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https://mathoverflow.net/questions/396840
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1
|
Fix $n\geq 2$ and let $$\mathbb{H}^{n}=\mathbb{R}\_{+}\times \mathbb{S}^{n-1}$$ be the hyperbolic space, so that any point $x\in \mathbb{H}^{n}$ can be represented in polar coordinates $x=(r, \theta)$, and equipped with the Riemannian metric $$g=dr^{2}+\sinh^{2}(r)d\theta^{2},$$ where $dr^{2}$ is the standard metric on $\mathbb{R}\_{+}$ and $d\theta^{2}$ is the standard metric on the sphere $\mathbb{S}^{n-1}$. Denote with $\mu$ the Riemannian measure on $\mathbb{H}^{n}$ and with $\sigma$ the Riemannian measure of co-dimension $1$ on hypersurfaces on $\mathbb{H}^{n}$. It is a known fact that $\mathbb{H}^{n}$ admits the Isoperimetric inequality $$\sigma(\partial \Omega)\geq f(\mu(\Omega)),$$ for all precompact open sets $\Omega\subset \mathbb{H}^{n}$ with smooth boundary, where $f$ is defined by $$f(v)=c\_{1}\left\{\begin{array}{cl}v, &v\geq 1\\ v^{\frac{n-1}{n}}, &v\leq 1,\end{array}\right.$$ for some small constant $c\_{1}>0$.
Let us fix some precompact open set $U\subset \mathbb{H}^{n}$ and consider $\mathbb{H}^{n}\setminus U$ as a manifold with boundary $\partial U$.
Now my **question**:
>
> Does the same Isoperimetric inequality now hold for precompact open sets $\Omega\subset \mathbb{H}^{n}\setminus U$ with smooth boundary (and possibly a smaller constant $c\_{2}>0$) and if yes, where can I find a reference for this statement?
>
>
>
**Specified question**:
>
> Let $U$ be a "nice" precompact open set, for example $U=B\_{1}(o)$ is an open ball of radius $1$ for some point $o\in \mathbb{H}^{n}$. Does the same Isoperimetric inequality now hold for precompact open sets $\Omega\subset \mathbb{H}^{n}\setminus B\_{1}(o)$ with smooth boundary (and possibly a smaller constant $c\_{2}>0$)?
>
>
>
Note that we assume that, when considering a precompact open set $\Omega\subset \mathbb{H}^{n}\setminus U$ with smooth boundary, we have $\partial U\cap \partial \Omega=\emptyset$.
If this is not known for the hyperbolic space $\mathbb{H}^{n}$, is a similar statement known for other Riemannian manifolds, for example $\mathbb{R}^{n}$?
Thanks in advance for your help!
|
https://mathoverflow.net/users/163368
|
Isoperimetric inequality for domains in the exterior of a precompact open set in Riemannian manifold
|
Some results are known under some restrictions. First, no uniform inequality can hold if $U$ is unrestrained: it could take the shape of (a neighborhood of) a bottle, and the bottleneck will make it possible to have $\Omega$ with large volume and small boundary. A natural restriction is to consider convex $U$ (so that the double of $\mathbb{H}^n$ still is negatively curved in the metric sense).
**To sum up:** assuming $U$ is convex, we even have sharp bounds. When $M=\mathbb{H}^n$ the dimensions $n=2,3,4$ are covered; when $M=\mathbb{R}^n$, you have a weaker inequality (the isoperimetric function has the form $f(v) = cv^{\frac{n-1}n}$, there is no linear asymptotic) but known in all dimensions.
**Added in edit:** I think that an equality $\sigma(\partial \Omega)\ge c\mu(\Omega)$ is true in all dimensions whenever $U$ is convex, and that the method in our paper with Kuperberg mentionned below can be used to prove this. The more precise bound for small $v$, I am not sure.
Note that $\mathbb{H}^n$ can even be replaced by a simply connected manifold $M$ of sectional curvature bounded above by some $\kappa\le 0$. In this setting:
1. Choe proved in 2003 gave the *sharp* inequality when $M=\mathbb{R}^n$ and $U$ is a ball, as well as several other restricted cases: **Relative isoperimetric inequality for domains outside a convex set.** *Archives Inequalities Appl 1 (2003): 241-250*.
2. In 2007, the general case of a convex set $U\subset \mathbb{R}^n$ was obtained by Choe, Ghomi and Ritoré. **The relative isoperimetric inequality outside convex domains in $\mathbf{R}^n$.** *Calculus of Variations and Partial Differential Equations 29.4 (2007): 421-429.*
3. In 2006, Choe treated the case when $M$ has variable (nonpositive) curvature and dimension $4$. **The double cover relative to a convex domain and the relative isoperimetric inequality.** *Journal of the Australian Mathematical Society 80.3 (2006): 375-382.*
4. In dimension $3$, the same was achieved by Choe and Ritoré in 2007. **The relative isoperimetric inequality in Cartan-Hadamard 3-manifolds.** *J. für die reine und angewandte Mathematic (2007): 179-191.* They also obtain the sharp inequality when $\kappa=-1$, in particular for $M=\mathbb{H}^3$.
5. With Kuperberg, we gave a new proof of Choe's result from 2006, treating dimensions $2$ and $4$ and including (sharp) results when $M$ has curvature bounded above by $\kappa$ (either $0$ or negative). **The Cartan–Hadamard conjecture and the Little Prince.** *Revista Matemática Iberoamericana 35.4 (2019): 1195-1258.* Our results can be used to tackle some finite union of convex sets (you need to ensure that geodesic rays reflecting on the boundary of $U$ can only bounce a bounded number of time).
I may have missed other relevant references, but you should catch them by looking at papers citing the above ones.
|
2
|
https://mathoverflow.net/users/4961
|
396900
| 163,875 |
https://mathoverflow.net/questions/396776
|
15
|
One of my colleagues gave me the following problem about 15 years ago:
Given three squares inside a 1 by 2 rectangle, with no two squares overlapping, prove that the sum of side lengths is at most 2. (The sides of the squares and the rectangle need not be parallel to each other.)
I couldn't find a solution or even a source for this problem. Does anyone know about it? I was told that this was like an exercise in combinatorial geometry for high school contests such as IMO, but I'm not sure if this problem was listed in such competitions. Has anyone heard of this problem, or does anyone know how to solve it?
Any kind of help will be appreciated.
|
https://mathoverflow.net/users/313754
|
Three squares in a rectangle
|
Since the squares are convex, we can draw lines which separate them. In particular, if two separating lines go from $(b-a,0)$ to $(b,1)$ and from $(c,1)$ to $(c+d,0)$, then we can prove the result in terms of those lines and those variables.
So: let the rectangle go from $(0,0)$ to $(2,1)$. Let $A$ be the leftmost square (or one such square). Let $C$ be the rightmost square (or one such square). Let $B$ be the other square.
Draw a line separating $A$ and $B$, and let $(b,1)$ and $(b-a,0)$ be its intersections with the lines $y=1$ and $y=0$.
Draw a line separating $B$ and $C$, and let $(c,1)$ and $(c+d,0)$ be its intersections with the lines $y=1$ and $y=0$.
Reasoning as in user21820's answer, we assume wlog that:
* $0<a$, so $A$ is left and above the line separating $A$ and $B$;
* $0<d$, so $C$ is right and above the line separating $B$ and $C$;
* $0<b$ and $c<2$ and $a-b+c+d>0$, so $B$ is below both lines.
(Since the lines may leave the rectangle, we do not assume $b-a>0$ or $b<2$ or $c>0$ or $c+d<2$.)
Lemma (proved at the end):
\begin{align}
\text{sidelength of }A &\le \min\!\left(\frac{b}{a+1},\,1\right)\\
\text{sidelength of }B &\le \min\!\left((a-b+c+d)u,\,1\right)\\
\text{sidelength of }C &\le \min\!\left(\frac{2-c}{d+1},\,1\right)
\end{align}
where
$$u=\max\left(
\frac{1}{a+d+1},
\frac{\sqrt{a^2+1}}{a^2+a+d+1},
\frac{\sqrt{d^2+1}}{d^2+d+a+1}
\right)$$
The factor $u$ satisfies $1/(a+1)>u$ and $1/(d+1)>u$ so long as $a<3.66$ and $d<3.66$ respectively. I will assume those inequalities for now to show that some functions are increasing or decreasing; I don't have a clean proof for those inequalities or without them yet.
In the corner case of $b=a+1$ and $c=1-d$, the side lengths are $1$, $0$ and $1$, and they sum to exactly $2$. We now use this in analyzing four cases.
**Case I**, $b\le a+1$ and $c\le 1-d$: The sum of the sidelengths is at most
$$\frac{b}{a+1}+(a-b+c+d)u+1$$ This is increasing in both $b$ and $c$, so its value is at most the corner value of $2$.
**Case II**, $b\le a+1$ and $c\ge 1-d$: The sum of the sidelengths is at most
$$\frac{b}{a+1}+(a-b+c+d)u+\frac{2-c}{d+1}$$ This is increasing in $b$ and decreasing in $c$, so its value is at most the corner value of $2$.
**Case III**, $b\ge a+1$ and $c\le 1-d$: The sum of the sidelengths is at most
$$1+(a-b+c+d)u+1$$
This is decreasing in $b$ and increasing in $c$, so its value is at most the corner value of $2$.
**Case IV**, $b\ge a+1$ and $c\ge 1-d$: The sum of the sidelengths is at most
$$1+(a-b+c+d)u+\frac{2-c}{d+1}$$
This is decreasing in both $b$ and $c$, so its value is at most the corner value of $2$.
So the sum of the sidelengths is at most $2$ in each case.
**Proof of Lemma:**
The sidelength of $A$ is clearly less than 1, and also clearly less than the maximum sidelength inscribed in the right triangle bounded by $x=0$, $y=1$, and the separator of $A$ and $B$. We use Polya's formula [here](https://math.stackexchange.com/a/532225) to calculate the sidelength in the triangle as the maximum of $b/(a+1)$ and $b\sqrt{a^2+1}/(a^2+a+1)$; since $a>0$, the maximum is just $b/(a+1)$.
A similar use of Polya's result bounds the sidelength of $C$. Yet another use of that result, now for a triangle which may be acute or obtuse, bounds the sidelength of $B$ by $|a-b+c+d|u$. Since we assumed $a-b+c+d>0$, we write this bound as $(a-b+c+d)u$.
|
5
|
https://mathoverflow.net/users/nan
|
396903
| 163,877 |
https://mathoverflow.net/questions/396849
|
4
|
When dealing with some hash functions that I was trying to speed up, I [toyed](https://mathoverflow.net/questions/394296/hamming-distance-between-ab-and-a-oplus-b-oplus-a-land-b-ll-1) with a binary operation with the goal to "approximate" the addition on $\{0,1\}^\*$ when seen as binary representation of the positive integers:
$$(a,b) \in (\{0,1\}^\*)^2 \mapsto a \oplus b \oplus ((a \land b) \ll 1)$$ where $\oplus$ is bitwise XOR and $\land$ is bitwise AND, and $\ll$ is left-shift by 1 position. (The purpose of $((a \land b) \ll 1)$ is to simulate the "carry-bit" operation.)
**Formal definition.** Let $\{0,1\}^\mathbb{N}$ denote the collection of functions $f:\mathbb{N}\to \{0,1\}$ and let $$\{0,1\}^\* = \{x \in \{0,1\}^\mathbb{N}: \exists N\in\mathbb{N}(\forall k\in\mathbb{N}(k\geq N\implies x(k)=0))\}.$$
Denote by $\ll 1$ the *shift* by one position, i.e. $\ll 1 : x \in \{0,1\}^\* \to x'\in \{0,1\}^\*$ where $x'(0) = 0$ and $x'(n+1) = x(n)$ for all $n\in \mathbb{N}$. We usually write $x \ll 1$ instead of $\ll 1(x)$.
For any $a,b\in\{0,1\}^\*$ let us write $$a +\_2 b := a \oplus b \oplus ((a\land b) \ll 1).$$
It is easily seen that $+\_2$ is not associative, and that $0$ is a neutral element for every $x\in \{0,1\}^\*$. Moreover, $+\_2$ is clearly commutative
**Question.** Given $a,b\in \{0,1\}^\*$, is there $x\in \{0,1\}^\*$ such that $a +\_2 x = b$? Is $x$ necessarily unique?
**Further question.** (Need not be answered for acceptance.) If the answer to the above question is positive, we would have a kind of "non-associative group". Do these have a proper name? Do they occur "naturally" somewhere in mathematics?
|
https://mathoverflow.net/users/8628
|
Non-associative commutative "group"
|
The answer to the first question is **yes**:
>
> **Claim.** Let $a, b \in \{0, 1 \}^{\ast}$ and let $n \ge 0$ be the least integer such that $a(i) = b(i) = 0$ for every $i > n$. Then the equation $$a +\_2 x = b$$ has a unique solution $x \in \{0, 1 \}^{\ast}$ which is recursively defined by $x(0) = a(0) + b(0)$, $x(i) = a(i - 1)x(i - 1) + a(i) + b(i)$ for $0 < i \le n$ and
> $x(n + 1) = a(n)x(n)$, $x(i) = 0$ for $i > n + 1$.
> Equivalently, the solution $x$ is given by
>
> $$
> \begin{array}{lll}
> x(0) &=& a(0) + b(0), \\
> x(1) &=& a(0)(a(0) + b(0)) + a(1) + b(1),\\
> ... & = & ..., \\
> x(n) & = & a(0) \cdots a(n - 1)(a(0) + b(0)) + a(1) \cdots a(n - 1)(a(1) + b(1)) + \cdots + a(n - 1)(a(n - 1) + b(n - 1)) + a(n) + b(n),
> \end{array}$$
> $x(n + 1) = a(n)x(n)$
> and $x(i) = 0$ for $n > n + 1$ where the operations $+$ and $\cdot$ refer to addition and multiplication in $\mathbb{Z} / 2 \mathbb{Z}$ which we identified with $\{0, 1 \}$.
>
>
>
>
> *Proof.* The equation is equivalent to $x \oplus ((a \wedge x) \ll 1) = a \oplus b$ which reads as a linear system over $\mathbb{Z} / 2 \mathbb{Z}$.
>
>
>
|
5
|
https://mathoverflow.net/users/84349
|
396908
| 163,880 |
https://mathoverflow.net/questions/396494
|
6
|
If $1\leq p<\infty$, it is easy to find nice necessary and sufficient equality conditions for the convolution inequality $$\lVert f\*g\rVert\_p\leq\lVert f\rVert\_1\lVert g\rVert\_p\qquad (f\in L^1(\mathbb{R}^n),\,g\in L^p(\mathbb{R}^n)),$$ for complex $f$ and $g$. But I cannot seem to determine one for $p=\infty$ nor find a reference that contains such a condition. Is there a natural equality condition in this case?
|
https://mathoverflow.net/users/306090
|
When is $\lVert f*g\rVert_\infty=\lVert f\rVert_1\lVert g\rVert_\infty$?
|
Here is a fairly simple condition.
It uses the following notion: A family of functions $f\_t\in L^1$ depending on a parameter $t$ in a measure space $X$ is said to *tend to $f$ somewhere* if the essential infimum of $\lVert f\_t-f\rVert\_1$ over $X$ is $0$.
Put $g\_s(t)=g(t-s)$. The condition is that $afg\_s\to\lvert f\rvert\lVert g\rVert\_\infty$ somewhere, for some constants $a=a(s)$ with $|a|=1$.
Clearly, this condition implies equality. For the converse, choose $a$ so that $\lvert\int fg\_s\rvert=\int\Re(afg\_s)$. Then equality implies that $\Re(afg\_s)\to\lvert f\rvert\lVert g\rVert\_\infty$ somewhere, and the condition follows via the inequality $\lVert\Im f\rVert\_1^2\leq2\lVert f\rVert\_1\lVert\lvert f\rvert-\Re f\rVert\_1$. ($\Re$ and $\Im$ denote real and imaginary parts.)
|
1
|
https://mathoverflow.net/users/306090
|
396910
| 163,881 |
https://mathoverflow.net/questions/396875
|
4
|
Considering the Jacobi theta: $\theta\_3(z) = \sum\_{n\in\mathbb{Z}} q^{n^2}$,
we can invert $\theta\_3-1$ in a small enough neighbourhood of 0.
Routine computation with Lagrange-Burmann inversion gives that the inverse have expansion
starting by:
$\frac{q}{2}-\frac{q^4}{16}+\frac{q^7}{32}-\frac{q^9}{512}-\frac{11 q^{10}}{512}+\frac{13 q^{12}}{4096}+ O(q^{13})$
Is there any known closed form (or recursive formula) giving the terms of this series?
(it is very closely related to OEIS A259938 suite: <https://oeis.org/A259938>, corresponding to $\theta\_3(q/2)$ but which doesn't seem very studied as well)
|
https://mathoverflow.net/users/70925
|
Closed formula for reversion of Jacobi theta series
|
Perhaps, the form given by [Lagrange inversion theorem](https://en.wikipedia.org/wiki/Lagrange_inversion_theorem) cannot be much simplified here. It expresses the $n$-th coefficient of a series reversion as the sum of $n-1$ values of exponential Bell polynomials.
From the practical perspective, since $\theta\_3-1$ contains nonzero coefficients only at square powers, computation of the $n$-th reversion coefficient amounts to iterating over the partitions of $n-1$ into squares $>1$ decreased by $1$ (almost [OEIS A243148](https://oeis.org/A243148) if we fix a number of squares). This yields the formula that Peter Taylor gave in the comments (modulo the corrections) for $n>1$:
$$A\_n = \frac{1}{n!2^n} \sum\_{(2^2-1)j\_2 + (3^2-1)j\_3 + \dots = n-1} (-1)^{j\_2+j\_3+\dots}\cdot \frac{(n-1+j\_2+j\_3+\dots)!}{j\_2!j\_3!\dots}.$$
|
3
|
https://mathoverflow.net/users/7076
|
396917
| 163,882 |
https://mathoverflow.net/questions/396922
|
1
|
There is a coloured operad $sOp$ such that $sOp$-algebras are single-coloured operads. This operad has a simple description in terms of generators and relations, say, as an operad $F(X)/R$. There is a more concrete description of $sOp$ as the operad $OpTrees$ of trees endowed with additional structure, see [1, section 4] or [2, Example 1.5.6]. The proof that the operads $F(X)/R$ and $OpTrees$ are isomorphic is not too hard, but not entirely trivial either.
Likewise, there should be a cartesian multicategory (or equivalently a multi-sorted Lawvere theory, or a multi-sorted abstract clone) $CartOp$ such that $CartOp$-algebras are single-coloured cartesian multicategories. Definition of $CartOp$ in terms of generators and relations seems fairly straightforward. However, concrete description of $CartOp$, description similar to that of $OpTrees$, is a bit more elusive.
Is there any reference that gives concrete description of $CartOp$?
[1] Pepijn van der Laan, Coloured Koszul duality and strongly homotopy operads <https://arxiv.org/abs/math/0312147>
[2] Clemens Berger, Ieke Moerdijk, Resolution of coloured operads and rectification of homotopy algebras <https://arxiv.org/abs/math/0512576>
---
Update. There is "Operads from the viewpoint of categorical algebra" of Tibor Beke, which gives description in terms of monads. I'll close the question. Any reference with a clean proof of isomorphism analogous to that of $F(X)/R\cong OpTrees$ is still welcome.
|
https://mathoverflow.net/users/78299
|
Lawvere theory of Lawvere theories
|
Viewing cartesian operads as algebraic theories, your question may be rephrased as: "How can we present the (multisorted) algebraic theory whose models are (monosorted) algebraic theories?"
This is given by the theory of [abstract clones](https://en.wikipedia.org/wiki/Clone_(algebra)#Abstract_clones). You can check that the category of monosorted algebraic theories is isomorphic to the category of abstract clones, the latter of which is defined as an $\mathbb N$-sorted universal algebra. It is then straightforward to translate back from algebraic theories to cartesian multicategories / cartesian coloured operads.
|
2
|
https://mathoverflow.net/users/152679
|
396926
| 163,886 |
https://mathoverflow.net/questions/396931
|
2
|
A graph $G$ is Hamiltonian if there is a [Hamiltonian cycle](https://mathworld.wolfram.com/HamiltonianCycle.html) in $G$.
Suppose $G$ is a [$k$-edge connected](https://en.wikipedia.org/wiki/K-edge-connected_graph) $k$-regular graph with $k>1$.
Does this ensure that $G$ is Hamiltonian?
If not, how about [vertex-transitive](https://en.wikipedia.org/wiki/Vertex-transitive_graph) $k$-regular graphs? (we know that vertex-transitive $k$-regular graphs are $k$-edge connected). We also know that $k$-regular hypercubes admit a Hamiltonian cycle (called Gray codes).
|
https://mathoverflow.net/users/114739
|
Is every $k$-edge connected $k$-regular graph Hamiltonian?
|
Assume $k \geq 4$, since, for $k=2$, the answer for both questions is **yes**, and for $k=3$, **no**, as there's the [Petersen graph](https://en.wikipedia.org/wiki/Petersen_graph).
The answer to the first question is **no**. To see this, we only need to prove that the constructions by Meredith in [1] give $k$-edge-connected graphs, as Meredith already proved that the graph is non-Hamiltonian and regular.
Let's recall Meredith's construction. Let $n,m$ be integers, $n=m+\alpha$, where $\alpha \in \{ -1,0,1\}$ such that $2m+n=k$. Take a Petersen graph $P$ and label the vertices $A, B, \cdots , J$, such that the edges are
* $AB,CD,EF,GH,IJ$;
* $AC,CE,EG,GI,IA$;
* $BF,FJ,JD,DH,HB$.
Then the $k$-regular Meredith graph $M\_k$ consists of vertices $X\_1,\cdots,X\_{k-1},x\_{11},\cdots,x\_{1n},x\_{21},\cdots,x\_{2m},x\_{31},\cdots,x\_{3m}$ (call them "vertices of type $X$") where $X$ denotes a letter in $\{A,B,C,D,E,F,G,H,I,J\}$ and $x$ the respective lowercase letter.
The edges are
* $X\_px\_{qr}$ where $X$ is a letter and $x$ its respective lowercase letter, and $p,q,r$ arbitrary;
* $x\_{1p}y\_{1p}$ where $xy \in \{ab,cd,ef,gh,ij\}$ and $p$ arbitrary;
* $x\_{2p}y\_{3p}$ where $xy \in \{ac,ce,eg,gi,ia\}$ and $p$ arbitrary;
* $x\_{3p}y\_{2p}$ where $xy \in \{bf,fj,jd,dh,hb\}$ and $p$ arbitrary;
[Wikipedia](https://en.wikipedia.org/wiki/Meredith_graph) contains some descriptions for $M\_4$.
Now we proof that $M\_k$ is $k$-connected. Let $\bar{M}\_k$ be a graph formed by removing $k-1$ edges from $M\_k$. We proceed to prove that $\bar{M}\_k$ is connected.
If the removal disconnects the subgraph induced by the vertices of type $X$ for some $X$, then the removed vertices must share some $x\_{qr}$ as an endpoint, since the induced subgraph is isomorphic to $K\_{k,k-1}$. But now we can check the structure of $\bar{M}\_k$ and discover that it's connected.
If the removal keeps connected the subgraphs induced by the vertices of type $X$ for all $X$, then we only need to ensure that the subgraphs are connected "in between". That is, we construct a subgraph $Q$ of $P$, where two vertices $M,N$ of $P$ are connected if and only if there's at least one edge between a vertex of type $M$ and a vertex of type $N$ in $\bar{M}\_k$. If $Q$ is connected, then $\bar{M}\_k$ is connected.
To see that $Q$ is connected, we recall that removing $3$ edges can only disconnect the Petersen graph into a vertex and the remaining subgraph; any other disconnection requires at least $4$ edges. Either way, there must be at least $k$ edges disconnected in $M\_k$ to make $Q$ disconnected.
So we have proven that $\bar{M}\_k$ is connected, and thus $M\_k$ is $k$-connected.
Lovasz [conjectured](https://en.wikipedia.org/wiki/Lov%C3%A1sz_conjecture) that every finite connected vertex-transitive graph contains a Hamiltonian cycle except five known counterexamples, i.e. $K\_2$, the Petersen graph, the Coxeter graph, and two graphs derived from the Petersen and Coxeter graphs by replacing each vertex with a triangle. All of them have vertex degree $\leq 3$. As the conjecture is neither proven or refuted, we can say that the answer to the second question is **unknown**.
[1] Meredith, G. H. J. (1973). Regular n-valent n-connected nonHamiltonian non-n-edge-colorable graphs. Journal of Combinatorial Theory, Series B, 14(1), 55-60.
|
5
|
https://mathoverflow.net/users/125498
|
396936
| 163,888 |
https://mathoverflow.net/questions/396924
|
0
|
given a bipartite graph $G(U,V,E\subseteq U\times V)$ with strictly positive edge-weights; is there an established name for the the task of calculating the lightest spanning subgraph and what is the best known algorithmic complexity?
Put more colloquially: if there are $p$ persons and $t$ tasks with possibly different cardinalities and a value that reflects how beneficial it is if person $p\_i$ takes over responsibility for task $t\_j$, which assignment is most beneficial under the constraint that each person must be responsible for at least task and every task must have at least one responsible person but not more than the number of tasks a person can handle or the number of persons that can collaborate on a task - these upper bounds are found in the vertex degrees.
|
https://mathoverflow.net/users/31310
|
Name for a type of assignment task
|
This is the [transportation problem](https://en.wikipedia.org/wiki/Transportation_theory_(mathematics)). The people correspond to supply nodes, with a lower bound of 1 and upper bound equal to the number of tasks that person can handle. The tasks correspond to demand nodes, with a lower bound of 1 and upper bound implied by the degree.
Not sure about the state-of-the-art, but the [network simplex algorithm](https://en.wikipedia.org/wiki/Network_simplex_algorithm) should be competitive.
|
1
|
https://mathoverflow.net/users/141766
|
396937
| 163,889 |
https://mathoverflow.net/questions/396897
|
11
|
Given a closed (perhaps irreducible) 3-manifold $M$ with an embedded surface $S$ and a knot $K$, what conditions allow the two to be isotoped to be disjoint? Obviously a necessary condition is that $[S] \cdot [K] = 0$, but when is this sufficient? [This](https://mathoverflow.net/questions/191585/knots-in-3-manifolds) answer gives a condition when $K$ is homotopically trivial, a case I am particularly interested in. Are there any other conditions in this case? How about other cases?
(Also would love references on the more general problem of making the algebraic and geometric intersections equal in a 3-manifold via isotopy)
|
https://mathoverflow.net/users/314845
|
When can a surface in a 3-manifold be isotoped off a knot?
|
Let’s assume that the manifold $M$ is irreducible and orientable and the surface $S$ is orientable. This is to avoid 1-sided surfaces.
First let’s assume that the surface $S$ is fully [compressible](https://en.wikipedia.org/wiki/Incompressible_surface?wprov=sfti1). That means that there is a sequence of compressions taking the surface to a collection of 2-spheres, for example a [Heegaard](https://en.wikipedia.org/wiki/Heegaard_splitting?wprov=sfti1) surface. Since the manifold is irreducible, these spheres lie inside of a ball (but they might be nested). Thus we can make $K$ disjoint from the balls by an isotopy. Reversing the process of compression, we can iteratively tube these spheres together to get back $S$ up to isotopy. In this process, we can assume that each tube misses $K$ by general position (it is a small regular neighborhood of an arc) and hence get an isotopic surface which misses $K$. So this takes care of the case of fully compressible surfaces, which will be the case in a non-Haken irreducible 3-manifold such as lens spaces and the Poincaré dodecahedral space.
In the general case, one compresses down the surface $S$ to a surface $S’$. By a similar argument, $K$ will miss $S$ up to isotopy if it misses $S’$ (although the converse is likely not true). In an irreducible 3-manifold, it must be Haken if $S’$ is not a collection of spheres (so a component is an incompressible surface of genus $>0$).
Thus assume $S$ is incompressible. In principle then one can determine if $S$ can be made disjoint from $K$. Let me sketch the case that $M$ and $M-K$ are [hyperbolic](https://en.wikipedia.org/wiki/Hyperbolic_3-manifold?wprov=sfti1). In this case, one can enumerate the finitely many disjoint incompressible surfaces of the same genus as $S$ in $M$ and $M-K$. There are algorithms using normal surface theory or geometric group theory. Then there is an algorithm to tell if these surfaces are isotopic (in the non hyperbolic case, there may be infinitely many surfaces of a fixed genus, and one has to work harder). If so, then $K$ can be isotoped disjoint from $M$, otherwise not. Of course, this only gives a sufficient condition in the compressible case by reducing to $S’$.
I suspect that there is an algorithm in general like Sam Nead suggests, but I expect that it isn’t written down in general.
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9
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https://mathoverflow.net/users/1345
|
396939
| 163,891 |
https://mathoverflow.net/questions/396918
|
2
|
Let $L/K$ be a field extension.
Let $R,S$ be two local commutative $K$-algebras and let $\varphi : R \to S$ be a homomorphism of $K$-algebras, not assumed to be local. Let's call a prime ideal $\mathfrak{p} \subseteq R \otimes\_K L$ *good* when $\mathfrak{p} \cap R = \mathfrak{m}\_R$. Notice that good prime ideals correspond to prime ideals in the tensor product of fields $R/\mathfrak{m}\_R \otimes\_K L$.
I wonder if the following is true:
**Question.** Is there some good prime ideal $\mathfrak{p} \subseteq R \otimes\_K L$ such that for all $f \in R \otimes\_K L$ with $f \notin \mathfrak{p}$ the image of $f$ in $S \otimes\_K L$ is not contained in every good prime ideal of $S \otimes\_K L$? Equivalently, the image of $f$ in $S/\mathfrak{m}\_S \otimes\_K L$ is not nilpotent.
In terms of the local $K$-schemes $X=\mathrm{Spec}(R)$, $Y=\mathrm{Spec}(S)$ and the morphism $Y \to X$, the question is the following: Is there some point in $X\_L$ over the closed point of $X$ such that every open neighborhood of it pulls back to an open subset in $Y\_L$ which contains a point over the closed point of $Y$?
I have checked some special cases, but either they were trivial or too hard to understand, since tensor products of fields can be nasty. Of course it is true when $L=K$, and it is also true when $\varphi$ is local. My feeling is that the statement is false in general, but I might be wrong. Maybe counterexamples can be constructed from localizations of $R$.
The context for this question is to prove a certain result for locally ringed spaces, and for locally ringed spaces with exactly two points it comes down to the question above.
|
https://mathoverflow.net/users/2841
|
Good prime ideals in tensor products of local rings
|
Let $x$, resp $y$ be the closed point of $X$, resp $Y$. Denote $f : Y \to X$ the given morphism. Then $f(y) \leadsto x$ (specialization). Let $x\_L$ be any point of $X\_L$ mapping to $x$. The morphism $X\_L \to X$ is flat. Hence there is a specialization $z \leadsto x\_L$ in $X\_L$ such that $z$ maps to $f(y)$. Since $Y\_L = Y \times\_X X\_L$, there is a point $w$ of $Y\_L$ which maps to $y$ in $Y$ and $z$ in $X\_L$. This proves what you want as $w$ will be in the inverse image of any open neighborhood of $x\_L$ you pick.
|
2
|
https://mathoverflow.net/users/152991
|
396942
| 163,892 |
https://mathoverflow.net/questions/396946
|
2
|
Let $X$, $Y$, and $Z$ be locally-compact, complete, and separable metric spaces and suppose that $X$ is compact; all non-empty.
Consider the spaces $C(X,C(Y,Z))$ and $C(X\times Y,Z)$ both equipped with their uniform convergence on compact topologies. Does the map:
$$
C(X\times Y,Z)\ni f \to (x\mapsto [y\mapsto f(x,y)]) \in C(X,C(Y,Z)),
$$
define a homomorphism; and if so, is it uniformly continuous?
**References?:** *Where can I read more about this map...surely this has been studied somewhere in the general topology literature; probably in the mid 1950-1960s I expect?*
|
https://mathoverflow.net/users/176409
|
Relationship between $C(X\times Y,Z)$ and $C(X,C(Y,Z))$
|
In the context of functions from one space to another, a key-word here might by "currying", for the process of converting $x,y\to f(x,y)$ to $x\to (y\to f(x,y)$. In the case of spaces of holomorphic functions with the natural topology, verification is not hard.
In other contexts, isomorphisms $\operatorname{Hom}(X\otimes Y,Z)\approx \operatorname{Hom}(X,\operatorname{Hom}(Y,Z))$ are instances of the Eilenberg-Maclane *adjunction*. In simple algebraic situations (e.g., for abelian groups) the verification is easy.
Those instances suggest that we might want to characterize the various topologies in a fashion so as to make the assertion be correct... unless there is a "counter-example" that is important *not* to exclude for your purposes. That is, in effect, do you need/want the isomorphism to be correct, or not necessarily. :)
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2
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https://mathoverflow.net/users/15629
|
396947
| 163,895 |
https://mathoverflow.net/questions/396954
|
0
|
Prove that $\sum\_{k=0}^n {k+1\choose 2}^R + \sum\_{k=0}^{-n-2} {k+1\choose 2}^R = 0.$ This can be shown using Faulhaber's formula but it's very long. Is there a nicer, shorter method? Any thoughts or ideas?
|
https://mathoverflow.net/users/265714
|
Prove that $\sum_{k=0}^n {k+1\choose 2}^R + \sum_{k=0}^{-n-2} {k+1\choose 2}^R = 0.$
|
The "correct" definition of $\sum\_{k=0}^{-n-2} f(k)$ is $-\sum\_{k=-n-1}^{-1}f(k)$. Now ${-a+1\choose 2}={a\choose 2}$. Thus the terms of $\sum\_{k=0}^{-n-2}{k+1\choose 2}^R$ cancel with the terms of $\sum\_{k=0}^n{k+1\choose 2}^R$.
|
4
|
https://mathoverflow.net/users/2807
|
396960
| 163,899 |
https://mathoverflow.net/questions/396362
|
2
|
For all $k,R \in \mathbb{N}$ fixed, prove that $ \sum\_{i=0}^{2k}( {n+R-1\choose R+i} + (-1)^{i+1}{ n+R+i\choose R+i } )\sum\_{j=0}^i {i\choose j}(-1)^j(i+1-j)^{2k}=0 $. I'm quite sure this is true but I'm not very sharp in seeing how to go about proving it. I was studying some stuff in some Pascal triangles is how I stumbled on this. Any ideas?
|
https://mathoverflow.net/users/265714
|
Prove that $ \sum_{i=0}^{2k}( {n+R-1\choose R+i} + (-1)^{i+1}{ n+R+i\choose R+i } )\sum_{j=0}^i {i\choose j}(-1)^j(i+1-j)^{2k}=0 $
|
~~First off, there should be $(-1)^{i+1}$ not $(-1)^{R+i}$ (now it's corrected in the question).~~
**UPDATE.** Argument below is simplified and streamlined.
The identity generalizes the [previous question](https://mathoverflow.net/q/395758), which essentially represents the case of $R=1$. In fact, this generalized identity has a somewhat easier proof (given below) than my [earlier proof for $R=1$](https://mathoverflow.net/q/395897).
Again, we start with noticing that the inner sum equals $i!\cdot S(2k+1,i+1)$, where $S(\cdot,\cdot)$ are Stirling numbers of second kind.
Let $k>0$ is fixed, and $LHS(2k,n,R)$ denote the identity left-hand side, i.e.
$$LHS(2k,n,R) := \sum\_{i=0}^{2k} \big( \binom{n+R-1}{R+i} + (-1)^{i+1} \binom{n+R+i}{R+i}\big) i! S(2k+1,i+1).$$
We start to prove the identity in two cases: $R=0$ and $n=0$.
In the case $R=0$, we have
\begin{split}
LHS(2k,n,0) &= \sum\_{i=0}^{2k} \big( \binom{n-1}i + (-1)^{i+1} \binom{n+i}i\big) i! S(2k+1,i+1) \\
&= \sum\_{i=0}^{2k} \big( (n-1)\_i - (-n-1)\_i \big) S(2k+1,i+1) \\
&= n^{2k} - (-n)^{2k} = 0.
\end{split}
In the case $n=0$ and $R>0$, we have
\begin{split}
LHS(2k,0,R) &= \sum\_{i=0}^{2k} (-1)^{i+1} i! S(2k+1,i+1) \\
&= -\sum\_{i=0}^{2k} (-1)\_i S(2k+1,i+1) \\
&= -0^{2k} = 0.\end{split}
Now, we are ready to prove the identity by induction on $n+R$. From above, it follows that $LHS(2k,n,R)=0$ when $(n,R)=(0,1)$ or $(n,R)=(1,0)$, i.e. when $n+R=1$. Now, when $n+R>1$, the cases $n=0$ or $R=0$ are addressed above, while in the case $n>0$ and $R>0$, we use [Pascal's rule](https://en.wikipedia.org/wiki/Pascal%27s_rule) to conclude that
$$LHS(2k,n,R) = LHS(2k,n,R-1) + LHS(2k,n-1,R) = 0$$
by the induction assumption.
|
2
|
https://mathoverflow.net/users/7076
|
396963
| 163,900 |
https://mathoverflow.net/questions/396941
|
9
|
Might there be a good reference on the interaction of number theory with statistical physics? I am particularly interested in innovations in number theory that have led to breakthroughs in statistical physics.
One such result that I am aware of is the Lee-Yang theorem. Marc Kac apparently used insights from Pólya's analysis of the Riemann Zeta function [2] in order to come up with an early proof of the Lee-Yang theorem(aka Lee-Yang Circle theorem) which states that the zeros of certain partition functions lie on the unit circle [1].
At present, I am more familiar with parallel developments such as the correspondence between Gaussian Unitary Ensembles(GUEs) and Montgomery's pair correlation conjecture [3].
References:
-----------
1. Knauf, Andreas (1999), "Number theory, dynamical systems and statistical mechanics", Reviews in Mathematical Physics, 11 (8): 1027–1060, CiteSeerX 10.1.1.184.8685
2. Pólya, G.: Collected Papers, Vol. II: Locations of Zeros (R.P. Boas, ed.).
Cambridge: M.I.T. Press 1974
3. Montgomery, Hugh L. (1973), "The pair correlation of zeros of the zeta function", Analytic number theory, Proc. Sympos. Pure Math., XXIV, Providence, R.I.: American Mathematical Society
|
https://mathoverflow.net/users/56328
|
Innovations in number theory leading to breakthroughs in statistical mechanics
|
1. The paper “The reasonable and unreasonable effectiveness of number theory in statistical mechanics” by George Andrews comes to mind. It is a nice survey that mentions some of the more striking appearances of number-theoretic results in statistical mechanics, such as the Rogers-Ramanujan identities and their connection to the hard hexagon model. It appeared in the book The Unreasonable Effectiveness of Number Theory (Ed. Stefan A. Burr, Amer. Math. Soc. 1993), which has a few other nice essays by different authors.
2. The journal [Communications in Number Theory and Physics](https://www.intlpress.com/site/pub/pages/journals/items/cntp/_home/_main/index.php) might have some relevant material.
|
10
|
https://mathoverflow.net/users/78525
|
396972
| 163,903 |
https://mathoverflow.net/questions/396952
|
4
|
In the paper of Rouquier on the dimension of triangulated categories (found [here](https://www.math.ucla.edu/%7Erouquier/papers/dimension.pdf)) lemma 3.5 says:
**Lemma** Let $\mathcal{T}$ be a triangulated category and let $\mathcal{T}\_1$ and $\mathcal{T}\_2$ be triangulated subcategories of $\mathcal{T}$ such that $\mathcal{T}=\mathcal{T}\_1\diamond\mathcal{T}\_2$. Then $dim\mathcal{T}\leq dim\mathcal{T}\_1+dim\mathcal{T}\_2+1$.
There is no proof given in the text and it says that the lemma is "clear". However, at the moment, I really can't see how one can proceed. My guess is that by assuming that $\mathcal{T}\_1=\langle M\_1\rangle\_{1+dim\mathcal{T}\_1}$ and $\mathcal{T}\_2=\langle M\_2\rangle\_{1+dim\mathcal{T}\_2}$, one can find an object $M$ such that $\mathcal{T}=\langle M\rangle\_{2+dim\mathcal{T}\_1+dim\mathcal{T}\_2}$.
Is there something trivial that I am missing?
Any help would be very much appreciated.
|
https://mathoverflow.net/users/171303
|
How to prove a lemma of Rouquier on the dimension of triangulated categories?
|
If $\mathcal{T}\_{1}=\langle M\_{1}\rangle\_{d\_{1}+1}$ and
$\mathcal{T}\_{2}=\langle M\_{2}\rangle\_{d\_{2}+1}$, then
$\mathcal{T}\_{1}\ast\mathcal{T}\_{2}\subseteq\langle M\_{1}\oplus
M\_{2}\rangle\_{d\_{1}+d\_{2}+2}$.
**Sketch proof:** Since $\mathcal{T}\_{i}\subseteq\langle M\_{1}\oplus
M\_{2}\rangle\_{d\_{i}+1}$, it suffices to show that
$$\langle M\_{1}\oplus
M\_{2}\rangle\_{s}\ast\langle M\_{1}\oplus M\_{2}\rangle\_{t}=
\langle M\_{1}\oplus M\_{2}\rangle\_{s+t}.$$
This follows by induction on $s$ and $t$, using the fact that $\ast$
is associative (i.e.,
$\mathcal{A}\ast(\mathcal{B}\ast\mathcal{C})=(\mathcal{A}\ast\mathcal{B})\ast
\mathcal{C}$ for triangulated subcategories
$\mathcal{A},\mathcal{B},\mathcal{C}$ of $\mathcal{T}$), which follows
from the octahedral axiom.
|
7
|
https://mathoverflow.net/users/22989
|
396987
| 163,905 |
https://mathoverflow.net/questions/396985
|
4
|
This is a simple question, just looking for a reference for a formula.
As far I understand the genus of a prime Fano $n$-fold is defined to be the genus of a complete intersection of $n-1$ smooth divisors in the system $|-K\_{X}|$ (see [https://www.math.ens.fr/~debarre/ExposePoitiers2013.pdf](https://www.math.ens.fr/%7Edebarre/ExposePoitiers2013.pdf)). For $n=3$ this number equals $$\frac{(-K\_{X})^3}{2} +1.$$
**Question:** What is the general formula for the genus of Fano variety?
Unfortunately I was not able to find this in the literature.
**Edit:** I thank Pop for the formula for the quantity given above. However, this does not agree with the use of term genus used in the literature (that was an incorrect guess on my part). By [https://www.math.ens.fr/~debarre/ExposePoitiers2013.pdf](https://www.math.ens.fr/%7Edebarre/ExposePoitiers2013.pdf) page 7 it seems that the most widely used definition of genus is $$\frac{(-K\_{X})^n}{2 \cdot \iota(X)^n } +1 ,$$ where $\iota(X)$ is the index of $X$ i.e. the maximal divisibility of $c\_{1}(X)$ in $H^{2}(X,\mathbb{Z})$.
I checked a few examples and this agrees with the way that other authors use the term genus in their papers.
**Question 2**: Why is this the accepted notion of genus, is there a geometric reason for this?
One could think that this is the genus of the intersection of $(n-1)$ smooth divisors in the system $$|\frac{-K\_{X}}{\iota(X)}|$$, but this is not the case. For example if the index is $1$ then the formula does not give the formula of Pop. Is the definition related to $K3$ surfaces?
**Edit 2**: Just to make it clear that the question is resolved I write a brief explanation.
As Tom Ducat pointed out, the formula I referenced made a strong assumption on the index. Indeed the genus of a Fano variety is the genus of a curve given by intersecting divisors in the class $A = -K\_{X}/\iota(X)$, where $\iota(X)$ is the index of $X$.Then by the computation of Pop's answer one can recover the formula $$g= 1+\frac{n-\iota(X)-1}{2}A^n$$, which agrees with Fujita's paper and the examples from the literature.
|
https://mathoverflow.net/users/99732
|
Formula for genus of a Fano variety
|
It's not clear from your question if you really want a reference (as per the first line) or the formula itself (as per what is written after "Question"). But the computation of the genus is not hard so let me write it here in case that answers the question.
If $C$ is such a smooth curve in $X$, then the normal bundle is $N\_{C/X} = (-K\_X)\_{|C}^{\oplus n-1}$. So adjunction says that
\begin{align\*}
\operatorname{deg} K\_C &= \operatorname{deg} (K\_X)\_{|C} + \operatorname{deg} \wedge^{n-1} N\_{C/X} \\
&= \operatorname{deg} (K\_X)\_{|C} + (n-1) \operatorname{deg} (-K\_X)\_{|C} \\
&= (n-2) \operatorname{deg} (-K\_X)\_{|C}\\
&= (n-2)(-K\_X)^n
\end{align\*}
Since $\operatorname{deg} K\_C = 2g(C)-2$ you get
\begin{align\*}
g(C) = \frac12 (n-2)(-K\_X)^n + 1
\end{align\*}
|
6
|
https://mathoverflow.net/users/121595
|
396988
| 163,906 |
https://mathoverflow.net/questions/396969
|
5
|
Suppose $\pi:\mathcal{X}\rightarrow S$ is a smooth family of complex affine varieties (to make things simpler, we can actually assume it is locally trivial). Let $P$ be a smooth projective variety, and assume there is an embedding $\mathcal{X}\hookrightarrow P\times S$ over $S$. Let $\overline{\pi}: \overline{\mathcal{X}}\rightarrow S$ be the projection for the closure of $\mathcal{X}$ in $P\times S$. Is it always true that for every (closed) point $s\in S$, $\overline{\pi^{-1}(s)}=\overline{\pi}^{-1}(s)$? Any references will be greatly appreciated.
|
https://mathoverflow.net/users/140928
|
When is the fiberwise compactification (not) equal to the compactification of the family?
|
No.
Let $S = \mathbb{A}^1\_{\mathbb{C}}$ and $\mathcal{X} = \mathbb{A}^1\_S \rightarrow S$ be the constant family. Let $\mathcal{Y}$ be the blow-up of the surface $\mathbb{P}^1\_S$ at the closed point over $0\in S$ lying at infinity. In other words: $\mathbb{P}^1\_S = \mathbb{P}^1\times \mathbb{A}^1$ and $\mathcal{Y}$ is the blow-up at the point $(\infty,0)$.
Since blowing-up is an isomorphism away from the exceptional locus, there is an open immersion $\mathcal{X} \hookrightarrow \mathcal{Y}$ of $S$-schemes. Since $\mathcal{Y}$ is irreducible, $\mathcal{X}$ is dense in $\mathcal{Y}$.
Since the morphisms $\mathcal{Y} \rightarrow \mathbb{P}^1\_S$ and $\mathbb{P}^1\_S \rightarrow S$ are projective, the composite $\mathcal{Y} \rightarrow S$ is projective as well. Choose $n \geq 0$ large enough such that there exists a closed embedding $\mathcal{Y} \hookrightarrow P\_S$, where $P = \mathbb{P}^n\_{\mathbb{C}}$. The closure $\bar{\mathcal{X}}$ of $\mathcal{X}$ in $P\_S$ is exactly $\mathcal{Y}$.
The fibre of $\mathcal{Y} \rightarrow S$ above $0$ is a union of two projective lines. However, the fibre of $\mathcal{X} \rightarrow S$ above $0$ is just $\mathbb{A}^1$. Therefore $\mathcal{X}\_0$ is not dense in $(\bar{\mathcal{X}})\_0$, as required.
|
4
|
https://mathoverflow.net/users/110362
|
396991
| 163,907 |
https://mathoverflow.net/questions/396883
|
3
|
$\DeclareMathOperator\SmallGroup{SmallGroup}$**Definition.** A subset $A$ of a group $G$ is called *product-1-free* if for any sequence of pairwise distinct elements $a\_1,\dots,a\_n$ of $A$ the product $a\_1\cdots a\_n$ is not equal to 1 in $G$.
For a finite group $G$, let $f\_1(G)$ be the largest cardinality of a product-1-free set in $G$.
**Example 1.** For every $n\ge 2$ the cyclic group $C\_n$ has $$f\_1(C\_n)\ge \left\lfloor\frac{\sqrt{8n-7}-1}2\right\rfloor.$$
This lower bound follows from the observation that $1+\dots+k<n$ for $k=\left\lfloor\frac{\sqrt{8n-7}-1}2\right\rfloor$.
**Example 2.** Each finite Boolean group $G$ has $f\_1(G)=\log\_2(|G|).$
>
> **Problem.** Is $\lfloor\log\_2(|G|)\rfloor\le f\_1(|G|)<\sqrt{2|G|}$ for any finite group $G$?
>
>
>
**Remark 1.** By a greedy algorithm mentioned in the comment of @Nick Gill, one can prove the following lower bounds:
$1)$ $f\_1(G)+2^{f\_1(G)}\ge |G|$ for every finite Abelian group $G$;
$2)$ $f\_1(G)+e\cdot f\_1(G)! \ge |G|$ for every finite group.
**Remark 2.** Calculations of $f\_1(G)$ in GAP show that a counterexample to the problem cannot be found among groups of cardinality $\le 50$ (see my partial answer below).
**Added in Edit.** After asking this question, I have found that it has been considered in the literature (see e.g. p.95 in [the book of Erdos and Graham](https://eu1lib.org/dl/447894/7c30e8)). In particular, the number $O(G)=f\_1(G)+1$ is known as [Olson's constant](https://en.wikipedia.org/wiki/Davenport_constant) of a group $G$. Below I write down some known non-trivial upper bounds for the number $f\_1(G)$.
1. By a result of [Olson](http://matwbn.icm.edu.pl/ksiazki/aa/aa28/aa2825.pdf) (1975), $f\_1(G)<3\sqrt{|G|}$ for any finite group $G$.
2. By a result of [Hamindoune and Zemor (1996)](http://matwbn.icm.edu.pl/ksiazki/aa/aa78/aa7823.pdf), $f\_1(C\_p)<\sqrt{2p}+5\ln(p)$ for any prime number $p$.
3. By a result of [Hamindoune and Zemor (1996)](http://matwbn.icm.edu.pl/ksiazki/aa/aa78/aa7823.pdf), $f\_1(G)\le \sqrt{2|G|}+O(|G|^{1/3}\ln(|G|))$ for any finite Abelian group $G$.
4. By a result of [Hoi Nguyen, E. Szemeredi, and Van Vu](https://people.math.osu.edu/nguyen.1261/cikk/zero.pdf) (2009), for every sufficiently large prime number $p$ we have
$f\_1(C\_p)=\big\lfloor\frac{\sqrt{8p-7}-1}2\big\rfloor$.
5. By a result of [Balandraud](https://link.springer.com/article/10.1007/s11856-011-0171-9) (2012), for every prime number $p$ we have $f\_1(C\_p)=\big\lfloor\frac{\sqrt{8p-7}-1}2\big\rfloor$.
6. By the observation of [Hoi Nguyen, E. Szemeredi, and Van Vu](https://people.math.osu.edu/nguyen.1261/cikk/zero.pdf), for every $n\ge 4$ and $m=\frac12n(n+1)-1$ the set $A=\{1,3,\dots,n,m-2\}\subset C\_m$ is product-1-free witnessing that $f\_1(C\_m)\ge n=1+\big\lfloor\frac{\sqrt{8m-7}-1}2\big\rfloor$.
7. It is easy to see that $f\_1(C\_n\oplus C\_n)\ge f\_1(C\_n)+n-1$ for every $n\in \mathbb N$. If $n>6000$ is prime, then $$f\_1(C\_n\oplus C\_n)=f\_1(C\_n)+n-1=n+\left\lfloor\tfrac{\sqrt{8n-7}-3}2\right\rfloor$$ according to the result of [Bhowmik and Schlage-Puchta](https://www.impan.pl/shop/en/publication/transaction/download/product/83035) (2010) who improved an earlier result of
[Gao, Ruzsa and Thangadurai](https://www.sciencedirect.com/science/article/pii/S009731650400038X?via%3Dihub) (2004).
|
https://mathoverflow.net/users/61536
|
Large product-1-free sets in finite groups
|
Let me sketch a strategy for proving the lower bound:
**Lemma**: Let $S\_1,\dots, S\_k$ be the composition factors of $G$. Then
$$ f\_1(G)\geq f\_1(S\_1)+\cdots +f\_1(S\_k).$$
**Sketch of proof**:
Take a series:
$$G=G\_0\rhd G\_1 \rhd \cdots \rhd G\_k=\{1\}$$
where $G\_{i-1}/G\_i\cong S\_i$. For each $i$ write $\ell\_i=f\_1(S\_i)$ and take $g\_{i,1},\dots g\_{i, \ell\_i}$ to be a set of elements in $G\_{i-1}$ such that $g\_{i,1}G\_i,\dots, g\_{i, \ell\_i}G\_i$ is a product-1-free set of $G\_{i-1}/G\_i=S\_i$. I claim that these elements will be a product-1-free set of $G$. To see this take $\Delta$ any subset of them and form a product in some order -- write this as $f\_1\cdots f\_r$. Let $i$ be the smallest integer such that $\Delta$ contains an element in $G\_{i-1}\setminus G\_i$. Now consider the product $(f\_1G\_i)\cdots (f\_rG\_i)$. A bunch of these will be equal to the identity (corresponding to elements $f\_j\in G\_i$). Those that aren't will be distinct and will correspond to a product-1-free set of $G\_{i-1}/G\_i$. Thus the product will not lie in $G\_i$ and so cannot equal $1$. QED
**Edit -- 8 Jul 2021 -- using comments of Sean Eberhard.**
The lemma reduces the problem to a question about simple groups. The original post showed that if $G$ is cyclic, then $f\_1(G)\sim \sqrt{|G|}>\log\_2|G|$. Combining this with the lemma gives the result for $G$ solvable. (And the answer by Taras Banakh does this in detail.)
So we must deal with $G$ non-abelian simple. These are dealt with by applying the bound in the original post to large cyclic subgroups.
**Suppose $G$ is of Lie type of rank $r$ over $\mathbb{F}\_q$.** Then one can check that $|G|<q^{8r^2}$. On the other hand there is typically a cyclic subgroup of order at least $q^{r-2}$ (I say "typically" because I haven't checked every case.)
Now $\sqrt{q^{r-2}}>\log\_2(q^{8r^2})$ unless $r$ is small. For $r$ small we can use the fact that $G$ has a large solvable subgroup (the Borel) for which the cyclic bound in the original post combined with the lemma gives a much better bound than that which is needed. This will give the bound for the whole group.
**Suppose $G=A\_n$ with $n\geq 5$.** In this case $G$ has an element of order $d$ where $d$ is the product of the first $k$ primes $p\_1,\dots, p\_k$ where $k$ is chosen to be as large as possible such that $p\_1+\cdots+p\_k\leq n$. Now $d\sim\exp(\sqrt{n\log n})$ and, again, we use the fact that $\sqrt{\exp(\sqrt{n\log n})}>\log(n!)$ provided $n$ is large enough.
**Suppose $G$ is sporadic**. This case looks more tricky -- element orders in the sporadics tend to be small compared to the size of the group. My strategy would be to choose a small index maximal subgroup for which one can prove a better lower bound (using the previous cases) and, with any luck, this will be sufficient to give the lower bound for the whole group.
**Edit 2 -- 8 July 2021**.
1. It seems to me that this same method should yield a similar lower bound even if you allow repeats in your product (but still bound the length of the product by the size of the set of course). There will be more exceptions of course, e.g. $C\_2^d$.
2. It also seems that one could obtain a similar lower bound for product-$g$-free sets for any $g\in G$. Defining $f\_g(G)$ in the obvious way, one could use the method of proof in the lemma to prove something like this:
>
> Let $g\in G$ and $N\lhd G$. If $g\in N$, then
> $$f\_g(G) \geq f\_{1}(G/N) + f\_g(N).$$
> If $g\in G\setminus N$, then
> $$f\_g(G) \geq f\_{gN}(G/N) + |N|.$$
>
>
>
This would again reduce the problem to a question about simple groups. If one could show something like $\sqrt{|G|}$ lower bound for cyclic groups as in the OP, then the same general bound follows.
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4
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https://mathoverflow.net/users/801
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397002
| 163,909 |
https://mathoverflow.net/questions/397005
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1
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Consider a subspace $V$ of $\mathcal{H}\_A \otimes \mathcal{H}\_B$, with $\mathcal{H}\_A$ and $\mathcal{H}\_B$ finite-dimensional Hilbert spaces, that is $1\_A \otimes U$ invariant for all unitary operators $U$ on $\mathcal{H}\_B$. Is $V = V\_A \otimes \mathcal{H}\_B$ for some subspace $V\_A$ of $\mathcal{H}\_A$?
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https://mathoverflow.net/users/69967
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Is a $1_A \otimes U$ invariant subspace of $\mathcal{H}_A \otimes \mathcal{H}_B$ a product $V_A \otimes \mathcal{H}_B$?
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The action of $1\_A\otimes U$ on $V$ defines the action of the unitary group $U(H\_B)$ on $V$. This action induces the action of Lie algebra $\mathfrak{u}(H\_b)$ on $V$. The action has the form $1\otimes T$ for skew-hermitian $T$. Every operator on $H\_B$ has the form $T\_1+iT\_2$ for $T\_1,T\_2$ skew-hermitian, so we deduce that the subspace $V$ closed under the action of $1\otimes S$ for any operator $S$. It follows that $V=V\_A\otimes H\_B$ for some $V\_A\subset H\_A$.
It seems like we can also directly use the fact, that unitary operators span the vector space of all operators.
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2
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https://mathoverflow.net/users/140292
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397010
| 163,910 |
https://mathoverflow.net/questions/396950
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4
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Suppose that $(M,g)$ is a smooth Lorentzian manifold, that $I\subset \mathbb R$ is a closed **finite** interval and that $\gamma: I \to M$ is a smooth null geodesic on $M$, that is to say, it satisfies
$$ \nabla\_{\dot{\gamma}(s)}\dot\gamma(s)=0 \quad \text{and} \quad g(\dot{\gamma}(s),\dot{\gamma}(s))=0\quad \text{for all $s \in I$}.$$
Assume also that $\beta: I\to M$ is a smooth curve such that
$$g(\dot{\beta}(s),\dot{\beta}(s)) =0 \quad \text{for all $s \in I$},$$
and additionally that $\beta$ is not completely overlapping with $\gamma$ at any point (this just means that there is no non empty **open** interval $J \subset I$ such that $\beta|\_{J}$ is a reparametrization of $\gamma$ on some interval of $I$).
Does it follow that $\gamma$ and $\beta$ can only intersect a finite number of times? If the answer is no in general, can one impose some soft assumptions to make the intersection points finite?
Thanks,
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https://mathoverflow.net/users/50438
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A question on null geodesics in Lorentzian geometry
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We argue by contradiction. Suppose there are infinitely intersections. By reversing time orientation if necessary, there exists a monotonically increasing sequence of times $s\_n \in I$ such that at each $\gamma(s\_n)$ the curves $\gamma$ and $\beta$ intersect. Since $I$ is compact $s\_n$ has a limit $s'\in I$. For convenience label the points $p\_n = \gamma(s\_n)$, and $p' = \gamma(s')$.
Since $I$ is compact, we have that the image $\beta$ is compact, and hence closed, in $M$. Since $p\_n \to p'$ and $p\_n \in \beta$, we have that $p'\in \beta$ also. By taking a subsequence if necessary, we can find a increasing sequence of times $t\_n$ such that $\beta(t\_n) = p\_n$ and $t\_n$ converges to $t'$, for which $\beta(t') = p'$.
**Case I**: suppose there exists some $n\_0$ such that for all $n > n\_0$, the causal segment $\beta|\_{[t\_n, t']}$ is geodesic (up to reparametrization).
* $\dot{\beta}(t')$ cannot be equal to $\dot{\gamma}(s')$: other wise by the uniqueness theorem for geodesics $\beta$ and $\gamma$ must overlap on some interval.
* Since the tangent vectors are not equal, this means that on every open neighborhood of the origin in $T\_{p'}M$, the exponential map will have to be non-injective, which contradicts the theorem on existence of normal neighborhoods.
So case I is ruled out.
**Case II**: the alternative is that $\beta|\_{[t\_n,t']}$ is not geodesic for any $n$. This implies that $p\_n$ is to the *chronological past* of $p'$ (See Theorem 8.1.2 and Corollary in Wald's *General Relativity*.)
Theorem 2.14 from <https://arxiv.org/abs/gr-qc/0609119> states that every point in a Lorentzian manifold has a causally convex globally hyperbolic neighborhood. Fix such a neighborhood $V'$ of $p'$. Since $\beta$ is compact, there is some $n\_0$ such that $\beta|\_{[t\_{n\_0}, t']}$ and $\gamma|\_{[s\_{n\_0}, s']}$ remain both in $V'$. For each $n > n\_0$, that $p\_n$ is to the chronological past of $p'$ implies there exists a sequence of *timelike* tangent vectors $v\_n\in T\_{p'}M$ with $\exp\_{p'}(v\_n) = p\_n$.
$v\_n$ must converge to zero, as the exponential map is a diffeomorphism on a sufficiently small neighborhood of zero, and $p\_n\to p'$. But this shows that for any neighborhood of zero there exists simultaneously a timelike and a null vector within that neighborhood both of which exponentiates to $p\_n$, contradicting the existence of normal neighborhoods.
This rules out case II.
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4
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https://mathoverflow.net/users/3948
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397011
| 163,911 |
https://mathoverflow.net/questions/397006
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4
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This is a question about marginals of probability measures, which seems unrelated to previous questions.
Let $\mathbb{S}^{d-1}\subset \mathbb{R}^d$ be the unit sphere. Assume that for each $\theta\in \mathbb{S}^{d-1}$ there is an associated probability measure $\mu\_\theta$ on $\mathbb{R}$.
**Question:** Under what conditions does there exist a probability measure $\mu$ on $\mathbb{R}^d$ such that
$$\mbox{if }X\sim \mu,\mbox{ then }\langle \theta,X\rangle \sim \mu\_\theta\mbox{ for all }\theta\in \mathbb{S}^{d-1}.$$
(Here $\langle\cdot,\cdot\cdot\rangle$ is the Euclidean inner product and $A\sim \nu$ means object $A$ has prob. law $\nu$.)
A **necessary condition** is that the map $\theta\mapsto \mu\_\theta$ be continuous under the weak topology on probability measures on $\mathbb{R}$. Is this condition sufficient?
(*Motivation* comes from the study of the Sliced Wasserstein distance. If the answer to the above question is "yes", then in principle one can "easily compute" barycenters for SW.)
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https://mathoverflow.net/users/8354
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Existence of measures with given 1d marginals
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$\newcommand\th\theta\newcommand\vpi\varphi\newcommand\R{\mathbb R}\newcommand\S{\mathbb S^{d-1}}$**Condensed version of the answer:** If $\th\cdot X\sim\mu\_\th$ for all $\th\in\S$, then for each nonzero $t\in\R^d$ the distribution, say $\mu\_t$, of $t\cdot X$ is determined by a simple rescaling of $\mu\_\th$ (the case $t=0$ is trivial). On the other hand, by [Bochner's theorem](https://en.wikipedia.org/wiki/Bochner%27s_theorem), a family $(\mu\_t)\_{t\in\R^d}$ of probability measures on $\R$ is the family of the distributions of the random variables $t\cdot X$ (for some random vector $X$ in $\R^d$) iff the function $\R^d\ni t\mapsto\vpi(t):=\int\_{\R}e^{ix}\mu\_t(dx)$ is positive definite and continuous. Thus, for any given family $(\mu\_\th)\_{\th\in\S}$ of probability measures on $\R$, your desired condition in question will hold if and only if $\vpi$ is positive definite and continuous.
**Detailed version of the answer:**
Suppose that
$$\text{there is a probability measure $\mu$ on $\R^d$ such that},\\
\text{if $X\sim\mu$, then $\th\cdot X\sim\mu\_\th$ for all $\th\in\S$.}\tag{0}$$
Then
$$E\_\mu f(\th\cdot X)=\int\_{\R}f\,d\mu\_\th \tag{1}$$
for all bounded continuous functions $f\colon\R\to\R$, where $E\_\mu$ denotes the expectation assuming that $X\sim\mu$. In particular, (1) implies
$$E\_\mu e^{ir\th\cdot X}=\int\_{\R}e^{irx}\,\mu\_\th(dx) \tag{2}$$
for all $r\in\R$ and all $\th\in\S$, so that
$$E\_\mu e^{it\cdot X}=\vpi(t)$$
for all $t\in\R^d$,
where
$$\vpi(t):=
\left\{
\begin{aligned}
\int\_{\R}e^{i|t|x}\,\mu\_{t/|t|}(dx) &\text{ if }t\in\R^d\setminus\{0\}, \\
1 &\text{ if }t=0.
\end{aligned}
\right.
\tag{3}$$
So, if the condition (0) holds, then the function $\vpi$ -- being the characteristic function (c.f.) of $X$ -- is positive definite and continuous.
Vice versa, by [Bochner's theorem](https://en.wikipedia.org/wiki/Bochner%27s_theorem), if $\vpi$ is positive definite and continuous, then there is a probability measure $\mu$ on $\R^d$ such that $E\_\mu e^{it\cdot X}=\vpi(t)$ for all $t\in\R^d$, so that (2) holds for all $r\in\R$ and all $\th\in\S$, that is, the c.f. of $\th\cdot X$ is the same as the c.f. of $\mu\_\th$, which means that $\th\cdot X\sim\mu\_\th$. So, (0) holds if the function $\vpi$, defined by (3), is positive definite and continuous.
Thus, for any given family $(\mu\_\th)\_{\th\in\S}$ of probability measures on $\R$, condition (0) will hold if and only if $\vpi$ is positive definite and continuous.
---
Somewhat related to this is the work by Shepp and his co-authors on probabilistic tomography, about partial restoration of the distribution of a random vector $X$ in $\R^d$ based on the known distributions of a finite number of linear functionals of $X$; see e.g. [this paper](https://core.ac.uk/download/pdf/214202814.pdf) and references there.
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5
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https://mathoverflow.net/users/36721
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397012
| 163,912 |
https://mathoverflow.net/questions/396533
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3
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I am new to Math Overflow, so I am not really sure whether this question fits the community standards. But, I posted this question in Stack Exchange and recieved no answers. Moreover, nothing even close to an answer to my question can be (easily) found online. So, I decided to post it among researchers to see whether I can get any significant explations.
I was studying Newton's Forward Interpolation and backward interpolation in a computer science course and the form that I got them in, is as follows-
**Forward Interpolation**
>
> $$f(x)=y=y\_0+\binom u1 \Delta y\_0+\binom u2 \Delta^2y\_0+\dots +\binom
> un \Delta^ny\_0$$ where \begin{align\*} x\_i&=x\_0+ih \;\text{ (equispaced
> points)}\\ u&=\frac{x-x\_0}{h}\\ \Delta y\_i &= y\_{i+1}-y\_i,
> \;i=0,1,\dots\\ \Delta^k y\_i &= \Delta^{k-1}y\_{i+1}-\Delta^{k-1}y\_{i}
> \end{align\*}
>
>
>
**Backward Interpolation**
>
> $$f(x)=y=y\_n+\binom u1 \Delta y\_{n-1} + \binom u2
> \Delta^2y\_{n-2}+\dots +\binom{u+n-1}n \Delta^n y\_0$$ where
> \begin{align\*} x\_i&=x\_0+ih \;\text{ (equispaced points)}\\
> u&=\frac{x-x\_n}{h}\\ \Delta y\_i &= y\_{n-1}-y\_{n-i-1}, \;i=0,1,\dots\\
> \Delta^k y\_i &= \Delta^{k-1}y\_{n-1}-\Delta^{k-1}y\_{n-i-1} \end{align\*}
>
>
>
Now, I understood polynomial approximation (that was taught just before these interpolations). But, I don't understand why and how these interpolations work.
I can guess that we have taken equispaced points, found the values of $f$ at those points, and tried to find a *better behaved* approximation that satisfies those values of $f(x)$. But, I don't have any intuition regarding how this approximation function behaves. I don't understand what role the binomials (that too with non integer values) play or what extra advantage the equispaced points give, and I have no idea of how the complex definitions of $\Delta^k$ help us to get this approximation. I tried to look into some of the expressions of $\Delta^k$ and these are what I calculated (about the forward part)-
\begin{align\*}
\Delta y\_0 &= y\_1-y\_0\\
\Delta^2y\_0 &= \Delta y\_1- \Delta y\_0\\
&=y\_2-2y\_1+y\_0
\end{align\*}
So, it's clear that these expressions wont simplify. They will just go on getting uglier.
As of now, I am completely confused about how this thing works. Also, what is the difference between Forward and Backward interpolation, and when to use which one? At this level of confusion, I don't think, rigorous proofs will be of much use. So, I would like to have a geometric interpretation, or simply a concrete intuition to arrive at such an expression.
Can somebody please help me with this? Thanks in advance.
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https://mathoverflow.net/users/311366
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About Newton's forward and backward interpolation
|
*The forward and backward finite differences and the derivative lower the degree of a polynomial by one.*
This property underlies the construction of series expansions of polynomials and, therefore, analytic functions with appropriate convergence properties in terms of diverse polynomial sequences, in particular, Sheffer sequences. In the terminology of the Sheffer operator calculus, these three operators are delta ops, or lowering ops.
The derivative acting on the power $x^n$ lowers the degree by one. In other words, the derivative is the lowering operator for the fundamental Sheffer sequence of polynomials $S\_n(x) = x^n$, the simplest such sequence. Specifically,
$$D \; x^n = n \; x^{n-1}.$$
Consequently,
$$\frac{D^k}{k!} \; x^n \; |\_{x=0} = \binom{n}{k} \; x^{n-k} \; |\_{x=0} = \delta\_{n-k}.$$
A polynomial of degree $n$ can be expanded as
$$p\_n(x) = \sum\_{k=0}^n \; c\_k \; x^k,$$
and the coefficients determined as
$$c\_k = \frac{D\_{x=0}^k}{k!} \; p\_n(x) ,$$
giving the Taylor series expansion
$$p\_n(x) = \sum\_{k \geq 0} [\; D\_{x=0}^k \; p\_n(x) \;] \; \frac{x^k}{k!}.$$
Now instead of the sequence of power monomials $x^n$, consider the polynomial falling factorials
$$(x)\_n = \frac{x!}{(x-n)!} = (x)(x-1) \cdots (x-n+1) = n! \; \binom{x}{n}.$$
The e.g.f. of a binomial Sheffer polynomial sequence $B\_n(x) = (B.(x))^n$ is
$$ e^{B.(x)t} = e^{xB(t)},$$
and the lowering operator defined by
$$L \; B\_n(x) = n \; B\_{n-1}(x)$$
is
$$L = B^{(-1)}(D),$$
where $B^{(-1)}(t)$ is the compositional inverse about the origin of $B(t)$. This is verified by
$$ B^{(-1)}(D) \; e^{B.(x) t} = B^{(-1)}(D) \; e^{x B(t)} = B^{(-1)}(B(t)) \; e^{x B(t)} = t \; e^{B.(x)t}.$$
The falling factorials have the e.g.f
$$e^{(x).t} = (1+t)^x = e^{x \ln(1+t)},$$
so their lowering op is
$$L = e^{D}-1 .$$
Check:
$$L \; (x)\_n = (e^{D}-1) \; (x)\_n = \frac{(x+1)!}{(x+1-n)} - \frac{x!}{(x-n)!}$$
$$= (x+1 -(x-n+1)) \; \frac{x!}{(x-n+1)!} = n \; (x)\_{n-1} . $$
Consequently,
$$\frac{(e^D-1)^k}{k!} \; (x)\_n \; |\_{x=0}= \binom{n}{k} \; (0)\_{n-k} = \delta\_{n-k}.$$
A polynomial of degree $n$ can be expanded as
$$p\_n(x) = \sum\_{k=0}^n \; c\_k \; (x)\_k$$
and the coefficients determined as
$$c\_k = \frac{(e^{D\_{x=0}}-1)^k }{k!} \; p\_n(x) ,$$
giving the series expansion
$$p\_n(x) = \sum\_{k \geq 0} [\; (e^D-1)^k \; p\_n(x) \; |\_{x=0} \;] \; \frac{(x)\_k}{k!}.$$
The lowering op is the forward difference op
$$(e^D -1) \; f(x) = f(x+1) - f(x), $$
and the $n$-th finite forward difference is
$$(e^D -1)^n \; f(x) = (-1)^n \; \sum\_{k=0}^n (-1)^k \; \binom{n}{k} \; e^{kD} \; f(x)$$
$$ = (-1)^n \; \sum\_{k=0}^\infty \; (-1)^k \; \binom{n}{k} \; f(x+k) := (-1)^n \; \nabla\_{k=0}^n \; f(x+k).$$
(For the purists: I use the convenient convention consistent with limits of the entire reciprocal of Euler's gamma function that $\binom{n}{m}$ vanishes for $m > n$. This allows invariance of the notation when $n$ is generalized to real or complex numbers.)
Then
$$p\_n(x) = \sum\_{k \geq 0} [\;(e^D-1)^k \; p\_n(x) \; |\_{x=0} \;] \; \frac{(x)\_k}{k!}$$
$$ = \sum\_{k \geq 0} \binom{x}{k} \; (-1)^k \; \nabla\_{m=0}^k \; p\_n(m)$$
$$ = \nabla\_{k=0}^{x} \; \nabla\_{m=0}^k \; p\_n(m),$$
the Newton series for the polynomial $p\_n(x)$.
The backward difference operator is
$$(1- e^{-D}) \; f(x) = f(x) - f(x-1).$$
The compositional inverse of $1-e^{-t}$ is $-\ln(1-t)$, so the backward difference operator is the lowering op of the binomial Sheffer sequence with the e.g.f
$$ e^{B.(x)t} = e^{-x \ln(1-t)} = (1-t)^{-x},$$
which is the e.g.f. for the rising factorials, a.k.a. the Pochhammer symbol,
$$(x)\_\bar{n} = (x+n-1)(x+n) \cdots (x) = n! \; \binom{x-1+n}{n} = (-1)^n\; n! \; \binom{-x}{n}.$$
The $n$-th order backward difference is
$$(1-e^{-D})^n \; f(x) = \sum\_{k=0}^n (-1)^k \; \binom{n}{k} \; e^{-kD} \; f(x)$$
$$ = \; \sum\_{k=0}^\infty \; (-1)^k \; \binom{n}{k} \; f(x-k) := \nabla\_{k=0}^n \; f(x-k),$$
and
$$p\_n(x) = \sum\_{k \geq 0} [\;(1-e^{-D})^k \; p\_n(x) \; |\_{x=0} \;] \; (-1)^k \; \frac{(-x)\_k}{k!}$$
$$ = \sum\_{k \geq 0} (-1)^k \; \binom{-x}{k} \; \nabla\_{m=0}^k \; p\_n(-m)$$
$$ = \nabla\_{k=0}^{-x} \; \nabla\_{m=0}^k \; p\_n(-m).$$
Check: $p\_n(-n) = \nabla\_{k=0}^{n} \; \nabla\_{m=0}^k \; p\_n(-m).$
These are the umbral relations cast by
$$y^x= (1-(1-y))^x = \nabla\_{k=0}^{x} \; \nabla\_{m=0}^k \; y^m$$
$$ = (1-(1-\frac{1}{y})^{-x} = \nabla\_{k=0}^{-x} \; \nabla\_{m=0}^k \; y^{-m}$$
where $=$ is understood to mean equivalence under analytic continuation, and in umbral notation, these Newton series may be expressed succinctly as
$$p\_n(x) = (p\_n(.))^x = (1-(1-p\_n(.)))^x =\nabla\_{k=0}^{x} \; (1-p\_n(.))^k $$
$$= \nabla\_{k=0}^{x} \; \nabla\_{m=0}^k \; p\_n(.)^m = \nabla\_{k=0}^{x} \; \nabla\_{m=0}^k \; p\_n(m)$$
$$ = (1-(1-\frac{1}{p\_n(.)}))^{-x} =\nabla\_{k=0}^{-x} \; (1-\frac{1}{p\_n(.)})^k $$
$$= \nabla\_{k=0}^{-x} \; \nabla\_{m=0}^k \; p\_n(.)^{-m} = \nabla\_{k=0}^{-x} \; \nabla\_{m=0}^k \; p\_n(-m).$$
(Info on connections between Mellin transform and Newton series interpolation and on the relations of the forward, backward, and central finite differences to the derivative and, therefore, to the slope of the tangent line to a point on a curve (and curvature, etc.) will appear in a couple of days on my blog.)
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3
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https://mathoverflow.net/users/12178
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397025
| 163,917 |
https://mathoverflow.net/questions/397014
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2
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Let $X$ be a projective variety over a field $K$. As a consequence of Kleiman's criterion, when $K$ is algebraically closed, we have that if $D$ is a nef divisor on $X$ then $D$ is pseudo-effective.
Does this hold also when $K$ is not algebraically closed?
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https://mathoverflow.net/users/14514
|
Nef and pseudo-effective divisors over non algebraically closed fields
|
I believe the answer is yes. I will use Keeler's paper [Kee03] (and the corrigendum [Kee18]) as a reference for these questions.
First, the theorem of the base holds over arbitrary fields by [Kee03, Theorem 3.6; Kee18, Theorem E2.2] (see also [Cut15, Proposition 2.3]), and hence one can talk about the Néron–Severi space as a finite-dimensional vector space, which Keeler denotes by $V(X)$ in [Kee03, p. 256]. The nef cone $\operatorname{Nef}(X)$ is closed in $V(X)$ [Kee08, p. 256], and we can define the pseudoeffective cone $\overline{\operatorname{Eff}}(X)$ as the closure of the big cone $\operatorname{Big}(X)$ (see, e.g., [Cut15, p. 8]).
Keeler shows that Kleiman's criterion holds over arbitrary fields in [Kee03, Theorem 3.9], and we have
$$\operatorname{Int}\bigl(\operatorname{Nef}(X)\bigr) = \operatorname{Amp}(X).$$
Note that $\operatorname{Nef}(X)$ and $\operatorname{Amp}(X)$ are denoted by $K$ and $K^\circ$ in [Kee03, p. 256].
Now we prove that nef divisors are pseudo-effective. First, by the asymptotic Riemann–Roch theorem [Kol96, Chapter VI, Theorem 2.15], we have
$$\operatorname{Amp}(X) \subseteq \operatorname{Big}(X).$$
Taking closures, we then have
$$\operatorname{Nef}(X) \subseteq \overline{\operatorname{Eff}}(X).$$
References
----------
[Cut15] Steven Dale Cutkosky, *Teissier's problem on inequalities of nef divisors,* J. Algebra Appl. **14** (2015), no. 9, 1540002, 37 pp. DOI: [10.1142/S0219498815400022](https://doi.org/10.1142/S0219498815400022). MR: [3368254](https://mathscinet.ams.org/mathscinet-getitem?mr=3368254).
[Kee03] Dennis S. Keeler, *Ample filters of invertible sheaves,* J. Algebra **259** (2003), no. 1, 243–283. DOI: [10.1016/S0021-8693(02)00557-4](https://doi.org/10.1016/10.1016/S0021-8693(02)00557-4). MR: [1953719](https://mathscinet.ams.org/mathscinet-getitem?mr=1953719).
[Kee18] Dennis S. Keeler, *Corrigendum to “Ample filters of invertible sheaves”,* J. Algebra **507** (2018), 592–598. DOI: [10.1016/j.jalgebra.2018.03.024](https://doi.org/10.1016/j.jalgebra.2018.03.024). MR: [3807062](https://mathscinet.ams.org/mathscinet-getitem?mr=3807062).
[Kol96] János Kollár, *Rational curves on algebraic varieties,* Ergeb. Math. Grenzgeb. (3), Vol. 32, Springer-Verlag, Berlin, 1996. DOI: [10.1007/978-3-662-03276-3](https://doi.org/10.1007/978-3-662-03276-3). MR: [1440180](https://mathscinet.ams.org/mathscinet-getitem?mr=1440180).
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4
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https://mathoverflow.net/users/33088
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397030
| 163,919 |
https://mathoverflow.net/questions/396951
|
7
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A rep-tile is a shape that can tile larger copies of the same shape.
**Question 1:** Are there any convex pentagons that are also rep-tiles?
**Remarks:** 15 convex pentagonal tiles of the plane are known and none of them *appears* to be a rep-tile. Assuming this observation is right, one can invoke a proof given in 2017 by Michel Rao - that these 15 are the *only* convex pentagonal tiles possible - to answer our question in the negative. However, I don't know if Rao's proof has been validated and if there is a simpler (elementary) proof that there are say no convex pentagonal rep-tiles. So, here one is actually asking if there is a simpler proof for a weaker claim.
**Definition:** Let us say a *multi-way rep-tile* is a polygon *P* with the property: if *P1* and *P2* are magnified copies of *P* and *P1* can be tiled with *m* units of *P* and *P2* can be tiled with *n* units of *P* with *n*> *m*, then, a layout of *n* units can form *P2* *without* *m* of the units in the layout together forming a *P1*. As shown on this page: <https://en.wikipedia.org/wiki/Rep-tile>, there are isosceles trapeziums with multi-way property (with *m*= 4 and *n* = 9). On the other hand, the square is obviously a rep-tile but *not* multi-way.
**Question 2:** Are there other convex polygons with this multi-way rep-tile property?
|
https://mathoverflow.net/users/142600
|
Are there any convex pentagonal rep-tiles?
|
Question 2 can be answered in the affirmative, at least: there are many triangles with the multi-way rep-tile property.
Every triangle has a simple tiling of itself with $k^2$ copies (just take the affine image of the standard equilateral tiling); these are called *quadratic tilings* by Michael Beeson in his 2010 [paper](http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.180.4908) on tiling triangles by congruent triangular tiles. So for a triangle to have this property, it suffices to have *any* non-quadratic tiling of size $m$, since we can choose a larger quadratic tiling with some $n>m$ tiles: the quadratic tiling has no subtriangles that are not themselves tiled quadratically, as should be obvious from looking at which straight lines can be formed. (This condition is of course also necessary.)
So we can then consult the linked paper for a variety of triangles with non-quadratic tilings: this is true of the $(30^\circ,60^\circ,90^\circ)$ triangle and any right triangle whose legs are in a rational ratio, which I *think* is exhaustive though I haven't read the full paper in a while to confirm this (it focuses mostly on the realizable numbers of tiles, which is related but not quite identical to our question here).
If you want to revise your definition to the stricter condition that there are two self-tilings, neither of which contains *any* other nontrivial self-tiling, there are still triangular examples: consider the unique $3$-tiling of the $(30^\circ,60^\circ,90^\circ)$ triangle by itself and its $k=2$ quadratic tiling.
---
As an addendum, the question of convex *hexagonal* rep-tiles ought to be much easier to resolve, since there are only [three](https://en.wikipedia.org/wiki/Hexagonal_tiling#Monohedral_convex_hexagonal_tilings) classes of convex hexagon which can tile the plane, and I believe none of them can even tile a half-plane (which is a prerequisite to being a polygonal rep-tile, though the proof involves a somewhat messy compactness argument). You can at least use the fact that any rep-tile must have an angle of at most $90^\circ$ to narrow the scope of possible tiles a little. Combined with a negative answer to Question 1, this would effectively reduce the classification problem of Question 2 to the convex quadrilaterals, which seems like it might prove rather difficult - it might end up being easiest to just classify all convex quadrilateral rep-tiles and then work out which ones are multi-way.
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6
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https://mathoverflow.net/users/89672
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397032
| 163,920 |
https://mathoverflow.net/questions/397023
|
15
|
A pair of distinguished generators of the fundamental group $\pi\_1(\partial(S^3 \setminus K))$ of the boundary torus of a knot complement are usually called the "meridian" and "longitude". However, this terminology has always seemed a bit odd to me: in geography a meridian is a line of longitude, so shouldn't the curve it intersects be a "latitude"? Does anyone know the origin of this language?
|
https://mathoverflow.net/users/113402
|
Why is the thing dual to a "meridian" called a "longitude"?
|
There is a fundamental asymmetry between latitude and longitude on a sphere, whereas on a torus, there is a symmetry between the two generators. This symmetry could motivate the use of nearly synonymous words.
The terminology may originate with the paper [On the homology invariants of knots](https://www.maths.ed.ac.uk/~v1ranick/papers/seifert2.pdf) by H. Seifert (*Quart. J. Math. Oxford (2)* **1** (1950), 23–32). The torus $T$ of interest to Seifert is the boundary of a closed tubular neighborhood $V$ of a knot in $\mathbb{R}^3$. Seifert says that a Jordan curve on $T$ that bounds on $V$ (respectively, on $\mathbb{R}^3 - V + T$) but not on $T$ is called a *meridian* (respectively, a *longitudinal circuit*). Seifert phrases these definitions in a way that highlights the symmetry, which makes the use of near synonyms seem natural.
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13
|
https://mathoverflow.net/users/3106
|
397035
| 163,922 |
https://mathoverflow.net/questions/397007
|
4
|
Let $A$ and $B$ be Hermitian matrices. Let $[A,B] = AB - BA$ be their commutator and let $[A,B]^+$ be the Moore-Penrose pseudoinverse of $[A,B]$.
Is it then true that $A$ and $B$ both commute with the projector $Q = [A,B] [A,B]^+ = [A,B]^+ [A,B]$ ?
$P = 1 - Q$ is effectively a projector onto the subspace where our two matrices commute and we should always have that $PAP$ commutes with $PBP$.
|
https://mathoverflow.net/users/174368
|
Do any two hermitian matrices A and B commute with the support of their commutator?
|
To attack this more theoretically: if A and B have a common eigenvector, then that must lie in $\ker Q$ and obviously $[Q,A]$ and $[Q,B]$ act trivially on this guy. Thus, we can take the perp to this eigenvector and ask the same question about the leftover matrices.
Thus, we should assume that $A$ and $B$ have no common eigenvectors; in this case, you get that $[Q,A]=[Q,B]=0$ if and only if $Q$ is the identity, i.e. if and only if $[A,B]$ is invertible, which is definitely not always the case (though as the other answer indicates, it is true generically).
**EDIT:** This is true because if $[Q,A]=[Q,B]=0$, then $A$ and $B$ preserve the kernel $\ker Q$, and as observed in the question, they would commute on this subspace, and thus have a common eigenvector. So if there are no such eigenvectors, $Q$ must be injective and thus the identity.
|
4
|
https://mathoverflow.net/users/66
|
397036
| 163,923 |
https://mathoverflow.net/questions/396978
|
3
|
The classical Busemann-Feller lemma in Euclidean space says the following.
Let $K\subset \mathbb{R}^n$ be a closed convex set.
Then
1. for any point $x\in \mathbb{R}^n$ there exists unique nearest point in $K$; let us denote it by $p(x)$.
2. the map $p\colon \mathbb{R}^n \to K$ does not increase distances, i.e. is 1-Lipschitz.
**Question. Does this result hold in an $n$-dimensional hyperbolic space? A reference would be helpful.**
Remark. I think part (1) is true for hyperbolic and spherical spaces; in the spherical space one should require in addition that $K$ is contained in an open half-sphere. I think part (2) is not true in the spherical space.
|
https://mathoverflow.net/users/16183
|
Busemann-Feller lemma in hyperbolic space
|
In any Hadamard space, projection into convex sets is non-expansive; see Proposition 2.4(4) in *[Metric spaces of non-positive curvature](https://www.springer.com/gp/book/9783540643241)* by Bridson and Haefliger.
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7
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https://mathoverflow.net/users/68969
|
397045
| 163,924 |
https://mathoverflow.net/questions/397042
|
5
|
Let $K$ be a field. Suppose $A$ and $B$ are $K$-algebra and there is a derived equivalence $F:D^b(A)\cong D^b(B)$ between their bounded derived categories. If we assume that $A$ has a tensor decomposition $A\cong A\_1\otimes\_K A\_2$ for two $K$-algebras $A\_1$ and $A\_2$.
Does $B$ also have a tensor decomposition $B\cong B\_1\otimes\_K B\_2$ such that $D^b(B\_1)\cong D^b(A\_1)$ and $D^b(B\_2)\cong D^b(A\_2)$.
|
https://mathoverflow.net/users/134942
|
Tensor decomposition under derived equivalence
|
Not in general.
For an easy example, let $C$ and $D$ be derived equivalent algebras. Then $A=C\times C$ and $B=C\times D$ are derived equivalent, and $A=C\otimes\_K(K\times K)$ has a tensor decomposition, but $B$ will usually not.
There are also connected examples. For example, $KA\_2\otimes\_K KA\_2$ is derived equivalent to $KD\_4$ (where $KA\_2$ and $KD\_4$ are path algebras of quivers of the indicated Dynkin type).
There are more examples in [this question](https://mathoverflow.net/questions/370535/when-is-an-algebra-derived-indecomposable). But also, examples are very easy to construct. If you take an algebra of the form $A=A\_1\otimes\_KA\_2$, then it will usually have many tilting complexes, and unless you choose one of the form $T\_1\otimes\_KT\_2$, there is no particular reason to expect its endomorphism algebra (which is derived equivalent to $A$) to have a tensor decomposition.
|
5
|
https://mathoverflow.net/users/22989
|
397049
| 163,925 |
https://mathoverflow.net/questions/397054
|
0
|
Suppose that I have a Jordan curve $J$ parametrized by the function $\phi$. Consider a sequence of parametric functions $\phi\_n$ parametrizing a sequence of Jordan curves $J\_n$, and denote by $H$ and $H\_n$ the mean curvature of $J$ and $J\_n$, respectively.
Suppose also that these Jordan curves are contained in a large open bounded ball $B \subset \mathbb{R}^2$. Is it enough to assume the convergence of $\phi\_n$ to $\phi$ in the $C^1$ topology to prove that $\|H\_n - H\|\_{L^{\infty}(B)} \to 0$? I think that with the additional assumptions that the parametric functions satisfy the conditions that their derivatives (up to second order) are bounded, then the convergence follows. Are these the minimum requirements for the said convergence to hold? Any lead or reference dealing with this kind of problem is very much welcome. (If $\phi\_n$ converges to $\phi$ in $C^2([0,1);\mathbb{R}^2)$, I think there is no problem with the desired convergence.)
|
https://mathoverflow.net/users/249354
|
convergence of the mean curvature under $L^\infty$ norm
|
The hypotheses you impose are not strong enough to ensure convergence of the curvature.
This is ultimately a local problem, so to find a counterexample we work with graphs of functions. Let $(f\_n \mid n \in \mathbf{N})$ be a sequence of functions in $C^2(-1,1)$ say with
$$
\sup \lvert f\_n \rvert + \lvert f\_n' \rvert \to 0 \text{ and}
\sup \lvert f\_n'' \rvert \leq 2,
$$
but $\lvert f\_n''(0) \rvert \geq 1$ for all $n$.
The (signed) curvature of the graph of $f\_n$ is
$$
k\_n = (1 + (f\_n')^2)^{-3/2} f\_n''.
$$
By construction we have that for large enough $n$,
$$
\lvert k\_n(0) \rvert \geq 1/2.
$$
**Edit.**
In fact the following is true: if $$\gamma\_n \to \gamma \text{ in $C^1$} \text{ and }k\_n \to k \text{ in $L^\infty$}$$ then $$\gamma\_n \to \gamma \text{ in $C^2$}.$$
This is essentially for the same reason as above: locally the curves can be written as graphs over a common line. After rotating and translating the graphs, we are reduced to the case where $f\_n,f \in C^2(-1,1)$ and $$f\_n \to f \text{ in $C^1$}.$$
But then also
$$ f\_n'' = (1 + (f\_n')^2)^{3/2} k\_n \to (1 + (f')^2)^{3/2} k = f''
\text{ in $L^\infty$}.$$ The convergence $\gamma\_n \to \gamma$ in $C^2$ follows by piecing together these local arguments.
|
1
|
https://mathoverflow.net/users/103792
|
397069
| 163,930 |
https://mathoverflow.net/questions/397072
|
1
|
Can someone point out links to Applied Topology/Topological Data Analysis conferences and journals?
Thank you!
|
https://mathoverflow.net/users/316155
|
Applied Topology/Topological Data Analysis conferences and journals
|
ATMCS (Algebraic Topology: Methods, Computation, and Science), Oxford University, Oxford, UK, June 20–24, 2022. (I just googled.)
International Conference on Computational Topology and Data Analysis ICCTDA on September 20-21, 2021 in Toronto, Canada
International Conference on Computational Topology and Topological Data Analysis ICCTTDA on December 09-10, 2021 in London, United Kingdom
International Conference on Computational Topology and Data Analysis ICCTDA on December 16-17, 2021 in Barcelona, Spain
ICCTTDA001 2022: 16. International Conference on Computational Topology and Topological Data Analysis
May 17-18, 2022 in Paris, France
|
0
|
https://mathoverflow.net/users/13268
|
397073
| 163,932 |
https://mathoverflow.net/questions/397076
|
2
|
Consider the following mixture model for a univariate density function
$$
(1) \quad f(x)=\int\_{(m, \sigma^2)\in D} g(x; m, \sigma^2) \mu(d(m, \sigma^2))
$$
where $D$ is a compact subset of $\mathbb{R}\times \mathbb{R}^+$, $(m, \sigma^2)$ denotes the pair of mean and variance, $g(\cdot; m, \sigma^2)$ is the univariate Normal density function
with mean $m$ and variance $\sigma^2$, $\mu$ is a probability measure over $D$.
I'm interested in understanding which classes of density functions can (or cannot) be "approximated" as in (1).
That is, can we characterise the class, $\mathcal{F}$, of densities, $f(\cdot)$, for which there exists $D,\mu$ such that the "distance" between $f(\cdot)$ and $\int\_{(m, \sigma^2)\in D} g(x; m, \sigma^2) \mu(d(m, \sigma^2))$ is small?
I found in various papers/books sentences along the lines of "There is an obvious sense in which the mixture of normals approach, given enough components, can approximate any density" (see [here](http://assets.press.princeton.edu/chapters/s10259.pdf) for instance).
I also found some papers proposing formalisation of this sentence within the finite mixture case. See also [here](https://mathoverflow.net/questions/27589/mixtures-of-gaussian-distributions-dense-in-distributions) for a related question. However, I could not find anything formally dealing with the infinite mixture case as (1).
Would you have some references to suggest? If I set $D$ very large, wouldn't (1) encompass a quite general class of densities?
|
https://mathoverflow.net/users/42412
|
About a mixture
|
$\newcommand\R{\mathbb R}\newcommand\si{\sigma}\newcommand\ep{\varepsilon}\newcommand\de{\delta}$Any probability distribution on $\R$ can be approximated by a discrete probability distribution on $\R$. Any discrete probability distribution on $\R$ is a mixture of Dirac probability distributions on $\R$. The Dirac probability distribution supported at a point $a\in\R$ can be approximated by the normal distribution with a small variance centered at $a$.
Thus indeed, if the set $D$ is large enough, then any probability distribution on $\R$ can be approximated by a mixture of normal distributions $N(m,\sigma^2)$ with $(m,\si^2)\in D$.
---
Another way: any probability distribution $\nu$ on $\R$ can be approximated by the convolution $\nu\*N(0,\si^2)$ of $\nu$ with the centered normal distribution $N(0,\si^2)$ with a small variance $\si^2$, and such a convolution is a mixture of normal distributions:
\begin{equation\*}
\nu\*N(0,\si^2)=\int\_\R N(y,\si^2)\nu(dy). \tag{0}
\end{equation\*}
If now the set $D$ is large enough to, say, contain a set of the form $C\_\de\times(0,\ep)$, where $\ep\in(0,\infty)$, $\de$ is a small positive number, and the set $C\_\de\subseteq\R$ is such that $\nu(C\_\de)>1-\de$, then for small $\si^2\in(0,\ep)$
\begin{equation\*}
\nu\approx\nu\*N(0,\si^2)\approx\int\_{C\_\de} N(y,\si^2)\nu(dy)
=\int\_D N(y,s^2)\mu(dy\times d(s^2)), \tag{1}
\end{equation\*}
where $\mu(dy\times d(s^2)):=\nu(dy)1(y\in C\_\de,s^2=\si^2)$ --
so that $\nu$ will be approximated by the mixture $\int\_D N(y,s^2)\mu(dy\times d(s^2))$ of normal distributions $N(m,s^2)$ with $(m,s^2)\in D$.
**Details:** The approximate equalities in (1) can be understood as follows. Suppose that $\si\_n^2\downarrow0$ and $\de\_n\downarrow0$ (as $n\to\infty$). Then, by [Slutsky's theorem](https://en.wikipedia.org/wiki/Slutsky%27s_theorem), $\nu\*N(0,\si\_n^2)\to\nu$ weakly; that is, for any bounded continuous function $g\colon\R\to\R$ we have
\begin{equation\*}
(\nu\*N(0,\si\_n^2))(g)\to\nu(g),
\end{equation\*}
where $\nu(g):=\int\_\R g\,d\nu$; this is how the first approximate equality in (1) can be understood.
Next, by (0), for any bounded continuous function $g\colon\R\to\R$ with $M:=\sup\_{x\in\R}|g(x)|<\infty$ we have
\begin{equation\*}
\begin{aligned}
&\Big|(\nu\*N(0,\si\_n^2))(g)-\int\_{C\_{\de\_n}} N(y,\si\_n^2)(g)\nu(dy)\Big| \\
=&\Big|\int\_{\R} N(y,\si\_n^2)(g)\nu(dy)-\int\_{C\_{\de\_n}} N(y,\si\_n^2)(g)\nu(dy)\Big| \\
\le&\int\_{\R\setminus C\_{\de\_n}} N(y,\si^2)(|g|)\nu(dy) \\
\le& M\nu(\R\setminus C\_{\de\_n})\le M\de\_n\to0;
\end{aligned}
\end{equation\*}
this is how the second approximate equality in (1) can be understood.
So, for any bounded continuous function $g\colon\R\to\R$,
\begin{equation\*}
\Big(\int\_{C\_{\de\_n}} N(y,\si\_n^2)\nu(dy)\Big)(g)
=\int\_{C\_{\de\_n}} N(y,\si\_n^2)(g)\nu(dy)\to\nu(g).
\end{equation\*}
That is, we have the weak convergence of the normal mixtures $\int\_{C\_{\de\_n}} N(y,\si\_n^2)\nu(dy)$ to $\nu$. (The equality in the latter display is an instance of the Fubini theorem.)
|
3
|
https://mathoverflow.net/users/36721
|
397079
| 163,935 |
https://mathoverflow.net/questions/397082
|
4
|
Let $[n]\_q=1+q+\dots +q^{n-1}$, $ {[n]\_q}! =[1]\_q [2]\_q \dots [n]\_q$ and $\binom{n}{j}\_q = \frac{[n]\_q!}{[j]\_q![n-j]\_q!}$ be the usual $q$-notation.
Consider the polynomials $p\_n(q,r,x)= \sum\_{j=0}^n q^{r\binom{j}{2}}\binom{n}{j}\_qx^j$ and let $d\_{n,r}(q)=\Delta\_{x} p\_n(q,r,x)$ be their [discriminants](https://en.wikipedia.org/wiki/Discriminant).
The well-known identity $p\_n(q,1,x) =(1+x)(1+qx)\dots(1+q^{n-1}x)$ implies that
$$d\_{n,1}(q)=\Delta\_{x} p\_n(q,1,x)=(1-q)^{2\binom{n}{2}}q^{2\binom{n}{3}}\Big(\prod\_{j=0}^{n-1}[j]\_q!\Big)^2.$$
This gives $ d\_{n,1}(q)=q^{n(n-1)^2} d\_{n,1}(q^{-1}).$
Computations suggest that more generally the following holds for integers $r$:
$$ d\_{n,1+r}(q)=q^{n(n-1)^2} d\_{n,1-r}(q^{-1}).$$
>
> **QUESTION.** Is this a known result? If yes, where can I find it, if no, any idea how to prove it?
>
>
>
|
https://mathoverflow.net/users/5585
|
Discriminants of some $q$-analogs of $(1+x)^n$
|
This is true.
We have
\begin{align\*}
p\_n (q^{-1}, 1-r, x) &= \sum\_{j=0}^n q^{ (r-1) \binom{j}{2}} \binom{n}{j}\_{q^{-1}} x^j \\
&= \sum\_{j=0}^n q^{ (r-1) \binom{j}{2}} q^{-j (n-j)} \binom{n}{j}\_{q} x^j \\
&=\sum\_{j=0}^n q^{ (r+1) \binom{j}{2}} q^{-j (n-j) - j (j-1)} \binom{n}{j}\_{q} x^j \\
&= \sum\_{j=0}^n q^{ (r+1) \binom{j}{2}} \binom{n}{j}\_{q} (x q^{-(n-1)} )^j \\
&= p\_n(q, 1+r, x q^{-(n-1)}).
\end{align\*}
and changing variables $x \to \lambda x$ in a polynomial multiplies the discriminant by $\lambda^{ n (n-1)}$.
|
7
|
https://mathoverflow.net/users/18060
|
397084
| 163,936 |
https://mathoverflow.net/questions/397024
|
3
|
I stumbled upon the following identity, which I have not tried to prove, but seems true:
the function $$f(t):={}\_2F\_1(1/2,2t;1-t;4)$$
is periodic of period 1, and more precisely
$$f(t)=\dfrac{1+2e^{-2\pi it}}{3}\;.$$
This begs several questions:
(1). Is this true? (2). Are there other (infinitely many?) formulas of this type (of course outside of trivial manipulations of this) ? (3) More generally, call a function $f(t)$ scale-periodic if there exists $A>0$ such that $A^{-t}f(t)$ is $1$-periodic (so periodic if $A=1$). I have found a number of such $f(t)$ of the form ${}\_2F\_1(a(t),b(t);c(t);z)$: is there some way to find them? (this last question was inspired by a paper of Beukers and Forsgard arXiv:2004.08117).
|
https://mathoverflow.net/users/81776
|
Periodic Gauss hypergeometric function
|
Gauss' contiguous relations provide a basis for finding a linear relationship between three functions of the form ${}\_2F\_1(a+k, b+l, c+m, z)$, henceforth $\mathbf{F}\left(\begin{matrix}a+k, b+l \\ c+m\end{matrix}\right)$. Various papers have been published on calculating these relationships; I'm using the notation of Vidūnas's paper [*Contiguous relations of hypergeometric series*](https://arxiv.org/abs/math/0109222):
$$\textbf{F}\left(\begin{matrix}a+k, b+l \\ c+m\end{matrix}\right) = \textbf{P}(k,l,m) \textbf{F}\left(\begin{matrix}a, b \\ c\end{matrix}\right) + \textbf{Q}(k,l,m) \textbf{F}\left(\begin{matrix}a+1, b \\ c\end{matrix}\right)$$
Vidūnas recommends finding the coefficients by first chaining up $k$ with his formula (9), but a naïve approach runs into bootstrap problems (as I discovered by trying it). Once $\textrm{Q}(2,0,0)$ is found, I think the approach works, but here we have such small shifts that we can do directly.
Rephrasing his (2), (3), (4) in terms of $\textbf{Q}$ we obtain
$$\mathbf{Q}(k+1,l,m) = \tfrac{(2a+2k-c-m+(b+l-a-k)z)}{(a+k)(1-z)}\mathbf{Q}(k,l,m) + \tfrac{(c+m-a-k)}{(a+k)(1-z)}\mathbf{Q}(k-1,l,m)$$
$$\mathbf{Q}(k,l,m-1) = \tfrac{a+k}{c+m-1} \mathbf{Q}(k+1,l,m) + \tfrac{c+m-a-k-1}{c+m-1} \mathbf{Q}(k,l,m)$$
$$\mathbf{Q}(k,l+1,m) = \tfrac{a+k}{b+l} \mathbf{Q}(k+1,l,m) + \tfrac{b+l-a-k}{b+l}\mathbf{Q}(k,l,m)$$
and by repeated application of these I calculate that
$$\mathbf{Q}(0,2,-1) = a \tfrac{(a+b-c+1)(a+b-c+2) + (c-a-1)(2a+b-c+1)(1-z) + (b-a+1)(c-a-1)(1-z)^2}{(c-1)b(b+1)(1-z)^2} \\
$$
$$\mathbf{P}(0,2,-1) = (c-a-1) \tfrac{a(a+b-c+2) + a(-2a+b+c-1)(1-z) + (b-a)(b-a+1)(1-z)^2}{(c-1)b(b+1)(1-z)^2}$$
Substituting your parameters $a=\tfrac12$, $b=2t$, $c=1-t$, $z=4$ I get respectively $0$ and $1$, which establishes the periodicity of your $f$.
---
To search systematically for scale-periodic hypergeometric functions, I propose the following approach:
1. Pick $k$, $l$, $m \in \mathbb{Z}$ (taking into account symmetries for efficiency).
2. Compute $\mathbf{Q}(k,l,m)$ and $\mathbf{P}(k,l,m)$.
3. Substitute $a = a\_0 + kt$, $b = b\_0 + lt$, $c = c\_0 + mt$ into $\mathbf{P}(k,l,m) = A$ and $\mathbf{Q}(k,l,m) = 0$. Equate coefficients of $t$ in each to get a collection of polynomial constraints over $a\_0,b\_0,c\_0,z,A$. Use Gröbner bases or other techniques to solve these constraints.
To take the obvious example of $k=0$, $l=2$, $m=-1$, and (for simplicity) applying the substitution $x = 1-z$, $\textbf{Q}(0,2,-1) = 0$ gives
$$
a\_0 (9-3x-2x^2) = 0 \\
6a\_0+6b\_0-6c\_0+9+(-5a\_0-b\_0+4c\_0-4)x+(-a\_0-b\_0+2c\_0-3)x^2 = 0 \\
(a\_0+b\_0-c\_0+1)(a\_0+b\_0-c\_0+2) +
(-a\_0+c\_0-1)(2a\_0+b\_0-c\_0+1)x +
(-a\_0+b\_0+1)(-a\_0+c\_0-1)x^2 = 0
$$
Ignoring the trivial $a\_0 = 0$, we see that $x \in \{-3, \tfrac32\}$.
### Case $x=-3$
$a\_0 = \tfrac12$ drops out of the next constraint, leaving the final constraint to reduce to $(b\_0+2c\_0-3)(b\_0+2c\_0-2) = 0$. If $b\_0+2c\_0-2 = 0$ then we're going to recover your $f$, so consider instead $b\_0+2c\_0-3 = 0$. This turns out to have no solutions: substituting into $\mathbf{P}$ and expanding we get a contradiction in the coefficients of $t^0$.
### Case $x = \tfrac32$
Comparing coefficients for $t^3$ in $\mathbf{P}$ gives $A = 1$; subsequently comparing coefficients for $t^2$ in $\mathbf{P}$ gives $a\_0 = 0$, so there is no non-trivial solution.
I therefore withdraw my blithe assertion in an earlier comment that question (2) is easily answered in the affirmative; it remains plausible, but with larger shifts we're going to have more coefficients of $t$ to impose constraints on a fixed number of variables and so there might not be very many sets of shifts which admit solutions.
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4
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https://mathoverflow.net/users/46140
|
397085
| 163,937 |
https://mathoverflow.net/questions/396990
|
7
|
Can the following uniformization statement be proved by $ZF+AD+DC$?
>
> For any binary relation $R\subseteq \mathbb{R}^2$ with the property that $\forall x (\{y\mid R(x,y)\}\mbox{ is at most countable and nonempty})$, then there is a function $f:\mathbb{R}\to \mathbb{R}$ uniformizing $R$.
>
>
>
|
https://mathoverflow.net/users/14340
|
Uniformization under AD
|
Here's an argument under the further assumption that $V=L(\mathbb{R})$. Something like this is presumably recorded somewhere. The main point comes from Steel's paper "Scales in $L(\mathbb{R})$", but I don't find it quite explicitly there.
Since $V=L(\mathbb{R})$, we can fix a real $x\_0$ such that our relation $R$ is $\mathrm{OD}\_{x\_0}$. Then for each $x\in\mathbb{R}$,
$R\_x=\{y\bigm|R(x,y)\}$ is $\mathrm{OD}\_{(x\_0,x)}$. It suffices
to see that $R\_x$ contains some element which is $\mathrm{OD}\_{(x\_0,x)}$,
since then we can set $f(x)=$ the least such (in the standard ordering of $\mathrm{OD}\_{(x\_0,x)}$).
In fact, for each real $x$, every countable $\mathrm{OD}\_x$ set of reals is $\subseteq\mathrm{OD}\_x$. For let's assume $x=\emptyset$; the relativization to other $x$ is routine. Fix a $\Sigma\_1$ formula $\varphi$ such that for some ordinals $\alpha<\beta$, the set $$A\_{\alpha\beta}=\{y\in\mathbb{R}\bigm|L\_\beta(\mathbb{R})\models\varphi(\alpha,\mathbb{R},y)\}$$ is countable; each OD set is of this form. We want to see that each such $A\_{\alpha\beta}\subseteq\mathrm{OD}$ (assuming it's countable). Let $A'\_{\alpha\beta}=A\_{\alpha\beta}$ if $A\_{\alpha\beta}$ is countable, and $A'\_{\alpha\beta}=\emptyset$ otherwise.
Claim: $A=\bigcup\_{\alpha\beta}A'\_{\alpha\beta}$ is countable.
Proof: Otherwise we can define a prewellorder $<^\*$ on $A$ such that the set of $<^\*$-predecessors of any given real is countable, and then argue like in the proof that there is no $\omega\_1$-sequence of pairwise distinct reals. QED (Claim).
Note now that $A$ is $\Sigma\_1^{L(\mathbb{R})}$ (here the standard notation allows the parameter $\mathbb{R}$ as default, but this is otherwise lightface, i.e. no other parameters). But by Proposition 2.11 of Steel's "Scales in $L(\mathbb{R})$" (applied at some sufficiently large and reflective level $\alpha\in\mathrm{OR}$), and since $A$ is countable, it follows that $A\subseteq\mathrm{OD}$, as desired.
|
7
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https://mathoverflow.net/users/160347
|
397096
| 163,941 |
https://mathoverflow.net/questions/396968
|
6
|
In section 35.1 of the book "Linear algebraic groups" by Humphreys, it is stated that the quasi-split but not split semisimple groups can only arise when the root system admits a nontrivial graph automorphism.
Moreover, it seems that the relative root system in this case is obtained by adjoining the vertices of Dynkin diagram which are sent to each other by the graph automorphism.
Also in the wikipedia page on quasi-split groups, it is stated that a quasi-split groups over a field correspond to actions of the absolute Galois group on a Dynkin diagram.
In both, there is no reference about this statement. In what paper can I find some theory about this?
|
https://mathoverflow.net/users/304053
|
Quasisplit but not split semisimple groups
|
The quasi-split forms of a split reductive group $G$ over a field $k$ are classified by the elements of the first Galois cohomology group of $k$ with values in $Out(G)$ (see Theorem 23.51 of Milne's book *Algebraic Groups*). So no outer automorphisms means no nonsplit quasi-split forms. When $G$ is semisimple, the outer automorphisms correspond to graph automorphisms of the dynkin diagram.
|
5
|
https://mathoverflow.net/users/316316
|
397098
| 163,942 |
https://mathoverflow.net/questions/394249
|
4
|
Let $X:=X\_{18}$ be an index one smooth prime Fano threefold of degree 18.
Consider its semi-orthogonal decomposition: $D^b(X)=\langle\mathcal{O}\_X,\mathcal{E}^{\vee},\mathcal{A}\_X\rangle=\langle\mathcal{Q}^{\vee},\mathcal{O}\_X,\mathcal{A}\_X\rangle$, where $\mathcal{E},\mathcal{Q}$ are tautological sub and quotient bundle on $X$ coming from Grassmannian $\mathrm{Gr}(2, 7)$. Note that $\mathcal{A}\_X\cong ^{\perp}\langle Q^{\vee},\mathcal{O}\_X\rangle$ or $\mathcal{A}\_X\cong ^{\perp}\langle \mathcal{O}\_X, \mathcal{E}^{\vee}\rangle$. It is known that $\mathcal{A}\_X\cong D^b(C\_2)$ where $C\_2$ is a smooth genus 2 curve(hyperelliptic curve). It is also known that the group of auto-equivalences of $D^b(C\_2)$ is generated by $Aut(C\_2), [1], -\otimes\mathcal{L}$(automorphism of the curve, shift functor and tensoring with line bundles).
There is a natural involution $\tau\in Aut(C\_2)$(hyperelliptic involution) inducing an auto-equivalence on $D^b(C\_2)$(still denoted by $\tau$). My question: Is there any way to write the auto-equivalence $\tau:\mathcal{A}\_X\rightarrow\mathcal{A}\_X$ purely in terms of composition of functors associated to the objects in $D^b(X)$?
For example, if $X:=X\_{10}$ is a special Gushel-Mukai threefold, its semi-orthogonal decomposition is given as $D^b(X)=\langle \mathcal{B}\_X, \mathcal{E},\mathcal{O}\_X\rangle$, where $\mathcal{B}\_X\cong\langle\mathcal{E},\mathcal{O}\_X\rangle^{\perp}$, the geometric involution $\tau$ on $X$ gives an auto-equivalence of $\mathcal{B}\_X$ and one can write $\tau^{-1}$ as $\mathrm{L}\_{\mathcal{E}}\circ\mathrm{L}\_{\mathcal{O}\_X}\circ(-\otimes\mathcal{O}\_X(H))[-1]$.
|
https://mathoverflow.net/users/41650
|
Auto-equivalences of non-trivial components of derived category of $X_{18}$
|
Let me answer the question by myself. After a intensively literature research, I found that the habilitation of Faenzi,Daniele contains everything I need, here is the link <http://dfaenzi.perso.math.cnrs.fr/publis/faenzi.hdr.pdf>
Section 3.2
|
1
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https://mathoverflow.net/users/41650
|
397112
| 163,945 |
https://mathoverflow.net/questions/397113
|
4
|
Assuming that for each fixed $k$, $(X\_{n,1},\ldots,X\_{n,k})\Longrightarrow(X\_1,\ldots,X\_k)$ where $X\_1,\ldots,X\_k$ are i.i.d. with mean zero and variance $\sigma^2$, will the array inherit the CLT from its limit? i.e. do I have (if $r\_n\to\infty$):
$\frac{1}{\sqrt{r\_n}}(X\_{n,1}+\cdots+X\_{n,r\_{n}})\Longrightarrow N(0,\sigma^2)$?
|
https://mathoverflow.net/users/316497
|
CLT for a converging array of random variables
|
No. Take any sequence $r\_n \to \infty$ and let $X\_{n,k}$ be iid $N(0,1)$ for $k \ne r\_n$, and $X\_{n,r\_n} = r\_n$. The hypothesis is satisfied because $(X\_{n,1}, \dots, X\_{n,k})$ are iid $N(0,1)$ as soon as $n$ is so large that $r\_n > k$. But
$$\frac{1}{\sqrt{r\_n}}(X\_{n,1} + \dots + X\_{n,r\_n}) \sim N(\sqrt{r\_n}, \frac{r\_n-1}{r\_n})$$
which is not even tight.
|
4
|
https://mathoverflow.net/users/4832
|
397118
| 163,948 |
https://mathoverflow.net/questions/397115
|
7
|
Recall that a set is **amorphous** iff it is infinite but has no partition into two infinite subsets. I'm interested in the possible structure (in the sense of model theory) which an amorphous set can carry. For example, if $T$ is a complete first-order theory with an amorphous model then $T$ must be strongly minimal; however, the converse fails badly (consider $Th(\mathbb{C};+,\times)$).
I'm curious whether second-order logic does a better job. For $T$ a complete second-order theory with no finite models, say that $T$ is **second-order strongly minimal** iff every second-order-definable subset of every model $\mathcal{A}\models T$ is either finite or cofinite in $\mathcal{A}$. For example, $Th(\mathbb{C};+,\times)$ is not second-order strongly minimal since $\mathbb{Z}\subset\mathbb{C}$ is second-order definable.
*EDIT: by "complete" I mean "maximally satisfiable:" a second-order theory $T$ is complete iff it is satisfiable and for each second-order sentence $\varphi$, either $\varphi\in T$ or $\neg\varphi\in T$.*
To be reasonably concrete, my main question is the following:
>
> **Question 1**: Is it consistent with $\mathsf{ZF}$ that every second-order strongly minimal theory has an amorphous model?
>
>
>
*(More broadly, I'm interested in whether there is a reasonably natural logic $\mathcal{L}$ such that consistently with $\mathsf{ZF}$ every "$\mathcal{L}$-strongly-minimal" theory has an amorphous model. Second-order logic just seems like a good candidate at the moment.)*
There's also a "virtual" version of the question which makes sense even in the presence of choice:
>
> **Question 2**: Is it consistent with $\mathsf{ZFC}$ that for every second-order strongly minimal theory $T$, there is some (set) symmetric extension in which $T$ has an amorphous model?
>
>
>
The exact relationship between these two questions isn't clear to me. That said, an affirmative answer to question 2 seems much more plausible than an affirmative answer to question 1.
|
https://mathoverflow.net/users/8133
|
Can second-order logic identify "amorphous satisfiability"?
|
Here's a monadic example. The second-order theory of the vector space $\mathbb F\_2^{\oplus\omega}$ as a vector space over $\mathbb F\_2$ is second-order strongly minimal because given a finite subspace $X$ **of any model of this theory**, any two points outside $X$ are related by an automorphism of the whole space fixing $X.$
There is a subtlety here pointed out by Emil Jeřábek in the comments. For Question 2, where we check strong minimality in ZFC, it is of course true that any automorphism of a subspace extends to the whole space. In ZF, this might not be true, but as Emil suggested we can use the axiom that there are complemented subspaces: $(\forall Y)(\exists Z)$ such that if $Y$ is a subspace then: $Z$ is a subspace with $Y\cap Z=\{0\},$ and $(\forall v)(\exists y\in Y)(\exists z\in Z)(v=y+z).$
This theory contains the second order axiom "($\exists P$) $P$ is a subspace of codimension 2", i.e. $P$ is closed under addition and there exists $v\notin P$ such that every $w\notin P$ satisfies $w+v\in P.$ So it can't have an amorphous model.
|
6
|
https://mathoverflow.net/users/164965
|
397121
| 163,951 |
https://mathoverflow.net/questions/396393
|
4
|
Let $\mathfrak{g}$ be a real semisimple Lie algebra with complexification $\mathfrak{g}\_\mathbb{C}$. Recall that a parabolic subalgebra in $\mathfrak{g}\_\mathbb{C}$ is one which contains a Borel subalgebra, and a parabolic subalgebra in $\mathfrak{g}$ is one which complexifies to a parabolic one in $\mathfrak{g}\_\mathbb{C}$.
**Question:** Given a Borel subalgebra $\mathfrak{b} \subset \mathfrak{g}\_\mathbb{C}$, is it contained in the complexification $\mathfrak{q}\_\mathbb{C}$ of some *minimal* parabolic subalgebra $\mathfrak{q} \subset \mathfrak{g}$?
---
Problem VII.13 in Knapp's "Lie Groups beyond an introduction" says this is true, under the additional condition that $\mathfrak{b}$ is built from the complexification of a maximally noncompact Cartan subalgebra of $\mathfrak{g}$, in the usual way via a root space decomposition and choice of positive roots. I am wondering to what extent we can drop this requirement -- or if this is a strong requirement at all?
Note that Knapp defines minimal parabolic subalgebras differently than I do here, but I believe their notion should be equivalent (this is related to the first part of my question [here on math.stackexchange,](https://math.stackexchange.com/questions/4172661/does-every-nilpotent-element-in-a-real-lie-algebra-lie-in-some-minimal-parabolic) for which I believe I have found an argument now, but I yet need to formalize it).
|
https://mathoverflow.net/users/126256
|
Complexifications of minimal parabolic subalgebras
|
NB: for brevity I'm going to refer to Borel subalgebras and parabolic subalgebra as Borels and parabolics, respectively
Even for a split (or quasi-split) Lie algebra not all complex Borels are the complexification of real Borels. Otherwise, for example, the real and complex flag manifolds would be identical. Naturally, a Borel cannot contain another one since they are supposed to be maximal solvable/minimal parabolic. So for these cases the condition would appear to be necessary as well as sufficient.
What is true is that a real parabolic $ \mathfrak{p} \leq \mathfrak{g} $ complexifies to a complex parabolic $ \mathfrak{p}^{\mathbb{C}} \leq \mathfrak{g}^{\mathbb{C}} $. Such a complex parabolic contains a family of Borels (of $\mathfrak{g}^{\mathbb{C}}$) and any one of these determines $ \mathfrak{p}^{\mathbb{C}} $ and thus $\mathfrak{p}$ uniquely.
To be more precise parabolics come in a number of conjugacy classes in correspondence with subsets of the Dynkin diagram and a given Borel subalgebra defines a unique parabolic in each conjugacy class that contains it. In particular we get a unique one in the class which contains the complexifications of the minimal parabolics in our real form. Because we chose it to be a complexification we can recover its real form but that wouldn't work otherwise. The uniqueness we mentioned above tells us that if a Borel $\mathfrak{b}$ is contained in some complex parabolic $\mathfrak{q}$ which is conjugate to $\mathfrak{p}^{\mathbb{C}}$ but is not a complexified parabolic itself then $\mathfrak{b}$ is not contained in any complexified minimal parabolic.
So now we have to ask if all Borel subalgebras contained within a given $ \mathfrak{p}^{\mathbb{C}} $ are of the form described by Knapp. I couldn't work that out off the top of my head but certainly there is some necessary condition.
|
2
|
https://mathoverflow.net/users/163024
|
397137
| 163,953 |
https://mathoverflow.net/questions/397089
|
7
|
Let $X$ be a topological space.
Let $\gamma:[a,b]\to X$ be continuous and injective.
$\gamma$ is said to be "openly extendable" if there is $[a,b]\subset (a',b')$ and a continuous and injective curve $\gamma':(a',b')\to X$ with $\gamma'\vert\_{[a,b]} = \gamma$.
Under what conditions on $X$, all continuous injective curves are openly extendable?
I'm specifically interested in the case where $X$ is metrizable.
|
https://mathoverflow.net/users/95265
|
Extending continuous injective curves both continuously and injectively
|
~~**EDIT 1:** Fixed the mistakes and added a full proof below~~
**EDIT 4: Found how to prove the existence of a connected neighborhood.
I am satisfied with this proof, so I'll accept the answer**
**Theorem:** If $X$ is connected, ~~locally connected~~ **EDIT 4: locally path connected**, locally compact and metrizable and $x=\gamma(a)$ and $y=\gamma(b)$ are not isolated within $X-(Im(\gamma)^\mathrm{o})$ then $\gamma$ is openly extendable.
**Proof:**
First, we want to find a neighborhood $U\_x$ with $W:=U\_x-Im(\gamma)^\mathrm{o}$ being connected.
Let $V\_x$ be a path connected neighborhood of $x$ separated from $y$.
There are three cases:
1. If $V\_x-Im(\gamma)$ is connected then so is $V\_x-Im(\gamma)^\mathrm{o}$ (and as $X$ is locally path connected, this neighborhood is path connected).
2. If for every path connected component $Y$ of $V\_x-Im(\gamma)$ we have $x\in\partial Y$.
Let $y\in Y$ and let $W\_x$ be a path connected neighborhood of $x$, by our assumption, $W\_x\cap Y\neq\emptyset$ and therefore $x$ is connected by a path to every point in $W\_x$, therefore $Y\cup\{x\}$ is path connected. But as this is true for all $Y$ then $$\bigcup\_{Y}(Y\cup\{x\}) = V\_x\cup\{x\} = V\_x-Im(\gamma)^\mathrm{o}$$ is path connected (as a union of path connected spaces with non empty pairwise intersection).
3. Let $Y$ be a path connected component of $V\_x$ with $x\notin \partial Y$. $Y$ is closed, and $X$ is $T\_3$ which means there is a closed neighborhood $W\_x$ of $x$ with $W\_x\cap Y = \emptyset$.
Let $U\_x$ be a path connected neighborhood of $x$ within $W\_x$.
We assumed $x$ is not isolated within $X-(Im(\gamma)^{\mathrm{o}})$ which means there is at least one point $x\neq w\in W\_x-Im(\gamma)$. As $x\notin\bar{Y}$ we have $w\notin Y$.
Now let $Z$ be the path connected component of $w$ in $V\_x-Im(\gamma)$. But now $Z\cup\{x\}$ is path connected as $Z$ is, and $x,w\in W\_x$ which is path connected. But now $Z\cup\{x\} = Z - Im(\gamma)^\mathrm{o}$ is path connected as we wanted.
~~Let $U\_x$ be compact, connected neighborhood of $x$ (it exists because $X$ is locally compact and locally connected, and we can refine one neighborhood to satisfy both conditions).
**EDIT 2:** The fact that $U\_x-(Im(\gamma)^{\mathrm{o}})$ is connected is not obvious, here is the explanation:
Because $X$ is $T\_3$ we can choose a neighborhood which is seperated from every $\gamma([t,s])$ where $t\geq t\_0>0$ which makes $U\_x-(Im(\gamma)^{\mathrm{o}})$ connected (otherwise there's a closed $K\subset Im(\gamma)^\mathrm{o}$ which separates $U\_x$ and we know $x\notin K$ therefore it contains the image of an interval $\gamma([t,s])$ with $t>0$ and we can choose $t\_0$ to be smaller than this $t$).
**End of EDIT 2**~~
~~**EDIT 3:** Finding a $U\_x$ for which $U\_x-Im(\gamma)^\mathrm{o}$ is connected is not obvious, and the argument doesn't work without it. It seems like it is true, but I'm not sure how to prove it.~~
Due to the properties of $X$, $W$ is also metrizable, connected and locally connected, and as $X$ is Hausdorff we can choose $U\_x$ to be compact which means so is $W$ (as $W = U\_x \cap (Im(\gamma)^\mathrm{o})^\mathrm{c}$ is closed).
Therefore the space $W$ is metrizable, compact, connected, and locally connected. Therefore it satisfies the conditions of the [Hahn Mazurkiewicz theorem](https://en.wikipedia.org/wiki/Space-filling_curve#The_Hahn.E2.80.93Mazurkiewicz_theorem). Therefore it is a Peano space which is arcwise connected. We have $$U\_x\cap Im(\gamma)^c\neq\emptyset$$ becuase we assume $x$ is not isolated. Let $z\in U\_x - Im(\gamma)$
Let $g:[0,1]\to U\_x-Im(\gamma)^\mathrm{o}$ be an arc between $x$ and $z$, i.e. a continuous bijection satisfying $g(0)=x, g(1) = z$. Now $g\cdot \gamma$ is continuous and injective (as both components are injective and their ranges do not overlap except at the common point).
$X$ is $T\_3$, so let $V\_y$ be a neighborhood of $y$ not intersecting $U\_x$ ($U\_x$ is compact and therefore closed). Repeating the same process for $y$ within $V\_y$ extends $\gamma$ into a continuous injective $\gamma\cdot g'$ with $Im(g')\subset V\_y$.
Now $\gamma':=g\cdot\gamma\cdot g'$ is injective because $V\_y\cap Im(g) = \emptyset$
$Im(g)^\mathrm{o}, Im(g')^\mathrm{o}\neq\emptyset$ bacuase $X$ is locally connected, and therefore $\gamma'$ satisfies the conditions of the lemma.
**Remark:** Among other things, this condition implies $\gamma$ cannot be surjective.
|
0
|
https://mathoverflow.net/users/95265
|
397139
| 163,955 |
https://mathoverflow.net/questions/397152
|
2
|
Recently I [asked a question](https://mathoverflow.net/questions/397115/can-second-order-logic-identify-amorphous-satisfiability) about whether a second-order analogue of strong minimality could correspond to amorphous satisfiability (= having a model whose underlying set cannot be partitioned into two infinite pieces). I missed a fairly obvious fact: roughly speaking, that in a second-order way we can assert the existence of a certain type of structure on the domain prohibiting amorphousness (e.g. a linear order or a non-surjective endofunction) without making any such structure definable.
I'd like to rephrase my original question to avoid this. One idea which seems promising is to demand something like a second-order witness property: whenever our theory asserts the existence of some structure on the domain there should be some definable such structure. The existence of a second-order sentence saying "The domain is amorphous" causes a bit of difficulty in making the question nontrivial, however (e.g. any maximally satisfiable second-order theory either proves that the domain is amorphous, in which case it obviously is amorphously satisfiable, or proves that there is a partition of the domain into two infinite piecs at which point the second-order witness property kills any type of minimality).
The easiest way to get around this, in my opinion, is to restrict attention to monadic theories. My question is whether it is consistent with $\mathsf{ZF}$ that every satisfiable monadic second-order theory $T$ with the following properties is amorphously satisfiable:
* $T$ is negation-complete: for each $\varphi$, either $\varphi\in T$ or $\neg\varphi\in T$.
* $T$ is "second-order strongly minimal:" no model of $T$ has a definable-with-(element-)parameters bi-infinite subset.
* $T$ has the "second-order witness property:" whenever $\exists X\varphi\in T$, there is some unary predicate symbol $U$ such that $\varphi[X/U]\in T$.
Note that something like the witness property is necessary here, per [Harry West's answer to my original question](https://mathoverflow.net/a/397121/8133).
|
https://mathoverflow.net/users/8133
|
Second-order strong minimality and amorphousness, take 2
|
I claim that if the language is countable, or merely well-orderable, then the witness property by itself prevents amorphous domains, and even infinite Dedekind-finite domains. Strong minimality has nothing to do with it.
**Theorem.** Any satisfiable complete theory $T$ in a well-ordered language with the witness property is never true on an amorphous domain, nor even on an infinite Dedekind-finite domain.
**Proof.** Suppose the language is well-ordered and complete theory $T$ has the witness property and is true in some infinite domain $M$. Since
$$\exists X(X\text{ has exactly one element })$$ is true, then by the witness property there is a predicate $U\_1$ whose extension is a singleton. Now,
$$\exists Y(Y\text{ has exactly two elements, one of them in }U\_1)$$
is true, so there is a predicate $U\_2$ having exactly one additional element beyond the element of $U\_1$. And further,
$$\exists Z(Z\text{ has exactly three elements, two of them in }U\_2)$$
is true, so there is a tripleton set $U\_3$. And so on.
In this way, we construct a countably infinite subset of the domain. So the domain is not Dedekind finite. $\Box$
We used the well-orderability of the language when picking the particular predicates $U\_k$, since there could be many witnessing predicates for the witness property.
The theorem shows that in order to have a theory with the witness property true on an amorphous domain, the language itself will have to be a little strange, perhaps even containing an amorphous set of unary predicate symbols.
|
3
|
https://mathoverflow.net/users/1946
|
397155
| 163,958 |
https://mathoverflow.net/questions/397140
|
1
|
Suppose that $A$ is real and symmetric matrix (or tensor) of dimension $3 \times 3$, with its spectral decomposition
$$A = \sum\_{i=1}^3 \lambda\_i\ n\_i\otimes n\_i$$
where $\lambda\_i$, $n\_i$ and $\otimes$ denote the eigenvalues, eigenvectors and dyadic product, respectively. Further, let $a$ be a vector of the form
$$ a = \sum\_{i=1}^3 f(\lambda\_i)\ n\_i$$
with an at least twice-differentiable function $f(\lambda\_i)$.
I need to compute the derivative (third-order tensor) $B$ with its components
$$B\_{ijk}=\frac{\partial a\_i}{\partial A\_{jk}}$$
also in the case that the eigenvalues are not distinct (e.g. $\lambda\_1=\lambda\_2\neq \lambda\_3$ or even $\lambda\_1=\lambda\_2= \lambda\_3$).
I know that this is possible for a matrix $C$
$$C = \sum\_{i=1}^3 f(\lambda\_i)\ n\_i\otimes n\_i$$
In this case, the derivative $\partial C/\partial A$ (fourth order tensor) is always computable, also if the eigenvalues are not distinct (see e.g. *R.W. Ogden, Non-Linear Elastic Deformations, 1997, p. 162*)
|
https://mathoverflow.net/users/160345
|
Derivative of eigenpair with respect to matrix
|
If the eigenvalues are distinct you simply fill in the [first order perturbation answer](https://en.wikipedia.org/wiki/Eigenvalue_perturbation) for the eigenvalues and eigenvectors,
$$\frac{\partial\lambda\_i}{\partial A\_{jk}}=(n\_i)\_{j}(n\_i)\_{k}(2-\delta\_{jk}),$$
$$\frac{\partial n\_{i}}{\partial A\_{jk}}=\sum\_{p\neq i}n\_p\frac{(n\_p)\_j(n\_i)\_k(2-\delta\_{jk})}{\lambda\_i-\lambda\_p},$$
and then
$$\frac{\partial a}{\partial A\_{jk}}=\sum\_i f'(\lambda\_i)n\_i\frac{\partial\lambda\_i}{\partial A\_{jk}}+\sum\_i f(\lambda\_i)\frac{\partial n\_{i}}{\partial A\_{jk}}.$$
If two (or more) eigenvalues are identical the eigenvectors $n\_i$ are not uniquely determined by the matrix $A$, and neither is the quantity $a=\sum\_i f(\lambda\_i)n\_i$. Now you may choose a particular set of eigenvectors and ask for the derivative $\partial a/\partial A\_{jk}$ for that particular choice, this will in general diverge when two eigenvalues $\lambda\_p$ and $\lambda\_q$ coincide, unless $(n\_p)\_j (n\_q)\_k=0$.
Note that this complication does not appear if you consider instead the quantity $C=\sum\_i f(\lambda\_i)n\_i\otimes n\_i$, because $C=f(A)$ is uniquely defined by $A$, even if the eigenvalues are not distinct.
|
1
|
https://mathoverflow.net/users/11260
|
397158
| 163,960 |
https://mathoverflow.net/questions/397141
|
0
|
Let us consider a matrix $A \in M\_{2 \times n}(\mathbb C)$ such that $rank A=2$. Let us denote by
$$
a\_1,\ldots,a\_d \in \mathbb C,
$$
where $d:=\binom{n}{2}$, the value of the $2 \times 2$ minors of $A$. We identify these values as a point $P=[a\_1:\ldots:a\_d] \in \mathbb P^{d-1}$, giving a map
$$
\psi: X:=\{A \in M\_{2 \times n}(\mathbb C): rank A=2\} \to \mathbb P^{d-1}
$$
such that
$$
A \mapsto [a\_1:\ldots:a\_d].
$$
Since $X$ is an open subvariety of a projective space parametrizing $2\times n$ matrices, my question is: given a point $P \in \mathbb P^{d-1}$, what is $\psi^{-1}(P)$?
|
https://mathoverflow.net/users/147236
|
How many matrices with given minors?
|
This answer is really just an expansion of Mohan's and Zach Teitler's comments. I have only provided more detail, that's all. I don't care much for the points and if you want, just give the points to Mohan and Zach Teitler, but I thought it may be useful to the OP and to other readers to provide a complete answer.
Let $\iota: Gr\_2(\mathbb{C}^n) \to \mathbb{P}^{d-1}$ be the Plücker embedding, and let
$$f: X \to Gr\_2(\mathbb{C}^n)$$
be the map which sends $A \in X$ to the span of its columns.
Then, it can be seen that $\psi = \iota \circ f$. In particular, the image of $\psi$ is the image of the Plücker embedding, which is described by the classical Plücker relations.
If $p \in \operatorname{Im}(\psi)$, how can we describe $\psi^{-1}(p)$? First note that $g \in GL(2,\mathbb{C})$ acts on $X$ by mapping $A$ to $Ag^{-1}$, preserving the fibers of $\psi$. Moreover, since $\iota$ is an embedding, then $\psi^{-1}(p) = f^{-1}(V)$, where $V \in Gr\_2(\mathbb{C}^n)$ is the unique point in the Grassmannian which maps to $p$ via the Plücker embedding $\iota$. And $f^{-1}(V)$ is the set of all bases of $V$, which is parametrized by $GL(2,\mathbb{C})$. Thus $\psi^{-1}(p)$ is a copy of $GL(2,\mathbb{C})$.
|
1
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https://mathoverflow.net/users/81645
|
397168
| 163,962 |
https://mathoverflow.net/questions/397162
|
7
|
It is well known that (small) coproducts [commute with connected limits](https://ncatlab.org/nlab/show/commutativity+of+limits+and+colimits#coproducts_commute_with_connected_limits) in $\mathbf{Set}$. With which class of limits do finite coproducts commute?
Ideally, we should furthermore like to know whether the class of finite coproducts is *closed* [1] in the sense that the class of finite coproducts is precisely the class commuting with the given class of limits in $\mathbf{Set}$.
[1] [Notes on Commutation of Limits and Colimits](https://arxiv.org/abs/1409.7860), Bjerrum–Johnstone–Leinster–Sawin (2015)
|
https://mathoverflow.net/users/152679
|
Finite coproducts commute with which limits in Set?
|
Indeed, let $D$ be a category. The canonical functor $D \to \pi\_0(D)$ is both cofinal and coinitial. Therefore, if finite coproducts commute with $D$-limits in a category $\mathcal C$, then finite coproducts commute with $\pi\_0(D)$-limits. And it is easily seen that the only discrete limit shapes with which finite coproducts commute in $Set$ are the singleton ones. So as Tom Goodwillie supposed, the only limit shapes with with finite coproducts commute in $Set$ are the connected ones.
Finite coproducts are not closed -- they don't include splitting of idempotents, which commutes with any limit whatsoever. But I believe that the finite disjoint unions of absolute colimit shapes do form a closed class.
|
11
|
https://mathoverflow.net/users/2362
|
397170
| 163,964 |
https://mathoverflow.net/questions/397146
|
2
|
Consider the power sum
$$S\_a(b)=1^{2b}+2^{2b}+\cdots+(3a-2)^{2b}.$$
Let $\nu\_3(x)$ denote the $3$-adic valuation of $x$.
>
> **QUESTION 1. (milder)** Is this true?
> $$\nu\_3\left(\frac{S\_a(b)}{S\_a(1)}\right)=0.$$
> **QUESTION 2.** Is this true? $\nu\_3(S\_a(b))=\nu\_3(2a-1)$.
>
>
>
|
https://mathoverflow.net/users/66131
|
Divisibility of (finite) power sum of integers
|
Notice that
$$2S\_a(b) \equiv 1^{2b} + 2^{2b} + \dots + (6a-4)^{2b} \pmod{6a-3}.$$
From Faulhaber's formula, we have
$$1^{2b} + 2^{2b} + \dots + (6a-4)^{2b} \equiv B\_{2b} (6a-3) \pmod{6a-3}.$$
It follows that
$$S\_a(b) \equiv \frac32B\_{2b} (2a-1)\pmod{3(2a-1)},$$
where $\nu\_3(\frac32 B\_{2b})=0$ by [von Staudt–Clausen theorem](https://en.wikipedia.org/wiki/Von_Staudt%E2%80%93Clausen_theorem).
Hence, $\nu\_3(S\_a(b))=\nu\_3(2a-1)$.
---
**ADDED.** Explanation of the [Faulhaber's formula](https://en.wikipedia.org/wiki/Faulhaber%27s_formula) implication:
**Lemma.** Let $n,m$ be positive integers, $m$ is even. Then
$$1^{m} + 2^{m} + \dots + (n-1)^{m} \equiv B\_{m} n \pmod{n}.$$
**Proof.** For $m=2$, the statement can be verified directly:
$$1^2 + 2^2 + \dots + (n-1)^2 = \frac{(n-1)n(2n-1)}{6} \equiv B\_2 n\pmod{n}.$$
For the rest assume $m\geq 4$.
Using Faulhaber's formula and taking into account that $B\_{m-1}=0$, we have
\begin{split}
& 1^{m} + 2^{m} + \dots + (n-1)^{m} \equiv 1^{m} + 2^{m} + \dots + n^{m} \\
&\equiv B\_{m} n + \frac{1}{m+1} \sum\_{k=3}^{m+1} \binom{m+1}{k} B\_{m+1-k} n^k \pmod{n}.
\end{split}
It remains to show that $\frac{1}{m+1} \binom{m+1}{k} B\_{m+1-k} n^k \equiv 0\pmod{n}$ for any integer $k\in [3,m+1]$.
Consider any prime $p\mid n$, and let $t:=\nu\_p(n)\geq 1$. Our goal is to show that $\nu\_p\big(\frac{1}{m+1} \binom{m+1}{k} B\_{m+1-k}\big) + t(k-1) \geq 0$.
It is enough to focus on the case $t=1$, from which the case $t>1$ follows instantly.
Since by von Staudt–Clausen theorem the denominators of Bernoulli numbers are square-free, we need to show that
$$\nu\_p\big(\frac{1}{m+1} \binom{m+1}{k}\big) + k - 2 \geq 0.$$
Noticing that $\nu\_p\big(\frac{1}{m+1} \binom{m+1}{k}\big) = \nu\_p\big(\frac1k\binom{m}{k-1}\big)\geq -\nu\_p(k)$, our goal reduces to showing that
$$k-2-\nu\_p(k)\geq 0.$$
For $k=3$ and $k=4$, this inequality is trivial, while for $k\geq 5$ it follows from the bound $\nu\_p(k)\leq \log\_2(k)$. QED
|
6
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https://mathoverflow.net/users/7076
|
397172
| 163,966 |
https://mathoverflow.net/questions/397174
|
2
|
Except for $\ p=2\ $ primes split into two disjoint classes, $\ p\equiv1\mod4\ $ and $\ p\equiv3\mod4.\ $ Squares respect this partition, odd prime $\ p=m^2+n^2\ \Leftrightarrow\ p\equiv1\mod4.\ $ On the other hand, *triangles* $\ \binom k2\ $ are oblivious to the $\mod4\ $ classification, as well as to the other classification $\mod6\ $ (each prime different from $2$ and $3$ is congruent to $1$ or $-1$).
**Question** What is a simple characterization of primes of the form $\ p=\binom m2+\binom n2\ ?\ $ or, in a sense, there are no simple characterizations (?).
|
https://mathoverflow.net/users/110389
|
Equation $\ p=\binom m2+\binom n2$
|
$p={m\choose 2}+{n\choose 2}$ is equivalent to $8p+2=(2m-1)^2+(2n-1)^2$. On the other hand, if $8p+2$ is a sum of two squares, these squares must be both odd (this is seen modulo 4). So, applying the criterion for representability as a sums of two squares, we get that *$p$ is a sum of two triangular numbers if and only if the prime factorization of $4p+1$ does not contain prime divisors $\equiv 3\pmod 4$ in odd power.* I do not think that there is a more explicit characterization.
|
15
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https://mathoverflow.net/users/4312
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397176
| 163,967 |
https://mathoverflow.net/questions/397185
|
0
|
Let $p$ be prime of the form $p=u^2+1$. For $a \in \mathbb{F}\_p,a \ne 0$,
define
$E\_a : x^3+a x z^2=y^2 z$
Let $B= \lfloor 2 \sqrt{p}\rfloor$
Must we have $(\#E\_a(\mathbb{F}\_p) -p - 1) \in \{2,-2,B,-B\}$?
|
https://mathoverflow.net/users/12481
|
Primes of the form $p=u^2+1$ and number of points on the elliptic curve $x^3+a x z^2=y^2 z$
|
Look at Section 18.4 in Ireland-Rosen "A classical introduction to modern number theory". Note $p\equiv 1 \pmod{4}$ and $p = \pi\cdot \bar{\pi}$ with $\pi = 1- iu \equiv 1 \pmod{2+2i}$. Let $\lambda: \mathbb{F}\_p \to \langle i \rangle$ be the character of order $4$, which is equal to $(\tfrac{\cdot}{p})\_4$.
Theorem 5 there shows that $$a\_p = \overline{\lambda(-a)} \, \pi + \lambda(-a)\,\bar{\pi}$$ where $a\_p$ is the negative of your expression $\# E(\mathbb{F}\_p)-p-1$.
If $\lambda(-a)=1$ then $a\_p= 2$. If $\lambda(-a)=i$ then $a\_p=-2u$. If $\lambda(-a) = -1$ then $a\_p=-2$. If $\lambda(-a)=-i$ then $a\_p=-2u$.
So, yes $a\_p\in\{-2u,-2,2,2u\}$.
|
7
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https://mathoverflow.net/users/5015
|
397186
| 163,970 |
https://mathoverflow.net/questions/397132
|
4
|
Let $R$ be a dvr and $f : X\to S$ a universally closed morphism of $R$-schemes.
Assume $X$ and $S$ are $R$-flat and universally closed.
>
> If the special fiber of $X\to S$ is a closed immersion, is $X\to S$ a closed immersion?
>
>
>
**remarks**
* My guess is "no", but I'm looking for a counterexample (I can imagine something where $f$ crashes a locus contained in the generic fiber to a point).
* Namely, I expect "no" for an answer because I see no reason why $f$ should have to be a monomorphism (unless one also assumes $f$ is a closed immersion on generic fibers, which I am not assuming). I'd expect some example such that for some $s\in S$ the fiber $f^{-1}(s)$ is nonempty and larger than $\kappa(s)$-point.
* I can believe $f$ is unramified. What I expect should fail is the monomorphism part.
* A silly example of a map that is a monomorphism on special fibers but not globally a monomorphism can already be, calling $K$ the fraction field of $R$ $$f : \text{Spec}(K)\coprod \text{Spec}(R) \to\text{Spec}(R)$$
induced by the ring map $R\to R\times K$, the identity on the first factor and the inclusion on the second. Clearly $f$ is the identity (hence a closed immersion) on special fibers, but it is not a monomorphism globally since the fiber over the generic point $\eta\in\text{Spec}(R)$ is two copies of $\eta$ and not one.
* **What I'm looking for** is a neat non-silly example of a geometric flavor (say $X$ and $S$ connected, of dimension $\ge 2$, perhaps $f$ of relative dimension $\ge 1$ with geometrically connected fibers). In the first example above $\text{Spec}(K)\coprod\text{Spec}(R)$ is not universally closed, so this is not an example directly relevant to the question. The second example above, if correct, should answer the question in the negative, as expected. I'd prefer some example for which one can (at least generically) "draw a picture".
|
https://mathoverflow.net/users/nan
|
Detecting closed immersions on fibers
|
I am just posting my comment as an answer, mostly to correct the mistake identified by @LaurentMoret-Bailly.
The property of being universally closed depends only on the underlying reduced scheme. Thus, there are counterexamples coming from a nilradical that is quasi-coherent, yet not coherent.
For one example, let $S$ be $\text{Spec}(R)$ and let $X$ be Spec of the $R$-algebra $R\oplus \text{Frac}(R)x$, where $x^2$ is zero. The morphism from $X$ to $S$ is a universal homeomorphism, thus universally closed. The morphism of closed fibers is an isomorphism, hence it is a closed immersion. Yet the morphism of generic fibers is projection from the dual numbers over Spec of $\text{Frac}(R)$, and this is not a closed immersion.
|
2
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https://mathoverflow.net/users/13265
|
397190
| 163,971 |
https://mathoverflow.net/questions/397187
|
3
|
$\newcommand{\sym}{\mathfrak{S}}$
$\newcommand{\rarr}{\rightarrow}$
Given a partition $\lambda$ of $n$ and a commutative ring $R$, writing $\sym\_n$ for the symmetric group, there is a Specht module $S^\lambda$ well-defined as an $R\sym\_n$-module. For example James does this in his book using polytabloids.
By a Specht filtration on an $R\sym\_n$-module $I$, I mean a finite filtration of $I$ where each cofactor is isomorphic to a Specht module.
Suppose there is an exact sequence $$0 \rarr M \rarr I^0 \rarr I^1 \rarr \cdots \rarr I^r \rarr 0$$
of $R\sym\_n$-modules where each $I^j$ for $0 \leq j \leq r$ has a Specht filtration. Does this imply $M$ should also have a Specht filtration in general? What if $R$ is a field whose characteristic is at least 5? (Hemmer and Nakano have papers showing Specht filtrations are better behaved in this situation)
|
https://mathoverflow.net/users/21848
|
Does a finite coresolution with Specht-filtered modules imply a Specht filtration?
|
No. Working with modules for $\mathbb{F}\_5S\_5$, take the Specht module $S^{(2,1,1,1)}$. Using $S^{(2,1,1,1)} \cong (S^{(4,1)} \otimes \mathrm{sgn})^\star$ and that $S^{(4,1)}$ has top $D^{(4,1)}$ and socle $\mathbb{F}\_5$, we get a short exact sequence
$$0 \rightarrow D^{(4,1)} \otimes \mathrm{sgn} \rightarrow S^{(2,1,1,1)} \rightarrow \mathrm{sgn}\rightarrow 0. $$
Here $S^{2,1,1,1)}$ and $\mathrm{sgn} \cong S^{(1^5)}$ are Specht modules, but the simple module $D^{(4,1)} \otimes \mathrm{sgn}$ has dimension $3$. Since the Specht modules for $\mathbb{F}\_5S\_5$ (other than $S^{(5)} \cong \mathbb{F}\_5$ and $S^{(1^5)} \cong \mathrm{sgn}$) all have dimension at least $4$, $D^{(4,1)} \otimes \mathrm{sgn}$ does not have a Specht filtration.
|
4
|
https://mathoverflow.net/users/7709
|
397191
| 163,972 |
https://mathoverflow.net/questions/397180
|
6
|
There is a well-know quotation of Euler in a letter from 1746 to Goldbach:
>
> Letztens habe ich gefunden, dass diese expressio $\sqrt{-1}^{\sqrt{-1}}$ einen valorem realem habe, welcher in fractionibus decimalibus
> $=0,2078795763$, welches mir merkwürdig zu seyn scheinet.
>
>
>
For the principal value of the logarithm the *expresssion* is $i^i=\exp(i \log(e^{i\pi/2}))= e^{-\pi/2}$.
My question is, how did Euler calculate this with such an accuracy?
|
https://mathoverflow.net/users/21051
|
How did Euler calculate $i^i$?
|
[On the numerical value of $i^i$](https://www.jstor.org/stable/2972387%0A) and [Historical notes on the relation $e^{-\pi/2}=i^i$](https://doi.org/10.2307/2972388) describe how these accurate computations can be performed with logarithmic tables.
Euler described how he arrived at the identity in a paper read at the Berlin Academy in 1746, giving more decimals (13) than in the letter to Goldbach. A later calculation by Gauss computed 35 decimal places. Euler did not present his computation, but Gauss did [[source](https://archive.org/details/werkecarlf03gausrich/page/n439/mode/2up)].
|
10
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https://mathoverflow.net/users/11260
|
397193
| 163,974 |
https://mathoverflow.net/questions/397178
|
8
|
Assume $X$ is a smooth projective variety over $\overline{\mathbf{F}}\_p$ and fix a prime $\ell\neq p$.
Let $F\_i$ be the geometric Frobenius on $\ell$-adic cohomology
$$H^i\_{\rm ét}(X,\overline{\mathbf{Q}}\_{\ell})$$
for fixed $i\ge 0$.
>
> * What relation is expected to hold between the minimal polynomial $m\_{F\_i}(T)$ of $F\_i$ and the characteristic polynomial $P\_i(T)$ of $F\_i$? (apart from $m\_{F\_i}\mid P\_i$)
> * Are $m\_{F\_i}(T)$ and $P\_i(T)$ conjectured to agree? If so, is this a known result?
> * Does $m\_{F\_i}(T)$ depend on the Weil cohomology theory chosen to compute $P\_i(T)$?
>
>
>
We know from Deligne's work on the Weil conjectures, that we have $P\_i(T)\in\mathbf{Z}[T]$, $P\_i(T)$ does not depend on the chosen Weil cohomology, and its roots are of the form $q^{-i/2}\rho$ for $\rho$ an algebraic number whose complex absolute value, for any complex embedding $\overline{\mathbf{Q}}\subset\mathbf{C}$, is one.
I'm mostly interested to understand to what extent $m\_{F\_i}(T)$ is, or expected to be, intrinsic.
|
https://mathoverflow.net/users/nan
|
Minimal vs characteristic polynomial of geometric Frobenius
|
It's conjectured -- see e.g. [this question](https://mathoverflow.net/questions/104102/semisimplicity-of-frobenius-operation-on-etale-cohomology?noredirect=1&lq=1) -- that the Frobenius is always semisimple, so its minimal polynomial is the radical of its characteristic polynomial (the product of its distinct linear factors, each with multiplicity 1). So $m\_{F\_i}$ should be independent of $\ell$.
This also shows that $m\_{F\_i}$ is different from $P\_i$ iff $P\_i$ has a root of multiplicity $> 1$. This can certainly occur, e.g. consider a supersingular elliptic curve over $\mathbf{F}\_{p^2}$.
|
6
|
https://mathoverflow.net/users/2481
|
397195
| 163,975 |
https://mathoverflow.net/questions/397150
|
9
|
Let $L/K$ be a (abelian, Galois) quadratic extension of number fields with $\text{Gal}(L/K)$ generated by $\sigma$ and $\mathfrak{p} = \alpha\mathcal{O}\_K$ a principal prime ideal of $\mathcal{O}\_K$. Assume $\mathfrak{p}$ splits as $\mathfrak{P} \sigma(\mathfrak{P})$ in $\mathcal{O}\_L$ and that $\alpha = \beta \sigma(\beta)$ for $\beta \in L^\times$ (so $\beta$ may not be integral though $\alpha$ is). I would like to show that $\mathfrak{P}$ is principal (and possibly that $\mathfrak{P} = \beta \mathcal{O}\_L$).
$\textbf{My attempt}$: It seems clear that if $\beta$ is not integral it must generate a fractional ideal of the form $$\beta\mathcal{O}\_L = \frac{\mathfrak{P}I}{\sigma(I)}$$ for some $I \subset \mathcal{O}\_L$.
We can assume $\mathfrak{P}I \cap \sigma(I) = \mathcal{O}\_L$, i.e. that the numerator and denominator are simplified: we can cancel common factors of $I$ and $\sigma(I)$ so that $I \cap \sigma(I) = \mathcal{O}\_L$, and if $\mathfrak{P} \cap \sigma(I) = \mathfrak{P}$ then $\sigma(\mathfrak{P})$ divides $I$ and we get that $$\beta\mathcal{O}\_L = \frac{\sigma(\mathfrak{P}) I'}{\sigma(I')}$$ where $I' = I/\sigma(\mathfrak{P})$. So, WLOG we can take the first expression for $\beta\mathcal{O}\_L$.
If $\beta$ is not integral we can find an integer $d \ne 1$ such that $d\beta$ is integral ($\beta$ is just a linear combination over the basis for $L$ and $d$ is the lcm of the denominators of the coefficients). Then $$ d\beta\mathcal{O}\_L = (d)\frac{\mathfrak{P}I}{\sigma(I)} \subset \mathcal{O}\_L.$$
Since this is integral, $\sigma(I)$ divides $(d)\mathfrak{P}I$. As $\mathfrak{P}I \cap \sigma(I) = \mathcal{O}\_L$, $\sigma(I)$ divides $(d)$. But so does $I$ (as $\sigma$ fixes $d$). If $I$ is nontrivial then this contradicts $I \cap \sigma(I) = \mathcal{O}\_L$, so either $I$ is trivial or $d = 1$. In either case we have $\beta\mathcal{O}\_L = \mathfrak{P}$ as desired.
This proof feels awkward and I suspect either it is wrong or just overly complicated. I'd appreciate any feedback!
|
https://mathoverflow.net/users/106850
|
Is it true that this ideal must be principal? (proof verification)
|
I claim that under the given circumstances $\mathfrak{P}$ is not necessarily principal, i.e., the statement claimed in the question is wrong.
Here is a counterexample. Consider $K = \mathbb{Q}$ and $L = \mathbb{Q}[\sqrt{-47}]$. Let $\alpha = -3$ and $\beta = \frac{1}{4}(1 \pm \sqrt{-47})$. Then the minimal polynomial of $\beta$ is $X^2 - \frac{1}{2} X + 3$. In particular, the norm of $\beta$ is $\beta\sigma(\beta) = -3$. (Note that $\beta$ is not integral, as its minimal polynomial is not integral.)
Then with the help of SAGE one computes the factorization in $L$: $3\mathcal{O}\_L =(3, \frac{\sqrt{-47}}{2} - \frac{1}{2})(3, \frac{\sqrt{-47}}{2} + \frac{1}{2})$. Say $\mathfrak{P}=(3, \frac{\sqrt{-47}}{2} - \frac{1}{2})$. Also, SAGE tells that the class group of $L$ is of order $5$ with generator $\mathfrak{a} = (2, \frac{\sqrt{-47}}{2} + \frac{1}{2})$, and that $\mathfrak{P} = \mathfrak{a}^2$ in the class group. Thus $\mathfrak{P}$ is not principal.
---
I believe that a mistake in the posted proof is contained in the sentense "If $I$ is non-trivial, then ...". Indeed, that $I$ and $\sigma(I)$ divide (d) only means that all $p$-adic valuations of $I$ and $\sigma(I)$ are smaller than those of $(d)$. Why this should imply that $I \cap \sigma(I) = \mathcal{O}\_L$?
|
10
|
https://mathoverflow.net/users/148992
|
397200
| 163,976 |
https://mathoverflow.net/questions/397078
|
3
|
In some urn model with parameter $p$, the generating function
$$
f\_p(z) \;=\; \frac{1+p\,z}{1-(1-p)\,z\,(1+p\,z)}
$$
is such that $[z^n]f\_p(z)$ is the probability that an $n$-urn configuration has a particular property.
It is known that this probability tends to zero if $p\gg n^{-1/2}$ and tends to one if $p\ll n^{-1/2}$, so $n^{-1/2}$ is a threshold for the property.
The following approach appears to give the correct probability for the property when $p=c\,n^{-1/2}$:
Let $\rho(p)$ be the dominant singularity of $f\_p$ (a simple pole). Then,
$$
\lim\_{n\to\infty}\rho\big(c\,n^{-1/2}\big)^{-n} \;=\; e^{-c^2} .
$$
Obviously, for fixed $p$, we have $[z^n]f\_p(z)\sim c\_p\,\rho(p)^{-n}$, for some (computable) constant $c\_p$, but substituting $p=c\,n^{-1/2}$ and then taking limits seems rather dubious.
**Question:** Can singularity analysis be used in this way? If so, under what conditions, and where can I find the appropriate justification?
[Of course, the specific details of this urn model and property aren't relevant; it's the general method I'm asking about.]
|
https://mathoverflow.net/users/34341
|
Using singularity analysis for probability at a threshold?
|
There are a few things working in this particular question that allow for that to occur. This is a rational function and the denominator is a quadratic in $z$ meaning that we can write $$[z^n] f\_p(z) = c\_1 \rho\_1^{-n} + c\_2 \rho\_2^{-n}$$ where $\rho\_j$ are the roots of the denominator and $c\_j$ are functions of $p$.
In the case at hand, we can substitute $p = c/\sqrt{n}$ and then see if the limit of this exact expression exists. Yes, $\rho\_2$ is larger than $\rho\_1$, but it could *a priori* be the case that $c\_2 \rho\_2^{-n}$ does not tend to $0$ as $n \to \infty$. In this concrete example, you can compute that $c\_2 \to 0$ and $\rho\_2^{-n} \to 0$ for $p = c/\sqrt{n}$ and $n \to \infty$.
Similarly, we need to understand the limiting behavior of $c\_1$. *A priori*, it could be the case that $c\_1 \to 0$ or something strange, but here it can be shown that $c\_1 \to 1$ in the regime listed.
It could be possible that there are more general conditions that guarantee that $c\_1 \not\to 0$ and similarly to deduce behavior on $c\_2 \rho\_2^{-n}$; nothing I've written---or you used in your post---takes into account that this is actually a bivariate rational function, just that it's rational in $z$. Regardless, some info about $c\_1$ has be taken into account, as we could just multiply the generating function by $2$ and alter the asymptotics by a constant factor.
|
2
|
https://mathoverflow.net/users/69870
|
397205
| 163,978 |
https://mathoverflow.net/questions/396867
|
9
|
>
> Is $\textrm{Spin}(8)$ a direct product of $\textrm{Spin}(7)$ and $S^7$?
>
>
>
I met this statement in the literature, but without a reference. If it is true, where is it explicitly written?
|
https://mathoverflow.net/users/309203
|
Is $\operatorname{Spin}(8)$ a direct product of $\operatorname{Spin}(7)$ and $S^7$?
|
As I suspected, the statement that the bundle $\mathrm{SO}(8)\to S^7$ is a product bundle, i.e., that
$$
\mathrm{SO}(8)\simeq S^7\times\mathrm{SO}(7)\tag1
$$
as bundles over $S^7$ is in N. Steenrod's [The topology of fibre bundles](https://mathscinet.ams.org/mathscinet/search/publdoc.html?arg3=&co4=AND&co5=AND&co6=AND&co7=AND&dr=all&pg4=AUCN&pg5=AUCN&pg6=TI&pg7=ALLF&pg8=ET&review_format=html&s4=Steenrod%2C%20N%2A&s5=&s6=The%20topology%20of%20fibre%20bundles&s7=&s8=All&sort=Newest&vfpref=html&yearRangeFirst=&yearRangeSecond=&yrop=eq&r=1&mx-pid=1688579) (as part of Theorem 8.6).
Note that Steenrod uses the notation $R\_n$ for what is more commonly notated $\mathrm{SO}(n)$, these days. (The '$R$' is for 'rotation group'. He uses $O\_n$ for the orthogonal group.) His proof, based on the octonions (which he calls 'the Cayley numbers'), is exactly what I outlined in my comment above. He does not give a reference to an earlier statement of the result, but refers to L. E. Dickson's famous *Linear Algebra* for the properties of the Cayley numbers that are used in the proof.
He does not discuss the spin groups, but, obviously, the equivalence of bundles
$$
\mathrm{Spin}(8)\simeq S^7\times\mathrm{Spin}(7)\tag2
$$
follows from (1) by passing to the respective double covers.
|
6
|
https://mathoverflow.net/users/13972
|
397209
| 163,979 |
https://mathoverflow.net/questions/397210
|
22
|
Bertrand Russell, I believe, somewhere presents a joke (if I remember correctly). Someone is shown the boat of another, and the first says: "I thought that your boat is longer than it is." The owner replies: "No, my boat is not longer than it is."
Does someone know the reference?
|
https://mathoverflow.net/users/37385
|
"The boat is not longer than it is."
|
Bertrand Russel, [**On denoting**](http://web.mnstate.edu/gracyk/courses/web%20publishing/russell_on_denoting.htm)
*Mind*, October 1905, pages 479-493.
>
> When we say: "George IV wished to know whether so-and-so", or when we
> say "So-and-so is surprising" or "So-and-so is true", etc., the
> "so-and-so" must be a proposition. Suppose now that "so-and-so"
> contains a denoting phrase. We may either eliminate this denoting
> phrase from the subordinate proposition "so-and-so", or from the whole
> proposition in which "so-and-so" is a mere constituent. Different
> propositions result according to which we do. I have heard of a touchy
> owner of a yacht to whom a guest, on first seeing it, remarked, "I
> thought your yacht was larger than it is"; and the owner replied, "No,
> my yacht is not larger than it is". What the guest meant was, "The
> size that I thought your yacht was is greater than the size your yacht
> is"; the meaning attributed to him is, "I thought the size of your
> yacht was greater than the size of your yacht".
>
>
>
|
19
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https://mathoverflow.net/users/11260
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397211
| 163,980 |
https://mathoverflow.net/questions/397060
|
3
|
I already asked this question on the Math Stack Exchange but did not get an answer.
I am currently working through Geiges proof of the Martinet-Lutz theorem, which can be
found [here](http://www.mi.uni-koeln.de/%7Egeiges/contact05.pdf), and am trying to figure out the effect of a half Lutz twist on the Euler class
of the contact structure. During the proof of Propositioin 3.15, he basically claims that given a contact structure $\xi$
on $M$ and a transverse link $L$, then performing a half Lutz twist along the components
of $L$ yields a contact structure $\eta$ such that
$$e(\eta) = e(\xi) - 2PD[L]$$
where $PD: H\_1(M)\to H^2(M)$ is Poincaré-Duality.
I am however fairly certain, that I have proven the same result with $2PD[L]$ replaced
by $PD[L]$, here is my proof:
* Choose a tubular neighbourhood $V$ of $L$, such that on each component the contact structure is given by $\ker d\theta + r^2d\phi$.
* Start with a vector field $X$ tangent to $\xi$, which is in general position with the zero section,
does not vanish over $V$ and is given by $\partial\_r$ on the boundary of $V$, then $e(\xi) = PD[X\cap M]$.
* After the Lutz twist the $X$ I have in mind is not tangent to $\eta$, however we
can replace $X$ on $V$ by $\partial\_r$, such that $X$ now vanishes along $L$.
The new Euler class is thus given by
$$e(\eta) = PD[X\cap M] = e(\xi) + PD[L].$$
Can anyone spot a mistake in my argument, or is the statement in Geiges wrong?
|
https://mathoverflow.net/users/316096
|
Effect of a Lutz twist on Euler number
|
I have actually solved this by now, his statement is indeed correct but
the proof is a bit misleading. I will only discuss the case where $K$ is a knot.
Instead of choosing a field which does not vanish in a tubular neighbourhood of $K$, one simply chooses a generic vector field which is given by the radial vector field $\partial\_r$ in a tubular neighbourhood of $K$, which vanishes along $K$.
After the Lutz twist the same vector field is also tangent to $\eta$
and still vanishes along $K$, however, the Lutz twist has made $K$ a negatively transverse knot such that the difference $e(\eta) - e(\xi)$ is
given by $-2PD[K]$.
|
1
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https://mathoverflow.net/users/316096
|
397216
| 163,981 |
https://mathoverflow.net/questions/397043
|
3
|
I am currently trying to understand the properties of Deligne-Lusztig induction, following Carter's *Finite groups of Lie type* and Digne-Michel's *Representations of finite groups of Lie type*. I am reasonably satisfied with the construction, but I am having difficulty understanding the proofs of the properties of the Deligne-Lusztig induction functor (it is worth noting that these two books have very different approaches to this theory, and as I am bouncing back-and-forth to fill in gaps, I may be missing something obvious). I am specifically interested in the simple case of inducing from a torus to a Borel, so I will try to simplify definitions and statements where possible.
In Digne-Michel (Definition 11.1), the DL functor $R\_{T \subset B}^G$ is the generalised induction functor associated to the $G^F$ -module-$T^F$ afforded by $H^\*\_{c}(L^{-1}(U))$. Here $B=TU$ is the Levi decomposition of the Borel subgroup $B$, $L : G \to G$ is the Lang map and $H^\*\_c$ is $l$-adic cohomology with compact support. If (like me) you have difficulty visualising this representation, we can concretely realise the DL functor on the level of characters by the formula
$$R\_{T \subset B}^G(\theta)(g) = \frac{1}{|T^F|} \sum\_{t \in T^F} \theta(t^{-1}) \mathcal{L}((g, t), L^{-1}(U))$$
where $\mathcal{L}((g,t), L^{-1}(U))$ denotes the Leftschetz number of the right-left multiplication action of $(g,t)$ on $L^{-1}(U)$. This is Proposition 11.2 in Digne-Michel.
The authors then remark that there is an adjoint functor $^\* R^G\_{T \subset B}$, called DL restriction, which can be given explicitly by the formula
$$^\* R^G\_{T \subset B}(\psi)(t) = \frac{1}{|G^F|} \sum\_{g \in G^F} \psi(g^{-1}) \mathcal{L}((g,t), L^{-1}(U)).$$
**Question 1:** Why does $R^G\_{T \subset B}$ admit an adjoint functor? This is not (to my knowledge) elaborated upon at all in either of the books I am following.
**Question 2:** I guessed that one can prove $R^G\_{T \subset B}$ admits an adjoint functor by using some abstract category theoretic condition, and then simply define $^\* R^G\_{T \subset B}$ as the adjoint. If this is the approach, then we get Frobenius reciprocity for free, but I don't know where the above formula comes from. If, on the other, we *define* $^\* R^G\_{T \subset B}$ by the above formula (which certainly seems like a very reasonable guess for an adjoint), how does one prove Frobenius reciprocity? Which of these is the easier/standard approach?
**Question 3:** Regardless of how one defines $^\* R^G\_{T \subset B}$, is $(R^G\_{T \subset B}, ^\* R^G\_{T \subset B})$ a bi-adjoint pair? I know that standard induction agrees with co-induction for finite groups, but I have very little intuition for DL induction and I cannot make a guess either way.
|
https://mathoverflow.net/users/175051
|
Frobenius reciprocity for Deligne-Lusztig induction/restriction
|
**Answer to Questions 1 and 2:** At least in Digne--Michel the functor ${}^\ast R\_{T \subseteq B}^G$ is rigorously defined. The reference is [Digne--Michel, p.47]. What they do is the following: If $G$, $H$ are any two finite groups and $M$ is a $G$-module-$H^{\rm opp}$, then we have the functor $R\_H^G \colon E \mapsto M \otimes\_{\mathbb{C}[H]} E$. This allows to define the DL-induction $R\_{T \subseteq B}^G$.
Then, the dual $H$-module-$G^{\rm opp}$ $M^\ast = {\rm Hom}(M,\mathbb{C})$ defines the adjoint functor. This can be verified using the Hom-$\otimes$-adjunction. Finally, [Digne--Michel, Prop. 4.5] applies to give the traces of both, $R\_H^G$ and ${}^\ast R\_H^G$, which answers your Q2.
|
3
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https://mathoverflow.net/users/148992
|
397222
| 163,982 |
https://mathoverflow.net/questions/397201
|
2
|
Let $E$ be the elliptic curve over $\mathbf{Q}\_3$ with Weierstrass equation $y^2 = x^3-x$.
It has complex multiplication by $\mathbf{Z}[\sqrt{-1}]$, with $\sqrt{-1}$ acting as $(x,y)\mapsto(-x,iy)$.
Call $R:=\mathbf{Z}\_3[\sqrt{-1}] = W(\mathbf{F}\_9)$.
Let $\mathcal{E}$ be the Néron model over $\mathbf{Z}\_3$, an abelian scheme (e.g. by Néron-Ogg-Shafarevich). Then the endomorphism $[-3] : \mathcal{E}\otimes\_{\mathbf{Z}\_3}R\to\mathcal{E}\otimes\_{\mathbf{Z}\_3}R$ is a lift of the $9$-power map on $\mathcal{E}\_{\mathbf{F}\_9}\to \mathcal{E}\_{\mathbf{F}\_9}$.
In the example by Serre discussed in the [answer](https://mathoverflow.net/q/106246) to this [question](https://mathoverflow.net/q/106238), a conjectural $\mathbf{Q}$-linear cohomology theory is discussed for varieties over $p$-adic fields, and it is argued that such theory cannot exist in light of the fact that a Frobenius action $F$ on $H^1(E\_{\overline{\mathbf{Q}}\_3})$ would satisfy $F^2 = -3=[-3]^\*$, $F[\sqrt{-1}]^\*=-[\sqrt{-1}]^\*F$, and of course $([\sqrt{-1}]^\*)^2=-1$, and these equations cannot be solved in $2\times 2$ matrices over $\mathbf{R}$ (at least, this is how I understood the question and answer).
My question only asks for a clarification on the linked question and answer.
>
> **question**
>
> It seems to me the linked question asks for a cohomology theory for varieties over $p$-adic fields. How does $F$ on $H^1$ arise, then?
>
>
>
I would expect such theory to be functorial *only with respect to morphisms of varieties over $p$-adic fields* (as it seems to be required in the linked question), and so $H^1(E\_{\overline{\mathbf{Q}}\_3})$ would not necessarily carry the effect of endomorphisms of $\mathcal{E}\_{\overline{\mathbf{F}}\_3}$, except those that are liftable.
The $3$-rd power map on $\mathcal{E}\_{\mathbf{F}\_3}$ is not liftable, I believe. Otherwise I'd expect the example to apply to the rational Betti cohomology of $E(\mathbf{C})$ for any $\sigma :\overline{\mathbf{Q}}\_3\simeq\mathbf{C}$.
**expectation** I expect the linked question asks for a theory that is *assumed* to carry some such $F$ (probably in the last sentence of the linked question - "taking value in Weil-Deligne representations over $\mathbf{Q}$"), and Serre's example shows $F$ cannot exist on any such $\mathbf{Q}$-linear theory. I'd just like to double-check this understanding is correct, to be sure. I'll leave the question here for other's benefit.
|
https://mathoverflow.net/users/nan
|
An example of Serre on the cohomology of some CM elliptic curves
|
Yes. There exists a "nice" cohomology theory for $p$-adic varieties, taking values in vector spaces over $\mathbb Q$, defined by fixing an embedding $\mathbb Q\_p \to \mathbb C$, base-changing along this embedding, and taking singular cohomology. Every morphism of varieties over $\mathbb Q\_p$ induces a map on cohomology groups in this theory (by the usual property of singular cohomology). So no argument of this type can rule out such a theory.
It is a cohomology theory valued in $\mathbb Q$-rational Weil-Deligne representations, which by assumption admit an action of Frobenius, that this argument rules out.
|
2
|
https://mathoverflow.net/users/18060
|
397225
| 163,984 |
https://mathoverflow.net/questions/397217
|
0
|
The Dirichlet inverse of the Euler totient function is:
$$\varphi^{-1}(n) = \sum\_{d \mid n} \mu(d)d \tag{1}$$
and the von Mangoldt function can be expanded/computed as:
$$\Lambda(n) = \sum\limits\_{k=1}^{\infty}\frac{\varphi^{-1}(\gcd(n,k))}{k} \tag{2}$$
Consider the sequences:
$$a(n)=\sum \_{k=1}^{j} \frac{\varphi^{-1}(\gcd (n,k))}{k}$$
$$b(n)=\sum \_{k=1}^{j+1} \frac{\varphi^{-1}(\gcd (n,k))}{k}$$
**Question:**
>
> For what values of $j$ does $b(n)$ eventually correlate better than $a(n)$ with the von Mangoldt function $\Lambda(n)$ as $n \rightarrow \infty$?
>
>
>
The exceptions of $j$ when $b(n)$ correlates worse than $a(n)$ appear to be:
$$j=7, 15, 24, 26, 31$$ which when added with $1$ gives initially a sequence of powers of some sort:
$$j+1=8, 16, 25, 27, 32$$
A very slow Mathematica program that computes the sequences $a(n)$ and $b(n)$ and compares their Pearson correlation with the von Mangoldt function, is:
```
Clear[a, n, k, start, end]
nnn = 200;
a[n_] := Total[Divisors[n]*MoebiusMu[Divisors[n]]];
Do[column = j;
earlier =
Table[Correlation[
Table[Sum[a[GCD[n, k]]/k, {k, 1, column}], {n, 1, nn}],
Table[N[MangoldtLambda[n]], {n, 1, nn}]], {nn, 2, nnn}];
later = Table[
Correlation[
Table[Sum[a[GCD[n, k]]/k, {k, 1, column + 1}], {n, 1, nn}],
Table[N[MangoldtLambda[n]], {n, 1, nn}]], {nn, 2, nnn}];
sign = Sign[later - earlier];
Print[If[Last[sign] == -1, j, 0]], {j, 2, 32}]
```
`nnn = 200;` is considered a large number that serves as the substitute for $n \rightarrow \infty$ in the program.
---
Edit 10.7.2021:
Correlations of partial sums:
```
Clear[a, n, k, start, end]
nnn = 400;
a[n_] := Total[Divisors[n]*MoebiusMu[Divisors[n]]];
Do[column = j;
earlier =
Table[Correlation[
Accumulate[
Table[Sum[a[GCD[n, k]]/k, {k, 1, column}], {n, 1, nn}]],
Accumulate[Table[N[MangoldtLambda[n]], {n, 1, nn}]]], {nn, 2,
nnn}];
later = Table[
Correlation[
Accumulate[
Table[Sum[a[GCD[n, k]]/k, {k, 1, column + 1}], {n, 1, nn}]],
Accumulate[Table[N[MangoldtLambda[n]], {n, 1, nn}]]], {nn, 2,
nnn}];
sign = Sign[later - earlier];
Print[{earlier, Count[sign, 1], sign, j}], {j, 2, 32}]
```
|
https://mathoverflow.net/users/25104
|
Correlating the von Mangoldt function with periodic sequences
|
TL;DR: This is not a full answer. I compute asymptotically the correlation of $b(n)$ and $\Lambda (n)$ (it turns out to be $0$) and the correlation of $b(n)$ and $a(n)$ in a relatively closed form sum (I don't think there's really a simpler expression). Although I do not show when the correlation of $b(n)$ and $a(n)$ is larger or smaller than $0$, I show that it is never equal exactly to $0$ except for $j = 1$.
---
For the rest of the answer let us denote by $\mathbb{E}\_N$ the average of a function on the numbers $1, \dots, N$. We will always look at correlations up to $N$, and then let $N \to \infty$. We also assume that $j > 1$.
First notice that the correlation of $b(n)$ and $\Lambda(n)$ tends to $0$ as $N \to \infty$. This is simply because $b(n)$ is a bounded sequence and $\Lambda (n)$ fluctuates wildly. More precisely, the correlation is equal to
$$\frac{\mathbb{E}\_N \left[ \Lambda (n) b(n) \right] - \mathbb{E}\_N \left[ \Lambda (n) \right] \mathbb{E}\_N \left[ b(n) \right]}{\left( \mathbb{E}\_N \left[ \Lambda (n)^2 \right] - \mathbb{E}\_N \left[ \Lambda(n) \right]^2 \right)^{\frac{1}{2}} \left( \mathrm{Var}\_N \left( b(n) \right) \right)^{\frac{1}{2}}}$$
As $b(n)$ is bounded, the numerator is at most a constant times $\mathbb{E}\_N \left[ \Lambda (n) \right]$, which by the Prime Number Theorem is asymptotically equal to $1$. Also by the Prime Number Theorem we have $\mathbb{E}\_N \left[ \Lambda (n)^2 \right] \sim \log N$, and as $b(n)$ is a non constant periodic sequence (with period $\mathrm{lcm} \left( 1, \dots, j + 1 \right)$) its variance is bounded by below. Therefore the correlation of $b(n)$ and $\Lambda (n)$ is bounded in absolute value by a constant times $\frac{1}{\log N}$, which tends to $0$.
---
Now, let us compute the correlation of $a(n)$ and $b(n)$. As they are periodic non-constant sequences, their variance is bounded by below. We will see that their covariance is asymptotically equal to a certain number, and whether this number is greater or lesser than $0$ exactly determines whether $b(n)$ is better correlated with $a(n)$ or with $\Lambda (n)$. First, we will compute their averages.
$$\mathbb{E}\_N \left[ a(n) \right] = \frac{1}{N} \sum\_{n = 1}^{N} \sum\_{k = 1}^{j} \frac{1}{k} \sum\_{d | k, n} d \mu (d) = \frac{1}{N} \sum\_{d = 1}^{j} d \mu(d) \left[ \frac{N}{d} \right] \sum\_{k \leq \frac{j}{d}} \frac{1}{d k}$$
As $N \to \infty$ this is is asymptotically equal to
$$\sum\_{d = 1}^{j} \frac{\mu (d)}{d} H\_{\left[ \frac{j}{d} \right]}$$
where $H\_m = \sum\_{k = 1}^{m} \frac{1}{k}$ is the Harmonic number. Replacing $j$ with $j + 1$ we get that $\mathbb{E}\_N \left[ b(n) \right]$ is asymptotically equal to $\sum\_{d = 1}^{j + 1} \frac{\mu (d)}{d} H\_{\left[ \frac{j + 1}{d} \right]}$ as $N \to \infty$. Now, we compute
$$\mathbb{E}\_N \left[ a(n) b(n) \right] = \frac{1}{N} \sum\_{n = 1}^{N} a(n) b(n) = \frac{1}{N} \sum\_{n = 1}^{N} \sum\_{k\_1 = 1}^{j} \sum\_{k\_2 = 1}^{j + 1} \frac{1}{k\_1 k\_2} \sum\_{d\_1 | n, k\_1} \sum\_{d\_2 | n, k\_2} d\_1 d\_2 \mu(d\_1) \mu(d\_2)$$
Notice that $d\_1 | n$ and $d\_2 | n$ if and only if $\mathrm{lcm} \left( d\_1, d\_2 \right) | n$. Interchanging the order of summation, and applying the fact that $\frac{1}{N} \left[ \frac{N}{m} \right] \sim \frac{1}{m}$ for $m \in \mathbb{N}$ we see that this sum is equal to
$$\sum\_{d\_1 = 1}^{j} \sum\_{d\_2 = 1}^{j + 1} \frac{\mu(d\_1) \mu(d\_2)}{\mathrm{lcm} \left( d\_1, d\_2 \right)} H\_{\left[ \frac{j}{d\_1} \right]} H\_{\left[ \frac{j + 1}{d\_2} \right]}$$
---
To sum it up, the covariance of $b(n)$ and $a(n)$ is asymptotically
$$\sum\_{d\_1 = 1}^{j} \sum\_{d\_2 = 1}^{j + 1} \frac{\mu(d\_1) \mu(d\_2)}{\mathrm{lcm} \left( d\_1, d\_2 \right)} H\_{\left[ \frac{j}{d\_1} \right]} H\_{\left[ \frac{j + 1}{d\_2} \right]} - \left( \sum\_{d\_1 = 1}^{j} \frac{\mu(d\_1)}{d\_1} H\_{\left[ \frac{j}{d\_1} \right]} \right) \left( \sum\_{d\_2 = 1}^{j + 1} \frac{\mu (d\_2)}{d\_2} H\_{\left[ \frac{j + 1}{d\_2} \right]} \right)$$
Let us show that this number is non-zero. For all $j > 1$, there exists a prime $\frac{j + 1}{2} < p < j + 1$. It turns out that the covariance has a factor of $p^2$ in the denominator. To prove this, let us see what terms could contribute such a factor. In order for $p$ to appear in the denominator of $H\_m$, we must have $m \geq p$, and since $p > \frac{j + 1}{2}$, in order for $p$ to appear in $H\_{\left[ \frac{j}{d\_1} \right]}$ or $H\_{\left[ \frac{j + 1}{d\_2} \right]}$ we must have $d\_1 = 1$ or $d\_2 = 1$.
In order for $p$ to appear in $d\_1, d\_2$ or $\mathrm{lcm} \left( d\_1, d\_2 \right)$ at least one of $d\_1, d\_2$ must be equal to $p$. Going over the very limited options, we see that
$$\sum\_{d\_1 = 1}^{j} \sum\_{d\_2 = 1}^{j + 1} \frac{\mu(d\_1) \mu(d\_2)}{\mathrm{lcm} \left( d\_1, d\_2 \right)} H\_{\left[ \frac{j}{d\_1} \right]} H\_{\left[ \frac{j + 1}{d\_2} \right]}$$
contains $\frac{1}{p^2} - \frac{2}{p^2} = - \frac{1}{p^2}$ and
$$\left( \sum\_{d\_1 = 1}^{j} \frac{\mu(d\_1)}{d\_1} H\_{\left[ \frac{j}{d\_1} \right]} \right) \left( \sum\_{d\_2 = 1}^{j + 1} \frac{\mu (d\_2)}{d\_2} H\_{\left[ \frac{j + 1}{d\_2} \right]} \right)$$
contains in total $\frac{0}{p^2}$, thus proving the claim.
|
1
|
https://mathoverflow.net/users/88679
|
397226
| 163,985 |
https://mathoverflow.net/questions/397154
|
3
|
Let $G$ be a reductive group over the finite field $\mathbb{F}\_q$. Then a *regular* embedding of $G$ is an $\mathbb{F}\_q$-rational embedding $\iota \colon G \rightarrow G'$ into a second reductive group over $\mathbb{F}\_q$, such that $\iota$ maps the derived group of $G$ isomorphically onto that of $G'$ and the center of $G'$ is connected. Moreover, a regular embedding is called *smooth regular*, if additionally the center of $G'$ is smooth. It is well-known that smooth regular embeddings always exist (see for example [1,Theorem 4.5],[2, Lemma 6.5]).
**Question.** Is it true that for any $G$ a smooth regular embedding $\iota \colon G \rightarrow G'$ exists, with the additional property that the center of $G'$ is an induced torus?
It seems quite plausible, but I could not find any reference (which is a bit annoying).
Note that for the somewhat dual notion of a *z-extension* $\widetilde G \rightarrow G$ it can be arranged -- and is even part of the definition of a z-extension (cf. [3,§1]) -- that ${\rm ker}(\widetilde{G} \rightarrow G)$ is induced.
References:
[1] Martin, B.: Étale slices for representation varieties in characteristic p.
[2] Taylor, J.: The Structure of Root Data and Smooth Regular Embeddings of Reductive Groups.
[3] Kottwitz, R.: Rational conjugacy classes in reductive groups.
|
https://mathoverflow.net/users/148992
|
Regular embeddings of a reductive groups with induced center
|
The answer is **Yes**.
Let $G\hookrightarrow G'$ be a smooth regular embedding. We write $Z(G')$ for the center of $G'$, which is an $F$-torus, where $F={\Bbb F}\_q$.
We construct a regular embedding $G'\hookrightarrow G''$ such that $Z'':=Z(G'')$ is an induced torus. We write $S=(G,G)=(G',G')$ for the derived group of $G$ and $G'$.
Consider the natural surjective homomorphism
$$\varphi\colon Z'\times\_F S\to G',\quad (z,s)\mapsto z^{-1}\cdot s.$$
Set $K=\ker\varphi=\big\{(z^{-1},z)\mid z\in Z'\cap Z(S)\big\}$. Then we have a canonical isomorphism
$$(Z'\times\_F S)/K\to G'.$$
Since $Z'=Z(G')$, we see that $Z'\supset Z(S)$, whence
$K=\big\{(z^{-1},z)\mid z\in Z(S)\big\}$.
Write $X'$ for the character group of $Z'$. Then $X'$ is a ${\rm Gal}(\overline F/F)$-module.
Clearly, there exists a surjective homomorphism $\alpha\colon X''\twoheadrightarrow X'$, where $X''$ is a permutation ${\rm Gal}(\overline F/F)$-module. Let $Z''$ denote the $F$-torus with character group $X''$, which is an induced torus. Then we have an injective homomorphism $\alpha^\*\colon\, Z'\hookrightarrow Z''$.
Consider the induced injective homomorphism
$$\alpha^{\*\*}\colon\, Z'\times\_F S\hookrightarrow Z''\times\_F S$$
and the induced injective homomorphism
$$\alpha^\vee\colon\, G'= (Z'\times\_F S)/K\hookrightarrow
(Z''\times\_F S)/\alpha^{\*\*}(K)=:G''.$$
Then $$\alpha^{\*\*}(K)=\big\{(\alpha^\*(z)^{-1},z)\mid z\in Z(S)\big\}.$$
It follows that the homomorphism $Z''\to Z(G'')$
is an isomorphism and that $\alpha^\vee$ is a regular embedding.
The composite injective homomorphism
$$ G\hookrightarrow G'\hookrightarrow G''$$
is a desired smooth regular embedding for which $Z(G'')$ is an induced torus.
|
2
|
https://mathoverflow.net/users/4149
|
397229
| 163,987 |
https://mathoverflow.net/questions/397221
|
3
|
Suppose $Y$ is an $(n-1)$-connected space, $n>2$, so we have Hurewicz isomorphisms $\pi\_n(Y)\cong H\_n(Y)$ and $\pi\_{n-1}(\Omega Y)\cong H\_{n-1}(\Omega Y)$. Let a map $\alpha\colon X\to\Omega Y$ be given. Naturally it induces a map $\beta\colon X\times S^1\to Y$. I want to show the following diagram is commutative:
$$\require{AMScd}
\begin{CD}
H\_{n-1}(X) @>\times[S^1]>> H\_n(X\times S^1)\\
@V\alpha\_\*VV @V\beta\_\*VV \\
H\_{n-1}(\Omega Y) @<\cong<< H\_n(Y).
\end{CD}
$$
Since $\Omega Y$ is path connected, we may homotope $\alpha$ to a based map. Then $\beta$ factors though the reduced suspension $\Sigma X$. If $X=S^{n-1}$ is a sphere, the commutativity would then follow from tracking down the definition of $\pi\_n(Y)\xrightarrow{\cong}\pi\_{n-1}(\Omega Y)$. However I don't know how this helps for the general case.
One can also phrase the question in cohomology in the obvious way. (In particular the cross product $\times[S^1]$ will be replaced by the slant product $/[S^1]$.)
|
https://mathoverflow.net/users/310606
|
Induced map in homology for a map to a loop space
|
The general case does follow from the case $Y=S^n$.
Without loss of generality $X=\Omega Y$ and $\alpha$ is the identity. Now the statement comes down to the assertion that the composition
$$
\pi\_{n-1}\Omega Y\to H\_{n-1}\Omega Y\to H\_n(\Omega Y\times S^1)\to H\_n Y
$$
corresponds to the Hurewicz map $\pi\_nY\to H\_nY$ via a natural isomorphism between $\pi\_{n-1}\Omega Y$ and $\pi\_nY$. This is true for all $Y$, and by naturality it is enough to prove it for $Y=S^n$.
|
7
|
https://mathoverflow.net/users/6666
|
397231
| 163,988 |
https://mathoverflow.net/questions/396816
|
3
|
If one wanted to obtain a fan for a toric variety of dimension $ n>1 $ whose Cox ring is $ \mathbb{Z}^{2} $ graded with weights $ \{(a\_{i},b\_{i})\}\_{i=1}^{n+2} $, then one could let $ B $ be the $ n \times (n+2) $ matrix whose $ (i,j) $-th entry is $ \delta(i,j) $ if $ 1 \le i, j \le n $, $ a\_{i} $ if $ j $ is equal to $ n+1 $ and $ b\_{i} $ if $ n $ is equal to $ n+2 $. After performing row reduction over the integers, one ends up with an $ n \times (n+2) $ matrix $ A $ with entries in $ \mathbb{Z} $. The $ n+2 $ columns of $ A $ are the rays for a fan in $ \mathbb{R}^{n} $. If the rays are $ \{u\_{\rho\_{1}},\dots,u\_{\rho\_{n+2}} \} $, then maximal dimensional cones $ \sigma $ are of the form $ \operatorname{Cone}(u\_{\rho\_{i\_{1}}},\dots,u\_{\rho\_{i\_{n}}}) $. The fan $ \Sigma $ is then obtained from the maximal cones and their faces. From $ \Sigma $ one obtains an ideal $ B(\Sigma) = \langle x^{\widehat{\sigma}} \rangle\_{\sigma \in \Sigma} $ where $ x^{\widehat{\sigma}} $ is $ \prod\_{i \mid \rho\_{i} \notin \sigma} x\_{i} $. From here the quotient of $ \mathbb{A}^{n+2}\_{\mathbb{C}} \setminus Z(B(\Sigma)) $ by the $ \mathbb{G}\_{m}^{2} $ action which sends $ x\_{i} $ to $ z\_{1}^{a\_{i}}z\_{2}^{b\_{i}}x\_{i} $ is isomorphic to the variety $ X\_{\Sigma} $ obtained from the fan $ \Sigma $. As a result, the Cox ring of $ X\_{\Sigma} $ has the desired grading.
What if instead of wanting to find an explicit fan of a toric variety of dimension $ n>1 $ whose Cox ring is $ \mathbb{Z}^{2} $ graded, one wants to find an explicit fan of a toric variety of dimension $ n>1 $ whose Cox ring is $ \operatorname{Hom}(\mathbb{Z}/\langle M \rangle \mathbb{Z}, \mathbb{C}^{\ast}) \times \operatorname{Hom}(\mathbb{Z}/\langle N \rangle \mathbb{Z}, \mathbb{C}^{\ast}) $ graded with weights $ (\overline{a\_{i}}, \overline{b\_{i}})\_{i=1}^{n} $? Is there a similar algorithm for obtaining the fan for such a variety?
|
https://mathoverflow.net/users/113893
|
How to create a toric variety whose Cox ring has a specific grading?
|
I realized an answer to this. Let $ B $ be the $ n \times (n+2) $-matrix with entries $ \delta(i,j) $ if $ 1 \le i,j \le n $, $ a\_{i} $ if $ j $ is equal to $ n+1 $ and $ b\_{i} $ if $ j $ is equal to $ n+2 $. Now perform row reduction (with operations strictly in the integers) on the matrix $ B $ until all entries of $ B $ in the $ n+1 $-st column are $ c \delta(1,i) $ and are equal to $ d \delta(2,j) $ for some integers $ c $ and $ d $. If $ \ell\_{1} $ is equal to $ \operatorname{LCM}(c,M) $ and $ \ell\_{2} $ is equal to $ \operatorname{LCM}(d,N) $, then multiply the first row by $ M/\ell\_{1} $ and the second row by $ N/\ell\_{2} $. The $ i $-th column is $ u\_{\rho\_{i}} \in \mathbb{N}^{n} $. If $ \rho\_{i} $ is equal to $ \mathbb{R}\_{+} u\_{\rho\_{1}} $, then the cone $ \sigma $ equal to $ \operatorname{Cone}(\rho\_{1},\dots,\rho\_{n}) $ is an affine toric variety.
The fan for the affine toric variety $ \operatorname{Spec}(\mathbb{C}[\sigma^{\vee} \cap M]) $ determines an ideal $ B(\Sigma) $. If $ \mathbb{A}^{n}\_{\mathbb{C}} $ is isomorphic to $ \operatorname{Spec}(\mathbb{C}[y\_{1},\dots,y\_{n}]) $ he variety $ \left(\mathbb{A}^{n}\_{k} \setminus Z(B(\Sigma))\right)//\left(\operatorname{Hom}(\mathbb{Z}/\langle M \rangle \mathbb{Z}, \mathbb{C}^{\ast}) \times \operatorname{Hom}(\mathbb{Z}/\langle N \rangle \mathbb{Z}, \mathbb{C}^{\ast})\right) $ is isomorphic to this variety and therefore has the desired grading. Here the action sends $ y\_{i} $ to $ z\_{1}^{\overline{a\_{1}}} z\_{2}^{\overline{b\_{i}}} y\_{i} $. If others up-vote this answer, then I will accept it. If someone sees something wrong with this, then let me know.
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1
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https://mathoverflow.net/users/113893
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397240
| 163,990 |
https://mathoverflow.net/questions/397245
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2
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Let $k$ be a number field and denote by $H^i(k,-)$ the Galois cohomology functor $H^i(\mathrm{Gal}(\bar{k}/k),-)$. Let $X$ be a smooth geometrically integral curve over $k$. One can easily show that the map $H^1(k,\mathrm{Pic}^0(X\_{\bar{k}})) \rightarrow H^1(k,\mathrm{Pic}(X\_{\bar{k}}))$ is surjective. Indeed, applying the functor $H^1(k,-)$ to the exact sequence $$0 \rightarrow \mathrm{Pic}^0(X\_{\bar{k}}) \rightarrow \mathrm{Pic}(X\_{\bar{k}}) \rightarrow \mathrm{NS}(X\_{\bar{k}}) \rightarrow 0,$$ we get the exact sequence $$H^1(k,\mathrm{Pic}^0(X\_{\bar{k}})) \rightarrow H^1(k,\mathrm{Pic}(X\_{\bar{k}})) \rightarrow H^1(k,\mathbb{Z}).$$ The last term is zero because $\mathbb{Z}$ has trivial Galois action and so any 1-cocycle $f \in H^1(k,\mathbb{Z})$ is simply an element of $\mathrm{Hom}(\mathrm{Gal}(\bar{k}/k),\mathbb{Z})$. But $\mathbb{Z}$ is torsion-free and thus all such maps are zero maps.
**Question 0.** Here I did not assume that $X$ is projective, but would all of the above still hold without this assumption? For example, I've never seen the Neron-Severi group of an affine curve discussed in any literature.
Now moving on, we have an exact sequence of Galois modules $$0 \rightarrow \mathrm{Pic}(X\_{\bar{k}})\_{\mathrm{tor}} \rightarrow \mathrm{Pic}(X\_{\bar{k}}) \rightarrow \mathrm{Pic}(X\_{\bar{k}})\_{\mathrm{free}} \rightarrow 0$$ where $\mathrm{Pic}(X\_{\bar{k}})\_\mathrm{tor}$ denotes the maximal torsion subgroup of $\mathrm{Pic}(X\_{\bar{k}})$ and $\mathrm{Pic}(X\_{\bar{k}})\_\mathrm{free} = \mathrm{Pic}(X\_{\bar{k}})/\mathrm{Pic}(X\_{\bar{k}})\_\mathrm{tor}$, the maximal free quotient.
**Question 1.** In this case $\mathrm{Pic}(X\_{\bar{k}})\_\mathrm{free}$ certainly does not have trivial Galois action, so we cannot reduce 1-cocycles to group homomorphisms $\mathrm{Gal}(\bar{k}/k) \rightarrow \mathrm{Pic}(X\_{\bar{k}})\_\mathrm{free}$. Therefore how do we go about computing $H^1(k,\mathrm{Pic}(X\_{\bar{k}})\_\mathrm{free})$?
I have thought about first studying $H^1(k,\mathrm{Pic}(X\_{\bar{k}}))$ using a wild idea as follows:
We apply the (étale) cohomology functor $H^i(X\_{\bar{k}},-)$ to the Kummer sequence $$0 \rightarrow \mu\_n \rightarrow \mathbb{G\_m} \rightarrow \mathbb{G}\_m \rightarrow 0$$ to obtain the long cohomology sequence $$0 \rightarrow \mu\_n(\bar{k}) \rightarrow \bar{k}^\* \rightarrow \bar{k}^\* \rightarrow H^1(X\_\bar{k},\mu\_n) \rightarrow \mathrm{Pic}(X\_{\bar{k}}) \rightarrow \mathrm{Pic}(X\_{\bar{k}}) \rightarrow H^2(X\_{\bar{k}},\mu\_n) \rightarrow H^2(X\_{\bar{k}},\mathbb{G}\_m)=0.$$
Then we apply $H^1(k,-)$ to $$H^1(X\_\bar{k},\mu\_n) \rightarrow \mathrm{Pic}(X\_{\bar{k}}) \rightarrow \mathrm{Pic}(X\_{\bar{k}}) \rightarrow H^2(X\_{\bar{k}},\mu\_n)$$ and study the result. But I'm not sure if there are any spectral sequences we can use, or if this approach is even feasible.
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https://mathoverflow.net/users/172132
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Computing $H^1$ with coefficients in a torsion-free abelian group
|
I will focus attention on smooth projective varieties $X$ over $k$ with $\mathrm{Pic}(X\_{\bar{k}})$ a free finitely generated abelian group, as they illustrate all the essential behaviour relevant to your question.
Here $\mathrm{Pic}^0(X\_{\bar{k}})$ is trivial so it is certainly *not true* in general that the map $H^1(k,\mathrm{Pic}^0(X\_{\bar{k}})) \rightarrow H^1(k,\mathrm{Pic}(X\_{\bar{k}}))$ is surjective. Your argument has problems as the Neron-Severi group is just equal to the Picard group here, and it is not isomorphic to $\mathbb{Z}$ with trivial Galois action in general.
As for your second question on how to calculate $H^1(k,\mathrm{Pic}(X\_{\bar{k}}))$. It depends a lot on the geometry of your specific situation. There are no cheap tricks which can help you using e.g. Kummer theory. If you are able to write down explicit generators for $\mathrm{Pic}(X\_{\bar{k}})$ and you know how Galois acts on them, then you can actually compute this cohomology group using standard commands for group cohomology in Magma. This is a tried and tested approach which appears in many papers; particularly common examples which appear in the literature include del Pezzo surfaces and K3 surfaces. You could have a look at the PhD Thesis of Martin Bright for example to see a detailed treatment of this method.
In general I should add that the Hochschild-Serre spectral sequence identifies $H^1(k,\mathrm{Pic}(X\_{\bar{k}}))$ with the so-called algebraic part of the Brauer group of $X$. It is for this reason people are usually interested in computing $H^1(k,\mathrm{Pic}(X\_{\bar{k}}))$, and any paper which calculates the Brauer group of a specific variety over a number field will almost certainly start by calculating $H^1(k,\mathrm{Pic}(X\_{\bar{k}}))$.
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4
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https://mathoverflow.net/users/5101
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397249
| 163,996 |
https://mathoverflow.net/questions/397257
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2
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Let $k=m+\sum^{m+1}\_{j=1} a\_j$ such that $a,m,k\in\mathbb{N}$ and $a\_1$ or $a\_{m+1}\geq 0$ with all other $a\geq1$. Note that we assume natural numbers start from $0$ and we have the restriction that $\sum^{m+1}\_{j=1} a\_j\geq m-1$. Why do there exist $F\_{k+2}$ solution sets for values of $m$ and $a\_\zeta$, $\forall1\leq\zeta\leq m$? How would this be proven?
For example, when $k=4$, we have $8$ solution sets,
\begin{equation}
\begin{split}
\{m,A\}&=\{0,\{4\}\},\\
&=\{1,\{3,0\}\},\\
&=\{1,\{0,3\}\},\\
&=\{1,\{1,2\}\},\\
&=\{1,\{2,1\}\},\\
&=\{2,\{0,1,1\}\},\\
&=\{2,\{1,1,0\}\},\\
&=\{2,\{0,2,0\}\},\\
\end{split}
\end{equation}
where $A=\bigcup^{m+1}\_{j=1} a\_j$. Note that $\{2,\{1,0,1\}\}$ is invalid since only the first or final $a$'s may be $0$. Also, $\{3,\{0,1,0\}\}$ is invalid since $\sum^{m+1}\_{j=1} a\_j\geq m-1$.
Any help would be much appreciated.
**Update**: Let $b\_j=a\_j-1$ for $1<j<m+1$ and $b\_j=a\_j$ when $j=1$ or $j=m+1$:
$$k=2m-1+\sum \_{j=1}^{m+1}b\_j.$$
Hence, for fixed $m$, there exist
$$\binom{k-2m+1+m+1-1}{m+1-1}=\binom{k-m+1}{m},$$
solutions for $k-2m+1=b\_1+\cdots +b\_{m+1}$. Summing over $m$ yields
$$F\_{k+2}=\sum \_{m=0}^{k+1}\binom{k+1-m}{m}.$$
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https://mathoverflow.net/users/167822
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Is this case of a generalised partition equivalent to Fibonacci numbers?
|
Without mistake on my behalf, a proof can be given as follows:
Denote the set of all solutions (for a given value of $k$)
by
$\mathcal F\_k$. Every element of $\mathcal F\_k$ ending
with a last coefficient $\geq 1$ corresponds to an element
of $\mathcal F\_{k-1}$ after decreasing its last element
(of the corresponding sequence $(a\_1,\ldots)$) by $1$.
Elements of $\mathcal F\_k$ ending with a last coefficient $0$
correspond similarly to elements of $\mathcal F\_{k-2}$:
Remove the last coefficient and decrease the (originally second-last)
remaining last coefficient by $1$. This proves the result by
induction after checking the initial conditions.
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3
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https://mathoverflow.net/users/4556
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397262
| 163,998 |
https://mathoverflow.net/questions/396342
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11
|
In Lurie's `Higher Algebra`, construction 4.4.2.7 presents a Bar construction in the setting of $\infty$-categories. The construction in 4.4.2.7 takes as input an $\mathcal{O}$-monoidal $\infty$-category $\mathcal{C}^\otimes \to \mathcal{O}^\otimes$ and a suitable pair of bi-modules in $\mathcal{C}^\otimes$ and gives a simplicial object in $\mathcal{C}$.
In the proof of the Barr-Beck theorem, this construction is used in lemma 4.7.3.13, where Lurie uses a simplicial object $\mathrm{Bar}\_T(T, M)\_\bullet$ with values in $\mathrm{LMod}\_T(\mathcal{C})$. All of the Bar constructions previously used in Higher-algebra are only defined as taking value in the underlying $\infty$-category.
Looking at the construction, it *feels* obvious that the $\mathcal{C}$-valued simplicial object $\mathrm{Bar}\_T(T, M)\_\bullet$ should lift to a $\mathrm{LMod}\_T(\mathcal{C})$-valued simplicial object, since the bar construction essentially comes from some functor $\mathbf{\Delta}^{op} \to \mathrm{Tens}\_{\succ}$ in which the objects should have structure of modules and the maps should be linear, but I am unable to formally write down a lift along the forgetful functor $\mathrm{LMod}\_T(\mathcal{C}) \to \mathcal{C}$ from this definition.
Is there a way to see how to to produce such a lift? More generally, given an $(A,B)$-bimodule $M$ in $\mathcal{C}$ and a $(B,C)$-bimodule $N$ in $\mathcal{C}$, the simplicial object $\mathrm{Bar}(M,N)\_\bullet$ should take value in the $\infty$-category of $(A,C)$-bimodules, rather than in $\mathcal{C}$. This extension to the bimodule case is also used without proof in HA 5.2.2.6.
[This question](https://mathoverflow.net/questions/290035/bar-construction-and-the-infty-categorical-barr-beck-theorem) asks about the definition of $\mathrm{Bar}\_T(T, M)\_\bullet$ but seems to be only about it as a $\mathcal{C}$-valued object (that's what the answer in the commens gives), and do not address the issue of making it $\mathrm{LMod}\_T(\mathcal{C})$-valued.
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https://mathoverflow.net/users/nan
|
Making the ($\infty$-categorical) Bar construction valued in (bi)-modules
|
In Lurie's construction of the relative tensor product, there's an $\infty$-operad $\mathrm{Tens}\_{[2]}^{\otimes}$ whose algebras in a monoidal $\infty$-category $\mathcal{C}$ are given by 3 associative algebras (say A,B,C) and two bimodules (say M for (A,B) and N for (B,C)). The simplicial bar construction is obtained by first restricting an algebra for this $\infty$-operad along a certain functor $\Delta^{\mathrm{op}} \to \mathrm{Tens}\_{[2]}^{\otimes}$, so that the composite to $\mathcal{C}^\otimes$ takes $[n]$ to $(M, B,\ldots,B,N)$ (with $n$ copies of $B$), and then taking the cocartesian pushforward to the fibre $\mathcal{C}$ over $\langle 1 \rangle$ (to get the simplicial diagram in $\mathcal{C}$ that takes $[n]$ to $M \otimes B^{\otimes n} \otimes N$).
To extend this to a simplicial diagram of bimodules, you want to first define a functor $\mathrm{Tens}\_{[1]}^{\otimes} \times \Delta^{\mathrm{op}} \to \mathrm{Tens}\_{[2]}^{\otimes}$ (where $\mathrm{Tens}\_{[1]}^{\otimes}$ is just the operad for bimodules). If we call the 3 objects of the bimodule operad $a,b$ (the two algebras) and $m$, then the composite with our algebra in $\mathcal{C}^\otimes$ should take $((a,\ldots,a,m,b,\ldots,b), [n])$ (with $i$ $a$'s and $j$ $b$'s) to
$(A,\ldots,A,M,B,\ldots,B,N,C,\ldots,C)$ with $i$ $A$'s, $n$ $B$'s and $j$ $C$'s.
Next you again take a cocartesian pushforward, so that you get a diagram in $\mathcal{C}^\otimes$ of the same shape, but which now takes this object in the source to $(A,\ldots,A,M \otimes B^{\otimes n} \otimes N,C,\ldots,C)$. This new diagram is then adjoint to a simplicial object in algebras for the bimodule operad, and indeed factors through the fibre of $(A,C)$-bimodules.
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5
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https://mathoverflow.net/users/1100
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397265
| 164,000 |
https://mathoverflow.net/questions/397233
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5
|
In this question I'd like to examine some properties of universally closed morphisms.
The question is self-contained. It can also be seen as a follow-up to this [question](https://mathoverflow.net/q/397132).
Let $R$ be a discrete valuation ring, $X$ and $S$ two $R$-flat and universally closed $R$-schemes.
Let $f : X\to S$ be a universally closed morphism. Assume the special fiber of $f$ is an isomorphism.
>
> * If $X$ and $S$ are reduced, is $f$ an isomorphism?
> * If $X$ and $S$ have reduced fibers, is $f$ an isomorphism?
>
>
>
Without either of the reducedness assumptions, the answer is "no", as shown by the example provided [here](https://mathoverflow.net/q/397190).
My expectation is that the answer is still "no" to both questions, but an example seems subtle to produce.
**Remarks**
In these remarks we further assume $f$ is flat, to see what more can be said in this case. After all, should the questions have positive answer, $f$ would have to be flat. The remarks are listed increasing the strength of the additional assumptions, to see what can be said in each ever-more-special case.
* From flatness we deduce $f$ is surjective, since generizations lift along $f$ (by flatness), and every point in $S$ specializes to a point in the special fiber (by universal closedness of $S$, if $s\in S$ is contained in the generic fiber, it maps to the generic point in $\text{Spec}(R)$. This specializes to the closed point in $\text{Spec}(R)$, and specializations lift along $S\to \text{Spec}(R)$ by universal closedness, so there is some $s\_0\in S$ contained in the special fiber, to which $s$ specializes).
* If $f$ is, in addition, *separated* and *locally of finite type*, then it is also proper, since $f$ is quasi-compact by universal closedness. It is therefore an isomorphism (it is quasi-finite, hence finite, hence finite flat of degree $1$). The question is interesting only if $f$ is *not* locally of finite type.
* To summarize, the case when $f$ is flat and separated is already unanswered but one has access to more info on $f$. A non-separated example, flat or not, to answer the question in the negative, would be already what I'm looking for. A flat separated example, if any, would be the best way to settle this puzzle, and I expect it is very subtle to find.
|
https://mathoverflow.net/users/nan
|
On universally closed morphisms of reduced schemes
|
Let $R$ be $\mathbb Z\_p$ (or any other dvr).
Let $S$ be obtained by gluing two copies of $\mathbb P^1\_{\mathbb Z\_p}$ away from the $0$-point in the special fiber, i.e. away from the vanishing locus of the ideal $(p,x)$ in local coordinates away from $\infty$.
Let $X$ be obtained by gluing two copies of $\mathbb P^1\_{\mathbb Z\_p}$ away from the $0$-section, i.e. away from the vanishing locus of the ideal $x$.
These are obtained by gluing two flat and universally closed schemes along an open set, hence flat and universally closed.
Then we can map $X \to S$ by mapping each copy of $\mathbb P^1\_{\mathbb Z\_p}$ to a different copy of $\mathbb P^1\_{\mathbb Z\_p}$, which is compatible with the gluing since we glue along a larger open set in $S$ than in $X$.
This map is an isomorphism in the special fiber since the constructions of $X$ and $S$, restricted to the special fiber, are identical.
But it is manifestly not an isomorphism, e.g. because it makes a non-separated generic fiber to a separated one.
|
2
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https://mathoverflow.net/users/18060
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397266
| 164,001 |
https://mathoverflow.net/questions/397237
|
0
|
Consider the space $\mathcal{M}\_{loc} (\mathbb{R}^d)$ of locally finite signed Radon measures, equipped with the weak\* topology in duality with $C\_b (\mathbb{R}^d)$. It is known that this is space is not metrizable, nor first countable (although I believe it is a Souslin space?).
On the other hand, in practice one often works not with the weak\* topology directly, but with weak\* *convergence*. So it would be interesting to look at the *sequential* weak\* topology on $\mathcal{M}\_{loc} (\mathbb{R}^d)$ (that is, a set is declared to be closed provided it is sequentially weak\* closed).
My question is, is this topology something that has already been studied in the literature? The only thing I can find is this [MO question](https://mathoverflow.net/questions/382906/borel-sigma-algebra-on-measures-generated-by-distance-inducing-weak-convergence) from several months ago.
EDIT: As user @memorial points out, the original statement of this question contains a basic error, since there is not actually a duality pairing between $C\_b (\mathbb{R}^d)$ and $\mathcal{M}\_{loc} (\mathbb{R}^d)$, rather, between $C\_b (\mathbb{R}^d)$ and $\mathcal{M} (\mathbb{R}^d)$, and between $C\_c (\mathbb{R}^d)$ and $\mathcal{M}\_{loc} (\mathbb{R}^d)$. Still, the question remains, since (as far as I know) the resulting weak\* topologies are not themselves sequential.
|
https://mathoverflow.net/users/100163
|
Reference request: sequential weak* topology on the space of signed Radon measures
|
This is not an answer since I cannot give the references you are requesting for reasons that will soon be apparent but am not entitled to comment. Firstly, there is something fundamentally wrong with the framework of your question--there is no duality between spaces of bounded, continuous functions on euclidean space and locally bounded measures thereon. (Try integrating the sine function on the line with respect to Lebesgue measure). What one does have are natural dualities
a) between the bounded, continuous functions and the finite Radon measures;
and
b) between the space of continuous functions with compact support and the locally finite measures.
Note that it has long been known that in case a) one cannot use the norm topology for this duality since it leads to a much larger dual space. This was remedied in the 50´s by R.C. Buck who introduced a complete l.c. topology (the strict topology) on the bounded, continuous functions which has the bounded Radon measures as dual. He worked in the context of functions and measures on locally compact spaces but this was soon extended to that of general completely regular spaces by various authors.
In b) one uses the natural l.c. topology on the functions of compact support as a so-called strict $LF$-space (Dieudonne and Schwartz). This was the basis of the Bourbaki approach to measure spaces, defining them as the dual spaces of suitable locally convex spaces of continuous functions (a precursor to Schwartz´ treatment of distributions).
With regard to your question, let me start with the case of functions and measures on a compact subset. In this case there is a natural l.c. topology on the space of measures which is complete and such that the associated convergent sequences are the ones in your question. This is the finest topology which agrees with the weak star one on bounded sets. Despite this description, it is l.c. I imagine that it coincides with the one you describe but I haven´t sat down to check this. This applies to any compact metrisable space, by the way.
At the moments, due to covid, I am not in a position to give precise references but the relevant themes are the Banach-Dieudonne theorem and the bounded weak star topology. There is a reasonable article on the latter in Wikipedia but it is, rather curiously, in German (with no English version).
Memory suggests that they are discussed in Koethe´s monograph and Schaefer´s standard text.
In the non-compact case, the natural topology on the locally bounded measures would be a corresponding projective limit of such space. This is then isomorphic to a closed subspace of a product of spaces of the type described above which means that it will inherit many properties from those of the compact case. It is even a complemented subspace (in the l.c. sense) which makes the situation even more tractable. I haven´t thought about whether it can be described as in your query. I am not aware of any source which deals with this directly but it is standard manipulation of inductive and projective limits of locally convex spaces and their duals.
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6
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https://mathoverflow.net/users/317800
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397268
| 164,002 |
https://mathoverflow.net/questions/397232
|
3
|
Let $C\_n=\frac1{n+1}\binom{2n}n$ be the well-known Catalan numbers. Here is a curiosity.
>
> **QUESTION.** Are there infinitely many $C\_n$ that are "radical", i.e. that are square-free?
>
>
>
|
https://mathoverflow.net/users/66131
|
"Radical" Catalan numbers?
|
If $s(n)$ is the largest integer such that $s(n)^2 \mid \binom{2n}{n}$, then the main result of Sárközy, A. (1985). On divisors of binomial coefficients, I. Journal of Number Theory, 20(1), 70-80 is
>
> If $\varepsilon > 0$, $n > n\_0(\varepsilon)$ then we have $$e^{(c-\varepsilon) \sqrt{n}} < s(n)^2 < e^{(c+\varepsilon) \sqrt{n}}$$ where $$c = \sqrt{2} \sum\_{k=1}^\infty \frac{1}{\sqrt{2k-1}} - \frac{1}{\sqrt{2k}}$$
>
>
>
Numerically, $c \approx 0.855$.
Define $s'(n) = \frac{s(n)}{\textrm{gcd}(s(n), n+1)}$. Then $s'(n)^2 \mid C\_n$. Noting that $e^{0.81 \sqrt{x}} > x + 1$ for all $x > 0$, $C\_n$ cannot be square-free for any $n > n\_0(c - 0.81)$, whence the number of square-free $C\_n$ is finite.
|
3
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https://mathoverflow.net/users/46140
|
397270
| 164,003 |
https://mathoverflow.net/questions/395393
|
4
|
Let $R=\mathbb{Z}/1024\mathbb{Z}$ and $G=GL(3,R)$. Let $H$ be the subgroup of $G$ consisting of all matrices with determinant $1$ which are congruent to the identity matrix modulo the ideal $4R$. Let
$A=\begin{pmatrix} 1 &0&512\\ 0& 1&0\\0&0 &1\end{pmatrix}$ and $B=\begin{pmatrix} 1 &4&2\\ 8& 1&4\\16 &8 &1\end{pmatrix}$.
Can one use GAP (or other software) to find out if $A\in B^H$? (Here $B^H$ is the subgroup of $G$ generated by the set $\{ B^h\mid h\in H\}$.)
|
https://mathoverflow.net/users/104638
|
A question about the possibilities of GAP
|
It is contained. Here is the way I did the calculation with GAP. It ought to work nicer, but there is a stupid technical issue in the way that makes it hard to implement a membership test in a subgroup. (The issue is basically that we cannot guarantee that elements will always lie in one big parent group.)
Thus, one needs to do things in a somewhat pedestrian way.
You will need my `matgrp` package for some calculations in the large groups over residue class rings to work:
```
gap> LoadPackage("matgrp");
true
gap> A:=[[1,0,512],[0,1,0],[0,0,1]];
[ [ 1, 0, 512 ], [ 0, 1, 0 ], [ 0, 0, 1 ] ]
gap> B:=[[1,4,2],[8,1,4],[16,8,1]];
[ [ 1, 4, 2 ], [ 8, 1, 4 ], [ 16, 8, 1 ] ]
gap> r:=Integers mod 1024;
(Integers mod 1024)
gap> Ar:=A*One(r);
[ [ ZmodnZObj( 1, 1024 ), ZmodnZObj( 0, 1024 ), ZmodnZObj( 512, 1024 ) ],
[ ZmodnZObj( 0, 1024 ), ZmodnZObj( 1, 1024 ), ZmodnZObj( 0, 1024 ) ],
[ ZmodnZObj( 0, 1024 ), ZmodnZObj( 0, 1024 ), ZmodnZObj( 1, 1024 ) ] ]
gap> Br:=B*One(r);
[ [ ZmodnZObj( 1, 1024 ), ZmodnZObj( 4, 1024 ), ZmodnZObj( 2, 1024 ) ],
[ ZmodnZObj( 8, 1024 ), ZmodnZObj( 1, 1024 ), ZmodnZObj( 4, 1024 ) ],
[ ZmodnZObj( 16, 1024 ), ZmodnZObj( 8, 1024 ), ZmodnZObj( 1, 1024 ) ] ]
gap> G:=GL(3,r);
GL(3,Z/1024Z)
gap> FittingFreeLiftSetup(G);; # build a data structure for group order
gap> Size(G);
406199075390515402701275136
```
Next we look at the image modulo 4. We construct it also as a permutation group (so we can use permutation group functionality for homomorphisms:
```
gap> rf:=Integers mod 4;
(Integers mod 4)
gap> gens:=List(GeneratorsOfGroup(G),m->List(m,r->List(r,x->Int(x)*One(rf))));;
gap> q:=Group(gens);
<matrix group with 5 generators>
gap> Size(q);
86016
gap> Size(GL(3,rf));
86016
gap> isop:=IsomorphismPermGroup(q);;
gap> p:=Image(isop);
```
We now construct a map from this permutation group to the matrix group and collect generators for its co-kernel -- these are evaluated relators for the factor that together will generate the kernel of the reduction map on $G$:
```
gap> reverse:=GroupGeneralMappingByImagesNC(p,G,GeneratorsOfGroup(p),
> GeneratorsOfGroup(G));;
gap> it:=CoKernelGensIterator(reverse);
<iterator>
gap> hg:=[];;
gap> for i in it do
> if not IsOne(i) then
> Add(hg,i);
> fi;
> od;
gap> Length(hg);
448
```
This number of generators is a bit too large. We just pick 20 random ones and verify they still generate the kernel. (Would iterate/try more generators if not):
```
gap> preH:=Group(List([1..20],x->Random(hg)));
<matrix group with 20 generators>
gap> FittingFreeLiftSetup(preH);;
gap> Size(G)/Size(preH);
86016
```
Now for determinant 1. We use the same idea for the homomorphism onto the determinant, and get $H$ as kernel
```
gap> dets:=List(GeneratorsOfGroup(preH),DeterminantMat);;
gap> d:=Group(det); # GAP will issue a harmless warning that it assumes the elements indeed are invertible
<group with 20 generators>
gap> Size(d);
256
gap> isop:=IsomorphismPermGroup(d);;
gap> p:=Image(isop);;
gap> reverse:=GroupGeneralMappingByImagesNC(p,preH,GeneratorsOfGroup(p),
> GeneratorsOfGroup(preH));;
gap> it:=CoKernelGensIterator(reverse);;
gap> hg:=[];;
gap> for i in it do
> if not IsOne(i) then
> Add(hg,i);
> fi;
> od;
gap> Length(hg);
4873
gap> H:=Group(List([1..20],x->Random(hg)));;FittingFreeLiftSetup(H);;
gap> Size(preH)/Size(H);
256
```
Now we are ready to calculate $B^H$. We use a standard closure algorithm, starting with $B$ and then take conjugates of generators with generators of $H$ until no new conjugates arise, that is the group is normal.
To test membership $x\in S$, we check (this is the kludge I mentioned) whether $|S|=|\langle S,x\rangle|$. (Indeed this ought to be better, but it still beats hand-calculations.)
```
gap> bco:=[Br];;
gap> sub:=Group(bco);;
gap> FittingFreeLiftSetup(sub);;
gap> Size(sub);
512
gap> for i in bco do
> for j in GeneratorsOfGroup(H) do
> x:=i^j;
> t:=Group(Concatenation(bco,[x]));
> FittingFreeLiftSetup(t);
> if Size(t)<>Size(sub) then
> Add(bco,x);
> sub:=t;
> fi;
> od;
> od;
```
We now use the same kludgy element test to see whether $A\in B^H$:
```
gap> Size(sub);
4503599627370496
gap> t:=Group(Concatenation(bco,[Ar]));
<matrix group with 11 generators>
gap> FittingFreeLiftSetup(t);;
gap> Size(t);
4503599627370496
```
The order stays the same after adding $A$, thus $A\in B^H$.
|
5
|
https://mathoverflow.net/users/59303
|
397274
| 164,004 |
https://mathoverflow.net/questions/397286
|
23
|
Why do we have two theorems one for the density of $C^{\infty}\_c(\mathbb{R}^n)$ in $L^p(\mathbb{R}^n)$ and one for the density of $C^{\infty}\_c(\Omega)$ in $L^p(\Omega)$? with $\Omega$ an open subset of $\mathbb{R}^n$.
Why not just the second one?
I was asked by the prof what is the difference between the density of $C(\Omega)$ and $C(\mathbb{R}^n)$ but all I found when checking the demonstrations is that we take $\Omega \neq \mathbb{R}^n$ when giving the proof for the second theorem because for $\Omega = \mathbb{R}^n$ we already have the first theorem.
|
https://mathoverflow.net/users/144902
|
Why do we have two theorems when one implies the other?
|
Some mathematicians seem to agree with you, and strive only to state and prove the most general versions of their theorems. I've had co-authors express that view. And I've sometimes had referee reports on my papers state this philosophical perspective explicitly, objecting to a warm-up theorem that I stated and proved early in the paper, even though later I proved harder, more general results. Earlier in my career, against my own judgement I would dutifully remove the objectionable warm-up presentations (and I did so even in what became one my most highly cited papers), but no longer.
I strongly disagree with the objection. I don't agree that one should seek to present only the most general forms of one's theorems. Rather, there is a definite value in proving easier or more concrete results first, even when one intends to move on to prove more encompassing results later. Indeed, I would say that often the main value of a theorem is concentrated in an easier, less general principal case.
The simpler results often aid in mathematical insight. Unencumbered with unnecessary generality or abstraction, they are often simply easier to understand, yet still illustrate the main idea clearly. Removing even a small generalization, such as restricting to $n=2$ or simplifying from an arbritrary real-like space $\Omega$ to the reals $\mathbb{R}$, can dramatically improve understanding, especially on your reader's first engagement with your argument. The reason is that every generalization, even very small ones, contributes yet another layer of difficulty and abstraction, contributing to the cognitive load that can make a difficult proof impenetrable.
On the first pass, it can often be best to focus on a simple, main case, which highlights the core ideas without unnecessary distractions. Once one has mastered such a case, then one has often thereby developed a familiarity of understanding of the core idea or technique of the argument, a framework of understanding capable of supporting a deeper understanding of the more general result. Having the easy case first makes the difficult case much easier to master. Indeed, often the key ideas of an argument have only to do with the special case in the first instance, and the generalizing steps are routine — all the more reason to omit them at first.
So this is not just for pedagogy, although certainly students new to a topic will appreciate mastering the easier versions of a theorem first. My point is that the practice is also important for experts, at every level of expertise. One gains ultimately a deeper understanding of the general result, when one sees how the core ideas and methods generalize those in a simpler case.
The same goes for mathematics talks. At conferences or seminars, please consider begining your talk with an easier special case that illustrates the theme or methods of your more general, advanced results. Your audience will definitely appreciate it.
So I have no problem with having two theorems, one of them implying the other, and I would find that to be a very sound way of proceeding in mathematics.
|
135
|
https://mathoverflow.net/users/1946
|
397289
| 164,008 |
https://mathoverflow.net/questions/396953
|
10
|
Let $G$ be a reductive algebraic group with choices of Borel subgroup and maximal torus $B \supseteq T$ and unipotent radical $U$ over an algebraically closed field $k$ of characteristic zero. Is the affine closure $\overline{G/U} = \operatorname{Spec}(A)$ Cohen–Macaulay? If so, is there a reference to this fact? If not, is there a reference to a specific counterexample?
Note: We have that $A$ is Cohen—Macaulay when $G = \operatorname{SL}\_2$ since $\overline{G/U} \cong \mathbb{A}^2$ is smooth. However, in the paper [Popov - Contractions of Actions of Algebraic Groups](https://www.researchgate.net/publication/252438955_Contractions_of_actions_of_reductive_algebraic_groups), the author seems to suggest that $A$ is not Cohen–Macaulay in general. Specifically, in section 6, Popov introduces the notion of a stable property of local rings of points of schemes, and explicitly states that "Among the examples of properties (P) of open type given in section 5 [which includes the property of being Cohen–Macaulay], the following properties are stable:" and goes on to give a list of properties which does not include the property of being Cohen–Macaulay. Furthermore, following Popov, an open property is stable if and only if it is closed under the invariants of a reductive group and the property is stable under tensor product of regular functions on the basic affine space of reductive groups, and the invariants of a reductive group acting on a Cohen–Macaulay ring is Cohen–Macaulay by the Hochster–Roberts theorem. If $A$ is not Cohen–Macaulay in general, I am interested in a specific counterexample.
Edit: For $\operatorname{SL}\_3$, we have $A \cong k[a,b,c,x,y,z]/(ax + by + cz)$ which is Cohen–Macaulay. To see this, note that using [this problem set](https://www.mit.edu/%7Efengt/Braverman_1.pdf) ([Wayback Machine](https://web.archive.org/web/20200328172550/https://www.mit.edu/%7Efengt/Braverman_1.pdf)), the algebra of functions $A$ is the quotient of the free polynomial algebra in six variables (three coming from the standard representation $V(\omega\_1)$ and three coming from the dual $V(\omega\_2)$, respectively) modulo the quadratic relation given by the multiplication rule $V(\omega\_1) \otimes V(\omega\_2) \cong V(\omega\_1 + \omega\_2) \oplus V(0) \to V(\omega\_1 + \omega\_2)$, where the rightmost arrow is the quotient arrow. One can check (for example, this computation is carried out on p178 in Fulton and Harris's book Representation Theory: A First Course) that the kernel of this map is one dimensional and contains $e\_1 \otimes e\_1^{\ast} + e\_2 \otimes e\_2^{\ast} + e\_3 \otimes e\_3^{\ast}$.
(Note the relations from the multiplication $V(\omega\_1) \otimes V(\omega\_1) \to V(2\omega\_1) \cong Sym(V(\omega\_1))$ are precisely expressing that the variables $a,b,c$ commute, and similarly for $V(\omega\_2) \cong V(\omega\_1)^{\ast}$.)
|
https://mathoverflow.net/users/104690
|
Is the affine closure of the basic affine space of a reductive algebraic group Cohen–Macaulay?
|
Update: It seems that every affine closure of such a quotient $G/U$ *is* Cohen-Macaulay. In the same list of properties as above, the author states that, by a theorem of Brion, the ring of functions of $\overline{G/U}$ has rational singularities. Immediately before this statement, the author also cites [Elkik - Singularites rationnelles et deformations](https://link.springer.com/content/pdf/10.1007/BF01578068.pdf), which in particular explicitly states (Definition 1, p141) that varieties with rational singularities are in particular Cohen-Macaulay.
Unfortunately, I have not been able to track down the referenced theorem of Brion cited.
EDIT: One can also explicitly compute that, when $G := \operatorname{Sp}\_4$ and $A$ is the ring of global functions on $G/U$, we have an isomorphism
$$A \cong k[a,b,c,d,e,w,x,y,z]/(2ad - 2bc - e^2, 2bz - 2dx + ey, 2bw + ay - ex, 2dw + cy - ez, ew + cx - az).$$
where the first five variables correspond to the highest weight representation of dimension 5 and the next four variables correspond to the highest weight representation of dimension 4. This was computed according to the exercise sheet above but, as some justification for this claim (as earlier posted relations contain a typo), notice that (as can be checked in, for example, Macaulay2) this ring has dimension 6 and is Gorenstein, as expected--see [Remark 5.6 here](https://arxiv.org/pdf/1010.1606.pdf) for the Gorenstein assertion, which also gives another answer to the above question.
|
4
|
https://mathoverflow.net/users/104690
|
397293
| 164,010 |
https://mathoverflow.net/questions/293118
|
3
|
I'm looking for the PDF version/scan of Henry P McKean Jr.'s paper on propagation of chaos. The reference is as follows -
**Propagation of chaos for a class of non-linear parabolic equations., In Stochastic Differential Equations (Lecture Series in Differential Equations, Session 7, Catholic Univ., 1967) (1967): 41-57.**
The paper/chapter has been cited numerous times (being one of the first papers in the field) but nowhere online can I find a pdf version of it. It seems to me that most people who cite it haven't actually seen/read it.
|
https://mathoverflow.net/users/106281
|
Looking for access to McKean's original paper?
|
I am not sure whether there is still a quest for McKean's original paper, but as a nice substitute, you can take a look at the very recent [review paper](https://arxiv.org/pdf/2106.14812.pdf) on "propagation of chaos". Theorem 5.1 there is a result due to McKean and the original proof of him is presented (starting from page 86).
|
1
|
https://mathoverflow.net/users/163454
|
397295
| 164,012 |
https://mathoverflow.net/questions/397285
|
4
|
I'd like to prove the following conjecture.
Let $x = \frac{p}{q}\pi$ be a rational angle ($p,q$ integers, $q \geq 1$).
Then
$f(x) = \frac{2}{\pi} \arccos{\left(2\cos^4(2x)-1 \right)}$
is irrational if $x$ is not an integer multiple of $\frac{\pi}{8}$. Is this true? Is the other direction true too?
I've tried to use the standard manipulations, Chebyshev polynomials, etc. (see <https://math.stackexchange.com/questions/398687/how-do-i-prove-that-frac1-pi-arccos1-3-is-irrational>), but am otherwise quite stuck.
Some evaluations just to see a trend (we can restrict domain of $x$ to be $[0,\pi/2]$):
\begin{align}
x & = \frac{\pi}{2} \implies f(x)= 0 \text{ (Rational) }, \\
x & = \frac{\pi}{3}, \implies f(x) = 2\arccos(-7/8)/\pi, \\
x & = \frac{\pi}{4}, \implies f(x) = 2 \text{ (Rational) }, \\
x & = \frac{\pi}{5}, \implies f(x) = 2\arccos(-3(3+\sqrt{5})/16)/\pi, \\
x & = \frac{2\pi}{5}, \implies f(x) = 2\arccos(3(-3+\sqrt{5})/16)/\pi, \\
x & = \frac{\pi}{6}, \implies f(x) = 2\arccos(-7/8)/\pi, \\
x & = \frac{\pi}{7}, \implies f(x) = 2\arccos(-1+2\sin^4(3\pi/14))/\pi \\
x & = \frac{2\pi}{7}, \implies f(x) = 2\arccos(-1+2\sin^4(\pi/14))/\pi \\
x & = \frac{2\pi}{7}, \implies f(x) = 2\arccos(-1+2\sin^4(\pi/7))/\pi \\
x & = \frac{\pi}{8}, \frac{3\pi}{8}, \implies f(x) = 4/3 \text{ (Rational) }\\
\cdots
\end{align}
|
https://mathoverflow.net/users/317106
|
Irrationality of this trigonometric function
|
**Lemma.** Let $n$ be a positive integer such that each number in the open interval $(n/4,3n/4)$ is not coprime with $n$. Then $n\in \{1,4,6\}$.
**Proof.**
* If $n=2m+1$ is odd, and $m\geqslant 1$, then $m\in (n/4,3n/4)$.
* If $n=4m+2$ and $m>1$, then $2m-1\in (n/4,3n/4)$
* If $n=2$, then $1\in (n/4,3n/4)$
* If $n=4m$ and $m>1$, then $2m-1\in (n/4,3n/4)$
Now assume that $f(x)=4t$ with rational $t$. Then $2\cos^42x-1=\cos (2\pi t)$ and $2\cos^42x=2\cos^2(\pi t)$, $\cos \pi t=\pm \cos^2 2x$. Replacing $t$ to $t+1$ if necessary, we may suppose that $\cos \pi t=\cos^22x$. Let $\pi t=2\pi a/b$, $2x=2\pi c/b$ for coprime (not necessarily mutually coprime) integers $a,b,c$. In other words, let $b$ be a minimal positive integer for which $\pi tb$, $2xb$ are both divisible by $2\pi$. Denote $w=\exp(2\pi i/b)$. In these notations we get $$\frac{w^a+w^{-a}}2=\frac{(w^c+w^{-c})^2}4.$$
This is polynomial relation for $w$ with rational coefficients. Thus, all algebraic conjugates of $w$ satisfy it aswell. These algebraic conjugates are all primitive roots of unity of degree $b$ (here I use the well known fact that the cyclotomic polynomial $\Phi\_b$ is irreducible). So, we are allowed to replace $w$ to $w^m$, where $\gcd (m,b)=1$. RHS remains non-negative, therefore so does LHS, and we get
$\cos (2\pi am/b)\geqslant 0$ for all $m$ coprime to $b$.
This is rare. Namely, denote $a=da\_1$, $b=db\_1$ where $d=\gcd(a,b)$ and $\gcd(a\_1,b\_1)=1$. Let $k$ be arbitrary integer coprime with $b\_1$. Denote
$m=k+Nb\_1$, where $N$ equals to the product of those prime divisors of $d$ which do not divide $k$. Now each prime divisor of $d$ divides exactly one of the numbers $k$ and $Nb\_1$, therefore it does not divide their sum $m$. Clearly $m$ is also coprime with $b\_1$, and totally $\gcd(m,b)=1$. Therefore $\cos (2\pi a\_1k/b\_1)=\cos(2\pi a\_1m/b\_1)=\cos (2\pi am/b)$ is non-negative. Next, $a\_1k$ takes all residues modulo $b\_1$ which are coprime to $b\_1$. It follows that there are no residues coprime with $b\_1$ in the interval $(b\_1/4,3b\_1/4)$. Thus by Lemma we get $b\_1\in \{1,4,6\}$, and $\cos \pi t=\cos 2\pi a/b=\cos 2\pi a\_1/b\_1\in \{0,1/2,1\}$ and $\cos 2x=\pm \sqrt{\cos \pi t}\in \{0,\pm \sqrt{2}/2,\pm 1\}$. This indeed implies that $2x$ is divisible by $\pi/4$, in other words, $x$ is divisible by $\pi/8$.
|
6
|
https://mathoverflow.net/users/4312
|
397304
| 164,017 |
https://mathoverflow.net/questions/397316
|
4
|
In short, the question is for any references describing how to use the Hardy-Littlewood circle method to find an asymptotic for the number of solutions to $F(x\_1, ..., x\_s) = k$ for $(x\_1, ..., x\_s) \in \mathbb{Z}^s$, where $F$ is some indefinite integral quadratic form, and $k\neq 0$ is a fixed integer.
An old paper of Sarnak et. al. (A Proof of Siegel's Weight Formula, see [here](https://collaborate.princeton.edu/en/publications/a-proof-of-siegels-weight-formula) or [here, for a PDF](https://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.543.8732&rep=rep1&type=pdf)) mentions that solving such a problem can be done with the Hardy-Littlewood circle method if $s\geq 5$ (the paper mentions this in section 2, just before formula 2.2). They mention the $s=4$ case is harder, and provide a reference to a paper; I tracked down the paper they refer to, although again it seems to only cover the diagonal case. I am not very intereseted in this case; I just want to see how to solve it for $s\geq 5$ for non-diagonal forms. I found some [lecture notes](https://web.math.princeton.edu/%7Egyujino/Sarnak%202020.pdf) of a course given by Sarnack covering the case of diagonal forms for $s\geq 5,$ but I do not see how to generalize. Is there some easy way to go from knowing this result for diagonal forms to non-diagonal ones, or some other source which covers solutions to indefinite forms? The proof given by Sarnack in those notes seems to use in a fundamental way the fact that the form is diagonal, so that a certain factorization can work.
|
https://mathoverflow.net/users/318125
|
Hardy-Littlewood circle method for non-diagonal quadratic forms
|
The "best" way to deal with quadratic forms using the circle method is via Heath-Brown's delta symbol method.
You can read about this in detail in the paper:
Heath-Brown - A New Form of the Circle Method, and its application to Quadratic Forms
Theorem 4 in particular gives an asymptotic formula for the problem you mention. Note that Heath-Brown is also able to obtain results for 4 variables.
|
4
|
https://mathoverflow.net/users/5101
|
397318
| 164,022 |
https://mathoverflow.net/questions/397314
|
4
|
Let $k$ be a number field. We have the well-known Kummer exact sequence of etale sheaves on $\mathrm{Spec}\, k$: $$1 \rightarrow \mu\_n \rightarrow \mathbb{G}\_m \rightarrow \mathbb{G}\_m \rightarrow 1.$$
**Question 0.** Applying the etale cohomology functor $H\_{et}^i(k,-)$, I know that $H\_{et}^i(k,\mathbb{G}\_m) = H^i(k,\bar{k}^\*)$, where the latter is a Galois cohomology group. What is the Galois cohomological equivalent for $H^i\_{et}(k,\mu\_n)$?
Let $\mu\_\infty := \mathrm{colim}\_n\mu\_n$, this group can be interpreted as the set of all $n$-th roots of unity, i.e., $\mu\_\infty \cong \mathbb{Q}/\mathbb{Z}$. We work only in Galois cohomology.
**Question 1.** I would like to know if we could obtain a similar Kummer sequence involving $\mu\_\infty$ instead of $\mu\_n$ for a fixed $n$. The reason for this is that I came across an exact sequence $$0 \rightarrow \mathrm{Br}(k) \otimes \_\mathbb{Z} \mathbb{Q}/\mathbb{Z} \rightarrow H^3(k,\mu\_\infty) \rightarrow H^3(k,\bar{k}^\*) \rightarrow 0$$ but I am unable to derive it from the usual Kummer sequence.
**Question 2.** The idea for this exact sequence is to prove that the middle term is 0. It is known that we have $H^3(k,\bar{k}^\*)=0$ in our setting and for any non-archimedean place $v$, we have $\mathrm{Br}(k\_v) \subset \mathbb{Q}/\mathbb{Z}$. But I'm also unsure how does it imply that $\mathrm{Br}(k) \otimes \_\mathbb{Z} \mathbb{Q}/\mathbb{Z}= 0$.
|
https://mathoverflow.net/users/172132
|
A Kummer exact sequence involving $\mu_\infty$
|
Regarding you question 1, it suffices to take the short exact sequence
$$1\to \mu\_\infty\to \mathbb{G}\_m\to \mathbb{G}\_m\otimes\_{\mathbb{Z}}\mathbb{Q}\to 1\,.$$
This sequence is exact in the étale topos of any scheme over $\mathbb{Q}$. Indeed to show surjectivity it is enough to show that for every ring $R$, any $x\in R^\times=\mathbb{G}\_m(R)$ and any $n\ge1$ integer, there is an étale extension $R\to R'$ and $y\in \mathbb{G}\_m(R')$ such that $y\otimes 1=x\otimes \frac{1}{n}$. But it is enough to take $y=\sqrt[n]{x}$ in $R'=R[y]/(y^n-x)$. Moreover $\mu\_\infty$ is just the kernel of the second map by definition.
Another way of thinking about the above short exact sequence is that it is the colimit of the short exact sequences
$$ 1\to \mu\_n \to \mathbb{G}\_m\xrightarrow{n}\mathbb{G}\_m\to 1$$
as $n$ goes to infinity along the divisibility poset (where the arrows on the central term are all the identities, while the arrows on the second term are multiplications by suitable integers).
Once you have this short exact sequence, you can look at the long exact sequence in cohomology
$$ H^2(k;\mathbb{G}\_m)\to H^2(k;\mathbb{G}\_m\otimes\mathbb{Q})\simeq H^2(k;\mathbb{G}\_m)\otimes\mathbb{Q}\to H^3(k;\mu\_\infty)\to H^3(k;\mathbb{G}\_m) \to H^3(k;\mathbb{G}\_m)\otimes\mathbb{Q}$$
and you can deduce the vanishing of $H^3(k;\mu\_3)$ from the vanishing of $H^2(k;\mathbb{G}\_m)\otimes\mathbb{Q}$ and $H^3(k;\mathbb{G}\_m)$.
---
Regarding your second question, the simplest reason I can think of is that $\operatorname{Br}(k)\otimes\_{\mathbb{Z}}\mathbb{Q}/\mathbb{Z}=0$ because the Brauer group of a field is torsion (e.g. Example II.2.22 in Milne's *Étale cohomology*) and $\mathbb{Q}/\mathbb{Z}$ is divisible. I don't see much point in passing through the completions of $k$.
|
2
|
https://mathoverflow.net/users/43054
|
397335
| 164,024 |
https://mathoverflow.net/questions/397136
|
3
|
Let $U\_q(\frak{sl}\_2)$ denote the quantum universal enveloping algebra of $\frak{sl}\_2$, and consider the adjoint action
$$
\mathrm{ad}\_X: U\_q({\frak sl}\_2) \to U\_q({\frak sl}\_2), ~~ Y \mapsto S(X\_{(1)})YX\_{(2)},
$$
where we have used sumless Sweedler notation. This gives $U\_q({$\frak sl}\_2)$ the structure of a $U\_q({\frak sl}\_2)$-module. What is the structure of this module? Does it decompose into a direct sum of finite-dim irreps, or are there infinite-dim reps in there? If decomposes into a direct sum of finite-dim irreps, then does every irrep appear, and are there multiplicities. How does it compare to the classical situation?
|
https://mathoverflow.net/users/153228
|
The adjoint representation of $U_q({\frak sl}_2)$ on itself
|
Unlike what happens in the classical case, it is not locally finite dimensional, basically because it has invertible elements. However its ad-locally finite part $U'$ is very large and it decomposes as a direct sum
$$U'=\bigoplus\_{V} V^\* \otimes V$$
where the sum is over the irreducible finite dimensional modules (note this is a purely 'quantum' phenomena).
A standard reference is Joseph-Letzter "[Local finiteness of the adjoint action for quantized enveloping algebras](https://www.sciencedirect.com/science/article/pii/002186939290157H)".
|
8
|
https://mathoverflow.net/users/13552
|
397337
| 164,025 |
https://mathoverflow.net/questions/397330
|
22
|
I am a PhD student in algebraic / arithmetic geometry and I never took a formal course in algebraic topology, even though I have some basic knowledge.
In algebraic geometry we deal exclusively with sheaf cohomology since we care about non-constant sheaves. But I feel, maybe in my naivety, that a lot of the important results (for usual topological spaces) are only true for singular and simplicial cohomology when they coincide with sheaf cohomology (Alexander duality and "$H^i=0$ for $i>$ the covering dimension" come to mind).
With that in mind, I wonder if it is worth for someone with a similar background to study the details of a first course in algebraic topology. (Perhaps on the level of Hatcher's book.) Do I lose something by just thinking in terms of sheaf cohomology?
|
https://mathoverflow.net/users/131975
|
Why should an algebraic geometer care about singular / simplicial (co)homology?
|
Sheaf cohomology is a powerful tool, but it isn't a replacement for all of basic algebraic topology. For example, fundamental groups and homology some topics that would get lost. And these topics are certainly relevant to algebraic geometry. Also, as pointed out in the comments, you would lose valuable intuition if you just stuck to the sheaf cohomology viewpoint.
Let me expand my original answer a bit. Let me focus on the simplest example, where $X$ is a smooth complex projective curve of genus $g$. One learns in topology that $X$ is obtained by identifying the sides of a $2g$-gon in the standard way. One can use this to extract 2 things about $X$.
1. One gets the homology $H\_1(X,\mathbb{Z})=\mathbb{Z}^{2g}$, with its intersection pairing equal to the standard symplectic form. For an algebraic geometer, this corresponds to the lattice of the Jacobian of $X$ together with its Riemann form. In particular, this is a principal polarization.
2. Also one gets the familiar presentation of the fundamental group
$$\pi\_1(X)= \langle a\_1\ldots a\_{2g}\mid [a\_1,a\_2]\ldots[a\_{2g-1}, a\_{2g}]\rangle$$
But why should an *algebraic* geometer care about this? Answer: because it tells us what etale covers of $X$ look like. The etale fundamental group of $X$ is the profinite completion of the above group. This is also true for the prime to $p$ part if $X$ lives in positive characteristic by lifting. Note that Grothendieck uses this reduction to the topological case in SGA1.
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24
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https://mathoverflow.net/users/4144
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397339
| 164,027 |
https://mathoverflow.net/questions/396687
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1
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I posted [this question](https://math.stackexchange.com/questions/4187132/reference-request-on-the-relationship-between-inscribed-polytopes-and-shadows-of) thinking that the response would be two or three answers that say "Counterexamples to this are found in every textbook—for example this one and this one and this one." But the result is that after two days there are 13 views and nobody has said anything except in a private email with a comment that probably misses the mark.
Two chords of a circle of unit radius have equal lengths if their corresponding arcs have equal lengths.
Suppose a polytope is the convex hull of finitely many points on the unit $(n-1)$-sphere in $\mathbb R^n.$ Each of its $(n-1)$-dimensional facets casts a shadow on the sphere, the light-source being at the center. Suppose a proposition says two such facets have equal $(n-1)$-dimensional volumes if their shadows have equal volumes. I suspect that that is *false* in general, even if both facets have equally many vertices. (Unlike with chords of circles, their shapes can differ—for example not all triangles are equilateral.)
Is a proof or a counterexample known?
|
https://mathoverflow.net/users/6316
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The relationship between facets of an inscribed polytope and those facets' shadows
|
The following is a detailed description of the counterexample given in the comments. The construction is the same, but with slightly more notation and detail for clarity. Also, the construction there gave two triangles with different areas but equal-area shadows; this answer constructs one polytope of which both of these triangles are faces (this is inessential, but it's unclear to me whether the original question asked about two facets of one polytope or of two different ones).
1. Let $T$ be any triangle inscribed in a unit circle in the $xy$-plane. By $``$the lift of $T$ to height $\varepsilon$ in $S^2$$"$ I mean the triangle
$$\varepsilon \hat{z} + \sqrt{1-\varepsilon^2}T,$$
which is similar to $T$, inscribed in $S^2$, and lies in the horizontal plane $\{z=\varepsilon\}$. Denote this lift by $L\_\varepsilon(T)$.
2. If $T$ is inscribed in the unit circle in the $xy$ plane and contains the origin in the interior of its convex hull, the shadow of $L\_\varepsilon(T)$ on $S^2$ approaches the entire upper hemisphere as $\varepsilon$ approaches $0$ from above, and it approaches the north hemisphere as $\varepsilon \to 1$ from below (these limits of shadows can be taken with respect to the Hausdorff metric. We care only about the areas of the shadows and not about the shadows themselves).
3. Let $T\_1$ be an equilateral triangle inscribed in the unit circle in the $xy$-plane and let $T\_2$ be a very narrow triangle inscribed in the same circle and containing the origin in its interior, such that $T\_2$ has small area $s$ (we will find a bound on $s$ later, but any $s<0.01$ will certainly do). Fix $\varepsilon\_1 = 0.1$. By part (2), we can choose $\varepsilon\_2>0$ such that the shadows of $L\_{\varepsilon\_1}(T\_1)$ and of $L\_{\varepsilon\_2}(T\_2)$ have equal area on $S^2$. It is clear that the areas of these lifts are different (the area of $L\_{\varepsilon\_2}(T\_2)$ is much smaller, as it is bounded above by $s$, which we may modify to be smaller than $A(L\_{\varepsilon\_1}(T\_1))$ if necessary).
4. The triangles $L\_{\varepsilon\_1}(T\_1)$ and $L\_{-\varepsilon\_2}(T\_2)$ are inscribed in $S^2$, have shadows of equal area on $S^2$, and have different areas. The planes they lie in do not intersect. Therefore the convex hull of their six vertices is a convex polytope inscribed in $S^2$ which has them as faces. This is a counterexample.
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2
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https://mathoverflow.net/users/75344
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397343
| 164,028 |
https://mathoverflow.net/questions/397332
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2
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Are there any reference for the classification of orientable disk bundle over a closed surface? I am particularly interested in the case if the surface is $S^2,RP^2,T^2$ or the Klein bottle.
Many thanks!
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https://mathoverflow.net/users/280895
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Classification of disk bundle over surfaces
|
Fix a base space $B$. Taking boundaries gives an equivalence from the category of (isomorphisms of topological) disk bundles over $B$ to the category of (isomorphisms of topological) circle bundles over $B$. When $B$ is a surface the latter are also called “Seifert fibered spaces”. These are described in many different references. One very nice exposition is given by Allen Hatcher in “Notes on basic three-manifold topology” - you can download this from his webpage.
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5
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https://mathoverflow.net/users/1650
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397350
| 164,030 |
https://mathoverflow.net/questions/397250
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5
|
Alice and Bob have $N\_A$ and $N\_B$ warriors under their command, numbered $1$~$N\_A$ and $1$~$N\_B$ respectively. Alice has $1$ fighting power at her disposal, and Bob has $b$ ($b\gt 0$). Before the game, they privately distribute their power between their warriors. When the game begins, both send their warrior #1 to a 1-to-1 fight. If a warrior with power $x$ fights one with power $y$, the former wins with probability $\frac{x}{x+y}$, and the latter with $\frac{y}{x+y}$. If #1 is defeated, #2 is sent to continue the next round of fight, so on and so forth until one side has all of their warriors defeated and loses the game. There's a small twist though: after a warrior defeats an $x$ power opponent, his power will increase by $cx$ ($c\geq 0$).
Clearly, a player's strategy is their method of distributing fighting power.
**Question 1**: if Alice and Bob are equally matched ($N\_A=N\_B$ and $b=1$), does Alice have a strategy that can guarantee her no less than 50% winning probability, no matter how Bob plays?
**Question 2**: is there a [dominant strategy](https://en.wikipedia.org/wiki/Strategic_dominance#Dominance_and_Nash_equilibria) for Alice?
---
**Note**: question 2 is a stronger claim, and implies question 1. For $c=1$, I know the answer is yes for both questions, because in that case the game is essentially the gambler's ruin problem, so every strategy gives the same winning probability for Alice. I suspect but can't prove an equal distribution is dominant for $c=0$ and give-it-all-to-one is dominant for $c\gt1$. A simulation for $N\_A=N\_B=2$, $b=1$ and $c=1/2$ suggests that giving 0.420341... to the first warrior guarantees 50% winning probability for Alice.
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https://mathoverflow.net/users/75935
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Is there a dominant strategy for this game?
|
This answer expands on my previous answer. It is too long for a comment on comment.
The outcome of the game to a player is 1 if he/she wins and 0 if he/she lose. Every pair of strategies induces an expected outcome, which is a number between 0 and 1 - the probability that the player wins. The min-max value of the game is (the infimum over all strategies of Bob of the supremum over all strategies of Alice of the probability that Alice wins). Similarly, the max-min value of the game is (the supremum over all strategies of Alice of the infimum over all strategies of Bob of the probability that Alice wins). If the two quantities coincide, the common number is called the value of the game. In the game you describe the two players are symmetric: when one player wins, the other loses; every strategy available to Alice is also available to Bob; and if each player adopts the strategy of their opponent, the probabilities that they win change as well. In symmetric games, the value, if it exists, is 0.5.
The question is, then, whether the value exists. There are theorems that guarantee the existence of the value, and they require:
1. The set of pure strategies should be compact (in a nice enough space; the space here is Euclidean, which is nice).
2. The payoff function should be continuous.
If I am not mistaken, in your example both conditions hold.
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1
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https://mathoverflow.net/users/64609
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397359
| 164,036 |
https://mathoverflow.net/questions/397317
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8
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Are there smooth closed manifolds $M^n$ in every dimension $n \geq 3$ with trivial mapping class groups and with $H^1(M^n;\mathbb{Z}/2\mathbb{Z})$ arbitrarily large?
I am under the impression that "generically" a manifold will have no mapping class group (but maybe I am totally mistaken). There are presumably lots of constructions of such manifolds with trivial mapping class groups. So I guess I'm hoping to hear of some such construction where $H^1(M^n;\mathbb{Z}/2\mathbb{Z})$ can get very large.
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https://mathoverflow.net/users/99414
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Manifolds with trivial mapping class group and large $H^1$?
|
I think that the construction in Belolipetsky--Lubotzky, Finite Groups and Hyperbolic Manifolds (<https://arxiv.org/abs/math/0406607>) provides a more algebraic construction of such manifolds for any $n$.
One of the results in this paper (Theorem 3.1) is the following : given groups $\Gamma \triangleright \Delta \triangleright M$ with $\Gamma/\Delta$ finite and $F = \Delta/M$ a nonabelian free group, there exists $\Gamma \ge D \ge \Delta$ and $B \le D$ of finite index with $N\_\Gamma(B) = B$ (in fact they prove that for any finite group $G$ there are infinitely many $B$ with $N\_\Gamma(B)/B$ isomorphic to $G$).
They use this to construct closed hyperbolic manifolds with trivial isometry group ; by Mostow rigidity the mapping class group of $X$ is isomorphic to the isometry group of $X$ (every homeomorphism is isotopic to an isometry), so these manifolds will have trivial mapping class group. Without too much details their construction uses a specific lattice $\Gamma$ in the isometry group of hyperbolic $n$-space $\mathbb H^n$, and subgroups $D, \Delta, M, B$ as in the previous paragraph ; the manifold is the quotient of $\mathbb H^n$ by $B$ (their construction ensures that $B$ is torsion-free though $\Gamma$ will likely not be).
Now to look at Betti numbers, for which we need to dive a bit more into the details of this paper. Since $X$ is aspherical we have
$$b\_1(X, \mathbb Q) = b\_1(B, \mathbb Q) \ge b\_1(D, \mathbb Q).$$
On the other hand $b\_1(D, \mathbb Q)$ is the dimension of the fixed subspace of $D/\Delta$ in $H^1(\Delta, \mathbb Q)$ (conjugation action on the normal subgroup $\Delta$ and its morphisms to $\mathbb Z$). By the definition of $D$ in the paper of Belolipetsky--Lubotzky (p. 4 in the arxiv version) we have that it acts trivially on the part of the cohomology coming from the surjective morphism $\Delta \to F$. So we have that $b\_1(X, \mathbb Q)$
is larger than the rank of $F$. It remains to remark that in the data for Theorem 3.1 the quotient $\Delta/M$ can be taken to have arbitrarily large rank (since a non-abelian free group contains free subgroups of finite index with arbitrarily large rank), so we get examples with $b\_1(B, \mathbb Q)$ arbitrarily large.
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5
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https://mathoverflow.net/users/32210
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397362
| 164,037 |
https://mathoverflow.net/questions/397356
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3
|
Let $M$ be a smooth manifold. A Lie algebroid over $M$ is a vector bundle $E\rightarrow M$ over $M$, with a Lie bracket on $\Gamma(M,E)$, a morphism of vector bundles $\rho:E\rightarrow TM$, such that, the following conditions are satisfied:
1. the map $\rho:E\rightarrow TM$ induce a morphism of Lie algebras $\Gamma(M,E)\rightarrow \Gamma(M,TM)$,
2. the Lie algebra structure on $\Gamma(M,E)$ is “compatible” with the $C^\infty(M)$-algebra structure on $\Gamma(M,E)$, upto a correction; this goes by the name Leibniz condition.
A generalized complex structure on a manifold $M$ is a morphism of vector bundles $J:TM\oplus TM^\*\rightarrow TM\oplus TM^\*$ such that it is compatible with some bracket operation and some ''inner product'' on $\Gamma(M,TM\oplus TM^\*)$.
In most of the references about generalized complex structures, they introduce the notion of Lie algebroid. I could not see detailed justification of introducing Lie algebroid over $M$ when discussing generalized complex structure on $M$. So, I am thinking of the following question:
>
> How is the theory of Lie algebroids useful in (better) understanding of generalized complex structures?
>
>
>
Any pointers are welcome. I have done google search for ''Lie algebroids and generalised complex structure" but could not find anything specific.
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https://mathoverflow.net/users/118688
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Use of theory of Lie algebroids in (better) understanding of generalised complex structures
|
The compatibility conditions that you mention in the definition of a generalized complex structure are equivalent to the statement that the $+i$-eigenbundle $L$ of $J$ is a complex Dirac structure:
1. compatibility of $J$ with the inner product is equivalent to $L$ being 'isotropic', and
2. compatibility of $J$ with the Courant bracket is equivalent to the sections of $L$ being closed under this bracket.
A Dirac structure automatically inherits the structure of a Lie algebroid over $M$. So $L$ in this case is naturally a Lie algebroid. Many of the properties and constructions for $J$ can be stated in terms of the Lie algebroid $L$. For example:
1. The Lie algebroid cohomology of $L$ is a fundamental invariant of $J$. When $J$ comes from a symplectic form, it is the usual de Rham cohomology, whereas when $J$ comes from a complex structure, it is a sum of Dolbeault cohomology groups.
2. A generalized holomorphic bundle for $J$ can be defined to be a representation of $L$. When $J$ comes from a symplectic form, this is a bundle with flat connection, and when $J$ comes from a complex structure, this is a co-higgs bundle.
Note that the Lie algebroid structure on $L$ is somewhat less information than $J$. On the other hand, if you view $L$ as a Dirac structure in $(TM \oplus T^{\*}M)\otimes \mathbb{C}$, then it is equivalent data to $J$. So it's often better to work with it in this way, and many constructions are very naturally stated using the formalism of Dirac geometry.
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3
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https://mathoverflow.net/users/69713
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397364
| 164,039 |
https://mathoverflow.net/questions/397373
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5
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The classical Artin-Rees lemma tells the following. Let $R$ be a Noetherian commutative ring and $I\subset R$ be an ideal. Let $M$ be a finitely generated $R$-module and $N\subset M$ be a submodule. Then there exists an integer $m\ge0$ such that for all $n\ge0$ the following equality of two submodules in $N$ holds:
$$
I^{n+m}M\cap N=I^n(I^mM\cap N).
$$
The usual proof is based on the Hilbert Basis Theorem: essentially, one uses the fact that the Rees ring $\bigoplus\_{n=0}^\infty I^n$ is graded Noetherian (since it is a quotient ring of the ring of polynomials over $R$ spanned by some finite set of generators of the ideal $I$).
I would like to specialize the Artin-Rees lemma to principal ideals $I=(s)\subset R$, and then extend it from multiplicative subsets of the form $\{1,s,s^2,s^3,\dotsc\}\subset R$ to other multiplicative subsets $S\subset R$.
So let $R$ be a Noetherian commutative ring and $S$ be a multiplicative subset in $R$. Assume for simplicity that $S$ is (at most) countable and all the elements of $S$ are regular (nonzero-divisors) in $R$.
Let $M$ be a finitely generated $R$-module and $N\subset M$ be a submodule. Does there exist an element $t\in S$ such that, for every $s\in S$, the equality of two submodules in $N$
$$
stM\cap N=s(tM\cap N)
$$
holds?
Or, at least, can one find an element $t\in S$ such that the above equality holds for a cofinal subset of elements $s\in S$ (in the divisibility preorder)?
A straightforward attempt to argue similarly to the Hilbert Basis Theorem proof of the Artin-Rees lemma does not seem to work, as the ring $\bigoplus\_{s\in S}sR$ does not have to be graded Noetherian when the multiplicative set $S$ is not finitely generated.
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https://mathoverflow.net/users/2106
|
Artin-Rees lemma for multiplicative subsets?
|
This is straightforward . Define the submodule $P$ as $N\subset P\subset M$, the set of all elements $m\in M$ such that $sm\in N$ for some $s\in S$. Then, choose $t\in S$ such that $tP\subset N$.
Now for what you need, clearly the right hand side is contained in the left. So, let $x\in stM\cap N$ for $s\in S$. Then, $x=stm$ and then, $m\in P$. Then $tm\in N$ by our choice of $t$. So, $tm\in tM\cap N$ and thus $x\in s(tM\cap N)$.
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3
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https://mathoverflow.net/users/9502
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397382
| 164,042 |
https://mathoverflow.net/questions/397380
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1
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Let $G$ be a simple graph (finite or infinite), $[n]\mathrel{:=}\{1,...,n\}$. Define the function:
$$\varepsilon\_n(G)\mathrel{:=}\min\_\phi{\lvert{\operatorname{dom} (\phi)}\rvert},$$
where $\phi$ is the partial function $\phi:V(G)\to[n]$ such that $\forall x,y\in[n]$, $x\neq y$, $\exists u,v\in V(G)$: $\phi(u)=x$, $\phi(v)=y$, $uv \in E(G)$ and $\forall u,v\in \operatorname{dom}(\phi)$, $u\neq v$, $\phi(u)=\phi(v) \implies uv\notin E(G) $.
I need information about this function, but I don't know where to search.
Intuitively, one can think of this function as the minimum number of characters we can write to express the inequality of $n$ numbers if we place them on the nodes of a graph and place symbols of inequality on the edges. It's obvious that $0 \leq \varepsilon\_n(G) \leq v(G) $,
$$\varepsilon\_n(K\_m)=\begin{cases}n,\ m\geq n\\0,\ m < n. \end{cases}$$
In particular, I am interested in the values of $\varepsilon\_n(\mathbb{Z}\_G)$, where $\mathbb{Z}\_G $ is the undirected infinite graph such that $V(\mathbb{Z}\_G )= \mathbb{Z} $, and $E(\mathbb{Z}\_G)=\{(i,j)\in \mathbb{Z}^2\mathrel: i+1=j \}$.
It's clear that $\varepsilon\_n(\mathbb{Z}\_G) \neq 0 $ for all $n$, but I have no rigorous proof of this statement. I am also interested in the complexity of calculating $ \varepsilon\_n(G) $ in the general case, but I do not understand how not to iterate over many $\phi$. I think that $\varepsilon\_n(G) \in \mathrm{FNP}$.
|
https://mathoverflow.net/users/175589
|
What is this invariant graph?
|
Rephrasing, given $G$ and $n$ you're looking for the smallest subset $V' \subseteq V$ of the vertices of $G$ such that the subgraph $G'$ induced by $V'$ has a vertex colouring $\phi$ on $n$ colours for which every pair of colours $c \neq d$ has an edge between those colours and there is no edge between two vertices of the same colour.
For your particular case of interest, when $n$ is odd we can colour a subchain of $\mathbb{Z}\_G$ in the sequence of colours of an Eulerian path on $K\_n$ to find that $\varepsilon\_n(\mathbb{Z}\_G) = \binom{n}{2} + 1$. When $n$ is even, $\varepsilon\_n(\mathbb{Z}\_G) = \binom{n}{2} + \frac{n-2}{2} + 1$ by a similar argument, as bof correctly pointed out in a comment improving the earlier claim of this answer: the addition of $\frac{n-2}{2}$ edges is necessary and sufficient for an Eulerian trace to exist.
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2
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https://mathoverflow.net/users/46140
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397383
| 164,043 |
https://mathoverflow.net/questions/397381
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1
|
I'm reading a book and have encountered a relation which seems to me to be impossible to prove, I would like to be sure if this is the case. The author gives a probability function as
$$p\_n = \frac{e^{-c\_1 n - c\_2/n}}{Z},$$
where $c\_1$ and $c\_2$ are constants and Z is a normalization factor and $n \geq 3$. Then by considering $\langle n \rangle = 6$ and defining $\alpha$ (the second moment) as $\alpha = \sum\_{n = 3}^{\infty} p\_n (n - 6)^2$, the author claims one can show that
\begin{equation}
\alpha + p\_6 = 1, \quad \quad \quad 0.66 < p\_6 < 1,
\end{equation}
\begin{equation}
\alpha p\_6^2 = 1 / 2 \pi, \quad \quad \quad 0.34 < p\_6 < 0.66.
\end{equation}
How is such a thing possible in the first place as these relations are not even dependent on $c\_1$ and $c\_2$?
|
https://mathoverflow.net/users/nan
|
To prove a relation involving a probability distribution
|
The equality $\alpha + p\_6 = 1$ can be rewritten as
$$\sum\_{n=3}^\infty g(n)p\_n=1,$$
where
$$g(n):=(n-6)^2+1(n=6)\ge1(n\in\{5,6,7\})+4\times1(n\notin\{5,6,7\}).$$
Therefore and because $p\_n>0$ for all $n\ge3$, we have
$$\sum\_{n=3}^\infty g(n)p\_n>\sum\_{n=3}^\infty p\_n=1,$$
so that the equality $\alpha + p\_6 = 1$ is always false.
---
As for the equality
\begin{equation}
\alpha p\_6^2 = 1 / 2 \pi, \quad \quad \quad 0.34 < p\_6 < 0.66,
\end{equation}
numerics suggest that it is also false in general. In particular, using Mathematica's numerical summation NSum[] command, for $c\_1=12$ and $c\_2=500$ we get $p\_6=0.50995\ldots\in(0.34,0.66)$ but $2\pi\alpha p\_6^2=0.90785\ldots\ne1$.
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2
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https://mathoverflow.net/users/36721
|
397384
| 164,044 |
https://mathoverflow.net/questions/397368
|
10
|
I came upon this statement in a stack answer.
Statement :
If $f\_n$ is a sequence of real valued functions (not necessarily continuous or measurable) on $[0,1]$ such that $f\_n$ converges point-wise to $0$, then there exists an infinite subset of $[0,1]$ where the convergence is uniform.
I couldn't prove it. I believe the claim is true because $[0,1]$ is uncountable but the set of sequences is countable only.
Any help would be appreciated in assistance to how to prove it.
|
https://mathoverflow.net/users/318305
|
Pointwise convergence imples uniform convergence in an infinite subset
|
For every sequence $(F\_n)\_{n \in \omega}$ with $F\_n:[0,1] \rightarrow \mathbb{R}$ converging pointwise to $0$, we can associate to every $x \in [0,1]$ an $f\_x \in \mathbb{\omega}^\omega$ in the following way: Set $f\_x(m):= \min\{n \in \mathbb{\omega}\,\, \colon \,\, \forall n' \geq n \,\,\,\, \vert F\_{n'}(x) \vert < \frac{1}{m}\}$.
Now the proof works in two steps.
In the **first step** we show that there exists $f^\* \in \omega^\omega$ such that for every $k \in \omega$ the set $\{x \in [0,1] \,\, \colon \,\, f\_x \restriction k \leq f^\* \restriction k\}$ is uncountable.
We will construct such an $f^\* \in \omega^\omega$ by induction on $k \in \omega$: Assume that $f^\* \restriction k$ has already been constructed, and we have that the set $\{x \in [0,1] \,\, \colon \,\, f\_x \restriction k \leq f^\* \restriction k\}$ is uncountable. Since $$\{x \in [0,1] \,\, \colon \,\, f\_x \restriction k \leq f^\* \restriction k\}= \bigcup\_{l \in \omega}\,\, \{x \in [0,1] \,\, \colon \,\, f\_x \restriction (k+1) \leq (f^\* \restriction k)^\frown l\}$$ we find $l \in \omega$ such that $\{x \in [0,1] \,\, \colon \,\, f\_x \restriction (k+1) \leq (f^\* \restriction k)^\frown l\}$ is uncountable. Now set $f^\*(k)=l$ and we see that $f^\*\restriction (k+1)$ has the required properties.
In the **second step** we inductively construct $(x\_k)\_{k \in \omega} \subseteq [0,1]$ injective and $(f\_k)\_{k \in \omega} \subseteq \omega^\omega$ increasing such that for every $k \in \omega$ we have $f^\* \leq f\_k$ , $f\_{x\_k} \leq f\_k$ and $f\_k \restriction (k+1) = f\_{k+1} \restriction (k+1)$. Once we have shown this, we can define $g \in \omega^\omega$ such that $g(k):=f\_k(k)$ and see that $f\_{x\_k} \leq g$ for every $k \in \omega$. But this proves that $(F\_n)\_{n \in \omega}$ converges uniformly on $(x\_k)\_{k \in \omega}$.
To this end assume that $x\_0,...,x\_{k-1}$ and $f\_0,...f\_{k-1}$ with the required properties have already been constructed. Since $\{x \in [0,1] \,\, \colon \,\, f\_x \restriction k \leq f^\* \restriction k\}$ is uncountable, we can find $x\_k \in \{x \in [0,1] \,\, \colon \,\, f\_x \restriction k \leq f^\* \restriction k\}$ different from $x\_0,...,x\_{k-1}$. Set $f\_k(m):=\max\{f\_{k-1}(m), f\_{x\_k}(m)\}$, and note that since $f\_{x\_k} \restriction k \leq f^\* \restriction k \leq f\_{k-1} \restriction k$, we have $f\_{k-1}\restriction k =f\_k \restriction k$. This finishes the proof.
|
6
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https://mathoverflow.net/users/134910
|
397385
| 164,045 |
https://mathoverflow.net/questions/397135
|
6
|
I am considering the following ODE
\begin{equation}
\begin{split}
&\frac{d^2}{dy^2}u + \frac{\alpha}{(1+y^2)^{\frac{r}{2}}}u = \delta(y)\\
&\lim\_{|y|\to \infty}u(y) = 0.
\end{split}
\end{equation}
Here $\alpha$ is a constant and $r > 0$ and $\delta(y)$ is the Dirac delta function. Is it possible to solve this ODE? I expect the solution to be singular in $\alpha$ as $\alpha$ approaches zero. If we can't get the explicit expression of the solution, can we analyze its singularity in $\alpha$ asymptotically as $\alpha \to 0$? For example, when $r = 0$, the solution behaves like $\frac{e^{-|y|\sqrt{-\alpha}}}{\sqrt{-\alpha}}$ which is singular in $\alpha$.
|
https://mathoverflow.net/users/114951
|
How to solve the following ODE with a parameter?
|
This is essentially an elliptic second-order ODE in $(0,\infty)$:
$$ (1 + y^2)^{r/2} u''(y) = (-\alpha) u(y) , $$
with boundary conditions $u'(0) = -1$ and $u(\infty) = 0$. This kind of problems are well-studied, the corresponding theory is known as *Krein's spectral theory of strings*.
The value $h(\alpha)$ of $u$ at $0$ for the given parameter $\alpha$ is the corresponding *spectral function*, and $h(-\alpha)$ is a Stieltjes function of $\alpha$ (other names: Nevanlinna–Pick function, Herglotz function). The relation between the coefficient (here: $(1+y^2)^{-r/2}$, roughly known as the *string*) and the spectral function $h$ is not explicit, but some comparison results are known.
I do not have time now to search for the relevant results, but I would start by looking at the following paper:
* S. Kotani and S. Watanabe, *Krein's spectral theory of strings and generalized diffusion processes*, in: Functional Analysis in Markov processes (Katata/Kyoto, 1981), Lecture Notes in Math. 923, Springer, Berlin, 1982, 235–259, [DOI:10.1007/BFb0093046](https://doi.org/10.1007/BFb0093046).
Another excellent reference with a chapter on Krein's theory is
* R. Schilling, R. Song, Z. Vondraček, *Bernstein Functions: Theory and Applications*. De Gruyter, Studies in Math. 37, Berlin, 2012, [DOI:10.1515/9783110269338](https://doi.org/10.1515/9783110269338).
---
*Edit: some additional details.*
Strictly speaking, our case corresponds to the *string* $m(dy) = (1 + y^2)^{-r/2} dy$, which is commonly identified with its distribution function $$m(y) = m([0, y)) = \int\_0^y (1 + s^2)^{-r/2} ds.$$
The spectral function can be defined in a number of equivalent ways: one of them inolves the integral of $(\phi\_N)^{-2}$, where $\phi\_N$ is the solution with "Neumann" initial condition $\phi\_N(0) = 1$, $\phi\_N'(0) = 0$. The one I am most familiar with defines it to be the reciprocal of $-\phi'(0)$, where $\phi$ is the solution with boundary conditions $\phi(0) = 1$ and $\phi(\infty) = 0$. This is clearly equivalent to the definition as $-u'(0)$ with $u$ as in the statement of the problem.
I failed once to find a simple reference for the above calculation in analytical terms, so together with Jacek Mucha we included a very brief discussion in the appendix to our paper (see Section A.3 therein):
* J. Mucha, M. Kwaśnicki, *Extension technique for complete Bernstein functions of the Laplace operator*, J. Evol. Equ. 18(3) (2018): 1341–1379 [DOI:10.1007/s00028-018-0444-4](https://doi.org/10.1007/s00028-018-0444-4).
|
7
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https://mathoverflow.net/users/108637
|
397387
| 164,046 |
https://mathoverflow.net/questions/397371
|
3
|
Suppose $(M,g)$ is a three dimensional smooth compact simply connected Riemannian manifold with boundary and suppose that $\Sigma$ is a smooth simply connected hypersurface in $M$ with a smooth boundary $\partial \Sigma \subset \partial M$. Let $D\_X Y$ denote the Levi-Civita connection on $(M,g)$. Let $Z$ be a smooth vector field on $\Sigma$ and let $f:\Sigma \to \mathbb R$ be a smooth function on $\Sigma$. Does there exist a smooth function $\phi$ in $M$ such that $\Sigma=\{\phi=0\}$ and such that
$$ (D\_{\nabla \phi} \nabla \phi)\big|\_{\Sigma} = f\, Z.$$
Note that the left hand side is just the restriction of $D\_{\nabla \phi}\nabla \phi$ to $\Sigma$ and that $\nabla \phi$ is the gradient of $\phi$ with respect to $g$.
If the answer is no, would it make a difference if the latter condition is replaced with
$$(D\_{\nabla \phi} \nabla \phi)\big|\_{\Sigma} - f\, Z \in \textrm{Span}\{(\nabla \phi)|\_{\Sigma}\}$$
|
https://mathoverflow.net/users/50438
|
A question on Levi-Civita connection and a fixed hyper surface
|
Not a full answer, but there definitely should be some integrability constraint.
Take the simplest case where $M = \mathbb{R}^3$ (the boundary is unimportant for the discussion here) and $\Sigma$ is the $x$-$y$ plane.
If $\phi$ is a defining function of $\Sigma$, then restricted to $\Sigma$ we have $\nabla \phi|\_{\Sigma} = \eta \partial\_z$ for some function $\eta: \Sigma \to \mathbb{R}$.
Symmetry of the Hessian requires then
$$ \nabla^2 \phi|\_{\Sigma} = \begin{pmatrix}
0 & 0 & \partial\_x \eta \\
0 & 0 & \partial\_y \eta \\
\partial\_x \eta & \partial\_y \eta & \* \end{pmatrix} $$
This shows that
$$ D\_{\nabla\phi} \nabla\phi = \eta (\partial\_x \eta, \partial\_y \eta, \*) $$
So a necessary condition for your equation to hold (setting $f \equiv 1$ since as you agreed it is unimportant) in the flat case is that $Z$ is the gradient of some scalar function on $\Sigma$, and that allowing there to be normal components (as in your modified question) has no impact.
---
The conclusion is unchanged in the Lorentzian setup, if you require $\Sigma$ to be non-degenerate. In the case where $\Sigma$ is null, however, since $\nabla\phi$ now lies in the tangent space of $\Sigma$ and $\Sigma$ is two dimensional, it may in fact be possible for the second variant to hold (but I don't have the energy to run through the analysis now).
|
4
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https://mathoverflow.net/users/3948
|
397392
| 164,048 |
https://mathoverflow.net/questions/395866
|
16
|
The following statement is a direct consequence of the Continuum Hypothesis:
>
> There exists a sequence $\langle f\_\alpha:\omega\_1\rightarrow\omega\_1 ~ \vert ~ \alpha<\omega\_1\rangle$ of functions such that there is no function $f:\omega\_1\rightarrow\omega\_1$ with the property that the sets $\{\xi<\omega\_1 ~ \vert ~ f(\xi)=f\_\alpha(\xi)\}$ are finite for all $\alpha<\omega\_1$.
>
>
>
Moreover, since a failure of this statement can be used to obtain an $\omega\_2$-sequence of subsets of $\omega\_1$ with pairwise finite intersection, results of Baumgartner in
*Baumgartner, James E.*, [**Almost-disjoint sets, the dense set problem and the partition calculus**](http://dx.doi.org/10.1016/0003-4843(76)90018-8), Ann. Math. Logic 9, 401-439 (1976). [ZBL0339.04003](https://zbmath.org/?q=an:0339.04003).
show that the statement is not equivalent to CH.
**Question:** Can the above statement consistently fail?
|
https://mathoverflow.net/users/90412
|
Diagonalizing against $\omega_1$-sequences of functions mod finite
|
The arguments in Section 6 of my paper "The nonstationary ideal in the $\mathbb{P}\_{\mathrm{max}}$ extension" show that there is a proper forcing adding a function from $\omega\_{1}$ to $\omega\_{1}$ which agrees with each such ground model function in only finitely many places. I say there that Todorcevic had done something similar in his "A note on the Proper Forcing Axiom". It follows that under PFA, and in the $\mathbb{P}\_{\mathrm{max}}$ extension, the statement above fails. You don't need any large cardinals, however.
|
6
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https://mathoverflow.net/users/31807
|
397394
| 164,049 |
https://mathoverflow.net/questions/397395
|
-6
|
Indeed,
**Conjecture** Let $\ p\ $ be a prime, $\ n>1\, $ -- a natural number, and let the largest prime divisor $\ q\ $ of $\ s(p^n) :=\sum\_{k=0}^n\,p^k\ $ satisfy $ q<p.\ $ Then $\ p^n\ =\ 7^3.$
The unique(?) exception would be
$$ s(7^3)\ =\ 400\ = 2^4\cdot5^2 $$
>
>
**EDIT** A micro-observation: when primes
$\ p\ q\ r\ $ satisfy $\ q|s(p^{r-1})\ $ then
$\ p>r\ $ and $\ q\ge r$.
More generally, $\ p>\rho $ and $\ q\ge \rho\ $ when $\ \rho\ $ is the smallest prime divisor of $\ r\,\ $ where this time $\ r\ $ is **not** assumed to be a prime.
|
https://mathoverflow.net/users/110389
|
(not so) Spectacular $\ 7^3$
|
There's also
$$s(67^2) = 1 + 67 + 67^2 = 4557 = 3 \cdot 7^2 \cdot 31 \tag{1}\label{eq1A}$$
In this case, $q = 31 \lt 67$ meets your criteria. I haven't checked further, but I'm fairly certain there are additional exceptions.
|
4
|
https://mathoverflow.net/users/129887
|
397397
| 164,050 |
https://mathoverflow.net/questions/397347
|
9
|
>
> Is a (finitely generated) torsion-free subgroup of a right-angled Coxeter group isomorphic to a subgroup of a right-angled Artin group?
>
>
>
It is well-known from the theory of special cube complexes that a subgroup $H$ of a right-angled Coxeter group $G$ contains a finite-index subgroup isomorphic to a subgroup of a right-angled Artin group (e.g. $H \cap [G,G]$), but is it true for the whole group if we assume in addition that $H$ is torsion-free?
The difficulty when applying the theory of special cube complexes comes from the fact that the induced action of $H$ on the usual CAT(0) cube complex $X$ of $G$ may invert hyperplanes (i.e. some isometries may stabilise a hyperplane and switch its halfspaces). This can be avoided by replacing $X$ with its barycentric subdivision, but then hyperplane-inversions yield self-osculations (i.e. there exist an element $h \in H$ and an oriented hyperplane $\vec{J}$ such that $\vec{J}$ and $g \vec{J}$ contain two intersecting oriented edges pointing to their common vertex and that do not span a square).
**NB:** In all the question, right-angled Artin/Coxeter groups are finitely generated. Recall that, given a (finite) simplicial graph $\Gamma$, the associated right-angled Coxeter group is defined by the presentation
$$\langle u \in V(\Gamma) \mid u^2=1 \ (u \in V(\Gamma)), \ [u,v]=1 \ (\{u,v\} \in E(\Gamma)) \rangle$$
and the associated right-angled Artin group by the presentation
$$\langle u \in V(\Gamma) \mid [u,v]=1 \ (\{u,v\} \in E(\Gamma)) \rangle$$
where $V(\Gamma)$ and $E(\Gamma)$ denote respectively the vertex- and edge-sets of $\Gamma$.
|
https://mathoverflow.net/users/122026
|
Subgroups of RAAGs vs. subgroups of RACGs
|
The fundamental group of the (non-orientable) closed surface of Euler characteristic -1 provides a counterexample.
On the one hand, it’s a subgroup of index 4 in the reflection group on the right-angled pentagon, so it embeds in a RACG, and of course it’s torsion-free.
On the other hand, [Crisp—Wiest](https://arxiv.org/abs/math/0303217v2) proved that it never embeds in a RAAG.
|
11
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https://mathoverflow.net/users/1463
|
397400
| 164,051 |
https://mathoverflow.net/questions/397396
|
4
|
Let $A$ and $B$ denote two *countably infinite* sets of ordinals.
Let $W\_A$ denote the supremum of ordinals *writable* by Ordinal Turing Machines with the set $A$ given as the source of parameters. That is, for any element $\tau$ of $A$, a $\tau$-th cell of the input tape is marked with $1$ before the start of a computation.
Similarly, let $W\_B$ denote the supremum of ordinals *writable* by Ordinal Turing Machines with the set $B$ given as the source of parameters. That is, for any element $\tau$ of $B$, a $\tau$-th cell of the input tape is marked with $1$ before the start of a computation.
Here “an ordinal writable by Ordinal Turing Machines” implies a countable ordinal $\alpha$ such that an Ordinal Turing Machine halts with an infinite binary sequence (written in the initial segment of length $\omega$ on the output tape) which encodes a well-ordering of natural numbers of order-type $\alpha$.
Question: do there exist a pair of $A$ and $B$ such that the supremum of all elements of $A$ is greater than the supremum of all elements of $B$, yet $W\_A < W\_B$? If no, why? The question separates into two possible situations: (i) an element in $A$ or $B$ may be uncountable; (ii) all elements in $A$ and $B$ are countable.
|
https://mathoverflow.net/users/122796
|
Countably infinite sets of ordinals as parameters for Ordinal Turing Machines
|
The answer is yes. In fact, you can make the difference enormous.
First, let me construct a certain set of ordinals.
**Lemma.** There is a countable set $X$ of ordinals such that:
1. Every OTM program that halts with $X$ as input, halts strictly before $\sup X$.
2. If a program does not halt with $X$ as input, then there is no countable end-extension $Y\supset X$ for which the program would halt on input $Y$.
**Proof.** We can construct $X$ in stages. Enumerate the programs $p\_0$, $p\_1$, and so on. To begin, start with any countable set $X\_0$. If $X\_n$ is defined, ask whether there is some countable end-extension of it to $X\_{n+1}$, extending only above the space used by the earlier programs, in such a way that $p\_n$ halts with this new input oracle. Let $X=\bigcup\_n X\_n$ be the limit oracle. This set is countable, and it has the two properties by construction. The computation of any program that halts was preserved, since the new ordinals were added only beyond the use of that program. And if a program does not halt with $X$, then no end-extension would make it halt, since otherwise we would have done so at the stage that that program was considered. $\quad\Box$
To form the desired sets $A$ and $B$, let $\alpha$ be any ordinal larger than $\sup X$. Let $A=X\cup\{\alpha+n\mid n\in\omega\}$. And let $B=\omega\cup\{\alpha\}$.
Notice that $\sup B<\sup A$. But by the lemma, the possible outputs of computations with $A$ are the same as the possible outputs from $X$, since $A$ end-extends $X$. Basically, the extra stuff we added on top of $A$ was no help in any computation, since all the computations used only the $X$ part and never got out to $\alpha$. But with $B$ as an oracle, we can search for the input $\alpha$ and thereby use $\alpha$ in effect as an input. So anything that $\alpha$ can write is also $B$-writable. In particular, with $B$ we can give $\alpha$ itself as output, but all ordinal outputs with $A$ are less than $\alpha$.
(If you want to consider reals coding ordinals, then we can find an $\alpha$ such that from $\alpha$ we can write a real coding an ordinal bigger than any ordinal coded by output from $X$. For this, we do the whole construction inside $L$, so $X$ is in $L$, and then pick $\alpha$ so that in $L\_\alpha$ we can define a real coding an ordinal larger than the countably many ordinals coded by reals output by $A$. Now, from $B$ we can compute $\alpha$ and therefore $L\_\alpha$ and therefore give this real as output.)
By choosing $\alpha$ suitably, therefore, we shall have $B<A$ but $W\_B>W\_A$, as desired.
**Moral.** The main idea is that we hid $\alpha$ from $A$, since it got lost in the complexities of $X$, and no program could find $\alpha$ at the top. But $\alpha$ is not hidden in $B$, because it is the only infinite ordinal in $B$, and so a program can easily find it.
|
6
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https://mathoverflow.net/users/1946
|
397406
| 164,053 |
https://mathoverflow.net/questions/397404
|
2
|
A [hypergraph](https://en.wikipedia.org/wiki/Hypergraph) $H=(V,E)$ is *sane* if $V$ is finite, $E \neq \emptyset$, and $\emptyset \notin E$, and $e\not\subseteq e'$ whenever $e\neq e' \in E$. Moreover, we call $H$ *summable* if there is a map $f:V\to \mathbb{Z}\_{\geq 0}$ such that for all $e, e' \in E$ we have $$\sum\_{v\in e}f(v) = \sum\_{w\in e'}f(w) > 0.$$
(The $>0$ requirement is there to exclude $f$ being the constant $0$-map.)
**Question.** Is every sane hypergraph summable?
|
https://mathoverflow.net/users/8628
|
Summable hypergraphs
|
No, not every sane hypergraph is summable. To see this, let $G$ be a star with five leaves $\ell\_1, \dots, \ell\_5$ all adjacent to a vertex $u$. Then turn $G$ into a sane hypergraph $H$ by adding the hyperedges $\{\ell\_1,\ell\_2\}$ and $\{\ell\_3, \ell\_4, \ell\_5\}$. Towards a contradiction, suppose that $f: V(H) \to \mathbb{Z}\_{\geq 0}$ is a function which shows that $H$ is summable. Since $\{u,\ell\_i\}$ is an edge of $H$ for all $i \in [5]$, we must have $f(\ell\_i)=f(\ell\_j)$ for all $i,j \in [5]$. But now the hyperedges $\{\ell\_1, \ell\_2\}$ and $\{\ell\_3, \ell\_4, \ell\_5\}$ have different sums.
|
2
|
https://mathoverflow.net/users/2233
|
397411
| 164,055 |
https://mathoverflow.net/questions/397370
|
1
|
What is the expression for the modular S-matrix of (p,q) minimal model? The Wiki <https://en.wikipedia.org/wiki/Minimal_model_(physics)> does not provide S-matrix
|
https://mathoverflow.net/users/17787
|
Modular S-matrix of (p,q) minimal model
|
The modular S-matrix appears in Section 10.6 of the Big Yellow Book by di Francesco, Mathieu and Sénéchal. For a freely available reference, there are my lecture notes <https://arxiv.org/abs/1609.09523> , eq. (A.28).
The Wikipedia article on minimal models was mostly written by someone who does not like the modular bootstrap. But you are free to complete it.
|
2
|
https://mathoverflow.net/users/12873
|
397421
| 164,059 |
https://mathoverflow.net/questions/397283
|
2
|
Let $P(n)$ be an irreducible polynomial of degree $2$ over the positive integers. Do there exist infinitely many positive integers $n$ such that $P(n)$ divides $n!$?
Edit: motivation by examples:
A) $p(n)=n^2+1$ (true, $21^2+1$ divides $21!$).
B) $p(n)=n^2+n+1$ (true, $74^2+74+1$ divides $74!$).
|
https://mathoverflow.net/users/174530
|
Polynomial whose values divide $n!$
|
Here is a completely elementary proof, inspired by Pasten's comments.
Let $P(n)=an^2+bn+c$.
Take $n=a^5x^4+2a^3(ab+2a+1)x^3+a(2a^3c+a^2b^2+6a^2b+3ab+6a^2+5a+1)x^2+(ab+2a+1)(2a^2c+2ab+b+2a+1)x+a^3c^2+2a^2bc+abc+2a^2c+ac+c+ab^2+b^2+2ab+2b+a+1$
Then
$P(n)=P\_1(x)P\_2(x)P\_3(x)$
where
$$\begin{align\*}
P\_1(x)=&a^2x^2+abx+2ax+ac+b+1\\
\\
P\_2(x)=&a^4x^2+a^3bx+2a^3x+2a^2x+a^3c+a^2b+ab+a^2+2a+1\\
\\
P\_3(x)=&a^5x^4\\
&+2a^4bx^3+4a^4x^3+2a^3x^3\\
&+2a^4cx^2+a^3b^2x^2+6a^3bx^2+3a^2bx^2+6a^3x^2+4a^2x^2+ax^2\\
&+2a^3bcx+4a^3cx+2a^2cx+2a^2b^2x+ab^2x+6a^2bx+4abx+bx+4a^2x+2ax\\
&+a^3c^2+2a^2bc+abc+2a^2c+c+ab^2+b^2+2ab+b+a\\
=&n-(a^2x^2+abx+2ax+x+ac+b+1)
\end{align\*}$$
Clearly for $x$ large enough $P\_1(x), P\_2(x), P\_3(x)$ are distinct and less than $n$ in absolute value.
Since the product of 3 distinct numbers $\le n$ divides $n!$, the result follows.$\qquad\qquad\blacksquare$
The idea behind this proof is that always $P(x)\;|\;P(P(x)+x)$ and applying this idea twice allows to factor $P(\text{some polynomial})$ into 3 factors all small enough.
Here is the relevant Maxima code:
```
(%i1) a*x*x+b*x+c;
(%o1) ...
(%i2) a*x^2+(b+1)*x+c;
(%o2) ...
(%i3) subst(a^2*x^2+a*b*x+2*a*x+x+a*c+b+1, x, (%o2));
(%o3) ...
(%i4) expand((%o3));
(%o4) ...
(%i5) subst(%o4, x, (%o1));
(%o5) ...
(%i6) expand((%o5));
(%o6) ...
(%i7) factor((%o6));
(%o7) (a^2*x^2+a*b*x+2*a*x+a*c+b+1)*(a^4*x^2+a^3*b*x+2*a^3*x+2*a^2*x+a^3*c+a^2*b+a*b+a^2+2*a+1)*(a^5*x^4+2*a^4*b*x^3+4*a^4*x^3+2*a^3*x^3+2*a^4*c*x^2+a^3*b^2*x^2+6*a^3*b*x^2+3*a^2*b*x^2+6*a^3*x^2+4*a^2*x^2+a*x^2+2*a^3*b*c*x+4*a^3*c*x+2*a^2*c*x+2*a^2*b^2*x+a*b^2*x+6*a^2*b*x+4*a*b*x+b*x+4*a^2*x
+2*a*x+a^3*c^2+2*a^2*b*c+a*b*c+2*a^2*c+c+a*b^2+b^2+2*a*b+b+a)
```
|
8
|
https://mathoverflow.net/users/2480
|
397434
| 164,063 |
https://mathoverflow.net/questions/397407
|
1
|
Let $X$ be a compact subset of the Euclidean space. Also let $Y$ be a separable Frechet space with the seminorms $\lVert \cdot \rVert\_n$'s.
Then let $C(X,Y)$ be the space of continuous mappings from $X$ into $Y$. It can be given the topology induced from the seminorms $\mid f \mid\_n := \sup\_{x \in X}\lVert f(x) \rVert\_n$ where $f \in C(X,Y)$.
My question is that, is $C(X,Y)$ Frechet and separable? If so, how to prove this?
Could anyone please help me?
|
https://mathoverflow.net/users/56524
|
Showing that $C(X,Y)$ is separable when $X$ is compact Hausdorff but $Y$ is just a separable Frechet space
|
If $d$ is a translation-invariant metric on $Y$ then
$$
D(f,g)=\sup\{d(f(x),g(x)):x\in X\}
$$
defines a translation-invariant metric on $C(X,Y)$, and $D$ is complete if $d$ is complete.
The topology induced by $D$ is the compact-open topology ([Arens, A topology for spaces of transformations, Ann. Math. 47 (1946), 480-495](https://doi.org/10.2307/1969087)).
Both $X$ and $Y$ are second-countable, hence the compact-open topology is separable ([E. A. Michael, On a theorem of Rudin and Klee, Proc. Amer. Math. Soc. 12 (1961), 921](https://doi.org/10.1090/S0002-9939-1961-0133108-4)).
|
2
|
https://mathoverflow.net/users/5903
|
397436
| 164,064 |
https://mathoverflow.net/questions/397435
|
36
|
We all know that the complex field structure $\langle\mathbb{C},+,\cdot,0,1\rangle$ is interpretable in the real field $\langle\mathbb{R},+,\cdot,0,1\rangle$, by encoding $a+bi$ with the real-number pair $(a,b)$. The corresponding complex field operations are expressible entirely within the real field.
Meanwhile, many mathematicians are surprised to learn that the converse is not true — we cannot define a copy of the real field inside the complex field. (Of course, the reals $\mathbb{R}$ are a *subfield* of $\mathbb{C}$, but this subfield is not a definable subset of $\mathbb{C}$, and the surprising fact is that there is no definable copy of $\mathbb{R}$ in $\mathbb{C}$.) Model theorists often prove this using the core ideas of stability theory, but I made a blog post last year providing a comparatively accessible argument:
* [The real numbers are not interpretable in the complex field](http://jdh.hamkins.org/the-real-numbers-are-not-interpretable-in-the-complex-field/).
The argument there makes use in part of the abundance of automorphisms of the complex field.
In a [comment](http://jdh.hamkins.org/the-real-numbers-are-not-interpretable-in-the-complex-field/#comment-10799) on that blog post, Ali Enayat pointed out that the argument therefore uses the axiom of choice, since one requires AC to get these automorphisms of the complex field. I pointed out in a [reply comment](http://jdh.hamkins.org/the-real-numbers-are-not-interpretable-in-the-complex-field/#comment-11161) that the conclusion can be made in ZF+DC, simply by going to a forcing extension, without adding reals, where the real numbers are well-orderable.
My question is whether one can eliminate all choice principles, getting it all the way down to ZF.
**Question.** Does ZF prove that the real field is not interpretable in the complex field?
I would find it incredible if the answer were negative, for then there would be a model of ZF in which the real number field was interpretable in its complex numbers.
|
https://mathoverflow.net/users/1946
|
Can one show that the real field is not interpretable in the complex field without the axiom of choice?
|
An interpretation of $(\mathbb R,+,\cdot)$ in $(\mathbb C,+,\cdot)$ in particular provides an interpretation of $\DeclareMathOperator\Th{Th}\Th(\mathbb R,+,\cdot)$ in $\Th(\mathbb C,+,\cdot)$. To see that the latter cannot exist in ZF:
* The completeness of the theory $\def\rcf{\mathrm{RCF}}\rcf$ of real-closed fields is an arithmetical ($\Pi\_2$) statement, and provable in ZFC, hence provable in ZF. Its axioms are clearly true in $(\mathbb R,+,\cdot)$, hence $\Th(\mathbb R,+,\cdot)=\rcf$.
* Similarly, ZF proves completeness of the theory $\def\acfo{\mathrm{ACF\_0}}\acfo$ of algebraically closed fields of characteristic $0$, hence $\Th(\mathbb C,+,\cdot)=\acfo$.
* The non-interpretability of $\rcf$ in $\acfo$ is again an arithmetical statement ($\Pi\_2$, using the completeness of $\acfo$), hence its provability in ZFC automatically implies its provability in ZF.
Of course, common proofs of some or all the results above may already work directly in ZF (e.g., if you take syntactic proofs of completeness, or if you make it work using countable models, etc.). My point is that it is not necessary to check the proofs, as the results transfer from ZFC to ZF automatically due to their low complexity.
|
34
|
https://mathoverflow.net/users/12705
|
397444
| 164,067 |
https://mathoverflow.net/questions/397445
|
7
|
Let $M$ be an $n$-dimensional topological closed manifold. Suppose $K$ is a compact subset of $M$ which is contractible in the sense that there exists a continuous map $F:K \times [0,1] \to M$ with $F(\cdot,0)=id$ and $F(\cdot, 1)=q \in M$.
Can we find an open neighborhood $U$ of $K$ such that $U$ is homeomorphic to $\mathbb R^n$?
|
https://mathoverflow.net/users/280895
|
Contractible set in a manifold
|
This is not true. Let K be one component (it doesn't matter which one) of the [Whitehead link](https://en.wikipedia.org/wiki/Whitehead_link), which has two components. Then K is contractible in the complement M of the other component. But K is not contained in a 3-ball in M. This can be seen in many ways; for instance if K were contained in a 3-ball, then each of its lifts to the universal cover of M would have trivial linking numbers. But you can readily draw the picture of the cover and see that some of those linking numbers are $\pm 1$.
|
9
|
https://mathoverflow.net/users/3460
|
397447
| 164,068 |
https://mathoverflow.net/questions/397424
|
2
|
I have the following question:
Let $u$ be a smooth subharmonic function on the unit disc $\mathbb{D}:=\left\{ z\in\mathbb{C}:\left|z\right|<1\right\} $.
Assume that $u=0$ on the boundary of $\mathbb{D}$ and
$$
\int\_{\mathbb{D}}\Delta u=1.
$$
Here $\Delta u$ is the Riesz measure of $u$.
Is there any chance to show that
$$
\Delta u\leq\frac{M}{1-\left|z\right|^{2}}dV
$$
as measures, for some positive constant $M$ (possibly an absolute constant, i.e. independent of $u$)?
Here $dV$ is the standard Lebesgue measure on $\mathbb{D}$.
Thanks for any suggestions.
|
https://mathoverflow.net/users/318721
|
A question on subharmonic functions on the unit disc
|
The first comment already shows that you cannot have $M$ independent of $u$.
But in fact you can construct a function smooth in the closed unit disk for
which $\Delta u$ has no uniform bound as $|z|\to 1$ whatsoever.
Just take
$$u\_1(z)=\sum\delta\_j\left(\log|z-a\_j|-\log|1-a\_jz|\right)=-\sum\delta\_jG(z,a\_j),$$
where $a\_j\in(0,1), a\_j\to 1$ is a fixed sequence, and $G$ is the Green function. Then smoothen near $a\_j$
by replacing this $u$ near $a\_j$ by a convolution with an infinitely smooth
radial $\phi\_j(z)$ with very small support. That these convolutions will match
away from the points $a\_j$ follows from the average property. By taking $\delta\_j$ and supports of $\phi\_j$ very small you ensure that the function is smooth in the closed disk, while $\Delta u$ tends to infinity arbitrarily fast.
Another way to do the same is to construct $u$ in the form $u(z)=f(\log|z|)$, where $f$ is a convex function. It is pretty clear that a convex function on $(-\infty,0]$ required properties exists,
and $\Delta u(z)=f''(\log|z|)$.
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2
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https://mathoverflow.net/users/25510
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397449
| 164,070 |
https://mathoverflow.net/questions/149867
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10
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Let $M$ be a (Hausdorff) smooth compact manifold and $G$ a Lie group acting smoothly on $M$. If $G$ is compact then, by [Mostow-Palais theorem](http://en.wikipedia.org/wiki/Mostow-Palais_theorem), there exists an equivariant smooth embedding $M\to {\mathbb R}^n$ (for some $n$) sending $G$ to a subgroup of $O(n)$. I am interested in generalizations of this theorem in the case when the group $G$ is noncompact. In this setting, linear representations of $G$ should be replaced with projective representations, as it is done in the context of algebraic group actions on algebraic varieties:
*Question:*
1. Suppose that $G$ is a connected noncompact linear Lie group acting smoothly on a smooth compact manifold $M$. Is there a smooth equivariant embedding $M\to {\mathbb R}P^{n-1}$ (for some $n$) sending $G$ to a subgroup of $PGL(n, {\mathbb R})$? (If it helps, let's assume that $G$ is semisimple; I would be also happy assuming that $M$ and $G$-action on $M$ are real-analytic.)
2. Is there a topological analogue of such equivariant embedding theorem, where $M$ is a *reasonable* compact topological space (metrizable, of finite covering dimension) and $G$ is still a connected linear Lie group?
The only results in this direction I am aware of are in algebro-geometric setting and my guess is that there is a similar construction in the context of smooth manifolds, involving choice of a $G$-vector bundle $\xi: E\to M$ and a projective embedding of $M$ using smooth sections of $\xi$. I checked math.sci.net for papers referring to the ones by Mostow and Palais and got nothing interesting; google search also returned nothing of value.
Addendum: I realized that even an infinitesimal version of this question is unclear. Here is the infinitesimal question in holomorphic setting:
Question 3. Let ${\mathfrak g}$ be a finite-dimensional (complex) Lie subalgebra of the Lie algebra of holomorphic vector fields on $B^n$, the open complex $n$-ball. Is there (for some $m$) a holomorphic embedding $f: B^n \to {\mathbb C}P^{m-1}$ so that $f\_\*({\mathfrak g})$ is the restriction (to the image of $f$) of a subalgebra of the Lie algebra of linear vector fields $psl(m, {\mathbb C})$?
Somehow, this version seems much more plausible than the similar question about $C^\infty$ vector fields.
Note that, because of lack of compactness, the standard argument of constructing an embedding using sections of a suitable line bundle does not work. (Maybe one can impose some restrictions on holomorphic sections which would be preserved by the action of ${\mathfrak g}$ and would imply finite-dimensionality and separation of points.)
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https://mathoverflow.net/users/21684
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Is there an analogue of Mostow-Palais equivariant embedding theorem for noncompact groups
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There cannot be such an equivariant embedding in general.
For a non-compact semisimple Lie group $G$ one can always find a cocompact lattice $\Gamma<G$, so $M=G/\Gamma$ will be a compact real analytic manifold on which $G$ acts real analytically.
However, any homomorphism $G\to \text{PGL}\_n(\mathbb{R})$ is algebraic, so the stabilizers of the corresponding action of $G$ on $\mathbb{P}^{n-1}(\mathbb{R})$ would be algebraic, thus you could not find a non-constant equivariant map $M\to \mathbb{P}^{n-1}(\mathbb{R})$, as $\Gamma$ is Zariski dense in $G$.
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2
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https://mathoverflow.net/users/89334
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397458
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