| |
Mon, 10/01/2018 - 13:25
One well-known, extremely slick proof of Cayley's tree enumeration theorem is the use of Prüfer sequences. Cayley also proved a version for forests, namely that the number of forests with $n$ labelled vertices that consist of $s$ distinct trees such that $s$ specified vertices belong to distinct trees is $sn^{n-s-1}$; see https://core.ac.uk/download/pdf/82105567.pdf . Is there a generalization of Prüfer sequences that corresponds to this quantity? Or more generally, is there a nice way to enumerate all $sn^{n-s-1}$ such forests, given $s$ and $n$?
| |
Mon, 10/01/2018 - 13:20
Suppose one has a non-negative square matrix $A$ where not all row-sums are equal. Furthermore, it has the following property:
There do not exist rows $i$ and $j$ such that $a_{ik} \geq a_{jk} \forall k$ with strict inequality at least once. Call such a matrix diverse.
Note that $A$ might have identical rows. Let $R$ be the principal submatrix obtained from $A$ by deleting all duplicate rows (except one exemplar, of course).
My conjecture:
If $R$ is singular, then $Rx = j$ has no solution where $j$ is the all-one vector.
The other alternative would be that it has infinitely many solutions. The possible source of duplicate rows has been eliminated by going from $A$ to $R$. Rows which are multiples of each other are forbidden by diversity. It has been proven that, if $A$ is the adjacency matrix of a simple graph, the conjecture holds without diversity (see source below).
Supposing someone can show the conjecture is wrong: Is it true if we require $A$ to be symmetric? Then the diversity property must also hold for the columns.
Huang, Liang-Hao, Bit-Shun Tam, and Shu-Hui Wu. "Graphs whose
adjacency matrices have rank equal to the number of distinct nonzero
rows." Linear Algebra and its Applications 438.10 (2013):
4008-4040.
| |
Mon, 10/01/2018 - 12:56
Consider two groups $G$ and $G'$, where $G$ is the direct product of groups $A$ and $B$, with $B$ abelian, and $G'$ is a nontrivial central extension of $A$ by $B$. Suppose that as groups, $H^1(G,M) \cong H^1(G',M)$ where $M$ is a trivial module in both cases. Furthermore, it can be verified that 1-cocycle representatives in $H^1(G',M)$ are related to 1-cocycle representatives in $H^1(G,M)$ by a bijection $f$- for example, taking $M = \mathbb Z$, the former function could be an integral multiple of the latter.
Given this information, under what conditions can we claim that the cohomology rings $H^*(G,A)$ and $H^*(G',A)$ are isomorphic? I want to use the cup product structure on cohomology, but I don't have a good understanding of what needs to be proved. I would appreciate any help/references!
| |
Mon, 10/01/2018 - 12:36
Let Z1,…,Zn be dependent gaussian random variables. Is it true that X=max{Z1,…,Zn} has a log-concave distribution function? This is true for the independent case, but is it true in general?
| |
Mon, 10/01/2018 - 12:31
As the title suggests, I'm trying to find motivation on the definition of the Thom spectrum of $-\xi$, or more generally on the definition of the Thom spectrum of a virtual bundle.
In this paper by S. Bauer (middle of page 7) he defines the Thom spectrum for a virtual bundle of the form $\xi - \Bbb R^m\times X$ as the desuspension:
$$ Th(\xi-\Bbb R^m\times X) = \sum^{-m}\Bbb S\wedge X^{\xi}$$
with this definition, let $\eta$ be a bundle over $X$ such that $\xi\oplus \eta\cong X\times \Bbb R^N$, then we have we have that $$Th(-\xi) = \sum^{-N}\Bbb S\wedge X^{\eta}$$
I'm wondering what is the reason of this definition and if it was somehow related with the concept of duality morphism (Rudyak, page 47, def 2.3)
My intuition is that with this definition we should have that the two spectra are dual in the sense above. This would explain for example why Bauer in the paper mentioned above, or Crabb & Knapp in this paper (page 90, Lemma 1.1) claim that there is a pairing given by cap product: $$\widetilde{h}^0(X^{-\xi})\times \widetilde{h}^r(X^{\xi})\to \widetilde{h}^r(X)$$ since we could interpret the first one as an homology group and then cap product is well defined.
So I tried proving that we have this aforementioned duality. In fact we have something resembling a duality between $Th(-\xi)$ and $Th(\xi)$: $$Th(-\xi)\wedge Th(\xi)=\sum^{-N}\Bbb S\wedge X^{\eta}\wedge \Bbb S\wedge X^{\xi}$$ $$ = \sum^{-N}\Bbb S\wedge X^{\eta}\wedge X^{\xi}$$ $$= \sum^{-N}\Bbb S\wedge (X\times X)^{\eta \times \xi}$$
but my problem is that $(X\times X)^{\eta \times \xi}$ is not quite $X_+\wedge \sum^N\Bbb S$, instead we have $$X_+\wedge \sum^N\Bbb S = \Delta^*(X\times X)^{\eta \times \xi}\wedge \Bbb S$$ where $\Delta \colon X \to X\times X$ is the diagonal map, and this prevents me for proving that we have an actual duality.
I'm slightly confused here so I apologise in advance if something is not super precise.
| |
Mon, 10/01/2018 - 12:05
Say we have r red balls, b blue balls, and g green balls in a bag. I want to know the probability of red balls being depleted, then blue balls being depleted, and finally green balls being depleted if we continually draw a random ball from the bag until we run out of balls. I know that the probability of red being depleted before blue is: $b/(r+b)$, and likewise probability of blue being depleted before green is $g/(b+g)$.
Are these two events (red being depleted before blue and blue being depleted before red) independent (in which case I just have to multiply the two probability together)? If they are not how could I go about calculating the probability of blue being depleted before green given red depleted before blue?
| |
Mon, 10/01/2018 - 10:57
Suppose I have a finite group $G$. With this group, I can associate an ortho-normal Hilbert space spanned by elements of the group $$\mathcal{H} = \{|g\rangle: g \in G \}$$.
Along with this Hilbert space, I can also define raising/lowering and projection operators:
$$
\begin{align}
T_+^g |k\rangle = & \delta_{g,k} |k\rangle \\
T_-^g |k\rangle = & \delta_{g^{-1},k} |k\rangle \\
L_+^g |k\rangle = & |gk\rangle \\
L_-^g |k\rangle =&|kg^{-1}\rangle
\end{align}
$$
which may look familiar from quantum double models for topological order in 2+1D.
I could alternatively work in the representation basis, related to the group element basis by the Fourier transform:
$$
|\mu,a,b\rangle = \sqrt{\frac{n_\mu}{|G|}} \sum_{g\in G}\Gamma^{ab}_\mu(g)|g\rangle
$$
where $\mu$ labels irreps of the group, $n_\mu$ is the dimension of $\mu$, and $\Gamma^{ab}_\mu$ are the irrep matrices, with $a,b = 1,\dots,n_\mu$.
My question is the following: suppose I consider instead the group $G \times G$ with basis $\{|\mu i j\rangle|\nu k l\rangle\}$. Does there exist a projection operator which projects onto the subspace where irreps are tied together i.e. take the form $\{|\mu i j\rangle|\mu k l\rangle\}$?
I know the orthogonality relation
$$
\sum_{g\in (G)_{cj}} |C| \chi_\mu(C) \bar{\chi}_\nu(C) = |G| \delta_{\mu,\nu}
$$
where $(G)_{cj}$ is the set of conjugacy classes $C$ of $G$. This seems to do exactly what I want but I don't see that there's a way to express this in terms of the operators $L_{\pm}, T_{\pm}$.
In other words, I'd like to find an operator $P$ such that
$$
\begin{equation}
P |\mu i j\rangle |\nu k l\rangle = \delta_{\mu,\nu} \sum_{a,b}\sum_{c,d} A_{ab}^{ij}B_{cd}^{kl} |\mu a b\rangle|\mu cd\rangle
\end{equation}
$$
in terms of the raising and lowering operators, where I am allowing for the fact that there may be some mixing between states within the same irrep (if it has dimension >1).
| |
| |
Mon, 10/01/2018 - 08:35
Let $n\ge 1$ be an integer. $\mathcal P_n$ be the vector space of all polynomial functions over $[a,b]$, of degree at most $n$.
My question is : Is it true that
$\inf_{x_0,x_1,...,x_n\in[a,b], x_0<x_1<...<x_n} \sup_{x\in [a,b]} \prod_{i=0}^n |(x-x_i)|=\inf_{P\in \mathcal P_n } \sup_{x\in [a,b]} |x^{n+1}-P(x)|$ ?
I can easily see that $\inf_{x_0,x_1,...,x_n\in[a,b],x_0<x_1<...<x_n} \sup_{x\in [a,b]} \prod_{i=0}^n |(x-x_i)| \ge \inf_{P\in \mathcal P_n } \sup_{x\in [a,b]} |x^{n+1}-P(x)|$ , but I don't know about the other reverse inequality.
| |
Mon, 10/01/2018 - 00:55
(Sorry for my poor english..)
Let $N$ be a positive integer and $\chi$ be a Dirichlet character modulo 4N. I already know that the $\mathbb{C}$-vector space $S_{k}(\Gamma_1(N))$ has a basis $\{F_1,\cdots F_n\}$ with the Fourier coefficients of $F_i$ (at infinity) are rational.
My questions are..
Does the $\mathbb{C}$-vector space $S_{k}(\Gamma_0(N),\chi)$ also has a basis $\{G_1,\cdots G_m \}$ with the Fourier coefficients of $G_i$ (at infinity) are rational?
Does the $\mathbb{C}$-vector space $S_{k+\frac{1}{2}}(\Gamma_0(4N),\chi)$ also has a basis $\{G_1,\cdots G_m \}$ with the Fourier coefficients of $G_i$ (at infinity) are in $K$? ($K$ is a number field.)
(Thanks for reading)
EDIT : How about number field $K$ instead of rational number? In other words, does the $\mathbb{C}$-vector space $S_{k}(\Gamma_0(N),\chi)$ also has a basis $\{G_1,\cdots G_m \}$ with the Fourier coefficients of $G_i$ (at infinity) are in $K$? ($K$ is a number field.)
| |
Mon, 10/01/2018 - 00:50
Consider the discrete monoid $M$ of nondecreasing continuous maps from $[0,1]$ to itself preserving the extremities. Note that the monoid is right-cancellative ($x.z=y.z$ implies $x=y$, since $z$ is always onto).
What do we know about the homotopy type of this monoid (viewed as a
one-object category) ? In particular, about its homotopy groups ?
My background on this subject is very small. By a paper from Dusa McDuff (On the classifying space of discrete monoids), every path-connected space has the same homotopy type as the classifying space of some monoid, and the fundamental group of $BM$ is the groupification of $M$.
EDIT: I am a bit confused between the English meaning and the French meaning of nondecreasing. The monoid I am talking about is not a group because a nondecreasing map preserving extremities is not necessarily one-to-one. I hope that this clarification will be helpful.
| |
Sun, 09/30/2018 - 17:10
Assume that $k$ is an infinite field. Let $G$ be a finite (constant) group, let $V$ be a faithful $G$-representation over $k$, $U$ a non-empty open subset of $V$ where the action is free. The map $\pi:U\to U/G$ is a $G$-torsor, and moreover, every $G$-torsor over $k$ arises as a fiber of $\pi$, and the set of $p\in (U/G)(k)$ such that $\pi^{-1}(p)$ is isomorphic to a given torsor is dense in $U/G$. This is actually true over every field extension $K/k$.
Let $\alpha,\beta$ be two $G$-torsors over $k$, and assume that there exists a $G$-torsor $P\to Z$ where $Z$ is an open subset of $\mathbb{P}^1$, and assume that $\alpha$ and $\beta$ both arise as fibers of $P\to Z$. Can I find a map from some open subset $Z'$ of $\mathbb{P^1}$ to $U/G$ such that the induced torsor above $Z'$ admits $\alpha$ and $\beta$ as fibers? The converse is of course obvious.
Note that if $U/G$ is retract rational (condition which does not depend on $V$) but only on $G$), then any two points on $U/G$ may be connected by an open subset of a rational curve, so everything is trivial in this case. Unfortunately the only examples that I have so far belong to this trivial class.
I assumed that $G$ is finite because it seems to me to already be interesting, but one might of course ask a similar question for an arbitrary linear algebraic group $G$ (in this case $V$ must be chosen generically free and not just faithful, and the quotient will be a rational quotient).
| |
Sun, 09/30/2018 - 10:21
Suppose $X$ is a Riemann surface and $ a\in\ X $ suppose $ \phi\in\mathcal O_a $ is a holomorphic function germ at $a.$ According to the theorem 7.8 of Forster's book Lectures on Riemann surfaces on page 47 $ \phi $ has a maximal analytic continuation $(Y,p,f,b).$
Now suppose $ \phi\in\mathcal O_a $ admits analytical continuation along every curve in $X$ which starts at $a$.Is it always true that $ p:Y\rightarrow X $ has path lifting property.
If I assume another extra condition that , $X$ is simply connected then Monodormy theorem imply that $p:Y\rightarrow X $ has path lifting property.
Another question is that whether $p:Y\rightarrow X $ is covering or not?
Obviously path lifting property of $p:Y\rightarrow X$ implies that $p:Y\rightarrow X$ is a covering since $p:Y\rightarrow X$ is unbranched holomorphic according to the definition. Conversely any covering map has path lifting property.
Definition of path lifting property of a map given in the book of Otto Forster's book on page 25 is that - A continuous map $p:Y\rightarrow X$ is said to have path lifting property if for every path $u:[0,1]\rightarrow X$ and for every point $y_0 \in Y$ with $p(y_0)=u(0)$ there exists a lifting $\tilde u :[0,1]\rightarrow Y$ of $u$ such that $\tilde u(0)=y_0$.
| |
Sun, 09/30/2018 - 01:18
Let $K\subset R^d$ be a compact set. It is well known that its Fourier dimension is defined by
$$\dim_F K=\sup\{s\ge 0: \exists \mu \in M_1(K) s.t. \hat{\mu}(x)=O(|x|^{-s/2})\}(|x|\to\infty),$$
where $M_1(K)$ denotes the set of probability measures with support in $K.$
My question is: Does the condition $\dim_FK=0$ imply that $K$ can not support a measure $\mu$ with decaying Fourier transformation? In other word, there is no Rajchman measure spported on $K?$
Maybe there exists measure with decaying Fourier transformation according other ``speed" rather than polynomial speed. Can somebody give an example? Many thanks.
| |
Sat, 09/29/2018 - 08:31
Asked this on stack exchange and got no response, so I'll try here.
An idea popped into my head awhile ago while doing a project on non-linear effects of systems of coupled oscillators, but I'm not an expert on this subject so I don't know if it's any good. From cursory searching I couldn't seem to find anything, which most likely means it doesn't work, or I don't know the correct jargon.
From my understanding, the foundation of the theory of linear systems can be attributed to the fact you can construct a basis in the space of solutions and you can use that basis to find other solutions subject to boundary conditions. This all rests on the fact that for some linear operator $L$ and two functions $f$ and $g$ that satisfy:
$$L(f)=0$$
$$L(g)=0$$
Then we have:
$$L(f+g)=0$$
The problem with non-linear systems is that you can't construct a basis this way. Suppose we have a linear operator $L$, but also a non-linear operator $N$, and two functions $f$ and $g$ that satisfy:
$$L(f)=N(f)$$
$$L(g)=N(g)$$
Then:
$$L(f+g)\ne N(f+g)$$
We can't construct a basis by addition. But perhaps there is still a way in general to construct a basis? Can we find some operator $\phi(f, g)$ that satisfies:
$$L(\phi(f, g))=N(\phi(f, g))$$
Where $\phi$ could be how to "combine" solutions into other solutions. For linear systems we'd simply have $\phi(f,g)=f+g$, but I thought that in general for this idea to be consistent you'd want to make sure $\phi$ obeys the following conditions:
$$\phi(f, g)=\phi(g, f)$$
$$\phi(f, 0)=f$$
$$\phi(f, \phi(g, h))=\phi(\phi(f, g), h)$$
The most general way I see to do this is to define $\phi$ as:
$$\phi(f,g)=F^{-1}(F(f)+F(g))$$
Where $F$ is some transformation. However, this implies:
$$F(\phi(f, g))=F(f)+F(g)$$
So that the system becomes linear under a suitable transformation $F$. I didn't attempt to figure out if a transformation $F$ of this type always exists, and it seems like in most practical cases trying to find $F$ requires you to solves another non-linear equation, so I don't know if it can be called progress.
Is there literature I can read that explores the gist of this idea, or is it simply considered trivial in the cases where a system can be made linear? Can it be shown that in some systems there is no such $F$?
| |
Sat, 09/29/2018 - 05:56
Problem
Let $c \in \mathbb{N}$ $;$ $\exists$ a prime $p$ for which:
$$p^c \mid (p-1)!+1$$
Does $\exists$ $M$ $\in$ $\mathbb{N}$ $;$ $\forall$ $c \geqslant M$
$;$ $\nexists$ $p$ satisfying the above?
When $c$ = $1$
The statement is equivalent to Wilson's Theorem. For every prime $p$:
$$p \mid (p-1)!+1$$
Proof:
$\forall$ $x \in {1,2,...,p-1}$ $\exists!$ $ x' \in {1,2,...,p-1}$ ;
$x \cdot x'\equiv 1 \pmod{p}$
$x=x' \iff p \mid x^2-1 \iff x = 1, x=p-1$
$(p-1)! = 1 \cdot (p-1) \cdot \prod{(x \cdot x')} \equiv 1^n \cdot (p-1) \equiv p-1 \pmod{p}$
$\implies p \mid (p-1)!+1$
QED
When $c$ = $2$
We have the statement:
$$p^2 \mid (p-1)!+1$$
The only known primes that satisfy this are $5$, $13$ and $563$.
$(5-1)!+1 = 25 = 5^2$
$(13-1)!+1 = 479001601 = 13^2 \cdot 2834329$
$563^2 \mid (563-1)!+1 \approx 1.128 \cdot 10^{1303}$
Such primes $p$ are known as Wilson Primes. It is conjectured that there are infinitely many Wilson Primes. However, if there exists a fourth Wilson prime $p$, then $p>2 \cdot 10^{13}$.
When $c \geqslant 3$
There are no known primes for which $p^3 \mid (p-1)!+1$ as if there is, then $p$ also has to be a Wilson Prime.
$(5-1)!+1 = 25 \equiv 25 \pmod{5^3}$
$(13-1)!+1 = 479001601 \equiv 676 \pmod{13^3}$
$(563-1)!+1 \equiv 91921010 \pmod{563^3}$
It is most likely due to following evidence that there exists an upper bound $M$ for which:
$$c \geqslant M \implies p^c \nmid (p-1)!+1$$
where $M \geqslant 3$.
- We consider $(p-1)!+1 \pmod{p^c}$
- We assume that every remainder divisible by $p$ (Wilson's Theorem) is equally probable.
- Thus, the probability of required remainder $0$ is $\frac{1}{p^{c-1}}$
- Thus, probable number of primes for given constant $c$ is:
$$\sum{\frac{1}{p^{c-1}}} = P(c-1)$$
where $P(x)$ is the Prime Zeta Function
When $c=2$, the expected number of Wilson primes is $P(1)$.
$$P(1)=\sum{\frac{1}{p}}$$
This sum is divergent. Thus, it is probable that there exist infinitely many Wilson primes.
Proof:
Define $N(x)$ to be the number of positive integers $n \leqslant x$ for which $p_i \nmid n$, where $i > j$ for constant $j$ and $p_i$ is the $i$th smallest prime. Then, we write:
$$n=k^2m$$
where $m$ is square-free.
As $m$ is square-free, and the only primes that divide it are $p_i$ for $1 \leqslant i \leqslant j$, it has $2^j$ possibilities.
$n^2 \leqslant x \implies n \leqslant \sqrt{x}$, thus giving $n$ a maximum of $\sqrt{x}$ possibilities.
$$\implies N(x) \leqslant 2^j\sqrt{x}$$
Assume the contrary, then for some $j$:
$$\sum_{i=j+1}^\infty \frac{1}{p_i} < \frac{1}{2}$$
We also have $x-N(x)$ is the number of numbers less than or equal to $x$ divisible by one or more of $p_i$ for $i>j$.
$$x-N(x) \leqslant \sum_{i=j+1}^\infty \frac{x}{p_i} < \frac{x}{2}$$
$$\implies 2^j\sqrt{x} > \frac{x}{2}$$
which is untrue for $x \geqslant 2^{2j+2}$
Thus the sum diverges. The divergence is similar to $\log{\log{x}}$ (Which is very slow).
Probable Answer To Problem
When $c \geqslant 3$, the sum converges and is less than $1$.
When $c=3$, $P(c-1) \approx 0.45$
When $c=4$, $P(c-1) \approx 0.17$
When $c=5$, $P(c-1) \approx 0.07$
When $c=6$, $P(c-1) \approx 0.03$
When $c=7$, $P(c-1) \approx 0.002$
We now go on to show why there most probably exists a constant $M$ such as the one in the problem. Consider:
$$\sum_{i=3}^\infty P(i-1)$$
We have:
$$\sum_{i=3}^\infty P(i-1) < \sum{\frac{1}{n(n+1)}} = \sum{\biggl(\frac{1}{n}-\frac{1}{n+1}\biggl)} = 1$$
Thus, the probable sum of the number of primes that satisfy the statement for $c \geqslant 3$, including a prime $p$, $n-2$ times, if the maximum $c$ satisfied is $n$, is less than $1$. However, if the answer to the problem is false, then, the sum would be infinite.
Thus, it is highly unlikely for there to be a solution for $c \geqslant 3$ as the probable answer is less than $1$ but the actual answer would be a positive integer. However, it is almost impossible for the answer to the problem to be false, as the probable answer is less than $1$ but the actual answer would be infinite!
Any of the following:
- Any progress or insight
- Answers conditional on conjectures
- Polynomial or logarithmic non-trivial bounds on $M$
will be accepted and appreciated.
P.S. This question is also in Mathematics Stack Exchange. Link:
https://math.stackexchange.com/questions/2651733/stronger-versions-of-wilsons-theorem
| |
Fri, 09/28/2018 - 16:27
Consider the normal product distribution, which is the distribution of the product of two or more independent normal variables. Particulary, focus in the case where the multiplied normal variables are $\mathcal{N}(0, 1)$.
Considering the following definition of subgaussian tail (taken from [1]):
Does the normal product distribution have subgaussian tail?
I have been doing some numerical experiments which suggest an affirmative answer ($a=0.1$):
[1] Matoušek, J. (2008). On variants of the Johnson–Lindenstrauss lemma. Random Structures & Algorithms, 33(2), 142-156.
| |
Fri, 09/28/2018 - 16:15
I am looking for an upper-bound of the Euler's totient function $\varphi$ which would be equivalent to the Riemann hypothesis. There is the following Nicolas' criterion about primorial numbers $N_k$ (product of the $k$ first prime numbers):
Nicolas' criterion (MR724536, Théorème 2)
- If the Riemann hypothesis is true, then for all $k \ge 1$ $$ N_k / \varphi(N_k) > e^{\gamma}\ln \ln N_k $$
- If the Riemann hypothesis is false then there are infinitely many $k$ such that the above inequality holds, and infinitely many $k$ such
that it does not.
In particular, the Riemann hypothesis is equivalent to the following upper-bound: $$\varphi(n)< \frac{n}{e^{\gamma}\ln \ln n}, \text{ for } n=N_k \text{ and } k>1.$$
Now, the above inequality does not hold for $n$ prime greater than $7$. So an upper-bound for all $n$ requires a modification. Let $\omega(n)$ be the number of distinct prime factors of $n$. Then $N_k$ is the smallest number $n$ satisfying $\omega(n)=k$. Consider the following step function $$s=\sum_{k=1}^{\infty}k\chi_{(N_{k-1},N_k]}.$$
Question: Is the Riemann hypothesis equivalent to the following upper-bound?
$$\varphi(n)< \frac{n2^{s(n)-\omega(n)}}{e^{\gamma}\ln \ln n}, \text{ for } n>2.$$
Because $s(N_k) = \omega(N_k)$, this upper-bound implies the Riemann hypothesis by Nicolas' criterion. What about the converse?
Remark: By the computation below, this upper-bound holds for $n<10^6$.
Computation
sage: %time test(1000000)
CPU times: user 23min 10s, sys: 20.8 s, total: 23min 31s
Wall time: 23min 58s
Code
# %attach SAGE/EulerRH.spyx
from sage.all import *
cpdef g(int x):
return x/(exp(euler_gamma)*ln(ln(x)))
cpdef f(int n, int a, int b, int c):
if n<=a:
return c
else:
return f(n,a*b,next_prime(b),c+1)
cpdef s(int n):
return f(n,2,3,1)
cpdef omega(int n):
return len(list(factor(n)))
cpdef test(int m):
cdef int n
for n in range(3,m):
if euler_phi(n)*(2**omega(n))>=g(n)*(2**s(n)):
print(n)
| |
Fri, 09/28/2018 - 14:07
Let $C\subset \mathbb{P}^2$ be a smooth conic without $k$-points.
Call the Chow $k$-motive zero-dimensional if it is a sum of $M\mathbb{L}^n$ where $M$ is an Artin motive, i.e. a part of a motive of zero-dimensional scheme.
Q. How to see that a motive of $C$ is not (or is) zero-dimensional?
| |