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most recent 30 from mathoverflow.net 2018-02-23T10:14:37Z

Forcings predicted by core model theory $+$$ZFC$ results proved by the method of core model theory

Thu, 02/22/2018 - 01:58

I have two unrelated question.

First question. To motivate the question, let me explain an example. The natural way to force the failure of singular cardinals hypothesis ($SCH$), is to start with a large cardinal $\kappa$, and make it singular while blowing up its power. However, results of core model theory show that if $SCH$ fails at $\kappa,$ then either $\kappa$ is a large cardinal, or it is a singular limit of large cardinals. Motivated by this fact, the long and short extender forcings were developed, showing that the second case can be forced as well.

Now my question is the following:

Question 1. Are there any other example of forcing notions, whose existence is first predicted using core model techniques and then they are discovered?

Second question. Forcing is a powerful tool to prove $ZFC$ results as well, see for example Forcing as a tool to prove theorems, and Examples of ZFC theorems proved via forcing and Proving results about complete Boolean algebras in ZFC using Boolean valued models and Producing finite objects by forcing!.

Surprisingly, one can also use the technique of core model theory to prove $ZFC$ results. One example that I know, is the following result of Woodin: Suppose that $V=L[s]$, where $s$ is an $ω$-sequence of ordinals. Then $GCH$ (and in fact much more) holds. See The universe constructed from a sequence of ordinals.

Question 2. Are there any other examples of $ZFC$ results whose proof uses techniques of core model theory. The same question for theories like $ZFC+\phi$, where $\phi$ is the assertion that some large cardinal(s) exist, for which we know a core model exists (so that we can apply the core model techniques).

Remark. I am not interested in results which use some kind of covering in the absence of large cardinals to get some results, like, if there is no measurable cardinal (or even larger cardinals), then square holds at singular cardinals or so on.

If a variety $X$ has finite automorphism group, is the same true for its $n$-fold self-products?

Wed, 02/21/2018 - 03:34

Let $X$ be an algebraic variety over $\mathbb{C}$. Let $n\geq 1$ be an integer and let $X^n$ be the $n$-fold self product of $X$.

Q. Is there an integer $n\geq 1$ and an algebraic variety $X$ over $\mathbb{C}$ such that $\mathrm{Aut}(X)$ is finite and $\mathrm{Aut} (X^n)$ is infinite?

Evaluating the sum of $kl^2$ over $p,q,k,l$ such that $pk +ql = n$

Tue, 02/20/2018 - 14:16

I have come across the following sum:

$$\sum_{\substack{p, q, k, l \in \mathbb{N} \\ k > l \\ pk + ql = n}}kl^2$$

and I am trying to simplify it, hoping to get a nice formula in terms of $n$ and some arithmetic functions of $n$.

For instance, the following sum can be evaluated as follows:

\begin{align*} \sum_{\substack{p, q, k, l \in \mathbb{N} \\ pk + ql = n}}kl &= \sum_{\substack{\alpha, \beta \in \mathbb{N}\\ \alpha+ \beta = n}}\sum_{k | \alpha}k \sum_{l | \beta}l \\ &= \sum_{\substack{\alpha, \beta \in \mathbb{N}\\ \alpha+ \beta = n}} \sigma_1(\alpha) \sigma_1(\beta)\\ &= (\sigma_1 \Delta \sigma_1)(n) \end{align*} where $\sigma_1(n)$ is the sum of divisors of $n$ and $\Delta$ is the discrete convolution. Ramanujan has a formula for the discrete convolution of $\sigma_1$ with itself given by $$(\sigma_1 \Delta \sigma_1)(n) = \frac{5}{12}\sigma_3(n) + \frac{1}{12}\sigma_1(n) - \frac{1}{2} n \sigma_1(n)$$ where $\sigma_3(n)$ is the sum of the cubes of the divisors of $n$.

Any thoughts on how one might proceed with the sum in the beginning of the question? The asymmetry between $k$ and $l$ is causing some problems so the same approach as for the second sum does not quite work.

I would be very much grateful for any suggestions. Thanks!

Rigid Non-archimedean Real Closed Fields

Tue, 02/20/2018 - 12:40

Question. Is there a countable rigid non-Archimedean real closed field?

Background:

  1. As usual, a structure is said to be rigid if the only automorphism of the structure is the identity map.

  2. It is well-known that any Archimedean ordered field is rigid; the argument is the same as that used by Darboux in 1880 to prove that the field of real numbers is rigid. Note that the Archimedean ordered fields are precisely those ordered fields that are isomorphic a subfield of the field of real numbers.

  3. In 1983 Shelah showed that its consistent with the axioms of set theory (ZFC) that there is an uncountable rigid non-Archimedean closed field. Here is the reference: Shelah, Saharon, Models with second order properties. IV. A general method and eliminating diamonds. Ann. Pure Appl. Logic 25 (1983), no. 2, 183–212.

  4. Using machinery from the metamathematics of arithmetic, one can build (many) countable non-Archimedean ordered fields. More specifically: it is well-known that countable nonstandard rigid models of PA exist (this is independently due to Gaifman and Ehrenfeucht). Let $\cal{M}$ be such a model, and let $\mathbb{Q}^{\cal{M}}$ be the field of rationals as computed in $\cal{M}$. By a classical theorem of Julia Robinson, $\cal{M}$ and $\mathbb{Q}^{\cal{M}}$ are bi-interpretable, therefore $\mathbb{Q}^{\cal{M}}$ is also rigid. One can use the same type of argument to show that there are (many) rigid non-Archimedean ordered fields in every infinite cardinality.

Do Sums of powers determine the numbers?

Tue, 02/20/2018 - 01:40

Let $x_1,x_2,...,x_n$ be complex numbers. Denote $S_k=\sum_{i=1}^{n}x_i^k$. Do values of $S_1,S_2,...,S_n$ determine $x_1,x_2,...,x_n$? What if $x_i\in Z/pZ$ for prime $p$? Would $S_1,S_2,...,S_n$ determine $x_1,x_2,...,x_n$ in that case?

Gradients of the Dominant Eigenvalue and Eigenvector

Tue, 02/20/2018 - 00:25

How can I compute the partial derivatives of the dominant eigenvalue and eigenvectors of a real symmetric matrix $\mathbf{A}$?

In particular, given

$ \mathbf{v}^* = \arg\max_{\mathbf{v}} \mathbf{v}^{\top}\mathbf{A}\mathbf{v},\quad \text{subject to } ~ \|\mathbf{v}\|_2 = 1, $

how can I find $\partial v_{i}^*/\partial A_{jk}$?

I know the power method is the usual way to compute the dominant eigenvalue and eigenvector. Is their any similar algorithm for computing the gradient?

Unlike another question, I am interested in an efficient computational solution rather than an analytical one.

Cofree conilpotent (cocommutative) coalgebra for $\infty$-categories

Tue, 02/20/2018 - 00:00

Let $\mathrm{K}$ be a field.

Denote $ \mathrm{Vect}_{\mathrm{K}} $ the category of K-vector spaces, $\mathrm{Coalg }^{\mathrm{conil } } $ the category of conilpotent, coaugmented, coassociative coalgebras over K and $\mathrm{Cocoalg }^{\mathrm{conil } } $ the category of conilpotent, coaugmented, coassociative and cocommutative coalgebras over K.

Then the forgetful functors $\mathrm{Coalg }^{\mathrm{conil } } \to \mathrm{Vect}_{\mathrm{K}} $ and $\mathrm{Cocoalg }^{\mathrm{conil } } \to \mathrm{Vect}_{\mathrm{K}} $ that take the cokernel of the coaugmentation admit right adjoints $\mathrm{T}: \mathrm{Vect}_{\mathrm{K}} \to \mathrm{Coalg }^{\mathrm{conil } }$ respectively $\mathrm{R}: \mathrm{Vect}_{\mathrm{K}} \to \mathrm{Cocoalg }^{\mathrm{conil } }$ and we have natural isomorphisms $\mathrm{T}(\mathrm{V}) \cong \oplus_{ i \geq 0} \mathrm{V}^{\otimes i}$ and $\mathrm{R}(\mathrm{V}) \cong \oplus_{ i \geq 0} (\mathrm{V}^{\otimes i})^{ \Sigma_i}$ of K-vector spaces.

Denote $ \mathrm{Calg} $ the category of augmented, associative and commutative algebras over K and $\mathrm{S} : \mathrm{Vect}_{\mathrm{K}} \to \mathrm{Calg}, \ \mathrm{V} \mapsto \oplus_{ i \geq 0} (\mathrm{V}^{\otimes i})_{ \Sigma_i} $ the free commutative algebra functor with its canonical augmentation.

If the field K has char. 0, we have a canonical isomorphism $(\mathrm{V}^{\otimes i})_{ \Sigma_i} \cong (\mathrm{V}^{\otimes i})^{ \Sigma_i}$ of K-vector spaces and so a canonical isomorphism $\mathrm{S}(\mathrm{V}) \cong \mathrm{R}(\mathrm{V})$ of K-vector spaces that makes $\mathrm{S}(\mathrm{V}) $ to a commutative and cocommutative bialgebra.

My question is: Does this generalize to $\infty$-categories in the following sense?

Let $\mathcal{C}$ be a stable presentable symmetric monoidal $\infty$-category.

Then one has $\infty$-categories $\mathrm{Calg}(\mathcal{C})^\mathrm{ }$ and $\mathrm{Cocoalg}(\mathcal{C})^\mathrm{ }:= (\mathrm{Calg}(\mathcal{C^\mathrm{op}})^\mathrm{})^\mathrm{op}$ of augmented, associative and commutative algebras respectively of coaugmented, coassociative and cocommutative coalgebras and one can define conilpotent coalgebras in the following way:

The forgetful functor $\psi: \mathrm{Cocoalg}(\mathcal{C})^\mathrm{ } \to \mathcal{C}$ that takes the cokernel of the coaugmentation admits a unique section $\mathrm{E} : \mathcal{C} \to \mathrm{Cocoalg}(\mathcal{C})^\mathrm{ }$ that sends an object $\mathrm{X}$ of $\mathcal{C}$ to its co-square-zero extension, whose coaugmentation has cocernel $\mathrm{X}$ with 0-comultiplication $0: \mathrm{X} \to \mathrm{X} \otimes \mathrm{X}.$

Denote $ \mathrm{Cocoalg}(\mathcal{C})^\mathrm{conil }$ the smallest full subcategory of $ \mathrm{Cocoalg}(\mathcal{C})^\mathrm{ }$ that contains the essential image of $\mathrm{E}$ and closed under small colimits. $ \mathrm{Cocoalg}(\mathcal{C})^\mathrm{conil } \subset \mathrm{Cocoalg}(\mathcal{C})^\mathrm{ } $ is a colocalization.

By the adjoint functor theorem $\psi: \mathrm{Cocoalg}(\mathcal{C})^\mathrm{ } \to \mathcal{C}$ admits a right adjoint and so the restriction $ \mathrm{Cocoalg}(\mathcal{C})^\mathrm{conil } \to \mathcal{C}$ of $\psi$ admits a right adjoint $\mathrm{R}.$

Similarly one defines $ \mathrm{Coalg}(\mathcal{C})^\mathrm{conil }$ and shows that the forgetful functor $ \mathrm{Coalg}(\mathcal{C})^\mathrm{conil } \to \mathcal{C}$ admits a right adjoint $\mathrm{T}.$

  1. Question: Given an object $\mathrm{X}$ of $\mathcal{C} $ do we have a canonical equivalence $\mathrm{T}(\mathrm{X}) \simeq \oplus_{ i \geq 0} \mathrm{X}^{\otimes i}$ in $\mathcal{C} $?

The forgetful functor $\mathrm{Calg}(\mathcal{C})^\mathrm{ } \to \mathcal{C} $ that takes the kernel of the augmentation admits a left adjoint $\mathrm{S}$ and we have a canonical equivalence $\mathrm{S}(\mathrm{X}) \simeq \oplus_{ i \geq 0} (\mathrm{X}^{\otimes i})_{ \Sigma_i} $ in $ \mathcal{C} $ for $\mathrm{X} \in \mathcal{C}.$

One can promote $\mathrm{S} $ to an oplax symmetric monoidal functor $\mathcal{C}^\times \to \mathrm{Calg}(\mathcal{C})^\otimes $ that yields a functor $$\bar{\mathrm{S}} : \mathcal{C} \simeq \mathrm{Cocoalg}(\mathcal{C}^\times) \to \mathrm{Cocoalg}(\mathrm{Calg}(\mathcal{C})^\otimes)=: \mathrm{Bialg}(\mathcal{C})$$ lifting $\mathrm{S} $.

The symmetric monoidal forgetful functor $\mathrm{Calg}(\mathcal{C}) \to \mathcal{C}$ (not the functor that takes the augmentation ideal) induces a forgetful functor $\alpha: \mathrm{Bialg}(\mathcal{C}) \to \mathrm{Cocoalg}(\mathcal{C})$.

This way for every $\mathrm{X} \in \mathcal{C}$ the object $\mathrm{S}(\mathrm{X} )$ carries the structure of a coaugmented conilpotent cocommutative coalgebra in $\mathcal{C}.$

Assume that $\mathcal{C} $ is additionally $\mathbb{Q}$-linear (i.e. receives a left adjoint symmetric monoidal functor from $\mathrm{H}(\mathbb{Q})$-module spectra.)

In this case the norm map $(\mathrm{X}^{\otimes i})_{ \Sigma_i} \to (\mathrm{X}^{\otimes i})^{ \Sigma_i}$ is an equivalence for every $\mathrm{X} \in \mathcal{C} $ and $i \geq 0.$

  1. Question: Is the composition $\mathcal{C} \xrightarrow{\bar{\mathrm{S}}} \mathrm{Bialg}(\mathcal{C}) \xrightarrow{\alpha} \mathrm{Cocoalg}(\mathcal{C})$ equivalent to $\mathrm{R}?$

By the definition of $\mathrm{Cocoalg}(\mathcal{C})^\mathrm{conil }$ and that for every $\mathrm{X} \in \mathcal{C}$ the coalgebra $\mathrm{S}(\mathrm{X} )$ is conilpotent, the statement of 2. is equivalent to the following condition:

For every $\mathrm{X}, \mathrm{Y} \in \mathcal{C} $ the map $$\mathrm{Cocoalg}(\mathcal{C})(\mathrm{E}(\mathrm{X}), \mathrm{S}(\mathrm{Y})) \to \mathcal{C}(\mathrm{X}, \psi(\mathrm{S}(\mathrm{Y}))) \to \mathcal{C}(\mathrm{X}, \mathrm{Y}) $$ induced by $\psi: \mathrm{Cocoalg}(\mathcal{C}) \to \mathcal{C}$ and composition with the projection to the first factor $ \psi(\mathrm{S}(\mathrm{Y})) \simeq \oplus_{ i > 0} (\mathrm{Y}^{\otimes i})_{ \Sigma_i} \to \mathrm{Y}$ is an equivalence.

Being a section of $\psi$ the functor $\mathrm{E}: \mathcal{C} \to \mathrm{Cocoalg}(\mathcal{C})^\mathrm{ }$ admits a right adjoint $\mathcal{P}$ by the adjoint functor theorem with $\mathcal{P} \circ \mathrm{R} \simeq \mathrm{id}.$

One can show that $\bar{\mathrm{S}} : \mathcal{C} \to \mathrm{Bialg}(\mathcal{C})$ is left adjoint to the composition $\bar{\mathcal{P}}: \mathrm{Bialg}(\mathcal{C}) \xrightarrow{\alpha} \mathrm{Cocoalg}(\mathcal{C}) \xrightarrow{\mathcal{P} } \mathcal{C}$.

If 2. would hold, we would have $\mathrm{id} \simeq \mathcal{P} \circ \mathrm{R} \simeq \mathcal{P} \circ \mathrm{\alpha} \circ \bar{\mathrm{S}} \simeq \bar{\mathcal{P}} \circ \bar{\mathrm{S}}$ being the unit of the adjunction $\bar{\mathrm{S}} : \mathcal{C} \rightleftarrows \mathrm{Bialg}(\mathcal{C}): \bar{\mathcal{P}}$.

So $\mathcal{C}$ would embed into $\mathrm{Bialg}(\mathcal{C})$ via $\bar{\mathrm{S}}.$ Can one expect this?

"Being in char 0" one can think of $\bar{\mathcal{P}}$ as an abelian version of primitive elements and of algebras over the monad $\bar{\mathcal{P}} \circ \bar{\mathrm{S}}$ associated to the adjunction $\bar{\mathrm{S}} : \mathcal{C} \rightleftarrows \mathrm{Bialg}(\mathcal{C}): \bar{\mathcal{P}}$ as abelian Lie algebras that are classically uniquely determined by their underlying object.

Brauer group of global fields

Mon, 02/19/2018 - 23:23

Is the Brauer group $\text{Br}(K)$ of a global field $K$

  • an $\ell$-divisible group for some prime $\ell$? If so, what $\ell$?

  • Is $\text{Br}(K)[n]$ finite, for $n$ integer?

I know from class field theory that it fits into an exact sequence

$$0\to \text{Br}(K)\to\bigoplus_v\text{Br}(K_v)\xrightarrow{\sum_v \text{inv}_v} \mathbf{Q}/\mathbf{Z}\to 0$$

with $v$ running over all places of $K$, and $K_v$ the completion of $K$ at $v$.

but I can't conclude.

Thanks very much.

Why limit for this function exists as x approaches 0? [on hold]

Mon, 02/19/2018 - 22:35

$$f(x) = \frac{\sqrt{x^{2}+100} - 10}{x^{2}} $$

As $\lim_{x \to 0.01} f(x)=0.05$ but, as we keep decreasing the value of $x$ to $0$, $\lim_{x \to 0.000001} f(x) = 0.000000$. So why this function's Limit exist?

Solving minimization problem $L_2$ IRLS

Mon, 02/19/2018 - 22:10

In the article ''' Chartrand, Rick, and Wotao Yin. "Iteratively reweighted algorithms for compressive sensing." Acoustics, speech and signal processing, 2008. ICASSP 2008. IEEE international conference on. IEEE, 2008. '''

It is given that one iteration of iterative reweighed least squares problem can be written as $\min \sum_{i=1}^N w_i u_i^2 $ subject to $\Phi u = b $ which has a closed form solution given as

$u^{(n)} = Q_n \Phi^T (\Phi Q_n \Phi^T)^{-1} b$. where $Q_n$ is the diagonal matrix containing entries $1/w_i$ And the closed form equation can be derived from the Euler Lagrange equation. Can someone please help me to get the derivation of the iteration

Is this a proof of Fermat's Last Theorem? [on hold]

Mon, 02/19/2018 - 21:26

There is no solution to $a^p +b ^p + c^p = 0$ where $p>3$ is prime and $a$, $b$, and $c$ are pairwise co-prime nonzero integers.

Lemma (A).

If ${d^p- e^p}/(d-e)=\prod_j{q_j}$ then $q_j=1\bmod{p}$ for all $j$ where $p$ and $q_j$ are odd primes given $p\nmid{d-e}$ and $\gcd(d,e) =1 $.

Assume $a^p +b ^p + c^p = 0$ where $p$ is an odd prime and $a$, $b$, and $c$ are pairwise co-prime nonzero integers.

Let

$x$ be such that $a \bmod{p} = x \bmod{p}$ and without loss of generality $-p < x < p,$

$y$ be such that $b \bmod{p} = y \bmod{p}$ and without loss of generality $-p <y < p,$

and

$z$ be such that $c \bmod{p} = z \bmod{p}$ and $-p \leq{z}\leq{p}.$

It should be obvious that $-2p<x+y+z<2p$. $x + y+ z$ can have any one of the values $-p$, $0$, or $p$ and that there may be two cases (w.o.l.o.g.):

(case 1) $z = -p$

(case 2) $p\nmid{zxy}.$

Let $x+y+z=0$ for each case.

Next, let $I_x$, $I_y$, and $I_z$ be integers such that

$a = x + pI_x$

$b = y + pI_y$

$c = z + pI_z.$

Adding the above three equations yields $$a +b + c = x + y+ z + p(I_x +I_y + I_z).$$

We know $$p^2 \| (a +b +c)$$

Therefore, we can show this one set of $x$, $y$, and $z$ such that $$x+y+z=0$$ yields $$p^2\|p(I_x +I_y + I_z)$$ or $$p\|(I_x +I_y + I_z).$$

Raising the set of the three equations all to the $p^{th}$ power yields by expansion

$a^p \bmod{p^3} = x^p + p^2I_x \bmod{p^3}$

$b^p \bmod{p^3} = y^p + p^2I_y \bmod{p^3}$

$c^p \bmod{p^3} = z^p + p^2I_z \bmod{p^3}$.

Adding these three equations yields $$p^3 \|x^p + y^p + z^p$$ where $$x+y+z=0.$$

(case 1) $z = -p.$

$x=p-y$ and taking this equation to the $p^{th}$ power yields by expansion $$x^p\bmod{p^3} = -y^p + p^2y^{p-1} - p^rR_0\bmod{p^3}$$ where $r\ge{3}$ and $R_0$ is some integer. That is $$x^p\bmod{p^3} = -y^p + p^2\bmod{p^3}.$$

We also have $$x^p\bmod{p^3} = -y^p + p^p\bmod{p^3}.$$

A contradiction. Therefore it can be shown that $p\nmid{abc}$.

(case 2) $p\nmid{xyz}$ that is $p\nmid{abc}$.

Lemma (A) shows, that for a given $d$ and $e$ and where $\gcd(d,e)=1$ and $p\nmid{d-e}$, ${d^p- e^p}/(d-e)=\prod_{j=1}^{m}(pQ_j +1)$ for all $j$ where ${Q_j}'s$ are integers.

We have $${x^p+ y^p}/(x+y)=\prod_{r=1}^{u}(pQ_r +1)^{p-1}=z^{p-1}$$ $${y^p+ z^p}/(y+z)=\prod_{s=1}^{v}(pQ_s +1)^{p-1}=x^{p-1}$$ $${z^p+x^p}/(z+x)=\prod_{t=1}^{w}(pQ_t +1)^{p-1}=y^{p-1}$$ or $${x^p+ y^p}=-\prod_{r=1}^{u}(pQ_r +1)^{p}=-z^{p}$$ $${y^p+ z^p}=-\prod_{s=1}^{v}(pQ_s +1)^{p}=-x^{p}$$ $${z^p+x^p}=-\prod_{t=1}^{w}(pQ_t +1)^{p}=-y^{p}.$$

Therefore $$x^p=1\bmod{p}$$ $$y^p=1\bmod{p}$$ $$z^p=1\bmod{p}$$ and so $$a^p=1\bmod{p}$$ $$b^p=1\bmod{p}$$ $$c^p=1\bmod{p}.$$

Since $a^p+b^p+c^p=0$, we now have $$1+1+1=3=0\bmod{p}$$ for all $$p>3.$$

A contradiction.

It is not the (case 2) that $p\nmid{z}$ and then $p\|{abc}.$

It is not the (case 1) that $p\|{xyz}$ and then $p\nmid{abc}.$

Contradiction; there is no solution to $a^p +b ^p + c^p = 0$ where $p>3$ is prime and $a$, $b$, and $c$ are pairwise co-prime nonzero integers.

Proof of Lemma (A).

result: If ${d^p- e^p}/(d-e)=\prod_j{q_j}$ then $q_j=1\bmod{p}$ for all $j$ where $p$ and $q_j$ are odd primes given $p\nmid{d-e}$ and $\gcd(d,e) =1 $.

proof.

(0.) $(d^p - e^p)/(d-e)$ is a positive odd integer and I leave this conclusion for the reader.

(1.) $(d^p - e^p)/(d-e) = d^{(p-1)} + ed^{(p-2)} + e^2d^{(p-3)} + ... + d^2e^{(p-3)} + de^{(p-2)} + e^{(p-1)}$.

(2.) $(d^p - e^p)/(d-e) - pd^{(p-1)} = d^{(p-1)} - d^{(p-1)} + (e-d)d^{(p-2)} + (e^2-d^2)d^{(p-3)} + ... + d^2(e^{(p-3)}-d^{(p-3)}) + d(e^{(p-2)}-d^{(p-2)} + (e^{(p-1)}-d^{(p-1)}).$

$(d-e)$ divides the RHS of (2.).

Therefore,

(3.) $(d-e) \| [(d^p - e^p)/(d-e) - pd^{(p-1)}]$.

But,

(4.) $\gcd{( (d^p - e^p)/(d-e), pd^{(p-1)})} = 1$.

Therefore,

(5.) $\gcd{((d^p - e^p)/(d-e), (d-e))} = 1$.

(6.) $(d^p - e^p) = (d-e) \prod_{j}[q_j] = \prod_{i}[r_i]\prod_j[q_j]$ where $d- e = \prod_i[r_i]$, $r_i$ are primes and $(d^p - e^p)/(d-e) = \prod_j[q_j]$, $q_j$ are odd primes.

(7.) For any $j$ and for any $i$, $q_j \neq{r_i}.$ This follows from

(5.) $\gcd{((d^p - e^p)/(d-e), (d-e))} = 1$.

(8.0) $d \neq{e} \bmod q_j$ for all $q_j$.

(8.1) $d^p = e^p\bmod{q_j}$ for any $q_j$.

(8.2.1) From (8.0 and 8.1) $d^{(p+1)}\neq{e^{(p+1)}} \bmod q_j$ for all $j$.

(8.2.2) From (8.0 and 8.1) $d^{(p-1)} \neq{e^{(p-1)}}\bmod q_j$ for all $j$.

(8.3) $d^2 \neq e^2 \bmod q_j$ for all $q_j$.

This follows from (8.2.1) and/or (8.2.2). That is, $2$ divides either or both $(p-1)$ or $(p+1)$.

(8.3.1) $d^{(p-2)} \neq{e^{(p-2)}} \bmod q_j$ for all $j$ (this follows from (8.3) and (8.1)).

(8.3.2) $d^{(p+2)} \neq e^{(p+2)} \bmod q_j$ for all $j$ (this also follows from (8.3) and (8.1)).

Now, $3$ divides $(p-2)$ or $(p-1)$ or $(p+1)$ or $(p+2)$. $3$ does not divide $p$ unless $p =3$. Then proof would be near complete.

Therefore,

(8.3.3) $d^3 \neq e^3 \bmod q_j$ for all $j$.

Induction.

Assume,

(8.m) $d^m\neq{e^m}\bmod q_j$ for all $q_j$ and $m < p$.

Also, it is also assumed, at this point $d^k \neq{e^k}\bmod{q_j}$ for $k: 1 =< k =< m < (p-1)$.

(8.m.1) $d^{(p-k)}\neq{ e^{(p-k)}} \bmod q_j$ for all $j$ (this follows from (8.m) and (8.1)).

(8.m.2) $d^{(p+k)} \neq {e^{(p+k)}}\bmod q_j$ for all $j$ (this follows from (8.m) and (8.1)).

(8.m.3) $[(m+1)\|(p-k)] or [(m+1)\|(p+k)]$

for exactly one $k: 1 =< k =< m < (p-1)$;

$(p-k)$ to $(p+k)$ has $2m + 1$ values in succession from $(p-m)$ to $(p+m)$, including $p$. $(m+1)$ must divide exactly one of the values in this range.

Therefore,

(8.[m+1]) $d^{(m+1)} \neq e^{(m+1)} \bmod q_j$ for all $j$.

end induction.

(9.) $d^n \neq e^n \bmod q_j$ for all $j$ $n: 1 \leq n < p$ and $n \neq Ip$ where $I$ is any positive integer.

Recall,

(6.) $(d^p - e^p) = (d-e) \prod_{j}[q_j] = \prod_{i}[r_i]\prod_j[q_j]$ for all $j$.

(10.) $d^{q_j-1} = e^{q_j-1}\bmod q_j$ for all $j$ using Fermat's Little Theorem.

(11.) $q_j - 1 \neq n$; Because if $q_j - 1 = n$ then $d^n = e^n \bmod q_j$ for all $j$ then there would be a contradiction.

(12.) $n$ is equal to any positive integer other than $Ip$ where $I$ is any positive integer.

This implies

(13.) $q_j = 1 \bmod{p}$ for all $j$.

(14.) If ${d^p- e^p}/(d-e)=\prod_j{q_j}$ then $q_j=1\bmod{p}$ for all $j$ where $p$ and $q_j$ are odd primes given $p\nmid{d-e}$ and $\gcd(d,e) =1 $.

End of Proof of Lemma (A).

Redundant boundary condition of a $1$st order ODE?

Mon, 02/19/2018 - 20:08

Consider the following $1$st order eigen ODE system of 2 components $(\alpha,\beta)(x)$ defined for $x\in[0,L]$ $$ F_{l+1}\beta+m\alpha=\lambda\alpha\\ F_{-l}\alpha-m\beta=\lambda\beta $$ where $\lambda$ is the eigenvalue, $l$ is integer, $m(0)$ is finite and $m(L)\rightarrow\infty$, and $F_k=-i(\frac{d}{dx}+\frac{k}{x})$.

Since it is just $1$st order, I thought only 2 boundary conditions (b.c.) are necessary, which can be most easily taken as $(\alpha,\beta)(L)=0$ since $m(L)$ diverges. So one doesn't need or just shouldn't involve b.c. at $x=0$?

However, on the other hand, if I do the very simple asymptotic analysis at $x\rightarrow0,x\rightarrow\infty$ for these equations, I find that $$ l\neq0,-1\qquad (\alpha,\beta)(0)=(\alpha,\beta)(\infty)=0\\ l=-1\qquad (\alpha,\beta')(0)=(\alpha,\beta)(\infty)=0 $$ with $l=0$ similar to $l=-1$. What I did is basically just noting that, when $k\neq0$, any component acted by $F_k$ is $0$ at $x=0$. And using this, we notice the other component's derivative vanishes. It seems that b.c. at $x=0$ is natural as well.

How to reconcile these two considerations? Where am I wrong? And if $m(x)$ is always finite and smooth enough, is the asymptotic analysis the correct b.c.?

Is every algebraic space a 1-geometric stack?

Mon, 02/19/2018 - 19:46

In many references (Toen, Higher and derived stacks: a global overview, Toen, Vezzosi, Homotopical algebraic geometry II, and so on), the definition of $n$-geometric stack appears.

In the non-derived case, the definition starts by declaring affine schemes as $(-1)$-geometric stacks and inductively defines $n$-geometric stacks by some procedure.

Toen-Vezzosi, HAG II, Remark 2.1.1.5 says that algebraic spaces and schemes are automatically 1-geometric stacks. I can check that schemes are 1-geometric. (It does not depend on whether a scheme is separated or not.) But I can't check easily that algebraic spaces are 1-geometric stacks.

Properties of codimension under pull back

Mon, 02/19/2018 - 19:14

If I pull back a cycle of codimension $c$ along a morphism of schemes I can easily see that the codimension can stay the same or the codimension can drop to all the way to $0$. But intuitively it seems the codimension can never increase. Is there a precise statement of this form?

To state it more carefully suppose $X,Y$ are irreducible schemes and $f:X \to Y$ is a morphism and $Z\subset Y$ is an irreducible subscheme of codimension $c$. Is it always true that if $T\subset f^{-1}(Z)$ is an irreducible component then $codim(T) \leq codim(Z)$?

If this is not true then are there reasonable hypothesis under which it is true, $X,Y$ Noetherian, finite type, smooth?

If $f$ is flat the codimension should be preserved but the inequality above seems to be true more generally.

Geometric inequality in a triangle [on hold]

Mon, 02/19/2018 - 17:28

Let $O$ be the orthocenter of isosceles triangle $ABC$, $AB=AC=c$. Let $OC$ meet the line segment $AB$ at point $P$. If $p=PO$, how can I prove that $c^4 \ge p^4+11p^2 c^2$?

Sphere equation with higher power

Mon, 02/19/2018 - 17:02

The shape described by $x^2 + y^2 +z^2 = 1$ is a sphere. But what do you call the shape described by $x^4 + y^4 +z^4 = 1$? Does it have a name?

Integral with inverse tangent [on hold]

Mon, 02/19/2018 - 16:44

If $a$ a positive number, calculate the value of the following integral $$\int_0^a (x^2-ax+a^2)\arctan(e^x-1)\,dx$$

Co-finite type abelian groups

Mon, 02/19/2018 - 16:15

Suppose $B$ is an abelian group such that for every integer $n\ge 1$, the $n$-torsion subgroup $B[n]$ is finite.

Let $B_{\rm tor} = \varinjlim_{n\ge 1} B[n]$ be the torsion subgroup of $B$.

Is it true that, necessarily, there exists an integer $d\ge 0$ such that

$$B_{\rm tor} \simeq (\mathbf{Q}/\mathbf{Z})^d\oplus F,$$ for $F$ a finite group?

What if we replace $B_{\rm tor}$ by $B[\ell^{\infty}] = \varinjlim_{n\ge 1} B[\ell^n]$ for a single prime $\ell$, and $(\mathbf{Q}/\mathbf{Z})^d\oplus F$ by $(\mathbf{Q}_{\ell}/\mathbf{Z}_{\ell})^d\oplus F_{\ell}$ for $F_{\ell}$ a finite $\ell$-group?

A "direct proof" that $\mathrm{Spa}(A,A^+)$ is quasicompact?

Mon, 02/19/2018 - 15:49

When Huber introduces affinoid rings in his paper titled Continuous Valuations, he shows that for an affinoid ring $(A,A^+)$, now more commonly called a Huber pair, the spectrum $\mathrm{Spa}(A,A^+)$ is spectral, and quasicompactness follows. He does the former in (approximately) the following steps:

(i) show that $\mathrm{Spv}(A)$ is spectral for any ring $A$,

(ii) show that $\mathrm{Spv}(A,I)$ is spectral for certain ideals $I$, by showing it is a pro-constructible subset of $\mathrm{Spv}(A)$,

(iii) show that $\mathrm{Spa}(A,A^+)$ is a pro-constructible subset of $\mathrm{Spv}(A,I)$ where $I$ is the ideal in $A$ generated by topologically nilpotent elements.

I'm wondering if anybody knows of a more direct way to show the quasicompactness of $\mathrm{Spa}(A,A^+)$, if one didn't want to define the spaces $\mathrm{Spv}(A)$ and $\mathrm{Spv}(A,I)$? Possibly even avoiding arguments involving constructible subsets (though this seems less likely)?

Question about the existence a line which separates a portion of points

Mon, 02/19/2018 - 15:01

Suppose we have a point set $S_1$ of size $n$ above $x$-axis and another point set $S_2$ of size $n$ below $x$-axis. No three points are on the same line. Let $n_1, n_2 \leq n$.Is there a line containing a point in $S_1$ and a point in $S_2,$ so that it has $n_1$ points of $S_1$ and $n_2$ points of $S_2$ on are it’s left side?

I think there is a line such that it has $n_1$ points of $S_1$ and $n_2$ points of $S_2$ on it’s left side, but I do not know how to prove this line contains a point of $S_1$ and a point of $S_2.$ Is there any theorem that I can use it for this problem? I appreciate any help/hint!

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