Quantum cup product and Dolbeault cohomology

Let $ X$ be a smooth projective variety over $ \mathbb{C}$ . We consider the small quantum cup product $ \star$ on the deRham cohomology ring $ \displaystyle H^*(X;\mathbb{C})=\bigoplus_{p,q}H^{p,q}(X)$ . Let $ H^{(i)}$ denote $ \displaystyle\bigoplus_{p-q=i}H^{p,q}(X)$ .

In ”Gamma Conjecture via Mirror Symmetry”, Galkin and Iritani claimed in the Appendix that $ H^{(i)}\star H^{(j)}\subseteq H^{(i+j)}$ in order to draw a conclusion about the spectrum of the operator $ c_1(X)\star$ acting on various subspaces of $ H^*(X)$ . They said this claim follows from the motivic axiom of the Gromov Witten theory. How should one argue following this line?

Thanks.

Remark: If $ X$ is convex, then $ \overline{\cal M}_{0,3}(X,\beta)$ is a smooth projective variety so the claim should be obvious. But in general we have to use virtual fundamental cycle to define $ \star$ . I don’t know what we should do then.

Factorizations as a product of primes minus one

Let $ x$ be a positive rational number. I am interested in factorizing $ x$ as a product of primes minus one. In fact, I would also like make sure the primes in the decomposition are distinct, and I want to impose congruence conditions on the primes (the application is too far afield to describe). Let me break down my main question (Q3) into some warm-ups. There are also some special cases following Q3.

Q1. Let $ x$ be a positive rational number. Are there collections

$ $ P = \{p_1, \dots, p_m\} $ $ $ $ Q = \{q_1, \dots, q_n\} $ $

of $ n+m$ odd prime numbers with $ P \cap Q = \emptyset$ such that

\begin{equation} \prod_{i = 1}^m (p_i – 1) = x \prod_{j = 1}^n (q_j – 1)? \tag{1} \end{equation}

We can call each such instance of (1) a factorization of $ x$ into a product of primes minus one. If the answer is yes, can $ x$ have infinitely many such factorizations?

Q2. Does Q1 still have a yes answer if we require the sets to be distinct primes, i.e., that $ P \cup Q$ is $ m+n$ distinct prime numbers? In other words, can we take each prime in the factorization to appear with multiplicity one? (Unlike the case of factorization into a product of primes, this is actually quite natural.)

My main question:

Q3. Does Q2 still have a yes answer if we put “legal” congruence constraints on the sets $ P$ and $ Q$ ? By “legal”, I mean that the congruence conditions don’t trivially make the equation impossible (e.g., a condition mod $ N$ that makes the equation unsolvable modulo N or some divisor/multiple of $ N$ ).

Example. Take $ x = 4$ with the condition that the primes in $ P$ are $ 3$ or $ 5$ mod $ 8$ and the primes in $ Q$ are $ 1$ or $ 7$ mod $ 8$ . Then $ (5-1) = 4$ , but also

$ $ (11-1)(13-1) = 4(31-1) $ $

so $ P = \{11, 13\}$ , $ Q = \{31\}$ works.

Moreover, suppose there is an integer $ k$ so that $ 13+16k$ and $ 31+40k$ are simultaneously prime. Then $ P = \{11, 13+16k\}$ , $ Q = \{31+40k\}$ also works. It follows from Dickson’s conjecture that infinitely many such $ k$ exist, but I believe this particular case is open. There are $ 27768$ such $ k$ between zero and a million, $ 211502$ up to ten million, and $ 1665924$ up to a hundred million. It’s pretty reasonable to guess this isn’t a counterexample to Dickson’s conjecture…

One could also perhaps motivate this question by recent work of Tao–Ziegler on polynomial patterns in the primes (and preceding results found in its references).

Some experimentation suggests a special case:

Q4. Can we always take $ P$ and $ Q$ to be of bounded cardinality to produce infinitely many distinct factorizations of $ x$ ?

For example, all positive integers $ x$ between $ 1$ and $ 100$ are of the form

$ $ (p-1)(q-1)=x(r-1) $ $

for $ p,q,r$ distinct.

Q5. If the answers to any of Q1-Q4 end up being no, can we characterize the positive rational numbers (or integers) $ x$ for which the answer is yes?

Honestly, I would even be happy if $ x=4$ admits infinitely many factorizations subject to the congruence conditions given in the example.

How to show that the product of two binary numbers cannot be determined in $AC^{0}$?

Input $ x = x_{0} … x_{n-1}$ for determine the xor over n-bits xi it is suffcient to multiply the following two n2-bits binary numbers: $ $ a = 0^{n-1} \hspace{2mm} x_{n-1} \hspace{2mm} 0^{n-1} \hspace{2mm} x_{n-2} \hspace{2mm} 0^{n-1} \hspace{2mm} … \hspace{2mm} 0^{n-1} \hspace{2mm} x_{1} \hspace{2mm} 0^{n-1} \hspace{2mm} x_{0} $ $

$ $ b = 0^{n-1} \hspace{3mm} 1 \hspace{5mm} 0^{n-1} \hspace{5mm} 1 \hspace{5mm} 0^{n-1} \hspace{2mm} … \hspace{2mm} 0^{n-1} \hspace{5mm} 1 \hspace{5mm} 0^{n-1} \hspace{5mm} 1 $ $

The product of two binary numbers can be determined in $ AC^{1}$ ?

unable to resolve product type ‘com.apple.product-type.system-extension’ for platform macosx (in target SimpleFirewallExtension))

I am trying to build and run FirewallExtension sample app from https://developer.apple.com/documentation/networkextension/filtering_network_traffic on catalina beta version, but getting error

unable to resolve product type ‘com.apple.product-type.system-extension’ for platform macosx MacOS Deployment Target is set to 10.15, but range of supported deployment target versions for this platform is from 10.6 to 10.14.99

Tried with Xcode 11. Does catalina needs some other setup to build network Extension? or anything is missed here?

magento 2 : How we can load product by sku in root file

i’m using magento 2 and i’ve to update particular product by sku,

i’ve used below code to load but it is taking too much time

$  objectManager = $  bootstrap->getObjectManager(); $  productRepository = $  objectManager->get('\Magento\Catalog\Model\ProductRepository'); $  product = $  productRepository->get(trim($  sku)); 

is there any other solution to load product from root which take less then above code,

Thank you,

Magento 2 : Add product to cart programmatically does not populate quote item id in validate method

How to add product to cart programmatically in Magento 2.2.0 ?

I have tried below code but when I do not get Quote item id in Validate method of below file.

\Magento\AdvancedSalesRule\Model\Rule\Condition\Product.php 

Code:

protected $  formKey;    protected $  cart; protected $  product;  public function __construct( \Magento\Framework\App\Action\Context $  context, \Magento\Framework\Data\Form\FormKey $  formKey, \Magento\Checkout\Model\Cart $  cart, \Magento\Catalog\Model\Product $  product, array $  data = []) {     $  this->formKey = $  formKey;     $  this->cart = $  cart;     $  this->product = $  product;           parent::__construct($  context); }  public function execute()  {    $  productId = 32;   $  params = array(                 'form_key' => $  this->formKey->getFormKey(),                 'product' => $  productId,                  'qty'   => 2                             );                    $  _product = $  this->product->load($  productId);        $  this->cart->addProduct($  _product, $  params);      $  this->cart->save();  } 

Sales Rule Validate Method Code:

File: \Magento\AdvancedSalesRule\Model\Rule\Condition\Product.php

Method:

public function validate(\Magento\Framework\Model\AbstractModel $  model)     {         //@todo reimplement this method when is fixed MAGETWO-5713         /** @var \Magento\Catalog\Model\Product $  product */         $  product = $  model->getProduct();         if (!$  product instanceof \Magento\Catalog\Model\Product) {             $  product = $  this->productRepository->getById($  model->getProductId());         }          if ($  model instanceof \Magento\Quote\Model\Quote\Item) {             die($  model->getItemId()); //NULL 

Thanks

Magento 2-Override FinalPriceBox using Preference or Plugin class of Configurable Product module

Anyone has idea how to make preference or plugin on Magento\ConfigurableProduct\Pricing\Render\FinalPriceBox class of Magento\ConfigurableProduct.

Basically, new function i wanted to add in the FinalPriceBox class and to use new function in final_price.phtml file.

I see code in the di.xml of module-configurableproduct module. But do not know how to use in new custom module?

<type name="Magento\ConfigurableProduct\Pricing\Price\FinalPrice">         <arguments>             <argument name="priceResolver" xsi:type="object">ConfigurableFinalPriceResolver</argument>         </arguments>     </type>