Saturday, May 28, 2022

Recent Questions - Mathematics Stack Exchange

Recent Questions - Mathematics Stack Exchange

Homological Algebra - Prove the $\ker g^* \neq g^*(M\otimes_R A)$

Posted: 28 May 2022 06:08 AM PDT

Let $M$ be a right $R$-module and consider a left exact sequence or left $R$-modules $$0\longrightarrow A\stackrel{f}\longrightarrow B\stackrel{g}\longrightarrow C.$$

Show that in general $\ker g^* \neq g^*(M\otimes_R A)$ that is, the sequence

$$M\otimes _RA\xrightarrow{g^*}M\otimes_RB \xrightarrow{g^*}M\otimes_RC$$

is not exact.

Hint: Apply $\mathbb Z_2\otimes_\mathbb Z-$ to the left exact sequence $0\longrightarrow \mathbb Z \longrightarrow\mathbb Q \longrightarrow \mathbb Q/\mathbb Z$

About Chern classes of symmetric powers

Posted: 28 May 2022 06:02 AM PDT

I have the following question

I'm training in computation of Chern classes. Let $\xi$, $\eta$ are complex vector bundles of rank two ($r=2$)

I'm trying to find $c_1(S^2 \xi \otimes \eta)$, $c_2(S^2 \xi \otimes \eta)$, $c_3(S^2 \xi \otimes \eta)$ where $S$ is a symmetric power in terms of Chern classes of $\eta$ and $\xi$.

My attempt is the following: in general case $c_1(E\otimes F) = rc_1(F) + sc_1(E)$ where $r$ and $s$ are ranks of vector bundles. In our case we have $c_1(S^2 \xi \otimes \eta)=2c_1(S^2 \xi)+2c_1(\eta)$ and I don't know how to define $c_1(S^2 \xi)$ in terms of chern classes of $\xi$ vector bundle.

For the second chern class in general case for vector bundles of ranks $s$ and $n$ we have the following equation

$c_2(E\otimes F) = rc_2(F) + sc_2(E) + \binom{r}{2}c_1(F)^2 + \binom{s}{2}c_1(E)^2 + (rs-1)c_1(E)c_1(F)$.

In our case we have obtained that

$c_2(S^2 \xi \otimes \eta)=2c_2(S^2 \xi)+2c_2(\eta)+c_1(\eta)^2+c_1(S^2 \xi)^2+3c_1(S^2 \xi)c_1(\eta)$

For the third Chern class we have that

$$c_3(E\otimes F)=3\textstyle\binom{s}{3}c_1(E)^3+ 3\binom{r}{3}c_1(F)^3 + 6\binom{s}{2}c_1(E)c_2(E)+ 6\binom{r}{2}c_1(F)c_2(F)\\ + 3sc_3(E) + 3rc_3(F)+3(rs-2)c_2(E)c_1(F)+3(rs-2)c_1(E)c_2(F)\\ +\left(\tfrac{3}{2}rs - 1\right)(s-1)c_1(E)^2c_1(F) + \left(\tfrac{3}{2}rs-1\right)(r-1)c_1(E)c_1(F)^2.$$

In our case for ranks $r=s=2$ we have obtained that

$$c_3(S^2 \xi \otimes \eta)=\textstyle c_1(S^2 \xi)^3+ c_1(\eta)^3\\ +6c_1(S^2 \xi)c_2(S^2 \xi) +6c_1(\eta)c_2(\eta) + 6c_3(S^2 \xi) +6c_3(\eta)+6c_2(S^2 \xi)c_1(\eta)+6c_1(S^2 \xi)c_2(\eta) +5c_1(S^2 \xi)^2c_1(\eta) + 5c_1(S^2 \xi)c_1(\eta)^2.$$

I hope that i'm right in these computations. I don't know how to define $c_1(S^2\xi)$, $c_2(S^2\xi)$, $c_3(S^2\xi)$ in terms of Chern classes of $\xi$ bundle. Can you help me please and explain it in more details if you don't mind. I just know from Bott Tu book that $c(S^{p} E)=\prod_{1 \leq i_1 \leq i_2 \leq \ldots \leq i_p \leq n} (1+x_{i_1}+\ldots+x_{i_p})$ and I don't know how to apply this formula for computation these classes. Also I want to know what happened in the case $c_1(\Lambda^2\xi)$, $c_2(\Lambda^2\xi)$, $c_3(\Lambda^2\xi)$ where $\Lambda$ is an exterior power

Thank you!

Is the ideal of a subscheme of a toric variety B-saturated?

Posted: 28 May 2022 05:54 AM PDT

Let $X$ be a normal toric variety over $\mathbb{C}$ with no torus factors. Let $S$ be the Cox ring of $X$, and $B \subset S$ be the irrelevant ideal. Given an ideal sheaf $\mathcal{I} \subseteq \mathcal{O}_X $, we can define the $S$-module $\Gamma_{\star}(\mathcal{I})$ by

$$ \Gamma_{\star}(\mathcal{I}) = \bigoplus_{\alpha \in \operatorname{Cl} X} \Gamma(X, \mathcal{I}(\alpha)) $$

Then we can define the ideal $I$ corresponding to $\mathcal{I}$ as the image of the standard map

$$ \varphi : \Gamma_{\star}(\mathcal{I}) \to \Gamma_{\star}(\mathcal{O}_X) = S $$ My question is: is the ideal $I$ always $B$-saturated? In other words, is it always true that $I = I: B$? This is definitely true for projective space, and, more generally, for smooth toric varieties. But is it true for singular toric varieties $X$? The problem is that the map $\varphi$ does not have to be injective.

Prove Bézout's lemma for $\mathbb{Z}[\sqrt{-19}]$

Posted: 28 May 2022 06:00 AM PDT

It is well-known that the Bézouts lemma can be established in any Euclidean domain. (The proof is based on Euclidean division.)

In some domains, Bézout's lemma still appears to be true despite being non-Euclidean.

Clearly, $\mathbb{Z}[\sqrt{-19}]$ is not Euclidean since it is not a UFD: $20=2*10=(1+\sqrt{-19})(1-\sqrt{-19})$, or even in general: Let $d\equiv 1\pmod{2}$ be a non-square; then $d+1=2*(d+1)/2=(1+\sqrt{-d})(1-\sqrt{-d}), d-1=2*(d-1)/2=(\sqrt{d}-1)(\sqrt{d}+1)$, so neither $\mathbb{Z}[\sqrt{-d}]$ nor $\mathbb{Z}[\sqrt{d}]$ are Euclidean.

How do you establish Bézout's lemma without referring to Euclidean division?

In particular, how do you prove Bézout's lemma for $\mathbb{Z}[\sqrt{-19}]$?

True location parameter in case of known data, does trimmed mean estimate the mean?

Posted: 28 May 2022 05:41 AM PDT

So, I've got confused about the case where there is a known dataset. Does sample mean unambiguously define the location parameter of the data? And if so, does sample trimmed mean estimate the mean? Because in case of simulations, the sample trimmed mean estimates the population trimmed mean, but in case the data is known, what it is, that the trimmed mean estimates?

$\mathbb{Q}/\mathbb{Z}$ decomposition and possible generalization

Posted: 28 May 2022 05:38 AM PDT

I was wondering whether $\mathbb{Q}/\mathbb{Z}$ can be decomposed into a direct sum of its subgroups and I thought of the following decomposition: $$\mathbb{Q}/\mathbb{Z} = \bigoplus_{p \text{ prime}}\left(\mathbb{Z}\left[\frac{1}{p}\right]\Big/\mathbb{Z}\right).$$

The isomorphism is quite obvious as via CRT any fraction $\frac{m}{p_1^{\alpha_1}...p_k^{\alpha_k}}$ can be uniquely written as a sum $\frac{m_1}{p_1^{\alpha_1}} + ... + \frac{m_k}{p_1^{\alpha_k}} \:\mathrm{mod}\,\mathbb{Z}$. Firstly, I want to ask: is this reasoning correct and this is indeed how $\mathbb{Q}/\mathbb{Z}$ is decomposed?

Afterwards, I got a feeling that this somehow seems like something more general and not linked to this particular example. I figured that each summand in the decomposition above is in fact a localization of $\mathbb{Q}/\mathbb{Z}$ at a prime ideal $(p) \subset \mathbb{Z}$. So, if we consider $M = \mathbb{Q}/\mathbb{Z}$ as an $R$-module where $R = \mathbb{Z}$, we get the following decomposition: \begin{align}M = \bigoplus_{\mathfrak{p} \in \mathrm{Spec}R \setminus \{(0)\}} M_\mathfrak{p}.\end{align}

So, my question is quite vague but I am too curious not to ask: $$\textbf{under which conditions can a module be decomposed in the above fashion?}$$

For finitely generated modules over principal ideal domains to me this feels very much Structure Theorem-ish. That is, we know that if $M$ is a f.g. module over a principal domain $R$ it can be decomposed as $$M = R^{\oplus s} \oplus R/(p_1^{\alpha_{11}}) \oplus ... \oplus R/(p_1^{\alpha_{1r_1}}) \oplus ... \oplus R/(p_k^{\alpha_{k1}}) \oplus ... \oplus R/(p_k^{\alpha_{kr_k}}).$$

Then the localization of $M$ at a prime ideal $(p_i)$ is $R_{(p_i)}^{\oplus s} \oplus R_{(p_i)}/(p_i^{\alpha_{i1}})R_{(p_i)} \oplus ... \oplus R_{(p_i)}/(p_i^{\alpha_{ir_i}})R_{(p_i)}$.

So to me it seems that the necessary condition for the decomposition to work is

1. that the module is equal to its torsion submodule;

also it seems that we also have to require that

2. for each prime ideal $\mathfrak{p}$ in $R$ and each nonnegative integer $\alpha$ we have equality $R/\mathfrak{p}^\alpha = R_\mathfrak{p}/\mathfrak{p}^\alpha R_\mathfrak{p}$.

Under which conditions does the second assumption hold? Will this two conditions be sufficient for a desired decomposition to occur, not necessarily if $M$ is f.g. and $R$ is a PID? Under which conditions such decomposition is possible?

I would really appreciate any help or advice. Thank you in advance!

And I apologize if my question is flawed in some way. I tried to express my understanding of the problem as much as possible and I hope I haven't missed anything.

Chebyshev inequality for poisson

Posted: 28 May 2022 05:38 AM PDT

A mail order company reviews telephone orders at a constant rate of thee per hour.

Let S be the total number of telephone orders in a 75 hour period, assuming each number of orders in each hour is independent, calculate the following:

  1. Mean and variance
  2. Chebyshev's inequality to calculate a lower bound for $P(200 \le S \le 250)$

I am slightly caught aback by the wording, but from what I understand the expectation of a poisson distribution is $\text{E(Y)} = \lambda$ and the variance of a poisson is $\text{Var(Y)}=\lambda^2$.

Therefore, $E(X) = 75\cdot Y \implies E(X) = 75\cdot E(Y) \implies E(X) = 75 \cdot \lambda \text{ }$ and $Var(X) = 75 \cdot Y \implies Var(X)=75^2\cdot Var(Y) \implies Var(X) = 75^2 \cdot \lambda^2$

In this case we have $\lambda = 3$ this implies that $E(X) = 225$, and $Var(X) = 50625$

For the chebyshev inequality we have the following:

$$P(|X-\mu| \ge k\sigma) \le\frac{1}{k^2}$$

We know that $\mu = 225$ and $\sigma = \sqrt{50625}$

$$\begin{align} &P(-25 \le S-225 \le 25)\\ &P(|S-225| \le 25) \end{align}$$

Then we calculate $k\sigma = 25$, therefore $$\frac{1}{k^2} = \frac{\sigma^2}{25^2} = \frac{50625}{625} = 81$$

However this does not seem right to me as $k$ is way too high. Perhaps I went wrong with my calculating with the variance?

What is the ramification index of $K(\zeta_m, \sqrt[n]{a})/\mathbb{Q}_p$?

Posted: 28 May 2022 05:38 AM PDT

Let $K$ be an unramified extension of degree $n$ over $\mathbb{Q}_p$.

Then $K=\mathbb{Q}_p(\zeta_n)$ such that $ p \nmid n$, where $\zeta_n$ is $n^{th}$ root of unity.

I want to calculate the ramification index of $K(\zeta_{p^m}, \sqrt[p^m]{p^2})$ over $K$.

Since $\zeta_{p^m}$ is $p^m$-th root of unity, $K(\zeta_{p^m})$ is totally ramified of $\phi(p^m)$, Euler phi-function.

On the other hand, $K(\sqrt[p^m]{p^2})$ is totally ramified over $K$ of degree $p^m$.

So $K(\zeta_{p^m}, \sqrt[p^m]{-p^2})$ is totally ramified over $K$ of degree $\phi(p^m)p^m$ and this is the ramification index, I think.

But there is another aspect, $K=\mathbb{Q}_p(\zeta_n)$ might contain some $p^m$-th roots of unity.

This is why I think the above calculation of ramification index of $K(\zeta_{p^m}, \sqrt[p^m]{p^2})$ over $K$ is not accurate.

I appreciate your help.

Does the following differential equation have a unique solution?

Posted: 28 May 2022 06:04 AM PDT

In a sufficiently small neighborhood around $x=2$ the differential equation

$\dfrac{dy}{dx} = \dfrac{y}{\sqrt{x}} , y(2)=4$ has a unique solution.

How do I prove this?

My attempt has so far been to fin the solution of the differential equation as

$$\log(y^2) = 2\sqrt{x} + 2\log(4) - 2\sqrt{2}$$

Is this solution unique? In the neighborhood of $2$ and what happens if we choose some other value of $x$?

$(u_n,v)_H\to C_v$ implies weak convergence?

Posted: 28 May 2022 06:04 AM PDT

Suppose $(u_n,v)_H$ converges for all $v\in H$, where $H$ is a Hibert space. Does this imply the existance of $u\in H$ such that ${u_n}_\overrightarrow{w}u$?

$U \subseteq W$ >> $ W^\perp \subseteq U^\perp$

Posted: 28 May 2022 05:33 AM PDT

$U \subseteq W$ >> $ W^\perp \subseteq U^\perp$ Question regarding this:
My exerciser at the study did this proof while assuming this:
$U+W=R^n$, so, need to prove: $U^\perp+W^\perp=R^n$. and that prove the problem at the post.
Why is it like that? and how should I think that if I see such a thing, I need to prove something else for it? and why proving this is equal?

What is the probability that randomly selected $2$ pets from this vet are cat or sick? Why doesnt addition rule for probability work?

Posted: 28 May 2022 06:06 AM PDT

There are $7$ dogs and $6$ cats in a vet. $2$ of the dogs and $2$ of the cats are sick.What is the probability that randomly selected $2$ pets from this vet are cat or sick ?

My approach : Let the set of cats (sick and normal cats)called "A" and the set of sick pets (sick dogs and cats) called "B" ,then $$P(A \cup B)=P(A)+P(B)-P(A \cap B)=\frac{\binom{6}{2}}{\binom{13}{2}}+\frac{\binom{4}{2}}{\binom{13}{2}}-\frac{\binom{2}{2}}{\binom{13}{2}}=20/78$$

My friends'approach : The set of $A \cup B$ have $8$ elements in it , so the answer is $$\frac{\binom{8}{2}}{\binom{13}{2}}=28/78$$

Who is right here ? .. Can you help me ?

Addentum : By comments of @L.F. ,I saw that my calculation does not contain the probability of selecting a normal cat and a sick dog. However , i cannot understand why , i applied addition rule of probability. Moreover , related question in this page confirms me.

What are some common methods to check if a sums over a subset of natural numbers have a closed form?

Posted: 28 May 2022 05:58 AM PDT

I have found many questions from people here which involve some crazy sums or expressions. Sometimes they are solvable, sometimes they aren't. If the expression is infact an integral then it's solvability can be easily checked on online sites like Wolfram Alpha or Integral Calculator. But, how about discrete expression, perhaps one involving binomial sums? How could one check if it has a closed form expression or an identity relating to it?

Limit of sequence with n! and nth root

Posted: 28 May 2022 05:59 AM PDT

Can anyone give a hint for solving this: $\lim\limits_{n\rightarrow\infty}\frac{1}{n^2}\sqrt[n]{\frac{(3n)!}{n!}}$. This sequence should converge to $\frac{27}{\mathrm{e}^2}$. I have no clue how to prove that in simple calculus. Is there anyone who can advise?

Is this regression equation consistently estimated?

Posted: 28 May 2022 05:47 AM PDT

Consider the structural system of equations where the where Y variables are endogenous, and the X variables are exogenous. The errors may be correlated contemporaneously between equations but not over time either within or between equations. The errors have a mean of 0 but different variances. \begin{equation}Y_{1t}= \beta_{11} + \beta_{12}Y_{2t} +\beta_{13}Y_{3t}+\beta_{14}X_{1t} +\beta_{15}X_{2t} + \epsilon_{1t} \end{equation}. \begin{equation}Y_{1t}= \beta_{21} + \beta_{22}Y_{3t} +\beta_{23}X_{1t}+\epsilon_{2t} \end{equation}. \begin{equation}Y_{1t}= \beta_{31} + \beta_{32}Y_{1t}+ \epsilon_{3t} \end{equation} where you can rewrite the structural equations as: \begin{equation}Y_{1t} - \beta_{12}Y_{2t} -\beta_{13}Y_{3t}= \beta_{11} +\beta_{14}X_{1t} +\beta_{15}X_{2t} + \epsilon_{1t} \end{equation}. \begin{equation}Y_{1t}- \beta_{22}Y_{3t} = \beta_{21} +\beta_{23}X_{1t}+\epsilon_{2t} \end{equation}. \begin{equation}Y_{1t}-\beta_{32}Y_{1t}= \beta_{31}+ \epsilon_{3t} \end{equation} Question: Please explain why/why not each 3 of these equations can be estimated consistently by OLS.

$f:R\to S$ is finite injective and $S$ is of dim $1$

Posted: 28 May 2022 05:34 AM PDT

Let $R$ be a local noetherian domain of dim $3$. I'm asked to construct examples $f:R\to S$ with the following properties:
1- $f$ is onto and $S$ is of dim $1$.
2- $f$ is injective and $S$ is of dim $1$.
3- $f$ is finite (as R-algebra) injective, and $S$ is of dim $1$.

  • For the 1st, I take $f:R\to S=R/p$ , where $p$ is of codim one.
  • For the 2nd, I take $f:R\to S=k[x]$ , where $k$ is quotient field of $R$.
  • For the 3rd, I dont know if it is possible or not? can you give a hint please?

Many thanks.

Given Array A. Let B be approximate array of A. Prove that If $A'[n+1]\ge B[n]$, then $B'[i] = B[i] (1\le i\le n)$ & $B'[n+1] = A'[n+1]$ hold.

Posted: 28 May 2022 05:50 AM PDT

Given Arrays of rational numbers. Let A[i] denote the $i^{th}$ element of an array $A$. For an array $A$ of length $n$, a monotone non-decreasing array $B$ of length $n$ that minimizes $\sum(A[i]-B[i])^2$ is called an approximate array of $A$. And $B$ is monotone non-decreasing. Let $A'$ be the array made by appending an element at the tail of array $A$ of length $n$. That is, $A'[i] = A[i](1\le i\le n)$. Let $B$ and $B'$ be approximate arrays of $A$ and $A'$.

Prove that If $A'[n+1]\ge B[n]$, then $B'[i] = B[i]\;\; (1\le i\le n)$ and $B'[n+1] = A'[n+1]$ hold.

Assume that B[1]=B[2]=...=B[n] and A'[n+1] < B[n]. Show that B'[1]=B'[2]=...=B'[n+1] and B'[n+1] < B[n], and find the value of B'[n+1].

Can this be proved by induction? I have no idea about $A'[n+1]\ge B[n]$

Are there functions which take themselves as input?

Posted: 28 May 2022 05:52 AM PDT

Are there functions which takes itself as an argument? Like: $f(x,f(x_{n\pm1}))$

A functions which value at a given point depends on its value on the next,or previous intake value. Or: $f(x,f(x))$.

A function which value at a given point depends on itself at that given point. This seems like nonsense because it should be: $f(x,f(x,f(x,f(x....))$

Number of functions mapping consecutive numbers to consecutive letters .

Posted: 28 May 2022 06:00 AM PDT

Let $N=\{1,2, \ldots, 9\}$ and $L=\{a, b, c\}$ which if the following is correct ?

  • $L \cup N$ is arranged on a line with the letters appearing consecutively (in any order). The number of such arrangements are less than $10 ! \times 5$.
  • More than half of the functions from $N$ to $L$ have $b$ in their range.
  • The number of one-to-one functions from $L$ to $N$ is less than 512 .
  • The number of functions $N$ to $L$ that do not map consecutive numbers to consecutive letters is greater than $512 .$

i have checked the first three parts as such for first one its false since it should be haviing 3! = 6 instead of 5 there for total permutations of L elements . Next one is simply $3^9 - 2^9$ by complementary couting . Third one is the $\binom{9}{3} 3!$ (less than 512) , next the final option i am not getting the statement at all , i think(for complementary counting) it means we must have all terms having consecutive letters as output but there are only 3 letters so how can it be possible ? Or is that means answer is same as total number of functions itself ?

Learning to translate natural-language phrases to formal logic

Posted: 28 May 2022 05:48 AM PDT

In natural language, we often use phrases like:

  1. "fix arbitrary"
  2. "such that"
  3. "we have"

In general, how to translate those phrases into formal logic?

One example condition is:

Fix arbitrary $x,y$. For all $f\in A$ such that $f(x)=0$, we always have $f(y)\geq0$, or we always have $f(y)\leq 0$.

Extending a certain bound to the von Neumann tensor product

Posted: 28 May 2022 05:39 AM PDT

I am wondering whether the following is true: Let $H_1, H_2$ denote two Hilbert spaces and let $$\phi_i\colon B(H_1)\overline{\otimes}B(H_2)\to B(H_2)$$ be ultraweakly continuous, completely positive linear maps for $i\in \{1,2\}$.

Assume now that there exists $\varepsilon>0$ such that $\Vert \phi_1(x)-\phi_2(x)\Vert\leq \Vert x\Vert \varepsilon$ for every $x$ in the algebraic tensor product $B(H_1)\odot B(H_2)$. Can we then conclude that $\Vert \phi_1(y)-\phi_2(y)\Vert\leq \Vert y\Vert \varepsilon$ for every $y\in B(H_1)\overline{\otimes}B(H_2)$?

My guess: by boundedness of $\phi_1-\phi_2$, we can extend the above inequality to hold true for elements in the minimal tensor product $B(H_1)\otimes B(H_2)$. Then, ultraweak continuity and a possible application of Kaplansky (?) takes us home.

I am unsure about all of this and would love some help.

Need Help in Reasoning with Extremal Points

Posted: 28 May 2022 05:40 AM PDT

There is a common logic I found in quite a few books and papers, which sounds intuitive, but I am unable to comprehend conceretely. The statement considers the following set-up in general:

Let $f:K\rightarrow \mathbb{C}$ be a continuous function on a locally compact Hausdorff space $K$ such that $|f(x)|\leq 1$ for all $x\in K$, and $\mu$ is a regular Borel probability measure on $K$ (i.e, $\mu\geq 0, \mu(K)=1$), whose support is compact. It is given that $\int_K f(x) d\mu(x) = \alpha$, where $|\alpha|=1$.

Then the author(s) state that "Since any point on the unit circle is an extreme point, and $|\int_K f \ d\mu|=1$, we must have that $f(x)=\alpha$ for each $x$ in the support of $\mu$."

I do not see it immediately, any help is greatly appreciated :)

Estimating two unknows from a set of linear equations

Posted: 28 May 2022 05:36 AM PDT

I have a set of approximate linear equations: $$P_i = A * F_i + B * T_i$$ And want to estimate constants A & B.

Some actual values are given below:

  50.0 = A * 266.0 + B * 1020.3  50.0 = A * 247.0 + B * 957.8  50.0 = A * 215.0 + B * 903.2  50.0 = A * 182.0 + B * 855.4  50.0 = A * 162.0 + B * 813.4  50.0 = A * 147.0 + B * 775.8  50.0 = A * 126.0 + B * 744.2  50.0 = A * 109.0 + B * 715.5  50.0 = A * 99.0 + B * 691.6  50.0 = A * 81.0 + B * 670.9  50.0 = A * 72.0 + B * 652.1  50.0 = A * 64.0 + B * 637.1  50.0 = A * 50.0 + B * 623.2  50.0 = A * 45.0 + B * 609.9  50.0 = A * 44.0 + B * 598.7  50.0 = A * 38.0 + B * 589.6  50.0 = A * 32.0 + B * 582.1  50.0 = A * 28.0 + B * 574.6  50.0 = A * 25.0 + B * 568.7

From this answer: Estimating two constants from fitted equation, I believe I can use a least squares approach, however I do not know how to go about this. I hope someone could point me in the right direction.

Much appreciated

Douglas Bayford

How to construct the Green's function solution for the barotropic Rossby wave eq with Dirichlet b.c.?

Posted: 28 May 2022 06:01 AM PDT

The linear differential equation I have been working on is the barotropic Rossby wave equation: $$ L\psi=(\frac{\partial}{\partial t}\nabla^2+\beta\frac{\partial}{\partial x} )\psi=\delta(\vec{x}-\vec{x}_0)\delta(t) $$ where $\psi(x,y,t)$ is the two-dimensional stream function varying with time $t$ and we are going to solve for it, $\beta$ is a real constant, $\nabla^2=\partial^2/\partial x^2+\partial^2/\partial y^2$ is two-dimensional Laplace operator, $\delta$ is the Dirac-delta function representing an impluse located at $\vec{x}_0$ in a 2D-plane $(x,y)$ at the initial moment $t=0$ and it forces the barotropic Rossby wave equation.

Actually, the green function solution for the barotropic Rossby wave equation (in a compact form) has been found by

Kloosterziel, R. C. & Maas, L. R. M. Green's functions for Rossby waves. J. Fluid Mech. 830, 387–407 (2017).

Here I put their solution (their eq. 1.3) down: $$\psi(x,y,t)=G_\delta=\frac{H(t)}{4}(J_0(z^+)Y_0(z^-)+J_0(z^-)Y_0(z^+))$$ where $H(t)$ the Heaviside unit step function, $J_0$ and $Y_0$ are the first- and second- kind Bessel functions of zero-order respectively, and $z^{\pm}=\sqrt{x\pm yi}$ with $i^2=-1$. This solution is however only for a infinitely large horizontal plane without considering any boundary conditions, so one may call it the fundamental Green's function solution.

When a solid boundary(impenetrable) is placed at the y axis, $x=0$, the imperetrable boundary condition(no normal flow into/out of the boundary) requests $\psi(x=0,y,t)=0$, i.e. the Dirichlet boundary. My question is how to construct the Dirichlet Green function solution using the above fundamental Green-function solution for the positive-x half plane: $\{ (x,y,t)\in \mathbb{R}^3 : x\in[0,\infty),y\in(-\infty,\infty),t\in[0,\infty) \}$, which satisfies the Dirichlet b.c.?**

I am wondering about using the method of mirror image, as inspired by

Haine, T. W. N. & Fuller, A. Boundary β-plumes and their vorticity budgets. Q. J. R. Meteorol. Soc. 142, 2758–2767 (2016). They solved for a steady beta-plume bounded by a west wall and their governning equation $$\beta\psi+\lambda\nabla^2\psi=-\delta(\vec{x}-\vec{x}_0)$$ contains no time derivative terms, thus it is very different case from ours. Here, $\lambda$ is a real constant representing friction coefficient.

I guess the Dirichlet Green's function solution can be written as some kind of $G_\delta-G_\delta^{\dagger}$, with $G^{\dagger}_\delta$ denoting the image (adjoint) Green's function solution for $$L^{\dagger}G_\delta^{\dagger}=\delta(\vec{x}-\vec{x}_0)\delta(t)$$ Here $L^\dagger=\frac{\partial}{\partial t}\nabla^2-\beta\frac{\partial}{\partial x}$ is the adjoint operator of $L$. Obviously, $L$ is not self adjoint since $L\ne L^{\dagger}$; the $\beta$-term changes sign in $L^{\dagger}$. Physically, the sign change means that $G_\delta^{\dagger}$ actually describes the Rossby waves radiated from the image source but propagating to the east(positive-x direction, while the original Rossby waves are westward). If the image source/Dirac function locates at $\vec{x}_0=(-x_0,y_0)$ such that it is a mirror reflection to the true souce at $\vec{x}_0=(x_0,y_0)$, the waves radiated from the two point sources will collide at $x=0$ and "compensate" for each other, then the boundary condition $\psi(x=0,y,t)=0$ is satisfied.

However, $$L(G_\delta-G_\delta^\dagger)=\delta(\vec{x}-\vec{x}_0)\delta(t) $$ cannot be satisfied in the right half plane, since $L(G_\delta^\dagger) \ne 0$ there. $G_\delta-G_\delta^\dagger$ cannot be the Dirichlet Green's function solution to our problem.

Then how to crack the problem? any help would be very appreciated!

Explanation for a part of this proof about the existence of a subgroup isomorphic to the quotient group

Posted: 28 May 2022 05:57 AM PDT

My question is about an exercise and its solution from a Chinese textbook Introduction to Modern Algebra (《近世代数引论》), which I translate into English as follows so that everyone will understand:

1.10.7 $\,$ Assume that $H$ is a subgroup of a finite abelian group $A$, prove that there exists a subgroup of $A$ isomorphic to $A/H$.
Proof: Write $A=G_1\oplus\dots\oplus G_m$, where $G_i$ is a Sylow $p$-subgroup of $A$ for $i=1,\dots,m$. By Exercise 1.9.5$^{[1]}$ we know $$H=H_1\oplus\dots\oplus H_m,\,H_i=H\cap G_i,\,1\leq i\leq m.$$ $\\$Since$$A/H\cong G_1/H_1\oplus\dots\oplus G_m/H_m,$$ it suffices to prove the existence of a subgroup of $G_i$ isomorpic to $G_i/H_i$. That is, without loss of generality, from now we assume $|A| = p^n,n\geq 1,|H|=p^m,m\leq n$.
By the structure theorem of abelian $p$-groups we know $$A=\mathbb{Z}_{p^{r_{\,1}}}\oplus\dots\oplus\mathbb{Z}_{p^{r_{\,t}}},\,\,\,\,H=\mathbb{Z}_{p^{s_{\,1}}}\oplus\dots\oplus\mathbb{Z}_{p^{s_{\,t}}},$$ $\\$where $\mathbb{Z}_{p^{r_{\,i}}}\leq\mathbb{Z}_{p^{s_{\,i}}},1\leq i\leq t,r_1+\dots+r_t=n,s_1+\dots+s_t=m$, where $r_i$ are positive integers and $s_i$ are nonnegative integers with $s_i\leq r_i$. Thus $$A/H\cong \mathbb{Z}_{p^{r_{\,1}-s_{\,1}}}\oplus\dots\oplus\mathbb{Z}_{p^{r_{\,t}-s_{\,t}}}\leq A.$$ (End of proof)$^{[2]}$

My question is about the part where $A$ is assumed to be a $p$-group. I realized that, given an arbitrary decomposition of $A$, there does not always exist a decomposition of $H$ compatible to that of $A$.
As a counterexample, set $A=\mathbb{V}=\{(1),(12)(34),(13)(24),(14)(23)\}$ and $H=\{(1),(12)(34)\}$. Denote by $A_1$ and $A_2$ respectively the subgroups $\{(1),(13)(24)\}$ and $\{(1),(14)(23)\}$, we know $A=A_1\oplus A_2$, but there cannot be $H_1\leq A_1$ and $H_2\leq A_2$ such that $H=H_1\oplus H_2$ because both $H\cap A_1$ and $H\cap A_2$ are $\{(1)\}$.

I wonder if we can choose a particular decomposition of $A$ such that a compatible decomposition of $H$ always exists? Or maybe the author actually mean $H\cong\mathbb{Z}_{p^{s_{\,1}}}\oplus\dots\oplus\mathbb{Z}_{p^{s_{\,t}}}$ instead of $H=\mathbb{Z}_{p^{s_{\,1}}}\oplus\dots\oplus\mathbb{Z}_{p^{s_{\,t}}}$ in order that $\mathbb{Z}_{p^{r_{\,i}}}\leq\mathbb{Z}_{p^{s_{\,i}}}$? It seems that if any of the assertions above is true then the conclusion follows, but I can prove neither of them. Can anyone show me if the above two assertions can be proved or disproved? If the above two assertion is not the case, is there a way to correct or improve the proof so that it will work out? Thanks in advance!

Remark
$[1]$ which asserts that if $G=G_1\oplus\dots\oplus G_m$, and if given any pair of $i,j,1\leq i\lt j\leq n,|G_i|$ and $|G_j|$ are relatively prime, then any subgroup $H$ of $G$ is direct product of its subgroups $H\cap G$.
$[2]$ The original Chinese texts of the exercise and the proof are as follows:

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Edit:

I try to elaborate on Brauer Suzuki's hints, but stuck somewhere. Here are my attempts, which is made by counting the number of direct summands of each type $C_{p^{k}}$ respectively for the decomposition of $H$ and $A$, so that we can attribute to each summand of $H$ a summand of the same type in $A$ without repetition.

Suppose that $G$ is a finite abelian $p$-group with decomposition $$G= C_{p^{k_1}}\oplus\dots\oplus C_{p^{k_n}}.$$ Denote $\{x^{p^n}:x\in G\}$ by $p^n G$ for $i=1,\dots,n$, and use the notation $U_{p}(m,G)$ for the number of direct summands of the type $C_{p^{m}}$ in the decomposition.

Suppose $$A= C_{p^{a_1}}\oplus\dots\oplus C_{p^{a_n}}.$$ By rearranging $C_{p^{a_{\,i}}}$ we may assume $a_1\leq a_2\leq\dots\leq a_n$. Similarly, for $H\leq A$ we have the decomposition into $$H= C_{p^{b_1}}\oplus\dots\oplus C_{p^{b_l}}$$ with $b_1\leq\dots\leq b_l$. Since the element of largest order in $A$ must have order $p^{a_n}$, we see $b_i\leq a_n$ for all $i=1,\dots,l$.

We count the number of direct summands of the type $C_{p^{a_{n}}}$, i.d. $U_p(a_n,H)$, in $$H= C_{p^{b_{1}}}\oplus\dots\oplus C_{p^{b_{l}}}.$$ Consider $$p^{a_n-1}H= p^{a_n-1}C_{p^{b_1}}\oplus\dots\oplus p^{a_n-1}C_{p^{b_l}}.$$ Any summand of type $p^{a_n-1}C_{p^{b_i}}$ with $b_i$ strictly less than $a_n$ will be $\{1\}$, while $p^{a_n-1}C_{p^{b_j}}\cong C_{p}$ for those $b_j=a_n$. Thus, $$U_p(a_n,H)=U_p(1,p^{a_n-1} H),$$ and $U_p(1,p^{a_n-1} H)$ can be calculated as $|p^{a_n-1}H|/p$, because all the non-trivial summands in the decomposition of $H$ must be of the same type $C_p$.

Similarly, we have $$U_p(a_n,A)=U_p(1,p^{a_n-1}A)=|p^{a_n-1}A|/p.$$ Since $p^n H\leq p^n A$, we have $$U_p(a_n,H)=|p^{a_n-1}H|/p\leq |p^{a_n-1}A|/p=U_p(a_n,A).$$ That is, for each direct summands of type $C_{p^{a_n}}$ (if there is any) in the decomposition of $H$, we have correspondently a summand of the same type in the decomposition of $A$.

Next we consider $U_p(a_n-1,H)$. Analogously, we multiplicate $H$ by $p^{a_n-2}$. Any summand which has order $p^{b_i}$ with $b_i\lt a_n-1$ in the original decomposition is transformed into $\{1\}$ by the left multiplication, and summands of order $p^{a_n-1}$ and $p^{a_n}$ become respectively isomorphic to $C_p$ and $C_{p^2}$.

Compared to the circumstance of $p^{a_n-1}H$, we construct the quotient group $p^{a_n-2}H/p^{a_n-1}H$, where both types become isomorphic to $C_p$, allowing us to calculate the number of type $p^{a_n-1}$ and $p^{a_n}$ by $|p^{a_n-2}H/p^{a_n-1}H|/p$. Similarly the number of type $p^{a_n-1}$ and $p^{a_n}$ in $A$ equals to $|p^{a_n-2}A/p^{a_n-1}A|/p$.

Here is exactly where I stuck. To compare the order of $p^{a_n-2}H/p^{a_n-1}H$ with the order of $p^{a_n-2}A/p^{a_n-1}A$, I try to construct an injection from the former into the later. It turns out this does not always work.

Now that we have an embedding $\iota: p^{a_n-2}H \to p^{a_n-2}A$, and since $p^{a_n-1}H \lhd p^{a_n-1}A$, there is an induced homomorphism $\phi: p^{a_n-2}H/p^{a_n-1}H \to p^{a_n-2}A/p^{a_n-1}A$, given by $p^{a_n-2}h+p^{a_n-1}H \mapsto p^{a_n-2}h+p^{a_n-1}A$.

But in order to make $\phi$ an injection, it seems $p^{a_n-2}H$ has to be a pure subgroup of $p^{a_n-2}A$, i.d. if an element of $p^{a_n-2}H$ can be expressed as $p^{a_n-1}a$ for some $a\in A$, then it can also be expressed as $p^{a_n-1}h$ for some $h\in H$. This does not always hold.

Unitors in star-autonomous categories

Posted: 28 May 2022 06:09 AM PDT

1.Context
Let $(C, \otimes, I, a, l,r)$ be a monoidal category. Suppose $S: C^{op} \xrightarrow{\sim} C$ is an equivalence of categories with inverse $S'$. Assume that there are bijections $\phi_{X,Y,Z}: Hom_C(X \otimes Y,SZ) \xrightarrow{\sim} Hom_C(X, S(Y \otimes Z))$ natural in $X,Y,Z$. (This makes $C$ a star-autonomous category.) Note that we do not assume that $S$ is a monoidal equivalence. For simplicity suppose that the associator $a$ is the identity and that $S$ and $S'$ are strict inverses.

2.Questions
Is the morphism $\phi_{B,I,S'B}^{-1}\big(S(l_{S'B})\big):B \otimes I \rightarrow SS'(B)=B$ equal to the right unitor $r_B$ for every $B \in C$?

Some ideas and follow-up questions:

  • Does $\phi$ map isomorphisms to isomorphisms? Maybe one can show that $\phi_{B,I,S'B}(r_B): B \rightarrow S(I \otimes S'B)$ is not in general an isomorphism. This would give a negative answer to the above question. The Yoneda lemma (covariant version) tells us something about the form of natural transformations between functors $F,Hom(-,X): C^{op} \rightarrow Set$. Can it be modified to cover natural transformations between functors $C^{op} \times C^{op} \times C^{op} \rightarrow Set$ as above?
  • Does there exist a property (for example the requirement that the unitor makes a certain diagram commute) that characterizes the left/right unitor uniquely? This would indicate a strategy for giving a positive answer: One could try to verify that $\phi_{B,I,S'B}^{-1}\big(S(l_{S'B})\big)$ satisfies the characteristic property.
  • What is a good place to look for counterexamples to the equality? I tried rigid monoidal categories – to no avail. For instance, consider the category of finite-dimensional vector spaces over a field $\mathbb F$ with usual tensor product. Set $S=S':=(-)^*$ the duality functor. Define $\phi$ as $\Big( \big(\phi_{X,Y,Z}(k)\big)(x) \Big)(y\otimes z):=\big(k(x\otimes y)\big)(z)$ for $k:X \otimes Y \rightarrow Z^*$. Then $\phi$ is natural in all three components. Denote by $i_X: X \rightarrow X^{**}$ the canonical identification of a finite-dimensional vector space $X$ with its double dual. One shows that $l_{X^*}^* \circ i_x=\phi_{X,\mathbb F, X^*}(i_x \circ r_x)$ for all finite-dimensional vector spaces $X$ over $\mathbb F$ by evaluating both sides first on an element $x \in X$ and then on simple tensors in $\mathbb F \otimes X^*$.

EDIT:
Using the naturality of $r$ and $\phi$ I was able to show that the above statement is equivalent to the claim that the following equality $$\phi_{S(I\otimes S'B), I,S'B}\big( S(l_{S'B}^{-1})\circ r_{S(I \otimes S'B)}\big)=id_{S(I\otimes S'B)}$$ holds. Maybe this observation is of help.

Possibly false theorem in a book. Quine's theorem proof. A.k.a. Blake's method for reducing boolean functions.

Posted: 28 May 2022 06:09 AM PDT

This is what I found in one text on discrete mathematics:

Theorem: given a full DNF of a boolean function, if we first simplify it using all identities of the type 1: $(x\land y)\lor (\bar{x}\land y) = y$ and $(x\lor y)\land (\bar{x}\lor y) = y$, and then apply all absorptions $(x\land y)\lor y = y$ and $(x\lor y)\land y = y$, we get a reduced DNF, namely the disjunction of all its prime implicates.

The text contained a "proof", but it was way too handwavy and actually similar to what I`ll describe below.

Here is my intuition of the proof: if we multiply a prime implicant by the expression of the type $(x\lor \bar{x})$, we can turn it into a disjunction of minterms. This way we can get a full DNF. Since then by applying to our full DNF operations of the type 1, which are inverse operations to multiplying by $(x\lor \bar{x})$, we should get all prime implicants. But there also will be some additional unwanted terms, that our absorbtions should take care of.

My question is how can I justify all the above steps and turn it into a complete rigorous proof?

Actually, I don`t even think the theorem as stated above is true at all. We probably need a more advanced identity - $$(a\land c)\lor (b\land \bar{c})=(a\land c)\lor (b\land \bar{c})\lor ab$$ for the first step instead of those of type 1 described in the theorem.

How to determine two matrices are conjugate.

Posted: 28 May 2022 05:58 AM PDT

Which of the following statements are true?

  • The matrices $ A=\left[ {\begin{array}{cc} 1 & 1 \\ 0 & 1\\ \end{array} } \right]$ and $ B=\left[ {\begin{array}{cc} 1 & 0 \\ 1 & 1\\ \end{array} } \right]$ are conjugate in $GL_2(\mathbb{R})$
  • The matrices $ A=\left[ {\begin{array}{cc} 1 & 1 \\ 0 & 1\\ \end{array} } \right]$ and $ B=\left[ {\begin{array}{cc} 1 & 0 \\ 1 & 1\\ \end{array} } \right]$ are conjugate in $SL_2(\mathbb{R})$
  • The matrices $ C=\left[ {\begin{array}{cc} 1 & 0 \\ 0 & 2\\ \end{array} } \right]$ and $ D=\left[ {\begin{array}{cc} 1 & 3 \\ 0 & 2\\ \end{array} } \right]$ are conjugate in $GL_2(\mathbb{R})$

I know the conjugate matrices have the same eigenvalues. But all of this matrices have the same eigenvalues. I am not sure how to determine that which matrices are conjugate to each other.

Discontinuity of Dirichlet function

Posted: 28 May 2022 05:34 AM PDT

Define $$f(x)= \begin{cases} 1, & \text{if }x\in\mathbb{Q}, \\ 0, & \text{if }x\in\mathbb{R}\setminus\mathbb{Q}. \end{cases}$$Then $f$ has a discontinuity of the second kind at every point $x$, since neither $f(x+)$ nor $f(x-)$ exists.

Proof: We'll consider only for $f(x+)$.

Case 1. If $x_0\in \mathbb{Q}$ then we can take $t_n=x_0+\frac{1}{n}$ at that $t_n\to x_0,t_n>x_0$ and $t_n\in \mathbb{Q}$. Hence $f(t_n)=1\to 1$ as $n\to \infty$.

Also we can take $t_n=x_0+\dfrac{\sqrt{2}}{n}$ at that $t_n\to x_0,t_n>x_0$ and $t_n\in \mathbb{R}\setminus\mathbb{Q}$. Hence $f(t_n)=0\to 0$ as $n\to \infty$

Case 2. For $x_0\in\mathbb{R}\setminus\mathbb{Q}$ we apply a similar argument.

We can take $t_n=x_0+\dfrac{1}{n}$ and in this case $f(t_n)\to 0$. Taking $t_n\in \mathbb{Q}$ such that $x_0<t_n<x_0+\dfrac{1}{n}$ we get $f(t_n)\to 1.$

Hence $f(x_0+)$ does not exists at any point $x_0\in \mathbb{R}$. Also $f(x_0-)$ does not exists at any point $x_0\in \mathbb{R}$.

Hence Dirichlet function has discontinuity of the second kind at every point of $\mathbb{R}^1$.

Is my proof true?

Rectilinear Motion (Calculus)

Posted: 28 May 2022 06:05 AM PDT

Problem: A body move vertically up from the earth according to $s= 64t-16t^2.$ Show that it has lost one-half its velocity in its first $48$ ft of rise.

My Answer: I really don't understand the problem and totally clueless on what to do but I try this:

$$V=64-16t$$

I let, $48=64t-16t^2,$ and I get $t=3.$

When $t=3,$

$$V= -32 \mbox{ ft./sec.}$$

But it's not $1/2.$ Please I really need to know this because tomorrow is the deadline of our homework and maybe I will be called for presentation of answer. This is the last problem on our homework set.
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