Your privacy

By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy.

Skip to content
# Tag: torsion

## dg.differential geometry – Do exist constant curvature manifolds (hyperbolic or elliptic) with torsion?

## vector bundles – The uniqueness of some semistable torsion free sheaves on Fano threefold

## nt.number theory – Finding torsion subgroups of Mordell curves and their isogenies

## at.algebraic topology – $pi_{2n-1}(operatorname{SO}(2n))$ element represents the tangent bundle $TS^{2n}$, not torsion and indivisible?

## general topology – Is there any difference between invariant factors and torsion coefficients?

## Projective cellular varieties with singular cohomology admitting torsion different from two

## Implication of generators being torsion elements.

## arithmetic geometry – When are marked points torsion

## Serge Lang’s Algebra Theorem 7.5 proof(about decomposition of finitely generated torsion module)

## abstract algebra – uniqueness of torsion coefficients

New and Fresh Private + Public Proxies Lists Everyday!

Get and Download New Proxies from NewProxyLists.com

Your privacy

By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy.

Let $X$ be a prime Fano threefold of index one and even genus $ggeq 6$, one can show that the moduli space of torsion free semistable sheaves $M(2,1,m_g)$ with $m_g=left lceil{frac{g+2}{2}}right rceil$ consists of single point, this means that the only semistable torsion free sheaf $F$ with $mathrm{ch}(F)=mathcal{ch}(mathcal{E})$ is $mathcal{E}$, where $mathcal{E}$ is the tautological sub-bundle on $X$ coming from the one in Grassmannian. If one assume that $F$ is a vector bundle, then the argument to prove the above statement is even easier. My question is

that does this statement also hold of the semistable torison free sheaf $F$ with $mathrm{ch}(F)=mathrm{ch}(mathcal{Q})$ such that $mathcal{Q}$ is the tautological quotient bundle on $X$. If we assume $F$ is a vector bundle, it seems that I can figure out the proof by using Borel-Weil-Bott theorem on Grassmannian, but if I only assume that $F$ is a semistable torsion free sheaf, is this argument correct and how to argue it?

If one considered the Mordell curve $E_{n}:y^2=x^3+n$ over $mathbb{Q}$ where $n$ is some non-zero integer. Is there a congruence condition on $n$ that could tell you exactly what the torsion subgroup of $E_{n}$ is? Moreover, are you able to tell what torsion subgroups appear in the isogeny class of $E_{n}$?

Question: Is the element $alpha$ in $pi_{2n-1}(operatorname{SO}(2n))$ representing the tangent bundle $TS^{2n}$ of the sphere $S^{2n}$ indivisible and not torsion?

My understanding so far —

An $operatorname{SO}(2n)$ bundle over $S^{2n}$ corresponds to an element in $pi_{2n}operatorname{BSO}(2n) =pi_{2n-1}operatorname{SO}(2n)$.

Not torsion: There does not exist any integer $m > 0$ such that $malpha$ is a trivial element.

Indivisible: There does not exist any integer $k > 1$ and any element $beta$ in $pi_{2n-1}operatorname{SO}(2n)$ such that $alpha=kbeta$.

Ref: Mimura, Toda: Topology of Lie groups. Chapter IV Corollary 6.14.

I’m working on Elements of Algebraic Topology of Munkres. Exercise 3(d) in page 26 requires to find the torsion coefficients and invariant factors of a finite abelian group $G$. I’ve never seen definitions of invariant factor, but according to some references like this, torsion coefficients and invariant factors seem to be exactly the same things, but I’m not sure.

Is there a simple enough example of a projective cellular variety $X$ (cellular decomposition by affine spaces) such that $H^*(X(mathbb{R}),mathbb{Z})$ has an element with torsion different from 2? (3,4,5, for example)

Why if the generators of a module are torsion elements then the module is a torsion module?

Is there a proof to this statement?

We consider a family of genus one curves defined by

$$displaystyle d_0 z^2 = F_{a,b}(u,v) text{ (1)},$$

where

$$F_{a,b}(u,v) = a(u^2 – v^2)^2 + 4bu^2 v^2$$

and $d_0 = F_{a,b}(u_0, v_0)$, where $u_0, v_0$ are fixed integers independent of $a,b$. Then, provided that $a,b$ are integers, the curve defined by (1) is a genus one curve with a marked rational point, namely $(u_0, v_0, 1)$.

How does one determine whether the marked point $(u_0, v_0, 1)$ is torsion? As $a,b$ vary over a box, say $max{|a|, |b|} leq X$, can one say anything about the distribution of pairs $(a,b)$ for which $(u_0, v_0, 1)$ is torsion/not torsion?

I’m reading Lang’s *Algebra* to learn about moduled over PIDs, and I don’t understand his proof of the first assertion of Theorem 7.5.

The assertion is:

Let E be a finitely generated torsion module $neq$ 0. Then E is the direct sum $$E=bigoplus_p E(p)$$, taken over all primes $p$ such that $E(p) neq 0$.

It is proved that if $a$ is an exponent for $E$, and suppose that $a=bc$ with $(b,c) = (1)$, then $E=E_a oplus E_b$.

Then Lang says that “our first assertion(the one above) then follows by induction, expressing $a$ as a product of prime powers.

I’m confused here. If $a=prod_{i=1}^r p_i^{k_i}$, then we will have $E=oplus_{i=1}^r E_{p_i^{k_i}}$, but how can we use this to prove $E=bigoplus_p E(p)$? Generally, $E(p)neq E_{p^r}$, right?

I’m studying Fraleigh Abstract Algebra section 11. Exercise 44 defines torsion coefficients as follows:

Finitely generated abelian group $G$ can be written in the form $Z_{m_1} times Z_{m_2} times ….

times Z_{m_r}$, where $m_i$ divides $m_{i + 1}$ for $i = 1, 2, …, r-1 $. The numbers $m_i$ can be shown to be unique, and are thetorsion coefficientsof $G$

How can one prove that the torsion coefficients are unique? I’m on a first course of abstract algebra, so I do not know about complex theorems.

DreamProxies - Cheapest USA Elite Private Proxies
100 Private Proxies
200 Private Proxies
400 Private Proxies
1000 Private Proxies
2000 Private Proxies
5000 Private Proxies
ExtraProxies.com - Buy Cheap Private Proxies
Buy 50 Private Proxies
Buy 100 Private Proxies
Buy 200 Private Proxies
Buy 500 Private Proxies
Buy 1000 Private Proxies
Buy 2000 Private Proxies
ProxiesLive.com
Proxies-free.com
New Proxy Lists Every Day
Proxies123.com
Buy Cheap Private Proxies; Best Quality USA Private Proxies