do a tiny bit better than $H^1$ by hole filling, you can get $C^\alpha$ without any ideas of De Giorgi. The simple hole filling computation already gives an exponent that scales like $\Lambda^{-1/2}$. What the Piccinini & Spagnolo paper does is optimize this computation, replacing an iteration in dyadic balls by a more careful monotonicity computation.
So $d=2$ should be expected to be much easier, and in hindsight it is maybe not so surprising that we can get the optimal exponent in this case.
|
9
|
https://mathoverflow.net/users/5678
|
414151
| 168,926 |
https://mathoverflow.net/questions/414150
|
2
|
Let $S$ be a smooth projective surface. We denote $S^{[n]}$ the Hilbert scheme of artinian subschemes in $S$ of length $n$, which is a smooth projective variety of dimension $2n$ by Fogarty. Let $I\subset S^{[n]}\times S$ be the universal correspondance, and let $p: I\to S^{[n]}$ and $q: I\to S$ be projections. To be precise, the fiber of $p$ over a point in $S^{[n]}$ is the corresponding subscheme of $S$.
We fix the following convention (which is quite common in literature):
Let $F$ be a vector bundle on $S$. $F^{[n]}$ is the locally free coherent sheaf on $S^{[n]}$ defined by $p\_\*q^\*F$.
It seems that the tangent bundle of $S^{[n]}$ coincides with $T\_S^{[n]}$ where $T\_S$ is the tangent bundle on $S$. The reason why I guess so is the following
The tangent space of $S^{[n]}$ at a point $z\in S^{[n]}$ representing $Z\subset S$ is given by $Hom(I\_Z/I\_Z^2, \mathcal O\_Z)$. In the case where $Z$ is non reduced, $Hom(I\_Z/I\_Z^2, \mathcal O\_Z)\cong H^0(Z,T\_{X|Z})$. Thus, $T\_{S^{[n]},z}=H^0(Z,T\_X|Z)=p\_\*q^\*F|\_z$. I do not see if these isomorphisms still hold when $Z$ is a reduced subscheme.
Any comments, responses and references are more than welcome !
|
https://mathoverflow.net/users/119184
|
tangent bundle of Hilbert schemes of points on a projective surface
|
The tangent bundle on $S^{[n]}$ is not quite isomorphic to $(T\_S)^{[n]}$, rather by Theorem B of Stapleton's paper listed below there is an injection of the former into the latter.
*Stapleton, David*, [**Geometry and stability of tautological bundles on Hilbert schemes of points**](http://dx.doi.org/10.2140/ant.2016.10.1173), Algebra Number Theory 10, No. 6, 1173-1190 (2016). [ZBL1359.14040](https://zbmath.org/?q=an:1359.14040).
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4
|
https://mathoverflow.net/users/6263
|
414152
| 168,927 |
https://mathoverflow.net/questions/414119
|
4
|
For a non recursive $x \in 2^{\omega}$, define $C\_x = \{y \in 2^{\omega}: x \leq\_T y\}$. Note that $y \in C\_x$ iff $(\exists e)(\forall n)(\Phi^y\_e(n) = x(n))$ where $\Phi\_e$ is the $e$th Turing functional. So $C\_x$ is a $G\_{\delta \sigma}$-set (countable union of countable intersection of open sets). Can we show that it isn't an $F\_{\sigma \delta}$-set?
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https://mathoverflow.net/users/475361
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Borel ranks of Turing cones
|
Assume $x$ is noncomputable, as otherwise it isn't true. Fix a $G\_{\delta\sigma}$ set $\bigcup\_i \bigcap\_j A\_{i,j}$, where the $A\_{i,j}$ are open. We'll show that this is not the complement of $C\_x$ by building a real $y$ which is either in both sets or in neither. This will be a finite extension argument; I think it would be more elegant to rework it into a forcing argument, but that would be more complicated to write.
We're going to identify $2^{\omega}$ with $2^{\omega\times\omega\times\omega}$ by pairing, so we can think of $y$ as a binary function on triples. The general shape of $y$ is the following: if $i$ is least with $y \in \bigcap\_j A\_{i,j}$, then there will be an $n\_0^i$ used to code $x(0)$. Specifically, there will be a unique value $b$ with $y(i, n\_0^i, b) = 1$, and the parity of $b$ will code $x(0)$, and we define $n\_1^i = \lfloor b/2\rfloor$. Then $n\_1^i$ codes $x(1)$ in the same way. In this way, each column codes a value of $x$ and tells you the next column to look at. Note that if this happens, then $y$ computes $x$ by starting at $n\_0^i$, searching for the $b$, and continuing.
If $y \not \in \bigcup\_i \bigcap\_j A\_{i,j}$, then we need to diagonalize against all functionals $\Phi\_e$ to ensure $\Phi\_e^y \neq x$.
So we have requirements $R\_i$ and $N\_e$ for attempting to meet these, which we order $R\_0, N\_0, R\_1, N\_1, \dots$, and we proceed one at a time.
An $R\_i$ requirement starts by picking a large $n\_0^i$ and looking for an extension into $A\_{i,0}$ that only puts $0$s on the $(i, n\_0^i)$ column (and also only 0s on the $(i', n\_{j\_{i'}}^{i'})$ columns, for $i' < i$; see later). If there is such an extension, it extends to that, then extends further to put a 1 on $(i, n\_0^i, b)$ for an appropriate large $b$, thus defining $n\_1^i$, and repeats with $n\_1^i$.
If it gets stuck at some $j\_i$ without an appropriate extension, we require that the $(i, n\_{j\_i})$ column contain only 0s, and then we move on to the $N\_i$ requirement.
The $N\_e$ requirement looks for an extension $\tau$ and $m$ with $\Phi\_e^{\tau}(m)\!\!\downarrow \neq x(m)$, and with $\tau$ containing only 0s on the $(i, n^i\_{j\_i})$ columns for $i \le e$. If there is such an extension, we take it. Otherwise, we must have already forced that $\Phi\_e^y$ is partial, as otherwise $x$ is computable by searching all extensions $\tau$ with $0$s on the appropriate columns and believing any computation $\Phi\_e^\tau(m)$ we see.
Either way, we then move on to $R\_{e+1}$.
If $y \in \bigcup\_i \bigcap\_j A\_{i,j}$, then fix the least $i$ with $y \in \bigcap\_j A\_{i,j}$. There is always an extension of the sort $R\_i$ is looking for (in particular, some initial segment of $y$ suffices), so we stay forever at $R\_i$, defining every $n\_j^i$. As discussed above, this means $y \in C\_x$.
If $y \not \in \bigcup\_i \bigcap\_j A\_{i,j}$, then for every $i$ there is a least $j\_i$ with $y \not \in A\_{i, j\_i}$. Then $R\_i$ must get stuck at this $j\_i$ and proceed to $N\_i$. So every $N\_e$ is addressed, ensuring $y \not \in C\_x$.
Either way, $y$ witnesses that $2^\omega \setminus C\_x \neq \bigcup\_i \bigcap\_j A\_{i, j}$, so $C\_x$ is not $F\_{\sigma\delta}$.
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3
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https://mathoverflow.net/users/32178
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414155
| 168,929 |
https://mathoverflow.net/questions/414153
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6
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The Pólya–Vinogradov inequality asserts that a non-principal Dirichlet character $\chi$ with modulus equal to $q$ satisfies
$$\displaystyle \left \lvert \sum\_{N < n < N+M} \chi(n) \right \rvert = O \left(\sqrt{q} \log q \right)$$
for all positive integers $N$, $M$.
My question concerns changing the summation condition to run over Eisenstein integers having bounded norm, and the character to a primitive cubic character given by the cubic residue symbol of some element $w \in \mathbb{Z}[\zeta\_3]$, where $\zeta\_3$ is a primitive third root of unity. Specifically, I want to understand the sum
$$\displaystyle \left \lvert \sum\_{A < N(z) < A + B} \left(\frac{z}{w} \right)\_3 \right \rvert$$
where $\left(\frac{\cdot}{\cdot} \right)\_3$ is the cubic residue symbol over the Eisenstein integers and $N(\cdot)$ is the norm on $\mathbb{Z}[\zeta\_3]$. Can a bound of $O(\sqrt{N(w)} \log N(w))$ be obtained as in the case of running over rational integers?
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https://mathoverflow.net/users/10898
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Pólya–Vinogradov inequality for Eisenstein integers
|
No.
Such a bound would imply a similar bound on
$$\displaystyle \left \lvert \sum\_{N(z) = M} \left(\frac{z}{w} \right)\_3 \right \rvert.$$
If $M$ is a product of distinct primes $p\_1,\dots p\_n$ congruent to $1$ mod $3$, the norms of primes $\pi\_1,\dots,\pi\_n$ then $$\sum\_{N(z) = M} \left(\frac{z}{w} \right)\_3 = \left( \left(\frac{1}{w} \right)\_3 + \left(\frac{-1 }{w} \right)\_3 \right) \left( \left(\frac{1}{w} \right)\_3 + \left(\frac{\omega }{w} \right)\_3 + \left(\frac{\omega^2 }{w} \right)\_3 \right) \prod\_{i=1}^n \left( \left(\frac{\pi\_i}{w} \right)\_3 + \left(\frac{\overline{\pi\_i} }{w} \right)\_3 \right)$$
so if we choose $w$ such that $\left(\frac{-1 }{w} \right)\_3 =\left(\frac{\omega }{w} \right)\_3 =1$ and then choose $n$ different primes $\pi$ such that $ \left(\frac{\pi\_i}{w} \right)\_3 =\left(\frac{\overline{\pi\_i} }{w}\right)\_3$ then this will have size $6 \cdot 2^n$.
Taking $n$ sufficiently large, we contradict any bound like the one you request.
---
To sketch a proof of a bound, let's first understand
$$\left \lvert \sum\_{z} \phi\left( \frac{ \sqrt{N(z)} - R}{T} \right) \left(\frac{z}{w} \right)\_3 \right \rvert$$
where $\phi$ is a smooth bump function. The trivial bound is $O(RT)$ as we are summing over an annulus of radius $R$ and thickness $T$. Another bound is provided by integrating the Fourier transform of $\phi\left( \frac{ \sqrt{N(z)} - R}{T} \right) $ against the Fourier transform of $\left(\frac{z}{w} \right)\_3 $. Since the Fourier transform of $\left(\frac{z}{w} \right)\_3 $ is a bunch of Gauss sums, the same at every point, the bound we get this way will be $\sqrt{N(w)}$ times the $L^1$ norm of the Fourier transform of $\phi\left( \frac{ \sqrt{N(z)} - R}{T} \right) $.
For $T$ small, the Fourier transform of $\phi\left( \frac{ \sqrt{N(z)} - R}{T} \right) $ will look like the integral of a phase over a circle, i.e. a Bessel function, thus, viewed as a function of a complex variable $x$, approximately constant for $|x| < R^{-1}$ and proportial to $1/\sqrt{x}$ for larger values of $x$. The bump function will give us a rapidly decreasing cutoff at a radius of $1/T$.
Since the value at $0$ should be $RT$, the Fourier transform will take the value $\approx RT$ for $|x| <R^{-1}$ and $\approx R^{1/2} T / |x|^{1/2}$ for $ R^{-1} < |x| < T^{-1}$. The $L^1$ norm of this is $$ \int\_{ R^{-1} }^{T^{-1} } (R^{1/2} T/ r^{1/2} ) \cdot 2 \pi r \cdot dr \approx R^{1/2} T (T^{-1})^{3/2} = (R/T)^{1/2} $$ so we get
$$\left \lvert \sum\_{z} \phi\left( \frac{ \sqrt{N(z)} - R}{T} \right) \left(\frac{z}{w} \right)\_3 \right \rvert \ll \min ( RT, (R N(w) /T)^{1/2} )$$
Now in the sharp cutoff case we can express the characteristic function interval of radius $\sqrt{A}$ and thickness $B/ \sqrt{A}$ as a sum of smoothed characteristic functions of the same radius and dyadically decreasing thickness. So the bound we get will be obtained by summing the bounds for (dyadically) different values of $T$ from $0$ to $B/\sqrt{A}$, and thus will be comparable to the bound for the worst case of $T$.
The worst case occurs when the two terms in the minimum are equal, $ RT= (R N(w) /T)^{1/2}$, or $T = (N(w)/ R)^{1/3}$, giving a bound of $$ R^{2/3} N(w)^{1/3} = (A N(w))^{1/3},$$ in other words, by working out the details, I believe it should be possible to prove a bound of the form
$$\displaystyle \left \lvert \sum\_{A < N(z) < A + B} \left(\frac{z}{w} \right)\_3 \right \rvert \ll (A N(w))^{1/3}$$
This gives cancellation for $A \gg N(w)^{1/2}$, so we still obtain cancellation in intervals of greater than square-root size, which makes sense as this is possible in the classical setting (<https://arxiv.org/abs/1508.00512>), but not as much cancellation for larger intervals, which also makes sense as the boundaries in this setting are less smooth than in the classical one.
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10
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https://mathoverflow.net/users/18060
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414156
| 168,930 |
https://mathoverflow.net/questions/414116
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7
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Let $X\to S$ be a family of smooth projective complex threefolds over a connected base $S$, could it happen that for some $a, b\in S(\mathbb{C})$, the fiber $X\_a$ is birational to $\mathbb{P}^3\_{\mathbb{C}}$, while the fiber $X\_b$ is not birational to $\mathbb{P}^3\_{\mathbb{C}}$?
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https://mathoverflow.net/users/nan
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Is rationality a deformation invariant property for smooth threefolds?
|
I believe this is still an open question in dimension $3$.
There is an example due to Hassett, Kresch and Tschinkel ([Stable rationality in smooth families of threefolds](https://arxiv.org/abs/1802.06107)) of a smooth projective family of threefolds over a connected base where some fibers are *stably* rational and others are not.
Also in dimension at least $4$, there are several examples of families over a connected base where $X\_a$ is rational and $X\_b$ is not even stably rational (you can find references in the linked paper).
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6
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https://mathoverflow.net/users/12402
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414157
| 168,931 |
https://mathoverflow.net/questions/414019
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3
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$c\_{0}$, the space of the scalar sequence that converges to $0$ endowed with the sup norm, has two well-known bases: the unit vector basis $(e\_{n})\_{n}$, where $e\_{n}(k)=1$ if $k=n$ and $0$ otherwise, and the summing basis $(s\_{n})\_{n}$, where $s\_{n}=\sum\_{k=1}^{n}e\_{k}$. It is known that $c\_{0}$ has no boundedly complete basis. Recall that a basis $(x\_{n})\_{n}$ for a Banach space $X$ is boundedly complete if for every scalar sequence $(a\_{n})\_{n=1}^{\infty}$ with $\sup\_{n}\|\sum\_{i=1}^{n}a\_{i}x\_{i}\|<\infty$, the series $\sum\_{n=1}^{\infty}a\_{n}x\_{n}$ converges.
We introduce a quantity measuring (non-)bounded completeness of a basis as follows:
Let $(x\_{n})\_{n=1}^\infty$ be a bounded sequence in a Banach space $X$. We set $$\textrm{ca}((x\_{n})\_{n=1}^\infty)=\inf\_{n}\sup\_{k,l\geq n}\|x\_{k}-x\_{l}\|.$$ Then $(x\_{n})\_{n=1}^\infty$ is norm-Cauchy if and only if $\textrm{ca}((x\_{n})\_{n=1}^\infty)=0$.
Let $(x\_{n})\_{n=1}^\infty$ be a basis for a Banach space $X$. We set $$\textrm{bc}((x\_{n})\_{n=1}^\infty)=\sup\Big\{\textrm{ca}((\sum\_{i=1}^{n}a\_{i}x\_{i})\_{n=1}^\infty)\colon
(\sum\_{i=1}^{n}a\_{i}x\_{i})\_{n=1}^\infty\subseteq B\_{X}\Big\},$$ where $B\_{X}$ is the closed unit ball of $X$.
Since we have proved that $\operatorname{bc}((e\_{n})\_{n})=\operatorname{bc}((s\_{n})\_{n})=1$, we have the following natural question:
Question. $\textrm{bc}((x\_{n})\_{n=1}^\infty)=1$ for every basis $(x\_{n})\_{n}$ in $c\_{0}$ ?
To answer the question, I want to know if there are other bases in $c\_{0}$ besides these two important classes of bases. Are there more references about bases in $c\_{0}$? Thank you.
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https://mathoverflow.net/users/41619
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Bases in $c_{0}$
|
You have answered your own question modulo a lemma that is basically obvious:
Lemma. Let $(x\_n)$ be a basis for a Banach space $X$ and let $Y=(\sum\_{n=0}^\infty E\_n)\_0$ be the $c\_0$-sum of $E\_n := \text{span} \{ x\_0,\dots,x\_n\}$. Let $(y\_k)$ be the obvious basis for $Y$ induced by concatenating the given bases for $E\_n$; $n=1,2,\dots$. Then
$\text{bc}((y\_n) \ge \text{bc} (x\_n)$.
In your other post about defining a quantitative version of non bounded completeness you pointed out that there is a basis $(x\_n)$ for $c$ for which $\text{bc} (x\_n)=2$. For this basis, $E\_n := \text{span} \{ x\_0,\dots,x\_n\}$ is obviously isometrically isomorphic to $\ell\_\infty^{ n+1}$ and hence $Y:= (\sum\_{n=0}^\infty E\_n)\_0$ is isometrically isomorphic to $c\_0$.
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1
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https://mathoverflow.net/users/2554
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414158
| 168,932 |
https://mathoverflow.net/questions/414159
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1
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Let define $F(s)=\int\_0^\infty f(u)e^{-su}du$. If $f$ is bounded and $\lim\_{t\to 0}f(t)$ exists. Then we can get $\lim\_{t\to 0}f(t)=\lim\_{s\to\infty}sF(s)$.
Can we use upper limits or lower limits to replace all the above limits? i.e. we only know $\limsup\_{t\to 0}f(t)$ or $\liminf\_{t\to0}f(t)$ exist. Can we get the following equality? $$\limsup\_{t\to 0}f(t)=\limsup\_{s\to\infty}sF(s), $$ $$\liminf\_{t\to 0}f(t)=\liminf\_{s\to\infty}sF(s)? $$
If it is correct, is there someone can give me some reference or if it is incorrect is there someone can give me a counterexample?
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https://mathoverflow.net/users/147009
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Initial and final Theorem for upper and lower limits?
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The answer is no.
Let $f(u)$ be defined piecewise. On intervals of the form $(2^{k},2^{k+1}]$, where $k\in \mathbb{Z}$, you set $f(u) = (-1)^k$. Then you have that $\limsup\_{t\to 0} f(t) = +1$ and $\liminf\_{t\to 0} f(t) = -1$.
The integral $s F(s) = \int\_0^\infty f(u/s) e^{-u} ~du$ is:
* Continuous in $s$
* For every $s$ strictly less than 1 in absolute value
* satisfies $s F(s) = 4s F(4s)$.
And hence you must have $|\limsup s F(s)|, |\liminf s F(s)| < 1$.
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2
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https://mathoverflow.net/users/3948
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414162
| 168,934 |
https://mathoverflow.net/questions/410104
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4
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Consider two independent continuous random walks on a graph $G$ with adjacency matrix $A$. I am interested in the probability that the two walkers will ever meet.
When the graph is a $k$-regular graph with $k=1,2$ then the probability is one, see this question for example: [What is the probability that two random walkers will meet?](https://mathoverflow.net/questions/35469/what-is-the-probability-that-two-random-walkers-will-meet)
I tried to derive a simple equation but I am doing something wrong and I cannot determine where it goes wrong, which is the reason of this post. Here is my reasoning:
Let $p\_i(t)$ (res. $q\_i(t)$) be probability that walker $1$ (res. $2$) is at node $i$ at time $t$. Let $\gamma(t)$ denote the probability that the two walkers have already met (at least one time) in the interval $[0,t]$. The probability that they have already met at $t+\mathrm{d}t$ is:
\begin{align}
\gamma(t+\mathrm{d}t)&=1\times \text{"probability they have already met by time t"} \quad + \text{"probability they have NOT already met by time t"}\times \text{"probability they meet in next increment"}\\
&=\gamma(t)+ (1-\gamma(t))\mathrm{d}t \underbrace{\left( \sum\_{ij}2\frac{A\_{ij}}{k}p\_i(t)q\_j(t) \right)}\_{\text{pr of meeting in next time increment}}.
\end{align}
In the last line I assumed regular graphs for this example, but this equation could be generalised to general graphs. (Conditioned on the event that walker $1$ is on vertex $i$ and walker $2$ is on vertex $j$ the probability that walker $2$ moves to vertex $i$ in the next time increment is $\frac{A\_{ij}}{k\_j}\mathrm{d}t$)
Taking the limit $dt\to 0$ gives me the master equation:
\begin{align}
\dot\gamma(t)&=(1-\gamma(t))\left( \sum\_{ij}2\frac{A\_{ij}}{k}p\_i(t)q\_j(t) \right)\\
&=(1-\gamma(t))f(t).
\end{align}
But if I solve this equation it gives me clearly wrong results, (first the integral of $f(t)$ diverges, and even if I take care of this divergence, the rate at which they meet is wrong). However I cannot see at which step I am taking a false assumption? Any remark or reference is always appreciated!
(I am NOT interested in the case where we reduce this problem to a single random walk, because I am interested in the generalization where the space is not homogeneous (e.g., not a regular graph, adding weights on the edges etc.).)
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https://mathoverflow.net/users/142153
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Probability that two walkers will meet on a graph
|
This is equivalent to the question whether two walks will necessarily meet infinitely often (with probability one) or not. This question has been studied in some detail, see in particular [2] where some general criteria are given.
[1] Krishnapur, Manjunath, and Yuval Peres. "Recurrent graphs where two independent random walks collide finitely often." Electronic communications in probability 9 (2004): 72-81.
[2] Barlow, Martin T., Yuval Peres, and Perla Sousi. "Collisions of random walks." In Annales de l'IHP Probabilités et statistiques, vol. 48, no. 4, pp. 922-946. 2012.
[3] Chen, XinXing, and DaYue Chen. "Two random walks on the open cluster of ℤ 2 meet infinitely often." Science China Mathematics 53, no. 8 (2010): 1971-1978.
[4] Hutchcroft, Tom, and Yuval Peres. "Collisions of random walks in reversible random graphs." Electronic Communications in Probability 20 (2015): 1-6.
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2
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https://mathoverflow.net/users/7691
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414163
| 168,935 |
https://mathoverflow.net/questions/414081
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0
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My question is about the compatibility and consistency between two definitions of cohomology in two books.
I asked this question about 10 days ago [on MathSE](https://math.stackexchange.com/questions/4351968/compatibility-and-consistency-between-two-definitions-of-cohomology-in-two-books)
and I set a bounty on it, but I didn't receive an answer.
I was reading cohomology from Neukirch's book, and there he referenced Hall's book. The two approaches are almost the same (are they not?), and they should give us the same results (cohomology groups). But I can not see their compatibility and consistency, and I can not recover them from each other. My problem is that: **I can not propose the compatibility between these two books.**
---
Hall's book:
**15.7. Cochains, Coboundaries, and Cohomology Groups.**
Given a double $\Omega$-modulo $A$ we define $C^n = C^n(A, \Omega)$ to be the additive group of all functions $f$ of $n$ variables which range independently over $\Omega$ and taking values in $A$, subject to the condition
$$ (15.7.1) \ \ \ \ \ \ \ \ \ \ \ \ \ \
f(\xi\_1, \dots, \xi\_n) = 0,$$
whenever at least one of the $\xi\_i = 1$. The elements of $C\_n$ are called
*$n$-dimensional cochains*. $C^0 = A$ by definition and a zero dimensional
cochain is simply any element of A.
The coboundary operator $\delta$ maps each $C\_n$ into the next, $C\_{n+1}$ in
accordance with the rule
$$(\delta f)(\xi\_0, \xi\_1, \dots, \xi\_n)\\
=\xi\_0f(\xi\_1, \dots, \xi\_n) \\
+\sum\_{t=1}^{n} (-1)^t (\xi\_0, \xi\_1, \dots, \xi\_{t-2}, \xi\_{t-1}\xi\_{t}, \xi\_{t+1}, \dots, \xi\_{n}) \\
+f(\xi\_0\xi\_1, \dots, \xi\_{n-1})\xi\_n . \textit{ something that I cannot read},
$$
And let $Z^n=Z^n(A, \Omega)= \ker (C^n\longrightarrow C^{n+1})$, and let $B^n=B^n(A, \Omega)= Im (C^{n-1}\longrightarrow C^{n})$. The quotient group $Z^n/B^n$ is called the $n$-dimensional cohomology group of the double $\Omega$-module $A$. We write it $$H^n(A, \Omega) = Z^n/B^n.$$
In the part *something that I can not read*, **in the above formula**, probably it is written $(-1)^{n+1}$, but I am not sure.
---
---
I have some confusions about coboundary operators and 1-cocycles and computing cohomology, and I can not find where the bugs are.
In Neukirch's Class field theory, page 13, does the last summand of $d\_q(\sigma\_1, \dots, \sigma\_q)$ written truly? Or should it be equal to
$$d\_q(\sigma\_1, \dots, \sigma\_q)=
\sigma\_1(\sigma\_2, \dots, \sigma\_{q})
+
\sum\_{i=1}^{q-1} (-1)^i (\sigma\_1, \dots, \sigma\_{i-1}, \sigma\_{i}\sigma\_{i+1}, \sigma\_{i+2}, \dots, \sigma\_{q})
+
(-1)^q(\sigma\_1, \dots, \sigma\_{q-1})\sigma\_q ?$$
I've seen something closely similar to these relations in Hall's book (The theory of groups), but I think at least one of them should have a typo. I know that they are not exactly the same, but I think they are not compatible unless for instance we replace the definition of Neukirch as above.
Also I have ambiguity about cocycles and 1-cocycles. If the things in Neukirch's book are correct, then I expected the cocycle should be $f(\sigma\tau)=\sigma f(\tau)+f(\sigma)$, and I am confused with $\delta(f(\sigma, \tau))=\sigma f(\tau)-f(\sigma\tau)+f(\sigma)\tau$, and I can not see the compatibility.
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https://mathoverflow.net/users/166540
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Coboundary operators, 1-cocycles and computing cohomology
|
Let me just put my comments as an answer: (i) To compute the cohomology groups $H^n(G,A)$, for $A$ a (left) $G$-module and where
$$H^n(G,A)=\text{Ext}^n\_{{\mathbb Z}G}({\mathbb Z},A)$$
one needs a projective resolution of the $G$-module ${\mathbb Z}$, say
$$\cdots\stackrel{d\_3}\rightarrow P\_2\stackrel{d\_2}\rightarrow P\_1\stackrel{d\_0}\rightarrow P\_0\stackrel{\varepsilon}\rightarrow {\mathbb Z}\rightarrow 0$$
such that
$$H^n(G,A)=\text{Ext}^n\_{{\mathbb Z}G}(
{\mathbb Z},A)=\ker(d^\*\_{n+1})/\text{Im}(d\_n^\*).$$
(ii) One choice of projective resolution of ${\mathbb Z}$ is given by the free ${\mathbb Z}$-modules $P\_n$ with basis
the $(n+1)$-tuples $(g\_0,\ldots,g\_n)$ of elements of the group $G$ and where the action of $G$ is by (left) translation: $g(g\_0,\ldots,g\_n)=(gg\_0,\ldots,gg\_n)$. It follows that the $P\_n$ are free $G$-modules with basis the $(n+1)$-tuples of the form $(1,g\_1,\ldots,g\_n)$. The $G$-morphisms
$$d\_n:P\_n\rightarrow P\_{n-1}$$
are defined on the generators by the formula
$$d\_n(g\_0,\ldots,g\_n)=\sum\_{j=0}^n(-1)^n(g\_0,\ldots,\widehat{g\_j},\ldots,g\_n)$$
where $\widehat{g\_j}$ means that this term is omitted.
The morphism $\varepsilon$ sends a generator $(g\_0)$ to $1$. One shows that the $(P\_n,d\_n)$ form indeed an exact sequence and upon appliying the functor $\text{Hom}\_{G}(-,A)$ one obtains a complex whose cohomology groups are $H^n(G,A)$. All of these is pretty standard fare in any homological algebra textbook. The important point is now to observe that an $n$-cochain $f\in \text{Hom}\_G(P\_n,A)$ is determined by its values in the generators $(g\_0,\ldots,g\_n)$ of $P\_n$ and since the cochain is a $G$-module morphism (we say that it is $G$-covariant), then
$$f(g\_0,g\_1,\ldots,g\_n)=f(g\_0(g\_0^{-1}g\_1,\ldots,g\_0^{-1}g\_n))=g\_0f(1,g\_0^{-1}g\_1,\ldots,g\_0^{-1}g\_n)$$
i.e., $f$ is determined by its values on the generators of $P\_n$ of the form $(1,g\_1,\ldots,g\_n)$ and then one may also consider another projective resolution of ${\mathbb Z}$ given by the free $G$-modules $Q\_n$ with generators $[g\_1,\ldots,g\_n]$ (and where $Q\_0$ is the free $G$-module generated by a symbol $[\;]$. One then shows that $P\_n\simeq Q\_n$ as $G$-modules and defining
$$\delta\_n:Q\_n\rightarrow Q\_{n-1}$$
by
$$\delta\_n[g\_1,\ldots,g\_n]=g\_1[g\_2,\ldots,g\_n]+\sum\_{j=1}^{n-1}(-1)^j[g\_1,\ldots,g\_jg\_{j+1},\ldots,g\_n]+(-1)^n[g\_1,\ldots,g\_{n-1}]$$
the $(Q\_n,\delta\_n)$ form a free $G$-resolution of ${\mathbb Z}$ and there is an isomorphism of chain complexes $\varphi:(P\_n,d\_n)\rightarrow (Q\_n,\delta\_n)$, answering your question. Let just add some details: First, the morphisms $\varphi\_n:P\_n\rightarrow Q\_n$ are given by $\varphi\_n(g\_0,\ldots,g\_n)=g\_0[g\_0^{-1}g\_1,g\_1^{-1}g\_2,\ldots, g\_{n-1}^{-1}g\_n]$ and the corresponding $\psi\_n:Q\_n\rightarrow P\_n$ are $\psi\_n[g\_1,\ldots,g\_n]=(1,g\_1,g\_1g\_2,g\_1g\_2g\_3,\ldots, g\_1g\_2\cdots g\_n)$. A lenghty, but straightforward, computation shows that $\delta\_n=\varphi\_{n-1}\circ d\_n\circ \psi\_n$.
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4
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https://mathoverflow.net/users/3903
|
414166
| 168,936 |
https://mathoverflow.net/questions/412957
|
7
|
$\DeclareMathOperator\SL{SL}\DeclareMathOperator\SO{SO}\DeclareMathOperator\SU{SU}\DeclareMathOperator\O{O}\DeclareMathOperator\Iso{Iso}$Let $ g $ be the round metric on the sphere $ S^n $. Since $ S^n $ is compact the isometry group $ \Iso(S^n,g) $ is also compact. And every compact group can be realized as the real points of some (reductive) linear algebraic group. Indeed, $ \Iso(S^n,g) = \O\_{n+1}(\mathbb{R}) $. The complex points of this group are $ \O\_{n+1}(\mathbb{C}) $. And $ \O\_{n+1}(\mathbb{C}) $ acts transitively on the tangent bundle of the sphere $ T(S^n) $.
Does this generalize from the round sphere to other compact homogeneous Riemannian manifolds?
In other words, Let $ (M,g) $ be a compact Riemannian homogeneous space. Then $ \Iso(M,g) $ is a compact Lie group. So there exists some (reductive) linear algebraic group whose real points are isomorphic to $ \Iso(M,g) $. The question is, does there always exist a linear algebraic group $ G $ such that the real points of $ G $ are isomorphic to the isometry group
$$
G\_\mathbb{R} \cong \Iso(M,g)
$$
and, in addition, the complex points of $ G $ act (transitively, smoothly) on the tangent bundle $ T(M) $?
Note that this question is equivalent to a question which does not a priori in involve any geometry:
**The manifold
$
G\_\mathbb{C}/H\_\mathbb{C}
$
is the tangent bundle to
$
G\_\mathbb{R}/H\_\mathbb{R}
$
where $ G\_\mathbb{R}$, $H\_\mathbb{R} $ are compact real forms of $ G\_\mathbb{C}$, $H\_\mathbb{C} $**
Note that while the action of $ G\_\mathbb{R} $ is by isometries, the action of $ G\_\mathbb{C} $ on $ T(M) $ can only be by isometries if $ M $ is parallelizable. So in particular the action of $ \O\_{n+1}(\mathbb{C}) $ on $ T(S^n) $ can only be by isometries in the cases $ n=1,3,7 $.
|
https://mathoverflow.net/users/387190
|
Does complexified isometry group act transitively on tangent bundle of compact Riemannian manifold?
|
Maybe these arguments are of interest to you. It is known that for any compact symmetric space $M$ the tangent bundle $TM$ possesses a canonical structure of a complex
manifold. Multiplication by $-1$ on $TM$ is an antiholomorphic involution; its set of fixed points is $M$, when identified with the zero section
of $TM$ (see e.g. Thm 2.5a of Szőke, R.: Complex structures on tangent bundles of Riemannian manifolds, Math. Ann. 291 (1991), 409-428). It
follows that the action of the Lie group $G\_{\mathbf R}$ on $M$ extends to an action on the complex manifold $TM$ by holomorphic
automorphisms. One proves that it extends to a holomorphic action of the complexification $G\_{\mathbf C}$ of $G\_{\mathbf R}$ on $TM$. This can be
done by considering the induced morphism from the real Lie algebra ${\mathfrak g}\_{\mathbf R}$ of $G\_{\mathbf R}$ into the complex Lie algebra
${\mathcal A}(TM)$ of holomorphic vector fields on $TM$. It naturally extends to the complexification ${\mathfrak g}\_{\mathbf C}$ of ${\mathfrak
g}\_{\mathbf R}$, which gives rise to a holomorphic action of the complexification $G\_{\mathbf C}^0$ of the neutral component $G\_{\mathbf R}^0$ of
the real Lie group $G\_{\mathbf R}$. The latter action can easily be extended to a holomorphic action of $G\_{\mathbf C}$ on the complex
manifold $TM$. It is not hard to see that the action is transitive (for more details see e.g. Thm C of Szőke, R.: Automorphisms of certain Stein manifolds,
Math. Z. 219 (1995), 357-385)
|
3
|
https://mathoverflow.net/users/85592
|
414174
| 168,939 |
https://mathoverflow.net/questions/414083
|
12
|
Roughly speaking, say that a logic $\mathcal{L}$ is **self-equivalence-defining** (SED) iff for each finite signature $\Sigma$ there is a larger signature $\Sigma'\supseteq\Sigma\sqcup\{A,B\}$ with $A,B$ unary relation symbols and an $\mathcal{L}[\Sigma']$-sentence $\eta$ such that the following are equivalent for all $\Sigma$-structures $\mathfrak{A},\mathfrak{B}$:
* $\mathfrak{A}\equiv\_\mathcal{L}\mathfrak{B}$.
* There is a $\Sigma'$-structure $\mathfrak{S}$ such that $\mathfrak{S}\models \eta$, $A^\mathfrak{S}\upharpoonright\Sigma\cong\mathfrak{A}$, and $B^\mathfrak{S}\upharpoonright\Sigma\cong\mathfrak{B}$.
For example, Fraisse showed that $\mathsf{FOL}$ is SED *(this is used crucially in the proof of Lindstrom's theorem - it's [a very happy construction](https://www.target.com/p/mathematical-logic-undergraduate-texts-in-mathematics-2nd-edition-by-h-d-ebbinghaus-j-flum-wolfgang-thomas-hardcover/-/A-85200040))*. That same argument gives as a corollary that $\mathsf{SOL}$ is also SED, roughly because $(i)$ "$X$ is the powerset of $Y$" is second-order expressible and $(ii)$ $\mathsf{SOL}$-elementary equivalence between two structures amounts to $\mathsf{FOL}$-elementary equivalence between their "power-structures."
The nicest logic whose SED status I don't know is $\mathcal{L}\_{\omega\_1,\omega}$. On the one hand, this logic isn't powerful enough to perform the same sort of "cheat" as $\mathsf{SOL}$. On the other hand, the direct game-theoretic attack analogizing the situation for $\mathsf{FOL}$ results in a game which is a bit too complicated for $\mathcal{L}\_{\omega\_1,\omega}$ to handle appropriately; see [Vaananen/Wang, *An Ehrenfeucht-Fraisse game for $\mathcal{L}\_{\omega\_1,\omega}$*](https://arxiv.org/abs/1212.0108), and note that this was also an issue in [this earlier question of mine](https://mathoverflow.net/questions/350778/improving-a-lindstrom-y-fact-about-mathcall-omega-1-omega). So my question is:
>
> Is $\mathcal{L}\_{\omega\_1,\omega}$ SED?
>
>
>
I'm separately interested in the general situation of which infinitary logics are SED. It's not hard to show that for $\kappa$ satisfying appropriate large cardinal properties we have that $\mathcal{L}\_{\kappa,\omega}$ is SED, but I don't see how to extract large cardinal strength from SEDness.
Note that looking at a single sentence $\eta$, as opposed to a theory $E$, is needed to avoid every (regular) logic being trivially SED; I messed this up in the original version of this question, and this was pointed out by Peter LeFanu Lumsdaine. Separately, I've added the forcing tag since, despite not being part of the question, general principles about forcing turn out to form a key component of the answer.
|
https://mathoverflow.net/users/8133
|
Can $\mathcal{L}_{\omega_1,\omega}$ detect $\mathcal{L}_{\omega_1,\omega}$-equivalence?
|
(Working in ZFC.)
No (re $\mathcal{L}\_{\omega\_1,\omega}$). Suppose it is. Consider the signature $\Sigma$ with just one binary relation symbol $<$. Let $\Sigma',\eta$ witness SED-ness for $\Sigma$.
Let $\pi:M\to V\_\theta$ be elementary,with $\theta$ some sufficiently large limit ordinal, $M$ transitive, $M$ of cardinality $\kappa=2^{\aleph\_0}$, with $\mathbb{R},2^{\aleph\_0}\subseteq M$. So $\mathrm{crit}(\pi)=\kappa^{+M}<\kappa^+$ and $\pi(\kappa^{+M})=\kappa^+$. Let $\mathfrak{A}=(\kappa^{+M},{\in}\upharpoonright\kappa^{+M})$ and $\mathfrak{B}=(\kappa^+,{\in}\upharpoonright\kappa^+)$. Note that $\mathfrak{A}\equiv\_{\omega\_1,\omega}\mathfrak{B}$, since $\pi$ is elementary and all the sentences in $\mathcal{L}\_{\omega\_1,\omega}$ are in $M$, since $\mathbb{R}\subseteq M$. So by hypothesis, there is some $\Sigma'$-structure $\mathfrak{G}$ such that $\mathfrak{G}\models\eta$, $A^{\mathfrak{G}}\upharpoonright\Sigma\cong\mathfrak{A}$ and $B^{\mathfrak{G}}\upharpoonright\Sigma\cong\mathfrak{B}$.
Now let $G$ be $V$-generic for $\mathrm{Coll}(\omega,\max(\kappa^{+},|\mathfrak{G}|))$.
In $V[G]$, there is an $\mathcal{L}\_{\omega\_1,\omega}$ sentence $\varphi$ in the signature $\Sigma$ asserting that the model is (isomorphic to) $\kappa^{+M}$. So in $V[G]$, $\mathfrak{A}\models\varphi$ but $\mathfrak{B}\models\neg\varphi$. So $V[G]$ models the statement "there are countable structures $\mathfrak{A}\_0,\mathfrak{B}\_0$ in the signature $\Sigma$ and an $\mathcal{L}\_{\omega\_1,\omega}$-sentence $\varphi\_0$ such that $\mathfrak{A}\_0\models\varphi\_0$ and $\mathfrak{B}\_0\models\neg\varphi\_0$
and there is a countable structure $\mathfrak{G}\_0$ in the signature $\Sigma'$ such that $\mathfrak{G}\_0\models\eta$ and $A^{\mathfrak{G}}\upharpoonright\Sigma\cong\mathfrak{A}\_0$ and $B^{\mathfrak{G}}\upharpoonright\Sigma\cong\mathfrak{B}\_0$". But this statement is $\Sigma^1\_2$ in a real coding $\eta$. (I don't see that it is $\Sigma^1\_1$ in such a real, because to assert that $\varphi\_0$ is really a sentence of $\mathcal{L}\_{\omega\_1,\omega}$ involves saying that it is built along a real ordinal.) Since $\eta\in V$ and by Shoenfield absoluteness, $V$ models the same statement. But this contradicts our assumptions.
|
13
|
https://mathoverflow.net/users/160347
|
414179
| 168,940 |
https://mathoverflow.net/questions/414182
|
1
|
I have a question that seems very basic, and yet I have not managed to find an answer after probably several hours of Google-searching.
Fix $0<a<b<\infty$, and let $\mathcal{P}\_{[a,b]}$ be the set of all probability distributions on $[a,b]$. For each $p \in \mathcal{P}\_{[a,b]}$, let
\begin{align\*}
\mu(p) &= \int\_{[a,b]} \! x \, p(dx) \\
\sigma(p) &= \sqrt{ \int\_{[a,b]} \! x^2 \, p(dx) \ - \ \mu(p)^2 } \\
\gamma(p) &= \frac{\sigma(p)}{\mu(p)}\text{.}
\end{align\*}
We call $\gamma(p)$ the *coefficient of variation* of $p$.
>
> What is $\max\_{p \in \mathcal{P}\_{[a,b]}} \gamma(p)$?
>
> And is there any paper or textbook that can be cited for the answer to this question?
>
>
>
---
**Remarks.**
I believe that this maximum must exist, as $\mu$ and $\sigma$ are continuous with respect to the topology of weak convergence, and $\mathcal{P}\_{[a,b]}$ is compact under the topology of weak convergence (since $[a,b]$ is compact).
It should be easy to show that any maximising $p$ has the following three properties:
* At least one of the intervals $(a,\mu(p)]$ and $[\mu(p),b)$ is a $p$-null set; hence in particular, at least one of the singletons $\{a\}$ and $\{b\}$ is a $p$-positive measure set.
* $\mu(p) \leq \frac{a+b}{2}$.
* The support of $p$ (i.e. the smallest $p$-full measure closed set) includes both $a$ and $b$.
The first of these must hold, otherwise one can spread some of the mass of $p$ away from $\mu(p)$ in such a manner as to increase $\sigma(p)$ while preserving $\mu(p)$.
The second of these must hold, otherwise one can reflect the mass of $p$ in the point $\frac{a+b}{2}$, resulting in a decreased $\mu(p)$ while $\sigma(p)$ stays the same.
Assuming the first of these, the third of these must hold: otherwise, one can linearly spread the mass of $p$ away from $a$ or $b$ as appropriate, and then $\sigma(p)$ will increase by a factor that is greater than the factor of increase of $\mu(p)$. (Specifically, either $\mu(p)$ will decrease, or else $\mu(p)-a$ will increase by the same factor that $\sigma(p)$ increases.)
My not-very-confident intuition is that any maximising $p$ must be fully supported on the boundary points. Of course, maximising $\gamma(p)$ over the set of probability measures supported on $\{a,b\}$ is an easy problem, as there is only one variable being varied (namely, $p(\{a\})$). Plugging the question into Wolfram Alpha (because I'm too lazy to do it by hand), we get that the maximisation is achieved by
$$ p(\{a\}) = \frac{b}{a+b}\text{,} \quad\quad p(\{b\}) = \frac{a}{a+b}\text{,} $$
for which we have
$$ \gamma(p) = \frac{b-a}{2\sqrt{ab}}\text{.} $$
But whether this remains the maximum over the whole of $\mathcal{P}\_{[a,b]}$ I do not know for sure.
|
https://mathoverflow.net/users/15570
|
What is the maximum possible coefficient of variation for data taking values within a specified range?
|
Indeed, for any random variable (r.v.) $X$ with values in $[a,b]$ and mean $\mu\in[a,b]$,
$$Var\,X\le Var\,X\_{a,b;\mu}=(b-\mu)(\mu-a), \tag{1}$$
where $X\_{a,b;\mu}$ is any r.v. with the unique distribution with mean $\mu$ supported on the two-point set $\{a,b\}$. So, the result that you conjectured follows, because
$$\max\_{\mu\in[a,b]}\frac{(b-\mu)(\mu-a)}{\mu^2}$$
is attained when $\mu$ is the harmonic mean of $a,b$ and equals $\dfrac{(b-a)^2}{4ab}$.
(However, the maximizing distribution, on the two-point set $\{a,b\}$, is discrete and thus does not have a density function.)
---
One way to prove inequality (1) is as follows. By [Vieta's formula](https://en.wikipedia.org/wiki/Quadratic_equation#Vieta%27s_formulas), for all $x\in[a,b]$ we have $x^2-(a+b)x+ab\le0$ and hence
$$x^2\le(a+b)x-ab, \tag{2}$$
with the equality iff $x\in\{a,b\}$. So,
$$EX^2\le(a+b)\mu-ab=EX\_{a,b;\mu}^2. \tag{3}$$
Since $Var\,X=EX^2-\mu^2$, we do now get (1).
Moreover, since the inequality in (2) is strict for $x\notin\{a,b\}$, the inequality in (3) is strict unless $X$ equals $X\_{a,b;\mu}$ in distribution. Thus, the inequality in (1) is strict unless $X$ equals $X\_{a,b;\mu}$ in distribution.
|
1
|
https://mathoverflow.net/users/36721
|
414183
| 168,941 |
https://mathoverflow.net/questions/414096
|
4
|
Let $b:\mathbb N\to \{0,1\}$ be the indicator function of integers that are a sum of two non-zero integer squares. Let $f(t)\in \mathbb Z[t]$ be an irreducible polynomial of degree $2$ with positive leading coefficient and not of the form $(at+b)^2+c^2$ for some rationals $a,b,c$. Then standard sieve heuristics would suggest that $$ \frac{x}{\sqrt{\log x} }\ll \sum\_{n\leq x } b(|f(t)|) \ll \frac{x}{\sqrt{\log x}}.$$ My question is: has this problem been studied before and are there any results confirming the lower bound, even for at least one polynomial $f$? The upper bounds are directly proved by the Brun or the Selberg sieve.
|
https://mathoverflow.net/users/9232
|
How often is the value of a quadratic polynomial equal to a sum of two integer squares?
|
This problem has been addressed in a paper of Friedlander and Iwaniec in Acta Mathematica 1978, called [Quadratic polynomials and quadratic forms](https://projecteuclid.org/journals/acta-mathematica/volume-141/issue-none/Quadratic-polynomials-and-quadratic-forms/10.1007/BF02545740.full). Under general conditions they count the number of integers $n\le x$ for which the values $g(n) = an^2+ bn+c$ (with $a>0$, and $a$, $b$, $c$ integers) may be represented by a given quadratic form $\phi(u,v) = Au^2+Buv+Cv^2$. The main result gives a lower bound of order $x/\sqrt{\log x}$ as may be expected, provided a (necessary) local condition is met.
|
7
|
https://mathoverflow.net/users/38624
|
414184
| 168,942 |
https://mathoverflow.net/questions/413906
|
9
|
Iwaniec [1] proved that
$$
\pi(x+x^\theta)-\pi(x) < \frac{(2+\varepsilon)x^\theta}{\eta(\theta)\log x},\ x>x\_0(\varepsilon,\theta).
$$
with
$$
\eta(\theta)=\frac{15\theta-2}{9}.
$$
(Actually, he proves that a function $\eta(\theta)>\theta$ exists, and that this is an admissible choice. This choice gives nontrivial information for $\theta>1/3.$ He gives others like $\eta\_1(\theta)=(1+\theta)/2$ for $\theta>1/2.$)
Two other questions have asked about the primes in this interval
* [Prime Power Gaps](https://mathoverflow.net/q/111029/6043)
* [Prime powers between $x$ and $x+x^\theta$](https://mathoverflow.net/q/262162/6043)
but neither asks about upper bounds, nor do answers give information.
Fundamentally, I'd like information on
$$
f(\theta) := \limsup\_{x\to\infty} \frac{\pi(x+x^\theta)-\pi(x)}{x^\theta/\log x}
$$
What modern results are available? The result above is $f(\theta) \le 18/(15\theta-2)$ for $1/3<\theta<1.$
[1] Henryk Iwaniec. [On the Brun–Titchmarsh theorem](https://doi.org/10.2969/jmsj/03410095). *Journal of the Mathematical Society of Japan*, **34**:1, pp. 95–123, 1982.
|
https://mathoverflow.net/users/6043
|
Primes between $x$ and $x+x^\theta$
|
Montgomery (1) gives a list of 40 exponent pairs $(\kappa,\lambda)$ which can be plugged into Iwaniec's formula
$$
\eta(\theta)=\left(1+\frac{1-\lambda+2\kappa}{3-\lambda-\kappa/2}\right)\theta - \frac{\kappa}{3-\lambda-\kappa/2}
$$
to yield bounds for $0<\theta\le1/2.$ Of these, 36 are optimal in some interval; adding the zeta function value for $\theta>1/2$ yields
$$
\eta(\theta)=\begin{cases}
\theta,&\text{ if }0<\theta\le1/9\\
\frac{1047\theta-2}{1029},&\text{ if }1/9\le\theta\le514/3597\\
\frac{531\theta-2}{515},&\text{ if }514/3597\le\theta\le322/2061\\
\frac{369\theta-2}{354},&\text{ if }322/2061\le\theta\le194/1101\\
\frac{444\theta-4}{417},&\text{ if }194/1101\le\theta\le130/669\\
\frac{189\theta-2}{176},&\text{ if }130/669\le\theta\le46/219\\
\frac{498\theta-8}{451},&\text{ if }46/219\le\theta\le2/9\\
\frac{228\theta-4}{205},&\text{ if }2/9\le\theta\le362/1569\\
\frac{1161\theta-22}{1037},&\text{ if }362/1569\le\theta\le182/743\\
\frac{1167\theta-26}{1027},&\text{ if }182/743\le\theta\le52/201\\
\frac{921\theta-22}{805},&\text{ if }52/201\le\theta\le342/1291\\
\frac{254\theta-8}{215},&\text{ if }342/1291\le\theta\le226/833\\
\frac{669\theta-22}{563},&\text{ if }226/833\le\theta\le118/415\\
\frac{116\theta-4}{97},&\text{ if }118/415\le\theta\le186/641\\
\frac{589\theta-22}{487},&\text{ if }186/641\le\theta\le94/303\\
\frac{587\theta-26}{473},&\text{ if }94/303\le\theta\le310/959\\
\frac{1439\theta-66}{1153},&\text{ if }310/959\le\theta\le286/855\\
\frac{1327\theta-66}{1049},&\text{ if }286/855\le\theta\le127/376\\
\frac{351\theta-22}{265},&\text{ if }127/376\le\theta\le74/217\\
\frac{393\theta-26}{293},&\text{ if }74/217\le\theta\le45/128\\
\frac{325\theta-22}{241},&\text{ if }45/128\le\theta\le62/171\\
\frac{347\theta-26}{251},&\text{ if }62/171\le\theta\le37/98\\
\frac{281\theta-22}{201},&\text{ if }37/98\le\theta\le66/173\\
\frac{823\theta-66}{585},&\text{ if }66/173\le\theta\le426/1093\\
\frac{317\theta-26}{224},&\text{ if }426/1093\le\theta\le682/1733\\
\frac{1243\theta-114}{851},&\text{ if }682/1733\le\theta\le438/1103\\
\frac{1601\theta-150}{1089},&\text{ if }438/1103\le\theta\le36/89\\
\frac{1651\theta-162}{1107},&\text{ if }36/89\le\theta\le486/1181\\
\frac{659\theta-66}{439},&\text{ if }486/1181\le\theta\le474/1141\\
\frac{1465\theta-150}{969},&\text{ if }474/1141\le\theta\le1626/3865\\
\frac{595\theta-66}{383},&\text{ if }1626/3865\le\theta\le120/281\\
\frac{1427\theta-162}{911},&\text{ if }120/281\le\theta\le422/973\\
\frac{1249\theta-150}{781},&\text{ if }422/973\le\theta\le846/1921\\
\frac{923\theta-114}{571},&\text{ if }846/1921\le\theta\le1542/3469\\
\frac{470\theta-60}{287},&\text{ if }1542/3469\le\theta\le34/75\\
\frac{15\theta-2}{9},&\text{ if }34/75\le\theta\le1/2\\
\frac{\theta+1}{2},&\text{ if }1/2<\theta\le7/12\\
2,&\text{ if }7/12<\theta<1.
\end{cases}
$$
This is hardly a satisfactory answer, giving only 20-year-old results and neglecting Vinogradov's method for small $\theta,$ but it may be useful so I'll leave it here.
I’ve added Huxley’s result
$$
\pi(x)-\pi(x-y) \sim \frac{y}{\log x}\text{ for }x^\theta \le y \le x/2
$$
with $\theta>7/12$.
1. Hugh L. Montgomery, [Harmonic Analysis as found in Analytic Number Theory](https://www.nato-us.org/analysis2000/papers/montgomery.pdf), in Twentieth Century Harmonic Analysis—A Celebration, Springer, 2001
2. M. N. Huxley, On the difference between consecutive primes, Invent. Math. 15 (1972), pp. 164-170.
|
6
|
https://mathoverflow.net/users/6043
|
414189
| 168,944 |
https://mathoverflow.net/questions/414103
|
2
|
Is the following analogue of Fermat's Little Theorem for Bernoulli numbers true?
>
> Let $D\_{2n}$ be the denominator of $\frac{B\_{2n}}{4n}$ where $B\_n$ is
> the $n$-th [Bernoulli number](http://oeis.org/A006863). If $\gcd(a, D\_{2n}) = 1$ then
>
>
> $$ a^{2n} \equiv 1\pmod{D\_{2n}}.$$
>
>
>
This question was [posted in MSE](https://math.stackexchange.com/questions/4343529/conjectured-analogue-of-fermats-little-theorem-for-bernouli-numbers) 3 weeks back but it is still open. Hence posting in MO.
|
https://mathoverflow.net/users/23388
|
Analogue of Fermat's little theorem for Bernoulli numbers
|
A proof is essentially given in Section 5.1 of [Notes on primitive lambda-roots](http://www.maths.qmul.ac.uk/%7Epjc/csgnotes/lambda.pdf) by P. J. Cameron and D. A. Preece.
|
5
|
https://mathoverflow.net/users/7076
|
414205
| 168,947 |
https://mathoverflow.net/questions/414209
|
15
|
**Motivation.** In a coin game, a player flips all their coins every turn, starting with just one coin. If the coins all land *heads* then the game stops; otherwise, the number of coins is doubled for the following turn. While the game may clearly terminate on any turn, there is in fact a positive probability that it will never terminate: this is the infinite product $$p = \prod\_{i=1}^\infty \left(1-\frac{1}{2^{(2^i)}}\right).$$
**Questions.** Do we have $p \in \mathbb{R}\setminus \mathbb{Q}$? Is $p$ transcendental?
|
https://mathoverflow.net/users/8628
|
Is $\prod_{i=1}^\infty (1-\frac{1}{2^{(2^i)}})$ transcendental?
|
I can prove that the number actually described by the word problem, which is $$ \prod\_{i=0}^{\infty} \left( 1- \frac{1}{2^{2^i}} \right),$$ is irrational, by a method similar to David Speyer's.
Expanding out the product, we get $$\sum\_{j=0}^{\infty} \frac{(-1)^{t\_j} }{2^j}= 1+\sum\_{j=1}^{\infty} \frac{(-1)^{t\_j} }{2^j} = \sum\_{j=1}^{\infty} \frac{1}{ 2^j} + \sum\_{j=1}^{\infty} \frac{(-1)^{t\_j} }{2^j} = \sum\_{j=1}^{\infty} \frac{1+ (-1)^{t\_j} }{2^j} = \sum\_{j=1}^{\infty} \frac{1+ (-1)^{t\_j} }{2} \cdot 2^{1-j} $$ where $t\_j$ is the number of $1$s in the binary expansion of $j$.
Thus the binary digit in the $j-1$th place is $1$ if $t\_j$ is even and $0$ if $t\_j$ is odd.
Since this sequence is not periodic, the number is irrational.
|
18
|
https://mathoverflow.net/users/18060
|
414214
| 168,952 |
https://mathoverflow.net/questions/414206
|
8
|
It is well known that in $\mathbb{Z}\_2$-valued simplicial cohomology (and other cohomologies)
$$ Sq^1 = \beta\;,$$
where $Sq^1$ is the first Steenrod square and $\beta$ is the Bockstein homomorphism for the short exact sequence
$$ \mathbb{Z}\_2 \overset{\cdot 2}{\rightarrow} \mathbb{Z}\_4 \overset{\mod 2}{\rightarrow} \mathbb{Z}\_2\;.$$
I was wondering whether there is a way to also interpret $Sq^2$ or even higher Steenrod squares in a similar fashion. One could think that instead of
$$\beta = g^{-1} d h^{-1}$$
for a short exact sequence of abelian groups
$$A\overset{g}{\rightarrow} B\overset{h}{\rightarrow} C$$
one could try something like
$$g^{-1}dh^{-1}di^{-1}$$ for an exact sequence
$$A\overset{g}{\rightarrow} B\overset{h}{\rightarrow} C\overset{i}{\rightarrow} D\;.$$
However, if I'm not mistaken, the long exact sequence splits into two short exact sequences, and the above is just the product of the two corresponding Bockstein homomorphisms. E.g., when applied to the long exact sequence
$$ \mathbb{Z}\_2 \overset{\cdot 2}{\rightarrow} \mathbb{Z}\_4 \overset{\cdot 2}{\rightarrow} \mathbb{Z}\_4 \overset{\mod 2}{\rightarrow} \mathbb{Z}\_2$$
we get $\beta^2$ for the short exact sequence given in the beginning, which is trivial.
Crossed module extensions are a more non-trivial analogue to ordinary group extensions which also come with an action of $C$ on $B$. It is known that crossed module extensions give rise to group 3-cocycles in $H^3(BD, A)$ just as ordinary extensions give rise to 2-cocycles. If $D$ acts trivially on $A$, the group 3-cocycle can be used to map a (simplicial) 1-cocycle to a 3-cocycle, and the "higher Bockstein" should generalize this to a map from $i$-cocycles to $i+2$-cocycles. However, I'm not sure how to use the action of $C$ on $B$ to define sich a higher Bockstein.
So I basically have two questions: 1) Is there a non-trivial analogue of a Bockstein operation for some sort of longer exact sequences/crossed module extensions? 2) If yes, is any of these operations equal to $Sq^2$ (in general, or maybe at least when applied to a specific degree)?
|
https://mathoverflow.net/users/115363
|
Analogue of Bockstein for crossed module extensions and higher Steenrod square
|
$\DeclareMathOperator{\Sq}{Sq}\newcommand{\Z}{\mathbb{Z}}$The short version is that every cohomology operation can be interpreted as a Bockstein operator for an "exact sequence" (read: fiber sequence) of grouplike $E\_\infty$ spaces.
Any cohomology operation $\delta:H^\*(-;A)\Rightarrow H^{\*+k}(-;B)$ such as $\Sq^i$ gives a morphism of Eilenberg-MacLane spectra $f\_\delta:HA\to \Sigma^k HB$. You can take the fiber of $f\_\delta$ to obtain a spectrum $F$ which defines a generalized cohomology theory $E\_F^\*:X\mapsto \pi\_{-\*} F^X$, and there is a resulting long exact sequence
$$
\dots\to E\_F^i(X)\to H^i(X;A)\xrightarrow{\delta}H^{i+k}(X;B)\to E\_F^{i+1}(X)\to\dots
$$
If $\delta = \Sq^1$, the fiber is again an Eilenberg-MacLane spectrum (namely $F\simeq H\Z/4$). If the degree of $\delta$ is bigger than $1$, it will have two non-zero homotopy groups, namely $A$ in degree $0$ and $B$ in degree $k-1$.
The connection to the crossed modules you mention is that applying the functor $\Omega^{\infty-1}$ to the fiber of $\Sq^2$ gives rise to a $2$-group, i.e. a homotopy type $X$ whose homotopy groups vanish outside degrees $1$ and $2$. As you mention, these are classified by the two groups $A = \pi\_1(X),B = \pi\_2(X)$, the action of the former on the latter, and a $k$-invariant in $H^3(A;B)$, which together can be packaged into the datum of a crossed module. However, these deloop once if and only if the action is trivial and the $k$-invariant vanishes. It is still possible to find reasonably easy algebraic models for spaces whose homotopy groups vanish outside degrees $k$ and $k+1$ (given by braided ($k=2$) and symmetric ($k\ge 3$) monoidal Picard (every object has a tensor inverse) groupoids), although I do not know a definition of $\Sq^2$ in this language.
Questions in the comments
=========================
1. You already discuss the resulting cohomology operation in the case that the action of $A$ on $D$ is trivial. For the general case, one first construts a natural transformation from $H^1(-;A)$ to local systems: $H^1(-;A)$ are isomorphism classes of $A$-principal bundles, and this natural transformation sends a principal bundle $P$ to $P\times\_A D$. A cohomology class in $H^k(A;D)$ then gives a natural transformation from $H^1(-;A)$ to $H^k$ of this local system, by the same construction as in the trivial case.
2. Yes, for a $(2,3)$-type $X$ the $2$-group $\Omega X$ is split, i.e. the action of $\pi\_1(\Omega X)\cong \pi\_2(X)$ on $\pi\_2(\Omega X)\cong \pi\_3(X)$ is trivial (this can be shown by the Eckmann-Hilton argument) and the $k$-invariant vanishes.
3. For degree $2$ cohomology operations, there is in fact a complete classification (compare [(Co)homology of the Eilenberg-MacLane spaces K(G,n)](https://mathoverflow.net/questions/24754/cohomology-of-the-eilenberg-maclane-spaces-kg-n) and the cited references): operations $H^2(-;A)\to H^4(-;B)$ are given by quadratic functions $q:A\to B$ (i.e. such that $q(x+y) - q(x) - q(y)$ is bilinear and $q(kx) = k^2q(x)$), and the resulting cohomology operation is given by a suitable version of the [Pontryagin square](https://encyclopediaofmath.org/wiki/Pontryagin_square). For $k\ge 3$, operations are in bijection with linear maps from $A\otimes \Z/2$ to $B$ (observe that such a *linear* map is also quadratic by the "freshman's dream"), and the resulting cohomology operation is the composition
$$
H^\*(-;A)\to H^\*(-;A\otimes Z/2)\xrightarrow{\Sq^2} H^{\*+2}(-;A\otimes\Z/2)\to H^{\*+2}(-;B)
$$
The relation to Picard groupoids is a consequence of the [Homotopy hypothesis](https://en.wikipedia.org/wiki/Homotopy_hypothesis), and given a braided Picard groupoid $C$, you can associate to it its abelian group $\pi\_0 C$ of isomorphism classes and the (abelian!) group $\pi\_1 C$ of automorphisms of the unit $1$, together with the map $q: \pi\_0 C\to \pi\_1 C$ which sends $x$ to the composition
$$
1\cong x\otimes x^{-1}\xrightarrow{\beta\_{x,x^{-1}}} x^{-1}\otimes x\cong 1
$$
It's a fun exercise to show that $q$ is quadratic in the above sense. It is not straightforward to give an inverse to this construction, i.e. construct the braided Picard groupoid from the quadratic map; for a reference in the symmetric setting, see [Cegarra, A. M.; Khmaladze, E., Homotopy classification of graded Picard categories, Adv. Math. 213 (2007)](https://doi.org/10.1016/j.aim.2007.01.015).
A chain level representative of $\Sq^2$ (and higher Steenrod squares) can be found in [Ralph M. Kaufmann, Anibal M. Medina-Mardones. Cochain level May-Steenrod operations](https://arxiv.org/abs/2010.02571v4).
|
11
|
https://mathoverflow.net/users/35687
|
414221
| 168,953 |
https://mathoverflow.net/questions/414186
|
8
|
**Definition:** *Highly composite prime gap*
The three composite numbers between the consecutive primes $643$ and $647$ each have at least three distinct prime factors. This is the first occurrence of prime gap of length $> 1$ where each composite number in the gap has at least $3$ distinct prime factors. We call prime gap between $643$ and $647$ as the highly composite prime gap of order $3$. We have the highly composite prime gaps for of order $k$ for $k = 3,4,5,6$ and $7$ as follows:
* $k = 3; p = 643$
* $k = 4; p = 51427$
* $k = 5; p = 8083633$
* $k = 6; p = 1077940147$
* $k = 7; p = 75582271489$
**Question 1**: Are there infinitely many highly composite prime gaps of order $k \ge 3$?
**Question 2**: Given $k$ is there always a highly composite gap of order $k$?
An ordinary linear regression between $k$ and $\log p$ gives a surprisingly strong fit with $R^2 \approx 0.99915$. Although it is based on only six data points, this suggests a relationship of the form $p \sim ab^k$ forsome fixed $a$ and $b$.
**Definition**: *Maximal highly composite gap*
The maximal highly composite gap is defined as a prime gap which is longer than any previous gap and each composite in the gap has at least $3$ distinct prime factors. The longest such gap I have found is of $75$ consecutive composite between the primes $535473480007$ and $535473480083$.
**Question 3**: Are there arbitrarily long prime gaps in which each composite number in the gap has at least three distinct prime factors?
**Note**: This question was [posted in MSE six months ago](https://math.stackexchange.com/questions/4182509/are-there-arbitrarily-long-prime-gaps-in-which-each-number-has-at-least-three-di); it got many votes but not answer hence posting in MO.
|
https://mathoverflow.net/users/23388
|
Are there highly composite prime gaps?
|
Assuming the [prime tuples conjecture](https://en.wikipedia.org/wiki/Dickson%27s_conjecture), all of these questions have affirmative answers. For instance, one can use the Chinese remainder theorem to find $a,b$ such that the tuple
$$ an+b, \frac{an+b+1}{2^2 \times 5}, \frac{an+b+2}{3 \times 7}, \frac{an+b+3}{2 \times 11}, an+b+4$$
(for instance) are an admissible tuple of linear forms (in that they have integer coefficients, and for each prime $p$ there exist a choice of $n$ in which all forms are simultaneously coprime to $p$); concretely, one can take $a=2^2\times 3 \times 5 \times 7 \times 11 = 4620$ and $b=19$. The tuples conjecture then shows that there are infinitely many $n$ in which these forms are all simultaneously prime, giving infinitely many gaps of order $3$ and length $4$. Similarly for other gap orders and lengths.
On the other hand, unconditionally not much can be said. I think at our current level of understanding we cannot even rule out the ridiculous scenario in which every sufficiently large prime gap $(p\_n,p\_{n+1})$ contains a semiprime. (Given that semiprimes are denser than the primes, it is likely that *most* prime gaps contain a semiprime, but it is highly unlikely (and inconsistent with the prime tuples conjecture) that *all* of them do.) The "bounded gaps between primes" technology of Goldston-Yildirim-Pintz, Zhang, Maynard, and Polymath does provide prime gaps of short length, but the method also inevitably produces several almost primes in the vicinity of such gaps, and with our current techniques we cannot prevent one of these almost primes being a semiprime inside the prime gap.
|
12
|
https://mathoverflow.net/users/766
|
414230
| 168,956 |
https://mathoverflow.net/questions/414224
|
0
|
Let $S(x, y, m, n) = \sum\limits\_{i=0}^m \binom{n}{i}x^i y^{n-i}$, where $0 < m < n$. I want to derive the relation between $S(x, y, m, n)$ and $S(x, y, m, n-1)$.
Is there any formulas I can use?
|
https://mathoverflow.net/users/475457
|
Sum of the first m terms of the expansion $(x+y)^n$
|
One can construct the relation using the identity
$$
\binom{n}{i} = \binom{n-1}{i} + \binom{n-1}{i-1}
$$
Then, writing the $i=0$ term separately,
\begin{eqnarray}
S(x,y,m,n) &=& y^n + \sum\_{i=1}^{m} \binom{n-1}{i} x^i y^{n-i}
+ \sum\_{i=1}^{m} \binom{n-1}{i-1} x^i y^{n-i} \\
&=& y^n + y\sum\_{i=1}^{m} \binom{n-1}{i} x^i y^{n-1-i}
+ \sum\_{i=0}^{m-1} \binom{n-1}{i} x^{i+1} y^{n-i-1} \\
&=& y S(x,y,m,n-1) + x\left[ S(x,y,m,n-1) - \binom{n-1}{m} x^m y^{n-1-m} \right] \\
&=& (x+y) S(x,y,m,n-1) \ - \ \binom{n-1}{m} x^{m+1} y^{n-1-m}
\end{eqnarray}
|
2
|
https://mathoverflow.net/users/134299
|
414232
| 168,958 |
https://mathoverflow.net/questions/414233
|
-2
|
Consider the number of integer partitions $p(n)$ of $n$ whose generating function is
$$\sum\_{n\geq0}p(n)\,x^n=\prod\_{k\geq1}\frac1{1-x^k}.$$
Also, the number of partitions into distinct parts $Q(n)$ of $n$ whose genertaing function is
$$\sum\_{n\geq0}Q(n)x^n=\prod\_{k\geq1}(1+x^k).$$
Expand the ratio of these two generating functions so that
$$\sum\_{n\geq0}a(n)x^n=\prod\_{k\geq1}\frac{1+x^k}{1-x^k}.$$
>
> **QUESTION.** Why is $a(n)\equiv 2\,\, (\text{mod}\, 4)$ iff $n$ is a perfect square, for $n\geq1$?
>
>
>
|
https://mathoverflow.net/users/66131
|
Congruence modulo 4 for a generating function leads to perfect squares?
|
$$\frac{1+x^k}{1-x^k}=(1+x^k)(1+x^k+x^{2k}+\dots)=1+2x^k+2x^{2k}+\dots$$
Multiplying all these together and looking at terms contributing to the coefficient of $x^n$, we see that the term will be contributing something divisible by $4$ unless it comes from multiplying a bunch of $1$s together with $2x^{k\cdot d}$ for $k\mid n$. This means $a(n)\equiv 2d(n)\pmod 4$, where $d(n)$ is the number of divisors of $n$. Since $d(n)$ is odd iff $n$ is a square, we get the result.
|
6
|
https://mathoverflow.net/users/30186
|
414235
| 168,959 |
https://mathoverflow.net/questions/361421
|
2
|
(**UPDATED** for rapid decay considerations + new question)
In dimension 2, the Radon transform range theorem states that a *rapidly decaying* (Schwartz) function $g(t,\theta)$ can be represented as a Radon transform of some function $f(x,y)$ (i.e. $g=R[f]$) if and only if, for all integers $n\geq0$
$$P\_n(\theta) := \int\limits\_{-\infty}^{\infty} t^n g(t,\theta) dt$$ is a homogeneous polynomial *of degree* $n$ in $\cos\theta$ and $\sin\theta$. This is often referred to as the *moment* or *Cavalieri conditions*. See e.g. Helgason's book p.5, Lemma 2.2 (2011 ed.) for the property that $P\_n$ must be a homogeneous polynomial of degree $n$.
**Question 1**: If $f$ is a radial function then its Radon transform $g=R[f]$ is known to be independent from $\theta$. Therefore all moments $P\_n(\theta)$ are also independent from $\theta$, in apparent violation of the property that $P\_n$ is a homogeneous polynomial in $\cos\theta, \sin\theta$ of degree $n$ when $n\geq1$. What am I missing?
For example, consider $f(x,y) = e^{-x^2-y^2} / \sqrt\pi$. Then $g(t,\theta) = e^{-t^2}$ which is independent from $\theta$ as expected of radial functions. The first moment is 0 which is NOT a homogeneous polynomial of degree 1, the second moment is $\sqrt\pi/2$ which is NOT a homogeneous polynomial of degree 2, and so forth.
**Question 2:** What happens when the above integral does not converge? Usually this happens when there is no solution to the Radon transform inverse problem, but consider $g(t,\theta) = (1-e^{-1/t^2})/|t|$ which is independent from $\theta$. After calculations, the inversion formula for radial function gives $$f(x,y) = f\_0\!\left(\sqrt{x^2+y^2}\right), \qquad f\_0(r) = \frac2{\pi r^3}\mathfrak D\!\left(\frac 1r\right)$$ where $\mathfrak D(x) := e^{-x^2}\int\_0^x e^{t^2} dt$ is Dawson's function. So there exists $f$ such that $g=R[f]$, and yet
$$
P\_n(\theta) = \int\limits\_{-\infty}^{\infty} t^n \frac{1-e^{-1/t^2}}{|t|} dt
$$
does not converge for $n\geq 2$.
**My partial answer to Q2**: This specific example is not a Schwartz function. Any references to range theorems for non-Schwartz functions appreciated. I found "A Range Theorem for the Radon Transform" (Madych and Solmon, 1988), other suggestions very appreciated.
Thanks!
p.
|
https://mathoverflow.net/users/158671
|
Radon transform range theorem and radial functions
|
To answer Q1: There are trig identities at play. First, 0 is usually accepted under the definition of "homogeneous polynomial" (i.e., it's a polynomial whose coefficients are all zero) so there is no contradiction there. But for the case of the second moment being constant, we have the identity $\sin^2(\theta) + \cos^2(\theta) = 1$, so actually a constant is representable by a homogeneous trig polynomial of degree 2 polynomial, so again, no contradiction.
**Edit:** for Q2, The moment conditions can indeed be violated when the function is not Schwartz. A good reference for this is the paper:
Solmon, D. C. (1987). *Asymptotic formulas for the dual Radon transform and applications*. Mathematische Zeitschrift, 195(3), 321-343.
The main theorem of this paper shows that the inverse Radon transform maps any even Schwartz function $\phi$ over $\mathbb{S}^{d-1}\times \mathbb{R}$ to a $C^\infty$-smooth function on $\mathbb{R}^d$ that decays like $O(\|x\|^{-d})$ as $\|x\|\rightarrow \infty$, i.e., absolutely integrable along hyperplanes, but not necessarily absolutely integrable over all of $\mathbb{R}^d$. Here $\phi$ does not have to satisfy any of the moment conditions, even though the defining integrals are convergent since $\phi$ is Schwartz.
|
2
|
https://mathoverflow.net/users/61024
|
414256
| 168,961 |
https://mathoverflow.net/questions/414261
|
0
|
Let $k,m$ and $\rho$ be positive integers. In "Zur Approximation der Exponentialfunktion und des Logarithmus. Teil II"
Mahler considered the complex integrals $A\_k(x)=\frac1{2i\pi}\int\_{\mathcal C\_0}\frac{e^{zx}\mathrm dz}{\prod\_{h=0}^m(z+k-h)^\rho}$ where $\mathcal C\_0$ is a circle centered in $0$ with radius lesser than $1$. He asserted that $A\_k=\sum\_{l=0}^{\rho-1}\frac{c\_l}{l!}x^l$ with $c\_l=\frac1{2i\pi}\int\_{\mathcal C\_0}\frac{z^{l-\rho}\mathrm dz}{\prod\_{\substack{0\le h\le m\\ h\ne k}}(z+k-h)^\rho}$. I do not manage to prove this. Any answer would be welcome.
|
https://mathoverflow.net/users/33128
|
It is obvious for Mahler, not for me!
|
These are just the details of prets's comment. Note that
\begin{align\*}
\frac{e^{zx}}{z^\rho}&=\sum\_{l\ge0}\frac{(zx)^l}{z^\rho l!}=
\sum\_{l\ge0}\frac{z^{l-\rho}x^l}{l!}
=\sum\_{l=0}^{\rho-1}\frac{z^{l-\rho}x^l}{l!}+\sum\_{l\ge\rho}\frac{z^{l-\rho}x^l}{l!}\\
&=\sum\_{l=0}^{\rho-1}\frac{z^{l-\rho}x^l}{l!}+\sum\_{n\ge0}\frac{z^{n}x^{\rho+n}
}{(\rho+n)!}.
\end{align\*}
The left sum is what you want and the second is an entire function. Then you may apply Cauchy's integral theorem.
|
3
|
https://mathoverflow.net/users/109085
|
414262
| 168,964 |
https://mathoverflow.net/questions/414271
|
4
|
Let $P\_{H}(G, t)$ be the number of vertex colorings of a graph $G$ in $t$ colors that avoid having a monochromatic subgraph $H$. In particular, for $H$ given by a single edge we recover the usual chromatic polynomial $P\_{H}(G, t) = P(G, t)$.
Question: Are there easy proofs that $P\_{H}(G, t)$ is a polynomial for $t \geq 0$ ?
|
https://mathoverflow.net/users/141327
|
Subgraph avoiding colorings
|
Fix a partition onto non-empty color classes (there are finitely many ways to do so, denote by $k$ the number of distinct color classes) without monochromatic $H$. After that, there is $t(t-1)\ldots(t-k+1)$ ways to assign colors. Sum up, you still get a polynomial.
|
6
|
https://mathoverflow.net/users/4312
|
414273
| 168,968 |
https://mathoverflow.net/questions/413996
|
7
|
Recall that a topological space $X$ is *scattered* if and only if every non-empty subset $Y$ of $X$ contains at least one point which is isolated in $Y$. Consider the statement: "Every scattered hereditarily separable space is countable".
In this [book](https://books.google.com.mx/books?id=k8uYhaIfHuAC&pg=PA229&lpg=PA229&dq=scattered+hereditarily+separable+is+countable&source=bl&ots=ys3y82RVM3&sig=ACfU3U33DBeHh6Eg6x44b-s3ivoOsjYN8Q&hl=en&sa=X&ved=2ahUKEwiO5tPO-rb1AhVuD0QIHdD_B00Q6AF6BAgaEAM#v=onepage&q=scattered%20hereditarily%20separable%20is%20countable&f=false) it is mentioned that, under $\mathsf{CH}$, K. Kunen constructed a compact, scattered, hereditarily separable space of size $\aleph\_1$. My question is if a "real" example of a scattered, hereditarily separable, uncountable space is known, or if this statement is independent of $\mathsf{ZFC}$.
|
https://mathoverflow.net/users/146942
|
Scattered hereditarily separable does not imply countable in ZFC
|
As clarified in the comments, the existence of a regular S-space is independent of ZFC, but a "real" $T\_2$ example can be constructed by taking a well-ordering of a set of reals in order type $\omega\_1$ and refining the Euclidean topology by declaring initial segments open. This example is $T\_2$ since it refines a $T\_2$ topology and is still hereditarily separable (any subset has an initial segment that is dense in the Euclidean topology and this initial segment is still dense in the stronger topology). It is not Lindelof since the cover by initial segments has no countable subcover, and it is scattered since any subset has a minimal element which is isolated in that subspace. This example is described in Mary Ellen Rudin's excellent book "Lectures on Set Theoretic Topology" published by the AMS.
|
8
|
https://mathoverflow.net/users/121994
|
414276
| 168,969 |
https://mathoverflow.net/questions/414265
|
11
|
$\DeclareMathOperator\PSU{PSU}$Let $ \PSU\_n $ be the projective unitary group. Let $ A\_m $ be the alternating group on $ m $ letters.
$ A\_5 $ is a maximal closed subgroup of $ PSU\_2 \cong SO\_3(\mathbb{R}) $.
The references in [The finite subgroups of SU(n)](https://mathoverflow.net/questions/17072/the-finite-subgroups-of-sun) show that $ A\_6 $ is a maximal closed subgroup of $ \PSU\_3 $.
The reference <https://arxiv.org/abs/hep-th/9905212> from the same MO question shows that $ A\_7 $ is a maximal closed subgroup of $ \PSU\_4 $. That leads me to ask: Is $ A\_{n+3} $ always a maximal closed subgroup of $ \PSU\_n $?
To see maximality in these first three cases it is enough to realize that these subgroups are all 2-designs (in fact all 3-designs) and are maximal finite.
|
https://mathoverflow.net/users/387190
|
Alternating subgroups of $\mathrm{SU}_n $
|
The comment of @abx is part of a general picture. For $n >7$ the alternating group $A\_{n}$ has no non-trivial complex irreducible character of degree less than $n-1$, so that for $n > 7$, $A\_{n}$ is not isomorphic to any subgroup of ${\rm GL}(n-3, \mathbb{C}).$
Furthermore, if $n > 7$ is also even, then the double cover of $A\_{n}$ (which, by a theorem of I. Schur is a maximal perfect central extension of $A\_{n}$) has no faithful complex irreducible character of odd degree, so certainly not of degree $n-3$ ( as an involution in its centre would be represented by a matrix of determinant $-1$). Hence the double cover of $A\_{n}$ has no non-trivial irreducible complex representation of degree $n-3$, faithful or not. Thus $A\_{n}$ does not embed as an irreducible subgroup of ${\rm PSU}(n-3, \mathbb{C})$ (and it is clear that a maximal closed subgroup has to be irreducible).
(A few points for clarity: for finite groups, all finite dimensional complex representations are equivalent to unitary ones. Also, there is no real distinction between projective representations (in Schur's sense) and genuine representations of covering groups. The fact that the minimal degree of an non-triival irreducible complex representation of $A\_{n}$ is $n-1$ ( for $n > 7$) can be found in almost any text on representation theory of $S\_{n}$ ( eg that of G.D. James)).
Later edit: I am not sure of the smallest degree of a faithful projective (in Schur's sense) complex representation of $A\_{n}$, though I believe this must be well-known to symmetric group experts. However, results such as Zsygmondy's theorem provide a lower bound. If $p$ is a prime less than or equal to $m-2$, then $A\_{m}$ contains a $p$-cycle which is conjugate to all its non-identity powers, from which it follows that $A\_{m}$ can have no non-trivial complex projective representation of degree less than $p-1.$ Hence if $n$ is an even integer such that there is a prime $p$ with $\frac{n}{2} < p < n-2$, we see that $A\_{n}$ has no complex irreducible projective representation of degree less than $\frac{n}{2}$ (and we may find similar bounds for odd $n$).
|
15
|
https://mathoverflow.net/users/14450
|
414287
| 168,972 |
https://mathoverflow.net/questions/414269
|
0
|
In this [question](https://mathoverflow.net/questions/412729/can-we-invoke-almost-supermartingale-theorem-for-deterministic-sequences), there is a proof for deterministic version of "Almost Supermartingale"
**Question: Can we extend [[1](https://mathoverflow.net/questions/412729/can-we-invoke-almost-supermartingale-theorem-for-deterministic-sequences)] as following? If yes, can we prove it?**
>
> Let the non-negative sequences be $\{V\_1^k\}$, $\{V\_2^k\}$, $\{S\_{1,2}^k\}$, and $\{ U\_{1,2}^k \}$ for $k=0,1,2,\ldots$. Let $\alpha\_0, \alpha\_1, \ldots$ and $\beta\_0, \beta\_1, \ldots$ be non-negative scalars satisfying $\sum\_{k=0}^\infty \alpha\_k < \infty$ and $\sum\_{k=0}^\infty \beta\_k < \infty$. Assume
> $$V\_1^{k+1} + V\_2^{k+1} \leq \left( 1 + \alpha\_k \right) V\_1^k + \left( 1 + \beta\_k \right) V\_2^k - S\_{1,2}^k + U\_{1,2}^k$$
> and
> $$\sum\_{i=1}^\infty U\_{1,2}^i < \infty.$$
> Then,
>
>
> 1. $V\_1^k \rightarrow V\_1^\infty$
> 2. $V\_2^k \rightarrow V\_2^\infty$
> 3. $\sum\_{k=0}^\infty S\_{1,2}^k < \infty$.
>
>
>
P.S.: Question: Is there any "name" of such convergence theorem in [[1](https://mathoverflow.net/questions/412729/can-we-invoke-almost-supermartingale-theorem-for-deterministic-sequences)] besides stating non-random version of (almost) supermartingale?
---
ATTEMPT:
We can split the above inequality
\begin{align}
V\_1^{k+1} \leq \left( 1 + \alpha\_k \right) V\_1^k - {\color{red}{\frac{1}{2}}}S\_{1,2}^k + {\color{red}{\frac{1}{2}}}U\_{1,2}^k \tag{1}
\end{align}
and
\begin{align}
V\_2^{k+1} \leq \left( 1 + \beta\_k \right) V\_2^k - {\color{red}{\frac{1}{2}}}S\_{1,2}^k + {\color{red}{\frac{1}{2}}}U\_{1,2}^k. \tag{2}
\end{align}
From [[1](https://mathoverflow.net/questions/412729/can-we-invoke-almost-supermartingale-theorem-for-deterministic-sequences)], we know that both $(1)$ and $(2)$ can have $V\_1^k \to V\_1^\infty$ and $V\_2^k \to V\_2^\infty$. Also, $\sum\_{k=0}^\infty S\_{1,2}^k < \infty$ follows directly from both $(1)$ and $(2)$.
**Does this above approach make sense?**
|
https://mathoverflow.net/users/123235
|
Proof of yet another extension of deterministic variant of "(Almost) Supermartingale" convergence theorem
|
No. E.g., for $j=1,2$ and $k=0,1,\dots$, take $V\_j^k=2+(-1)^{j+k}$, $S\_{1,2}^k=0$, $U\_{1,2}^k=0$, $\alpha\_k=0$, and $\beta\_k=0$.
|
2
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https://mathoverflow.net/users/36721
|
414296
| 168,975 |
https://mathoverflow.net/questions/413805
|
10
|
The Clausen-Scholze theory of condensed mathematics offers an abelian category with enough projective objects that embraces the study of arbitrary locally compact (and Hausdorff) groups. The behaviour of the tensor product is managed by restricting to a subcategory of solid abelian groups within the category of condensed abelian groups and there is a solidification functor that is left adjoint to this inclusion. Putting my first toe in the water, I am restricting to F\_p vector spaces where p is prime: in other words, condensed abelian groups of prime exponent. The Clausen-Scholze theory already provides sufficient in this special case to be able to solve problems that were inaccessible with classical Galois cohomology. My question: is every condensed vector space over a finite field automatically solid? And if not, exactly what would be the advantage of solidification in this context?
|
https://mathoverflow.net/users/124943
|
Are condensed vector spaces over finite fields always solid?
|
Peter Scholze's comment gives a good answer to the main direct question and tells us that the condensed mod p vector space with basis a compact Hausdorff space S is solid if and only if S is finite. The advantages of solidification are going to become more apparent as we use the condensed maths more widely. Solidification is particularly important when using tensor product. One situation where I believe there is no need to invoke solidification is when tensoring a solid vector space (over F\_p) with a finite vector space: such a tensor is automatically solid from the get-go.
|
1
|
https://mathoverflow.net/users/124943
|
414301
| 168,976 |
https://mathoverflow.net/questions/412066
|
7
|
Consider the map
$$f:\mathbb C^2\to\mathbb C^2$$
$$(x,y)\mapsto(x^2y,xy^2)$$
We can view $f$ as induced by the map of monoids $g:\mathbb Z^2\_{\geq 0}\to\mathbb Z^2\_{\geq 0}$ given by the matrix $(\begin{smallmatrix}2&1\cr 1&2\end{smallmatrix})$. Thus $f$ is a map of log schemes.
>
> Is $f$ log smooth?
>
>
>
If I understand Proposition 6.1 from <https://arxiv.org/abs/alg-geom/9406004> correctly, the answer is supposed to be yes, since $\ker g$ and $(\operatorname{coker} g)\_{\mathrm{tors}}$ are both finite.
On the other hand, log smooth maps are supposed to be flat, and flat maps are open. Thus if $f$ were log smooth it would have to be open. But it is not open: $(0,0)$ is in the image of $f$, but $(0,a)$ is not whenever $a\ne 0$.
|
https://mathoverflow.net/users/35353
|
Is $(x^2y,xy^2)$ log smooth?
|
It is smooth. For a log map to have good "fiber bundle properties", one needs more than smoothness. Rather, one needs the relevant maps of monoids to be **exact** in the sense of Kato. A reference is Nakayama--Ogus "Relative rounding in toric and logarithmic geometry" [Mathscinet](https://mathscinet.ams.org/mathscinet-getitem?mr=2740645) [Journal](https://msp.org/gt/2010/14-4/p09.xhtml).
|
0
|
https://mathoverflow.net/users/35353
|
414307
| 168,977 |
https://mathoverflow.net/questions/414164
|
9
|
Let $f\_i=f\_i(x\_1,x\_2,\ldots, x\_n),i=0,1,2, \ldots $ be a family of symmetric polynomials. For the partition $\lambda=(\lambda\_1,\lambda\_2, \ldots, \lambda\_n)$ consider the determinant
$$
D\_\lambda(f)=\left | \begin{array}{lllll}
f\_{\lambda\_1} & f\_{\lambda\_1+1} & f\_{\lambda\_1+2} & \ldots & f\_{\lambda\_1+n-1}\\
f\_{\lambda\_2-1} & f\_{\lambda\_2} & f\_{\lambda\_2+1} & \ldots & f\_{\lambda\_2+n-2}\\
\vdots & \vdots & \vdots & \ldots & \vdots \\
f\_{\lambda\_n-(n-1)} & f\_{\lambda\_n-(n-2)} & f\_{\lambda\_n-(n-3)} & \ldots & f\_{\lambda\_n}
\end{array} \right|.
$$
It is well known (Jacobi−Trudi formulas) that for the elementary symmetric polynomials $e\_i=e\_i(x\_1,x\_2,\ldots, x\_n)$ and for the complete homogeneous symmetric polynomials $h\_i=h\_i(x\_1,x\_2,\ldots, x\_n)$ we have
$$
D\_\lambda(h)=s\_{\lambda}(x\_1,x\_2,\ldots, x\_n) \text{ and } D\_\lambda(e)=s\_{\lambda'}(x\_1,x\_2,\ldots, x\_n),
$$
where $s\_{\lambda}(x\_1,x\_2,\ldots, x\_n)$ is the Schur polynomial and $\lambda'$ is the conjugate partition.
**Question.** Is there a similar expression for $D\_\lambda(p)$ where $p\_i=p\_i(x\_1,x\_2,\ldots, x\_n)$ is the power sum symmetric polynomials?
By direct calculation for $n=2, \lambda\_2\geq 1$ I got that
$$D\_{(\lambda\_1,\lambda\_2)}(p)=-s\_{(\lambda\_1-1,\lambda\_2-1)} V(x\_1,x\_2)^2$$ and for $n=3,\lambda\_3\geq 2$
$$D\_{(\lambda\_1,\lambda\_2,\lambda\_3)}(p)=-\frac{s\_{(\lambda\_1-1,\lambda\_2-1,\lambda\_3-1)}}{s\_{(1,1,1)}} V(x\_1,x\_2,x\_3)^2$$
and so on.
Here $V$ is the Vandermonde determinant.
I hope that must be a nice formula and for any $n$.
|
https://mathoverflow.net/users/40637
|
Determinant connection between Schur polynomials and power sum polynomials
|
I don't know a fully general result, but your pattern for partitions $\lambda$
of length $\leq n$ with $n$-th entry $\lambda\_{n}\geq n-1$ and with $n$
indeterminates persists:
>
> **Theorem 1.** Let $n$ be a positive integer. Let $\lambda=\left( \lambda
> \_{1},\lambda\_{2},\ldots,\lambda\_{n}\right) $ be an integer partition with at
> most $n$ parts. Assume that $\lambda\_{n}\geq n-1$. Consider polynomials in $n$
> indeterminates $x\_{1},x\_{2},\ldots,x\_{n}$. For each nonnegative integer $k$,
> we set
> \begin{align\*}
> p\_{k}:=x\_{1}^{k}+x\_{2}^{k}+\cdots+x\_{n}^{k}.
> \end{align\*}
> (This is the $k$-th power-sum symmetric polynomial in $x\_{1},x\_{2}
> ,\ldots,x\_{n}$ when $k>0$. We have $p\_{0}=n$.) Define the $n\times n$-matrix
> \begin{align\*}
> P:=\left( p\_{\lambda\_{i}-i+j}\right) \_{1\leq i\leq n,\ 1\leq j\leq
> n}=\left(
> \begin{array}
> [c]{cccc}
> p\_{\lambda\_{1}} & p\_{\lambda\_{1}+1} & \cdots & p\_{\lambda\_{1}+n-1}\\
> p\_{\lambda\_{2}-1} & p\_{\lambda\_{2}} & \cdots & p\_{\lambda\_{2}+n-2}\\
> \vdots & \vdots & \ddots & \vdots\\
> p\_{\lambda\_{n}-n+1} & p\_{\lambda\_{n}-n+2} & \cdots & p\_{\lambda\_{n}}
> \end{array}
> \right) .
> \end{align\*}
> Let $\mu=\left( \mu\_{1},\mu\_{2},\ldots,\mu\_{n}\right) $ be the partition
> defined by
> \begin{align\*}
> \mu\_{i}=\lambda\_{i}-\left( n-1\right) \ \ \ \ \ \ \ \ \ \ \text{for each
> }i\in\left\{ 1,2,\ldots,n\right\} .
> \end{align\*}
> (This is indeed a partition, since $\mu\_{n}=\underbrace{\lambda\_{n}}\_{\geq
> n-1}-\left( n-1\right) \geq0$.) Let $s\_{\mu}$ be the corresponding Schur
> polynomial in the $n$ indeterminates $x\_{1},x\_{2},\ldots,x\_{n}$. Furthermore,
> let
> \begin{align\*}
> V\_{n}:=\prod\_{1\leq i<j\leq n}\left( x\_{i}-x\_{j}\right)
> \end{align\*}
> be the Vandermonde determinant. Then,
> \begin{align\*}
> \det P=\left( -1\right) ^{n\left( n-1\right) /2}s\_{\mu}\cdot V\_{n}^{2}.
> \end{align\*}
>
>
>
*Proof.* Let $A\_{\mu}$ be the $n\times n$-matrix
\begin{align\*}
\left( x\_{j}^{\mu\_{i}+n-i}\right) \_{1\leq i\leq n,\ 1\leq j\leq n}=\left(
\begin{array}
[c]{cccc}
x\_{1}^{\mu\_{1}+n-1} & x\_{2}^{\mu\_{1}+n-1} & \cdots & x\_{n}^{\mu\_{1}+n-1}\\
x\_{1}^{\mu\_{2}+n-2} & x\_{2}^{\mu\_{2}+n-2} & \cdots & x\_{n}^{\mu\_{2}+n-2}\\
\vdots & \vdots & \ddots & \vdots\\
x\_{1}^{\mu\_{n}+n-n} & x\_{2}^{\mu\_{n}+n-n} & \cdots & x\_{n}^{\mu\_{n}+n-n}
\end{array}
\right) .
\end{align\*}
It is then well-known that
\begin{equation}
s\_{\mu}=\dfrac{\det\left( A\_{\mu}\right) }{V\_{n}}
.
\label{darij1.eq.slam=frac}
\tag{1}
\end{equation}
Indeed, this is the alternant formula for Schur polynomials. For a proof, see,
e.g., Corollary 2.6.7 in the lecture notes [Darij Grinberg and Victor Reiner,
*Hopf Algebras in Combinatorics*,
arXiv:1409.8356v7](https://arxiv.org/abs/1409.8356v7). (The notations in those
notes are not quite ours. Namely, our matrix $A\_{\mu}$ is the transpose of the
matrix whose determinant is $a\_{\mu+\rho}$ in the notes, whereas our $V\_{n}$
is $a\_{\rho}$ in these notes. Corollary 2.6.7 has to be applied to $\mu$
instead of $\lambda$.)
Let $B$ be the $n\times n$-matrix
\begin{align\*}
\left( x\_{i}^{j-1}\right) \_{1\leq i\leq n,\ 1\leq j\leq n}=\left(
\begin{array}
[c]{cccc}
1 & x\_{1} & \cdots & x\_{1}^{n-1}\\
1 & x\_{2} & \cdots & x\_{2}^{n-1}\\
\vdots & \vdots & \ddots & \vdots\\
1 & x\_{n} & \cdots & x\_{n}^{n-1}
\end{array}
\right) .
\end{align\*}
The Vandermonde determinant formula yields
\begin{align\*}
\det B & =\prod\_{1\leq i<j\leq n}\underbrace{\left( x\_{j}-x\_{i}\right)
}\_{=-\left( x\_{i}-x\_{j}\right) }=\prod\_{1\leq i<j\leq n}\left( -\left(
x\_{i}-x\_{j}\right) \right) \\
& =\left( -1\right) ^{n\left( n-1\right) /2}\underbrace{\prod\_{1\leq
i<j\leq n}\left( x\_{i}-x\_{j}\right) }\_{=V\_{n}}=\left( -1\right) ^{n\left(
n-1\right) /2}V\_{n}.
\end{align\*}
However, we have
\begin{equation}
A\_{\mu}B=P.
\label{darij1.eq.AB=P}
\tag{2}
\end{equation}
(Indeed, for any $i,j\in\left\{ 1,2,\ldots,n\right\} $, the $\left(
i,j\right) $-th entry of the matrix $A\_{\mu}B$ is
\begin{align\*}
\sum\_{k=1}^{n}\underbrace{x\_{k}^{\mu\_{i}+n-i}x\_{k}^{j-1}}\_{\substack{=x\_{k}
^{\mu\_{i}+n-i+j-1}=x\_{k}^{\lambda\_{i}-i+j}\\\text{(since }\mu\_{i}=\lambda
\_{i}-\left( n-1\right) \text{ and}\\\text{thus }\mu\_{i}+n-i+j-1=\lambda
\_{i}-\left( n-1\right) +n-i+j-1=\lambda\_{i}-i+j\text{)}}} & =\sum\_{k=1}
^{n}x\_{k}^{\lambda\_{i}-i+j}\\
& =x\_{1}^{\lambda\_{i}-i+j}+x\_{2}^{\lambda\_{i}-i+j}+\cdots+x\_{n}^{\lambda
\_{i}-i+j}=p\_{\lambda\_{i}-i+j},
\end{align\*}
which happens to be precisely the $\left( i,j\right) $-th entry of the
matrix $P$. Thus, \eqref{darij1.eq.AB=P} follows.)
Now, the two matrices $A\_{\mu}$ and $B$ are square matrices. Hence,
\begin{align\*}
\det\left( A\_{\mu}B\right) & =\underbrace{\det\left( A\_{\mu}\right)
}\_{\substack{=s\_{\mu}V\_{n}\\\text{(by \eqref{darij1.eq.slam=frac})}}
}\cdot\underbrace{\det B}\_{=\left( -1\right) ^{n\left( n-1\right) /2}
V\_{n}}\\
& =s\_{\mu}V\_{n}\cdot\left( -1\right) ^{n\left( n-1\right) /2}V\_{n}=\left(
-1\right) ^{n\left( n-1\right) /2}s\_{\mu}\cdot V\_{n}^{2}.
\end{align\*}
In view of \eqref{darij1.eq.AB=P}, we can rewrite this as
\begin{align\*}
\det P=\left( -1\right) ^{n\left( n-1\right) /2}s\_{\mu}\cdot V\_{n}^{2}.
\end{align\*}
This proves Theorem 1. $\blacksquare$
The claim of Theorem 1 can further be rewritten by observing that (in $n$
indeterminates $x\_{1},x\_{2},\ldots,x\_{n}$) we have
\begin{align\*}
s\_{\lambda}=s\_{\mu}\cdot\left( x\_{1}x\_{2}\cdots x\_{n}\right) ^{n-1}
\end{align\*}
(because the entries of $\lambda$ are the respective entries of $\mu$ plus
$n-1$). The product $x\_{1}x\_{2}\cdots x\_{n}$ can also be rewritten as
$s\_{\left( 1^{n}\right) }$, where $\left( 1^{n}\right) $ is the partition
$\left( 1,1,\ldots,1\right) $ with $n$ entries.
|
4
|
https://mathoverflow.net/users/2530
|
414316
| 168,981 |
https://mathoverflow.net/questions/414302
|
10
|
Let $R$ be a complete local $\mathbf{Z}\_p$-algebra, for some prime $p$. In the 1970 paper [Group schemes of prime order](https://doi.org/10.24033/asens.1186) by Oort and Tate, they write down an explicit finite flat group scheme $G\_R(a, b)$ of rank $p$, for each pair of elements $a,b \in R$ satisfying $ab = p$, and show that every FFGS over $R$ of rank $p$ is isomorphic to one of these.
>
> What are the homomorphisms (of $R$-group schemes) from $G\_R(a, b)$ to $G\_R(a', b')$?
>
>
>
The underlying scheme of $G\_R(a, b)$ is $\{ X : X^p - aX = 0\}$. If we look for homomorphisms which are "linear", $X \mapsto \lambda X$ for some $\lambda$, then (after a bit of unravelling) we conclude that $\lambda$ has to be a point of the $R$-scheme
$$\{ Z : Z(aZ^{p-1} - a') = Z(b - b'Z^{p-1}) = 0\}.$$
(This is a group scheme, but neither finite nor flat over $R$ in general, although it is a FFGS if $\{a, b'\}$ generate the unit ideal of $R$.)
**Are these all the homomorphisms** $G\_R(a, b) \to G\_R(a', b')$?
(The answer is clearly "yes" for $p = 2$. It is also "yes" for $p = 3$ via a deeply nasty polynomial computation.)
|
https://mathoverflow.net/users/2481
|
Homomorphisms between Oort–Tate group schemes
|
Yes. The Tate-Oort description is an equivalence of categories between finite flat group schemes (over a $\Lambda$-scheme $S$ where $\Lambda$ is a certain ring decribed in the paper) and the category of triples $(L , a, b)$ where $L$ is an invertible sheaf over $S$ and $a\in \Gamma(S,L^{\otimes (p−1)})$, $b\in\Gamma(S,L^{\otimes (1− p)})$ satisfy $a\otimes b=w\_p\cdot 1$. The morphisms between $(L,a,b)$ and $(L',a',b')$ are the morphisms of invertible
sheaves $f:L\to L'$, viewed as global sections of $L^{\otimes -1}\otimes L'$, such that $a\otimes f^{\otimes p}=f\otimes a'$ and $b'\otimes f^{\otimes p}=f\otimes b$. In the case where the base is a local scheme, this is exactly what you wrote. All I wrote is almost verbatim from the Tate-Oort paper.
|
8
|
https://mathoverflow.net/users/17988
|
414318
| 168,982 |
https://mathoverflow.net/questions/414292
|
1
|
Very specific question. We work over $\mathbb{C}$, although really just want alg. closed of char. 0.
Suppose that $G$ is an algebraic group and $V$ is a finite-dimensional $G$-module, meaning that we have a comodule map
$$
a:V\to\mathbb{C}[G]\otimes V.
$$
Let $v\in V$ be a non-zero vector; then we may choose a basis $v\_1=v, v\_2,\dots,v\_n$ of $V$, and write
$$
a(v\_1)=\sum\limits\_if\_i\otimes v\_i.
$$
Now it is clear that the ideal $(f\_1-1,f\_2,\dots,f\_n)$ cuts out exactly the subgroup of $G$ given by the stabilizer of $v$ in $G$.
My question is as follows: how could I go about computing the cohomology of the Koszul complex defined by the elements $f\_1-1,f\_2,\dots,f\_n$ in $\mathbb{C}[G]$?
**Edit:** I should add that I have an idea of what the answer 'should' be: let $\mathfrak{g}$ denote the Lie algebra of $G$, and write $H$ for the stabilizer of $v$ in $G$. Write $W\subseteq V$ for the subspace $\mathfrak{g}\cdot v$. Then this is a representation of $H$, so $V/W$ in particular is also a representation of $H$. I expect that the cohomology of the Koszul complex is isomorphic to the ring:
$$
\mathbb{C}[H]\otimes \bigwedge(V/W)
$$
Here I'm viewing the Koszul complex as an operator on the supercommutative superalgebra $\mathbb{C}[G]\otimes\bigwedge V$, where $f\_i$ 'corresponds to' the basis vector $v\_i\in V$.
|
https://mathoverflow.net/users/97652
|
Koszul complex of equations defining a stabilizer
|
See arXiv:1411.7994, Proposition 1.28 for a description of the cohomology dg-algebra of a Koszul complex in the case when the zero locus has codimension smaller than the number of equations.
|
3
|
https://mathoverflow.net/users/4428
|
414319
| 168,983 |
https://mathoverflow.net/questions/414242
|
5
|
The following definition should be standard, but let me state it just in case there is some ambiguity:
If $\mathscr{L}$ is a set of subsets of a set $X$ that is closed under finite unions and intersections and contains $\varnothing,X$ (or more generally, if $\mathscr{L}$ is a distributive lattice with top and bottom elements), let us say that $\mathscr{F} \subseteq \mathscr{L}$ is a **filter** in $\mathscr{L}$ provided it contains $X$ (the top element), is closed under finite intersections (meets), and is an up-set (meaning that if $A \subseteq B$ with $A \in \mathscr{F}$ and $B \in \mathscr{L}$ then in fact $B \in \mathscr{F}$). Let us say that such a filter is **proper** iff it does not contain $\varnothing$; and that it is an **ultrafilter** iff it is proper and maximal for inclusion among proper filters.
If $\operatorname{Ult}(\mathscr{L})$ denotes the set of ultrafilters in $\mathscr{L}$, then for every $A\in\mathscr{L}$ we define a set $Z(A) := \{\mathscr{U}\in\operatorname{Ult}(\mathscr{L}) : \mathscr{U} \ni A\}$ of ultrafilters containing $A$ as an element. It is almost trivial that $Z(A\cap B) = Z(A) \cap Z(B)$, and it is also true, though slightly less trivial, that $Z(A\cup B) = Z(A) \cup Z(B)$ (sketch of proof: the inclusion $\supseteq$ is clear, so let us see $\subseteq$: if $\mathscr{U} \ni A\cup B$ and $\mathscr{U} \not\ni A$ then the filter generated by $\mathscr{U}$ and $\{A\}$ contains $\varnothing$, so $A\cap U = \varnothing$ for some $U\in\mathscr{U}$, and then $\mathscr{U} \ni (A\cup B)\cap U = B\cap U \subseteq B$ so $\mathscr{U}\ni B$).
In particular, the $Z(A)$ form a basis of closed sets for a topology on $\operatorname{Ult}(\mathscr{L})$ called the **Zariski topology**. (Digression: note that they *also* form a basis of open sets for another topology, the *Stone topology*; I mention this in passing, because this confused the hell out of me: part of the confusion comes from the fact that, if $\mathscr{L}$ is actually a Boolean algebra, — e.g., if we consider all subsets of $X$, or more generally the clopen subsets of a topological space $X$, — then the complement of $Z(A)$ is $Z(X\setminus A)$ so the two topologies coincide.)
As an example, if $X$ is a topological space and $\mathscr{Z}$ is the lattice of zero-sets of continuous real-valued functions on $X$ (“**z-sets**”), then $\operatorname{Ult}(\mathscr{Z})$, with its Zariski topology, is the Stone-Čech compactification of $X$ (Gillman & Jerison, *Rings of Continuous Functions* (1960), points (a) and (b) in the proof of theorem 6.5).
This example has long bothered me because z-sets seem to make a fairly incongruous appearance, and I wondered why we don't take closed sets instead, which seem more “fundamental”, and what happens if we do. Let me ask precisely that:
**Question:** if $X$ is a topological space, $\mathscr{C}$ is the lattice of closed sets of $X$, can we better describe the space $\operatorname{Ult}(\mathscr{C})$ of ultrafilters in $\mathscr{C}$, endowed with its Zariski topology? Does it have a name? What is “wrong” with it that makes it less interesting than the Stone-Čech compactification?
(Note that if $X$ is metric, then closed sets and z-sets coincide, so the above space *is* the Stone-Čech compactification.)
Maybe assume that $X$ is $T\_1$, which ensures that we have a continuous map $X \to \operatorname{Ult}(\mathscr{C})$ taking $x\in X$ to the set of $F \subseteq X$ closed such that $F\ni x$.
**Remark:** For the description of the space of ultrafilters of *open* sets (with the Stone topology), see [this answer on MSE](https://math.stackexchange.com/questions/592357/can-the-ultrafilters-in-the-poset-of-open-subsets-be-made-into-a-topological-spa/4361689#4361689), where I prove that it is the “Gleason space” of $X$.
|
https://mathoverflow.net/users/17064
|
Ultrafilters of closed sets
|
The construction you describe when $\mathscr{C}$ consists of all closed sets of $X$ is known as the Wallman compactification of $X$. I'll denote if $\omega(X)$. It is due to Wallman; *Lattices and topological spaces*, Ann. Math. 39 (1938) 112-126.
Of course some sort of techincal assumption is required.
>
> Let $X$ be a $T\_1$ space. Then $\omega(X)$ is a compact $T\_1$ space containing $X$ as a dense subspace. Moreover it has the property that every continuous map $X\rightarrow K$ into a compact Hausdorff space $K$ extends over $\omega(X)$. The space $\omega(X)$ is Hausdorff if and only if $X$ is normal, and in this case $\omega(X)\cong\beta(X)$.
>
>
>
Shanin later generalised Wallman's construction; *On special extensions of topological spaces*, Dokl. SSSR 38 (1943) 6-9, *On separation in topological spaces*, Dokl. SSSR 38 (1943) 110-113, *On the theory of bicompact extensions of topological spaces*, Dokl. SSSR 38 (1943) 154-156. The compactifications that Shanin constructed allowed for the ultrafilters to come from more general lattices $\mathscr{L}$ of closed subsets of $X$. Of course at the expense of added assumptions: $\mathscr{L}$ is required to be a so-called *$T\_1$-base* for the closed subsets of $X$. Denote by $\omega(X;\mathscr{L})$ the Wallman-Shanin compactification built using the $T\_1$-base $\mathscr{L}$.
Here are some examples to convince you that these compactifications are interesting.
>
> 1. $X$ is locally compact $T\_2$ and $\mathscr{L}$ consists of all $(i)$ compact subsets of $X$, and $(ii)$ all closed subsets $A\subseteq X$ for which there is a compact $K\subseteq X$ with $A\cup K=X$. Then $\omega(X;\mathscr{L})$ is the Alexandroff compactification of $X$.
>
>
>
>
> 2. $X$ is Tychonoff and $\mathscr{L}=\mathscr{Z}(X)$ is the collection of zero sets. Then $\omega(X;\mathscr{L})\cong\beta(X)$, as you have recognised.
>
>
>
>
> 3. $X$ is rim-compact $T\_2$ and $\mathscr{L}$ is the set of all finite intersections of regularly closed sets with compact boundaries. Then $\omega(X;\mathscr{L})=\mathfrak{f}(X)$ is the Freudenthal compactification of $X$.
>
>
>
|
4
|
https://mathoverflow.net/users/54788
|
414329
| 168,985 |
https://mathoverflow.net/questions/413705
|
1
|
Let $(x\_{n})\_{n=1}^\infty$ be a bounded sequence in a Banach space $X$. We set
$$\textrm{ca}((x\_{n})\_{n=1}^\infty)=\inf\_{n}\sup\_{k,l\geq n}\|x\_{k}-x\_{l}\|.$$
Then $(x\_{n})\_{n=1}^\infty$ is norm-Cauchy if and only if $\textrm{ca}((x\_{n})\_{n=1}^\infty)=0$.
Let $X$ be a Banach space with a basis $(x\_{n})\_{n=1}^\infty$. Recall that $(x\_{n})\_{n=1}^\infty$ is boundedly complete if for every scalar sequence $(a\_{n})\_{n=1}^{\infty}$ with $\sup\_{n}\|\sum\_{i=1}^{n}a\_{i}x\_{i}\|<\infty$, the series $\sum\_{n=1}^{\infty}a\_{n}x\_{n}$ converges.
Let $(x\_{n})\_{n=1}^\infty$ be a basis for a Banach space $X$. We set
$$\textrm{bc}((x\_{n})\_{n=1}^\infty)=\sup\Big\{\textrm{ca}((\sum\_{i=1}^{n}a\_{i}x\_{i})\_{n=1}^\infty)\colon \|\sum\_{i=1}^{n}a\_{i}x\_{i}\|\leq 1, \forall n\Big\}.$$
Clearly, $(x\_{n})\_{n=1}^\infty$ is boundedly complete if and only if $\textrm{bc}((x\_{n})\_{n=1}^\infty)=0$.
We consider some classical non-boundedly complete bases as follows:
1. $\operatorname{bc}((e\_{n})\_{n})=1$, where $(e\_{n})\_{n}$ is the unit vector basis of $c\_{0}$.
2. $\operatorname{bc}((s\_{n})\_{n})=1$, where $(s\_{n})\_{n}$ is the summing basis of $c\_{0}$.
3. $\operatorname{bc}((e\_{n})\_{n=0}^{\infty})=2$, where $(e\_{n})\_{n=0}^{\infty}$ is the unit vector basis of $c$ ($e\_{0}=(1,1,1,\ldots)$).
4. $\operatorname{bc}((e\_{n})\_{n})=1$, where $(e\_{n})\_{n}$ is the unit vector basis of the James space $\mathcal{J}$.
Question 1. $\textrm{bc}((x\_{n})\_{n=1}^\infty)\in \{0,1,2\}$ for every basis $(x\_{n})\_{n}$ ?
Since $c\_{0}, C[0,1]$ or $L\_{1}[0,1]$ has no boundedly complete basis, we have the following natural questions:
Question 2. $\textrm{bc}((x\_{n})\_{n=0}^\infty)=$ ?, where $(x\_{n})\_{n=0}^\infty$ is the Faber-Schauder basis for $C[0,1]$ (See I. Singer, Bases in Banach spaces I, pp. 11 for the Faber-Schauder basis).
Question 3. $\textrm{bc}((h\_{n})\_{n=1}^\infty)=$ ?, where $(h\_{n})\_{n=1}^\infty$ is the Haar basis for $L\_{1}[0,1]$.
PS: We have proved that if $(x\_{n})\_{n}$ is an $1$-unconditional basis, then $\textrm{bc}((x\_{n})\_{n=1}^\infty)=0$ or $1$.
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https://mathoverflow.net/users/41619
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Quantifications of boundedly complete bases
|
This is an answer to Dongyang's query in the comments. I show $\text{bc}(x\_n)\ge 1$ if $(x\_n)$ is a monotone basis for $X$ that is not boundedly complete. Let $(x\_n^\*)$ be the functionals biorthogonal to $(x\_n)$ and let $Y$ be the closed linear span in $X^\*$ of $(x\_n^\*)$, so that $(x\_n^\*)$ is a monotone basis for $Y$. By monotonicity of $(x\_n)$, the natural evaluation map from $X$ into $Y^\*$ is an isometric isomorphism and, since $(x\_n)$ is not boundedly complete, this map is not surjective. Therefore we can regard $X$ as being a proper subspace of $Y^\*$. Thus there is a unit vector $y^\*$ in $Y^\*$ whose distance to $X$ is arbitrarily close to one (say, larger than $1-\epsilon$).
You can write $y^\*=\sum a\_n x\_n$ with the understanding that the convergence of the sum is in the weak$^\*$ topology on $Y^\*$. By monotonicity, for every $N$,
$\|\sum\_{n=1}^N a\_n x\_n\| \le 1 = \|y^\*\|$. On the other hand, for every $N$, $\|y^\* - \sum\_{n=1}^N a\_n x\_n\|$ is at least as large as the distance from $y^\*$ to $X$, which implies that for every $N$, $\|\sum\_{n=N}^\infty a\_n x\_n\| > 1-\epsilon$. Now for any fixed $N$, $\sum\_{n=N}^M a\_n x\_n$ converges weak$^\*$ to $\sum\_{n=N}^\infty a\_n x\_n$ and hence for large $M$,
$\|\sum\_{n=N}^M a\_n x\_n\| > 1-\epsilon$. From this it follows that
$\text{bc}(x\_n)\ge 1$.
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0
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https://mathoverflow.net/users/2554
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414331
| 168,986 |
https://mathoverflow.net/questions/414313
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0
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If we know that there are two vectors $x,y\in\mathbb{R}^d$ satisfying
\begin{equation}
\|x\|\ge c\_1 \|y\|, \quad x^Ty\ge c\_2\|x\|^2,
\end{equation}
where $c\_1>0$ and $c\_2>0$ are some given constants, can we always find a positive definite matrix $M\in\mathbb{R}^{d\times d}$ such that
\begin{equation}
x=My?
\end{equation}
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https://mathoverflow.net/users/178204
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Find a matrix transformation for 2 vectors with conditions
|
Yes. We only need $x^T y > 0$, which is implied by your conditions if neither x nor y is zero. Let $s$ be the positive square root of $x^T y$, and $A$ a $d\times d$ real matrix who's first column equals $\frac1s x$. Let the remaining columns be a basis for the codimension-$1$ space perpendicular to y. Let $e\_1$ be the first standard basis vector.
As x is not perpendicular to y, $A$ is non-singular, and $A \left(s e\_1\right) = x$.
Also $A^T y = \left(\frac1s x^T y, 0, \dots, 0\right)^T = s e\_1$.
Therefore $A A^T y = A \left(s e\_1 \right) = x$. We let $M = A A^T$, which is positive definite.
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2
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https://mathoverflow.net/users/474034
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414334
| 168,987 |
https://mathoverflow.net/questions/414320
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8
|
In [1, example C.1.2.8], a locale $Y$ (dense in another locale
$X$) without any point is given. I fail to understand the point
of such point-less locale - **Why can't we identify those as the
trivial locales, and what's so great about considering locales
that have no points?**
>
> Anyway, here's the construction of $X$ and $Y$ (taken from
> [1]). Let $A$ be an uncountable nonempty set (e.g. $\mathbb{R}$)
> (equipped with the discrete topology), and let $X$ be the set of
> all functions $\mathbb{N} \to A$, equipped with the Tychonoff
> topology. For each $a \in A$, let $X\_a$ be the subspace $\{f \in
> X \,|\, a \in im(f)\}$, and let $$ Y = \bigcap\_{a\in A} X\_{a}.$$
> Now the point set $Y\_p$ of $Y$ is empty because there is no onto
> map from $\mathbb{N}$ to $\mathbb{R}$.
>
>
>
In [2, section 5], Johnstone demonstrates why considering such
locales could be useful. The main argument is that *topoi* are
nice things to consider. However, at the point of writing, the
(external) applications of topos theory seem lacking. Hopefully
the situation has changed in mathematics in recent years. Thus
the second question: **How does the consideration of pointless
locales help topos theory, and how does that in turn applies
(externally) to mathematics?**
### Reference
* [1] Sketches of an Elephant: A Topos Theory Compendium [Peter
T. Johnstone]
* [2] The point of pointless topology-[Peter T. Johnstone]
|
https://mathoverflow.net/users/124549
|
What's the point of a point-free locale?
|
A good answer to both questions is provided by the following [variant of the Gelfand duality for commutative von Neumann algebras](https://mathoverflow.net/questions/23408/reference-for-the-gelfand-duality-theorem-for-commutative-von-neumann-algebras/360088#360088),
which shows that the following categories are equivalent:
* The category CSLEMS of [compact strictly localizable enhanced measurable spaces](https://mathoverflow.net/questions/20740/is-there-an-introduction-to-probability-theory-from-a-structuralist-categorical/20820#20820);
* The category HStonean of hyperstonean spaces and open maps.
* The category HStoneanLoc of hyperstonean locales and open maps.
* The category MLoc of [measurable locales](https://ncatlab.org/nlab/show/measurable+locale), defined as the full subcategory of the category of locales consisting of complete Boolean algebras that admit sufficiently many continuous valuations.
* The opposite category CVNA^op of commutative von Neumann algebras,
whose morphisms are normal \*-homomorphisms of algebras in the opposite direction.
The first category, despite the rather complicated name,
is essentially *the* correct category for measure theory:
it incorporates equality almost everywhere,
a (generalized) σ-finiteness property,
and an abstract variant of the Radon measure property,
which eliminate pathological measurable (and measure) spaces
for which some of the most basic theorem of measure theory (such as the Riesz representation theorem or the Radon–Nikodym theorem) fail.
Of particular interest is the fourth category MLoc of mesurable locales.
It is a full subcategory of the category of locales,
which quite interesting: it demonstrates that both point-set general topology (as implemented by the category of topological spaces) and point-set measure theory (as implemented by the above category CSLEMS) are a part of pointfree general topology, implemented by *full* subcategory of the category of locales.
These parts (i.e., general topology and measure theory) are almost disjoint: locales corresponding to topological spaces are spatial, i.e., have enough points. On the other hand, points in a measurable locale are in a bijective correspondence with atoms in the original measure space.
In particular, atomless measure spaces (i.e., what is typically used in practice) correspond to locales that have no points at all.
Returning to topos theory: working in the topos of sheaves of sets on a measurable locale amounts to doing ordinary mathematics in *measurable families* over a measurable space. For example, doing internal linear algebra in such a topos corresponds to working with measurable vector bundles, etc.
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15
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https://mathoverflow.net/users/402
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414338
| 168,989 |
https://mathoverflow.net/questions/412686
|
11
|
I almost expect the answer to this question to be no, but I can't find it explicitly said anywhere.
Given a formal group law $f$ of height $n$ over a perfect field $k$ of characteristic $p$, we can construct the Morava E-theory $E(n)$. The resulting cohomology theory is
$$E(n)^m (X) = MP^m(X)\otimes \_{MP(\*)} R$$
where $R = W(k)[[v\_1,...,v\_{n-1}]]$ is the Lubin-Tate ring of $f$ and $MP$ is complex cobordism. Also
$$ \pi\_\bullet (E) = E(n)^\bullet (\*) = R[\beta, \beta^{-1}]$$
so at least the homotopy groups of the spectrum are independent of our choice of formal group law. This suggests that maybe the spectrum is the same, and the difference is the multiplicative structure on it. I think my confusion also might come from the fact that people usually refer to it as "Morava E-theory" rather than "a Morava E-theory", somewhat implying uniqueness.
However, the main problem I see with this is that the map $MP(\*) \to R$ from the Lazard ring classifying the universal deformation of $f$ (which makes $R$ a $MP(\*)$-module) will be different for different formal group laws, so I would expect different values (as a graded abelian group) of the cohomology theory depending on the formal group chosen.
Can someone please link an appropriate reference?
|
https://mathoverflow.net/users/170467
|
Does the spectrum of Morava E-theory depend only on height?
|
Here's an argument that Eric Peterson and I came up with showing that the homotopy type of Morava $E$-theory only depends on the choice of perfect char $p$ field $k$ and the height $n<\infty$.
$\textbf{Lemma}$: Let $R$ be a ring with two Landweber exact formal group laws $e,f:\text{Laz}\rightarrow R$ and let $E$ and $F$ be the spectra corresponding to these two formal group laws under the Landweber exact functor theorem. Then $E$ and $F$ have the same homotopy type if there is a ring extension $u: R\rightarrow S$ which is split as an $R$-module map, and such that the formal group laws $u\circ e$ and $u\circ f$ are isomorphic and Landweber exact.
$\textbf{Proof}$: First we show that $E\simeq F$ if the map $\eta:E\_\*\rightarrow F\_\*E$ (induced by $1\wedge id: \mathbb{S}\wedge E\rightarrow F\wedge E$) is split (as a map of $F\_\*$-modules). Note that $E\_\*$ is canonically an $F\_\*$ module by the \textit{equality} $E\_\*=R=F\_\*$. We claim that $F\_\*E$ is flat over $F\_\*$. That boils down to interpreting Landweber exactness as flatness of maps to $\mathcal{M}\_{fg}$, noting that flatness is preserved under pullback, and that (Spec of) $E\_\*$, $F\_\*$ $F\_\*E$ and $\mathcal{M}\_{fg}$ fit into a pullback square (with $\mathcal{M}\_{fg}$ in the bottom right and Spec of $F\_\*E$ in the top left). It follows that $F\wedge E$ represents the cohomology theory $X\mapsto F^\*X\otimes\_{F\_\*}F\_\*E$, and then Brown representability (and Strickland's result about the absence of phantom maps between Landweber exact spectra) produces a map $\phi: \text{Mod}\_{F\_\*}(F\_\*E,F\_\*)\rightarrow F^\*E$. Now suppose $s$ is a splitting of $\eta$. Then there is a map $\phi(s):E\rightarrow F$ which on homotopy groups (set $X=\text{pt}$ above) is the composite $s\circ\eta$ which is the identity map (recall that $E\_\*$ and $F\_\*$ are canonically isomorphic to $R$), so $E\simeq F$.
Now we show that $\eta:E\_\*\rightarrow F\_\*E$ is split if there is a ring extension $u: R\rightarrow S$ which is split as an $R$-module map, and such that the formal group laws $u\circ e$ and $u\circ f$ are isomorphic and Landweber exact. Write $g:=u\circ e$ and $h:=u\circ f$. Assume they are isomorphic and Landweber exact. Let $G$ and $H$ be the spectra associated to $g$ and $h$. Then we have the following commutative diagram of $F\_\*(=R)$-modules
$\require{AMScd}$
\begin{CD}
S=G\_\* @>1\otimes 1\otimes id>> H\_\*G= S\otimes\_{MU\_\*}MU\_\*MU\otimes\_{MU\_\*}S\\
@A u A A @AA u\otimes id\otimes u A\\
R=E\_\* @>>1\otimes 1\otimes id> F\_\*E=R\otimes\_{MU\_\*}MU\_\*MU\otimes\_{MU\_\*}R
\end{CD}
Since the right vertical map is split (as a map of $F\_\*$-modules), splitting the bottom horizontal map (as a map of $F\_\*$-modules) is equivalent to splitting the diagonal map (as a map of $F\_\*$-modules). Moreover, the top horizontal map is split (as an $S=H\_\*$-algebra (and hence $F\_\*$-algebra) map in fact) by first choosing an isomorphism of $g$ and $h$, which induces an $H\_\*$-algebra isomorphism $H\_\*G\simeq G\_\*G$ and then using the multiplication map $G\wedge G\rightarrow G$ to split $G\_\*\rightarrow G\_\*G$. Therefore splitting the diagonal map is equivalent to splitting the left vertical map (as a map of $F\_\*$-modules).
$\textbf{End of proof}.$
Now we apply the lemma to show that at height $n<\infty$ the homotopy type of Morava $E$-theory depends only on the choice of a perfect characteristic $p$ field. In other words, let $k$ be a perfect characteristic $p$ field and let $E\_1(n)$ and $E\_2(n)$ be two Morava $E$-theory spectra corresponding to two arbitrary height $n< \infty$ formal group laws over $k$. Then we'll show that as spectra, $E\_1(n)\simeq E\_2(n)$.
It suffices to check the conditions of the lemma. Let $\overline{k}$ be the algebraic closure of $k$. Let $R$ and $S$ be the (2-periodic) Lubin-Tate deformation rings $W(k)[[u\_1,...,u\_{n-1}]][\beta^\pm]$ and $W(\overline{k})[[u\_1,...,u\_{n-1}]][\beta^\pm]$. Let $\tilde{e}$ and $\tilde{f}$ be formal group laws over $R$ universally deforming $e\_1$ and $e\_2$. Then $\tilde{e}\_1$ and $\tilde{e}\_2$ are Landweber exact and the corresponding spectra are the Morava $E$-theory spectra $E\_1(n)$ and $E\_2(n)$. Let $u:R\rightarrow S$ be the map that extends coefficients by $W(k)\rightarrow W(\overline{k})$, sends $u\_i$ to $u\_i$, and $\beta$ to $\beta$. Then the formal group laws $\tilde{g}:=u\circ \tilde{e}\_1$ and $\tilde{h}:=u\circ \tilde{e}\_2$ over $S$ are isomorphic and Landweber exact, since they universally deform the isomorphic formal group laws $g:u\circ e\_1$ and $h:u\circ e\_2$ over $\overline{k}$. Finally, the map $u:R\rightarrow S$ is split as an $R$-module map, because the map $W(k)\rightarrow W(\overline{k})$ is split as a $W(k)$-module map (See (ill Sawin's comments at [Splitting the Witt vectors of $\overline{\mathbb{F}\_p}$](https://mathoverflow.net/questions/413773/splitting-the-witt-vectors-of-overline-mathbbf-p)))
|
11
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https://mathoverflow.net/users/163893
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414342
| 168,991 |
https://mathoverflow.net/questions/414293
|
8
|
Let $\mathcal C=(\mathcal O, \mathcal M)$ be a category internal to topological spaces. Thus $\mathcal O$ and $\mathcal M$ are topological spaces: the space of objects and the space of morphisms respectively. These spaces come endowed with structure maps: $i \colon \mathcal O \to \mathcal M$, $s, t\colon \mathcal M\to \mathcal O$, and $c\colon \mathcal M \times\_{\mathcal O} \mathcal M \to \mathcal M$, which satisfy well-known identities. Here $\mathcal M \times\_{\mathcal O} \mathcal M$ is the pullback $\mathcal M\xrightarrow{s} \mathcal O \xleftarrow{t} \mathcal M$.
Let Top be the category of topological spaces. Feel free to use your favorite ``convenient'' category of topological spaces. My question is
>
> What is the correct notion of a topological functor $F\colon \mathcal C\to $Top ?
>
>
>
Intuitively I feel that the following is a reasonable notion:
Proposed definition: A functor $F\colon \mathcal C\to $Top consists of a space over $\mathcal O$, that I will denote by $F\to \mathcal O$. This map is not necessarily a fibration. The fiber at a point $x\in \mathcal O$ corresponds to the value of $F$ at $x$. Furthermore, there has to be a structure map
$$F\times\_{\mathcal O} \mathcal M \to F$$
which satisfies certain more or less evident relations.
>
> Is this notion of a topological functor in the literature? Does it have a name? Where can I read about it?
>
>
>
Disclosure: I have actually used this definition in a couple of papers, but in an ad hoc manner. I want to know if other people used it, and if it has been developed systematically.
The following question is not of immediate practical consequence to me, but it is presumably important for the general picture:
>
> Is it possible to reinterpret this definition as an internal functor from $\mathcal C$ to Top, perhaps using the language of higher categories?
>
>
>
What I really want to know is whether people have studied the homotopy theory of such functors.
>
> Is there a projective model structure on the category of functors $\mathcal C \to $Top, where weak equivalences/fibrations are fiber homotopy equivalences/fibrations over $\mathcal O$? Has anyone studied homotopical notions, such as homotopy limits and colimits, derived Kan extensions, and so forth, in this setting?
>
>
>
|
https://mathoverflow.net/users/6668
|
What is the right notion of a functor from an internal topological category to topological spaces?
|
The definition you propose is that of a [$\mathsf{Top}$-internal diagram](https://nlab-pages.s3.us-east-2.amazonaws.com/nlab/show/internal+diagram) in $\mathcal{C}$. It comes from viewing categories as many-object monoids, and functors/presheaves on categories as the many-object generalisation of left or right modules over a monoid (and then internalising these notions to $\mathsf{Top}$). Similarly, the bimodule version of this is called an "[internal profunctor](https://nlab-pages.s3.us-east-2.amazonaws.com/nlab/show/internal+profunctor)".
You can find more about internal diagrams in the following references:
* Andrade, [From manifolds to invariants of Eₙ-algebras](https://arxiv.org/abs/1210.7909), Section 2.6.
* Borceux, [Handbook of Categorical Algebra, Volume
I](https://www.cambridge.org/br/academic/subjects/mathematics/logic-categories-and-sets/handbook-categorical-algebra-volume-1?format=PB),
Section 8.2.
* Jacobs, [Categorical Logic and Type Theory](https://people.mpi-sws.org/%7Edreyer/courses/catlogic/jacobs.pdf), Section 7.4.
* Johnstone, [Sketches of an Elephant](https://nlab-pages.s3.us-east-2.amazonaws.com/nlab/show/Sketches+of+an+Elephant), Volume 2, Section B.2.3.
* Johnstone, [Topos Theory](https://books.google.com.br/books?id=rLrFAgAAQBAJ&pg=PA49&lpg=PA49&dq=%22internal+diagram%22&source=bl&ots=J89ATihHOP&sig=ACfU3U3-VzKj5AnTfnTAXGlWV5-qKFooqw&hl=en&sa=X&ved=2ahUKEwiHjemp8MH1AhVZq5UCHXIrCP0Q6AF6BAggEAM#v=onepage&q=%22internal%20diagram%22&f=false), Section 2.2.
* Tomašić, [A topos-theoretic view of difference algebra](https://arxiv.org/abs/2001.09075), Section I.4.
* Wong, [The Grothendieck Construction in Enriched, Internal and ∞-Category Theory](https://digital.lib.washington.edu/researchworks/handle/1773/44365), Section 4.4. (This one treats the non-Cartesian case).
* Wong, [Smash Products for Non-cartesian Internal Prestacks](https://arxiv.org/abs/1907.09666), paper version of the above.
* [MO 199237](https://mathoverflow.net/questions/199237) and [MO 263927](https://mathoverflow.net/questions/263927).
---
(Incidentally, a second possible (non-standard) such definition would be via [locally internal categories](https://nlab-pages.s3.us-east-2.amazonaws.com/nlab/show/locally+internal+category), the "locally small version" of internal categories: we could pick a nice category of spaces $\mathsf{Spc}$ (one that is locally Cartesian closed; in particular [that of compactly generated Hausdorff spaces isn't](https://link.springer.com/article/10.1007/BF00872904); see also [this nLab page](https://nlab-pages.s3.us-east-2.amazonaws.com/nlab/show/nice+category+of+spaces)), and then view it as a locally $\mathsf{Spc}$-internal category, via [self-internalisation](https://mathoverflow.net/a/390348). Then a topological functor from $\mathcal{C}$ to $\mathsf{Spc}$ would be a locally $\mathsf{Spc}$-internal functor from the externalisation of $\mathcal{C}$ (i.e. $\mathcal{C}$ viewed as a locally internal category) to this locally $\mathsf{Spc}$-internal version of $\mathsf{Spc}$. I don't think these two definitions agree (**Edit:** *They actually do! See the comments below*), though in any case this second definition is definitely a hassle to work with! :/)
|
13
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https://mathoverflow.net/users/130058
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414345
| 168,992 |
https://mathoverflow.net/questions/414340
|
6
|
Let $V$ be the vertices of a regular $p$-agon in the plane, where $p$ is prime, and let $C$ be a set. Given two maps $f,g:V\rightarrow C,$ I believe it is true that either
(a) $f=g\circ r$ for some rotation $r:V\rightarrow V,$ or
(b) there is $h:V\rightarrow C$ such that for all $v\in V,$ $h(v)\neq f(v),$ but for all rotations $r:V\rightarrow V,$ there exists $v\in V$ (depending on $r$) such that $h(v)=g\circ r(v).$
Note these are mutually exclusive. (*Aside*: Calling the maps "colorings" of $V,$ I am saying if $f$ and $g$ are colorings, which are not just rotations of each other, then there is another coloring $h$ that doesn't match $f$ anywhere, but matches every rotation of $g$ somewhere. Perhaps this sounds more intuitive.)
My proof is of course elementary but involves some case by case analysis, so I would like to ask if it follows from known results, or has an easy proof that I have missed. I don't know if it's essential that $p$ be prime, but I need that for my proof.
|
https://mathoverflow.net/users/313687
|
A question about colorings of the vertices of a p-agon, where p is prime
|
We may identify $V$ with the field $\mathbb{F}\_p$ of size $p$, then denoting $g\_k(x):=g(x+k)$ for $k\in \mathbb{F}\_p$ the questions reads as:
if $f(x)\not\equiv g\_k(x)$ for each $k\in \mathbb{F}\_p$, prove that there exists $h(x)$ such that $h(x)\ne f(x)$ for all $x$; but for each $k$ there exists $x(k)$ for which $h(x(k))=g\_k(x(k))$.
If $g\equiv c$ is a constant coloring, we find a vertex $v\in \mathbb{F}\_p$ for which $f(v)\ne c$ and put $h(v)=c$, and put $h(x)\in C\setminus \{f(x)\}$ for $v\ne x$ arbitrarily (this is possible since $|C|\geqslant |\{c,f(v)\}|=2$.)
If $g$ is not constant, we additionally require that $x(k)\colon \mathbb{F}\_p\to \mathbb{F}\_p$ is a bijection. Then we need $f(x(k))\ne g\_k(x(k))$ for all $k$ (and $h$ is automatically defined by $h(x(k))=g\_k(x(k))$). Consider the $p\times p$ matrix with both rows and columns indexed by $\mathbb{F}\_p$ and put a star at the entry $(x,k)$ iff $f(x)\ne g(x+k)$. We want $p$ non-attacking rooks on the starred entries. Assume the contrary, then by Hall lemma (or Koenig theorem if you prefer) there exist sets $A,B\subset \mathbb{F}\_p$ such that $|A|+|B|=p+1$ and $f(x)=g(x+k)$ for $x\in A$, $k\in B$.
If $|B|=1$, then $|A|=\mathbb{F}\_p$ and $f(x)=g(x+k)$ for all $x$ where $B=\{k\}$. This contradicts to our assumption.
If $|A|=1$, then $B=\mathbb{F}\_p$ and we get that $g$ is constant, which also is not the case.
So, assume hereafter that $1<|A|,|B|<p$. Then we may also assume without loss of generality that $g$ takes colors from the set $\mathbb{F}\_p$. Replacing $f(x)$ to 0 for $x\notin A$, we get the same for $f$, and $f(x)=g(x+k)$ for $x\in A$, $k\in B$ still holds. Now the functions $f,g\colon \mathbb{F}\_p\to \mathbb{F}\_p$ are given by polynomials of degree at most $p-1$. Denote degree of $g$ by $d$, we have $d\geqslant 1$ since $g$ is not constant. Then we may write $d=d\_1+d\_2$ where $0\leqslant d\_1<|A|$, $1\leqslant d\_2<|B|$. Then the coefficient of $x^{d\_1}k^{d\_2}$ in the polynomial $h(x,k):=g(x+k)-f(x)$ is non-zero (it equals to a leading coefficient of $g$ times some non-zero binomial coefficient.) Any other monomial $x^a k^b$ in $h$ has either $a<d\_1$ or $b<d\_2$ (or both). Thus by generalized Combinatorial Nullstellensatz there exist $x\in A$, $k\in B$ with $h(x,k)\ne 0$, a contradiction.
Generalized Combinatorial Nullstellensatz is the following slight generalization of Alon's Combinatorial Nullstellensatz:
whenever $f(x\_1,\ldots,x\_k)$ is a polynomial with coefficients in the field $K$, the coefficient of the monomial $\prod x\_i^{d\_i}$ in $f$ is non-zero, and for each other monomial $\prod x\_i^{c\_i}$ in $f$ there exists at least one index $j$ such that $c\_i<d\_i$, then for arbitrary sets $A\_i\subset K$ for which $|A\_i|>d\_i$ there exist $a\_i\in A\_i$ such that $f(a\_1,\ldots,a\_k)\ne 0$.
UPDATE. Here is an alternative proof, due to Maxim Didin, of the more general fact:
**Theorem 1**. Let $(G,+)$ be a finite abelian group, and $f$, $g$ be two colorings (maps) from $G$ to a color set $C$ such that for every $k\in G$, $f(x)\not\equiv g(x+k)$ (in other words, $f$ is not a shifted $g$). Then there exists a coloring $h\colon G\to C$ such that $f(x)\ne h(x)$ for all $x\in G$, but for every $k$ there exists $x\in G$ such that $h(x)=g(x+k)$.
Denote $n=|G|$. For $k\in G$, denote by $g\_k(x):=g(x+k)$ a shift of $g$ by $k$. Some shifts may coincide. Denote by $\Phi$ the set of all distinct shifts of $g$. The following strengthens Theorem 1.
**Theorem 2**. In above notations and under assumptions of Theorem 1, there exists an injection $\eta:\Phi\to G$ such that for all $\varphi\in \Phi$ we have
$f(\eta(\varphi))\ne \varphi(\eta(\varphi))$.
(To deduce Theorem 1 from Theorem 2, define $h(\eta(\varphi)):=\varphi(\eta(\varphi))$ for $\varphi\in \Phi$, that is possible by injectivity of $\eta$; also define $h(x)=g(x)$ for $x\notin \eta(\Phi)$.)
To prove Theorem 2, we say that a shift $\varphi\in \Phi$ likes an element $x\in G$ if $f(x)\ne \varphi(x)$. Then by Hall marriage condition, to check the conclusion of Theorem 2 it is sufficient (and necessary) to prove that any set $M\subset \Phi$ of shifts of $g$ like in total at least $|M|$ elements $x\in G$. If $|M|=1$, this is true by our assumption. For $|M|>1$ this follows from the following
**Proposition**. For any set $M\subset \Phi$, $|M|>1$, there exist at least $|M|$ elements $x\in G$ such that $|\{\varphi(x)|\varphi\in M\}|\geqslant 2$.
(Obviously any such element $x$ is liked by some shift from $M$ that yields Hall's condition.) Proposition in turn follows from the key
**Lemma**. Let $A\subsetneq G$ be an arbitrary subset of $G$ which is not equal to $G$. Call a shift $g\_k\in \Phi$ of the coloring $g$ $A$-appropriate if $g\_k(x)=g(x)$ for all $x\in A$. Then there exist at most $n-|A|$ distinct $A$-appropriate shifts.
**Proof of Lemma**. If the only $A$-appropriate shift is $g\_0(x)=g(x)$, we are done as $1\leqslant n-|A|$. So, assume that $g\_k$ is $A$-appropriate but $g\_k\not\equiv g$. For every $x\notin A$ define $P(x)$ as the maximal set of the form $\{x,x-k,x-2k,\ldots\}$ for which $x-k,x-2k,\ldots$ all belong to $A$ (it may happen that $P(x)=\{x\}$). Since $g\_k$ is $k$-appropriate, the coloring $g$ is constant on each set $P(x)$. The whole $G$ is partitioned onto $n-|A|$ sets $P(x)$ and several orbits $\{a,a+k,a+2k,\ldots\}$ which are contained in $A$. Since $g\_k\not\equiv g$, there exists an element $\alpha\in G$ such that $g(\alpha)\ne g(\alpha-k)$. Assume that $g\_c\in \Phi$ is $A$-appropriate, that is, $g(x)=g(x+c)$ for all $x\in A$. If both $\alpha-c$, $\alpha-c-k$ belong to $A$, then $g(\alpha)=g(\alpha-c)=g(\alpha-c-k)=g(\alpha-k)$, a contradiction. Thus $\alpha-c$ must belong to some set $P(x)=\{x,x-k,x-2k,\ldots, x-(d-1)k\}$ for $x\notin A$. Also, either $\alpha-c=x$ or $\alpha-c=x-(d-1)k$. Let us show that for fixed $x$, at most one element of $P(x)$ may appear as $\alpha-c$ (when $c$ vary). Assume the contrary, then in particular $d>1$. If $\alpha-c=x$, we have $\alpha-c-k=x-k\in A$, thus $g(\alpha-k)=g(\alpha-c-k)=g(x)$. If $\alpha-c=\alpha-(d-1)k\in A$, then $g(\alpha)=g(\alpha-c)=g(x)$. Since both equations $g(\alpha-k)=g(x)$, $g(\alpha)=g(x)$ can not hold simultaneously, we see that at most one element of $P(x)$ may be equal to $\alpha-c$. Thus there exists at most $n-|A|$ possible $c$ for which $g\_c$ is $A$-appropriate.
It remains to deduce Proposition from Lemma that is rather tautological.
**Proof of Proposition**. Without loss of generality $g\_0=g\in M$. If the claim does not hold, there exist $n-k+1$ elements $x$ for which $|\{\varphi(x)|\varphi\in M\}|=1$, i.e., $\varphi(x)=g(x)$ for all $\varphi\in M$. But then by Lemma $|M|\leqslant k-1$, a contradiction.
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