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https://mathoverflow.net/questions/417195 | 2 | Let $\{X\_i \}\_{i \in \mathbb{N}}$ be a sequence of i.i.d. random variables satisfying $\mathbb{E} X\_1 = 0$ and $\mathbb{E} X\_1 ^2 < \infty$. Assume that $\{S\_n \}\_{n \in \mathbb{N}}$ is a non-lattice random walk, where $S\_n = X\_1+...+ X\_n$. I am wondering whether there is a 'local large deviation theorem' running along these lines:
**Theorem.** *Let $0 < a < b$. Then as $n \to \infty$ uniformly for $r \in [an, bn]$*
$$
\mathbb{P} \{ S\_n \in [r,r+1] \} \sim \int\limits\_{r} ^{r+1} \kappa \_n (x) dx,
$$
*where $\kappa \_n$ is a function given explicitly.*
Similar results are available for $r = o(n)$ in [1], at least for absolutely continuous random variables. This question can also be formulated in terms of the associated convolution operator. Let $a \in L^1$, $a \geq 0$. Define the operator $L$ on some function space by
$$
Lu (x) = \int\limits \_{\mathbb{R}} u(y)a(x-y) dy
$$
and set $u\_n = L ^n \delta \_0$ (or alternatively $u\_n = \frac{1}{2 \varepsilon}L ^n \mathbf{1} \_\varepsilon$, where $\mathbf{1} \_\varepsilon$ is the indicator of $[-\varepsilon, \varepsilon]$ for a small $\varepsilon > 0$). What can we say about
$$
\int\limits \_{r} ^{r+1} u\_n (x)dx.
$$
for $r \in [an, bn]$, possibly under some additional assumptions on $a$?
[1]: Richter, W. (1957). Local limit theorems for large deviations. Theory of Probability & Its Applications, 2(2), 206-220.
| https://mathoverflow.net/users/41071 | Equivalent of a local limit theorem in the large deviation region and asymptotics of a convolution operator | For general classes of bounded pdf's of $X\_1$, including pdf's with exponential-like, super-exponential, and sub-exponential tails, your Theorem follows from the considerations in [Sections 2.1 and 2.2](https://www.sciencedirect.com/science/article/abs/pii/009630039490197X), with
$$\kappa\_n(x)=p\_{S\_n}(r)e^{-s\_0(x-r)},$$
where $p\_{S\_n}$ is the pdf of $S\_n$,
$$s\_0:=\sup\{s\ge0\colon m(s)<r/n\},$$
and
$$m(s):=\frac{EX\_1e^{sX\_1}}{Ee^{sX\_1}}.$$
As can also be seen from those considerations, the asymptotics of $p\_{S\_n}(r)$ will very much depend on how heavy the right tail of the distribution of $X\_1$ is.
| 1 | https://mathoverflow.net/users/36721 | 417217 | 169,969 |
https://mathoverflow.net/questions/417214 | 9 | I'm reading Thurston's article "Shapes of polyhedra and triangulations of the sphere." In the introduction he claims the following:
>
> "${}^{(1)}$There are procedures to refine and modify any triangulation of a surface until every vertex has either 5, 6 or 7 triangles around it, or with more effort, ${}^{(2)}$so that there are only 5 or 6 triangles if the surface has positive Euler characteristic, only 6 triangles if the surface has zero Euler characteristic, or only 6 or 7 triangles if the surface has negative Euler characteristic..."
>
>
>
I divide the quote into two claims. I proved the first claim, but I'm stuck with the second one. There must be a global argument I'm not seeing. Does anyone know how to do it?
| https://mathoverflow.net/users/148805 | Refining a triangulation | Suppose that $S$ is a closed, connected surface with negative Euler characteristic. Suppose that $T$ is a triangulation of $S$.
Define "refine" to mean "replace each triangle by four triangles" (so that edge midpoints become vertices of valence six). This does not improve any of the vertices of "concentrated positive or negative curvature", but it does isolate them.
Suppose that there are no vertices of degree five or lower. Refine as above to isolate all vertices of degree greater than six. Suppose that the vertex $v$ has degree eight or higher. We "split" at $v$ - we choose two edges $e$ and $e'$ at $v$, cut $T$ along $e \cup e'$ and insert a pair of triangles. This increases the number of triangles by two, the number of vertices by one, and the number of edges by three. The new vertices $u$ and $u'$ have degree less than that of $v$. Also, two old vertices of degree six will now have degree seven. Repeat.
---
As a bit of motivation: the original triangulation gives a metric on $S$. Refinement has the effect of scaling the metric up, which "brings the curvature down". Cutting and inserting triangles disperses concentrated negative curvature (at a vertex) into adjacent (almost) flat regions.
---
In general, if there are vertices of degree less than six, we refine and then pair them with vertices of higher degree, and (carefully!) cancel curvature.
In the case of Euler characteristic zero, after pairing in this way, only vertices of degree six remain.
In the case of positive Euler characteristic, we have some vertices of degree five (either six or twelve) left at the end.
| 7 | https://mathoverflow.net/users/1650 | 417219 | 169,970 |
https://mathoverflow.net/questions/417101 | 17 | Hurwitz's theorem says that the only division composition algebras over the real numbers $\mathbb{R}$ are the real numbers themselves $\mathbb{R}$, the complex numbers $\mathbb{C}$, the quaternions $\mathbb{H}$, and the octonions $\mathbb{O}$. However, in pure constructive mathematics without any weak axiom of choice, the notion of the set of real numbers bifurcates into multiple incompatible notions, such as the Cauchy real numbers $\mathbb{R}\_C$, the Dedekind real numbers $\mathbb{R}\_D$, the Escardó-Simpson real numbers $\mathbb{R}\_E$, and the MacNeille real numbers $\mathbb{R}\_M$, and I'd imagine the same for $\mathbb{C}$, $\mathbb{H}$, and $\mathbb{O}$ (i.e. Cauchy complex numbers, Dedekind quaternions, etc). For which of these sets of real numbers, complex numbers, quaternions, and octanions, if any at all, does Hurwitz's theorem still hold true?
---
Edit: Swapped out "normed division algebra" for "division composition algebra" for the following reason:
Classically, Hurwitz's theorem is also expressed in terms of finite-dimensional normed division algebras over that set of real numbers. However, finite-dimensional normed divison algebras and division composition algebras over the real numbers do not coincide in constructive mathematics because there are multiple different types of real numbers in constructive mathematics.
In a division composition algebra, the norm $\lvert a \rvert := \langle a, a\rangle$ has a codomain of the ground field.
However, that is not necessarily true of finite-dimensional normed division algebras over some field of real numbers $\mathbb{R}\_X$ in constructive mathematics, because the notion of "multiplicative norm" bifurcates into multiple definitions based upon which set of real numbers $\mathbb{R}\_Y$ is used as the codomain of the norm. It is perhaps more appropriate to call them finite-dimensional $\mathbb{R}\_Y$-normed division $\mathbb{R}\_X$-algebras. One can have a finite-dimensional normed division $\mathbb{R}\_C$-algebra with a norm valued in $\mathbb{R}\_D$, where $\mathbb{R}\_C$ are the Cauchy real numbers and $\mathbb{R}\_D$ are the Dedekind real numbers, but such a finite-dimensional normed division algebra is not a composition algebra, and is not covered under Hurwitz's theorem.
Thus, for sets of real numbers $\mathbb{R}\_X$ and $\mathbb{R}\_Y$, every division composition $\mathbb{R}\_X$-algebra for some set of real numbers $\mathbb{R}\_X$ is a finite-dimensional $\mathbb{R}\_X$-normed division $\mathbb{R}\_X$-algebra, but a finite-dimensional $\mathbb{R}\_Y$-normed division $\mathbb{R}\_X$-algebra is only a division composition $\mathbb{R}\_X$-algebra if $\mathbb{R}\_Y$ is isomorphic to $\mathbb{R}\_X$.
| https://mathoverflow.net/users/nan | Is Hurwitz's theorem true in constructive mathematics? | There is a weakening of Hurwitz's theorem that is true constructively, with essentially the same proof:
Let $A$ be a division composition algebra. Then any chain of proper subalgebras $\mathbb{R} = A\_0 \subsetneq A\_1 \subsetneq \cdots \subsetneq A\_n = A$ has length $n \leq 3$ (where "proper" means "contains an element with positive distance from the previous algebra).
We can also show that in general, any inclusion of subalgebras generated by adding one element must come from (a quotient of) the [General Cayley-Dickson construction](https://en.wikipedia.org/wiki/Cayley%E2%80%93Dickson_construction#General_Cayley%E2%80%93Dickson_construction) with parameter $\cdot \gamma$, $\gamma \leq 0$.
| 3 | https://mathoverflow.net/users/100508 | 417226 | 169,972 |
https://mathoverflow.net/questions/413983 | 2 | It is well-known that an elliptic curve $E$ that has a point of order $2$ and is represented as $E=[0,a,0,b,0]$ has a *$2$-isogenous* curve $E^\prime=[0,-2a,0,a^2-4b,0]$, see e.g. p. 507 in
* A. Dujella, [*Number Theory*](https://web.math.pmf.unizg.hr/%7Eduje/numbertheorybook.html), University of Zagreb, Školska knjiga, Zagreb, 2021, ISBN: 978-953-0-30897-8, 621 pp.
>
> **Question:** Does a similar simple formula for $E^\prime$ exist for a curve with $2$-torsion expressed in the *Tate normal form* $E=[1-c,-b,-b,0,0]$? We may assume that all torsion points on $E$ are known.
>
>
>
**Rationale:** I am working on $\mathbb{Z}/16\mathbb{Z}$ curves over cubic fields. The curves are generated by the formulas on p. 584 in
* D. Jeon, C. H. Kim, and Y. Lee, *Families of elliptic curves over cubic number fields with prescribed torsion subgroups*, Mathematics of Computation, Volume 80, Number 273, January 2011, pp. 579–591, doi:[10.1090/S0025-5718-10-02369-0](https://doi.org/10.1090/S0025-5718-10-02369-0).
For $t=\frac{4}{7}$, Magma struggles to calculate the last generator.
The `DescentInformation` is very limited over number fields in Magma ($2$-descent only). `IsogenousCurves` and `IsIsogenous` are not implemented at all.
Sometimes, feeding a $2$-isogenous curve or a different model of the curve helps.
| https://mathoverflow.net/users/95511 | $2$-isogenous to a curve in the Tate normal form | John Cremona has some explicit code for calculating the 2-torsion points of curves in the general Weierstrass [a1,a2,a3,a4,a6] format, and part of that formula is finding rational roots to the cubic equation
(1) $ P(x,[W]) = 4(x^3+e\_{a2}x^2+e\_{a4}x+e\_{a6})+(e\_{a1}x+e\_{a3})^2)$
where [a1,a2,a3,a4,a6] are from the Weierstrass form W.
If you stick in a curve of the form [0,a,0,b,0] then (1) becomes
(2) $4x^3 + 4ax^2 + 4bx = 0$ (to find the 2-torsion points)
and it is easy to see that $x=0$ is one torsion point. The other 2 are the quadratic roots of $x^2 + 4ax + b$ which need to be rational, if three 2-torsion points are to be found.
However if we use the Tate form above of $[1-c,-b,-b,0,0]$ and stick it into (1) we derive
(2) $P(x,b,c) = 4x^3 + (c^2 - 2c + (-4b + 1))x^2 + (2bc - 2b)x + b^2$
and we want rational roots of this cubic for x.
Maxima tells us that the 3 roots of this cubic in [b,c] is
$x\_1=\left({{-1}\over{2}}-{{\sqrt{3}\,i}\over{2}}\right)\,\left({{b\,
\sqrt{-b\,\left(c^4-3\,c^3+\left(3-8\,b\right)\,c^2+\left(-20\,b-1
\right)\,c+16\,b^2+b\right)}}\over{8\,3^{{{3}\over{2}}}}}+{{-{{3\,b^
2}\over{4}}-{{\left(4\,b-c^2+2\,c-1\right)\,\left(\left(c-1\right)\,
b\right)}\over{8}}}\over{6}}+{{\left(4\,b-c^2+2\,c-1\right)^3}\over{
1728}}\right)^{{{1}\over{3}}}-{{\left({{\sqrt{3}\,i}\over{2}}+{{-1
}\over{2}}\right)\,\left({{\left(c-1\right)\,b}\over{6}}-{{\left(4\,
b-c^2+2\,c-1\right)^2}\over{144}}\right)}\over{\left({{b\,\sqrt{-b\,
\left(c^4-3\,c^3+\left(3-8\,b\right)\,c^2+\left(-20\,b-1\right)\,c+
16\,b^2+b\right)}}\over{8\,3^{{{3}\over{2}}}}}+{{-{{3\,b^2}\over{4}}
-{{\left(4\,b-c^2+2\,c-1\right)\,\left(\left(c-1\right)\,b\right)
}\over{8}}}\over{6}}+{{\left(4\,b-c^2+2\,c-1\right)^3}\over{1728}}
\right)^{{{1}\over{3}}}}}+{{4\,b-c^2+2\,c-1}\over{12}}$
$x\_2 = \left({{\sqrt{3}\,i}\over{2}}+{{-1}\over{2}}\right)\,\left({{b\,
\sqrt{-b\,\left(c^4-3\,c^3+\left(3-8\,b\right)\,c^2+\left(-20\,b-1
\right)\,c+16\,b^2+b\right)}}\over{8\,3^{{{3}\over{2}}}}}+{{-{{3\,b^
2}\over{4}}-{{\left(4\,b-c^2+2\,c-1\right)\,\left(\left(c-1\right)\,
b\right)}\over{8}}}\over{6}}+{{\left(4\,b-c^2+2\,c-1\right)^3}\over{
1728}}\right)^{{{1}\over{3}}}-{{\left({{-1}\over{2}}-{{\sqrt{3}\,i
}\over{2}}\right)\,\left({{\left(c-1\right)\,b}\over{6}}-{{\left(4\,
b-c^2+2\,c-1\right)^2}\over{144}}\right)}\over{\left({{b\,\sqrt{-b\,
\left(c^4-3\,c^3+\left(3-8\,b\right)\,c^2+\left(-20\,b-1\right)\,c+
16\,b^2+b\right)}}\over{8\,3^{{{3}\over{2}}}}}+{{-{{3\,b^2}\over{4}}
-{{\left(4\,b-c^2+2\,c-1\right)\,\left(\left(c-1\right)\,b\right)
}\over{8}}}\over{6}}+{{\left(4\,b-c^2+2\,c-1\right)^3}\over{1728}}
\right)^{{{1}\over{3}}}}}+{{4\,b-c^2+2\,c-1}\over{12}}$
and
$ x\_3 = \left({{b\,\sqrt{-b\,\left(c^4-3\,c^3+\left(3-8\,b\right)\,c^2+
\left(-20\,b-1\right)\,c+16\,b^2+b\right)}}\over{8\,3^{{{3}\over{2}}
}}}+{{-{{3\,b^2}\over{4}}-{{\left(4\,b-c^2+2\,c-1\right)\,\left(
\left(c-1\right)\,b\right)}\over{8}}}\over{6}}+{{\left(4\,b-c^2+2\,c
-1\right)^3}\over{1728}}\right)^{{{1}\over{3}}}-{{{{\left(c-1\right)
\,b}\over{6}}-{{\left(4\,b-c^2+2\,c-1\right)^2}\over{144}}}\over{
\left({{b\,\sqrt{-b\,\left(c^4-3\,c^3+\left(3-8\,b\right)\,c^2+
\left(-20\,b-1\right)\,c+16\,b^2+b\right)}}\over{8\,3^{{{3}\over{2}}
}}}+{{-{{3\,b^2}\over{4}}-{{\left(4\,b-c^2+2\,c-1\right)\,\left(
\left(c-1\right)\,b\right)}\over{8}}}\over{6}}+{{\left(4\,b-c^2+2\,c
-1\right)^3}\over{1728}}\right)^{{{1}\over{3}}}}}+{{4\,b-c^2+2\,c-1
}\over{12}}$
I really don't see a simple form here for rational roots. unless b and c take very specific values so that one of $x\_1$, $x\_2$ or $x\_3$ become rational.
I did also consider the elliptic curve invariants $c\_4$ and $c\_6$ equivalency, but it means moving from a sextic to a quartic equation for the two variables (b,c --> a,b)
(continuing) after reading up a bit more on 2-isogenies and Magma defining fields, the following Magma code and results is your answer, no, there is no simple elliptic curve:
>
> F<a,b,c>:=FunctionField(Rationals(),3);
>
>
> E:=EllipticCurve([1-c,-b,-b,0,0]);
>
>
> E;
>
>
> Elliptic Curve defined by y^2 + (-c + 1)*x*y - b*y = x^3 - b*x^2 over Multivariate rational function field of rank 3 over Rational Field
>
>
> E1, f := IsogenyFromKernel(E, DivisionPolynomial(E, 2));
>
>
> E1;
>
>
> Elliptic Curve defined by y^2 + (-c + 1)*x*y - b*y = x^3 - b*x^2 + (-5*b^2 + 5/2*b*c^2 + 5/2*b*c - 5*b - 5/16*c^4 + 5/4*c^3 - 15/8*c^2 + 5/4*c - 5/16)*x + (-3*b^3 + 9/4*b^2*c^2 + 7/2*b^2*c + 10*b^2 - 9/16*b*c^4 + 1/4*b*c^3 + 21/8*b*c^2 - 15/4*b*c + 23/16*b + 3/64*c^6 - 9/32*c^5 + 45/64*c^4 - 15/16*c^3 + 45/64*c^2 - 9/32*c + 3/64) over Multivariate rational function field of rank 3 over Rational Field
>
>
> f;
>
>
> Elliptic curve isogeny from: CrvEll: E to CrvEll: E1
>
>
> taking (x : y : 1) to ((x^4 + (-b + 1/4*c^2 - 1/2*c + 1/4)*x^3 + (b^2 - 1/2*b*c^2 + 1/2*b + 1/16*c^4 - 1/4*c^3 + 3/8*c^2 - 1/4*c + 1/16)*x^2 + (-1/2*b^2*c - 3/2*b^2 + 1/8*b*c^3 - 3/8*b*c^2 + 3/8*b*c - 1/8*b)*x + (3/4*b^3 + 1/16*b^2*c^2 - 1/8*b^2*c + 1/16*b^2)) / (x^3 + (-b + 1/4*c^2 - 1/2*c + 1/4)*x^2 + (1/2*b*c - 1/2*b)*x + 1/4*b^2) : (x^6*y + (-2*b + 1/2*c^2 - c + 1/2)*x^5*y + (b^2*c - b^2 - 1/2*b*c^3 + 3/2*b*c - b + 1/16*c^5 - 5/16*c^4 + 5/8*c^3 - 5/8*c^2 + 5/16*c - 1/16)*x^5 + (5/2*b*c - 5/2*b)*x^4*y + (-1/2*b^3*c + b^3 + 3/8*b^2*c^3 - 11/8*b^2*c^2 - 5/2*b^2*c + 7/2*b^2 - 3/32*b*c^5 + 1/2*b*c^4 - 17/16*b*c^3 + 9/8*b*c^2 - 19/32*b*c + 1/8*b + 1/128*c^7 - 7/128*c^6 + 21/128*c^5 - 35/128*c^4 + 35/128*c^3 - 21/128*c^2 + 7/128*c - 1/128)*x^4 + 5*b^2*x^3*y + (1/2*b^3*c^2 + 9/4*b^3*c - 5*b^3 - 1/4*b^2*c^4 + 11/16*b^2*c^3 - 9/16*b^2*c^2 + 1/16*b^2*c + 1/16*b^2 + 1/32*b*c^6 - 3/16*b*c^5 + 15/32*b*c^4 - 5/8*b*c^3 + 15/32*b*c^2 - 3/16*b*c + 1/32*b)*x^3 - 5*b^3*x^2*y + (-5/4*b^4*c + 7/2*b^4 + 1/8*b^3*c^3 - 9/16*b^3*c^2 + 3/4*b^3*c - 5/16*b^3 + 3/64*b^2*c^5 - 15/64*b^2*c^4 + 15/32*b^2*c^3 - 15/32*b^2*c^2 + 15/64*b^2*c - 3/64*b^2)*x^2 + (2*b^4 - 1/2*b^3*c^2 + 1/2*b^3*c)*x*y + (-b^5 + 5/8*b^4*c^2 - 7/8*b^4*c + 1/4*b^4 + 1/32*b^3*c^4 - 1/8*b^3*c^3 + 3/16*b^3*c^2 - 1/8*b^3*c + 1/32*b^3)*x - 1/2*b^4*c*y + (11/32*b^5*c - 1/16*b^5 + 1/128*b^4*c^3 - 3/128*b^4*c^2 + 3/128*b^4*c - 1/128*b^4)) / (x^6 + (-2*b + 1/2*c^2 - c + 1/2)*x^5 + (b^2 - 1/2*b*c^2 + 2*b*c - 3/2*b + 1/16*c^4 - 1/4*c^3 + 3/8*c^2 - 1/4*c + 1/16)*x^4 + (-b^2*c + 3/2*b^2 + 1/4*b*c^3 - 3/4*b*c^2 + 3/4*b*c - 1/4*b)*x^3 + (-1/2*b^3 + 3/8*b^2*c^2 - 3/4*b^2*c + 3/8*b^2)*x^2 + (1/4*b^3*c - 1/4\*b^3)*x + 1/16*b^4) : 1)
>
>
>
~
~
| 2 | https://mathoverflow.net/users/6046 | 417236 | 169,974 |
https://mathoverflow.net/questions/417246 | 2 | I am working on symplectic geometry and I have some questions about a degeneration of $\mathbb{P}^2$.
Question: Can we obtain the moment polytope (or the polytope associated with the anti-canonical divisor) of $\mathbb{P}(a^2,b^2,c^2)$ as a Newton-Okounkov body of $\mathbb{P}^2$? (Here $(a,b,c)$ is a Markov triple satisfying $a^2 + b^2 + c^2 = 3abc$.)
The motivation of this question is as follows: It is known by Hacking-Prokhorov that the weighted projective space $\mathbb{P}(a^2,b^2,c^2)$ admits a $\mathbb{Q}$-Gorenstein smoothing with a generic fiber $\mathbb{P}^2$ where $(a,b,c)$ is a Markov triple. I want to understand a $\mathbb{Q}$-Gorenstein smoothing of $\mathbb{P}(a^2,b^2,c^2)$ with generic fiber $\mathbb{P}^2$ as a toric degeneration of $\mathbb{P}^2$ with the central fiber $\mathbb{P}(a^2,b^2,c^2)$. Dave Anderson proved that if a polytope $P$ can be realized as a Newton-Okounkov body with a certain condition (finitely generatedness of a semigroup), then there exists a toric degeneration of $X$ with the central fiber $X\_0$ where $X\_0$ is a toric variety whose normalization is a normal toric variety associated with the polytope $P$.
I would really appreciate for any comment.
Thank you!
| https://mathoverflow.net/users/11705 | Markov triples and Newton-Okounkov bodies of $\mathbb{P}^2$ | The polytopes you are interested in are related by sequences of combinatorial mutations, as described [here](https://arxiv.org/abs/1212.1785) and [here](https://arxiv.org/abs/1302.1152). If two polytopes $P\_1$ and $P\_2$ are related by a combinatorial mutation, then there is a construction due to Ilten ([here](https://arxiv.org/abs/1205.4664)) of a flat family $\pi\colon\mathcal X\to \mathbb P^1$ such that $\pi^{-1}(0)=X\_{P\_1}$ is the toric variety defined by the spanning fan of $P\_1$ and $\pi^{-1}(\infty)=X\_{P\_2}$ is the toric variety defined by the spanning fan of $P\_2$. You can interpret the toric degeneration of $\mathbb P^2$ to $\mathbb P (a^2,b^2,c^2)$ as a following a sequence of 1-parameter families by moving along the relevant edges of the Markov tree.
| 3 | https://mathoverflow.net/users/104695 | 417255 | 169,981 |
https://mathoverflow.net/questions/417173 | 6 | I'm reading Milnor's [notes](http://www.math.stonybrook.edu/%7Ejack/DYNOTES/) on dynamical systems and in Lecture 3 he gives an example of an attractor with no natural measure, which he attributes to Mañé. I can find no other reference in which this example is discussed; no paper by Ricardo Mañé, no books or papers in which this example is mentioned. Milnor also states that an earlier paper by Zakharevich has a similar example, but again no reference is given, and I cannot find this paper anywhere. Has this or any other counterexample been published anywhere else? Any references would be highly appreciated.
| https://mathoverflow.net/users/477928 | Mañé's example of an attractor with no natural measure | **Q:** Has this or any other counterexample been published anywhere else?
**A:** In [A continuous Bowen-Mañé type phenomenon](https://www.aimsciences.org/journals/displayArticlesnew.jsp?paperID=3046) examples of vector fields without physical measure for certain parameters are discussed under the name "*Bowen-Mañé type phenomena*". No primary reference to either scientist is given.
The Bowen-Mañé example is also called "*Bowen's eye*" in the literature, see [arXiv: 2010.08945](https://arxiv.org/abs/2010.08945) and [arXiv:1609.05356](https://arxiv.org/abs/1609.05356). The attribution to Bowen is due to Takens, who writes in a [1994 paper:](https://www.emis.de/journals/em/docs/boletim/vol251/v25-1-a6-1994.pdf)
>
> "I attribute this example to Bowen: although he never published it, I
> learned this example through a paper by Ruelle who referred to Bowen
> (we have no reference: even Ruelle could not localize that paper)."
>
>
>
[[source]](https://www.stefanoluzzatto.net/uploads/2/6/7/1/26719559/2017-09_tunisia.pdf)
| 3 | https://mathoverflow.net/users/11260 | 417269 | 169,985 |
https://mathoverflow.net/questions/417261 | 1 | Let $X$ be a vector space equipped with a norm $p$ and a seminorm $q$. Denote the completion of $X$ with respect to $p$ with $X\_p$ and with respect to $p+q$ by $X\_{p+q}$. Then the induced map $\iota : X\_{p+q} \to X\_p$ is well-defined and continuous but not necessarily injective as can be seen in analogy to [this answer](https://math.stackexchange.com/a/901717) to [Kernel of the Extension of a Bounded Linear Operator](https://math.stackexchange.com/questions/901579/kernel-of-the-extension-of-a-bounded-linear-operator) by taking $X = l^2$,
\begin{equation}
p(x) = \sqrt{\sum\_{n = 1}^\infty \left( \frac{x\_n - x\_{n+1}}{n} \right)^2}, \qquad
q(x) = \sqrt{\sum\_{n = 1}^\infty \left( \frac{x\_n}{n} \right)^2} \, .
\end{equation}
Then setting $x^m\_n = m/(m+n)$ we have $\lim\_{m \to \infty} p(x^m) = 0$, $x^m$ being Cauchy in $q$ and $\lim\_{n \to \infty} q(x^m) = \sqrt{\pi^2/6}$ such that the equivalence class of $x^m$ in $X\_{p+q}$ lies in the kernel of $\iota$ while being nonzero in $X\_{p+q}$.
On the other hand we can take the Schwartz space $X = \mathcal{S}(\mathbb{R})$ and let $p = \lVert \cdot \rVert\_{L^1}$ and $q = \lVert \cdot \rVert\_{L^2}$.
Then $\iota$ is certainly injective because any sequence $f\_n$ in $\mathcal{S}(\mathbb{R})$ with $\lim\_{n \to \infty} \lVert f\_n \rVert\_{L^1} = 0$ has a subsequence converging to zero almost everywhere such that if $f\_n$ is Cauchy in $L^2 (\mathbb{R})$, the corresponding limit $f$ in $L^2 (\mathbb{R})$ (which exists by completeness) has to be the zero function. Thus $\lim\_{n \to \infty} \lVert f\_n \rVert\_{L^2} = 0$ as well proving that the equivalence class of $f\_n$ is zero in $X\_{p+q}$.
In the latter example we have much more structure: $X$ is a reflexive, complete and metrisable nuclear space and $p$ and $q$ are continuous norms on $X$.
But I suspect that these conditions are not sufficient.
What are some necessary and sufficient conditions for $\iota$ to be injective?
| https://mathoverflow.net/users/18936 | When is a natural map between completions injective? | A simple sufficient condition for two norms $p\le r$ on $X$ to induce an *injective* continuous linear map $i:X\_r\to X\_p$ between the completions is that the unit ball $B\_r=\{x\in X: r(x)\le 1\}$ is $p$-closed.
Indeed, for $x\in X\_r$ with $i(x)=0$ choose a sequence $x\_n\in X$ with $x\_n\to x$ in $X\_r$. This sequence is $r$-Cauchy so that, for every $\varepsilon>0$, there is $n\_\varepsilon\in \mathbb N$ with $x\_n-x\_m\in\varepsilon B\_r$ for all $m,n\ge n\_\varepsilon$. The continuity of $i$ and $i|\_X=id\_X$ yield $x\_m=i(x\_m)\to i(x)=0$ in $X\_p$ and hence in $(X,p)$. For every $n\ge n\_\varepsilon$, this implies $$x\_n=\text{$p$-}\lim\limits\_{m\to\infty} x\_n-x\_m+x\_m\in \overline{\varepsilon B\_r}^p= \varepsilon B\_r.$$ This shows that $x\_n$ converges to $0$ in $(X,r)$ so that $x=0$.
---
You can relax the assumption to $\overline{B\_r}^p \subseteq cB\_r$ for some constant. Moreover, this principle can be generalized to general uniform spaces. In one form or another you find it in most text books about locally convex spaces, for instance in Grothendieck's *Topological Vector Spaces*.
| 3 | https://mathoverflow.net/users/21051 | 417278 | 169,990 |
https://mathoverflow.net/questions/417276 | 5 | What's a good example of a simple algebra over a field of characteristic $0$ which has a non-inner derivation but also has the invariant basis number property (IBN)?
I'm under the impression that when an algebra is simple Artinian, all derivations are inner. If I'm missing something subtle about what can happen please let me know, because that would be a lot easier to use than what follows.
Here is what I've been pursuing. In this article,
>
> K. Goodearl. Simple self-injective rings need not be Artinian. (1974)
>
>
>
Two examples are given of self-injective, simple, non-Artinian rings that have the IBN property. Roughly recapitulating them, they are
1. The coordinatizing ring for an irreducible continuous geometry $L$ "in case
$\infty$" (i.e. fails the DCC) as constructed in von Neumann's Examples of Continuous Geometries.
2. The maximal right ring of quotients of an $AW^\ast$-factor type $\mathrm{II}\_f$
I'm new to non-inner derivations and AW\* algebras and what the coordinatizing ring for a continuous lattice looks like, so it's tough to make progress. If there is a way to demonstrate such a non-inner derivation, or alternatively a reason that neither of these algebras has such derivations, either one would settle this question.
| https://mathoverflow.net/users/19965 | Finding non-inner derivations of simple $\mathbb Q$-algebras | Let $K$ be any field. I will give a simple $K$-algebra with IBN and a noninner derivation. My example will be a contracted monoid algebra. These have IBN by @PaceNielsen's nice answer [here](https://mathoverflow.net/questions/248345/can-the-trivial-module-be-stably-free-for-a-monoid-ring).
Let $X$ be an infinite set and let $M$ be the monoid with zero given by the presentation $$\langle X\cup X^\*\mid x^\*y=\delta\_{x,y}, x,y\in X\rangle$$ where $X^\*$ is a bijective copy of $X$. Let $R=K\_0M$ be the contracted monoid algebra. So it has $K$-basis $M\setminus \{0\}$ and the product extends that of $M$, where we identify the zero of $M$ with the zero of $K$. This is the Leavitt path algebra of an infinite bouquet of circles and is well known to be simple. This was first proved by Douglas Munn (W. D. Munn. *Simple contracted semigroup algebras*. In Proceedings of the Conference on Semigroups
in Honor of Alfred H. Clifford (Tulane Univ., New Orleans, La., 1978), pages 35–43. Tulane Univ.,
New Orleans, La., 1979) to the best of my knowledge.
I claim that it has a noninner derivation. This [paper](https://arxiv.org/pdf/1509.05075.pdf) studies outer derivations of Leavitt path algebras and the derivation I am using is from there, although technically that paper does not allow digraphs with infinite out-degree.
It is easy to check that every nonzero element of $M$ can be written uniquely in the form $pq^\*$ where $p,q$ are words in $X$ (possibly empty) and you extend $\ast$ to words in the obvious way, so that $\*$ is an involution on $M$.
Now define $d\colon R\to R$ on the basis $M\setminus \{0\}$ as follows. Fix $x\in X$ and put $$d\_x(pq^\*) = (|p|\_x-|q|\_x)pq^\*.$$ Here $|w|\_x$ is the number of occurrences of the letter $x$ in $w$.
It is easy to check that $d\_x$ is a derivation. It is also not too hard to check that it is not inner using that $d\_x(x)=x$ and you cannot find $m\in R$ with $xm-mx=x$.
**Details added (update).**
I found a little time to add in some details. First of all the presentation of $M$ is a complete rewriting system (it is length reducing and there are no overlaps of rules) and the nonzero reduced elements are the elements $pq^\*$ with $p,q$ words in $X$.
To see that $R=K\_0M$ is simple, let $I$ be a nonzero ideal and let $$0\neq a=\sum\_{m\in M}c\_mm\in I.$$
We shall use the following *observation*. Use $|w|$ for the length of a word $w$ over $X$. If $p,q,w,a,b$ are words in $X$ and $0\neq pq^\*w=sb^\*w$ with $|b|,|q|\geq |w|$, then $pq^\*=sb^\*$. Indeed, we must have $q=wz$ and $b=wy$ for the product to be nonzero and then $pq^\*w = pz^\*$ and $sb^\*w = sy^\*$, whence $p=s$ and $z=y$, i.e., $q=b$.
First suppose that $supp(a)$ contains $1$. Since $X$ is infinite, we can find a letter $x\in X$ such that $x$ appears in no $p,q$ with $pq^\*\in supp(a)$. Then $x^\*ax=x^\*1x=1$ as $x^\*pq^\*x=0$ whenever $x$ does not appear in $p, q$ and at least one of $p$ or $q $ is nonempty. Next we show that we can modify $a$ to contain $1$ in its support.
Suppose that some word $p$ over $X$ is in the support of $a$ and assume $p$ has minimal length with this property. Then $p^\*a$ has $1$ in its support and it is nonzero since $p^\*wz^\*=1$ in $M$ implies $z=1$ and $w=p$, and so the coefficient of $1$ in $p^\*a$ is the coefficient of $p$ in $a$. Thus we are in the previous case, and hence done.
So suppose that all elements of the support of $a$ are of the from $pq^\*$ with $q$ nonempty. Choose $pq^\*$ in the support of $a$ with $|q|$ minimal. By our *observation* above, the coefficient of $p$ in $aq$ is the same as the coefficient of $pq^\*$ in $a$ because $|q|\leq |b|$ for any $sb^\*$ in the support of $a$ and so $sb^\*q\neq pq^\*q=p$ by the *observation*. This puts us in the previous case, and hence we are done.
It now follows that $R$ is simple.
Next lets check that $d$ is a derivation. Call a mapping $\theta\colon M\setminus \{0\}\to G$ with $G$ a group a *partial homomorphism* if $\theta(mn)=\theta(m)\theta(n)$ whenever $mn\neq 0$. For example, there is a partial homomorphism $\theta\colon M\to F\_X$ (where $F\_X$ is the free group on $X$) with $\theta(pq^\*) = pq^{-1}$, and this is in fact the universal partial homomorphism from $M$ to a group, although we don't need this. Now there is a homomorphism $\eta\colon F\_X\to \mathbb Z$ sending our fixed $x$ to $1$ and $X\setminus \{x\}$ to $0$. Then $\nu:=\eta\theta\colon M\to \mathbb Z$ is a partial homomorphism with $\nu(pq^\*) = |p|\_x-|q|\_x$.
Therefore, $d(m) = \nu(m)m$ for $m\in M$, where we consider $\nu(0)0=0$ (even though technically $\nu(0)$ is not defined). Then we easily check that, for $m,n\in M$, we have
$$d(mn) = \nu(mn)mn = (\nu(m)+\nu(n))mn = m\nu(n)n+\nu(m)mn = md(n)+d(m)n$$ where this works out even if $mn=0$.
Finally, we observe that $d$ is not inner. Indeed, if $d$ is inner then $x=d(x)=xr-rx$ for some $r\in R$. Note that since $1$ commutes with $x$, we may assume without loss of generality that $1\notin supp(r)$. Since left multiplication by $x$ is injective on $M$ (because $x^\*x=1$), we cannot have $x\in x\cdot supp(r)$ because $1\notin supp(r)$. Thus $x\in supp(r)x$. But the only way that can happen is if $xx^\*\in supp(r)$ since if $pq^\*x=x$ and $p\neq 1$, then $q=p=x$ (use the *observation*). Let $n\geq 1$ be maximal with $(x^n)(x^n)^\*\in supp(r)$. Then $xr$ has $(x^{n+1})(x^n)^\*$ in its support (since left multiplication by $x$ is injective) and hence $(x^{n+1})(x^n)^\*\in supp(r)x$ since $xr-rx=x$. But if $pq^\*\in supp(r)$ with $pq^\*x = (x^{n+1})(x^n)^\*$ with $n\geq 1$, then $|q|\geq 2$ and the only possibility is $p=x^{n+1}$, $q=x^{n+1}$ by the *observation*. Thus $x^{n+1}(x^{n+1})^\*\in supp(r)$, contradicting the choice of $n$. We deduce that $d$ is not inner.
| 4 | https://mathoverflow.net/users/15934 | 417282 | 169,992 |
https://mathoverflow.net/questions/417283 | 3 | I came across the following inequality, which should hold for any integer $k\geq 1$:
$$\sum\_{j=0}^{k-1}\frac{(-1)^{j}2^{k-1-j}\binom{k}{j}(k-j)}{2k+1-j}\leq
\frac{1}{3}.$$
I have been struggling with this statement for a while. It looks valid for small $k$, but a formal proof seems out of reach with my tools. Any suggestions on how to approach this?
| https://mathoverflow.net/users/478035 | An inequality involving binomial coefficients and the powers of two | For $j=0,\dots,k-1$,
\begin{equation\*}
\frac1{2k+1-j}=\int\_0^1 x^{2k-j}\,dx.
\end{equation\*}
So,
\begin{equation\*}
\begin{aligned}
s:=&\sum\_{j=0}^{k-1}\frac{(-1)^{j}2^{k-1-j}\binom{k}{j}(k-j)}{2k+1-j} \\
&=\int\_0^1 dx\,\sum\_{j=0}^{k-1}(-1)^{j}2^{k-1-j}\binom{k}{j}(k-j)x^{2k-j} \\
&=\int\_0^1 dx\,kx^{k+1}(2x-1)^{k-1}=I\_1+I\_2,
\end{aligned}
\tag{1}\label{1}
\end{equation\*}
where
\begin{equation\*}
I\_1:=\int\_0^{1/2} dx\,kx^{k+1}(2x-1)^{k-1}=(-1)^{k-1}\frac{k!(k+1)!}{(2k+1)!}\le
\frac1{2^{2k+2}}\le\frac1{16k^2}, \tag{2}\label{2}
\end{equation\*}
\begin{equation\*}
I\_2:=\int\_{1/2}^1 dx\,ke^{g(x)},
\end{equation\*}
\begin{equation\*}
g(x):=(k+1)\ln x+(k-1)\ln(2x-1).
\end{equation\*}
Next, $g(1)=0$, $g'(1)=3k-1$, and, for $x\in(1/2,1)$,
\begin{equation\*}
g''(x)=-\frac{k+1}{x^2}-\frac{4(k-1)}{(2x-1)^2}\le-(k+1)-4(k-1)=3-5k
\end{equation\*}
and hence $g(x)\le h(x):=
(3k-1)(x-1)+(3-5k)(x-1)^2/2$.
So,
\begin{equation\*}
I\_2\le\int\_{-\infty}^1 dx\,ke^{h(x)}=J(k):=\sqrt{\frac{\pi }{2}} e^{\frac{(1-3 k)^2}{10 k-6}} k\,
\frac{\text{erf}\left(\frac{1-3 k}{\sqrt{10 k-6}}\right)+1}{\sqrt{5 k-3}}.
\tag{3}\label{3}
\end{equation\*}
Let
\begin{equation\*}
H(k):=\text{erf}\left(\frac{1-3 k}{\sqrt{10 k-6}}\right)+1
-\left(\frac{1}{3}-\frac{1}{16k^2}\right)\frac{\sqrt{\frac{2}{\pi }} e^{-\frac{(1-3 k)^2}{10 k-6}} \sqrt{5 k-3} }{k}.
\end{equation\*}
Then
\begin{equation\*}
H'(k)=\frac{e^{-\frac{(1-3 k)^2}{10 k-6}} \left(160 k^4-647 k^3+75 k^2+456 k-162\right)}{48 \sqrt{2 \pi } k^4 (5 k-3)^{3/2}}>0
\end{equation\*}
for $k\ge4$ and $H(k)\to0$ as $k\to\infty$. So, for $k\ge4$ we have $H(k)<0$ or, equivalently,
\begin{equation\*}
J(k)<\frac{1}{3}-\frac{1}{16k^2}.
\end{equation\*}
Therefore and in view of \eqref{1}, \eqref{2}, and \eqref{3}, for $k\ge4$ we have
\begin{equation\*}
s<\frac13,
\end{equation\*}
as desired. Checking the latter inequality for $k=1,2,3$ is straightforward.
| 4 | https://mathoverflow.net/users/36721 | 417289 | 169,993 |
https://mathoverflow.net/questions/416648 | 17 | EDIT: immediately **after** bountying the question *(whoops ...)* I found, while looking for something else entirely, that Sauro Tulipani [gave an explicit algorithm](https://www.jstor.org/stable/2273991?refreqid=excelsior%3Abff6195c8445cbcfc1c488c4c2530272&seq=1#metadata_info_tab_contents) for producing a Horn sentence $\varphi\_p$ saying "There are not exactly $p$ objects" and $length(\varphi\_p)=O(p^5\log p).$ So my question should be rephrased as:
>
> Is Tulipani's solution optimal?
>
>
>
(From Tulipani's article I also learned that the construction mentioned below was originally due to Appel, and was rediscovered later by Blass independently.)
---
This question is motivated by Blass' paper [There are not exactly five objects](https://www.jstor.org/stable/2274177?seq=1#metadata_info_tab_contents). Briefly, Wilkie observed that a general semantic result implies that "There are not exactly $p$ objects" must be expressible by a Horn sentence$^\*$ in the equality-only language for each prime $p$, but actually finding such sentences is nontrivial. Blass' example is a bit long (with length exceeding $10^{10^{22}}$ in the case $p=5$); he goes on to say that a somewhat shorter example exists for $p=5$ (due to Morley) but that it doesn't obviously generalize to arbitrary primes.
I'm curious what the optimal length is, as a function of the index of the prime involved:
>
> For $i\in\mathbb{N}$ let $H(i)$ be the length of the shortest Horn sentence equivalent to "There are not exactly $p\_i$ objects," where $p\_i$ is the $i$th prime. What can we say about the growth rate of $H$?
>
>
>
In particular, it is consistent with my current understanding that Blass' construction is *asymptotically* optimal. I suspect that's not the case, but I don't have any actual evidence for that.
*(The tag "universal-algebra" is a bit tentative; my understanding is that, although not strictly equational, Horn logic is still of interest to UA, but I could easily be overestimating its relevance.)*
---
$^\*$That is: "a first-order sentence in prenex normal form whose matrix is a conjunction of conditionals in each of which the consequent is atomic or "false" and the antecedent is a conjunction of atomic formulas." The obvious first-order formulation $$\forall x\_1,...,x\_p[(\bigwedge\_{1\le i<j\le p}x\_i\not=x\_j)\rightarrow\exists y(\bigwedge\_{1\le i\le p}y\not=x\_i)]$$ is **not** a Horn sentence, so the problem is nontrivial.
| https://mathoverflow.net/users/8133 | How hard is it to say "not exactly $p$" with a Horn sentence? | I will show how to improve Tulipani’s construction from $O(p^5)$ symbols to $O(p^3)$ symbols, or $O(p^3\log p)$ bits.
Recall that Tulipani’s sentence is
$$H\_p=\forall\vec x\,\exists\vec s\,\exists\vec u\,\exists y\:\Bigl(G\_p(\vec x,\vec s,\vec u)\land\let\ET\bigwedge\ET\_i(y=x\_i\land u\_p=x\_0\to x\_1=x\_0)\Bigr),$$
where $G\_p(\vec x,\vec s,\vec u)$ is a Horn formula in variables $\{x\_i:0\le i<p\}$, $\{s\_{ij}:0\le i\le j<p\}$, $\{u\_i:2\le i\le p\}$, defined as the conjunction of
$$\begin{gather}
\tag1\label{assoc}\ET\_{i,j,k,h,r}(x\_h=s'\_{ij}\land x\_r=s'\_{jk}\to s'\_{hk}=s'\_{ir}),\\
\tag2\label{canc}\ET\_{i,j,k}(s'\_{ij}=s'\_{ik}\to x\_j=x\_k),\\
\tag3\label{zero}\ET\_is\_{0i}=x\_i,\\
\tag4\label{cyc}u\_2=s\_{11}\land\ET\_{2\le j<p}\ET\_i(u\_j=x\_i\to u\_{j+1}=s'\_{1i}),
\end{gather}$$
where $s'\_{ij}:=s\_{\min\{i,j\},\max\{i,j\}}$. The meaning of $G\_p$ is that given elements $\vec a,\vec b,\vec c$ of a model such that each $b\_{ij}$ belongs to the set $A=\{a\_i:i<p\}$, $G\_p(\vec a,\vec b,\vec c)$ is true iff the recipe
$$\tag+\label{add}a\_i+a\_j=a\_j+a\_i=b\_{ij},\qquad i\le j<p,$$
gives a well-defined operation $+$ on $A$ such that
* $(A,+)$ is an abelian group with zero element $a\_0$,
* $c\_j=ja\_1$ for each $j$.
The bottleneck here is the formula $\eqref{assoc}$ with $p^5$ conjuncts, which ensures (in view of $\eqref{zero}$) that $+$ is well defined, and associative.
First, I claim that one can replace $\eqref{assoc}$ with the two formulas
$$\begin{gather}
\tag{$1'$}\label{def}\ET\_{i,j,k}(x\_i=x\_j\to s'\_{ik}=s'\_{jk}),\\
\tag{$1''$}\label{nucl}\ET\_{i,j,h,r}(x\_h=s'\_{ij}\land x\_r=s'\_{j1}\to s'\_{h1}=s'\_{ir})
\end{gather}$$
with $O(p^4)$ symbols. Let $G'\_p$ denote $\eqref{def}\land\eqref{nucl}\land\eqref{canc}\land\eqref{zero}\land\eqref{cyc}$, and let $H'\_p$ be $H\_p$ with $G\_p$ replaced with $G'\_p$. It is easy to see that if $A=\{a\_i:i<p\}$ and $\vec b\in A$, then $G'\_p(\vec a,\vec b,\vec c)$ holds iff $\eqref{add}$ gives a well-defined operation on $A$ such that
* $(A,+)$ is a commutative loop with zero element $a\_0$,
* $a\_1$ belongs to the [nucleus](https://groupprops.subwiki.org/wiki/Nucleus)
$$N=\{z\in A:\forall x,y\in A\:(x+y)+z=x+(y+z)\},$$
* $c\_j=ja\_1$ for each $j$ (meaning a sum of $j$ copies of $a\_1$, whose bracketing does not matter because of the previous point).
Since $G\_p$ implies $G'\_p$, $H'\_p$ still holds in models of size $\ne p$.
The nucleus $N$ is associative, and it is a subgroup of $A$. If $\sim$ denotes the equivalence relation
$$a\sim b\iff N+a=N+b,$$
then the equivalence class of $a$ is $N+a$: on the one hand, if $a\sim b$, then $b\in N+b=N+a$ as $0\in N$. On the other hand, if $b=n+a$ with $n\in N$, then
$$N+b=N+(n+a)=(N+n)+a=N+a$$
using the definition of $N$ and the fact that it is a subgroup. Thus,
$$\{N+a:a\in A\}$$
is a partition of $A$ into sets of size $|N+a|=|N|$. As a consequence, $|N|$ divides $|A|$, and more generally, if $H$ is a subgroup of $N$, then $|H|$ divides $|A|$.
In particular, if $G'\_p(\vec a,\vec b,\vec c)$ where the $a\_i$ are pairwise distinct, then the order of the cyclic subgroup of $A$ generated by $a\_1$ divides $p=|A|$. This implies that $H'\_p$ fails in a model of size $p$ if we take for $\vec x$ a bijective enumeration of its elements, using the same argument as Tulipani.
This gives a sentence $H'\_p$ with $O(p^4)$ symbols. In order to further reduce it to $O(p^3)$, note that $\eqref{nucl}$ can be replaced with
$$\ET\_{i,j}\exists z\ET\_h\bigl((x\_h=s'\_{ij}\to s'\_{h1}=z)\land(x\_h=s'\_{j1}\to s'\_{ih}=z)\bigr).$$
If we want to keep $H'\_p$ in prenex normal form, we can introduce new variables $\{z\_{ij}:i,j<p\}$, expand the quantifier prefix of $H'\_p$ to $\forall\vec x\,\exists\vec s\,\exists\vec u\,\exists y\,\exists\vec z$, and replace $\eqref{nucl}$ with
$$\ET\_{i,j,h}\bigl((x\_h=s'\_{ij}\to s'\_{h1}=z\_{ij})\land(x\_h=s'\_{j1}\to s'\_{ih}=z\_{ij})\bigr).$$
This gives a sentence with $O(p^3)$ symbols, still using $O(p^2)$ variables.
A few more optimizations, which however do not change the $O(p^3)$ asymptotic:
* We can drop $\eqref{def}$. On the one hand, this only makes the sentence weaker, hence it continues to hold in models of size $\ne p$. On the other hand, the failure for models of size $p$ is achieved when the $x\_i$ are pairwise distinct, in which case $\eqref{def}$ holds automatically.
* We can drop $\eqref{zero}$ and the $s\_{0i}$ variables, putting $s'\_{0i}:=x\_i$ instead.
* We only use the $u\_j$ variables to compute $u\_p=px\_1$. If we count to $p$ using doubling rather than adding $1$ at each step, we only need the $O(\log p)$ variables $u\_j$ for $j=\lfloor p2^{-k}\rfloor$, $k\le\log\_2p-2$, and we can reduce the size of $\eqref{cyc}$ to $O(p\log p)$ symbols.
It’s hard to guess what the optimal size should be; the only lower bound I can think of is that using a straightforward pebble game argument, any sentence (Horn or otherwise) that distinguishes a set with $p$ elements from a set with $p+1$ elements needs to use more than $p$ distinct variables (and a fortiori needs quantifier rank more than $p$). This bound is tight for not-necessarily-Horn sentences, as exhibited in the question.
| 11 | https://mathoverflow.net/users/12705 | 417291 | 169,994 |
https://mathoverflow.net/questions/417253 | 2 | Suppose that $A,B$ are real analytic subsets of $\Omega\subseteq \mathbb{R}^n$ and $p\in A\cap B \neq \emptyset$. Does the intersection inequality from complex analysis still hold, i.e. does the following inequality hold:
$\mathrm{dim}^{\mathbb{R}}\_p(A\cap B) \geq \mathrm{dim}^{\mathbb{R}}\_p(A)+\mathrm{dim}^{\mathbb{R}}\_p( B)-n$?
By real analytic subset, I mean sets that are locally given as the zero set of finitely manly real analytic functions and by dimension, I mean the maximal dimension of regular points near $p$ as manifolds. The complex analytic proofs that I know use the "Active Lemma", i.e. $\mathrm{dim}(\left\{f=0\right\})=n-1$, for non-zero $f$, which does not hold over $\mathbb{R}$.
If not, are there explicit counterexamples?
| https://mathoverflow.net/users/109193 | Dimension of intersection of real analytic sets | For a counterexample take a sphere in 3 space and a plane tangent to it .
| 5 | https://mathoverflow.net/users/4696 | 417293 | 169,996 |
https://mathoverflow.net/questions/417284 | 2 | Lets work with Harvey's [Friedman](https://cpb-us-w2.wpmucdn.com/u.osu.edu/dist/1/1952/files/2014/01/TalkAxiomSetThy-ts05xn.pdf) theory ${\sf K}(W)$, formulated in the language of set theory with a primitive constant symbol $W$ added, i.e. in ${\sf FOL}(\in,W)$
**Axioms:**
**Extensionality:** $$\forall Z \, (Z \in X \iff Z \in Y) \implies \\\forall Z \, (X \in Z \iff Y \in Z)$$
**Subworld Separation:** $$\forall A \in W \, \exists X \in W \, \forall Y \, (Y \in X \iff Y \in A \land \phi)$$; where formula $\phi$ doesn't use the symbol "$X$".
**Reducibility:** if $\phi$ is a formula in ${\sf FOL} (\in)$, with all parameters among "$X,\vec{P}$ " then: $$\forall \vec{P} \in W [ (\exists X: \phi) \implies \exists X \in W: \phi]$$
/
Which is equiconsistent with ZFC.
Is it consistent to add the following principle:
**Generalization:** if $\phi$ is a formula in ${\sf FOL} (\in)$, with all parameters among symbols "$X,\vec{P}$ "; then: $$ \forall \vec{P} \in W [(\forall \operatorname {infinite} X \in W: \phi) \implies \phi(W)]$$
| https://mathoverflow.net/users/95347 | Is it consistent to add a generalization axiom on top of Ext.+Subworld Separation+Reduciblity? | Generalization holds in K(W). Suppose it did not, that ¬(W) holds and ∀infinite∈:.
Then ∃X((infiniteX)∧¬), since this is true when X is W. By Reducibility
∃X∈W((infiniteX)∧¬). But this contradicts the fact that ∀infinite∈:.
| 2 | https://mathoverflow.net/users/133981 | 417303 | 169,998 |
https://mathoverflow.net/questions/417304 | 3 | I have read in a paper about the following result:
Let $V$ be a separable Hilbert space and $(\Omega,A\_{\Omega},P)$ a probability space. Suppose that $Y\_1,Y\_2,...$ is a sequence of independent $V$-valued random variables. If $E\left(\Vert Y\_i\Vert\_{V}^m\right)\leq \frac{1}{2} m! B^2 L^{m-2}$ $\forall m\geq 2$, then, $\forall n\in N$ and $\epsilon>0$
$$ P\big(\big\Vert \frac{1}{n} \sum\_{i=1}^n Y\_i\big\Vert\_{V}>\epsilon\big)\leq 2 \exp\Big(-\frac{n\epsilon^2}{B^2+L\varepsilon+B\sqrt{B^2+2L\epsilon}}\Big).$$
There was no proof given but it was mentioned that the result is well known. I think I found some literature that shows the result if we would assume that $E(Y\_i)=0$ for all $i\in N$ holds (see [Theorem 3.3 of Pinelis](https://arxiv.org/pdf/1208.2200.pdf)). But I don't know much about martingales so maybe it wouldn't follow. Does anyone know if the result is actually true and if $E(Y\_i)=0$ is a necessary condition?
| https://mathoverflow.net/users/163533 | Concentration inequality for Hilbert space valued random variables | The condition $EY\_i=0$ cannot be dropped.
Indeed, if e.g. the $Y\_i$'s are iid with $\mu=EY\_i\ne0$ and $n\to\infty$, then, by the law of large numbers, the left-hand side of the inequality in question will go to $1$ for each $\epsilon\in(0,\|\mu\|\_V)$, whereas the right-hand side of the inequality will go to $0$.
| 2 | https://mathoverflow.net/users/36721 | 417306 | 169,999 |
https://mathoverflow.net/questions/414893 | 2 | Let $G$ be a finite group and $\lambda \in \text{Irr}(G)$ an irreducible complex character of $G$.
Let $m(\lambda) := \min \{ \vert G : H \vert \mid H \leq G, \lambda\vert\_H \text{ has a linear component}\}.$
Is there a universal constant $c$, independent of $G$ and $\lambda$, such that
$$ \frac{m(\lambda)}{\lambda (1\_G)} \leq c? $$
When we consider only monomial groups such a constant is given by 1.
Is there such a constant when we consider only solvable or simple groups?
| https://mathoverflow.net/users/173766 | Existence of universal bound related to characters | There is no such universal constant for solvable groups, at least. Essentially this is because $m(\chi)$ as defined in the question, is multiplicative. That is, if $G\_1$ and $G\_2$ have coprime orders, and if $\chi\_i\in\mathrm{Irr}(G\_i)$ for $i=1,2$ then $\chi\_1\times\chi\_2\in\mathrm{Irr}(G\_1\times G\_2)$ satisfies $m(\chi\_1\times\chi\_2)=m(\chi\_1)m(\chi\_2).$
This follows because the subgroups of $G\_1\times G\_2$ are just the $H\_1\times H\_2$ where $H\_i\subseteq G\_i,$ and $(\chi\_1\times\chi\_2)\_{H\_1\times H\_2}=(\chi\_1)\_{H\_1}\times (\chi\_2)\_{H\_2}$ contains a linear constituent if and only if $(\chi\_i)\_{H\_i}$ contains a linear constituent for $i=1,2.$ Hence if $c(\chi)=m(\chi)/\chi(1)$ then in the situation above, also $c(\chi\_1\times\chi\_2)=c(\chi\_1)c(\chi\_2).$
For odd primes $p>q$ with $q$ dividing $p+1,$ it is known that there exists a solvable group $G$ of order $|G|=p^3q$ and a primitive $\chi\in\mathrm{Irr}(G)$ of degree $p.$ If $H\subseteq G$ and $\chi\_H$ contains a linear constituent then $|G:H|>p$ since $\chi$ is not monomial, so $|G:H|\ge pq$ and $c(\chi)\ge q$ (in fact equality holds). For any $n\ge 1$ we can find $n$ such groups with pairwise coprime orders, for example as follows: Start with $|G\_1|=5^3\cdot 3,$ and for $n\ge 2,$ choose $q\_n$ odd and prime to $\prod\_{i=1}^{n-1}|G\_i|$ and $p\_n$ with $p\_n=-1$ mod $q\_n$ and $p\_n=1$ mod $\prod\_{i=1}^{n-1} |G\_i|.$ Setting $G=\prod\_1^n G\_i$ then $\chi=\prod\_1^n\chi\_i$ (where $\chi\_i\in\mathrm{Irr}(G\_i)$ is any primitive character of $G\_i$ as mentioned above) has $c(\chi)\ge \prod\_1^n q\_i\ge 3^n.$
Since simple groups can't have pairwise coprime orders this doesn't work there, though I guess that there is no universal bound in that case.
| 1 | https://mathoverflow.net/users/313687 | 417310 | 170,000 |
https://mathoverflow.net/questions/403366 | 4 | Let $f : X \to Y$ be projective and smooth morphism of complex algebraic varieties. Here we care about the algebraic topology of $X$ and $Y$, so use classical topology for simplicity.
I can take the constant sheaf $\mathbb{Q}\_X$ and (derived) push it forward to get $f\_\* \mathbb{Q}\_X \in D^b\_c(Y,\mathbb{Q})$. There is a celebrated theorem of Deligne that $f\_\* \mathbb{Q}\_X$ is *semi-simple*, i.e. isomorphic to a direct sum of its cohomology sheaves. The argument uses hard Lefschetz along the fibres. (It is also true that each summand is a semi-simple local system, as a polarizable VHS, however I want to ignore that extra piece of information below.)
Suppose I replace $\mathbb{Q}$ with $k := \mathbb{F}\_p$.
**Question:** Is it true that $f\_\* k\_X \in D^b\_c(Y,k)$ is always semi-simple? That is, does it always split as a direct sum of its (not-necessarily semi-simple) cohomology sheaves.
I had always assumed the answer was no, but woke up this morning feeling unusually optimistic. (I have tried several times to produce a counter-example.) I understand that this is deep water, and I am happy with a heuristic answer either way (potentially using motives).
| https://mathoverflow.net/users/919 | Decomposition of direct image of a smooth morphism, Deligne's theorem, motives | We can give many counterexamples to semisimplicity in positive characteristic using the observation that if the canonical morphism $k\_Y\rightarrow f\_\*k\_X$ in $D^b\_c(Y,k)$ is split, then the induced morphism in $k$ cohomology must be injective: $$f^\*:H^\*(Y,k)\rightarrow H^\*(X,k).$$
So we just need to cook up nontrivial elements in this kernel. For instance, take any nontrivial abelian cover $f:X\rightarrow Y$ of degree $p$. This is classified by an element $\alpha$ of $H^1(Y,\mathbb{Z}/p\mathbb{Z})$, and since this cover is the pullback of the universal degree $p$ cover along a morphism $Y\rightarrow B(\mathbb{Z}/p\mathbb{Z})$, we see that $\alpha$ vanishes in $H^1(X,\mathbb{Z}/p\mathbb{Z})$, since $E(\mathbb{Z}/p\mathbb{Z})$ is contractible.
For a more geometric example, let $E$ be an elliptic curve over $Y$, and $A$ an $E$ torsor over $Y$. Then this is classified up to homotopy by a map to $B(S^1\times S^1)\cong K(\mathbb{Z},2)\times K(\mathbb{Z},2)$. Since the total space $E(S^1\times S^1)$ is contractible, the associated classes $(\alpha,\beta)$ in $H^2(Y,\mathbb{Z})$ will lie in the kernel of $H^2(Y,\mathbb{Z})\rightarrow H^2(A,\mathbb{Z})$. Then one can reduce this mod $p$ and get counterexamples over a field if the orders of these elements aren't divisible by $p$.
Finally, one can use nonliftable projective bundles to give counterexamples, in line with Anonymous' suggestion. If $f:X\rightarrow Y$ is a (complex) dimension $n-1$ projective bundle, then its obstruction to being the projectivisation of a complex vector bundle is a class $\beta\_f\in H^2(X,\mathbb{C}^\*)\cong H^3(X,\mathbb{Z})$. Then one can check in the universal case of $u:U\rightarrow BPGL\_n(\mathbb{C})$, we have $u^\*(\beta\_u)=0$, so this class $\beta\_f$ is always in the kernel of the map:
$$H^3(Y,\mathbb{Z})\rightarrow H^3(X,\mathbb{Z}).$$
These examples have been written up in full detail in an upcoming paper, which should be coming to the arXiv soon.
| 2 | https://mathoverflow.net/users/128502 | 417327 | 170,005 |
https://mathoverflow.net/questions/405604 | 1 | In [[1](https://www.jstor.org/stable/pdf/2157207.pdf?refreqid=excelsior%3Ac8cb5873955a58f5a2eba65be8b3f13c)], it is shown in theorem 1.2 that for symmetric $n \times n$ matrices $A$, $B$, we have
$$
\min\_{Y \in Y^\*} \text{tr}(Y^TAY) =
\text{tr}(X^TAX) =
\sum\_{i=1}^p \lambda\_i,
$$
with
$$
\text{
$X^TBX = I^p$
and
$X^TAX = \mathrm{diag}(\lambda\_1,\dots,\lambda\_p)$,
}
$$
$Y^\*$ being the set of all $n\times p$ matrices for which $Y^TBY = I^p$, and the columns of $X$ correspond to the eigenvectors belonging to the first $p$ eigenvalues $\lambda\_1 \leq \dots \leq \lambda\_p$.
It is then stated that this is a direct consequence of theorem 1.1 above and the Courant-Fischer theorem. While I understand the proof for 1.1 (as given in [[2]](https://books.google.nl/books?id=eXQXwCDD9agC&lpg=PR2&ots=AiStMDbESI&dq=J.%20N.%20FRANKLIN%2C%20Matrix%20Theory%2C%20Prentice%20Hall%2C%20Englewood%20Cliffs&lr&hl=nl&pg=PA106#v=onepage&q=J.%20N.%20FRANKLIN,%20Matrix%20Theory,%20Prentice%20Hall,%20Englewood%20Cliffs&f=false)), I don't get how this is used together with C-F to yield the proof for 1.2. A lot of other papers refer to this proof, so that doesn't help figuring it out. Moreover, I read that the min-max theorem and the Cauchy interlacing theorem both get referred to as Courant-Fischer theorem sometimes, so I'm note sure which C-F the writers are referring to.
I'd be very thankful for any help or suggestions :)
| https://mathoverflow.net/users/401351 | Trace minimization for generalized eigenvalue problem | After some work, I figured out the proof, using -- indeed -- the Courant-Fischer theorem and parts of Cauchy's Interlacing and Poincaré's Separation theorems. I've carved out the part of my thesis proving this, which can be found [here](https://github.com/dnndbrkt/stratifyingautism/blob/6e4660b66c351cb83177736e963752e9d9f2c13f/th%203.8.pdf).
| 1 | https://mathoverflow.net/users/401351 | 417328 | 170,006 |
https://mathoverflow.net/questions/417324 | 5 | **Motivation**: The following is a theorem of Berrick-Hesselholt (essentially also due to Linnell, though not in this form):
>
> Let $G$ be a group. Suppose that for every subgroup of $G$ isomorphic to $\mathbb Q$, $G$ has a quotient in which the image of this subgroup is central and nontrivial. In this case the Bass trace conjecture holds for $G$.
>
>
>
I can add it for context, but for my question it is not important to know what this conjecture is - I just state this as motivation.
**Question** : What are some examples of *finitely presented* groups which *do not* have this property ?
That is, they have a subgroup isomorphic to $\mathbb Q$ such that for each quotient of $G$, its image is central only if it is trivial.
| https://mathoverflow.net/users/102343 | Potential counterexamples to Bass' trace conjecture | a) There are old results which directly imply the existence of such groups:
(1) Boone Higman 1972: every f.g. group with solvable word problem embeds into a simple subgroup of a finitely presented group.
(2) Every countable group with solvable word problem embeds into a f.g. group with solvable word problem (reference? the original HNN construction directly works since it consists of explicit amalgams — alternatively here Ph. Hall produced in the 50s an explicit 3-generated metabelian group with solvable word problem, with copies of $\mathbf{Q}$).
---
b) A more explicit example is the group $\tilde{T}$ obtained as the set of self-homeomorphisms of $\mathbf{R}/\mathbf{Z}$ commuting with $\sigma:n\mapsto n+1$, that are piecewise affine with dyadic slopes and breakpoints.
The center of $\tilde{T}$ is the infinite cyclic group $\langle\sigma\rangle$ and the quotient is naturally identified with Thompson's group $T$, which is a finitely presented simple group. The normal proper subgroups of $\tilde{T}$ are precisely the subgroups of $\langle\sigma\rangle$.
If $Q$ is any copy of $\mathbf{Q}$ in $\tilde{T}$, it follows that the image of $Q$ in $T$ is non-central (since $\mathbf{Q}$ has no nontrivial cyclic quotient). Hence the same holds in every nontrivial quotient of $\tilde{T}$.
Finally, that $\tilde{T}$ contains a copy of $\mathbf{Q}$ (and even continuum many such copies) is an original observation of Belk, Matucci, Hyde ([arXiv](https://arxiv.org/abs/2005.02036)).
([The other answer](https://mathoverflow.net/a/417332/14094) is closely related as it refers to a more complicated finitely presented simple group containing $\tilde{T}$. The group $\tilde{T}$ itself is not finitely presented but has few enough normal subgroups for the condition to hold.)
| 9 | https://mathoverflow.net/users/14094 | 417334 | 170,009 |
https://mathoverflow.net/questions/417074 | 2 | Let us say a set of $n$ rectangles is *rectifiable* if all $n$ rectangles together form a big rectangle without gaps or overlaps.
**Question:** How hard computationally is the question of deciding whether a set of $n$ rectangles with all dimensions integers is rectifiable?
If we further constrain the question by disallowing rotations of the tile rectangles OR by insisting that the longest sides of all rectangles should have same orientation in the layout, what happens?
Note: Analogous questions can be asked with sets of $n$ triangles and in higher dimensions.
Ref: [What rectangles can a set of rectangles tile?](https://mathoverflow.net/questions/285018/what-rectangles-can-a-set-of-rectangles-tile)
| https://mathoverflow.net/users/142600 | On sets of rectangles that can all together form at least one big rectangle | The problem is NP-complete. There is a simple reduction from the bin packing problem [1].
Suppose we want to determine items $a\_1, \dots, a\_n$ fit in $k$ bins of size $h$, $kh = \sum\_{i=1}^n a\_i$.
Let $w \gt 2kh$. Each item of size $a\_i$ is represented as a rectangle of size $w \times 2 a\_i$. One rectangle of size $k w \times 1$ is used as the base of the rectangle packing.
The set of rectangles is rectifiable if and only if the items will fit in the bins.
No item can have a different orientation than the base rectangle, and thus whether the rotation is allowed is irrelevant.
* [1]: Korf, Richard E., Michael D. Moffitt, and Martha E. Pollack. "Optimal rectangle packing." Annals of Operations Research 179.1 (2010): 261-295.
| 4 | https://mathoverflow.net/users/476793 | 417337 | 170,010 |
https://mathoverflow.net/questions/417188 | 1 | I am studying a recent paper in which the author worked on the rectangular, flat 3 tori. It can be realized, the author explained, as $\mathbb{R}^3 \over (L\_1 \mathbb Z \times L\_2 \mathbb Z \times L\_3 \mathbb Z)$ with $L\_j \in (0, \infty),j=1,2,3.$ For notational convenience, we use the coordinates for the standard torus $\mathbb{T}^3 :=
{\mathbb{R}^3 \over \mathbb{Z}^3}$ and incorporate the geometry of the torus into the Riemannian metric, using the corresponding Laplace-Beltrami operator
$$\triangle = \sum\_{j=1}^3 L\_j^{-2} \frac{\partial^2}{\partial x\_2^2}$$
We then define the Schrodinger propagator $e^{i t \triangle} $ by
$$\mathcal{F}(e^{i t \triangle } f)(\xi)=e^{- 2 \pi i t \sum\_{j=1}^3 L\_j^{-2} \xi\_j^2} \hat{f}(\xi), \,\,\, for\,\,\, \xi =(\xi\_1,\xi\_2,\xi\_3)\in \mathbb{Z}^3.$$
I have some difficulties understanding how he scaled the coordinates of the Laplace operator. Also, I can not get how he got the semigroup. Could you please explain for me in details. Thanks in advance.
| https://mathoverflow.net/users/471464 | The semigroup of Laplace-Beltrami operator on 3-flat torus | You are probably just overthinking it since this is basically just a multivariable calculus change of variables.
The transformation $\mathbb{R}^3 \to \mathbb{R}^3$ given by
$$ (x\_1, x\_2, x\_3) \mapsto (y\_1,y\_2,y\_3) = (L\_1 x\_1, L\_2, x\_2, L\_3 x\_3)$$
maps the torus $\mathbb{T}^3 = \mathbb{R}^3 / \mathbb{Z}^3$ to the rectangular torus $\mathbb{R}^3 / L\_1\mathbb{Z} \times L\_2 \mathbb{Z}\times L\_3 \mathbb{Z}$. In other words, this defines a *change of variables* between the standard torus and the rectangular torus.
The change of variables satisfies
$$ \frac{\partial}{\partial x\_i} = L\_i \frac{\partial}{\partial y\_i}$$
So the Laplace operator on the rectangular tori
$$ \sum \left(\frac{\partial}{\partial y\_i}\right)^2 = \sum \frac{1}{(L\_i)^2} \left( \frac{\partial}{\partial x\_i} \right)^2 $$
in the new coordinates.
| 2 | https://mathoverflow.net/users/3948 | 417345 | 170,011 |
https://mathoverflow.net/questions/417316 | 2 | Given an adjunction $F\dashv G:\mathcal{C}\rightleftarrows\mathcal{D}$ with unit $\eta$ and counit $\epsilon$, we naturally have a monad $(G\circ F,\eta,G\epsilon\_F)$ on $\mathcal{C}$ and a comparison functor $K:\mathcal{D}\to\mathcal{C}^{G\circ F}$ (where $\mathcal{C}^{G\circ F}$ is the [E-M category](https://ncatlab.org/nlab/show/Eilenberg-Moore+category) of this monad) given by $$K(D)=\big(G(D),G(\epsilon\_D)\big),$$ $$K(f:D\to D')=G(f).$$
This functor is unique satisfying $$?\circ K=G, \hspace{10mm} K\circ F=\widehat
{G\circ F},$$ as shown in e.g. [Mac Lane, p.142](http://www.mtm.ufsc.br/%7Eebatista/2016-2/maclanecat.pdf).
It seems like the proof that this functor is unique satisfying these equations should be way easier than it is in Mac Lane. He uses the fact that $$(1\_\mathcal{C},K):F\dashv G:\mathcal{C}\rightleftarrows\mathcal{D}\longrightarrow\widehat{G\circ F}\dashv\ ?:\mathcal{C}\rightleftarrows\mathcal{C}^{G\circ F}$$ is a morphism of adjunctions by the above equations and the fact that both adjunctions have the same unit, then looks at the equivalent counit condition for morphisms of adjunctions to conclude that any other functor $K':\mathcal{D}\to\mathcal{C}^{G\circ F}$ satisfying $?\circ K'=G$ and $K'\circ F=\widehat{G\circ F}$ agrees with $K$ on structure maps.
>
> Why is this not immediately true since $K$ and $K'$ agree on arrows?
>
>
>
In particular, any two functors $F,G:\mathcal{A}\rightrightarrows\mathcal{B}$ which agree on arrows immediately agree on objects since $$F(X)=F(dom(1\_X))=dom(F(1\_X))=dom(G(1\_X))=G(dom(1\_X))=G(X),$$ and the equation $?\circ K=G=\ ?\circ K'$ tells us that $K$ and $K'$ agree on arrows -- they're both just $G$ on arrows since $?$ leaves arrows unchanged.
>
> Can we just repeat the above argument with $F=K$ and $G=K'$ and be done?
>
>
>
I can't find anything wrong with this reasoning, but I suspect Mac Lane would have taken this route if it worked.
| https://mathoverflow.net/users/92164 | Uniqueness of comparison functors | Your argument that any two functors that agree on arrows must agree on objects depends on assuming that the homsets of a category are disjoint, so that every arrow has exactly one domain and codomain. But even if the homsets of $\mathcal{C}$ are disjoint, the homsets of $\mathcal{C}^{G\circ F}$ won't generally be: a given arrow $f:X\to Y$ in $\mathcal{C}$ can be a $(G\circ F)$-algebra morphism between more than one pair of $(G\circ F)$-algebra structures on $X$ and $Y$.
In general, it's dangerous to try to do category theory treating the arrows in a non-dependently typed manner. In particular, it doesn't really make sense to ask whether two arrows are equal unless you *already* know that their domains and codomains are equal (so that they have the same type).
| 5 | https://mathoverflow.net/users/49 | 417347 | 170,012 |
https://mathoverflow.net/questions/417287 | 4 | Consider the parabolic equation in $p: \mathbb R^2\to\mathbb R$
$$\partial\_t p + b(t)\partial\_x p + D(t,x)\partial^2\_{xx}p=0,$$
where $b$, $D$ are nice enough functions. I look for the continuity of the derivatives $\partial\_t p$, $\partial\_x p$ of the solution. It is known by Nash's paper ([Continuity of Solutions of Parabolic and Elliptic Equations](https://www.jstor.org/stable/2372841?seq=1#metadata_info_tab_contents)) that, under very reasonable conditions on $b$, $D$, we have the Hölder-type continuity of $p$. Is there any work concerning such continuity analysis of $\partial\_t p$, $\partial\_x p$?
PS: My idea is to consider $q\mathrel{:=}\partial\_x p$. Then
$$\partial\_t q + b(t)\partial\_xq + D(t,x)\partial^2\_{xx}q=-\partial\_x D(t,x)\partial\_{xx}p$$
is a similar parabolic equation for $q$ with an additional source. But the term $\partial\_x D(t,x)\partial\_{xx}p$ contains $\partial\_{xx}p$, which makes the estimation even harder….
| https://mathoverflow.net/users/261243 | Reference request: continuity of the derivatives of the (fundamental) solution to a parabolic equation | The equation for $q = p\_x$ can be written in divergence form as
$$q\_t + b(t)q\_x + (D(t,\,x)q\_x)\_x = 0,$$
so Nash's theorem (which applies to divergence-form equations) implies that $p\_x$ is Holder continuous under mild hypotheses on the coefficients (boundedness and measurability).
If the coefficients are more regular (e.g. in Holder classes) then parabolic Schauder estimates (e.g. in the book of Lieberman) give higher regularity of $p$, the general principle being that $p$ has twice as many spatial derivatives as temporal ones (by the scaling of the equation).
| 4 | https://mathoverflow.net/users/16659 | 417352 | 170,013 |
https://mathoverflow.net/questions/417144 | 1 | The process $X(t)=\int\_0^t B(s) ds+B(t)$ is a centered continuous Gaussian process. Therefore it defines a Gaussian measure on $C[0,T]$. Therefore there is a Cameron-Martin space with Cameron-Martin norm. I can compute the covariance and get some complicated expression.
Is there a clean expression for the Cameron-Martin norm associated to $X$?
| https://mathoverflow.net/users/341290 | What is the Cameron-Martin norm associated to $X(t)=\int_0^t B(s) ds+B(t)$? | I will show that for $x\in\mathcal C^1$, $x$ is in the Cameron-Martin space $\mathcal H$ and
$$ |x|\_\mathcal{H}^2 = \int\_0^T\left(x'(t)-\int\_0^t\mathbf e^{-(t-s)}x'(s)\mathrm ds\right)^2\mathrm dt. $$
Expanding the product and using Fubini's theorem, it can be rewritten as
$$ |x|\_\mathcal{H}^2 = \int\_0^T|x'(t)|^2\mathrm dt + \int\_0^T\int\_0^Tk(s,t)x'(s)x'(t)\mathrm ds\mathrm dt $$
for some explicit kernel $k$. I claim that with slightly more effort, one can show that $\mathcal H=\mathrm H^1$ *as vector spaces*, and that the above expression stays true for $x'\in\mathrm L^2$ the weak derivative. Hopefully this is clean enough for your purposes.
**The Cameron-Martin space.**
Define $\overline F:\mathcal C\to\mathcal C$ as the continuous map sending $B$ to $X$; formally, $F(b):t\mapsto\int\_0^tb\_sds + b\_t$. Set $i\_B:\mathrm H^1\to\mathcal C$ the usual Cameron-Martin inclusion for Brownian motion.¹ The important observation is that the continuous map $\overline F\circ i\_B:\mathrm H^1\to\mathcal C$ is injective. Let us postpone the proof to the end of the answer.
Define $\mathcal H$ as the image of $\mathrm H^1$ under $\overline F\circ i\_B$, together with the inner product coming from $\mathrm H^1$ (we use the injectivity here). In other words, $\mathcal H$ is the set of all $\overline F\circ i\_B(b)$ for $b\in\mathrm H^1$, and $|\overline F\circ i\_B(b)|\_\mathcal{H}^2=|b|\_{\mathrm H^1}^2$. It is obviously canonically isomorphic to $\mathrm H^1$, through some $F:\mathrm H^1\to\mathcal H$.
I claim that the inclusion $i\_X:\mathcal H\hookrightarrow\mathcal C$ is the Cameron-Martin space of $X$. Indeed, for any $\phi,\psi\in\mathcal C^\*$, and noting that $\overline F\circ i\_B=i\_X\circ F$,
$$ \begin{align\*}
\mathbb E[\phi(X)\psi(X)]
&= \mathbb E[\phi\circ\overline F(B)\psi\circ\overline F(B)] \\
&= \big\langle\phi\circ\overline F\circ i\_B,\psi\circ\overline F\circ i\_B\big\rangle\_{(\mathrm H^1)^\*} \\
&= \big\langle\phi\circ i\_X\circ F,\psi\circ i\_X\circ F\big\rangle\_{(\mathrm H^1)^\*} \\
&= \big\langle\phi\circ i\_X,\psi\circ i\_X\big\rangle\_\mathcal{H^\*}.
\end{align\*} $$
**Expression for the norm.**
Suppose $x\in\mathcal C$ is actually of class $\mathcal C^1$, and define
$$ b:t\mapsto
\int\_0^t\mathbf e^{-(t-s)}x'(s)\mathrm ds
= \mathbf e^{-t}\int\_0^t\mathbf e^sx'(s)\mathrm ds. $$
Then $b$ is clearly in $\mathrm H^1$, and in fact we will show $x=F(b)$, so $x\in\mathcal H$ and $|x|\_\mathcal{H}=|b|\_{\mathrm H^1}$ and we have the expected expression for the Cameron-Martin norm.
The fact that $x=F(b)$ follows from computing derivatives:
$$ \frac{\mathrm d}{\mathrm dt}\big(F(b)(t)-x(t)\big)
= b(t) + b'(t) - x'(t)
= 0. $$
**Injectivity of the transform.**
Since $i\_B$ is injective, we need only show that $\overline F$ is injective. Choose $b\in\mathcal C$ such that $\overline F(b)=0$. This means precisely that
$$ 0=\overline F(b)(t)=\int\_0^tb(s)\mathrm ds + b(t) $$
for all $t$. Then the function
$$ t\mapsto\mathbf e^t\int\_0^tb(s)\mathrm dt $$
is $\mathcal C^1$ with vanishing derivative hence always zero, and using the relation again,
$$ b(t) = -\int\_0^tb(s)\mathrm ds = -\mathbf e^{-t}\cdot0 = 0. $$
---
¹ $\mathrm H^1$ is the Sobolev space of functions $b$ that can be written as the integral of a function $\dot b\in\mathrm L^2$.
| 1 | https://mathoverflow.net/users/129074 | 417357 | 170,014 |
https://mathoverflow.net/questions/417351 | 0 | Let $E,F$ be two holomorphic vector bundles on a compact Kahler manifold $X$. Denote by $\mathbb{P}(E), \mathbb{P}(F)$ the associated projective bundles and $L\_E=\mathcal{O}\_E(-1), L\_F=\mathcal{O}\_F(-1)$ the tautological line bundles. Let $f:E\to F$ a bundle map.
Is there a "canonical" induced map
$$\tilde{f}:\tilde{L}\_E\to \tilde{L}\_F$$
where $\tilde{L}\_E,\tilde{L}\_F\to Y$ are some pull back of $L\_E,L\_F$ to some space $Y$?
| https://mathoverflow.net/users/102114 | Induced homomorphism on tautological line bundles $\mathcal{O}_E(1),\mathcal{O}_F(1)$ | Let $Z = \mathbb{P}(E) \times\_X \mathbb{P}(F)$. Consider the composition
$$
p\_1^\*L\_E \to p^\*E \to p^\*F \to p^\*F/p\_2^\*L\_F,
$$
where $p\_1 \colon Z \to \mathbb{P}(E)$, $p\_2 \colon Z \to \mathbb{P}(F)$, and $p \colon Z \to X$ are the natural projections,
and the middle arrow above is the pullback of $f$. Let
$$
Y \subset Z
$$
be the zero locus of the composition. Then on $Y$ the composition vanishes, hence the composition of the first two arrows factors (in a unique way) through a morphism
$$
(p\_1^\*L\_E)\vert\_Y \to (p\_2^\*L\_F)\vert\_Y
$$
as required.
| 2 | https://mathoverflow.net/users/4428 | 417358 | 170,015 |
https://mathoverflow.net/questions/417262 | 1 | Let $X$ be a smooth projective toric variety over $\mathbb{C}$. It is acted by the compact torus $T=(S^1)^n$.
The $T$-equivariant cohomology $H^\*\_T(X)$ (with coefficients in a field, say) is an algebra over the ring of the $T$-equivariant cohomology of the point $H^\*\_T(pt)$. The ideal $H^{>0}\_T(pt)\cdot H^\*\_T(X)$ is clearly two-sided.
**Is it true that the quotient algebra $H^\*\_T(X)/H^{>0}\_T(pt)\cdot H^\*\_T(X)$ is isomorphic to $H^\*(X)$ as a graded algebra?**
| https://mathoverflow.net/users/16183 | A connection between equivariant and non-equivariant cohomology of toric variety | Yes. The keyword here is equivariantly formal. More generally if $X$ is a (possibly singular) projective variety over $\mathbb{C}$ whose ordinary cohomology $H(X)$ vanishes in odd degrees, and if $X$ admits an algebraic action of a torus $T=(\mathbb{C}^\*)^n$ with compact torus $K=(S^1)^n$ then equivariant cohomology $H\_K(X)$ is a free module over $H\_K(pt)$ and can be obtained as extension of scalars from the ordinary cohomology, and
$$H(X)\cong H\_K(X)/H\_K^{>0}(X)\cdot H\_K(X).$$
This follows from Theorem 14.1 in [this paper](https://www.math.ias.edu/%7Egoresky/pdf/equivariant.jour.pdf) of Goresky-Kottwitz-MacPherson.
Also Proposition 2 from [this paper](https://www-fourier.ujf-grenoble.fr/%7Embrion/notesmontreal.pdf) of Brion applies to your situation.
| 4 | https://mathoverflow.net/users/66536 | 417360 | 170,016 |
https://mathoverflow.net/questions/417344 | 4 | Let $ \Omega $ be a bounded domain with smooth boundary. Consider the Poisson equation
\begin{eqnarray}
-\Delta u&=&f\text{ in }\Omega\\
u&=&0\text{ on }\partial\Omega
\end{eqnarray}
where $ f\in C\_0^{\infty}(\Omega) $. By using the Lax-Milgram theorem, we can find the solution in $ H\_0^1(\Omega) $ and then enhance the regularity for $ u $. I want to ask what condition should I assume for $ f $ can I obatin that $ \nabla u=0 $ on $ \partial\Omega $. More generaly we can replace the operator $ -\Delta $ by other elliptic operators. Can you give me some referances or hints?
| https://mathoverflow.net/users/241460 | The behavior of $ \nabla u $ on the boundary for Poisson equations | The first observation is that the $u$ above satisfies $\nabla u=0$ on $\partial \Omega$ if and only if $f$ is orthogonal to all harmonic functions $v$ in $\Omega$, continuous up the the boundary. In fact, $\int\_{\Omega} fv=\int\_{\Omega} (\Delta u) v=\int\_{\Omega} u \Delta v=0$, by the boundary conditions. Conversely, if this holds for $f$, then, since $u=0$ on $\partial \Omega$, $0=\int\_{\Omega} (\Delta u) v=\int\_{\partial \Omega} v \frac{\partial u}{\partial n}$ for every harmonic $v$. This gives $\nabla u=0$ at the boundary, since $v$ is arbitrary on $\partial \Omega$ and the tangential derivatives of $u$ are zero by the boundary conditions.
The second remark is that such $u$ vanishes in a neighborhood of the boundary where $f$ is zero (hence $u$ is harmonic). In fact, the equation and the boundary conditions imply that all the derivatives up to the second order vanish on the boundary. Continuing $u$ by zero across the boundary we obtain an harmonic function vanishing in an open set and hence in any connected set containing the boundary where $f=0$.
This observation allows to change $\Omega$ to a ball $B\_R$ containing it. In fact, if $u$ solves the problem above in $\Omega$, since it is zero in a neighborhood of the boundary, it solves the same probelm in $B\_R$ and conversely if it solves in $B\_R$ then it is 0 in $B\_R \setminus \Omega$, by the above argument and $u, \nabla u=0$ at $\partial \Omega$.
Let us take therefore $\Omega=B\_R$. By the above discussion and by density, the problem has a solution iff $f$ is orthogonal to all harmonic polynomials $P$, that is $\int\_{\mathbb R^n} P(x)f(x)=0$ or $\left (P(iD)\hat f\right )(0)=0$.
| 5 | https://mathoverflow.net/users/150653 | 417361 | 170,017 |
https://mathoverflow.net/questions/417330 | 3 | Assume that $f\_0,f\_1,f\_2$ are polynomial functions of degree two in two variables. This means that the $f\_i$ are linear combinations with real coefficients of $x^2,xy,x,y^2,y,1$.
Consider the function $f = f\_1^2-af\_0f\_2:\mathbb{R}^2\rightarrow\mathbb{R}$ where $a\in \mathbb{R}\_{>0}$. Is it true that for a "random" choice (whit a suitable definition of random) of the $f\_i$ there exists $(x\_0,y\_0)\in\mathbb{R}^2$ such that $f(x\_0,y\_0) \geq 0$.
Clearly, this does not hold for any choice of the $f\_i$. Take for instance $f\_1\equiv 0$, $f\_0 = x^2+1$, $f\_2 = y^2+1$. Then $f(x,y) = -a(x^2y^2+x^2+y^2+1) < 0 $ for all $(x,y)\in\mathbb{R}^2$.
Write
$$f\_0 = a\_1 x^2 + a\_2xy+ a\_3 x+ a\_4 y^2+ a\_5 y+ a\_6;$$
$$f\_1 = b\_1 x^2 + b\_2xy+ b\_3 x+ b\_4 y^2+ b\_5 y+ b\_6;$$
$$f\_2 = c\_1 x^2 + c\_2xy+ c\_3 x+ c\_4 y^2+ c\_5 y+ c\_6;$$
Then $f$ corresponds to the point $(a\_1,\dots,c\_6)\in\mathbb{R}^{18}$.
Taking
$$f\_0 = \epsilon\_0(x^2 + y^2 + 1);$$
$$f\_1 = \epsilon\_1(x^2 + y^2 + 1);$$
$$f\_2 = \epsilon\_2(x^2 + y^2 + 1);$$
we have $f = (\epsilon\_0^2-a\epsilon\_1\epsilon\_2)(x^2 + y^2 + 1)$ which is always negative when $\epsilon\_0^2-a\epsilon\_1\epsilon\_2 < 0$.
But the point $(f\_0,f\_1,f\_2)$ lies inside the linear subspace $a\_2 = a\_3 = a\_5 = b\_2 = b\_3 = b\_5 = c\_2 = c\_3 = c\_5 = 0$.
Thank you very much.
| https://mathoverflow.net/users/14514 | Positivity of real functions in two variables | $\newcommand\R{\mathbb R}\newcommand\c{\mathsf c}\newcommand\ep{\varepsilon}$The answer is no. Indeed, after clarifications given by the OP in comments and in the original post, the question can be stated as follows:
>
> For $i=0,1,2$, let
> $$f\_i(x,y)=a\_{i,0}+a\_{i,1}x+a\_{i,2}y+a\_{i,3}xy+a\_{i,4}x^2+a\_{i,5}y^2,$$
> where the $a\_{i,j}$ are real numbers.
> For a real $a>0$, let $M\_a$ be the set of all matrices $(a\_{i,j}\colon i=0,1,2,\,j=0,\dots,5)\in\R^{3\times6}$ such that $f(x,y):=f\_1(x,y)^2-af\_0(x,y)f\_2(x,y)\ge0$ for some $(x,y)\in\R^2$.
> Is it true that the dimension (in whatever appropriate sense) of the complement $M\_a^\c$ of $M\_a$ to $\R^{3\times6}$ is strictly less that $18$ (the dimension of $\R^{3\times6}$)?
>
>
>
Let $g(x,y):=1+x^2+y^2$. For real $\ep>0$, let $N\_\ep$ denote the set of all matrices $(a\_{i,j}\colon i=0,1,2,\,j=0,\dots,5)\in\R^{3\times6}$ such that
$|a\_{i,j}-b\_j|<\ep$ for all $i=0,1,2,\,j=0,\dots,5$, where $(b\_0,\dots,b\_5)=(1,0,0,0,1,1)$. Then for $i=0,1,2$ and all real $x,y$ we have
$$|f\_i(x,y)-g(x,y)|<\ep(1+|x|+|y|+|xy|+x^2+y^2)
\le2\ep g(x,y),$$
since $|x|\le(1+x^2)/2$, $|y|\le(1+y^2)/2$, and $|xy|\le(x^2+y^2)/2$.
So, taking now any $a\in(1,2)$ and any $\ep\in(0,\frac{\sqrt a-1}{2(\sqrt a+1)})$, we get ($\ep<1/2$ and) for all real $x,y$
$$f\_1(x,y)^2-af\_0(x,y)f\_2(x,y)\le g(x,y)^2[(1+2\ep)^2-a(1-2\ep)^2]<0.$$
So, $N\_\ep\subseteq M\_a^\c$ and the dimension of $N\_\ep$ is $18$. Thus, the dimension of $M\_a^\c$ is $18$, not $<18$.
| 1 | https://mathoverflow.net/users/36721 | 417367 | 170,019 |
https://mathoverflow.net/questions/417364 | 0 | I am interested in the determinant of $W = X \* X'$, where $X \in \mathbb{R}^{k \times n}$ is a matrix with each row drawn IID from some sub-Gaussian distribution on $\mathbb{R}^{n}$. (I am aware of some universality results, so happy to also consider a "standard" Wishart matrix with parameter $k/n$). Edit: say e.g. that the diagonal of $W$ is all ones, and the off-diagonal are of order $1/\sqrt n$.
Question (**short**): is there an LDP for the empirical spectral distribution of $W$ to Marchenko-Pastur when $k \approx \sqrt{n}$?
Question (**long**):
There is an argument by Ofer here <https://mathoverflow.net/q/372456> that finds the expected determinant of $W$, if $k/n \to c$ for some constant $c \in (0,1)$. The idea is to use an LDP for the convergence of the empirical spectral distribution of Wishart to Marchenko-Pastur.
I am interested in the case of $k/n \to 0$, and in particular, the scaling $k \approx \sqrt{n}$. Letting $\lambda := k/n$, and $\mu(\lambda)$ denote the corresponding M-P law, it is easy to check
$$ \mathbb{E}\_{X \sim \mu(k,n)}[\log x] \asymp -\lambda/2 + O(\lambda^2) $$
I would hope that this implies something like
$$ \mathbb{E}\det(W) = \exp(k \lambda + O(\lambda)) := exp(k^2/n + O(k/n))$$
But if $k/n \to 0$, the referenced LDP gives the trivial answer $\mathbb{E}\det(W) = o(k)$. Is there a more "quantitative" LDP known that allows me to take $k$ and $n$ jointly to $0$?
| https://mathoverflow.net/users/134361 | LDP for Marchenko Pastur with k/n tending to 0 | For standard Gaussians, and with the matrix $W/n$,
the proof of the LDP given by Ben Arous-Guionnet adapts
to the Wishart setup. However, you will have different scalings and so the non-commutative entropy term (of exponential scaling $k^2$) will disappear, and the proof more or less trivializes.
If I did not make a stupid computational mistake, the large deviations will have speed $kn$ and rate function
$I(\mu)= \frac12 \int (x-\log x) \mu(dx)$. You see that $I(\mu)=0$ iff $\mu=\delta\_1$.
| 1 | https://mathoverflow.net/users/35520 | 417370 | 170,020 |
https://mathoverflow.net/questions/417362 | 10 | Does there exists a triple $(G, X, \pi)$, where $G$ is a compact group, $X$ an infinite dimensional Banach space over $\mathbf{C}$, and $\pi : G \to B(X)$ a strongly continuous representation of $G$, such that if $Y$ is a non-zero closed invariant subspace of $X$ in the sense that $\pi(g)Y \subseteq Y$ for all $g \in G$, then $Y = X$ (we call nonzero strongly representations of $G$ satisfying this property topologically irreducible)?
I am not sure that if this is settled already, as the question seems natural, yet I am ignorant of the relevant literature, nor could I come up with a proof or disproof. Let me be (perhaps overly) precise about the question asked in the title. By compact group, I mean Hausdorff plus quasi-compact, and topologically irreducible means that the only invariant closed subspaces are either $0$ or the whole space. More generally, can we find such an example by replacing infinite dimensional Banach spaces by complete infinite dimensional locally convex Hausdorff spaces instead? While we are on the subject, I'd also appreciate some recommendation of books, papers/surveys on representations of topological groups on Banach spaces that is not focused on unitary representations on Hilbert spaces?
| https://mathoverflow.net/users/40789 | Existence of a strongly continuous topologically irreducible representation of a compact group on an infinite dimensional Banach space? | For Banach spaces, the question is no : there is a form of the Peter-Weyl theorem, due to Shiga, which implies that in every Banach space representation of a compact group, the finite-dimensional sub-representations span a dense subspace. In particular, strongly continuous irreducible representations on a Banach space are finite-dimensional. I am not sure about arbitrary locally convex topological vector space.
Shiga's paper is [here](https://projecteuclid.org/journals/journal-of-the-mathematical-society-of-japan/volume-7/issue-3/Representations-of-a-compact-group-on-a-Banach-space/10.2969/jmsj/00730224.full). If the link does not work, the precise reference is:
K. Shiga, Representations of a compact group on a Banach space, Journal of the Mathematical Society of Japan 7 (1955), 224–248.
| 9 | https://mathoverflow.net/users/10265 | 417371 | 170,021 |
https://mathoverflow.net/questions/417372 | 4 | To simplify the notation, assume $V=L$. We have $\lvert V\_{\omega\_{1}} \rvert=\aleph\_{\omega\_{1}}$ and $\lvert H(\aleph\_{1})\rvert=\aleph\_{1}$, so in particular $V\_{\omega\_{1}} \models \exists x \forall \alpha\; x \not\in L(\alpha) $ since $L(\omega\_{1})=H(\aleph\_{1})$.
Using the Löwenheim–Skolem theorem we have a transitive countable set $M\prec V\_{\omega\_{1}}$. In particular, $M\in H(\aleph\_{1})$. We have $H(\aleph\_{1})\prec\_{1}V\_{\omega\_{1}}$, so $M\prec\_{1}H(\aleph\_{1})$. By the condensation lemma $M=L(\alpha)$. $\alpha$ must be a limit ordinal, but $L(\beta)\in L(\beta+1)$, so $L(\alpha)\models \forall x\exists \beta \; x\in L(\beta)$ for all ordinal limit $\alpha$. This is a contradiction with $M\prec V\_{\omega\_{1}}$.
Where is the mistake? I didn't find it.
Edit: I think the Mostowski collapse lemma is not the core of problem.
We have that exist a $\alpha$ limit with $V\_{\omega\_{1}}\in L(\alpha)$, and we have $V\_{\omega\_{1}}\prec\_{1}L(\alpha)$, but as $V\_{\omega\_{1}}$ is transitive his transitive collapse is himself, but is not possible.
The heart of problem is the relation $V\_{\omega\_{1}}\prec\_{1} L(\alpha)$, but why is it wrong?
Edit: thanks the comentaries the solution is, despite $V\_{\omega\_{1}}$ and $L(\alpha)$ satisfies the same $\Sigma\_{1}$ sentences $\varphi$, we need to consider all formulas $\varphi(x)$ that are $\Sigma\_{1}$
| https://mathoverflow.net/users/472959 | A doubt about the Gödel condensation lemma | Your mistake is that taking the Mostowski collapse does **not** preserve elementarity.
We do have a countable transitive $A$ and a countable $M$ with $$A\cong M\preccurlyeq V\_{\omega\_1},$$ where $M$ comes from downward Lowenheim-Skolem applied to $V\_{\omega\_1}$ and $A$ is the collapse of $M$, but that does not imply $A\preccurlyeq V\_{\omega\_1}$. Elementary substructurehood is more than just agreement of theories, it also takes into account exactly how the smaller structure sits inside the larger one. Moving the smaller structure around (e.g. via the Mostowski collapse) may break elementarity.
| 13 | https://mathoverflow.net/users/8133 | 417377 | 170,023 |
https://mathoverflow.net/questions/416940 | 10 | $\DeclareMathOperator\Spec{Spec}\newcommand{\perf}{\mathrm{perf}}\DeclareMathOperator\SHC{SHC}$I have just finished reading the paper "The spectrum of prime ideals in tensor triangulated categories" in which Balmer proposes his notion of spectrum which nowadays is considered central in the understanding and classification of the homotopy categories which we want to study in the concrete mathematical practice (to name a few examples: the $G$-equivariant stable homotopy category for $G$ a compact Lie group, or the derived category of quasi-coherent sheaves on a scheme).
Since I am not familiar with this notion I wanted to ask here various questions about the underlying ideas of such concept.
(1) I noticed that all the examples proposed by Balmer in his paper deal with compact objects, in the sense that the proposed tensor triangulated categories (t.t. categories from now on) can be identified with the full-subcategories of compact objects in a larger t.t. category. And from what I remember every other example which I read in different sources does the same thing: we study the Balmer spectrum of compact objects in a larger t.t. category. Balmer does not explicitly state that this must be the case, indeed his definition does not require the involved objects to be compact a priori.
For this abstract machinery to work we only need the t.t. category to be essentially small. I could think that this is the problem: in general we cannot guarantee that the t.t. category we are interested in is essentially small so we restrict to the subcategory of its compact objects for this property to be more likely.
But I have other reasons to believe that this justification is not completely correct: if we indulge in the intuition suggested by the choice of words, we should think of the support of an object in our t.t. category as an higher categorical analogue of the usual support of a function. Fixing the domain of our functions to be compact spaces ensures that the support will also be compact. So if we consider also non-compact objects the support could be non "topologically small".
Thus I am inclined to believe for the complete t.t. categories either the Balmer spectrum is too big to be computed or its is not the correct notion we want to use to classify their tensor subcategories.
(2) Related to the previous question: if the proposed notion of Balmer spectrum should be applied only to categories of compact objects, what can we deduce about the whole category of possibly non-compact objects? Suppose we consider an essentially small t.t. category $\mathcal{T}$ and we manage to compute the Balmer spectrum of $\mathcal{T}^c$, can we deduce any information regarding the thick tensor ideals or localizing tensor ideals of $\mathcal{T}$?
Two classical examples of this are $D(R)$, the derived category of a commutative ring $R$, and $\SHC$, the stable homotopy category. For $D^{\perf}(R)$ this is homeomorphic to the usual Zariski spectrum $\Spec(R)$, while for $\SHC^\mathrm{c}$ we have the classification provided by the thick subcategory theorem from chromatic homotopy theory. But I have never seen a classification (even partial) of their thick tensor subcategories or thick localizing subcategories.
(3) What information does the Balmer spectrum encode? Balmer proves that there is a bijection between the Thomason subsets of this spectrum and the radical thick tensor ideals of the t.t. category. But other than this? At first I expected that if two t.t. categories had isomorphic spectrum then they would have a sufficiently compatible t.t. structure. Then I found the following interesting example: we have that the Balmer spectrum of the category of compact rational $S^1$-equivariant spectra is homeomorphic to $\Spec(\mathbb{Z})$. If $H \leq S^1$ is a closed subgroup then the kernel of $\phi^H$, the non-equivariant geometric $H$-fixed points, provides a Balmer prime. Then $\ker \phi^{S^1}$ corresponds to the generic point $(0)$, while $\ker \phi^{C\_n}$ can be mapped to $(p\_n)$ where we order the prime numbers $\{p\_n : n \geq 1 \}$.
Therefore $S^1\text{-}\SHC^\mathrm{c}\_{\mathbb{Q}}$ and $D^{\perf}(\mathbb{Z})$ have the same Balmer spectrum, but they are very different t.t. categories: for one, the latter has a compact generator given by the tensor unit, while this is not the case in the former category. I would have thought that the t.t. structure would have been more rigid with respect to the Balmer spectrum, but this seems not to be the case.
If you wanted a more precise question: if two t.t. categories have homeomorphic Balmer spectra, can we translate this to any information on the two categories? What if the homeomorphism is induced by a monoidal exact functor? Can we deduce it is fully faithful, essentially surjective or any other property?
I hope that my questions are not too vague or naïve.
| https://mathoverflow.net/users/131453 | Understanding Balmer spectra | I am not an expert in tt-geometry, but let me try to answer some of your questions.
(1) You are correct, the Balmer spectrum is typically not well-suited to study the "big" categories - this is because all definitions that appear only use "finitary" things : tensor products, cones/extensions, finite direct sums, retracts. This makes it, as defined, ill-suited for studying big categories where you also have interesting infinitary phenomena.
In a big category, you might be more interested in studying localizing ideals for instance, where you can take arbitrary (homotopy) colimits, but then you run into subtle issues about compact generation and telescope conjectures etc. (which are also studied !)
This is not a hopeless situation, though : a lot of work has been done (is probably being done) about finding suitable notions of support for "big" categories (see e.g. Balmer's paper *Homological support of big objects in tensor-triangulated categories* - this is far from the only one on the topic, see e.g. *Big categories, big spectra* by Balchin and Stevenson)
In fact, the place where the Balmer spectrum is somehow the best suited is when the monoidal structure interacts well with finiteness: namely in rigid situations (resp. rigidly compactly generated).
There was not a clear question here, so I hope this answers it.
(2) I think I've answered this partially in my answer to (1). Thick tensor ideals in the small world give rise to localizing ideals in the big world, but in general there is no way to go back, and even when there is the comparison is not perfect (a keyword here is telescope conjecture; but it's not the only thing, and the example of $SHC$ should be enlightening : say we look at a prime $p$, then the kernel of $K(n)\otimes -$ and $E\_n \otimes -$ are very different, as witnessed by the difference between $L\_{K(n)}$ and $L\_n$, but their kernels agree in $SHC^c$). The Balchin-Stevenson paper I mentioned earlier has a section "Comparison maps". Probably other papers that study this kind of thing raise the same kind of question, so you might want to look at that literature (if someone more knowledgeable wants to edit my answer and add some references about this, they would be most welcome !).
The moral is somehow that "big" things are harder to classify.
(3) An abstract homeomorphism of spectra is unlikely to give you any information, except that the "large scale" structure of the two tt-categories is the same (but that would be tautological : one could define this large scale structure by the Balmer spectrum) . This is the same thing with ordinary commutative rings: an abstract homeomorphism of spectra won't tell you much.
You can say much more if the homeomorphism is induced by a tt-functor between them f course, and somehow the functoriality of the Balmer spectrum is key to Balmer's approach, and to computations (e.g. the computation of the spectrum of the equivariant stable homotopy category relies heavily on leveraging the various geometric fixed points functors that one has). If you think in terms of rings, a morphism of commutative rings $R\to S$ that induces a homeomorphism of spectra doesn't tell you that they are isomorphic : indeed, there is some nilpotent business happening here. But you can think of it as some [going up/down theorem](https://en.wikipedia.org/wiki/Going_up_and_going_down).
Balmer has a paper about this kind of question, called *On the surjectivity of the map of spectra associated to a tensor-triangular functor*. He proves there for instance that if the map of spectra is surjective (on closed points), then the original functor is conservative. He further completely characterizes surjectivity in terms of detection of nilpotence (which is related to the nilpotent problem I mentioned earlier for rings).
Certainly, more things can be said if the map is a homeomorphism, and again someone more aware of the literature on the topic could probably say more than I did (if anyone wants to edit and add some references, it would be great, as before).
As a general rule of thumb, a good place to test ideas/conjectures about this stuff is with commutative rings or more generally schemes : the spectrum of the perfect derived category of a (nice) scheme $X$ is exactly (the underlying space of) $X$, in a way compatible with morphisms of schemes. This is to some extent not the most interesting case, but it's a good way to check intuitions.
| 4 | https://mathoverflow.net/users/102343 | 417386 | 170,026 |
https://mathoverflow.net/questions/417393 | 1 | It is known that cobordism provides a complete classification of surfaces as: a surface is cobordant to either $S^2$ or $\mathbb{R}P^2$. I am looking for a reference with contains a proof of this fact.
| https://mathoverflow.net/users/475761 | Reference request: full classification of surfaces as being cobordant to $S^2$ or $\mathbb{R}P^2$ | The orientable closed surface with genus $g$ (denoted as $F\_g$) can be realized as the boundary of a central-symmetric closed body $K$ in $\Bbb{R}^3$ (for example, $F\_1 \cong \Bbb{T}^2$ can be realized as the boundary of a solid torus centered at the origin). Choose a large closed ball $B$ centered at the origin with its interior containing $K$, then $M=B \backslash (K^\circ)$ is the coboundary of $F\_g$ and $\Bbb{S}^2$.
Quotient $M$ by the group action generated from $x \mapsto -x$, the resulting manifold is the coboundary of the non-orientable closed surface with genus $[\dfrac{g+3}{2}]$ and $\Bbb{RP}^2$.
| 3 | https://mathoverflow.net/users/166298 | 417395 | 170,028 |
https://mathoverflow.net/questions/417390 | 3 | Let $A=[a\_{ij}]$ be a $3\times 3$ matrix, where $a\_{ii}$ is a real number, and $a\_{ji}=\overline{a\_{ij}}$ is the complex conjugate of $a\_{ij}$ for all $1\leq i,j\leq 3$, i.e $A^t=[\overline{a\_{ij}}]$. Let $\lambda\_1,\lambda\_2,\lambda\_3$ be eigenvalues of $A$ (not necessarily distinct) and $u\_i=(a\_i,b\_i,c\_i)$ be an eigenvector of $A$ corresponding to $\lambda\_i$, $i=1,2,3$. Is this true that
$\sum\_{i=1}^3\frac{|x\_i\overline{y\_i}|}{||u\_i||^2}\leq 1$, where $x\_i,y\_i\in\{a\_i,b\_i,c\_i\}$ (not necessarily distinct)? Is this true in general for an $n\times n$ matrix by replacing $n$ with $3$?
| https://mathoverflow.net/users/36341 | An inequality relating to entries of eigenvectors | The inequality holds for any $n$ if all eigenvalues of the Hermitian matrix $A$ are distinct, so that the eigenvectors form a unitary matrix.$^\ast$
It does not hold if some eigenvalues are identical.
In that case the corresponding eigenvectors need not be orthogonal, you could choose them nearly parallel, say $u\_1=(1,0,0)$ and $u\_2=(\sqrt{1-\epsilon},\sqrt\epsilon,0)$, with $u\_3=(0,0,1)$; then take $x\_i=a\_i$, $y\_i=a\_i$, and you find that the inequality is violated:
$$\sum\_{i=1}^3\frac{|x\_i\overline{y\_i}|}{||u\_i||^2}=2-\epsilon>1.$$
---
$^\ast$ If the eigenvectors $u\_i$ are the rows of an $n\times n$ unitary matrix $U$, the inequality in the OP is the statement that
$$\sum\_{i=1}^n |U\_{ij}\bar{U}\_{ik}|\leq 1,$$
for any pair of integers $j,k\in\{1,2,\ldots n\}$. Define $U'\_{ij}=U\_{ij}\,\exp\bigl(-i\arg[U\_{ij}\bar{U}\_{ik}]\bigr)$,
then
$$\left(\sum\_{i=1}^n |U\_{ij}\bar{U}\_{ik}|\right)^2=\left(\sum\_{i=1}^n U'\_{ij}\bar{U}\_{ik}\right)^2\leq\left(\sum\_{i=1}^n|U'\_{ij}|^2\right)\left(\sum\_{i=1}^n|U\_{ij}|^2\right)=1.$$
| 4 | https://mathoverflow.net/users/11260 | 417396 | 170,029 |
https://mathoverflow.net/questions/417392 | 1 | I have noticed experimentally that the following question has a positive answer.
Let $p>5$ and $H$ be a subgroup of $(\mathbb Z/p\mathbb Z) ^\*$, with $a\in H$ and $a>2$.
Is it true that $$(a-1)\; | \; p\sum\limits\_{h \in H} (-h/p \mod a) ?$$
| https://mathoverflow.net/users/110301 | Strange result of divisibility | I assume that in the calculations, you are identifying the elements of $\mathbb Z/p\mathbb Z$ with $\{0,\dots,p-1\}\subseteq\mathbb Z$, and likewise, that the $\bmod a$ operation takes values in $\{0,\dots,a-1\}$. Then the result follows from
>
> **Lemma:** If $a,b>0$ are coprime and $-a<h<b$, then $$a(ha^{-1}\bmod b)-b((-hb^{-1})\bmod a)=h.$$
>
>
>
**Proof:** By negating $h$ and swapping the roles of $a$ and $b$ if necessary, we may assume $h\ge0$. Let $u=(ha^{-1}\bmod b)$. Then $au=h+bv$ for some $v$, and since $0\le au<ab$ and $0\le h<b$, we have $0\le v=\lfloor au/b\rfloor<a$. Also $bv\equiv-h\pmod a$, thus $v=((-hb^{-1})\bmod a)$. QED
Consequently,
$$\begin{align\*}
p\sum\_{h\in H}((-hp^{-1})\bmod a)
&=a\sum\_{h\in H}(ha^{-1}\bmod p)-\sum\_{h\in H}h\\
&=(a-1)\sum\_{h\in H}h,
\end{align\*}$$
as $\{ha^{-1}\bmod p:h\in H\}$ is just another enumeration of $H$.
| 4 | https://mathoverflow.net/users/12705 | 417398 | 170,031 |
https://mathoverflow.net/questions/417408 | 1 | I have an integral of the form
$$ \int\_R^{\infty} e^{-x} x^n \vert L\_m^{\alpha}(x) \vert^2 \ dx,$$
where $L\_m^{\alpha}$ is the generalized Laguerre polynomial and $n \ge 0.$
I would to get a nice explicit exponential bound on this integral in terms of $R$ (I intend to take $R$ large). However, most upper bounds on these polynomials that I could find involve exponentials themselves, see for instance here [on DLMF.](https://dlmf.nist.gov/18.14)
Does anybody see a way to get a nice estimate on this integral for $R$ large? The exponential decay is clear, as we are integrating an exponential against a polynomial. However, I would like to have a bound that is a good tradeoff between compact and efficient.
| https://mathoverflow.net/users/457901 | Exponential decay bound on integral | Let $a:=\alpha$. The integral in question is
\begin{equation\*}
I(R):=\int\_R^\infty e^{-x} x^n L\_m^a(x)^2\, dx,
\end{equation\*}
[where](https://en.wikipedia.org/wiki/Laguerre_polynomials#Explicit_examples_and_properties_of_the_generalized_Laguerre_polynomials)
\begin{equation\*}
L\_m^a(x)=\sum\_{i=0}^m b\_i,\quad b\_i:=(-1)^i \binom{m+a}{m-i}\frac{x^i}{i!},
\end{equation\*}
so that for $i=0,\dots,m-1$ and $x\ge R>0$
\begin{equation\*}
\frac{|b\_i|}{|b\_{i+1}|}=\frac{i+1}{(m-i)x}\,|i+1+a|\le\frac{c\_{m,a}}R,\quad
c\_{m,a}:=m\max(|m+a|,|a+1|)
\end{equation\*}
and hence for $R>c\_{m,a}$
\begin{equation\*}
L\_m^a(x)^2\le\frac{b\_m^2}{(1-c\_{m,a}/R)^2}
=\frac1{(1-c\_{m,a}/R)^2}\frac{x^{2m}}{(m!)^2}.
\end{equation\*}
Thus, for $R>\max(c\_{m,a},n+2m)$, by [Proposition 2.7](https://www.semanticscholar.org/paper/Exact-lower-and-upper-bounds-on-the-incomplete-Pinelis/df79633882e070bc48bda463ccbbc9919a256030),
\begin{equation\*}
\begin{aligned}
I(R)&\le J(R):=\frac1{(1-c\_{m,a}/R)^2}\frac1{(m!)^2}\,\frac{e^{-R}R^{n+2m}}{1-(n+2m)/R} \\
&\sim\frac{e^{-R}R^{n+2m}}{(m!)^2}
\end{aligned}
\tag{1}\label{1}
\end{equation\*}
as $R\to\infty$. So, we do get an exponential decrease in $R$.
The upper bound $J(R)$ on $I(R)$ in \eqref{1} is asymptotically exact (as $R\to\infty$), since $L\_m^a(x)^2\sim\dfrac{x^{2m}}{(m!)^2}$ as $x\to\infty$.
| 2 | https://mathoverflow.net/users/36721 | 417418 | 170,034 |
https://mathoverflow.net/questions/417414 | 1 | $T>0$ is a parameter.
Consider the linear Diophantine equation $ax+by=c$ where $a,b$ are coprime.
Suppose $a,b$ are of magnitude $T^{1+\epsilon}$ and $c$ is of magnitude $T^2$.
1. For how many such equations we can expect $x,y$ to be of magnitude $T^{1+\epsilon}$ for a fixed $c$ and we vary $a,b$ coprime of magnitude $T^{1+\epsilon}$? Call such solutions Fundamental.
Such solutions are also **mininum normed**.
2. The usual way of solving such equations is to solve $ax'+by'=1$ and choose $x=x'c$ and $y=y'c$. But this does not provide a polynomial time algorithm to find $x,y$ of magnitude $T^{1+\epsilon}$ if there exists one. How do we find such solutions when they exist without using integer programming and directly using number theory?
| https://mathoverflow.net/users/10035 | Fundamental solutions to linear Diophantine equations and their existence and computation | Solve $ax'+by'=1$, then take $$x = x'c - b \left\lfloor \frac{x'c}{ b} \right\rfloor$$
$$y = y'c +a\left\lfloor \frac{x'c}{ b} \right\rfloor$$
then we have $ax+by= ax'c +by'c = c$ and (if $b>0$ for simplicity) $0 \leq x < b$ so $$|x| < b$$ and $$|y| = \left| \frac{c-ax}{b} \right| \leq \frac{|c|}{|b|} + \frac{|a||x|}{|b|}< \frac{|c|}{|b|}+a\leq T^{1-\epsilon} + T^{1+\epsilon}.$$ So there is always a solution, and the algorithm to find it is clearly polynomial time.
We can optimize the norm by subtract one more copy of $b$ from $x$ if it helps.
| 4 | https://mathoverflow.net/users/18060 | 417419 | 170,035 |
https://mathoverflow.net/questions/417404 | 3 | I have noticed experimentally that the following question has a positive answer.
Is it true that for all even and convex functions $f$, $g$:
$$\int\_0^1 f(\sin(1/x)) \times g(\cos(1/x)) dx \leq \int\_0^1 f(\sin(1/x)) dx \times \int\_0^1 g(\cos(1/x))dx? $$
| https://mathoverflow.net/users/110301 | $\int_0^1 f(\sin(1/x)) \times g(\cos(1/x)) dx \leq \int_0^1 f(\sin(1/x)) dx \times \int_0^1 g(\cos(1/x))dx? $ | $\newcommand\abs[1]{\lvert#1\rvert}$It already suffices that $f$ and $g$ be even and nondecreasing on $[0,1]$ (which of course is the case if $f$ and $g$ are even and convex). Indeed, then the identity $\abs\sin^2+\abs\cos^2=1$ implies that for all real $u$, $v$ we have
$$(\abs{\sin u}-\abs{\sin v})(\abs{\cos u}-\abs{\cos v})\le0\tag{1}\label{1}$$
and hence
$$\begin{aligned}h(u,v)&:=[f(\sin u)-f(\sin v)][g(\cos u)-g(\cos v)] \\
&=[f(\abs{\sin u})-f(\abs{\sin v})][g(\abs{\cos u})-g(\abs{\cos v})]\le0,
\end{aligned}$$
so that the difference between the left-hand side of your inequality and its right-hand side is
$$\frac12\,\int\_0^1\int\_0^1 dx\,dy\,h\Big(\frac1x,\frac1y\Big)\le0.
$$
---
One may note that the above reasoning holds if in the inequality in question one replaces all instances of $1/x$ by $k(x)$, where $k$ is any Borel-measurable function from $(0,1)$ to $\mathbb R$. Also, one can replace $\sin$ and $\cos$ by any functions $S$ and $C$ from $\mathbb R$ to $\mathbb R$ that are Borel-measurable and (say) bounded and "negatively dependent" in the sense that \eqref{1} holds with $S$ and $C$ in place of $\sin$ and $\cos$:
$$(\abs{S(u)}-\abs{S(v)})(\abs{C(u)}-\abs{C(v)})\le0
\tag{2}\label{2}$$
for all real $u$, $v$.
In particular, inequality \eqref{2} will hold if $|S|$ is any increasing function and $|C|$ is any decreasing one (or vice versa).
| 7 | https://mathoverflow.net/users/36721 | 417421 | 170,036 |
https://mathoverflow.net/questions/417411 | 3 | $\DeclareMathOperator\Mod{Mod}$Let $S$ be a surface and $P=\{a\_1,...,a\_n\}$ be a pants decomposition of $S$. Denote by $\Mod(S)$ the mapping class group of $S$.
Define the stabilizer of $\Mod(S)$ on $P$ to be $$A=\{f\in \Mod(S), f(a\_i)=a\_i, i =1,...,n\}.$$
What is $A$?
I was once told that it is generated by Dehn twists and hyperelliptic involution, but I cannot find a reference for that. Is it true, and why? Could you give a reference for this?
I saw this on Lemma 3.2 [in this paper by U. Wolf](https://www.math.kit.edu/iag3/%7Ewolf/media/wolf-the-action-of-the-mapping-class-group-on-the-pants-complex.pdf), which says that $A$ is generated by Dehn twists and half twists But I didn't understand it yet.
| https://mathoverflow.net/users/nan | Stabilizer of the action of the mapping class group on a pants decomposition | Suppose that $S$ is closed (without boundary), connected, and oriented.
If $S$ has genus two then the stabiliser is generated by Dehn twists about the $a\_i$, the hyperelliptic, and a reflection.
If $S$ has genus greater than two, then there is no hyperelliptic symmetry, but there will still be a reflection symmetry.
This is because any mapping class that stabilises the "cuffs" of the pants (the curves $a\_i$) either permutes the pants (and so there are exactly two pants) or preserves each pants setwise (and thus is isotopic to the identity, or a reflection, off of a small neighbourhood of the cuffs).
| 2 | https://mathoverflow.net/users/1650 | 417432 | 170,039 |
https://mathoverflow.net/questions/417424 | 2 | My question is similar to [The mean of points on a unit n-sphere $S^n$](https://mathoverflow.net/questions/231501/the-mean-of-points-on-a-unit-n-sphere-sn).
I have a unit $n$-sphere $S^n$ and a set $P$ of points lying on its surface.
I use geodesic distance metric $d(p,q)=\arccos(pq^T)$.
Additionally I have a guarantee that all of my points are positive (unit vectors with all coordinates greater than or equal to 0).
The task is to find a centroid of those points, whose uniqueness is guaranteed because all points are positive.
In such case, does the centroid always coincide with the normalised arithmetic mean of the points? Formally
\begin{gather\*}
\mu\_\text{arithmetic} = \frac{1}{\lvert P\rvert}\sum\_{p\in P}p \\
\mu\_\text{geodesic}=\arg\min\_c\sum\_{p\in P}d(p,c)^2.
\end{gather\*}
Question: do we have
$$\mu\_\text{geodesic}=\frac{\mu\_\text{arithmetic}}{\lVert\mu\_\text{arithmetic}\rVert}?$$
| https://mathoverflow.net/users/163471 | The mean of positive points on a unit $n$-sphere $S^n$ | The answer is no. E.g., let $P$ be the set $\{1,e^{it},e^{i3t}\}$ of points on the unit circle in $\mathbb C=\mathbb R^2$, where $t$ is a small positive real number. Then the geodesic mean of $P$ is
$$e^{i(4/3)t}=1+\frac{4 i t}{3}-\frac{8 t^2}{9}-\frac{32 i t^3}{81}+O\left(t^4\right),$$
whereas the arithmetic mean of $P$ is
$$\frac{1+e^{i t}+e^{3 i t}}{\left| 1+e^{i t}+e^{3 i t}\right| }
=1+\frac{4 i t}{3}-\frac{8 t^2}{9}-\frac{14 i t^3}{27}+O\left(t^4\right),$$
so that the geodesic mean of $P$ differs from the arithmetic mean.
| 3 | https://mathoverflow.net/users/36721 | 417434 | 170,040 |
https://mathoverflow.net/questions/417439 | 2 | Let $p$ integer prime, $f$ a function of $A=\mathbb F\_p^n$ to $\mathbb F\_p$, with $n\geq p+1$.
Is it true that : for all $x\in A, \sum\limits\_{\sigma \in S\_n} s(\sigma) \times f(x\_\sigma) =0$?
$s$ the signature
$S\_n$ is the group of all bijection of $U\_n=\{1,...,n\}$ to $U\_n$.
If $x=(x\_1,...x\_n)$ then $x\_\sigma=(x\_{\sigma(1)},...,x\_{\sigma(n)})$.
>
> The answer is yes, if $p=2$ but what about the other case?
>
>
>
| https://mathoverflow.net/users/110301 | A new convolution, on function of $\mathbb F_p^n$ to $\mathbb F_p$ still zero? | As long as $n \geq p+1$, two of the entries of $x$ must be the same by the pigeonhole principle. Let $\tau$ be a transposition fixing those two entries. Then $s(\sigma \circ \tau) = -s(\sigma)$ but $x\_{\sigma \circ \tau} = x\_\sigma$ so $f(x\_{\sigma \circ \tau} ) =f(x\_\sigma)$. Thus the terms in your sum for $\sigma$ and $\sigma \circ \tau$ cancel.
Since ${}\circ \tau$ acts on $S\_n$ with $n!/2$ orbits of size 2, we can divide the sum into $n!/2$ pairs of terms which cancel, so the whole sum vanishes.
| 7 | https://mathoverflow.net/users/18060 | 417442 | 170,041 |
https://mathoverflow.net/questions/417444 | 6 | I have a question involving preservation of cofiltered limits. Ordinarily this would be a very boring question, but it comes up in condensed math in its analogue of the completeness concept.
The concept is that of "solid", which says that for each profinite set $S = \text{lim} S\_i$, maps of condensed abelian groups $\mathbb{Z}[S] \rightarrow M$ lift against the canonical map $\mathbb{Z}[S] \rightarrow \text{lim} \mathbb{Z}[S\_i]$.
One broad goal of condensed math is to extend commutative algebra to subsume some areas of functional analysis. The concept of solid plays a role similar to completion. But I don't as of yet understand the condition very well.
I don't understand why the free condensed abelian group $\mathbb{Z}[S]$ is not identically $\text{lim} \mathbb{Z}[S\_i]$. I am trying to come up with a broad class of examples as follows:
>
> Let $\mathcal{S}$ be a site. Consider the category of sheaves of sets on the site $\mathcal{S}$, $\text{Sh}(\mathcal{S}, \text{Set})$, and the category of sheaves of abelian groups $\text{Sh}(\mathcal{S}, \text{Ab})$ on the site $\mathcal{S}$. I am looking for a general class of examples showing that the free functor $\text{Sh}(\mathcal{S}, \text{Set}) \rightarrow \text{Sh}(\mathcal{S}, \text{Ab})$ does not preserve cofiltered limits.
>
>
>
| https://mathoverflow.net/users/30211 | Condensed math and cofiltered limits | It should be noted that already in $\mathbf{Set}$, the free functor $\mathbf Z^{(-)} \colon \mathbf{Set} \to \mathbf{Ab}$ does not preserve cofiltered limits. For a cofiltered diagram $D \colon \mathcal I \to \mathbf{Set}$, write $S\_i$ for its value at $i \in \mathcal I$, write $S$ for its limit, and write $\pi\_i \colon S \to S\_i$ for the canonical projection.
There is always a map
\begin{align\*}
\phi \colon \mathbf Z^{(S)} &\to \lim\_\leftarrow \mathbf Z^{(S\_i)}\\
\sum\_{k=1}^n n\_k s\_k &\mapsto \left( \sum\_{k=1}^n n\_k\pi\_i(s\_k) \right)\_i.
\end{align\*}
Denote its $i^{\operatorname{th}}$ component by $\phi\_i$. For a set $X$ and an element $z=\sum\_{k=1}^n n\_k x\_k \in \mathbf Z^{(X)}$ with $x\_k \neq x\_{k'}$ for $k \neq k'$, write $\operatorname{Supp}(z)$ for $\{x\_1,\ldots,x\_n\}$, and denote by $|z|$ its cardinality. If $f \colon X \to Y$ is a map, we denote the induced map $\mathbf Z^{(X)} \to \mathbf Z^{(Y)}$ by $f$ as well (by abuse of notation). For $z \in \mathbf Z^{(X)}$, we have $\operatorname{Supp}(f(z)) \subseteq f(\operatorname{Supp}(z))$, so $|f(z)| \leq |z|$ with equality if and only if $f\_\* \colon \operatorname{Supp}(z) \to \operatorname{Supp}(f(z))$ is a bijection.
**Lemma.** *The map $\phi$ is injective, and its image consists of those $(x\_i)\_{i \in \mathcal I}$ for which there exists $n \in \mathbf Z$ with $|x\_i| \leq n$ for all $i \in \mathcal I$.*
In other words, the image consists of the sequences $(x\_i)\_i$ of bounded support.
*Proof.* For injectivity, if $x = \sum\_{k=1}^n n\_ks\_k \in \ker(\phi)$ is such that $s\_k \neq s\_{k'}$ for $k \neq k'$, then there exists $i \in \mathcal I$ such that $\pi\_i(s\_k) \neq \pi\_i(s\_{k'})$ for $k \neq k'$ (here we use that the sum is finite and that $\mathcal I$ is cofiltered). Then $\phi\_i(x) = \sum\_{k=1}^n n\_k\pi\_i(s\_k)$ is zero by assumption, so all $n\_k$ are zero.
For the image, it is clear that $|\phi\_i(x)| \leq n$ for all $i \in \mathcal I$ if $x \in \mathbf Z^{(S)}$ has $|x| = n$. Conversely, if $|x\_i| \leq n$ for all $i \in \mathcal I$, then decreasing $n$ if necessary, we may assume $|x\_{i\_0}| = n$ for some $i\_0 \in \mathcal I$. Then $|x\_i| = n$ for all $f \colon i \to i\_0$ since $n = |x\_{i\_0}| = |D(f)(x\_i)| \leq |x\_i| \leq n$. Replacing $\mathcal I$ by the coinitial segment $\mathcal I/i\_0$ we may therefore assume $|x\_i| = n$ for all $i \in \mathcal I$.
Constancy of $|x\_i|$ means that for every morphism $f \colon i \to j$ in $\mathcal I$, the map $f\_\* \colon \operatorname{Supp}(x\_i) \to \operatorname{Supp}(x\_j)$ is a bijection. Setting $T = \lim\limits\_\leftarrow \operatorname{Supp}(x\_i)$, we see that each projection $\pi\_i \colon T \to \operatorname{Supp}(x\_i)$ is a bijection. Functoriality of the limit gives an injection $T \hookrightarrow S$, giving elements $s\_1,\ldots,s\_n \in S$ such that $\pi\_i(\{s\_1,\ldots,s\_n\}) = \operatorname{Supp}(x\_i)$ for all $i \in \mathcal I$. The coefficients must also be constant under the bijections $\operatorname{Supp}(x\_i) \to \operatorname{Supp}(x\_j)$, so we get an element $x = \sum\_{k=1}^n n\_ks\_k \in \mathbf Z^{(S)}$ with $\phi(x) = (x\_i)\_i$. $\square$
**Example.** An example where $\phi$ is not surjective: let $S = \mathbf Z\_3$ with $S\_i = \mathbf Z/3^i$. Define the element $(x\_i)\_i \in \prod\_i \mathbf Z^{(S\_i)}$ where $x\_i \in \mathbf Z^{(S\_i)} = \mathbf Z^{S\_i}$ has coordinates (for $k \in \{0,\ldots,3^i-1\}$) given by
$$x\_{i,k} = \begin{cases} 1, & \text{the first $3$-adic digit of } k \text{ is } 1, \\ -1, & \text{the first $3$-adic digit of } k \text{ is } 2, \\ 0, & k=0.\end{cases}$$
These form an element of $\lim\limits\_\leftarrow \mathbf Z^{(S\_i)}$: any fibre of $\mathbf Z/3^{i+1} \to \mathbf Z/3^i$ above $k \in \{0,\ldots,3^i-1\}$ consists of $\{k,3^i+k,2 \cdot 3^i+k\}$, of which $x\_{i+1,k} = x\_{i,k}$ and the others are $1$ and $-1$ since $3^i+k$ starts on $1$ and $2 \cdot 3^i+k$ starts on $2$.
Since $\operatorname{Supp}(x\_i) = S\_i \setminus \{0\}$, we see that $(x\_i)\_i$ does not have bounded support, hence is not in the image of $\phi$.
**Corollary.** *For any presheaf topos $\mathbf T = \mathbf{PSh}(\mathscr C) = [\mathscr C^{\operatorname{op}},\mathbf{Set}]$ on a small nonempty category $\mathscr C$, the free functor $\mathbf Z^{(-)} \colon \mathbf T \to \mathbf{Ab}(\mathbf T)$ does not preserve cofiltered limits.*
*Proof.* Let $f \colon \mathscr C \to \*$ be the map to the terminal category $\*$, on which $\mathbf{PSh}(\*) = \mathbf{Set}$. Note that $f$ has a section $g$ since $\mathscr C$ is nonempty. We saw above that in $\mathbf{Set}$, the free functor $\mathbf Z^{(-)}$ does not preserve cofiltered limits. The pullback $f^\* \colon \mathbf{Set} \to \mathbf{PSh}(\mathscr C)$ takes a set $S$ to the constant sheaf $\underline{S}$ on $\mathscr C$. Since limits, colimits, and free abelian group objects in presheaf categories are pointwise, $f^\*$ commutes with formation of limits, colimits, and free abelian group objects. Thus pulling back everything along $f^\*$ gives the natural map $\underline{\mathbf Z}^{(\underline S)} \to \lim\limits\_\leftarrow \underline{\mathbf Z}^{(\underline S\_i)}$, which is not an isomorphism since it isn't after applying the section $g^\*$ (this is just evaluation at an object $g(\*) \in \mathscr C$). $\square$
The class of topoi where the free functor does not commute with cofiltered limits is probably much larger still. For instance, my example does not include condensed sets, which is close to a presheaf topos but not quite. I'm not even sure what a topos would look like where these *do* commute!
| 12 | https://mathoverflow.net/users/82179 | 417452 | 170,043 |
https://mathoverflow.net/questions/417454 | 11 | Recently I learned a nice [constructive proof of the irrationality of $\sqrt{2}$](https://en.wikipedia.org/wiki/Square_root_of_2#Constructive_proof), which uses the 2-adic valuation of an integer: the count of how many times a number is divisible by 2. The valuation requires *some* induction to construct, and [this nice answer by François Dorais](https://mathoverflow.net/questions/19857/has-decidability-got-something-to-do-with-primes/20944#20944) talks about how Robinson's Arithmetic $Q$ isn't strong enough to prove $\sqrt{2}$ irrational.
By the question "how much induction...?" I mean what is the complexity of the statement that is used in the application of induction to prove the existence of the valuation (I think the particular case $p=2$ is not special here). Further, I think the only property really needed in this irrationality proof is that the *parity* of the valuation is well-defined, so in principle it is *this specific property* that I need to know the strength of:
>
> there is a well-defined multiplicative function $p\_2\colon \mathbb{N}\to \{\pm 1\}$ encoding the parity of the 2-adic valuation.
>
>
>
I can easily think of a recursion (say in some dependent type theory, or a proof assistent) that defines this function, but I don't know how to classify the precise strength of the induction principle needed, in the usual arithmetic hierarchy.
[As an aside, I really like this proof, not just because it gives a constructive lower bound on how far a rational is from $\sqrt{2}$, but also because it doesn't rely on more extensive factorisation properties of integers, like one of the most common proofs relying on fractions in 'lowest terms', or on the beautiful, but more subtle, use of infinite descent]
| https://mathoverflow.net/users/4177 | How much induction does a p-adic valuation need? | If you want to stick to theories in the basic language of arithmetic $\langle0,1,+,\cdot,<\rangle$, the irrationality of $\sqrt2$ can be easily proved in the theory $IE\_1$ (i.e., using induction for bounded existential formulas), since it proves the $\gcd$ property; or even more directly, you can just prove
$$\forall a,b<x\,(a^2=2b^2\to b=0)$$
by induction on $x$ (this is a bounded *universal* formula rather than existential, but $IE\_1=IU\_1$).
$IE\_1$ also proves that any number $x$ can be written uniquely as $yz$ where $y$ is odd and $z$ a power of two (meaning $z$ has no odd divisor apart from $1$), and you can define the parity of $v\_2(x)$ by saying that $v\_2(x)$ is even iff $z$ is a square. This will be multiplicative as required in the question. Alternatively (but leading to an equivalent definition), $IE\_1$ proves that any $x$ can be uniquely written as $y^2z$ with $z$ square-free, and then we can define that $v\_2(x)$ is even iff $z$ is odd.
Some larger fragment of $I\Delta\_0$ has a $\Delta\_0$-definition of the graph of exponentiation, which will enable to define $v\_2(x)$ itself.
However, even $IE\_1$ is essentially overkill. People usually study bounded arithmetic in languages that are in various ways more suitable than the basic language of arithmetic, and then the 2-adic valuation is definable in very weak fragments; usually not because of some clever trick, but since it more or less belongs to the language. E.g., whenever you have a well-defined bit predicate
$$\DeclareMathOperator\bit{bit}\bit(i,X)=\lfloor X2^{-i}\rfloor\bmod2,$$
you can define
$$v\_2(X)=i\iff\bit(i,X)=1\land\forall j<i\:\bit(j,X)=0.$$
In particular, you can do just that (with a $\Sigma^B\_0$ formula) for binary integers $X$ in the basic theory $V^0$ (see [1]), which cannot even define general *multiplication*, hence it has no meaningful way of stating (let alone proving) the irrationality of $\sqrt2$. The latter can be done in the extension $\mathrm{VTC}^0$ of $\mathrm V^0$ (see [1] again for the definition).
In the realm of one-sorted theories of arithmetic, which is perhaps more in line with the spirit of the question, you can do all this using $\Sigma^b\_0$ length-induction (i.e., in the theory $\Sigma^b\_0$-LIND as defined e.g. in [2]).
[1] Stephen A. Cook, Phuong Nguyen: *Logical foundations of proof
complexity*, Cambridge University Press, 2010. <https://doi.org/10.1017/CBO9780511676277>
[2] Chris Pollett: [*A propositional proof system for $R^i\_2$*](http://www.cs.sjsu.edu/faculty/pollett/papers/psystem.pdf). In: Proof complexity and feasible arithmetics (P. Beame, S. Buss, eds.), DIMACS Series in Discrete Mathematics and Theoretical Computer Science 39 39, AMS, 1997, pp. 253–278.
| 9 | https://mathoverflow.net/users/12705 | 417460 | 170,045 |
https://mathoverflow.net/questions/333536 | 11 | $\DeclareMathOperator\PGL{PGL}\DeclareMathOperator\Conj{Conj}$Let $G$ be the symmetric group $S\_n$ or the projective general linear group $\PGL\_2(n)$.
Let $X$ be a cyclically reduced word in the abstract variables
$x\_1, x\_2, \ldots,x\_k$, i.e. $X$ is a product containing $x\_1, x\_2, \ldots,x\_k$ and their inverses, without any element appearing next to its own inverse in any cyclic permutation. (Only words with length $4$, $6$, $8$ are needed in my research.)
Consider the probability $P$ that the word sums to $1$, with each $x\_i$ chosen uniformly and independently from $G$.
**Question:**
What are the upper bounds of $\log\_{|G|}P$?
If $\log\_{|G|}P$ converges when $n\to\infty$, what's the value?
Answers are acceptable for either $G=S\_n$ or $G=\PGL\_2(n)$.
**Known:**
If there's a variable occurring only once in $X$, then $P$ is exactly $1/|G|$.
If $X=x\_1^k$, then the limit is $-1/k$ for symmetric groups by David E Speyer's argument.
As Richard Stanley pointed out, if $X=x\_1x\_2x\_1^{-1}x\_2^{-1}$, then $P=|\Conj(G)|/|G|$. ($|\Conj(G)|$ is the number of conjugacy classes of $G$)
The formula $P=|\Conj(G)|/|G|$ holds for the words $x\_1x\_1x\_2x\_2$ and $x\_1x\_2x\_1x\_2^{-1}$ if all the characters of $G$ are real, and that's exactly the case for $S\_n$ and $\PGL\_2(n)$.
| https://mathoverflow.net/users/125498 | Probability of words summing to $1$ in $S_n$ or $\mathrm{PGL}_2(n)$ | $\DeclareMathOperator\PGL{PGL}$I believe the best result in this direction is due to M. Larsen & A. Shalev (2012); see [this](https://www.researchgate.net/publication/256735756_Fibers_of_word_maps_and_some_applications) paper. I'll summarize their results here. This doesn't answer the questions whether the limit exists or what its precise value is, though, so this is not a full answer.
Denote the length of the word $X$ by $\ell$ (while $k$ is the size of the alphabet).
As for $S\_n$, Proposition 2.3 gives $P < n!^{-\eta}$ for large enough $n$, if $\eta < \frac{2k-1}{4\left((2k)^{2\ell+1}-1\right)}$, so $\log\_{|S\_n|}P < -\eta. $
As for $\PGL\_n(\mathbb{F}\_q)$, Proposition 3.3 gives $\log\_{|G|}P \le -\eta$ for large enough $n$ (and every $q$), if $\eta < \frac{1}{1800\ell^2}$. (This holds for all classical groups of Lie type).
Shalev and Larsen handle all finite simple groups in their paper. They also refer to [this](https://people.math.carleton.ca/%7Ejdixon/Residual.pdf) paper, which gives a bound for simple groups $G$ of Lie type of bounded
rank $r$: $ P \le \frac{C}{q}$ where $C = C(r, X)$ is a constant, and $q \ge G^{\epsilon}$ for some $\epsilon = \epsilon(r)$, hence $\log\_{|G|}P \le -\eta$ for large enough $G$ if $\eta < \epsilon$.
EDITED: this last paragraph answers the $n=2, q\to\infty$ case.
| 5 | https://mathoverflow.net/users/474608 | 417466 | 170,047 |
https://mathoverflow.net/questions/417462 | 1 | I am looking for a copy of the paper "On some problems of Bellman and a theorem of Romanoff", P. Erdős, J. Chinese Math. Soc. 1951. Can someone help me by providing a link or copy of the paper?
| https://mathoverflow.net/users/160943 | Looking for a 1951 paper by Erdős titled "On some problems of Bellman and a theorem of Romanoff", published in J. Chinese Math. Society | I found a digital version of paper [1] in [the list of published papers of Paul Erdős](https://www.renyi.hu/%7Ep_erdos/Erdos.html): click on the title of the reference below and you'll see the same digital object.
**Reference**
[1] Paul Erdős, "[On some problems of Bellman and a theorem of Romanoff](https://www.renyi.hu/%7Ep_erdos/1951-08.pdf)", Journal of the Chinese Mathematical Society (N.S.) (1951), 409--421, [MR0072161](https://mathscinet.ams.org/mathscinet-getitem?mr=MR0072161).
| 3 | https://mathoverflow.net/users/113756 | 417467 | 170,048 |
https://mathoverflow.net/questions/417238 | 0 | Let $\mu$ be a probability distribution on $\mathbb R^d$ with "sufficiently regular" density $p$. Let $f:\mathbb R^d \to \mathbb R$ be a "sufficiently regular" function. Finally, for every $t \ge 0$, define
$$
s\_f(t) := \mu(f^{-1}((-\infty,t])) = \int\_{f^{-1}((-\infty,t])}p(x)dx
$$
>
> **Question.** *What is the derivative of $s\_f$ w.r.t $t$ ?*
>
>
>
It seems I should be able to solve my problem in principle using Lemma 3.1 of [Malliavin Calculus for non Gaussian differentiable measures and surface measures in Hilbert spaces](https://arxiv.org/abs/1608.05964). However, that paper is hard to parse for a non-expert like myself.
---
Now, suppose $f$ depends "smoothly" on a parameter $\theta$, i.e let $\Theta$ be a nonempty subset of some $\mathbb R^n$ and suppose $F:\mathbb R^d \times \Theta \to \mathbb R$ is "smooth", and for any $\theta \in \Theta$, define $s\_\theta := s\_{f\_\theta}$, where $f\_\theta(x):=F(x,\theta)$ for all $x \in \mathbb R^d$.
>
> **Question.** *For every $t \ge 0$, what is the gradient of $\theta \mapsto s\_\theta(t)$.*
>
>
>
Examples
--------
* *Linear:* $f(x) = w^\top x - c$, for some unit-vector $w \in \mathbb R^d$ and scalar $c \in \mathbb R$. Here $\theta = (w,c) \in \mathbb R^{d + 1}=:\Theta$.
* *Quadratic:* $f(x) = \pm (r^2-\|x\|^2)$, here $\theta = r^2 \in (0,\infty) =: \Theta$.
| https://mathoverflow.net/users/78539 | Time-derivative of integral over sub-level set $s(t) := \int_{f^{-1}((-\infty,t])}p(x)dx$ | This is a comment (but too long for that format) to suggest a simple and elementary approach: we can write
$$s\_f(t)=\int H(t-f(x))p(x)\,dx$$ and manipulate formally (but see below) to get
$$\frac{d}{dt}s\_f(t)=\int \delta(t-f(x))p(x)\,,dx$$ where $H$ and $\delta$ are the Heaviside function and the Dirac distribution respectively. This suggests using the elementary theory of distributions for your computations, which can then be carried out in the concrete situations you describe using well-established formulae.
Similarly, if $$s\_{\theta}(t)=\int H(t-f(x,\theta))p(x)\,dx,$$ we can compute its partial derivatives with respect to the parameters as
$$\int \delta(t-f(x,\theta))\frac{\partial}{\partial \theta\_i}f(x,\theta)\,p(x)\,dx.$$
Note that in order to make this rigorous, one requires, within the context of distribution theory, the notions of parametrised integrals, differentiation under the integral sign, composition of a distribution with a smooth function (and the chain rule) and the product of a smooth function and a distribution. All of these are standard fare (they were developed by a cohort of prominent mathematicians in the 50´s and 60´s, together with methods for explicit computations for concrete functions--all this at the level of a first year analysis course). The two examples for $f$ that you mention are particularly simple.
In case of interest, I would be happy to include references.
| 1 | https://mathoverflow.net/users/317800 | 417471 | 170,049 |
https://mathoverflow.net/questions/417263 | 11 | It seems there are many subtly different notions of the shape of a topological space (and, more generally, toposes).
For instance, Lurie [*Higher topos theory*] defines this one:
**Definition 1.**
The shape of a topos $\mathcal{E}$ is the pro-object in $\mathcal{S}$ representing the endofunctor $p\_\* p^\* : \mathcal{S} \to \mathcal{S}$, where $p$ is the unique geometric morphism from the $\infty$-sheaf topos $\mathcal{E}$ to the $(\infty, 1)$-category $\mathcal{S}$ of $\infty$-groupoids.
Here is another, perhaps closer to what is studied in classical shape theory:
**Definition 2.**
The shape of a topological space $X$ is the pro-object $\varprojlim\_{\mathcal{U} : \textrm{Cov} (X)} \mathrm{B} \mathcal{U}$ where $\mathcal{U}$ ranges over the poset $\textrm{Cov} (X)$ of covering sieves $\mathcal{U}$ of *non-empty* open subspaces of $X$ and $\mathrm{B} \mathcal{U}$ is the geometric realisation of the nerve of $\mathcal{U}$.
There is a canonical comparison from definition 1 to definition 2.
Roughly speaking, it comes from the Grothendieck plus construction for sheafification.
The sheaf $p^\* K$ is the sheafification of the constant presheaf with value $K$, so there is a comparison morphism $\varprojlim\_{U : \mathcal{U}} K \to p\_\* p^\* K$ for each covering sieve $\mathcal{U}$.
We have $\varprojlim\_{U : \mathcal{U}} K \cong \mathcal{S} (\mathrm{B} \mathcal{U}, K)$, and taking the colimit over all covering sieves $\mathcal{U}$ yields a morphism
$${\textstyle \varinjlim}\_{\mathcal{U} : \textrm{Cov} (X)^\textrm{op}} \mathcal{S} (\mathrm{B} \mathcal{U}, K) \to {\textstyle \varinjlim}\_{\mathcal{U} : \textrm{Cov} (X)^\textrm{op}} p\_\* p^\* K \cong p\_\* p^\* K$$
which yields a morphism from the pro-object representing $p\_\* p^\*$ to $\varprojlim\_{\mathcal{U} : \textrm{Cov} (X)^\textrm{op}} \mathrm{B} \mathcal{U}$.
**Question.** Is this an equivalence?
When $K$ is discrete, the constant presheaf with value $K$ is almost separated – there is a problem over $\varnothing$, but it can be ignored – so $\varinjlim\_{\mathcal{U} : \textrm{Cov} (X)^\textrm{op}} \mathcal{S} (\mathrm{B} \mathcal{U}, K) \to p\_\* p^\* K$ is in fact a bijection in this case.
I am less clear about the situation for non-discrete $K$ – do we need to iterate the plus construction even for constant presheaves?
Can we fix the problem by replacing $\textrm{Cov} (X)$ with some suitable category of hypercovers?
| https://mathoverflow.net/users/11640 | What is the connection between Lurie's definition of shape and Čech homotopy? | The plus construction has to be iterated, yes. The topological space from [this answer](https://mathoverflow.net/a/31386/20233) provides a simple counterexample. Let $X=\{a,b,c,d\}$ with opens $\{a\},\{b\},\{a,b\},\{a,b,c\},\{a,b,d\}$. Then the plus construction does not change the global sections of the constant presheaf on $X$ with fiber $K$. However, the global sections of the associated sheaf is the free loop space $\mathcal LK=K\times\_{K\times K}K$. So the shape of $X$ is a circle, and in this case Definition 2 "does not work".
The problem can more or less be fixed by using hypercovers, because the analogue of the plus construction with hypercovers always produces the hypersheafification (this is essentially Verdier's hypercovering theorem). So the analogue of Definition 2 gives the shape of the $\infty$-topos of hypersheaves on $X$.
| 3 | https://mathoverflow.net/users/20233 | 417481 | 170,052 |
https://mathoverflow.net/questions/417477 | 9 | Let $\mathbb{F}\_2=\{0,1\}$ be the field with two elements, and let
$u:\mathbb{F}\_2^n\rightarrow \mathbb{F}\_2$. *Suppose that $n$ is odd.*
>
> Is it possible that
> $$
> \sum\_{x \in \mathbb{F}\_2^n}(-1)^{u(x)+u(x+a)}= 0,
> $$
> for every $a \neq 0$ in $\mathbb{F}\_2^n$?
>
>
>
I treat the sum here as natural number, not as an element of $\mathbb{F}\_2$.
---
This question arose in the context of [this question](https://math.stackexchange.com/questions/4395191/every-boolean-function-is-multiplicative-with-probability-greater-than-1-2). (Trying to derive a lower bound on the approximate multiplicativity of a Boolean function). When $n$ is even [there are](https://math.stackexchange.com/a/4396116/104576) such functions; this is related to [Bent functions](https://en.wikipedia.org/wiki/Bent_function).
| https://mathoverflow.net/users/46290 | Are there functions $\mathbb{F}_2^n \to \mathbb{F}_2$ satisfying these special relations? | I think the answer is no. To see this, observe that
\begin{equation\*}
\left(\mathbb{E}\_{\mathbf{x}}[(-1)^{u(\mathbf{x})}]\right)^2=\mathbb{E}\_{\mathbf{x}}[(-1)^{u(\mathbf{x})}]\mathbb{E}\_{\mathbf{y}}[(-1)^{u(\mathbf{y})}]=\mathbb{E}\_{\mathbf{x},\mathbf{y}}[(-1)^{u(\mathbf{x})+u(\mathbf{y})}]=\mathbb{E}\_{\mathbf{x},\mathbf{a}}[(-1)^{u(\mathbf{x})+u(\mathbf{x}+\mathbf{a})}],
\end{equation\*}
where we use independence and then rewriting $\mathbf{y}=\mathbf{x}+\mathbf{a}$, and where all expectations are uniform over $\mathbb{F}\_2^n$. The assumption then is that for $\mathbf{a}=\mathbf{0}$, the expectation over $\mathbf{x}$ is clearly $1$, while for all other values of $\mathbf{a}$, the expectation over $\mathbf{x}$ is $0$. It follows that
\begin{equation\*}
\left(\mathbb{E}\_{\mathbf{x}}[(-1)^{u(\mathbf{x})}]\right)^2 = \frac{1}{2^n}\iff \mathbb{E}\_{\mathbf{x}}[(-1)^{u(\mathbf{x})}] = \frac{\pm 1}{2^{n/2}}
\end{equation\*}
However, it's not hard to see that this expectation must also be an integer multiple of $1/2^n$. But then this integer must be $\pm 2^{n/2}$, but this is integral if and only if $n$ is even.
| 12 | https://mathoverflow.net/users/170770 | 417489 | 170,054 |
https://mathoverflow.net/questions/417501 | 9 | What is an example of a Hopf algebra with a non-invertible antipode?
| https://mathoverflow.net/users/478224 | Hopf algebra with a non-invertible antipode | Theorem of Takeuchi (in *Free Hopf algebras generated by coalgebras, 1971*) asserts that free Hopf algebra $H(C)$ over a coalgebra $C$ has injective antipode, and it is bijective precisely (at least over alg. closed field) when $C$ is pointed.
On the other hand, in a paper *Faithful flatness over Hopf subalgebras - counterexamples, 2000* P. Schauenburg constructs a Hopf algebra with surjective, but non-injective antipode. Example is constructed as follows. There's a "free Hopf-with-bijective-antipode" functor from coalgebras to Hopf algebras; one can find a biideal in such free-with-bijective-antipode Hopf algebra over matrix coalgebra $M\_4(k)$ that is stable under antipode, but not under its inverse, and quotient will be the example. Paper can be found on author's homepage <http://schauenburg.perso.math.cnrs.fr/personnelle.html>
| 11 | https://mathoverflow.net/users/81055 | 417505 | 170,058 |
https://mathoverflow.net/questions/416734 | 4 | Let $(A,L)$ be a polarized abelian variety. I know that the degree of the polarization is the Euler characteristic of $L$, so that
$d = \chi(L) = \dim H^0(A,L)$
since $L$ is ample.
I've read in a lot of papers the sentence
>
> Let $(A,L)$ be a polarized abelian variety of dimension $g$ and of type $(d\_1, \dots, d\_g)$, etc.
>
>
>
I've seen the formal definition of type in Birkenhake-Lange book, but I cannot relate this definition with the degree of a polarization, although I am pretty sure they are related.
Indeed, when a statement says "type $(1,1,\dots,1,d)$", I think of "degree $d$", but it is just an intuition.
So my question is:
>
> what is exactly a type of an abelian variety? How do we relate type and degree?
>
>
>
Thanks for help!
| https://mathoverflow.net/users/152522 | Type vs degree of a polarized abelian variety | **To answer your first question**, i.e., what is type on a (polarized) abelian variety $A$? It is already answered [here](https://mathoverflow.net/q/177246). Roughly speaking, for a polarized abelian variety $(A,L)$, there is an integral basis $\{dx\_i,dy\_i\}\_{i=1}^{g}$ of $H^1(A,\mathbb Z)$, and positive integers $d\_1,\ldots, d\_g$ with $d\_i|d\_{i+1}$ such that
$$c\_1(L)=\sum\_i d\_idx\_i\wedge dy\_i.\tag{1}\label{1}$$
The $g$-tuple $(d\_1,\ldots,d\_g)$ is called the type of $(A,L)$.
For example, when $A=J(C)$ is the Jacobian variety of a smooth projective curve $C$, one can choose $L$ to be principal polarization ($d\_1=\cdots=d\_g=1$). This follows from the fact that there is a symplectic basis $\{\gamma\_i,\delta\_i\}\_{i=1}^g$ on $H\_1(J(C),\mathbb Z)=H\_1(C,\mathbb Z)$ such that $\gamma\_i\cdot\delta\_j=\delta\_{ij}$ and $\gamma\_i\cdot\gamma\_j=\delta\_i\cdot\delta\_j=0$.
**For your second question**, in fact, the book (Birkenhake-Lange, p.119) defines the *degree of a polarization* of $L$ to be the product $\prod\_id\_i$. So the relation you're looking for is probably
$$\chi(L)=\prod\_id\_i.\tag{2}\label{2}$$
*Proof of \eqref{2}.* First, by Hirzeburch-Riemann-Roch theorem, one obtains
$$\chi(L)=\int\_Ach(L)td(A)=\int\_Ach(L)=\int\_A\frac{c\_1(L)^g}{g!},\tag{3}\label{3}$$
where the second and the third identities follow from the fact that the tangent bundle of $A$ is trivial and the definition of the Chern character on a line bundle.
Now, by \eqref{1}, one has
$$c\_1(L)^g=(g!\prod\_id\_i)dx\_1\wedge dy\_1\wedge\cdots\wedge dx\_g\wedge dy\_g,$$ so $\int\_Ac\_1(L)^g=g!\prod\_id\_i$. Combine with identity \eqref{3}, the equality \eqref{2} follows. $\Box$
As a final remark, the proof of the identity \eqref{2} can be more concrete: By Kodaira vanishing theorem, all higher degree cohomology of $L$ vanish, so $\chi(L)=\dim H^0(A,L)$, one it reduces to show that there are exactly $\prod\_id\_i$ independent global sections on $L$. These sections correspond to theta functions on $A$. One can refer to Griffiths-Harris, p.317-320 for how these theta functions are found explicitly.
| 1 | https://mathoverflow.net/users/74322 | 417512 | 170,060 |
https://mathoverflow.net/questions/417490 | 2 | Let $\mathfrak{so}(p,q)$ be the real definite/indefinite orthogonal Lie algebra, $p,q\ge0$, $p+q=n\in\mathbb{N}$, and $L\subset\mathfrak{so}(p,q)$ a Lie subalgebra with non-trivial centre, $\mathrm{Z}(L)\neq0$.
**Question:**
Is there an upped bound $c(p,q)$ on the dimension of $L$, $\dim L\le c(p,q)$, better than the dimension of the maximal proper subalgebra?
**Discussion:**
I have tried to brute-force a solution through a classification of closed subgroups of classical groups as in
M. Liebeck, G. Seitz, "On the subgroup structure of classical groups", Invent. Math. 134, 1998.
But the option (ii) of Theorem 1 is not very explicit, and I cannot extract any dimension-relevant information from there.
Thank you.
| https://mathoverflow.net/users/478218 | An upper bound on the dimension of a subalgebra of $\mathfrak{so}(p,q)$ with non-trivial centre | [Answer completely rewritten.]
Indeed we have:
>
> $\DeclareMathOperator\so{\mathfrak{so}}$Fix $n\ge 4$ and $p,q\ge 0$ with $p+q=n$; write $r=\min(p,q)$. Then the minimal codimension for the centralizer of a nonzero element in $\so(p,q)$ is $2n-6$ if $r\ge 2$ and $2n-4$ if $r\in\{0,1\}$.
>
>
>
(The cases $n=2,3$ being straightforward: in this case $2n-4$ is the minimal codimension.)
To prove the result let's consider separately $r\ge 2$ and $r\le 1$. First we have
>
> For every nondegenerate quadratic form $q$ in dimension $n\ge 4$ over a field $K$ of characteristic zero, the codimension of the centralizer of every nonzero element of $\mathfrak{so}(q)$ is $\ge 2n-6$, which moreover is achieved if $r\ge 2$.
>
>
>
First let us check that it is achieved if $r\ge 2$. Indeed, we then choose a basis $(e\_1,\dots,e\_n)$ for which the bilinear form $b$ associated to $q$ satisfies $b(e\_1,e\_n)=b(e\_2,e\_{n-1})=1$, and such that the subspaces of bases $(e\_1,e\_2,e\_{n-1},e\_n)$ and $(e\_3,\dots,e\_{n-2})$ are orthogonal. Decomposing matrices with blocks of size $1+1+(n-4)+1+1$, consider the matrix
$$g=\begin{pmatrix}0 & 0 & \quad 0\quad & 1 & 0\\ 0 & 0 & 0 & 0 & -1 \\ \\0 & 0 & 0 & 0 & 0 \\ \\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\end{pmatrix}.$$
Then its centralizer in $\so(q)$ has codimension $2n-6$. Indeed, the latter verification can be performed over an algebraic closure, in which case we can suppose furthermore that $b(e\_i,e\_j)=\delta\_{i+j,n+1}$. So $\so(q)$ consists of matrices that are skew-symmetric with respect to the antidiagonal. In such a basis, the centralizer of $g$ consists of those matrices of the form
$$\begin{pmatrix}a & b & \quad x\quad & 1 & 0\\ c & -a & y & 0 & -1 \\ \\0 & 0 & h & -y^\mathrm{t} & -x^\mathrm{t} \\ \\ 0 & 0 & 0 & a & -b\\ 0 & 0 & 0 & -c & -a\end{pmatrix}$$
for $a,b,c,d$ arbitrary scalars, $x,y$ arbitrary $(n-4)$-vectors, and $h$ arbitrary square matrix of size $n-4$ that is skew-symmetric with respect to the antidiagonal. This has codimension $2n-6$.
In the other direction, we can directly pass to an algebraic closure. Now I borrow the argument from [this answer by Peter McNamara](https://mathoverflow.net/a/416962/14094): we can suppose that (the projection of) $g$ belongs to closed adjoint orbit in $\mathbf{P}(\so\_n)$ (indeed passing from a non-closed orbit to its boundary decreases the orbit dimension and does not decrease the dimension of the centralizer). And (see that same answer), this means that $g$ is, up to conjugation, a nonzero element in a root space. Fixing a Cartan subalgebra, we have 1 or 2 orbits of roots according to whether $n\ge 4$ is even or odd (call them long roots and short roots, assuming that for even $n$ there are only long roots — for $n=3$ there are only short roots). The long roots have a centralizer of codimension $2n-6$ (this is the computation above). Whence this lower bound on the codimension. (The short roots have a centralizer of codimension $2n-4$, but this is a priori only relevant for $n=3$.)
In order to deal with the case $r\le 1$, here's a result in general.
>
> Fix $n\ge 2$. For every nondegenerate quadratic form $q$ in dimension $n$ over a field $K$ of characteristic zero, the codimension of the centralizer of every nonzero *semisimple* element of $\mathfrak{so}(q)$ is $\ge 2n-4$, which is achieved.
>
>
>
To show that it is achieved, diagonalize $q$, and thus choose an orthogonal decomposition of the ambient space as $V=V\_2\oplus V'$ with $V\_2$ 2-dimensional. Choose a nonzero element $g$ of the 1-dimensional $\so(q|\_{V\_2})$. Extend it as zero on $V'$. So we can view $g$ as an element of $\so(q)(K)$. Then its centralizer has codimension $2n-4$. Indeed, to compute it, we can extend extend scalars to assume $K$ algebraically closed (this is inessential, just simplifies computations), and hence we can suppose that the basis is orthonormal. Then $\so(q)$ is the space of skew-symmetric matrices and the centralizer of $g$ consists of those with the (linearly independent) conditions $g\_{ij}=0$ for $i=1,2$, $j=3,\dots,n$.
To show that this is a lower bound, we can suppose that the field is algebraically closed. So, we can deal with the bilinear form $b(e\_i,e\_j)=\delta\_{i+j,n+1}$ as above, so that $\so(q)$ consists of skew-symmetric matrices with respect to the skew-diagonal. A Cartan subalgebra consists of those diagonal matrices therein, and we can restrict to elements in this Cartan subalgebra. Replacing each nonzero diagonal entry $x$ with $x/|x|$ will not decrease the centralizer, so we can suppose that $x$ has only the eigenvalues $\pm 1$ and possibly $0$. Then if $m\_0$ and $m\_1$ are the multiplicity of the eigenvalues $0$ and $1$, we have $n=m\_0+2m\_1$ and the centralizer has dimension $d=m\_1^2+m\_0(m\_0-1)/2$. Writing $m=m\_0$ this yields $d=(n-m)^2/4+m(m-1)/2$ (where $m<n$ is such that $n-m$ is even). One sees this is maximal for $m=n-2$, namely dimension $1+(n-2)(n-3)/2$, and this means codimension
$2n-4$.
To conclude the initial result: if the rank $r$ is zero all elements are semisimple and the result follows.
It remains to consider the case of $\so(1,n-1)$ (this might be true over more general fields in rank 1, but I haven't checked). One direction is already settled: there is a semisimple element whose centralizer has codimension $2n-4$. We have to see that every nonzero element $g$ has a centralizer of codimension $\ge 2n-4$. Passing from an element to its nilpotent and semisimple part in the additive Jordan-Dunford decomposition does not decrease the dimension of centralizer. So we can suppose that $g$ is nilpotent or semisimple. The semisimple case is already settled. So we can suppose $g$ nilpotent. All nonzero nilpotent elements in $\so(1,n-1)$ are indeed conjugate and for each of them, the centralizer indeed has codimension $2n-4$.
(At the group level, this centralizer is essentially a semidirect product $\mathbf{R}^{n-2}\rtimes\mathrm{O}(n-3)$, namely the centralizer of a nontrivial translation of $\mathbf{R}^{n-2}$ in its isometry group.)
| 3 | https://mathoverflow.net/users/14094 | 417519 | 170,062 |
https://mathoverflow.net/questions/417520 | 5 | Let $f = x^n + a\_{n-1}x^n + \cdots + a\_0$ be a monic polynomial of degree $n \geq 2$ with integer coefficients. By $\text{Gal}(f)$ we mean the Galois group over $\mathbb{Q}$ of the Galois closure of $f$. Define $H(f) = \max\{|a\_i|\}$ denote the naive or box height of $f$. Hilbert's irreducibility theorem asserts that for most integer $n$-tuples $(a\_0, \cdots, a\_{n-1})$ with $H(f) \leq B$ say, the corresponding polynomial $f$ has symmetric group $S\_n$. If we assume that one of the coefficients $a\_i$ of the polynomials are fixed, then is it still true that the number of such polynomials whose galois group is not symmetric group is $o(B^{n-1})?$
Thanks in advance for any assistance.
| https://mathoverflow.net/users/160943 | Galois groups of specific classes of polynomials with one coefficient fixed | This is equivalent (by Hilbert) to asking whether the partially specialized polynomial still has symmetric Galois group (over the respective function field). This holds unless you specialized the constant coefficient to $0$.
According to *Cohen, S. D.*, [**The Galois group of a polynomial with two indeterminate coefficients**](http://dx.doi.org/10.2140/pjm.1980.90.63), Pac. J. Math. 90, 63-76 (1980), [ZBL0408.12011](https://zbmath.org/?q=an:0408.12011), you may even prescribe all but two of the coefficients and still get the full symmetric group, unless you specialized such that the resulting polynomial is a polynomial in $x^r$ for some $r>1$.
| 9 | https://mathoverflow.net/users/127660 | 417527 | 170,064 |
https://mathoverflow.net/questions/417528 | 0 | This question is motivated by considerations on [conflict-free colorings](https://mathoverflow.net/questions/416793/conflict-free-coloring-of-mathbbr-with-the-euclidean-topology), which arose while studying assignment problems for frequencies in cellular networks.
A *[hypergraph](https://en.wikipedia.org/wiki/Hypergraph)* $H=(V,E)$ is said to be *linear* if for all $e\_1\neq e\_2 \in E$ we have $|e\_1\cap e\_2| \leq 1$.
Let $H=(\omega, E)$ be a linear hypergraph such that $e$ is infinite for all $e\in E$. Is there a necessarily a map $c: \omega\to \{0,1\}$ such that for all $e\in E$ there is $v^\*\in e$ such that $$c^{-1}\big(\big\{c(v^\*)\big\}\big) \cap e \;= \;\{v^\*\}\;\;?$$
(More informally, we want every edge $e\in E$ to have a vertex $v^\*\in e$ such that all the vertices in $e\setminus\{v^\*\}$ are colored with the (unique) color in $\{0,1\}\setminus\{c(v^\*)\}$.)
| https://mathoverflow.net/users/8628 | Conflict-free coloring of linear hypergraphs on $\omega$ | Let $V$ be the set of points in the affine plane $\mathbb Q^2$, and $E$ be the set of lines.
Assume that the coloring is possible. Say a line is *white* if it contains a unique black point, and *black* otherwise.
There exist two white points (on two parallel lines), hence a white line through them. There exists a white point outside that line, hence infinitely many white lines concurrent at that white point (and passing through white points on the found line). Similarly, take infinitely many black lines concurrent at a black point.
Take 5 white and 5 black lines out of those collections; they have at least 20 points of mutual intersection. But at most 5 of those points are black (one per white line), and at most 5 are black. A contradiction.
| 1 | https://mathoverflow.net/users/17581 | 417532 | 170,067 |
https://mathoverflow.net/questions/417388 | 6 | Let $S$ be a monoid. On p. xvii of P.M. Cohn's *Free Ideal Rings and Localization in General Rings* (CUP, 2006), one reads that
* an element $u \in S$ is *regular* if (quote) "[...] it can be cancelled, i.e. $ua = ub$ or $au = bu$ implies $a = b$";
* $S$ is a *conical* monoid if (quote) "$ab = 1$ implies $a = 1$ (and so also $b = 1$)".
On p. 53, one then reads that $S$ is an *invariant* monoid if each of its elements is invariant, an element $c \in S$ being *invariant* if $c$ is regular and $cS = Sc$ (so in particular, an invariant monoid is cancellative).
Now, with these definitions in mind, Problem 0.9.10 in the same book (p. 58) asks:
>
> Is every invariant conical monoid necessarily commutative?
>
>
>
The problem is numbered "10°" by Cohn, and on p. xiv one reads:
>
> [...] open-ended (or open) problems are marked °, though sometimes this may refer only to the last part; the meaning will usually be clear.
>
>
>
At first, I had mistakenly thought (see the first version of this post) that the problem had a trivial answer, due to the following (flawed) argument:
>
> Start with a semigroup $H$, pick an element $e \notin H$, and let $H^{(e)}$ be the (unique) magma obtained by extending the operation of $H$ to a binary operation on $H^{(e)}$ in such a way that $ex = xe := x$ for every $x \in H^{(e)}$. Clearly $H^{(e)}$ is a monoid (some people would call it an *unconditional unitization* of $H$); and it is cancellative, commutative, or invariant if and only if so is $H$, resp. On the other hand, $H^{(e)}$ is obviously conical (regardless of whether this is the case with $H$). Therefore, if $H$ is a non-commutative invariant monoid (such as the monoid of non-zero elements of a non-commutative skew-field), then $H^{(e)}$ provides a negative answer to Cohn's question.
>
>
>
The issue here is that $H^{(e)}$ is a cancellative monoid if and only if $H$ is a cancellative semigroup without unity: If $H$ is a monoid with identity $1\_H$, then $xe = x1\_H = x = 1\_H x = ex$ for every $x \in H$ (and, by construction, $e \ne 1\_H$). So, my (new) question is:
>
> **Q.** Is anyone aware of any progress on Cohn's problem since 2006?
>
>
>
**Update (12/03/2022).** Cohn's problem has been quickly solved by Pace Nielsen [below](https://mathoverflow.net/a/417478/16537). A second and, in some sense, much easier solution was communicated by George Bergman to Pace Nielsen and is now the bulk of the [accepted answer](https://mathoverflow.net/a/417555/16537) of this thread.
| https://mathoverflow.net/users/16537 | Problem 0.9.10 in Cohn's "Free Ideal Rings and Localization in General Rings" (CUP, 2006) | I mentioned this problem to George Bergman and he offered the following much simpler solution, which he has given permission for me to post. In his own words:
>
> Take any nonabelian group $G$ with a nontrivial homomorphism
> $f\colon G \to \mathbb{Z},$ and let $S$ be the submonoid of $G$ with
> underlying set $\{e\} \cup \{x\in G\, |\, f(x) > 0 \}.$
>
>
>
If $G$ is the free group on the two symbols $x,y$, and $f$ is the homomorphism determined by the rule $f(x)=f(y)=1$, then I believe that the monoid $M$ described in my other answer is the "duo-hull" of $x$ and $y$ inside the monoid $S$.
| 2 | https://mathoverflow.net/users/3199 | 417555 | 170,071 |
https://mathoverflow.net/questions/417550 | 4 | I need a series expansion to describe a general gaussian-like (bell shaped) function. I couldn't find a rigorous definition of "bell shaped" online but in essence the function should have the following:
1. $f(x)>0$ (positive)
2. $f(x)=f(-x)$ (symmetric)
3. $\int \_{-\infty}^{\infty}f(x)dx <\infty$
4. $f^{(n)}(x) = 0$ has exactly $n$ roots (bell shaped)
5. $f^{(n)}(x) \rightarrow 0 $ as $|x| \rightarrow \infty$ for all $n$
Example would be: Gaussian, Lorentzian curve, Voigt profile, Sech.
I need a series expansion that the gives approximants for all such functions. The series expansion should only give "bell shaped" functions for any choice coefficients (as with fourier series - for example - only producing periodic functions).
Does this exist?
| https://mathoverflow.net/users/478260 | Series expansion for gaussian-like function | You could try a series of Hermite functions,
$$f(x)=\sum\_{n=0}^N a\_n \frac{d^{2n}}{dx^{2n}}e^{-x^2}.$$
The function $f$ satisfies your conditions 1,2,3,5 by construction, and if the $a\_n$'s decay rapidly with $n$ it will look "bell-shaped".
Actually, for unconstrained $a\_n$'s, and including also odd derivatives, this expansion is dense in $L^2(\mathbb{R})$, see [Approximating with Gaussians.](https://arxiv.org/abs/0805.3795)
| 2 | https://mathoverflow.net/users/11260 | 417569 | 170,076 |
https://mathoverflow.net/questions/417132 | 0 |
>
> If $f\in C^1(\mathbb R)$ satisfies $f'(x)>f(f(x))$ for all $x\in\mathbb R$, then $f(f(f(x)))\leq0$ for all $x\geq0$.
>
>
>
I have some trouble to prove this. I wonder if there's some relations between this problem and the ODE $ f'(x)=f(f(x)) $. Could anybody provide a solution or some hints on this problem?
| https://mathoverflow.net/users/241460 | $f'(x)>f(f(x))$ implies $f(f(f(x)))\leq0$ for nonnegative $x$ | Well, this is Problem 4 of Day 1 of IMC 2012, proposed by Tomáš Bárta from Prague, see [here](https://www.imc-math.org.uk/imc2012/IMC2012-day1-solutions.pdf)
| 2 | https://mathoverflow.net/users/4312 | 417571 | 170,077 |
https://mathoverflow.net/questions/417493 | 1 | A copy of the Cantor set is a space homeomorphic to $2^{\omega}$.
Suppose that $X$ is a Hausdorff space that contains a copy $C^{\prime}$ of the Cantor set. Let $U$ be a nonempty subset open in $C^{\prime}$, also let $D$ be a countable set dense in $X$ such that $D\cap U$ is dense in $U$. Does anyone have any idea how to prove that the set of accumulation points of $D\cap U$ is infinite?
| https://mathoverflow.net/users/475617 | Subsets of the Cantor set | To answer the question: every point of $U$ is an accumulation point of $D\cap U$, hence there are continuum many accumulation points. The ambient space $X$ plays no role here; everything takes place in the Cantor set.
| 3 | https://mathoverflow.net/users/5903 | 417579 | 170,080 |
https://mathoverflow.net/questions/417574 | 6 | Let $(X,d,m)$ be a metric measure space. We say that it is *doubling in the sense of metric spaces* if for every:
$x\in X$ and every $r>0$ there exists some **(metric) doubling constant** $C\_d\geq 0$ such that
$$
Ball(x,r) \mbox{ can be covered by at-most $C\_d$ balls of radius $r/2$}.
$$
There is a different, [related and in some spirit "equivalent"](https://www.ams.org/journals/proc/1998-126-02/S0002-9939-98-04201-4/S0002-9939-98-04201-4.pdf), notion of *doubling in the sense of metric measure spaces*, which states that there is a constant $C\_m\geq 0$ such that: for every $x\in X$ and each $r>0$
$$
m(Ball(x,r)) \leq C\_m m(Ball(x,r/2)).
$$
If $(X,d,m)$ is doubling in the sense of metric measure spaces, with constant $C\_m$, then is it doubling in the sense of metric spaces? And if so, can we $C\_m$ to deduce an upper-bound for $C\_d$?
Note, I'm most interested in the case where $m$ is an $s$-dimensional Hausdorff measure.
| https://mathoverflow.net/users/469470 | Relationship between doubling constant of a metric space and of a metric measure space | Apart from the obvious counterexample of the measure being $0$, if $(X,d,m)$ is doubling in the sense of metric measure spaces it will be doubling in the sense of metric spaces.
Consider a ball $B(x,r)$. If for some $n$, $B(x,r)$ cannot be covered by $n$ balls of radius $\frac{r}{2}$, then we can obtain by recursion a sequence of points $x\_1,\dots,x\_n$ in $B(x,r)$ which are pairwise at distance $\geq\frac{r}{2}$. Thus the balls $B(x\_i,\frac{r}{4})$ are disjoint, and they are all contained in $B(x,2r)$. Suppose $m(B(x\_1,\frac{r}{4}))$ is the smallest of all the $m(B(x\_i,\frac{r}{4}))$.
Then $m(B(x\_1,4r))\geq m(B(x,2r))\geq\sum\_{i=1}^nm(B(x\_i,\frac{r}{4}))\geq nB(x\_1,\frac{r}{4})$, so either $n\leq C\_m^4$ or $m(B(x\_1,4r))=0$. If you only consider finite distances the second option implies that $m(X)=0$.
So an upper bound would be $C\_d\leq C\_m^4$.
| 10 | https://mathoverflow.net/users/172802 | 417580 | 170,081 |
https://mathoverflow.net/questions/417572 | 2 | On paper [A procedure for improving the upper bound
for the number of $n$-ominoes](https://www.cambridge.org/core/services/aop-cambridge-core/content/view/CA6D64DA91090FFAA49EBC83C098E3AF/S0008414X00050628a.pdf/a-procedure-for-improving-the-upper-bound-for-the-number-of-n-ominoes.pdf) by D. A. Klarner & R. L. Rivest, it is known that the number $t(n)$ of polyominoes with area $n$ satisfies $3.72^n<t(n)<6.75^n$.
What is the best current result?
| https://mathoverflow.net/users/174530 | Number of polyominoes with area $n$ | Wikipedia suggests that no-one has improved on Klarner, D.A.; Rivest, R.L. (1973) [A procedure for improving the upper bound for the number of n-ominoes](https://www.cambridge.org/core/services/aop-cambridge-core/content/view/CA6D64DA91090FFAA49EBC83C098E3AF/S0008414X00050628a.pdf/a-procedure-for-improving-the-upper-bound-for-the-number-of-n-ominoes.pdf), Canadian J of Math. 25 (3): 585–602, which gives $$\lim\_{n \to \infty} (t(n))^{1/n} \le 4.649551$$
However, that is not conclusive because it appears to overlook Barequet, G., Rote, G., & Shalah, M. (2016) [λ> 4: An improved lower bound on the growth constant of polyominoes](https://geometry.stanford.edu/papers/brs-lg4iubgcp-16/brs-lg4iubgcp-16.pdf), CACM, 59(7), 88-95, which gives a lower bound of $$t(n) \ge 4.00253176^n$$
Two of the same authors (Gill Barequet, Mira Shalah) have a [preprint on arxiv](https://arxiv.org/abs/1906.11447) which purports to show that the upper bound is $$t(n) \le 4.5252^n$$
| 6 | https://mathoverflow.net/users/46140 | 417581 | 170,082 |
https://mathoverflow.net/questions/417256 | 6 | Motivation: for any ring $R$ there is the natural monomorphism $\mathrm{in} \colon R \to \mathrm{End}(R\_{add}): r \mapsto (x \mapsto rx)$, where $R\_{add}$ is an additive abelian group ( rings are assumed to be associative with identity, but not necessarily commutative). So a ring is exactly an abelian group with a distinguished subgring of its endomorphisms (and a fixed bijection between elements and distinguished endomorphisms). Some rings are "complete" in the sense that they "contain" all endomorphisms of the underlying abelian group. For example, $\mathrm{in}$ is an isomorphism for $R = \mathbb{Z}, \mathbb{Q}, \mathbb{Z}\_n$.
1. What is known about the classification of abelian groups $A$ such that there is an isomorphism between the abelian groups $\mathrm{End}(A)$ and $A$?
Each such isomorphism gives some "complete ring" structure on $A$.
2. What is known about the uniqueness (up to isomorphism) of the "complete ring" structure on an abelian group?
I'm interested in the answers to these questions, with any additional assumptions that seem natural to you. I am especially interested in the answers for commutative rings.
| https://mathoverflow.net/users/148161 | Abelian groups such that $A \cong \mathrm{End}(A)$ and "complete rings" | The rings, you call ``complete'' are known as $E$-rings (as Ulrich Pennig mentioned in the comments).
Some comments on your questions
1. There are too many results on the $E$-rings to list them here and I'd rather direct you to the book by Göbel and Trlifaj *Approximations and endomorphism algebras of modules*. However to give you some sense that we have no hope to obtain any reasonable classification - every Abelian cotorsion-free group embeds in an $E$-ring. Formally a group $A$ is cotorsion-free if there are only null homomorphisms $\mathbb{Z}^\wedge\_p\to A$. This is equivalent to the claim that if another Abelian group $C$ admits a compact topology then every homomorphism $C\to A$ is null. To make long story short - in the absence of compactness you may add to $A$ new elements to get more endomorphisms and then add yet more elements to kill the unwanted endomorphisms.
2. Given an Abelian group $A$ the $E$-ring structure on $A$ is unique up to the choice ot the identity element. Every invertible element of an $E$-ring $A$ can be chosen as the identity element for another ring structure on $A$.
| 6 | https://mathoverflow.net/users/16678 | 417582 | 170,083 |
https://mathoverflow.net/questions/396372 | 2 | Let $K$ be a finite extension of $\mathbb{Q}\_p$. Let $\mathcal{O}$ be its ring of integers and $\mathfrak{m}$ the maximal ideal. Pick a uniformiser $\pi$. The construction using theory of Lubin--Tate formal groups yields an abelian Galois extension $K\_\pi$ of $K$ such that $\mathrm{Gal}(K\_\pi/K)\simeq\mathcal{O}^\times$. Note that it is well-known that $\mathcal{O}^\times\simeq\mu\_{q-1}\times\mathbb{Z}/p^a\times\mathbb{Z}\_p^d$ where $d=[K:\mathbb{Q}\_p]$ and $q$ the number of elements of the residue field of $\mathcal{O}$ and $a$ some integer.
>
> There should be Galois sub-extensions of $K$ in $K\_\pi$ corresponding to each copy of $\mathbb{Z}\_p$. Is it known what they are?
>
>
>
| https://mathoverflow.net/users/44005 | Field extension corresponding to a quotient of units of local fields | The isomorphism you write down involving the $\mathbb{Z}\_p$ is not canonical, so one cannot describe them without fixing additional data.
For $\mathbb{Q}\_p$ you have $a=1$ and this tower is the totally ramified (p-)cyclotomic tower.
| 3 | https://mathoverflow.net/users/478296 | 417585 | 170,084 |
https://mathoverflow.net/questions/417197 | 5 | Fix integers $1 \leq k \leq n$ and suppose $\mathbf{x} \in \mathbb{R}^n$ is such that $e\_j(x\_1,x\_2,\ldots,x\_n) \geq 0$ for all $1 \leq j \leq k$, where $e\_j$ is the $j$-th [elementary symmetric polynomial](https://en.wikipedia.org/wiki/Elementary_symmetric_polynomial).
>
> **Question 1:** Is it true that $x\_1 + x\_2 + \cdots + x\_{n-k+1} \geq 0$?
>
>
>
The answer is trivially "yes" in the edge cases $k = 1$ and $k = n$, but the intermediate cases seem much less obvious.
While Question 1 is the one that I'm really interested in, there is a natural generalization of it that is perhaps known, so I'll ask it now:
>
> **Question 2:** Is it true that $e\_j(x\_1,x\_2,\ldots,x\_{n-1}) \geq 0$ for all $1 \leq j \leq k-1$?
>
>
>
In other words, can we use non-negativity of elementary symmetric polynomials in $n$ variables to infer non-negativity of elementary symmetric polynomials in $n-1$ variables? If the answer to Question 2 is "yes" then we can use it $k-1$ times to see that the answer to Question 1 is "yes" too.
| https://mathoverflow.net/users/11236 | Dimension reduction for non-negativity of elementary symmetric polynomials | The answer to question 2, and therefore question 1, is "yes". We abbreviate $e\_k(x\_1, x\_2, \ldots, x\_{n-1})$ to $b\_k$ and $e\_k(x\_1, x\_2, \ldots, x\_{n-1}, x\_n)$ to $a\_k$, so $a\_k = x\_n b\_{k-1} + b\_k$.
We are trying to show that, if $a\_1$, $a\_2$, ..., $a\_k \geq 0$ then $b\_1$, $b\_2$, ..., $b\_{k-1} \geq 0$. We prefer to prove the contrapositive: If $b\_j<0$ then one of $a\_1$, $a\_2$, ..., $a\_j$, $a\_{j+1} <0$. We may assume that $j$ is minimal with $b\_j<0$, so $b\_1$, $b\_2$, ..., $b\_{j-1} \geq 0$.
**Case 1:** $x\_n<0$. Then $a\_j = x\_n b\_{j-1} + b\_j < 0$.
**Case 2:** $x\_n \geq 0$. First of all, if $b\_{j-1} \leq 0$, then $a\_j<0$ and, if $b\_{j+1} \leq 0$, then $a\_{j+1} < 0$. So we may assume that $b\_{j-1}$ and $b\_{j+1}>0$.
Now, [Newton's inequalities](https://en.wikipedia.org/wiki/Newton%27s_inequalities) give
$$\frac{b\_{j-1} b\_{j+1}}{\binom{n-1}{j-1} \binom{n-1}{j+1}} \leq \frac{b\_j^2}{\binom{n-1}{j}^2}$$
or
$$b\_{j-1} b\_{j+1} \leq \frac{j(n-j-1)}{(j+1)(n-j)} b\_j^2 < b\_j^2.$$
So
$$0<\frac{b\_{j+1}}{- b\_j} < \frac{-b\_j}{b\_{j-1}}.$$
We must either have $x\_n > \tfrac{b\_{j+1}}{- b\_j}$ or $x\_n < \tfrac{-b\_j}{b\_{j-1}}$. In the first case, $0 > x\_n b\_j + b\_{j+1} = a\_{j+1}$; in the second case, $a\_j = x\_n b\_{j-1} + b\_j < 0$. Either way, we have found a negative $a\_j$.
| 4 | https://mathoverflow.net/users/297 | 417597 | 170,087 |
https://mathoverflow.net/questions/417500 | 7 | Let $X$ be a compact Riemann surface of genus $g$, then $K^1\_{\mathrm{top}}(X)\cong\mathbb{Z}^{2g}$. Is there a explicit description of a set of basis of $K^1\_{\mathrm{top}}$? (e.g., For cohomology $H^1(X,\mathbb{Z})\cong\mathbb{Z}^{2g}$ we may take the 1-cochains ``around the holes'')
Furthermore, we define the Mukai vector of $\kappa\in K^1\_{top}(X)$ to be $v(\kappa)=\mathrm{ch}(\kappa)\sqrt{\mathrm{td}(X)}$, and the Euler pairing on $K^\*\_{top}(X)$ by $\langle a,b\rangle=(v(a^\vee),v(b))$ where $(-,-)$ is the pairing in $H^\*(X,\mathbb{Q})$. Do we know the pairing with respect to the basis? (The definition of $\mathrm{ch}\colon K^\*(X)\to \oplus H^\*(X)$ is given in [https://www.maths.ed.ac.uk/~v1ranick/papers/ahvbh.pdf](https://www.maths.ed.ac.uk/%7Ev1ranick/papers/ahvbh.pdf))
| https://mathoverflow.net/users/nan | Topological K-theory of Riemann surface | Following @Kiran's suggestion in the comments, I'll outline why the map $U(1)\to U$ induces an isomorphism between cohomology and K-theory in this setting. At the end I'll also explain a different perspective that might be helpful.
The inclusion maps $U(n)\to U(n+1)$ are $(2n-1)$-connected, so the map $U(1)\to U$ is 1-connected. This means that for any 2-dimensional CW complex X, the induced map Map$(X, U(1))\to$ Map$(X, U)$ is $-1$-connected, and in particular a surjection on $\pi\_0$ (I believe this fact about connectivity appears in May's Concise Course; it's proven by induction on skeleta of X). So $[X, U(1)]\to [X, U]$ is surjective. (This is a a borderline case of the result I'm quoting and I would recommend checking carefully that it does work... Note that there's no difference between based and unbased homotopy classes of maps in this setting, because the action of $\pi\_1 U(1) = \pi\_1 U$ is trivial on $\pi\_\* U(1)$ and on $\pi\_\* (U)$, as these are groups.) Now, say $X = M^g$, a closed Riemann surface of genus $g$. Knowing in advance that $[M^g, U(1)] = H^1 (M^g; \mathbb Z) = \mathbb{Z}^{2g} = K^1 (M^g) = [M^g, U]$, this surjection must be an isomorphism (note that the group structures are induced by the group structures on $U(1)$ and $U$, so the function between homotopy sets is a group homomorphism.)
Another way of thinking about $K^1 (M^g)$ is to consider vector bundles over the suspension $\Sigma M^g$. Since the attaching map of the 2-cell in $M^g$ is a commutator in $\pi\_1 (\bigvee\_{2g} S^1)$, the attaching map of the 3-cell in $\Sigma M^g$ is a commutator in $\pi\_2 (\bigvee\_{2g} S^2)$, and hence is nullhomotopic. This means $\Sigma M^g \simeq (\bigvee\_{2g} S^2) \vee S^3$, which gives $[\Sigma M^g, BU] \cong \pi\_2 (BU)^{2g} \oplus \pi\_3 (BU) = \mathbb{Z}^{2g}$ (and by Bott periodicity $[\Sigma M^g, BU] \cong [M^g, \Omega BU] \cong [M^g, U]$). See Tyrone's comment below for a way to make this explicit.
| 4 | https://mathoverflow.net/users/4042 | 417598 | 170,088 |
https://mathoverflow.net/questions/417602 | 2 | I'm reading a paper on the classical Gagliardo-Nirenberg interpolation inequality [arXiv link](https://arxiv.org/abs/1812.04281) and there is a inequality used
$$
|v-\overline{v}|\le \left\Vert v' \right\Vert\_{r,I} \ell^{1-\frac{1}{r}}, r\ge 1
$$
where $\overline{v}:=\frac{1}{\ell}\int\_I v(x)dx$, $I$ is an interval on $R$, $v'$ is the derivative.
It looks quite simple, quite similar to the Hölder inequality, but where does the derivative come from? And I know that using the Poincaré inequality there will be a constant bound $C$, but then where does the term $\ell^{1-1/r}$ come from?
I think it should be a quite simple question but I am just stuck at it. Thanks for your comments!
| https://mathoverflow.net/users/295572 | A simple 1-dimensional inequality, maybe Poincaré inequality or Hölder inequality? | By the mean value theorem, $\bar v=v(t)$ for some $t\in I$. So, for all $x\in I$,
$$|v(x)-\bar v|=|v(x)-v(t)|
=\Big|\int\_t^x v'\Big|
\le\int\_I|v'|\le\|v'\|\_r\, \ell^{1-1/r};$$
the latter inequality is an instance of Hölder's inequality.
| 3 | https://mathoverflow.net/users/36721 | 417606 | 170,091 |
https://mathoverflow.net/questions/416794 | 5 | I have several questions regarding proposition 2.3 in "Cherednik and Hecke algebras of varieties with a finite group action", by Pavel Etingof. Let $X$ be a complex affine algebraic variety and $g$ an automorphism of finite order. We define the $D(X)$-bimodule $D(X)g$ by $a(bg)c=abg(c)g$.
1. In the aforementioned proposition, it is stated that the standard equivalence of left and right $D$-modules on the second factor induces two left $D$-modules $\tilde{D}(X)$ and $\tilde{D}(X)g$ on $X\times X$ coming from $D(X)$ and $D(X)g$ respectively. I would like to understand how this equivalence can be used to induce $D$-modules on $X\times X$.
2. Let $i:X\rightarrow X\times X$ be the morphism given by $x \mapsto (x, g^{-1})$ and $\Delta$ be the diagonal. The proposition states that $\tilde{D}(X)=\Delta\_{\*}O\_{X}$ and $\tilde{D}(X)g=i\_{\*}O\_{X}$. Then, the following chain of equalities is stated:
$$Ext\_{D(X)\otimes D(X)^{op}}^{m}(D(X),D(X)g)=Ext\_{D(X\times X)}^{m}(\tilde{D}(X),\tilde{D}(X)g)= Ext\_{D(X\times X)}^{m}(\Delta\_{\*}O\_{X},i\_{\*}O\_{X})=Ext\_{D(X)}^{m}(O\_{X},\Delta^{!}i\_{\*}O\_{X})$$
I would like to understand the first and third equalities.
| https://mathoverflow.net/users/476832 | Two identities involving Ext functors in the context of D-modules | I asked Prof. Etingof and he answered the following:
1. A $D$-module on $X\times X$ is a module over $D(X)\otimes D(X)$, while initially, these are bimodules over $D(X)$, which are modules over $D(X)\otimes D(X)^{op}$. The equivalence I mentioned is a Morita equivalence between $D(X)$ and $D(X)^{op}$ which defines a Morita equivalence between $D(X)\otimes D(X)$ and $D(X)\otimes D(X)^{op}$.
2. The first equality follows from my answer to (1). The third equality follows because the functors $\Delta\_{ \* }$ and $\Delta^{ \* }$ are adjoint.
| 2 | https://mathoverflow.net/users/476832 | 417615 | 170,093 |
https://mathoverflow.net/questions/417440 | 3 | $\DeclareMathOperator\Aut{Aut}\DeclareMathOperator\Gl{Gl}$Let $\mathcal{D}=\mathbb{C}[[t\_{1},\dotsc , t\_{n}]]$ be a formal polydisc over $\mathbb{C}$, and $G$ be a finite group. On Lemma 7.8 of [Etingof and Ma - Lecture notes on Cherednik algebras](https://arxiv.org/abs/1001.0432) it is stated that every action $\rho : G\rightarrow \Aut(\mathcal{D})$ is linearizable. The proof given in the notes goes as follows:
We have a decomposition $G = \Gl\_{n}(\mathbb{C})\ltimes \Aut\_{U}(\mathcal{D})$, where $\Aut\_{U}(\mathcal{D})$ are the automorphisms such that their derivative at the origin is the identity.
We need to show that the image of $\rho$ can be conjugated into $\Gl\_{n}(\mathbb{C})$. The obstruction to this is in the cohomology group $H^{1}(G, \Aut\_{U}(\mathcal{D}))$, which is trivial as $\Aut\_{U}(\mathcal{D})$ is a pro-unipotent algebraic group over $\mathbb{C}$.
I have two questions regarding this proof:
1. This proof seems to be assuming that the group $\Aut\_{U}(\mathcal{D})$ is abelian. I would like to see why is this the case.
2. Why does the fact that $\Aut\_{U}(\mathcal{D})$ is a pro-unipotent algebraic group over $\mathbb{C}$ imply the vanishing of $H^{1}(G, \Aut\_{U}(\mathcal{D}))$?
| https://mathoverflow.net/users/476832 | On the linearizability of the action of a finite group on a formal polydisc | For $1 \to A \to B \to C \to 1$ an exact sequence of (not necessarily abelian) algebraic groups, we have a long exact sequence
$$ H^0(G, A) \to H^0(G, B) \to H^0(G, C) \to H^1(G, A) \to H^1(G, B) \to H^1(G, C) $$
It follows that if $H^1(G,A) = H^1(G, C) =0$ then $H^1(G,B)=0$.
We can now prove that for finite $G$, and $U$ a unipotent algebraic group $H^1(G,U)=0$, by induction on $\dim U$. There is always a surjection from $U$ to a vector space $V$, and the kernel $K$ is unipotent of smaller dimension. Now $H^1(G,K) =0$ by the induction hypothesis, and $H^1(G, V)=0$ by representation theory of finite groups over characteristic zero fields, so $H^1(G,U)=0$, completing the induction step.
Now for finite $G$ and pro-unipotent $U$, $H^1(G,U)=0$ by a limiting argument.
---
A limiting argument, probably not the best way:
Write $U$ as an inverse limit of unipotent groups $U\_n$. We want to show our cocycle in $C^1( G, U) $ is the coboundary of an element of $U$. We know for each $n$ there exists an element in $U\_n$ whose coboundary is the projection of our cocycle to $C^1(G, U\_n)$. In fact, the set of such elements of $U\_n$ form an algebraic variety, a coset of an algebraic subgroup.
The image of this variety in $U\_1$ is a coset of a subgroup in $U\_1$, thus a closed subvariety. As $n$ grows, these subvarieties form a descending chain, which stabilizes at some $n$. Pick a point $x\_1$ in the stabilization. Then for each $n$ the intersection of the subvariety with the inverse image of $x\_1$ is nonempty, and still a coset of a closed subgroup.
Look at the images of these intersections in $U\_2$. Choose a point $x\_2$ that's in all of them. Then repeat...
Then $(x\_1,x\_2,\dots)$ gives the desired point of $U$.
| 2 | https://mathoverflow.net/users/18060 | 417617 | 170,094 |
https://mathoverflow.net/questions/417631 | 1 | This is a question about central limit theorems when the dimension is increasing. Suppose now I have a random vector $X\_N = (X\_{N1}, \cdots, X\_{Np})\in\mathbb{R}^p$. For all $c\_p\in\mathbb{R}^p$ with $\|c\_p\|\_2 = 1$, suppose we have $c\_p^\top X\_N \xrightarrow{d} N(0,1)$ as $N\to\infty, p\to\infty$. What can we say about the asymptotic distribution of $\sum\_{i=1}^p X\_{Ni}^2$ (after normalization if needed)? Is it normal or there is a counterexample?
Several thoughts:
1. If $X\_N$ are i.i.d. $N(0,1)$, we have CLT.
2. If $p$ is fixed, by Cramer-Wood theorem we have asymptotic normality for the $p$-dimensional random vector; then by continuous mapping theorem we have chi-square distribution. But sum of chi-square(1) converge to a normal eventually!
So what I need here is probably a combination of CLT and CMT in increasing dimensions. But I have limited knowledge in this direction.
*Background information*: why would we care about this type of CLT? Suppose $X\_N$ is an aggregation of $p$ summary statistics. Typically we can show any finite linear combination of them are asymptotically normal. If we care about testing a global null to see whether the $p$ true estimands are zero, it is natural to consider a procedure involving a sum-of-square statistics.
More concretely, think about a multiple testing problem where we have $p$ different null hypothesis $H\_{0j},j=1,\cdots,p$. For each null hypothesis, we have some testing statistics, say $X\_{Nj} = N^{-1}\sum\_{i=1}^Nx\_{ij}$. Each of these testing statistics are asymptotically normal. If we consider whether they are jointly normal, it's likely to be true since we do linear combination of the $p$ coordinates we have a linear combination of all the samples involved in the $p$ studies, for which we could apply the general CLT. That's why we have a condition such as "for all $c\_p$...". If we test the global null(that is, all null hypotheses are true) we might consider aggregating the testing statistics by taking a summation of squares. But now, how do we rigorously justify this squared summation also has certain stable distibution?
| https://mathoverflow.net/users/465240 | Central limit theorem of random vectors when the dimension is increasing | In such generality, virtually nothing can be said about the asymptotic distribution of $V\_{Np}:=\sum\_{i=1}^p X\_{Ni}^2$ or even about the existence of such an asymptotic distribution. In particular, $V\_{Np}$ may have a non-normal asymptotic distribution or no asymptotic distribution at all.
Indeed, consider the following three simple settings.
**Setting 1:** All the $X\_{Ni}$'s are iid standard normal and $c\_p^\top c\_p=1$. Then $Y\_{Np}:=c\_p^\top X\_N\sim N(0,1)$ and $V\_{Np}\sim\chi^2\_p\approx N(p,2p)$ (as $p\to\infty$), so that $V\_{Np}$ is asymptotically normal.
**Setting 2:** $X\_{N1}\sim N(0,1)$, $X\_{N2}=\cdots=X\_{Np}=0$, and $c\_p=[1,0,\dots,0]^\top$. Then $Y\_{Np}=X\_{N1}\sim N(0,1)$ and $V\_{Np}=X\_{N1}^2\sim\chi^2\_1$, so that the asymptotic distribution of $V\_{Np}$ is not normal.
**Setting 3:** This is a combination of Settings 1 and 2: for odd $N$ we use Setting 1, and for even $N$ we use Setting 2. Then there is no asymptotic distribution at all.
| 2 | https://mathoverflow.net/users/36721 | 417633 | 170,100 |
https://mathoverflow.net/questions/398509 | 7 | The definition of $L\_\infty$-algebra is by now pretty standard. I gather that the sign conventions given in Lada–Markl's paper *Strongly homotopy Lie algebras*, Communications in Algebra **23** Issue 6 (1995) (arXiv:[hep-th/9406095](https://arxiv.org/abs/hep-th/9406095)) are widely used, and I will keep to them here. I will not rehash the definition of $L\_\infty$-algebra, because I'm sure the people who can answer this question will know it well-enough, or could look it up. I emphasise that in my case, I am **not** assuming anything is finite-dimensional, so that I am unwilling to dualise from dg coalgebras to dg algebras in case there is something funny going on.
What I want to know is what is the correct set of conditions on a weak map of 2-term $L\_\infty$-algebras. Such a thing is a dg-coalgebra map of the graded cocommutative cofree coalgebra equipped with the differential that is the sum of all the brackets of the $L\_\infty$-algebra.
I have found several published papers that all use the definition of Lada–Markl for $L\_\infty$-algebras, specialise to the case of 2-term $L\_\infty$-algebras or slight variations, then give conflicting definitions for a morphism. None of them have commented on potential errors in the others, or if there are conventions that differ leading to different signs. The coherence conditions have various signs flipped in various ways, and it's not clear to me, a complete uninitiate for dealing with cofree coalgebras of infinite-dimensional vector spaces, how to even arrive at even one of these sets. All of the sources I have seen so far do not actually calculate these coherence conditions, merely state them as an definition, rather than deriving them from the definition I just gave (if there is a published source working this stuff out, I'd *love* to see it).
Note: The SE software is suggesting to me [this question](https://mathoverflow.net/questions/139175/what-is-a-homotopy-between-l-infty-algebra-morphisms), except that uses an apparently different sign convention compared to Lada–Markl. However, that's exactly the sort of formula I'm after - just with the signs sorted (which is of course the whole difficulty in this game), and I don't trust myself to not miss something.
---
**Edit** I will point out that Lada and Markl do provide an example of an explicit formula for a class of weak maps, namely from a $L\_m$-algebra (including $m=\infty$) to a dg-Lie algebra (i.e. an $L\_\infty$-algebra with no brackets of arity higher than 2). This is Definition 5.2, and in Remark 5.3 they give their definition of weak map, merely as a map of dg coalgebras between the free coalgebras generated from the $L\_\infty$-algebras, with the differential coming from extending the collection of brackets as a derivation. The Remark then claims that the collection of formulas in Definition 5.2 defines a weak map.
What I would like is to know if this is indeed consistent. No derivation is given, and maybe it's obvious. However, the signs are the tricky part, and I have no idea how the particular combination of signs in Definition 5.2 are arrived at.
| https://mathoverflow.net/users/4177 | What is the correct definition of weak map between 2-term $L_\infty$-algebras? | In the article *Classification of 2-term $L\_\infty$-algebras* (arXiv:[2109.10202](https://arxiv.org/abs/2109.10202)), Kevin van Helden gives the definition of a morphism of 2-term $L\_\infty$-algebras (Definition 2.3), and he was kind enough to share with me some private calculations that go through and checks the definition agrees with the one given in Lada and Markl's Remark 5.3.
| 2 | https://mathoverflow.net/users/4177 | 417635 | 170,102 |
https://mathoverflow.net/questions/417640 | 10 | Given a finite field $\mathbb{F}\_q$ with $q=p^m$ where $p$ is the characteristic.
For any subset $S=\{a\_1,\dots,a\_n\}$ of $\mathbb{F}\_q$, if any partial sum (i.e. the sum of elements in a non-empty subset of $S$) is non-zero, then we may call $S$ a good subset.
My question is what's the maximal cardinality $f(q)$ of a good subset $S$?
Or are there any (lower) bounds for $f(q)$?
| https://mathoverflow.net/users/471160 | The maximal subset of a finite field where the sum of any subset is non-zero | I could trace this question down to the paper of Erdős and Heilbronn "[On the addition of residue classes mod $p$](https://doi.org/10.4064/aa-9-2-149-159)" (Acta Arith. 17 (1970), 227–229), where it is shown that for a prime $p$, if $A$ is a subset of the $p$-element group with $\lvert A\rvert>6\sqrt{3p}$, then the subset sum set of $A$ is the whole group.
It seems that the first to prove that $\lvert A\rvert>c\sqrt{\lvert G\rvert}$, with an absolute constant $c$, suffices to represent $0$ in any abelian group $G$, was Szemerédi ("[On a conjecture of Erdős and Heilbronn](https://doi.org/10.4064/aa-17-3-227-229)", Acta Arith. 17 (1970), 227–229).
Some tightly related problems are considered in a paper of Eggleton and Erdős
"[Two combinatorial
problems in group theory](https://old.renyi.hu/~p_erdos/1972-25.pdf)" (Acta Arith. 21 (1972), 111–116).
Three years later, Olson ("[Sums of sets of group elements](https://doi.org/10.4064/aa-28-2-147-156)", Acta Arith. 28
(1975), 147–156) proved that if $A$ is a zero-sum-free set in a finite
abelian group $G$, then $\lvert A\rvert<3\sqrt{\lvert G\rvert}$.
Hamidoune and Zémor ("[On zero-free subset sums](https://doi.org/10.4064/aa-78-2-143-152)", Acta Arith. 78 (1996), 143–152) showed that for the group of prime order $p$ it suffices to have $\lvert A\rvert>(1+o(1))\sqrt{2p}$.
The exact result for the prime-order groups was obtained by Nguyen,
Szemerédi, and Vu ("[Subset sums modulo a prime](https://doi.org/10.4064/aa131-4-1)", Acta Arith. 131
(2008), no. 4, 303–316).
There were several further improvements in the general case, with the current
record $\lvert A\rvert<\sqrt{6\lvert G\rvert}$ due to Gao, Huang, Hui, Li, Liu, and Peng ("[Sums
of sets of abelian group elements](https://doi.org/10.1016/j.jnt.2019.07.026)", J. Number Theory, 208 (2020),
208–229). In fact, they prove a stronger result: if $A$ is a finite
zero-sum-free subset of an abelian group, then $A$ spans at least as many as
$1+\lvert A\rvert^2/6$ pairwise distinct subset sums.
Two other important papers on this subject are mentioned in the [answer](https://mathoverflow.net/a/417687) of
[Le p'tit bonhomme](https://mathoverflow.net/users/171221/le-ptit-bonhomme).
| 16 | https://mathoverflow.net/users/9924 | 417646 | 170,107 |
https://mathoverflow.net/questions/417621 | 8 | Let $K \subset S^3$ be a slice knot. Then it bounds a smooth embedded disk $D \subset B^4$. Let $S^3\_{p/q}(K)$ denote a $3$-manifold obtained by $p/q$-surgery on $K \subset S^3$.
The following theorem is due to Gordon:
>
> Gordon, C. M. (1975). Knots, homology spheres, and contractible 4-manifolds. Topology, 14(2), 151-172.
>
>
>
**Theorem:** For a slice knot $K \subset S^3$, $S^3\_{\pm 1}(K)$ bounds a contractible $4$-manifold.
I wonder that Gordon's theorem can be generalized to $1/n$ surgeries on slice knots for all $n$?
| https://mathoverflow.net/users/475366 | Gordon's approach: slice knots and contractible $4$-manifolds | Yes, the generalisation is also true. This must be written somewhere, but I don't know where (any help from other users?), and finding such a statement is often hard.
So, here's the idea instead. Turn the surgery into an integral surgery, i.e. do 0-surgery on $K$ and $-n$-surgery on a meridian $L$ of $K$. 4-dimensionally, you're constructing a 4-manifold $X$ by attaching two 2-handles to $B^4$. $X$ contains a 0-sphere (the capped-off slice disc of $K$) and a $-n$-sphere (the capped off meridian disc of $L$) intersecting transversely once. A regular neighbourhood $N$ of the union of these two spheres is a plumbing (by definition) whose boundary is $S^3$ (this is certainly done in Gompf and Stipsicz's book). Now surger out $N$ and replace it with a 4-ball $B$, to get $W$. $W$ is an integral homology ball by excision ($\tilde H\_\*(W) = H\_\*(W,B) = H\_\*(X,N)$) and the long exact sequence of the pair $(X,N)$. The fundamental group hasn't changed with the surgery either: $X\setminus N$ is simply-connected because both $N$, $\partial N$ and $X$ are, and then $W$ is also simply-connected (both steps use Seifert-van Kampen).
| 6 | https://mathoverflow.net/users/13119 | 417648 | 170,109 |
https://mathoverflow.net/questions/417110 | 1 | Consider the Cauchy problem
$$\left\{\hspace{5pt}\begin{aligned}
&-\dfrac{\partial u }{\partial t}
+a\dfrac{\partial^2 u}{\partial x^2}
+b \dfrac{\partial u }{\partial x}
+c u
= f(u) \leq 0& \hspace {10pt} &\text{for $(x,t) \in \mathbb{R} \times (0,T]$}
;\\
&u(x,T) = g(x)\geq 0 & \hspace{10pt} &\text{for $x \in \mathbb{R}$.}
\end{aligned}\right.$$
Here we assume that $u$, $a>0$, $b$, $c<0$, $f \leq 0$ and $g\geq 0$ are smooth enough. Moreover, I have the local bound
$\max\_{t}\|u\|\_{L^2(-R,R)} \lesssim 1$ and $\|\partial\_x u\|\_{L^2((-R,R) \times [0,T])} \lesssim 1$ for any fixed $R >0$. Also, I know that $u\geq0$.
I am going to prove that $\sup\_{[a,b] \times [0,T]} u \lesssim 1$ for any fixed $a<b$. But in the parabolic PDE book by Gary Lieberman (Theorem 6.17 in Chapter VI.6), I only have $\sup\_{[a,b] \times [\delta,T-\delta]} u \lesssim 1$. Is there some theorem or method to extend it globally in time?
| https://mathoverflow.net/users/87922 | Local boundedness for Cauchy problem | I finally find a result in Theorem 11.17 in Chapter XI.6 in parabolic PDE book by Gary Lieberman. But it restricts on the case of one dimension space.
| 0 | https://mathoverflow.net/users/87922 | 417650 | 170,110 |
https://mathoverflow.net/questions/417187 | 2 | I encountered an example in a paper telling that $\underline{SM}(\mathbb{R}^{0|1},X)\cong \pi TX $, where $X$ is some fixed ordinary Riemannian manifold, $\pi TX $ is the supermanifold with base manifold $X$ and structural sheaf $\pi(\wedge^\*(TX)^\vee) $ according to my advisor, seen as a functor $\pi TX:=SM(-,\pi TX): SM^{\mathrm{op}} \to Set$, $S\mapsto SM(S,\pi TX)$ where $SM$ stands for the category of supermanifolds. $\underline{SM}(\mathbb{R}^{0|1},X)$ is the functor which acts on objects by $\underline{SM}(\mathbb{R}^{0|1},X)(S)= SM(S\times \mathbb{R}^{0|1},X) $, with the obvious action on morphisms. "$\cong$" means that there is a natural isomorphism between the two functors.
Now, by the theorem that supermanifolds are "affine" we have
$$\underline{SM}(\mathbb{R}^{0|1},X)(S) \cong SAlg\_{\mathbb{R}}(C^\infty(X), C^\infty(S)\otimes C^\infty(\mathbb{R}^{0|1})) $$
where $SAlg\_{\mathbb{R}}$ is the category of super $\mathbb{R}$-algebra in which the morphisms are parity-preserving $\mathbb{R}$-algebra homomorphisms. Let $\theta$ be the odd coordinate of $\mathbb{R}^{0|1}$ and we see that an element in $\underline{SM}(\mathbb{R}^{0|1},X)(S)$ is identified with a super $\mathbb{R}$-algebra morphism $\Phi^\*:C^\infty(X)\to C^\infty(S)\otimes C^\infty(\mathbb{R}^{0|1})\cong C^\infty(S) \oplus C^\infty(S)\cdot \theta $.
Decompose $\Phi^\*$ using the direct sum and we can write $\Phi^\*=f+\phi\theta$ where $f:C^\infty(X)\to C^\infty(S)$ is a super $\mathbb{R}$-algebra morphism and $\phi:C^\infty(X)\to C^\infty(S)$ is a parity-reversing map such that
$$\phi(ab)=\phi(a)f(b)+f(a)\phi(b)= \phi(a)f(b)+(-1)^{p(a)}f(a)\phi(b)$$
for any $a,b\in C^\infty(X)$, where $p(a)$ stands for the parity of $a$ which is always zero.
The paper refers to Deligne & Morgan's *Notes on Supersymmetry (following Joseph Bernstein)* and says that this is the standard description of $\pi TX$ in terms of its $S$-points so we get $\underline{SM}(\mathbb{R}^{0|1},X)\cong \pi TX $, but Deligne & Morgan's description is not clear to me at all. My advisor told me that $\pi TX$ is supposed to be as above and left the rest to me as an exercise.
I have no difficulty understanding $\pi TX$ for sure, but I don't have a clue how a morphism $\varphi: S\to \pi TX$, which can be identified with a super $\mathbb{R}$-algebra morphism $\varphi^\*:\pi(\wedge^\*(TX)^\vee) \to C^\infty(S) $, could be translated to some $\Phi^\*:C^\infty(X)\to C^\infty(S)\oplus C^\infty(S)\cdot \theta$ satisfying the axioms above.
For the simplest case, say $X=\mathbb{R}^1$, we have $\pi(\wedge^\*(TX)^\vee) =C^\infty(\mathbb{R})\oplus C^\infty(\mathbb{R})\cdot dx$ where $dx$ is considered to be odd. From this $\varphi^\*$ gives a parity-preserving $\mathbb{R}$-algebra $\bar f:C^\infty(X)\to C^\infty(S)$ which I guess is the candidate of $f$ (or not?). Put $s:=\varphi^\*(dx)$ and we see that $\varphi^\*$ is determined by $\bar f$ and $s$, as
$$ \varphi^\* (a+bdx)=\bar f(a)+\bar f(b)s.$$
Unless $\phi=\bar f\cdot s$, which doesn't make any sense, I cannot see where $\phi$ could possibly come from. I think I do need some help about this. Thanks in advance.
| https://mathoverflow.net/users/167862 | Parity reversed tangent bundle as a supermanifold | My advisor told me the answer; I was just one step away from it.
A morphism $\varphi:S\to \pi TX$ is determined by the natural transformation between the structural sheaves, which is locally determined, so it suffices to assume that $X$ has coordinates $x\_1,\cdots,x\_n$. Now $C^\infty(\pi TX)=\pi(\wedge^\*(TX)^\vee) $ is finitely generated by $d x\_1,\cdots,dx\_n$ over $C^\infty(X)$, the composition $C^\infty(X)\hookrightarrow \pi(\wedge^\*(TX)^\vee)\to C^\infty(S)$ gives our $f$ and the remaining data of $\varphi^\*:\pi(\wedge^\*(TX)^\vee)\to C^\infty(S)$ is given by the images of $d x\_1,\cdots,dx\_n$.
Let $s^i:= \varphi^\*(dx\_i)$, then $\varphi^\*$ on $dx\_i$'s can be expressed by (with abuse of notation) $\varphi^\*=\sum\_i \partial\_{x\_i}\otimes\_f s^i$ in light of $\left(\sum\_i\partial\_{x\_i}\otimes \_f s^i\right)(dx\_j)= \sum\_i dx\_j(\partial\_{x\_i})\otimes\_f s^i =s^j $.
Now, $\sum\_i \partial\_{x\_i}\otimes\_f s^i$ gives an element in $\text{Der}(C^\infty(X),C^\infty(X))\otimes\_f C^\infty(S)$, which is by definition (see [Wikipedia](https://en.wikipedia.org/wiki/Inverse_image_functor)) a global section of the pullback bundle $f^\*\pi TX$ on $S$. Finally, as $\text{Der}(C^\infty(X),C^\infty(X))\otimes\_f C^\infty(S)\cong \text{Der}\_f(C^\infty(M),C^\infty(N))$, where $\text{Der}\_f$ denotes derivations with respect to $f$, via $W\otimes s\mapsto s(f^\*\circ W)$ with inverse $V\mapsto \sum\_i \partial\_{x\_i}\otimes \_f V(x\_i) $, the above relates $\varphi$ with $(f,\phi)$ bijectively.
| 1 | https://mathoverflow.net/users/167862 | 417651 | 170,111 |
https://mathoverflow.net/questions/417587 | 2 | I need a (numerically) evaluable function for the number $N\_{n,k}$ of endofunctions $f: [n] \rightarrow [n]$ without fixed points that have exactly $k$ two-cycles, where $[n] := \{1,\dotsc,n\}$. In formal terms, what is
$$N\_{n,k} := \lvert \{ f:[n]\rightarrow [n] \mid \forall a: f(a) \neq a \land \lvert\{ a \in [n] \mid f^2(a) = a\}\rvert = 2k\ \}\rvert? $$
I have found a few sources on this, but have trouble to transfer the results to this specific question. An explicit formula would be greatly appreciated, while a proof or indication of a proof would be as well, though less importantly so.
| https://mathoverflow.net/users/119246 | Number of endofunctions in [n] without fixed points with exactly k two-cycles | First choose your 2-cycles, for a factor of $\binom{n}{2k}(2k-1)!!$. (Note that we require the convention that $(-1)!! = 1$). Then count functions $g: [n-2k] \to [n]$ with no fixed points or 2-cycles. There are $\binom{n-2k}{2j}(2j-1)!! (n-1)^{n-2k-2j}$ functions with no fixed points and at least $j$ 2-cycles, so an inclusion-exclusion gets $$\binom{n}{2k}(2k-1)!! \sum\_{j \ge 0} (-1)^j \binom{n-2k}{2j}(2j-1)!! (n-1)^{n-2k-2j}$$
Alternatively we can just start the whole thing as an inclusion-exclusion from $j = k$, but then the Möbius function adds a binomial coefficient:
$$\sum\_{j \ge k} (-1)^{j-k} \binom{j}{k} \binom{n}{2j}(2j-1)!! (n-1)^{n-2j}$$
| 3 | https://mathoverflow.net/users/46140 | 417653 | 170,112 |
https://mathoverflow.net/questions/417655 | 5 | Apart from the abstract types of the crystallographic groups, are there any other abstract groups that admit a proper, co-compact, uniformly bilipschitz action on $\mathbb{R}^3$?
| https://mathoverflow.net/users/159356 | Infinite groups that admit a discrete, co-compact, bilipschitz action on $\mathbb{R}^3$ |
>
> Fix $k\ge 0$. Let $\Gamma$ be a discrete group. Then $\Gamma$ (a) has a proper cocompact, uniformly bilipschitz action on the Euclidean space $\mathbf{R}^k$ if and only if (b) it has an isometric one, if and only if (c) it has a finite index subgroup isomorphic to $\mathbf{Z}^k$.
>
>
>
Proof: That (c) implies (b) is very standard and (b) trivially implies (a). Suppose (a): consider such an action $(g,x)\mapsto gx$, assume each map $(1/C,C)$-bilipschitz. Fix $x$ in $\mathbf{R}^k$ and a symmetric generating subset $S$ of $\Gamma$. Write $M=\max\_{s\in S}d(x,sx)$ and $C'=CM$.
Then $g\mapsto gx$ is a quasi-isometry. For $g,h\in\Gamma$, let $n=d\_S(g,h)$ be their word distance: consider elements $g\_0,\dots,g\_n$ with $g\_0=g$, $g\_n=h$, $d(g\_i,g\_{i+1})\le 1$. Then
$$d(gx,hx)\le \sum\_{i=0}^{n-1} d(g\_ix,g\_{i+1}x)\le C\sum\_{i=0}^{n-1} d(x,g\_i^{-1}g\_{i+1}x)\le nCM=C'd\_S(g,h),$$
and $d(gx,hx)\ge C^{-1}d(x,g^{-1}h)$ which tends to infinity when $d\_S(g,h)$ tends to infinity.
So $x\mapsto gx$ Lipschitz, uniformly proper, and has cobounded image and is a map between geodesic spaces. Hence it is a quasi-isometry.
We conclude by using that a group that is quasi-isometric to $\mathbf{R}^k$ satisfies (c). (This is a standard consequence of Gromov's theorem on groups with polynomial growth, e.g. using Pansu's theorem, and also admits proofs not appealing to Gromov's theorem.)
| 5 | https://mathoverflow.net/users/14094 | 417661 | 170,114 |
https://mathoverflow.net/questions/417570 | -1 | A simple, undirected graph $G=(V,E)$ is said to be *vertex-transitive* if for all $a,b\in V$ there is a graph isomorphism $\varphi:G\to G$ such that $\varphi(a) = b$.
If $G = (\omega, E)$ is vertex-transitive and connected, is there a bijection $p:\mathbb{Z}\to \omega$ such that $\{p(k), p(k+1)\} \in E$ for all $k\in\mathbb{Z}$?
| https://mathoverflow.net/users/8628 | Hamiltonian $\mathbb{Z}$-paths in connected countably infinite vertex-transitive graphs | If $G$ is a regular tree of degree $d \geq 3$, there's clearly no such $p$.
| 2 | https://mathoverflow.net/users/66104 | 417664 | 170,115 |
https://mathoverflow.net/questions/417436 | 6 | Cross Posting this from MSE since it's been there for almost a month and it got a couple upvotes but no answers. MSE link [Is every finite subgroup the integer points of a linear algebraic group?](https://math.stackexchange.com/questions/4375025/is-every-finite-subgroup-the-integer-points-of-a-linear-algebraic-group)
Let $ K $ be a compact connected Lie group. For every finite subgroup $ \Gamma $ of $ K $ does there exist a linear algebraic group $ G $ such that the integer points are
$$
G\_\mathbb{Z} \cong \Gamma
$$
and the real points are
$$
G\_\mathbb{R} \cong K.
$$
I'm interested in this because sometimes the integer points are cool like
$$
\operatorname{SO}\_3(\mathbb{Z}) \cong S\_4.
$$
EDIT: Here is an attempt to clarify what I am looking for.
Consider 3 by 3 matrices with complex entries. For a $ 3 \times 3 $ complex matrix the conditions
$$
I=MM^T
$$
and
$$
det(M)=1
$$
are polynomial in the entries of $ M $. The polynomials defining these conditions all have integer coefficients. The subset of matrices that satisfy these two constraints is the Lie group $ SO\_3(\mathbb{C}) $. Now if we restrict the entries to be real then we get exactly the group $ SO\_3(\mathbb{R}) $ which is a compact connected Lie group. Finally, if we restrict the entries to be integers we get $ SO\_3(\mathbb{Z}) $ which is a finite group with 24 elements isomorphic to the symmetric group $ S\_4 $.
So what I was really interested in was the idea that for any compact connected lie group $ K $ and finite subgroup $ \Gamma $ we can find a (finite collection of integer coefficient) polynomial constraints on the entries of a square matrix such that the complex matrices satisfying those constraints form a Lie group, the matrices satisfying those constraints and having real entries form a compact connected Lie group, and finally the matrices satisfying those constraints and having integer entries form a finite group isomorphic to $ \Gamma $.
| https://mathoverflow.net/users/387190 | Is every finite subgroup the integer points of a linear algebraic group? | The answer to this question is no.
There exist finite groups $\Gamma \subset K$ of a compact connected Lie group $K$, such that for any algebraic $\mathbb Q$-group $G$ with $G(\mathbb R)=K$, the finite group $\Gamma $ cannot be a subgroup of $G(\mathbb Z)$:
We take $K=SU(2)$ and $\Gamma $ to be a Dihedral group of the form $(\mathbb Z/l\mathbb Z)\rtimes {\mathbb Z}/2{\mathbb Z}$ where the nontrivial element of the group ${\mathbb Z}/2{\mathbb Z}$ operates by $x\mapsto -x$ on ${\mathbb Z}/l{\mathbb Z}$. Here $l$ is a large prime.
Suppose $G$ is an algebraic group defined over $\mathbb Q$ such that $G({\mathbb R})=K$, and $\Gamma \subset G(\mathbb Z)$. Then for almost all primes $p$, $\Gamma $ injects into $G({\mathbb F}\_p)=G(\mathbb Z/p\mathbb Z)$. Moreover, for almost all primes $p$, the order of $G({\mathbb F}\_p)$ is $(p^2-1)(p^2-p)/(p-1)=p(p^2-1)$. Further, $l$ divides this order since $\Gamma $ is a subgroup of $G({\mathbb F}\_p)$.
Therefore, for almost all primes $p$, we have $p(p^2-1)\equiv 0 \quad (mod \quad l)$. But by Dirichlet's theorem on primes in arithmetic progressions, the residue class of the generator of the unit group of ${\mathbb Z}/l{\mathbb Z}$ is represented by infinitely many primes $p$. Hence the order of such a $p$ (modulo $l$) is $l-1$. On the other hand $p(p^2-1)$ is divisible by $l$ which means that $l-1\leq 2$, and $l$ cannot be a large prime.
I am pretty sure that a much simpler proof can be found, but this is "a proof".
ADDED later: The proof shows that the "large prime" $l$ need only satisfy $l\geq 5$. Moreover, the gcd of the numbers $p(p^2-1)$ as $p$ varies over primes large enough, is just $24$. Hence the order of $\Gamma $ is $\leq 24$.
| 6 | https://mathoverflow.net/users/23291 | 417672 | 170,118 |
https://mathoverflow.net/questions/417669 | 4 | Are there $2^{\aleph\_0}$ pairwise non-isomorphic connected [vertex-transitive](https://en.wikipedia.org/wiki/Vertex-transitive_graph) graphs $G$ with $V(G) = \omega$?
| https://mathoverflow.net/users/8628 | Non-isomorphic connected vertex-transitive graphs on $\omega$ | Yes. In particular, there are continuously many pairwise non-quasi-isometric Cayley graphs with countably infinite vertex set. A proof of this is in the paper *Continuously many quasi-isometry classes of 2-generator groups* by Bowditch.
| 5 | https://mathoverflow.net/users/159356 | 417675 | 170,119 |
https://mathoverflow.net/questions/417676 | 3 | Let $R$ be a (not necessarily commutative) ring, $M$ a left $R$-module, and $N$ a right $R$-module. We say that a pairing
$$
\langle -,-\rangle:M \otimes\_R N \to R
$$
is non-degenerate if, for all $n \in N$ there exists an $m \in M$ such that $\langle m,n\rangle \neq 0$, \textbf{and} for all $m \in M$, there exists an $n \in N$ such that $\langle m,n\rangle \neq 0$.
Such a pairing will give embeddings
$$
M \hookrightarrow N^\*, ~~~~~ N \hookrightarrow M^\*,
$$
where $M^\*$ and $N^\*$ denote the dual modules of $M$ and $N$ respectively. In general (even for infinite-dimensional vector spaces) this will not give isomorphisms
$$
N \simeq M^\*, ~~~~ M \simeq N^\*.
$$
However, if we assume that $M$ and $N$ are finitely-generalted projective, then does non-degeneracy imply that we get isomorphisms?
If it fails in the general noncommutative setting, I would still be interested in a positive answer in the commutative setting.
| https://mathoverflow.net/users/478224 | Nondegenerate pairings versus perfect pairings for finitely generated projective modules | No: for $R=\mathbb{Z}$ and even for $M,N$ both free (i.e. free Abelian groups), non-degenerate doesn't imply perfect. (You get finite index sublattices, so torsion quotients, unlike the case of vector spaces.)
| 5 | https://mathoverflow.net/users/13215 | 417677 | 170,120 |
https://mathoverflow.net/questions/417666 | 0 | Is there a method to make a rep-n rep-tile for any number n, using only triangles? And if there is no such method, what's the smallest number for which there's no example?
I'm only considering rep-tiles which are all of the same size, as they make up the larger congruent rep-tile.
I first asked the more general question, can you do it for any shape, not just triangles? But then I felt confident that you can do it with rectangles, for any n.
| https://mathoverflow.net/users/478353 | Is there a method to make a rep-n rep-tile for any number n, using only triangles? | You can indeed do it for rectangles, by taking a ratio of sides equal to $\sqrt{n}$.
For triangles, M. Beeson (2012), [Triangle Tiling I: the tile is similar
to ABC or has a right angle](https://arxiv.org/pdf/1206.2231.pdf) (pre-print) gives the following results:
>
> When the tile is similar to $ABC$, we always have "quadratic tilings" when $N$ is a square. If the tile is similar to $ABC$ and is not a right triangle,
> then $N$ is a square. If $N$ is a sum of two squares, $N = e^2 + f^2$, then a right triangle with legs $e$ and $f$ can be $N$-tiled by a tile similar to $ABC$; these tilings are called "biquadratic". If the tile and $ABC$ are $30-60-90$ triangles, then $N$ can also be three times a square. If $T$ is similar to ABC, these are all the possible triples $(ABC, T, N)$.
>
>
>
$n = 6$ is not a square, a sum of two squares, or three times a square, so this is the first $n$ for which there is no example.
| 3 | https://mathoverflow.net/users/46140 | 417681 | 170,122 |
https://mathoverflow.net/questions/417641 | 1 | Let $X$ be a reflexive Banach space, $z, x, v \in X$. If $f: X \times X \rightarrow \mathbb{R}$ is continuous regarding its first argument and locally Lipschitz regarding its second argument. $\{z\_i\}, \{x\_i\}$ and $\{v\_i\}$ are arbitrary sequences converging to $z, x$ and $v$, respectively.
**I wonder** if generalized directional derivative $f^{0}(z, x; v)$ (in [Clarke sense](https://encyclopediaofmath.org/wiki/Clarke_generalized_derivative)) satisfies the following inequality:
$$
f^{0}(z, x ; v) \geqslant \limsup \_{i \rightarrow \infty} f^{0}\left(z\_{i}, x\_{i} ; v\_{i}\right)
$$
**And I also want to know** how to prove it if the inequality holds.
Thanks for your help in advance.
| https://mathoverflow.net/users/478336 | Proof that Clarke generalized directional derivative is upper continuous | $\newcommand\R{\mathbb R}$The answer is no.
Indeed, apparently, in your post the Clarke derivative is meant with respect to the second argument of $f$:
$$f^0(z,x;v):=\limsup\_{\substack{y\to x\\ t\downarrow0}}
\frac{f(z,y+tv)-f(z,y)}t.$$
Let now $X:=\R$ and
$$f(z,x):=xe^{-x/|z|}\,1(z\ne0,x>0)$$
for real $z,x$ -- that is, $f(z,x)=xe^{-x/|z|}$ if $z\ne0$ and $x>0$, and $f(z,x)=0$ otherwise. Then $f(z,x)$ is continuous in $z$ and $1$-Lipschitz in $x$.
However, $f^0(0,0;1)=1(z\ne0)$, so that $f^0(z,x;v)$ is not upper-semicontinuous at $(0,0,1)$.
| 0 | https://mathoverflow.net/users/36721 | 417683 | 170,123 |
https://mathoverflow.net/questions/417684 | 11 | Does anyone know anything about M. Meyniel? [According to zbMath](https://www.zbmath.org/authors/?q=ai%3Ameyniel.m), he published precisely one mathematics paper, in which he gave a sufficient condition for hamiltonicity of digraphs:
>
> "[Une condition suffisante d'existence d'un circuit hamiltonien dans un graphe oriente](https://doi.org/10.1016/0095-8956(73)90057-9)" (*JCTB* **14** (1973), 137–147).
>
>
>
In that paper, he was listed with an address but no affiliation. The address was 13, rue Poirier de Narçay, Paris 14ᵉ, which appears to be an apartment above a [game store](https://www.games-workshop.com/de-AT/Warhammer-Paris-14e?_requestid=11221417) for what it is worth.
I assume that M. Meyniel is distinct from Henri Meyniel of [Meyniel graphs](https://en.wikipedia.org/wiki/Meyniel_graph) and [Meyniel's conjecture](https://en.wikipedia.org/wiki/Cop_number#General_results). That said, the paper "[Sufficient conditions for a digraph to be Hamiltonian](https://doi.org/10.1002/(SICI)1097-0118(199606)22:2%3C181::AID-JGT9%3E3.0.CO;2-J)" (*J. Graph Th.* **22** (1996) 181–187) is dedicated to the memory of Henri Meyniel and lists Henri Meyniel as the author of M. Meyniel's paper (item [15] in the bibliography).
| https://mathoverflow.net/users/2663 | Who is M. Meyniel? | You can be quite sure that M. Meyniel means "Monsieur Meyniel" (a common usage in French).
Here is what I think is definite proof that M. Meyniel is H. Meyniel: The acknowledgement of the 1973 paper by M. Meyniel thanks J.C. Bermond, so evidently Bermond knew the author. In the article [Cycles in digraphs - a survey](https://hal.inria.fr/hal-02424283/file/44-BeTh81-surveycyclesdigraphs.pdf) Bermond and Thomassen cite the 1973 paper as follows:
>
> Meyniel, H., Une condition suffisante d'existance d'un circuit hamiltonien dans un graphe orienté, *J. Combinatorial Theory B* 14(1937), 137–147
>
>
>
| 29 | https://mathoverflow.net/users/11260 | 417685 | 170,124 |
https://mathoverflow.net/questions/417698 | 4 | Let $a$ be an element of $\mathbb{F}\_p$, which is not a quadratic residue.
Define $$f(x) = \frac{x + a}{x+1},$$ which is a rational function on $\mathbb{F}\_p$. In fact, if we set $f(-1)=\infty$ and $f(\infty)=1$, then $f:\mathbb{F}\_p\cup\{\infty\}\rightarrow \mathbb{F}\_p\cup\{\infty\}$ is a bijection.
What is the order of $f$ under the operation of composition?
I expect that $f$ has order $p+1$, but I don't know how to prove it.
**Edit:**
As pointed out rightfully below, the sign of $a$ was wrong at first!
| https://mathoverflow.net/users/166993 | Order of a rational function on $\mathbb{F}_p$ | As @DavidESpeyer [suggests](https://mathoverflow.net/questions/417698/order-of-a-rational-function-on-mathbbf-p#comment1072032_417698), I think you meant ${}+ a$ in place of ${}- a$. As @KevinCasto [says](https://mathoverflow.net/questions/417698/order-of-a-rational-function-on-mathbbf-p#comment1072031_417698), you are then looking for the order of $\begin{bmatrix} 1 & a \\ 1 & 1 \end{bmatrix}$ as an element of $\operatorname{PGL}\_2(\mathbb F\_p)$, i.e. (since its eigenvalues are $1 \pm \sqrt a$), the order of $1 \pm \sqrt a$ as an element of $\mathbb F\_{p^2}^\times/\mathbb F\_p^\times$. Certainly this order divides $p + 1 = \lvert\mathbb F\_{p^2}^\times/\mathbb F\_p^\times\rvert$; but, as @DavidESpeyer also [suggests](https://mathoverflow.net/questions/417698/order-of-a-rational-function-on-mathbbf-p#comment1072032_417698), it need not equal $p + 1$. Indeed, if $p = 5$ and $a = 2$, then the element has order $3$, since $\begin{pmatrix} 1 & 2 \\ 1 & 1 \end{pmatrix}^3 = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}$. (If you really did mean ${}- a$ instead of ${}+ a$, then the same example works; just pretend I took $a = 3$ instead of $a = 2$.)
EDIT: There was some discussion about what would happen if you had ${}- a$ in place of ${}+ a$. If $-a$ as well as $a$ is not a quadratic residue (i.e., if $-1$ *is* a quadratic residue), then the reasoning above shows that the order divides (but need not equal) $p + 1$. If $-a$ *is* a quadratic residue, then you are now looking for the order of $(1 + \sqrt{-a}, 1 - \sqrt{-a})$ as an element of the quotient of $\mathbb F\_p^\times \times \mathbb F\_p^\times$ by the diagonal copy of $\mathbb F\_p^\times$. The order thus now divides $p - 1$. Taking $p = 7$ and $a = 3$, and observing that $\begin{pmatrix} 1 & -3 \\ 1 & 1 \end{pmatrix}^3 = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$, shows that the order may be a proper divisor of $p - 1$.
| 7 | https://mathoverflow.net/users/2383 | 417700 | 170,131 |
https://mathoverflow.net/questions/417701 | 2 | Let $p > 2$ be a prime and $q = p^r$ for some $r \in \mathbb{Z}^+$. I will assume that all roots of unity lie in $\mathbb{C}\_p^{\times}$. Let $\zeta$ a primitive $p$-th root of unity. Let $Tr : \mathbb{F}\_q \to \mathbb{F}\_p$ be the trace. Also, denote $\pi$ to be the maximal prime in $\mathbb{Z}\_p[\zeta]$ such that $\pi^{p-1} = -p$.
For a multiplicative character $\psi: \mathbb{F}\_q \to \mathbb{C}\_p^{\times}$ ($\psi(0) = 0$), the Gauss sum for $\psi$ is defined to be
\begin{align\*}
G(\psi) = \sum\_{c \in \mathbb{F}\_q} \psi(c) \zeta^{Tr\_{\mathbb{F}\_q/\mathbb{F}\_p}(c)}
\end{align\*}
I feel that the following should be true:
If $\psi$ is order 2 and $r =1$ then $G(\psi) = \pi^{(p-1)/2}$ or $-i\pi^{(p-1)/2}$, where $i \in \mathbb{C}\_p$ is a solution to $X^2 + 1 = 0$.
My thought is that $\pi^{(p-1)/2}$ is playing the role of $i\sqrt{p}$ in the traditional value for the Gauss sum (where we view everything taking place in $\mathbb{C}$). Is this correct? Or is there something more subtle going on that I'm missing?
| https://mathoverflow.net/users/171396 | Value of the quadratic Gauss sum viewed in $\mathbb{C}_p$ | That $G(\psi,\zeta)^2 = \psi(-1)p$ is pure algebra, so it holds in $\mathbf C$ or $\mathbf C\_p$ or any other field not of characteristic $2$ that contains a nontrivial $p$th root of unity.
You could write down a $p$-adic formula for your quadratic Gauss sum using the Gross–Koblitz formula.
First let's normalize the link between your nontrivial $p$th root of unity and your choice of $\pi$ such that $\pi^{p-1} = -p$. To each $\pi$ there is a unique nontrivial $p$th root of unity $\zeta$ such that $\zeta \equiv 1 + \pi \bmod \pi^2$, where the congruence means $\lvert\zeta - (1 + \pi)\rvert\_p \leq \lvert\pi\rvert\_p^2$, or equivalently $\lvert\zeta - (1 + \pi)\rvert\_p < \lvert\pi\rvert\_p$ since $\mathbf Q\_p(\pi) = \mathbf Q\_p(\zeta)$. Write the $\zeta$ fitting that congruence mod $\pi^2$ as $\zeta\_{\pi}$.
Every character of $\mathbf F\_q^\times$ with values in $\mathbf C\_p$ is a power of the Teichmüller character $\omega\_q$ (interpret $\mathbf F\_q$ as $\mathbf Z\_p[\zeta\_{q-1}]/(p)$). For the Gross–Koblitz formula it is convenient to write characters of $\mathbf F\_q^\times$ as powers of $\omega\_q^{-1}$, say as $\omega\_q^{-k}$ for $0 \leq k < q-1$. The quadratic character $\psi$ of $\mathbf F\_q^\times$ is $\omega\_q^{(q-1)/2} = \omega\_q^{-(q-1)/2}$, so $k = (q-1)/2$.
Let the base $p$ expansion of $k$ be $d\_0 + d\_1p + \cdots + d\_{f-1}p^{f-1}$. When $k = (q-1)/2 = (p^f-1)/2$, all of its base $p$ digits are $(p-1)/2$, so the sum of the base $p$ digits is $f(p-1)/2$.
The Gross–Koblitz formula for the quadratic character $\psi$ says
$$
-G(\psi,\zeta\_\pi) = \pi^{f(p-1)/2}\Gamma\_p\left(\frac{(p-1)/2}{q-1}\right)^f,
$$
where $\Gamma\_p$ is Morita's $p$-adic Gamma-function. Note the minus sign on the left side: normalizing Gauss sums with an overall minus sign is reasonable for various purposes, like here and in the Hasse–Davenport relation. On the right side of the formula above,
$\pi^{f(p-1)/2}$ is a square root of $\pi^{f(p-1)} = (-p)^f = (-1)^fq$.
$\newcommand\sgn{\genfrac(){}{}}$In the special case $q = p$ (so $f = 1$), you're working with the classical quadratic Gauss sum for $\mathbf F\_p$ and the Legendre symbol. In this case
$$
-G\left(\sgn\cdot p,\zeta\_\pi\right) = \pi^{(p-1)/2}\Gamma\_p\left(\frac{1}{2}\right),
$$
where $\pi^{(p-1)/2}$ is a square root of $\pi^{p-1} = -p$. For $p > 2$ it is known that $\Gamma\_p(1/2)^2 = -\sgn{-1}p$, so if you square the right side above then you get $\pi^{p-1}\Gamma\_p(1/2)^2 = -p(-\sgn{-1}p) = \sgn{-1}p p$, which is the formula for the square of the mod $p$ quadratic Gauss sum that I mentioned at the start of this answer (when $q = p$).
| 7 | https://mathoverflow.net/users/3272 | 417710 | 170,135 |
https://mathoverflow.net/questions/417539 | 1 | I recall two definitions from Banach space theory
>
> **Definition 1.** Let $E$ be a Banach space, then a basis $(e\_n)\_{n\in\mathbb{N}}$ of $E$ is called $1$-spreading if $$\left\|\sum\_{i=1}^k a\_i e\_{m\_i}\right\|\le\left\|\sum\_{i=1}^k a\_i e\_{n\_i}\right\|$$ whenever $k$ is a positive integer, $(a\_i)\_{i=0}^k$ are scalars and $n\_1<\ldots<n\_k$ and $m\_1<\dots < m\_k$.
>
>
> **Definition 2.** Let $E$ be a Banach space and $(x\_n)\_{n\in\mathbb{N}}$ a basic sequence of $E$. A spreading model of $(x\_n)\_{n\in\mathbb{N}}$ is a normalized basic sequence $(y\_n)\_{n\in\mathbb{N}}$ in a Banach space $F$ such that for every $\epsilon > 0$ and $k\in\mathbb{N}$, there is an $N$ such that $$(1+\epsilon)^{-1}\left\|\sum\_{i=1}^k a\_i y\_i\right\|\le \left\|\sum\_{i=1}^k a\_i x\_{n\_i}\right\|\le (1+\epsilon)\left\|\sum\_{i=1}^k a\_i y\_i\right\|$$ for all $N<n\_1<\dots<n\_k$ and a sequence $(a\_i)\_{i=1}^k$ of scalars.
>
>
>
Now, looking at these two definitions I get the following:
* If a basis is $1$-spreading then, fixing $n\in\mathbb{N}$, all the subspaces of dimension $n$ generated by a subset of the basis are isometrically isomorphic.
* If $(y\_n)\subset F$ is a spreading model of $(x\_n)\subset E$ then every finite dimensional subspace generated by a subset of $(y\_n)$ is $(1+\epsilon)$-isomorphic to a finite dimensional subspace of $E$.
My two related question, that I supect being almost trivial (I'm relatively new to Banach space theory), are the following:
1. If a basis is $1$-spreading then, fixing $n\in\mathbb{N}$, are all subspaces of dimension $n$ isometrically isomorphic? What about infinite dimensional subspaces? Are them all isometrically isomorphic to the space $E$ itself?
2. If $(y\_n)\subset F$ is a spreading model of $(x\_n)\subset E$, is every finite dimensional subspace of $F$ $(1+\epsilon)$-isomorphic to a finite dimensional subspace of $E$? Again what about the infinite dimensional subspaces of $F$?
Thanks!
| https://mathoverflow.net/users/141146 | Definition of $1$-spreading basis and spreading model | It might be more useful to read first the structure of classical $\ell\_p$ spaces before starting on spreading models. These questions are about those and has very little to do with spreading models (only exception is the pull back mentioned below).
1. The only Banach space with any one of these properties is the Hilbert space. So for instance the unit vector basis of $\ell\_p$, $p\neq 2, p<\infty$ is 1-spreading but it contains $\ell\_2^n$'s almost isometrically as well as $\ell\_p^n$'s, and many other types depending on $p$. Every non-Hilbertian Banach space contains infinite dimensional subspaces not isomorphic to the space. So the answer is always No with only one exception of the Hilbert space.
2. The first assertion is true and trivial (or routine depending on background). The finite dimensional subspace can be seen almost isomoterically as a subspace (possibly much higher dimensional) of finite dimensional spanned by basis vectors, say, $(e\_i)\_1^m$. (Take $\varepsilon$-net in the unit ball, approximate each vector by a finitely supported vector, then the union of all basis vectors used in the process gives $(e\_i)\_1^m$.) Then you pull back $(e\_i)\_1^m$into $E$.
The answer to the second assertion is negative. For instance, Tsirelson space does not contain copies of $\ell\_1$ but all of its spreading models are isomorphic to $\ell\_1$. Spreading models are usually 'nicer' than the generating space, and they are usually the starting point when looking for nice finite dimensional subspaces. But the caveat is you can't always pull back infinite dimensional information from the spreading models.
| 7 | https://mathoverflow.net/users/3675 | 417714 | 170,136 |
https://mathoverflow.net/questions/417502 | 4 | I'm not sure this question fully qualifies as a research-level math question, but from my (limited) past experience on stackexchanged I feared this question might not get an answer there.
**Setting**: the category $\cal{O}$ associated to a complex semisimple Lie algebra $\frak{g}=\frak{n}\oplus\frak{h}\oplus\frak{n}^-$. The Verma with highest weight $\lambda$ is denoted $M\_\lambda$, and its irreducible quotients are $L\_\lambda$. The projective cover of $L\_\lambda$ is $P\_\lambda$, which is self-dual precisely when $\lambda$ is $\varrho$-antidominant. Let the contragredient dual be denoted $M^\dagger$, and let $\mathbb C\_\lambda$ denote the one-dimensional space with an $\mathfrak h$-action given by $\lambda$.
**Question summary**: for the tensor of a Verma with a finite-dimensional, take the spectral sequence associated to the filtered complex which is the Chevalley cochain complex filtered by the Verma filtration on the tensor we are interested in -- then we should get a spectral sequence $H^{p+q}(\mathfrak{n} : M\_{\lambda+\lambda\_p})\ {\Longrightarrow\_p}\ H^{p+q}(\mathfrak{n}:M\_\lambda\otimes V)$. However I am applying this to examples such as $M\_{-1}\otimes L\_1=P\_{-2}$ and $M\_0\otimes L\_2=M\_2\oplus P\_{-2}$ in $\frak sl\_2$ and getting the wrong answers. I reproduce my work below and would be very grateful if someone could point out where I went wrong.
**Details**: For a finite-dimensional object $V\in\cal O$, the tensor product $M\_\lambda\otimes V$ admits a "standard filtration" or "Verma flag", $M\_\lambda\otimes V=M(0)\supset M(1)\supset\cdots\supset 0$, whose quotients are $M(i)/M(i+1)=M\_{\lambda\_i}$, where $\{\lambda\_i\}\_i$ are the weights of $V$ labeled such that $\lambda\_i\ge\lambda\_j\implies i\le j$.
Let $C^\bullet$ be the Chevalley cochain complex $\operatorname{Hom}\_\mathbb{C}(\mathfrak{n}^{\wedge\bullet},M\_\lambda\otimes V)$, and consider the filtration $F^p C^{p+q}=\operatorname{Hom}\_\mathbb{C}(\mathfrak{n}^{\wedge(p+q)},M(p))$. This is a bounded decreasing filtration of a bounded cohomologically graded complex. Then this guy has a spectral sequence
$$E\_1^{p,q}=H^{p+q}(\operatorname{Hom}\_\mathbb{C}(\mathfrak{n}^{\wedge(p+q)},M(p))/\operatorname{Hom}\_\mathbb{C}(\mathfrak{n}^{\wedge(p+q)},M(p+1)))=H^{p+q}(\mathfrak{n}:M\_{\lambda+\lambda\_p}) \ {\Longrightarrow\_p}\ H^{p+q}(\mathfrak{n}:M\_\lambda\otimes V),$$
where $\deg \text{d}\_r=(r,1-r)$.
I tried to apply this to $M\_{-1}\otimes L\_1=P\_{-2}$. In that case we would get the first page $E\_1$ has (I will denote $H^k(\mathfrak{n}:M)$ as just $H^k(M)$) $H^0(M\_0)$ in the $(0,0)$ position, $H^1(M\_0)$ in the $(0,1)$ position, $H^1(M\_{-2})$ in the $(1,0)$ position, and $H^0(M\_{-2})$ in the $(1,-1)$ position, and nothing anywhere else. But $M\_{-2}=L\_{-2}$ is self-dual, and recall from category $\cal O$ that $M^\dagger$ has a standard filtration if and only if $H^{>0}(\mathfrak{n}:M)=0$ (the obstruction to the contragredient having a standard filtration is measured by higher Lie algebra cohomology out of $\mathfrak n$). Hence $H^1(M\_{-2})=0$. Then $E\_1$ just has three nonzero terms, each lying in its own row. The differential $\text d\_1$ has degree $(1,0)$ and points to the right. So $E\_2=E\_1$. It's also easy to see that the later differentials, of degree $(r,1-r)$, will never connect two nonzero terms. So actually $E\_1=E\_\infty$.
Let us compute the actual terms of this page. $H^0(M\_0)=\mathbb C\_0\oplus\mathbb C\_{-2}$ and $H^0(M\_{-2})=\mathbb C\_{-2}$ since $H^0(M)$ is just the $\mathfrak n$-invariants of $M$, and $H^1(M\_{0})=\mathbb C\_{-2}$ since in general there is a formula $\operatorname{ch}\_M=\sum\_\lambda \chi(\operatorname{Ext}^\bullet(M\_\lambda,M))\operatorname{ch}\_{M\_\lambda}$. This would imply, by the convergence of the spectral sequence, that
$$H^0(\mathfrak n:M\_{-1}\otimes L\_1)=\mathbb C\_0\oplus\mathbb C\_{-2}\oplus\mathbb C\_{-2},\qquad H^1(\mathfrak n:M\_{-1}\otimes L\_1)=\mathbb C\_{-2}.$$
HOWEVER this cannot be correct, since $M\_{-1}\otimes L\_1=P\_{-2}$ is both injective and projective, so in particular the $-2$-weight space of $H^1(\mathfrak n:P\_{-2})$, which is given by $\operatorname{Ext}^1(M\_{-2},P\_{-2})$, must vanish. The $H^0$ is also incorrect, since $\text{ch}\_{P\_{-2}}=\text{ch}\_{M\_0}+\text{ch}\_{M\_{-2}}$ then forces $\dim\operatorname{Ext}^0(M\_{-2},P\_{-2})=\dim\operatorname{Ext}^0(M\_{0},P\_{-2})=1$, i.e. $H^0(\mathfrak n:P\_{-2})=\mathbb C\_0\oplus\mathbb C\_{-2}$. So the right answer should be
$$H^0(\mathfrak n:P\_{-2})=\mathbb C\_0\oplus\mathbb C\_{-2},\qquad H^1(\mathfrak n:P\_{-2})=0.$$
I've tried a similar computation with $M\_0\otimes L\_2=M\_2\oplus P\_{-2}$ with similarly disastrously wrong answers.
What gives?
| https://mathoverflow.net/users/476521 | Spectral sequence from standard/Verma filtration/flag to compute Lie algebra cohomology of tensor product with respect to $\mathfrak{n}$ | I think the issue here is that the subquotients in the standard filtration have weights (your $\lambda\_i$'s) which are ordered the other way. To be clear, if the weights of $\nu\_0$, ..., $\nu\_n$ of $L$ are ordered so that $\nu\_i \le \nu\_j$ implies $i\le j$, then one obtains a standard filtration for $M\_\lambda \otimes L$ with subquotients $M(i)/M(i+1) = M\_{\lambda+\nu\_i}$. Explicitly, the filtration comes from applying the exact functor $\mathcal{U}(\mathfrak g) \otimes\_{\mathcal{U}(\mathfrak{b})}-$ to the filtration of $U(\mathfrak{b})$-modules $N(0)=\mathbb{C}\_\lambda \otimes L \supset N(1) \supset ... \supset N(n)=0$, where $N(k)=U(\mathfrak{b})\cdot \{v\_k,...,v\_n\}$. (Had you ordered the weights the other way, the $N(i)$ would not be $\mathfrak b$-submodules!)
In your first example, $\nu\_0=-1$ and $\nu\_2=1$, so $M=M\_{-1}\otimes L\_1$ has a filtration with $M(0)=M$, $M(1)=M\_{0}$ and $M(0)/M(1)=M\_{-2}$ (so $\lambda\_0=-2$, $\lambda\_1=0$). Since it only has two steps, the spectral sequence amounts to a single long exact sequence computation: the one associated to the short exact sequence of complexes $0 \rightarrow C^\bullet(\mathfrak n : M(1)) \rightarrow C^\bullet(\mathfrak n : M) \rightarrow C^\bullet(\mathfrak n : M\_{-2}) \rightarrow 0$. It is easy to see that this does agree with your character computations for $P\_{-2}$.
| 4 | https://mathoverflow.net/users/138150 | 417716 | 170,137 |
https://mathoverflow.net/questions/417690 | 148 | Teaching group theory this semester, I found myself laboring through a proof that the sign of a permutation is a well-defined homomorphism $\operatorname{sgn} : \Sigma\_n \to \Sigma\_2$. An insightful student has pressed me for a more illuminating proof, and I'm realizing that this is a great question, and I don't know a satisfying answer. There are many ways of phrasing this question:
**Question:** Is there a conceptually illuminating reason explaining any of the following essentially equivalent statements?
1. The symmetric group $\Sigma\_n$ has a subgroup $A\_n$ of index 2.
2. The symmetric group $\Sigma\_n$ is not simple.
3. There exists a nontrivial group homomorphism $\Sigma\_n \to \Sigma\_2$.
4. The identity permutation $(1) \in \Sigma\_n$ is not the product of an odd number of transpositions.
5. The function $\operatorname{sgn} : \Sigma\_n \to \Sigma\_2$ which counts the number of transpositions "in" a permutation mod 2, is well-defined.
6. There is a nontrivial "determinant" homomorphism $\det : \operatorname{GL}\_n(k) \to \operatorname{GL}\_1(k)$.
7. ….
Of course, there are many proofs of these facts available, and the most pedagogically efficient will vary by background. In this question, I'm not primarily interested in the pedagogical merits of different proofs, but rather in finding an argument where the existence of the sign homomorphism looks *inevitable*, rather than a contingency which boils down to some sort of auxiliary computation.
The closest thing I've found to a survey article on this question is a 1972 note ["An Historical Note on the Parity of Permutations" by TL Bartlow](https://doi.org/10.1080/00029890.1972.11993124) in the American Mathematical Monthly. However, although Bartlow gives references to several different proofs of these facts, he doesn't comprehensively review and compare all the arguments himself.
Here are a few possible avenues:
* $\Sigma\_n$ is a Coxeter group, and as such it has a presentation by generators (the adjacent transpositions) and relations where each relation respects the number of words mod 2. But just from the definition of $\Sigma\_n$ as the group of automorphisms of a finite set, it's not obvious that it should admit such a presentation, so this is not fully satisfying.
* Using a decomposition into disjoint cycles, one can simply compute what happens when multiplying by a transposition. This is not bad, but here the sign still feels like an *ex machina* sort of formula.
* Defining the sign homomorphism in terms of the number of pairs whose order is swapped likewise boils down to a not-terrible computation to see that the sign function is a homomorphism. But it still feels like magic.
* Proofs involving polynomials again feel like magic to me.
* Some sort of topological proof might be illuminating to me.
| https://mathoverflow.net/users/2362 | Conceptual reason why the sign of a permutation is well-defined? | (This is a variant of Cartier's argument [mentioned](https://mathoverflow.net/a/417727) by Dan Ramras.)
Let $X$ be a finite set of size at least $2$. Let $E$ be the set of edges of the complete graph on $X$. The set $D$ of ways of directing those edges is a torsor under $\{\pm1\}^E$. Let $G$ be the kernel of the product homomorphism $\{\pm1\}^E \to \{\pm1\}$. Since $(\{\pm1\}^E:G)=2$, the set $D/G$ of $G$-orbits in $D$ has size $2$. The symmetric group $\operatorname{Sym}(X)$ acts on $X$, $D$, and $D/G$, so we get a homomorphism $\operatorname{Sym}(X) \to \operatorname{Sym}(D/G) \simeq \{\pm 1\}$. Each transposition $(ij)$ maps to $-1$ because if $d \in D$ has all edges at $i$ and $j$ outward except for the edge from $i$ to $j$, then $(ij)d$ equals $d$ except for the direction of the edge between $i$ and $j$.
| 95 | https://mathoverflow.net/users/2757 | 417732 | 170,145 |
https://mathoverflow.net/questions/417691 | 0 | Let $X$ be a reflexive Banach space, $z, x, v \in X$. $\{z\_i\}, \{x\_i\}$ and $\{v\_i\}$ are arbitrary sequences converging to $z, x$ and $v$, respectively.
**I would like to know** under which **conditions of the function $f: X \times X \rightarrow \mathbb{R}$**, generalized directional derivative $f^{0}(z, x; v)$ (in [Clarke sense](https://encyclopediaofmath.org/wiki/Clarke_generalized_derivative)) satisfy the following inequality:
$$
f^{0}(z, x ; v) \geqslant \limsup \_{i \rightarrow \infty} f^{0}\left(z\_{i}, x\_{i} ; v\_{i}\right)
$$
**Conditions of the function $f$ I can accept:** For this $f$, I need it is locally Lipschitz regarding its second argument at least. Regarding its first argument, I want the function to have properties like Lipschitz continuous or something similar to continuous.
**And I also want to know** how to prove it if there are these conditions.
Thanks for your help in advance.
| https://mathoverflow.net/users/478336 | The conditions used to prove upper semicontinuous of generalized directional derivative (in Clarke sense) | $\newcommand\R{\mathbb R}$The conditions that you can accept are not enough for the conclusion that you desire.
Indeed, as in the [previous answer](https://mathoverflow.net/a/417683/36721), let $X:=\R$ and
$$f(z,x):=xe^{-x/|z|}\,1(z\ne0,x>0)$$
for real $z,x$ -- that is, $f(z,x)=xe^{-x/|z|}$ if $z\ne0$ and $x>0$, and $f(z,x)=0$ otherwise. Then $f(z,x)$ is $C$-Lipschitz in $z$ (with $C:=(2/e)^2$) and $1$-Lipschitz in $x$.
However, $f^0(0,0;1)=1(z\ne0)$, so that $f^0(z,x;v)$ is not upper-semicontinuous at $(0,0,1)$.
---
If we modify the above example a bit, as follows:
$$f(z,x):=xe^{-|x|/|z|}\,1(z\ne0)$$
for real $z$ and $x$, then we will see that $f(z,x)$ is, not only $C$-Lipschitz in $z$ and $1$-Lipschitz in $x$, but also continuously differentiable in $z$ and in $x$, and yet still with $f^0(0,0;1)=1(z\ne0)$, so that $f^0(z,x;v)$ is not upper-semicontinuous at $(0,0,1)$.
---
On the other hand, the conclusion you desire will obviously hold if $f(z,x)$ has a partial derivative in $x$ which is continuous in $(z,x)$.
| 0 | https://mathoverflow.net/users/36721 | 417734 | 170,147 |
https://mathoverflow.net/questions/417740 | -1 | What is an example of a simple graph $G = (\{1,\ldots,n\}, E)$, where $n\in\mathbb{N}$ is a positive integer, with the following properties?
1. There is a path in $G$ of length $n$,
2. every vertex has at least $2$ neighbors, and
3. $G$ does *not* have a [Hamiltonian cycle](https://en.wikipedia.org/wiki/Hamiltonian_path).
| https://mathoverflow.net/users/8628 | Path of length $n$ but no Hamilton cycle | Assuming you mean a graph with a Hamiltonian path but no Hamiltonian cycle,
* The Petersen graph is a standard example.
* Any pendant-free graph with a Hamiltonian path and a bridge is an easy example; e.g. take two pendant-free graphs $G\_1$ and $G\_2$ with Hamiltonian paths (and optionally with Hamiltonian cycles). Let $s\_1, t\_1$ be endpoints of a Hamiltonian path in $G\_1$ and similarly $s\_2, t\_2$ in $G\_2$. Join them either by merging $s\_1$ and $s\_2$ or by adding a single edge $s\_1 - s\_2$.
* The class of [maximally non-Hamiltonian graphs](https://mathworld.wolfram.com/MaximallyNonhamiltonianGraph.html) may be of particular interest to you.
| 3 | https://mathoverflow.net/users/46140 | 417749 | 170,151 |
https://mathoverflow.net/questions/417744 | 1 | Lets say I take an arbitrary closed and smooth $d$-manifolds $\mathcal{M}$. Now, it is a well-known fact that whenever I take two (sufficiently nice embedded) closed $d$-balls $B\_{1}$ and $B\_{2}$ in $\mathcal{M}$, then the manifold obtained by cutting out the interior of $B\_{1}$ is homeomorphic to the manifold obtained by cutting out the interior of $B\_{2}$ from $\mathcal{M}$. In other words, for a closed manifold, it does not depend on where I cut out a ball (up to homeomorphism). This is basically a consequence of the famous and highly non-trivial annulus theorem. For dimension 2 and 3, it follows from the famous triangulation theorems by Radó (1924) and Moise-Bing (1952,1959), but for higher dimension it was only proven in 1969 by Kirby ($d>4$) and in 1982 by Quinn (for d=4).
I am wondering if some similar statement is true for pseudomanifolds. By a pseudomanifold, I mean the following:
>
> Let $\Delta$ be a finite abstract $d$-dimensional simplicial complex.
> Its geometric realization $\vert\Delta\vert$ is $d$-dimensional
> **pseudomanifold** (without boundary), if the following conditions are
> fulfilled:
>
>
> 1. It is "pure", i.e. every simplex $\sigma\in\Delta$ of dimension $<d$ is the face of some $d$-simplex.
> 2. It is "non-branching", i.e. every $(d-1)$-simplex is face of exactly two $d$-simplices.
> 3. It is "strongly connected", i.e. for every two $d$-simplices $\sigma,\tau\in\Delta\_{d}$, there is a sequence of $d$-simplices
> $\sigma=\sigma\_{1},\sigma\_{2},\dots,\sigma\_{k}=\tau$ such that
> $\sigma\_{l}\cap\sigma\_{l+1}$ is a $(d-1)$-simplex $\forall l$.
>
>
>
Obviously, every PL-manifold is a pseudomanifold, but not vice versa. A famous example is the pinched torus, which is obtained by identifying two distinct points of the $2$-sphere.
So, **my question is**,
>
> when I remove a closed $3$-ball inside a closed pseudomanifold (for
> example, by deleting the interior of a single $3$-simplex or any
> subcomplex PL-homeomorphic to it), does the result depend on where I cut the ball (up to PL-homeomorphism)?
>
>
>
I am mostly interested in the $3$-dimensional case, which is usually a little bit easier than the higher-dimensional cases...
| https://mathoverflow.net/users/259525 | Annulus theorem for pseudomanifolds | The homeomorphism type of the space left after deleting the interior of the ball can depend on whether the ball intersects the singular locus (necessarily in the boundary of the ball) or not. If your ball intersects the singular locus, then once you remove the ball interior such points need not have neighborhoods homeomorphic to Euclidean half space, which will be the case if you remove the interior of a simplex in the interior of the manifold part of the pseudomanifold. On the other hand, if you restrict the balls to lie entirely disjoint from the singular locus, then your problem occurs entirely within the manifold obtained by removing the singular locus, in which case you can apply the classical results.
As a concrete example, suppose you triangulate the torus and then take your pseudomanifold to be the suspension of that torus. Let your ball be one of the cones on one of the 2-simplices of the torus. When you remove the interior of the ball, the cone point now has a neighborhood homeomorphic to the cone on a torus with an open disk removed. A local homology argument can then be used to show that the cone point does not have a neighborhood homeomorphic to Euclidean half-space.
| 3 | https://mathoverflow.net/users/6646 | 417751 | 170,152 |
https://mathoverflow.net/questions/417747 | 4 | Let $K=\mathbb{Q}(\sqrt{6})$. I am looking to determine all $K$-rational points on the curve
$$C: y^{2}=3x^6-24x^5+24x^4-54x^3+24x^2-24x+3.$$
More precisely, $C$ is a twist of the modular curve $X\_{0}(26)$. I know that Bruin and Najman (<https://arxiv.org/abs/1406.0655>) have determined all quadratic points on $X\_{0}(26)$ using the finiteness of the Jacobian of $X\_{0}(26)$ over $\mathbb{Q}$.
Let $J$ denote the Jacobian of $C$. Then $J$ has rank 1 over $\mathbb{Q}$. It's also interesting that $C(\mathbb{Q})=\emptyset$ (this can be seen using the TwoCoverDescent function on Magma).
I would really appreciate any pointers on how to proceed.
| https://mathoverflow.net/users/409515 | Finding the $K=\mathbb{Q}(\sqrt{6})$-rational points on the twist of $X_{0}(26)$ | I used Magma to point search on $C/K$ up to a height of $1000$ and it appears that $C(K) = \emptyset$. If that's true, then one can probably use the Mordell-Weil sieve to prove it. Here's a bit more detail.
The curve $C$ has four automorphisms defined over $\mathbb{Q}$ and for one of these (the map $(x,y) \mapsto (1/x,y/x^{3})$), the quotient curve has genus $1$. In particular, there is a map from $\phi : C \to E$, where $E : y^{2} = x^{3} - 651x - 12742$. This curve $E$ has rank $1$ over $K$.
For a finite set $S$ of prime ideals of $\mathcal{O}\_{K}$ (all of which are primes of good reduction for $E/K$), one can write down the commutative diagram
$$
\require{AMScd}
\begin{CD}
C(K) @>>> E(K)\\
@VVV @VV{\beta}V\\
\prod\_{\mathfrak{p} \in S} C(\mathbb{F}\_{\mathfrak{p}}) @>\alpha>> \prod\_{\mathfrak{p} \in S} E(\mathbb{F}\_{\mathfrak{p}})\\
\end{CD}
$$
The horizontal maps in this diagram use the map from $C$ to $E$, while the vertical maps use reduction modulo the prime ideals in $S$.
If $P \in C(K)$ is a point, then the image of $\alpha$ and the image of $\beta$ (as subsets of $\prod\_{\mathfrak{p} \in S} E(\mathbb{F}\_{\mathfrak{p}})$) have a non-trivial intersection. So if one can find a set $S$ of primes for which the image of $\alpha$ and the image of $\beta$ are disjoint, this proves that $C(K)$ is empty.
For more about this, I recommend [The Mordell-Weil sieve: Proving non-existence of rational points on curves](https://arxiv.org/abs/0906.1934) by Nils Bruin and Michael Stoll.
| 5 | https://mathoverflow.net/users/48142 | 417763 | 170,154 |
https://mathoverflow.net/questions/417759 | -1 | Given two compact oriented surfaces that have the same number of genus and boundary components. How to construct a homeomorphism that sends one to another?
| https://mathoverflow.net/users/nan | Construct a homeomorphism between two surfaces | A closed surface of genus g can be cut along 2g closed curves (all at one base point) to obtain a 4g-gon. Do this for both surfaces. Then choose a homeomorphism between the 4g-gons which matches the corresponding pairs of curves, so you get a well-defined homeomorphism of surfaces. (A 4g-gon is a cone over some interior cone point. You can first choose an appropriate homeomorphism of boundaries, then fix an interior „cone point“ for both 4g-goes, and then just take the cone map, that extends the chosen homeomorphism of boundaries into the interior, sending one cone point to the other.)
Similarly, surfaces with boundary can be cut into polygons, where now some edges of the polygon correspond to boundary curves.
| 2 | https://mathoverflow.net/users/39082 | 417764 | 170,155 |
https://mathoverflow.net/questions/417755 | 2 | Let $K(x)\_{n\times n}$ be a positive definite matrix defined on $x\in D$ and $K\_{i,j}(x)\in C^2(D)$ (or generally $C^k$) for any $1\le i,j\le n$. Of course for any $x$, there exists a invertable matrix $A(x)$ such that $A(x)A(x)^t=K(x)$, and $A(x)$ is not unique. What I want to ask is, whether $K(x)$ has $C^2$ (or $C^k$) decomposition. That is, there exists $A(x)\in C^2(D)$ (or $C^k$), such that $A(x)A(x)^t=K(x)$. If not, what conditions can we add to make this true?
Moreover, are there some references about this question? Thanks a lot!
| https://mathoverflow.net/users/176547 | Decomposition of a positive definite matrix | The [Cholesky decomposition](https://en.wikipedia.org/wiki/Cholesky_decomposition) does what you want. It depends smoothly on the input matrix, because every step in the algorithm is a smooth function. It's all just basic arithmetic and square roots.
| 3 | https://mathoverflow.net/users/3041 | 417771 | 170,159 |
https://mathoverflow.net/questions/417558 | 2 | It is known that the $T\_0$ and $T\_2$ axioms are not preserved under open, closed and continuous maps (for instance, see here: [An example of open closed continuous image of $T\_0$-space that is not $T\_0$](https://math.stackexchange.com/questions/1411751/an-example-of-open-closed-continuous-image-of-t-0-space-that-is-not-t-0) and here: [An example of open closed continuous image of $T\_2$-space that is not $T\_2$](https://math.stackexchange.com/questions/1412492/an-example-of-open-closed-continuous-image-of-t-2-space-that-is-not-t-2?rq=1%5D)). However, it is not difficult to verify that the Hausdorff property is preserved under perfect functions (closed with compact fibers).
Is it also true that a perfect image of a $T\_0$ space is $T\_0$ as well? It is not difficult to see that a perfect image of a finite $T\_0$ space is indeed $T\_0$. What about infinite $T\_0$ spaces?
| https://mathoverflow.net/users/146942 | Preservation of separation axioms under perfect functions | Yes, it seems to be true.
First, towards a counterexample. Let $f\colon X \to Y$ be a perfect map from a $T\_0$ space. If $Y$ is not $T\_0$, it contains a two-point indiscrete space $B$, and $f\colon f^{-1}[B] \to B$ is also a perfect map. So if there is a counterexaple at all, there is a counterexample realized by a $T\_0$ space $X$ consisting of two compact parts such that $(\*)$: every non-empty closed subset intersects both of them. Necessarily $X$ is compact. (In fact, given $(\*)$, $X$ is compact if and only if both parts are compact since every open cover of one part is an open cover of $X$.)
However, no compact $T\_0$ space satisfying $(\*)$ exists. Every compact space contains a minimal non-empty closed subset by Zorn's lemma. And in a $T\_0$ space, this minimal closed set has to be a singleton.
| 2 | https://mathoverflow.net/users/112373 | 417779 | 170,162 |
https://mathoverflow.net/questions/417484 | 6 | *This is a minor variation of a question originally [asked on MSE](https://math.stackexchange.com/questions/4349128/partial-consistency-of-peano-arithmetic) by user779130 and bountied by me, without success. Throughout, "length" refers to the number of symbols, not lines, in a proof.*
For $T$ an "appropriate" theory, let $p\_T(n)$ be the length of the shortest $T$-proof of "There is no proof of $0=1$ in $\mathsf{PA}$ of length $<n$." We always have an exponential upper bound on the length of $p\_T$ by brute-force-checking, and sufficiently strong $T$s have $p\_T=O(1)$. However, beyond this very little is clear to me:
* Is $p\_{\mathsf{PA}}$ sub-exponential?
* If so, is $p\_{\mathsf{PRA}}$ sub-exponential?
(When I first read the original version of this question, I thought I recalled a theorem that $p\_{\mathsf{PA}}$ is in fact exponential, but I was unable to track it down. In fact, I already can't rule out the seemingly-ridiculous possibility of $p\_\mathsf{PRA}$ having polynomial-bounded growth rate.)
| https://mathoverflow.net/users/8133 | Proving short consistency: can we do better than brute force search? | Since the concept involves two theories, but the notation in the question only indicates one of them, I will instead write $p\_{S,T}(n)$ for the shortest $S$-proof of $\DeclareMathOperator\con{Con}\con\_T(\def\ob{\overline}\ob n)$, expressing that there is no $T$-proof of contradiction of length $<n$.
Here, each of $T,S$ is a consistent theory extending a suitable base theory (say, the theory $\def\pv{\mathrm{PV}\_1}\pv$ of polynomial-time functions, which is a weak subtheory of $I\Delta\_0+\Omega\_1$) with a fixed polynomial-time enumeration of axioms. To avoid trivializing a part of the question, let $\ob n$ denote the binary numeral defined by $\ob 0=0$, $\ob{2n}=S(S(0))\cdot\ob n$, and $\ob{2n+1}=S(\ob{2n})$, so that the length of the sentence $\con\_T(\ob n)$ is $\Theta(\log n)$.
Consequently, there is a trivial lower bound
$$p\_{S,T}(n)=\Omega(\log n)\tag1\label{lb}$$
for all $S$ and $T$. On the other hand, there is a trivial upper bound
$$p\_{S,T}(n)=2^{O(n)},\tag2\label{ub}$$
as we can just enumerate all strings of length $n$, and check that none of them is a $T$-proof of a contradiction.
We can study $p\_{S,T}$ in three different regimes:
**1.** If $S$ is sufficiently stronger than $T$, that is: if $S$ proves $\con\_T=\forall x\,\con\_T(x)$, then
$$p\_{S,T}(n)=\Theta(\log n),$$
thus the lower bound $\eqref{lb}$ is tight. Indeed, we can prove $\con\_T(\ob n)$ by taking a fixed proof of $\con\_T$ and instantiating the universal quantifier to $\ob n$.
**2.** If $T=S$ (including the case $p\_{\def\pa{\mathrm{PA}}\pa}=p\_{\pa,\pa}$ in the question), the problem was answered by Pudlák [1], who proved
>
> **Theorem 1:** For any $T$, there are constants $c>\epsilon>0$ such that
> $$n^\epsilon<p\_{T,T}<n^c$$
> for sufficiently large $n$.
>
>
>
Notice that both the lower bound and the upper bound here are nontrivial.
To be precise, Pudlák proves the polynomial upper bound only for finitely axiomatized theories $T$, and more generally, for theories axiomatized by schemata of a certain restricted form, which applies e.g. to PA and ZFC. The bounds were improved (under some restrictions) in Pudlák [2].
**3.** If $S$ is weaker than $T$ (including the case $p\_{\def\pra{\mathrm{PRA}}\pra}=p\_{\pra,\pa}$ in the question), the growth rate of $p\_{S,T}$ becomes a difficult open problem. If $T\supseteq S$, then Theorem 1 implies the polynomial lower bound
$$p\_{S,T}(n)>n^\epsilon,$$
but that’s just about all we know for a fact.
We have every reason to expect that if $T$ is sufficiently stronger than $S$, then the trivial upper bound $\eqref{ub}$ cannot be significantly improved, i.e., $p\_{S,T}$ is exponential. It seems likely that this should hold already when $T\supseteq S+\con\_S$. However, we cannot even prove the much weaker statement
>
> **Conjecture 2:** For every $S$, there is $T$ such that $p\_{S,T}$ is not polynomially bounded.
>
>
>
In fact, Krajíček and Pudlák [3] proved that Conjecture 2 is equivalent to
>
> **Conjecture 3:** There is no optimal propositional proof system.
>
>
>
Here, a *propositional proof system* (pps) is a polynomial-time predicate $P(\pi,\phi)$ (meaning “$\pi$ is a $P$-proof of $\phi$”) such that
$$\phi\in\def\taut{\mathrm{TAUT}}\taut\iff\exists\pi\:P(\pi,\phi),$$
where $\taut$ is the set of all classical propositional tautologies.
A pps $P$ *simulates* a pps $Q$ if there exists a polynomial $p$ such that
$$Q(\pi\_Q,\phi)\implies\exists\pi\_P\:\bigl(|\pi\_P|\le p(|\pi\_Q|+|\phi|)\land P(\pi\_P,\phi)\bigr),$$
and a pps $P$ is *optimal* if it simulates any other pps. A pps $P$ is *polynomially bounded* if there is a polynomial $p$ such that
$$\phi\in\taut\implies\exists\pi\:\bigl(|\pi|\le p(|\phi|)\land P(\pi,\phi)\bigr).$$
Any polynomially bounded pps is clearly optimal, but the converse is not necessarily true.
It is generally assumed that *every* pps requires proofs of exponential size to prove some tautologies. However, we are far from proving this; the weaker statement that there are no polynomially bounded pps is equivalent to the famous $\mathrm{NP\ne coNP}$ problem. We can only prove unconditional superpolynomial (or even exponential) lower bounds on weak proof systems such as Resolution. No nontrivial lower bounds are known even for the *Frege* proof system (called Hilbert outside proof complexity), which is the textbook proof system using a finite set of axiom schemata and schematic rules (modus ponens).
The proof of the equivalence of Conjecture 2 and Conjecture 3 works roughly as follows. On the one hand, for any theory $S$, we can define a proof system $Q\_S$ by
$$Q\_S(\pi,\phi)\iff \pi\text{ is an $S$-proof of “$\phi\in\taut$”}$$
(this is called the *strong proof system of $S$* in Pudlák [4]) and we show that if $S$ violates Conjecture 2, then $Q\_S$ is optimal. On the other hand, if $P$ is an optimal pps, then the theory axiomatized by the reflection principle
$$S=\pv+\forall\pi,\phi\:(P(\pi,\phi)\to\phi\in\taut)$$
violates Conjecture 2.
As already mentioned, it is consistent with current knowledge that Frege is polynomially bounded, and therefore optimal. Since the reflection principle for Frege is provable already in the base theory $\pv$, it follows that we cannot at present even disprove the statement
>
> $p\_{S,T}$ is polynomially bounded for all $S$ and $T$.
>
>
>
Thus, for example, we cannot prove anything about $p\_{\pra,\pa}$ besides the $n^\epsilon$ lower bound.
See also [5] for a discussion of statements related to Conjectures 2 and 3.
**References:**
[1] Pavel Pudlák: [*On the length of proofs of finitistic consistency statements in first order theories*](http://math.cas.cz/%7Epudlak/fin-con.pdf), in: Logic Colloquium '84 (J. B. Paris, A. J. Wilkie, G. M. Wilmers, eds.), Studies in Logic and the Foundations of Mathematics vol. 120, 1986, pp. 165–196, doi [10.1016/S0049-237X(08)70462-2](https://doi.org/10.1016/S0049-237X(08)70462-2).
[2] Pavel Pudlák: [*Improved bounds to the length of proofs of finitistic consistency statements*](http://math.cas.cz/%7Epudlak/improved.pdf), in: Logic and Combinatorics (S. G. Simpson, ed.), Contemporary Mathematics vol. 65, AMS, 1987, pp. 309–331.
[3] Jan Krajíček, Pavel Pudlák: [*Propositional proof systems, the consistency of first order theories and the complexity of computations*](http://math.cas.cz/%7Epudlak/propositional.pdf), Journal of Symbolic Logic 54 (1989), no. 3, pp. 1063–1079, doi [10.2307/2274765](https://doi.org/10.2307/2274765), jstor [2274765](https://www.jstor.org/stable/2274765).
[4] Pavel Pudlák: *Reflection principles, propositional proof systems, and theories*, [arXiv:2007.14835](https://arxiv.org/abs/2007.14835) [math.LO].
[5] Pavel Pudlák: [*Incompleteness in the finite domain*](https://math.cas.cz/%7Epudlak/inco.pdf), Bulletin of Symbolic Logic 23 (2017), no. 4, pp. 405–441, doi [10.1017/bsl.2017.32](https://doi.org/10.1017/bsl.2017.32).
| 3 | https://mathoverflow.net/users/12705 | 417783 | 170,163 |
https://mathoverflow.net/questions/417780 | 8 | By Deligne's theorem, each coherent topos has enough points. What would be an example of a Grothendieck topos with enough points which is *not* coherent?
| https://mathoverflow.net/users/478438 | Topos with enough points but not coherent | Here are some examples :
1. For any topological space $X$, the topos of sheaf $\operatorname{Sh}(X)$ has enough points. In most cases this is not a coherent topos. If I remember correctly (for $X$ sober), this is only coherent if $X$ is a [spectral space](https://en.wikipedia.org/wiki/Spectral_space). In any case, spaces like $\mathbb{R}$ are definitely not coherent toposes.
2. For any topos $\mathcal{T}$, and any set of points $X$ of $\mathcal{T}$, you get geometric morphism $X \to \mathcal{T}$ (where by $X$ I mean the topos $\operatorname{Set}/X$), you can take its image factorisation $X \twoheadrightarrow I \hookrightarrow \mathcal{T}$ and $I$ is a subtopos of $\mathcal{T}$ with enough points. Generally, non coherent toposes don't have a lot of coherent subtoposes (though of course this can happen), so this will often gives example of non-coherent topos with enough topos. There is also a way to do this for $X$ the class of all points of $\mathcal{T}$ despite the size issues.
3. There is another completeness theorem like Deligne's which says that any "separable" Grothendieck topos has enough points. Separable essentially means that the topos can be defined by a site whose underlying category is countable and whose topology is generated by a countable family of basic covering Sieve. This is done in Makkai and Reyes's book "[First order categorical logic](https://marieetgonzalo.files.wordpress.com/2018/04/makkai-reyes-book.pdf)" (theorem 6.2.4).
There are many separable toposes that are not coherent. In terms of classifying toposes, coherent means you only use finitary logic, while separable means you can use infinitary logic but the theory should have a countable signature and a countable set of axioms. So none of the two class is included in the other, but I would personally consider that in term of which classical topos they contains, separable toposes form a much larger class.
| 17 | https://mathoverflow.net/users/22131 | 417793 | 170,166 |
https://mathoverflow.net/questions/417782 | 0 | Let S be a compact orientable surface. Let A and B be two subsurfaces of S that have the same signature. How to check if there is a homeomorphism of S that sends A to B and if so, find one?
Here a subsurface of $S$ is a component of S \ simple closed curves.
I saw this question [Mapping-Class Groups of Subsurfaces of a Hyperbolic Surface](https://mathoverflow.net/questions/283567/mapping-class-groups-of-subsurfaces-of-a-hyperbolic-surface) and <https://arxiv.org/pdf/math/9906122.pdf> but I didn't find the part related to my question yet.
| https://mathoverflow.net/users/nan | Construct a homeomorphism of a surface that sends a subsurface to another subsurface | We are assuming that $A$ and $B$ are both connected and contain their boundary points. (If we do not assume $A$ and $B$ are connected then the problem becomes much harder.) Let $(A\_i)$ be the closures of the components of $S - A$. Let $(B\_j)$ be closures of the components of $S - B$. Define $(g\_i, b\_i, s\_i)$ to be the genus of $A\_i$, the size $|\partial A\_i|$, and the size $|A \cap A\_i|$, respectively.
We now sort the list $(A\_i)$ using the complexities $(g, b, s)$. We do the same for the list $(B\_j)$. Finally, there is a homeomorphism of $S$ sending $A$ to $B$ if and only if the two lists of complexities are identical.
To produce an "explicit" homeomorphism is much more work. Any algorithm will be sensitive to the details of how the data (of $S$, $A$, and $B$) are presented. One technique is to carefully order the boundary components of $A$ and $B$, compute all pairwise intersection numbers, and then perform Dehn twists to reduce the intersection numbers to zero. This is morally similar to Lee Mosher's automatic structure for the mapping class group.
| 1 | https://mathoverflow.net/users/1650 | 417796 | 170,168 |
https://mathoverflow.net/questions/370283 | 17 | A set $X\subseteq\omega^\omega$ is **unravelable** iff there is a possibly larger set $A$ and a clopen set $Y\subseteq A^\omega$ (with respect to the product topology coming from the discrete topology on $A$) such that winning strategies for the game on $A$ with payoff set $Y$ can be converted to winning strategies for the game on $\omega$ with payoff set $X$ in a particularly simple way (see Martin's paper [*A purely inductive proof of Borel determinacy*](https://www.math.ucla.edu/%7Edam/booketc/purely_inductive.pdf) for the precise definition). By Gale–Stewart, unravelability implies determinacy; however, it is not the only route to proving the determinacy of a pointclass.
My question is whether there is a known upper bound on the complexity of an unravelable pointclass. To make this somewhat precise:
>
> Is it consistent with $\mathsf{ZFC}$ that every set in $\mathcal{P}(\mathbb{R})^{L(\mathbb{R})}$ is unravelable?
>
>
>
The strongest results I can find are *vastly* weaker than this, namely that under large cardinal assumptions $\Pi^1\_1$ sets are unravelable *(Neeman [1](https://www.sciencedirect.com/science/article/pii/S0168007200000221),[2](https://link.springer.com/article/10.1007/BF02771982); I remember seeing the same result for Borel-on-$\Pi^1\_1$ sets, but I can't find a reference for it at the moment)*. On the other hand, I don't see an easy proof that this is anywhere close to the most unravelability we can expect, nor can I find this stated in the literature.
| https://mathoverflow.net/users/8133 | What sets can be unraveled? | I emailed Itay Neeman, and he told me the following:
>
> As far as I know it's open. I don't think anything is known about
> unraveling beyond what you can get from my methods. These give the
> Suslin operation on $\Pi^1\_1$ sets, and various iterations of that. In
> terms of the large cardinal hierarchy it's still all just using
> measures. Anything above that is open I think.
>
>
> In particular I don't think it's known if $\Sigma^1\_2$ sets can be
> unraveled.
>
>
>
For now I think that settles this question. (Itay later said that it's unclear whether we "should" expect unravelability of $\Sigma^1\_2$ sets from large cardinals or not in the first place. So I think my main takeaway at this point is that unravelability is **not** best thought of as yet another tameness property, where the slogan "large cardinals make definable things tame" is expected to prevail.)
| 6 | https://mathoverflow.net/users/8133 | 417802 | 170,169 |
https://mathoverflow.net/questions/417797 | 4 | I am interested in the following Poincaré-type inequality,
$$ \int\_{S(r)} \lvert u-\bar{u}\rvert^2 d\sigma \leq C(N) \int\_{S(r)} |u\_{\theta}|^2 d\sigma$$
where $\bar{u} = \frac{1}{\lvert S(r)\rvert}\int\_{S(r)} u d\sigma$ and $u\_{\theta}$ denotes the tangential derivative of $u$. The domain $S(r)$ is just the $N$-dimensional sphere with radius $r$. The function $u$ can be assumed to be smooth for the purpose of this question.
In the case $S(r)$ is $1$ dimensional (i.e. a circle), then the constant $C(1)=1$ and this is the Wirtinger's inequality. Are there any references where I can find the best constant in higher dimensions?
| https://mathoverflow.net/users/100801 | Best constant for Poincaré inequality on spheres | The best constant is just the multiplicative inverse of the smallest positive eigenvalue of the Laplacian on the sphere. On $\mathbb{S}^N$ this the smallest eigenvalue is $N$, so $C(N)$ in that case equals $1/N$.
You can figure out the appropriate $r$ scaling yourself.
| 8 | https://mathoverflow.net/users/3948 | 417807 | 170,171 |
https://mathoverflow.net/questions/417804 | 9 | Do there exist integers $x$ and $y$ such that $\frac{x^3-x^2-2 x+1}{y^2-4}$ is an integer?
In other words, can any integer representable as $x^3-x^2-2 x+1$ have any divisor representable as $y^2-4$?
This is the simplest non-trivial example of my earlier question [Integer points of rational function in 2 variables](https://mathoverflow.net/questions/404703/integer-points-of-rational-function-in-2-variables) .
| https://mathoverflow.net/users/89064 | Can $y^2-4$ be a divisor of $x^3-x^2-2 x+1$? | No. The roots of $x^3 - x^2 - 2x + 1$ are $-(\zeta + \zeta^{-1})$ where $\zeta$ is a 7th root of unity; this soon implies [*see below*] that any prime factor is either $7$ or $\pm 1 \bmod 7$, and thus that all factors of $x^3 - x^2 - 2x + 1$ are congruent to $0$ or $\pm 1 \bmod 7$. In particular it is not possible for two factors to differ by $4$, so no number of the form $y^2 - 4 = (y-2) (y+2)$ can divide $x^3 - x^2 - 2x + 1$.
*added later*: To show that any prime factor $p$ of $x^3 - x^2 - 2x + 1$ is either $7$ or $\pm 1 \bmod 7$, let $k$ be the finite field of order $p^2$, and $\zeta \in k$ a root of the quadratic equation $\zeta^2 + x\zeta + 1 = 0$ (any quadratic equation with coefficients in the $p$-element field has a root in $k$). Then $\zeta^7 = 1$, so either $\zeta = 1$ or the multiplicative group $k^\times$ of $k$ has a subgroup of size $7$. In the former case, $x = -2$, and then $x^3 - x^2 - 2x + 1 = -7$ so $p=7$. In the latter case, [Lagrange's theorem](https://en.wikipedia.org/wiki/Lagrange%27s_theorem_(group_theory)#Applications) gives $7 \mid \#k^\times = p^2-1$, so $p \equiv \pm 1 \bmod 7$. **QED**
| 30 | https://mathoverflow.net/users/14830 | 417811 | 170,173 |
https://mathoverflow.net/questions/417787 | 2 | I'll begin with a broad question: if $M$ is a smooth manifold and $E \to M$ is a stably trivial bundle, can one determine lower bounds on the rank $k$ of the trivial bundle needed such that $E \oplus \underline{\mathbb{R}}^k$ is trivial?
An obvious example is for the tangent bundle of spheres: $TS^n \to S^n$. Here, $k=1$. If I'm not mistaken, if $\Sigma$ is a homotopy $n$-sphere and $f:S^n \to \Sigma$ is a homotopy equivalence, then $f^\*T\Sigma \cong TS^n$ which means that $T\Sigma$ is also stably trivial and $k=1$. Kervaire-Milnor showed that $\mathbb{Z}$-homology spheres also have stably trivializable tangent bundles. I do not really understand their proof but it doesn't seem to give any quantitative results on how large the rank needs to be in order to stabilize.
Refined question: In the broad question above, let's require $M$ to be a homology sphere and $E \to M$ to be the tangent bundle, and then ask the same question about lower bounds.
Related question: For a given positive integer $k$, does there exist a smooth manifold with stably trivial tangent bundle that requires at least a rank $k$ trivial bundle to stabilize?
| https://mathoverflow.net/users/121144 | Quantitative results for stabilizing tangent bundles of homology spheres | If $E \to X$ is a rank $r$ real vector bundle, then it is classified by a map $X \to BO(r)$. The existence of an isomorphism $E \cong E\_0\oplus\underline{\mathbb{R}}$ (equivalently, the existence of a nowhere-zero section of $E$), corresponds to lifting $X \to BO(r)$ through the map $BO(r-1) \to BO(r)$ induced by the inclusion $O(r-1)\hookrightarrow O(r)$. The obstructions to such a lift lie in $H^n(X; \pi\_{n-1}(S^{r-1}))$ as the homotopy fiber of $BO(r-1) \to BO(r)$ is $O(r)/O(r-1) = S^{r-1}$. Moreover, the obstructions to the uniqueness of such a lift, which corresponds to the uniqueness of $E\_0$ up to isomorphism, lie in $H^n(X; \pi\_n(S^{r-1}))$. This allows us to conclude the following:
* If $\operatorname{rank}E > \dim X$ (the cohomological dimension of $X$), then all the obstructions to existence vanish, so $E \cong E\_0\oplus\underline{\mathbb{R}}$ for some $E\_0$ with $\operatorname{rank}E\_0 = \operatorname{rank}E - 1$.
* If $\operatorname{rank}E > \dim X + 1$, then all the obstructions to uniqueness also vanish, so $E \cong E\_0\oplus\underline{\mathbb{R}}$ and $E\_0$ is unique up to isomorphism.
In particular, if $M$ is an $n$-dimensional smooth manifold with stably trivial tangent bundle, then $TM\oplus\underline{\mathbb{R}}^m \cong \underline{\mathbb{R}}^{n+m}$ for some $m \geq 0$. If $m \geq 2$, then $\operatorname{rank}(TM\oplus\underline{\mathbb{R}}^m) > \dim M + 1$, so there is a unique vector bundle $E\_0$ up to isomorphism with $TM\oplus\underline{\mathbb{R}}^m \cong E\_0\oplus\underline{\mathbb{R}}$. Now note that $TM\oplus\underline{\mathbb{R}}^m = (TM\oplus\underline{\mathbb{R}}^{m-1})\oplus\underline{\mathbb{R}}$ and $TM\oplus\underline{\mathbb{R}}^m \cong \underline{\mathbb{R}}^{n+m} \cong \underline{\mathbb{R}}^{n+m-1}\oplus\underline{\mathbb{R}}$, so by uniqueness, we have $TM\oplus\underline{\mathbb{R}}^{m-1} \cong \underline{\mathbb{R}}^{n+m-1}$. After finitely many applications of this argument, we see that $TM\oplus\underline{\mathbb{R}}$ is trivial - that is, $k = 1$ unless $M$ is parallelisable, in which case $k = 0$. More generally, if $E \to X$ is a stably trivial bundle with $\operatorname{rank}E = \dim X$, then $k = 1$ unless $E$ is trivial, in which case $k = 0$.
If $E \to X$ is stably trivial and $\operatorname{rank}E < \dim X$ then it is possible that a larger value of $k$ is needed. Such examples can be constructed as in [this answer](https://mathoverflow.net/a/279246/21564). First note that $TS^n$ is non-trivial for $n \neq 1, 3, 7$. On the other hand, if $n$ is odd, then $TS^n \cong E\_0\oplus\underline{\mathbb{R}}^{\rho(n+1)-1}$ where $\rho(n+1)$ denotes that $(n+1)^{\text{st}}$ Radon-Hurwitz number: if $n + 1 = 2^{4a+b}c$ where $a \geq 0$, $0 \leq b \leq 3$, and $c$ is odd, then $\rho(n+1) = 8a + 2^b$. Therefore $E\_0\oplus\underline{\mathbb{R}}^{\rho(n+1)} \cong TS^n\oplus\underline{\mathbb{R}} \cong \underline{\mathbb{R}}^{n+1}$; i.e. for the vector bundle $E\_0$, the value of $k$ is $\rho(n+1)$.
| 4 | https://mathoverflow.net/users/21564 | 417820 | 170,176 |
https://mathoverflow.net/questions/417790 | 3 | This question is about the argument for Lemma 3.7 in [Forcing axioms and stationary sets](https://doi.org/10.1016/0001-8708(92)90038-M) ([MSN](https://mathscinet.ams.org/mathscinet-getitem?mr=MR1174395)) by Boban Veličković.
He defines a game $G\_\alpha$ between two players, playing objects in $H\_\kappa$, depending on a function $F : [\kappa]^{<\omega} \to \kappa$. At stage $n$, Player I plays an interval of ordinals $I\_n$ below $\kappa$ and an ordinal $\xi\_n \in I\_n$. Player II plays an ordinal $\mu\_n$. $\min(I\_{n+1})$ must be above $\mu\_n$. Player I wins if the closure $M$ of $\{ \xi\_n : n < \omega \} \cup \alpha$ under $F$ is contained in $\bigcup\_n I\_n$, and $M \cap \omega\_1 = \alpha$.
Now I understand the argument for why player I has a winning strategy when we ignore the requirement on $\alpha$. For the $\alpha$-indexed games, he says: Let $A\_F$ be the set of $\alpha<\omega\_1$ such that player II has a winning strategy $\sigma\_\alpha$. "Since $A\_F$ has cardinality $\leq \aleph\_1$, there is a strategy $\sigma$ which dominates all the $\sigma\_\alpha$. It follows that $\sigma$ is a winning strategy for II in $G\_\alpha$ for every $\alpha \in A\_F$."
What does it mean to say that a strategy "dominates" a lot of other ones (for the same player), and why does such $\sigma$ exist?
| https://mathoverflow.net/users/11145 | Veličković's model game | What constitutes a partial play doesn't depend on $\alpha$. And $\kappa$ is assumed to have cofinality strictly larger than $\aleph\_1$. So he means: define $\sigma$ applied to a partial play to just be some ordinal below $\kappa$ that is above all the outputs of the $\aleph\_1$-many strategies $\sigma\_\alpha$.
(And it is easy to see that for any fixed $\alpha$, if $\sigma\_\alpha$ is a winning strategy for II in the game $G\_\alpha$, then so is any function from partial runs into $\kappa$ that is $\ge \sigma\_\alpha$.)
| 4 | https://mathoverflow.net/users/26319 | 417838 | 170,180 |
https://mathoverflow.net/questions/417800 | 25 | I am working on a paper which will extend a result in my thesis and have boiled one problem down to the following: show that the symmetric matrix $M\_p$, whose definition follows, is invertible for all odd primes $p$. Letting $p>3$ be prime and $\ell = \frac{p-1}{2}$, we define
$$M\_p = \begin{pmatrix} 2ij - p - 2p\left\lfloor\frac{ij}{p}\right\rfloor\end{pmatrix}\_{1\leq i,j\leq \ell}$$
**Examples:**
1. For $p=5$ we have $M\_5 = \begin{pmatrix} -3 & -1 \\ -1 & 3 \end{pmatrix}$ and $\det(M\_5) = -1\cdot 2\cdot 5$.
2. For $p=7$ we have $M\_7 = \begin{pmatrix} -5 & -3 & -1 \\ -3 & 1 & 5 \\ -1 & 5 & -3 \end{pmatrix}$ and $\det(M\_7) = 2^2 \cdot 7^2$.
3. For $p=11$ we have $M\_{11} =
\begin{pmatrix}
-9 & - 7 & -5 & -3 & -1 \\
-7 & -3 & 1 & 5 & 9 \\
-5 & 1 & 7 & -9 & -3 \\
-3 & 5 & -9 & -1 & 7 \\
-1 & 9 & -3 & 7 & -5
\end{pmatrix}$ and
$\det(M\_{11}) = -1\cdot 2^4\cdot 11^4$.
Though this (seemingly) nice formula that we see above fails for primes greater than 19, though the determinant has been checked to be non-zero for primes less than 1100. (My apologies if this question is not as motivated or as well discussed as is desired. If there are any questions or if further clarification is needed just let me know!)
| https://mathoverflow.net/users/477847 | Show that these matrices are invertible for all $p>3$ | Experimentally, we have the following formula for $p$ prime:
$$\det(M\_p)=(-1)^{(p^2-1)/8}(2p)^{(p-3)/2}h\_p^-\;,$$
where $h\_p^-$ is the minus part of the class number of the $p$-th cyclotomic
field, itself essentially equal to a product of $\chi$-Bernoulli numbers.
I have not tried to prove this, but since there are many determinant
formulas for $h\_p^-$ in the literature, it should be possible.
| 24 | https://mathoverflow.net/users/81776 | 417854 | 170,183 |
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