ct.category theory – Additivization of functors in an abelian monoidal category (crosspost from MSE)

I posted a question a week ago on math.stackexchange. As is sometimes the case, I got no answers. Considering that the question is about a research article, I hope that it might be relevant for MathOverflow.

Here is the original question:

I’m having trouble with the proof of Lemma 2.9 in “Cohomology of Monoids in Monoidal Categories” by Baues, Jibladze, and Tonks, and I was wondering if someone could clarify a detail. I’ll try to summarize the context of the lemma.

Context

Let $$(Bbb A,circ,I)$$ be an monoidal category where $$Bbb A$$ is abelian: in particular, $$circ$$ is not necessarily additive in both arguments. Suppose that $$circ$$ is left distributive, i.e. the natural transformation
$$(X_1circ Y)oplus(X_2circ Y)rightarrow (X_1oplus X_2)circ Y$$
is an isomorphism. For example, $$Bbb A$$ could be the category of linear operads (this is a motivating example of the article). Given an endofunctor $$F$$ of $$Bbb A$$, we define its cross-effect
$$F(A|B):=ker(F(Aoplus B)rightarrow F(A)oplus F(B)).$$
The additivization of $$F$$ is then the functor $$F^text{add}$$ defined by
$$F^text{add}(A):=text{coker}left(F(A|A)rightarrow F(Aoplus A)xrightarrow{F(+)}F(A)right).$$
The idea is that $$F^text{add}$$ is the additive part of $$F$$.

Let $$(M,mu,eta)$$ be an internal monoid in $$Bbb A$$, and let $$L_0$$ be the endofunctor of $$Bbb A$$ defined by $$L_0(A)=Mcirc(Moplus A)$$. Let $$L:=L_0^text{add}$$ be the additivization of $$L_0$$. (In the case of operads, represented as planar trees, I see $$L(A)$$ as the space of trees whose nodes are all labeled by elements of $$M$$ except for one leaf, which is labeled by an element of $$A$$.)

Suppose now that $$Bbb A$$ is right compatible with cokernels, i.e. that

for each $$AinBbb A$$, the additive functor $$Acirc-:Bbb ArightarrowBbb A$$ given by $$Bmapsto Acirc B$$ preserves cokernels.

Then, in the proof of Lemma 2.9, the authors claim the following:

By the assumption that $$Bbb A$$ is right compatible with cokernels it follows that $$L(L(X))$$ is the additivisation of $$L_0(L_0(X))$$ in $$X$$ (…).

Remarks

If anyone could provide an explanation of the last claim, I would be very grateful. However, my inability to understand how to show this might be related to two other issues I have:

1) Elsewhere in the literature, cross-effects are only defined when $$F$$ is reduced, i.e. $$F(0)=0$$ (e.g. here, section 2). But we can always reduce a functor by taking the cokernel of $$F(0)rightarrow F(X)$$, so I don’t think it’s much of a problem.

2) In the first quote, the authors state that $$Acirc -$$ is additive, which is quite the opposite of the initial hypothesis that $$circ$$ be left distributive, and not necessarily right distributive. How to resolve this apparent conflict?

nt.number theory – On the \$mathsf{LCM}\$ of a set of integers defined by moduli of powers

For integers $$a,b,t$$ define $$mathcal R_t(a,b)={qinmathbb Zcap(1,min(a^t,b^t)): a^tequiv b^tbmod q}$$ and $$mathsf{LCM}(mathcal R_t(a,b))$$ to be $$mathsf{LCM}$$ of all entries in $$mathcal R_t(a,b)$$.

Similar reasoning to On \$mathsf{LCM}\$ of a set of integers gives $$mathsf{LCM}(mathcal R_t(a,b))leqmathsf{LCM}(T_t(a,b))$$ where $$T_t(a,b)$$ is defined as $$T_t(a,b)=Big{qinmathbb Zcap(1,infty):q|Big((a-b)sum_{i=0}^{t-1}a^{t-1-i}b^iBig)Big}.$$

So $$mathsf{LCM}(mathcal R_t(a,b))=mathsf{LCM}(T_t(a,b))$$ holds.

If $$a,binbig(frac r2,rbig)$$ hold and are coprime then what is the probability $$mathsf{LCM}(mathcal R_t(a,b)) at some $$alphain(0,t)$$ and $$beta>0$$?

set theory – Why are quotient sets (types) called quotients — are they the inverse of some product?

There seems to be a beautiful relation between natural numbers and sets (and types),
as in the size of a discriminate union, cartesian product, and function type,
is described by the sum, product, exponential of the sizes of the components. (As I learned from type theory). This also makes it easy to see why the symbols + and x are used for discriminate union and cartesian product (sum type and product type).

$$forall A, B, C : text{sets} \ A + B = C ~ implies ~ |A| + |B| = |C|\ A times B = C ~ implies ~ |A| times |B| = |C|\ A → B = C ~ implies ~~~~~~~~ |B|^{|A|} = |C|$$

However, why are quotient sets (and quotient types) called quotients and use the symbol $$/$$?

That does not seem to make sense to me. At the very least, to deserve the name quotient, I would expect them to somehow be the inverse of some product. I first thought they should be the inverse of the cartesian product, I tried to google this, but I cannot find anything. Is there some relation between quotient (sets) and (cartesian) products, that I am missing?

nt.number theory – Reciprocity theorem with n >= 5

My equation is:

$$2^{(p-1)/n} equiv 1 pmod p$$

And this is always true: $$1 equiv p pmod n$$

I want to know which form $$p$$ must have to make equation 1 true.

I know for $$n=3$$ (cubic reciprocity) the form is $$p=x^2+27y^2$$.

For $$n=4$$ (biquadratic reciprocity) the form is $$p=x^2+64y^2$$

I know the solution has to deal with Gaussian integers and Artin’s reciprocity theorem.

Can someone please show me how to exactly do this maybe with the example $$n=5$$.

number theory – On quadratic form \$L^2+27M^2\$

If $$4q=L^2+27M^2$$, where $$qequiv 1 (mod 3)$$ is a power of $$p$$. How can I show that there is only one pair of integers (L,M) satisfying the condition where $$Lequiv 1 (mod 3)$$ and M up to sign.

I have proved by elementary number theory that there is at most one pair of POSITIVE integers. But what if $$4q=L’^2=L^2+27M^2$$, where $$Lequiv L’equiv 1$$?

measure theory – Prove that the composition of a measurable function with a continuous function results into a measurable function

Let $$Omega$$ be a measurable subset of $$Bbb R^n$$, and let $$W$$ be an open subset of $$Bbb R^m$$. If $$f:Omegato W$$ is measurable, and $$g:WtoBbb R^p$$ is continuous, then $$gcirc f:OmegatoBbb R^p$$ is measurable.

MY ATTEMPT

Let $$VsubseteqBbb R^p$$ be an open set. Then we have that $$g^{-1}(V)$$ is open. Since $$f$$ is measurable, $$f^{-1}(g^{-1}(V))$$ is also measurable.

In other words, for every open set $$VsubseteqBbb R^p$$, the set $$(gcirc f)^{-1}(V)$$ is measurable, and the result holds.

I am a little bit unsure about the wording of the proof. Could someone please check my arguments?

fa.functional analysis – A question about Schwartz-type functions used in analytic number theory

In analytic number theory we like to weigh our counting functions with a smooth function $$f$$, so that we may apply Poisson’s summation formula and take advantage of Fourier transforms. Typically the weight function $$f$$ will be a Schwartz type function with the following properties:

1) $$f(x) geq 0$$ for all $$x in mathbb{R}$$;

2) $$f(x) = 1$$ for $$x in (-X,X)$$ say;

3) $$f(x) = 0$$ for $$|x| > X + Y$$; and

4) $$f^{(j)}(x) ll_j Y^{-j}$$ for $$j geq 0$$.

In most applications the dependence on $$j$$ in the last condition does not matter, since $$j$$ would be bounded. However, in a problem I am considering it might be worthwhile to make $$j$$ a (slow growing) function of $$X$$ so it then becomes relevant to know how the bound depends on $$j$$. Is it possible to give an explicit example of a function $$f$$ satisfying the above properties for which the dependence can be made explicit?

nt.number theory – Number of distinct sum of two and four squares under sum condition

Given $$0leq mleq n$$ and $$x_1+dots+x_4=n$$ holds what is the

1. distribution of

2. maximum

number of distinct values of the sum of squares function $$x_1^2+dots+x_4^2$$ on conditions

a. $$0=x_1=x_2leq x_3,x_4leq m$$

b. $$0leq x_1,x_2,x_3,x_4leq m$$ holds?

(Note we do not force order).

complexity theory – Dividing students into 4 groups based on preferences is NP-complete

Given a set of students $$H$$ of size $$n$$, and a set $$E subseteq H times H$$ of pairs of students that dislike each other, we want to determine whether it’s possible to divide them into $$4$$ groups such that:

• no two students that dislike each other end up in the same group,
• the size of each group must be at least $$frac{n}{5}$$.

I want to prove that this problem is NP-complete. I suspect that I could use the NP-completeness of the independence set problem, yet I have some problems with finding an appropriate reduction.

Let $$G = (H, E)$$ an undirected graph – each edge represents two students that dislike each other.

For the groups to be of the required size, their size must be $$k in left (frac{n}{5}, frac{2n}{5} right ) cap mathbb{N}$$. I could then try checking whether there is an independence set of size $$k$$ (which would mean there are $$k$$ students that potentially like each other), remove its vertices, and repeat for the next $$k$$. However, I don’t think this would result in a polynomial number of size combinations.

Do you have any advice on constructing this reduction?

group theory – Elements of \$mathbb{F}_7^*/mathbb{F}_7^{*3}\$

I think I have forgotten some basic group theory, but I am having hard time representing the elements from $$mathbb{F}_7^*/mathbb{F}_7^{*3}$$, where $$mathbb{F}_7^{*3}$$ denotes all elements that are cubes in $$mathbb{F}_7^*$$. I have figured out that $$mathbb{F}_7^{*3} = {bar{1},bar{6}}$$ and hence $$mathbb{F}_7^*/mathbb{F}_7^{*3}$$ is isomorphic to $$mathbb{Z}/3mathbb{Z}$$. However, I am looking for representative elements from $$mathbb{F}_7^*$$. Any help would be appreciated.