text:
\section{Introduction}
There is a well-known technique for the construction of Vassiliev link
invariants: define a weight system (i.e.\ a linear form on the space of chord
diagrams respecting certain relations)
on the basis of some Lie algebraic data and pull it back by
the universal Vassiliev-Kontsevich invariant. But unfortunately, the latter
is not known explicitly enough to allow direct evaluation of these link
invariants.

Efforts have been made to handle the universal Vassiliev-Kontsevich
invariant by considering only
``elementary'' parts of links into which any link may be cut.
This approach has been successful in so far as one
may hope to find skein relations for the link invariants coming from Lie
algebras---a skein relation being an equation implying a recursive algorithm
for the computation of a link invariant, for example the one that
determines the famous Jones polynomial:
$$t^2 P(\psdiag{2}{6}{opkreuz})-t^{-2}P(\psdiag{2}{6}{onkreuz})
= (t^{-1}-t)P(\psdiag{2}{6}{owbog}).$$
\medskip

It has been shown that the link invariants obtained from
the classical simple Lie algebras ${\fraksl}_n$, ${\frakso}_n$, and ${\fraksp}_n$
satisfy certain
versions of the skein relation of the HOMFLY polynomial (${\fraksl}_n$; see [LM 1])
resp.\ the Kauffman polynomial (${\frakso}_n$, ${\fraksp}_n$; see [LM 2]).
But what about the exceptional simple Lie algebras?
\medskip

In this paper, we deal with the case of the exceptional Lie algebra
$\frakg_2$. By means of a generalization of the notion of links---since we have
to introduce branchings, we call them 3-nets---we manage to establish a
skein relation for the $(\frakg_2,V)$-invariant, $V$ being the 7-dimensional
``standard'' representation of $\frakg_2$. As a by-product, we obtain an extension
of the ($\frakg_2,V$)-invariant to closed 3-nets. Kuperberg's skein relation
for the quantum $\frakg_2$-invariant (see [K]) turns out to be a special case of
ours; it is not too surprising that there is a connection between these
skein relations since the restrictions to knots of the two invariants
coincide according to a result of Piunikhin's in [P].

We expect that our method can be adapted to the case of the other
exceptional Lie algebras.
\medskip

To the reader not familiar with Lie theory, we recommend [H]
and [FH]. For an introduction to Vassiliev invariants and weight systems,
see [BN 1]; a more general definition of weight systems is given in [V],
section 6.
\medskip

{\bf Overview} over the categories and functors appearing in this paper:
\vspace{-2mm}\\
\begin{center}
$\I$ \\
\hspace*{9mm}\psdiag{0}{14}{Pfeil} \vspace{-1.2cm}\\
\footnotesize the $(\frakg_2,V)$-invariant
\end{center}

\begin{tabular}{c c c c c}
\hspace{2.4cm} & \hspace{2.4cm} & \hspace{2.4cm} & \hspace{2.4cm}
& \hspace{2.4cm} \\
${\cal T}_{tpf}$ & $\stackrel{\hat Z_f}{\longrightarrow}$ & $\hat{\cal D}_3$ &
$\stackrel{\Psi}{\longrightarrow}$ & ${\cal C}(\frakg_2)$ \\
\end{tabular}

\begin{tabular}{p{2.4cm} p{2.4cm} p{2.4cm} p{2.4cm} p{2.4cm}}
\footnotesize the category of {\bf t}ri\-valent,
{\bf p}a\-ran\-the\-sized,
{\bf f}ramed tangles; eve\-ry 3-net corresponds to a morphism of this
category &
\footnotesize universal Vassi\-liev-Kontsevich in\-va\-riant &
\footnotesize the category of 3-diagrams (a generalization \newline of chord dia\-grams) &
\footnotesize $(\frakg_2,V)$-weight system &
\footnotesize a mo\-di\-fi\-ca\-tion of a sub\-ca\-te\-go\-ry of the ca\-te\-go\-ry of
re\-pre\-sen\-ta\-tions of $\frakg_2$
\end{tabular}
\vspace{3ex}

{\bf Acknowledgements} \hspace{0.3em} We would like to express our thanks to Ch.\ Riedtmann
for supervision, for encouragement, and for thoroughly reading this paper
and to P. Vogel for crucial suggestions and for patiently answering our
many questions; we also thank Ch.\ Blanchet for helpful discussions,
A. Beliakova for important references and K. Ott
for suggesting some linguistic improvement.


TITLE: Proportionality and Thales Theorem
QUESTION [1 upvotes]: let $k$ and $l$ be parallel lines. Let $A, B, C$ be points on $k$ and $A' B', C'$ be points on $l$ such that the lines $AA', BB'$ and $CC'$ intersect at a common point. Prove that $\frac{|A'B'|}{|B'C'|}=\frac{|AB|}{|BC|}$.
I tried using Thales's theorem on proportionality but I got confused
Any hints,
Thanks

REPLY [1 votes]: I think that you have to have two considerations, and they are, where is the point $P$ in which they intersect( inside or outside the parallels). I think (I am not completely sure) it depends on how your order the points. I´ll just give you one consideration, and it´s that the point $P$ is inside the parallels.

Here I just put five points ($A,C,A',C',P$),I know I haven´t put the other two points, but I can already say, that´s the point $P$, where they intersect. I can say that some angles, for example (angle $CAA'\cong AA'C'$, because they are angles between parallels) are equal to others. I´d like to say the name exact of them but, since I am not an english speaker I don´t know what they are(I´d like you to edit it, and put them). And now I can say that triangles $ACP \sim A'C'P$ by $AAA$.

So now drawing this second graphic I get some pairs of angles that are equal (angle that measure the same have the same colors)because they are angles between lines (Opossed to the vertex? is that the correct word? and also I get other similar triangles. And because they are similar, I can establish the following proportions:
$1)$ $$\frac{|AB|}{|A'B'|}=\frac{|BP|}{|B'P|}$$
$2)$ $$\frac{|BC|}{|B'C'|}=\frac{|BP|}{|B'P|}$$
Therefore
$$\frac{|AB|}{|A'B'|}=\frac{|BC|}{|B'C'|}=\frac{|BP|}{|B'P|}$$
So, we have:
$$\frac{|AB|}{|A'B'|}=\frac{|BC|}{|B'C'|}$$
And then if I multiply by $|A'B'|$ and $\frac{1}{|BC|}$, I get to:
$$\frac{|AB|}{|BC|}=\frac{|A'B'|}{|B'C'|}$$ which what you are asking for.
But whoever asked this question could say to you, that was not what they meant, because when they asked the question, they didn´t specify the order of the points. So here would be there other solution (Of course, it requires a new graphic):
(The difference is the order of the points)
So, here we also have similar triangles, which, technically, are the same as before.
So, for example, $ACP \sim PA'C'$ by $AA$. So, I can establish the following proportionS:
$1)$ $$\frac{|AC|}{|A'C'|}=\frac{|CP|}{|C'P|}$$
$2)$ $$\frac{|CB|}{|C'B'|}=\frac{|CP|}{|C'P|}$$
Therefore:
$$\frac{|AC|}{|A'C'|}=\frac{|CB|}{|C'B'|}$$
(I am going to solve this problem using a property of proportions that I´ll also demonstrate, but you could solve it geometrically.)

So, I´ll aply what I´ve just demonstrated:
$$\frac{|AC|+|CB|}{|A'C'|+|C'B'|}=\frac{|AC|}{|A'C'|}=\frac{|CB|}{|C'B'|}$$
$$\frac{|AB|}{|A'B'|}=\frac{|CB|}{|C'B'|}$$
And we get to:
$$\frac{|A'B'|}{|C'B'|}=\frac{|AB|}{|CB|}$$. And, I´d say, in this case, they are the two solutions. I haven´t done the cases where, $P$, where they intersect are outside the parallels, but I think (I am not completely sure), there should be two solutions as well (changing the order). =)


TITLE: Questions about a topological category
QUESTION [1 upvotes]: Given topological spaces $(X_i,\tau_i)$ with sets
$\mathbf S_i=\{\mathcal A\in \tau_i^2|\mathcal A\supseteq\Delta X_i^2\}$, where $\tau^2$ is the product topology and $\Delta$ is the diagonal. The sets S make it possible to judge how close to points $x,y\in X$ are to each other: the smaller open sets $\mathcal A\in \mathbf S$ such that $(x,y)\in\mathcal A$, the nearer $(x,y)$ is to the diagonal and the nearer $x$ and $y$ are.
Define function $f:X_1\to X_2$ to be uniform iff:
$\;\forall x,y\in \!X_1\, \forall \mathcal E_2\!\in \mathbf S_2\, \exists\mathcal D_1\in\mathbf
S_1: ((x,y)\in\mathcal D_1\implies(f(x), f(y))\in\mathcal E_2)$
$\qquad$ or equivalent
$\;\mathcal E_2\!\in \mathbf S_2\implies((f\times f)^{-1}(\mathcal E_2)\subseteq \mathbf S_1 $.
Apparently, topological spaces $X$ with "uniform" functions form a category since the identity and the composition of "uniform" functions are "uniform" functions, and it would be interesting to know things about that category, for example if the inverse of a bijective morphism is a morphism.

It should be pointed out perhaps, that this category is not the
  category of uniform spaces and as far as I know the morphisms in
  general don't have to be continuous.

Further I wonder if "uniformity" is equivalent with
$\mathcal B\subseteq X_2\times X_2\implies((x,y)\in\overline{(f\times f)^{-1}(\mathcal B)}\implies (f(x),f(y))\in\overline{\mathcal B})$?
See also: Is this a criterion for continuity? about a similar but not equal condition on a function.

REPLY [1 votes]: The fonction $x \mapsto 1/x$ defined on $[1,+\infty[$ is uniformly continuous. (Exercise !) The function $x\mapsto 1/x$ defined on $]0,1]$ is continuous but not uniformly continuous (exercise again !) and these two functions are inverse one to each other.
As on a metric space, with the uniform structure given by the metric, being uniform amouts to be uniformly continuous, the previous toy example shows that the inverse of a bijective morphism is not necessarily a morphism in the category of uniform spaces with uniform maps as morphisms.
For you other question, the answer is yes, and I also think that it is a proposition in Bourbaki, Topologie générale, Chapitre II, or maybe an exercise in the exercises for this chapter, but I don't have it next to me, so that I cannot confirm.
EDIT : after the edit of the question, what is meant is not leading to a uniform structure indeed, and is definitely not a uniform structure as well.
The also from your see also is misplaced, as the definitions of uniformity are not the same here and there. This question has horribly bad natations (i,j etc) and is too much unprecise and changing.


TITLE: show that $A$ is orthogonal if and only if its column vectors are orthonormal
QUESTION [0 upvotes]: can someone help me with this proof
Question :
$A$ is orthogonal if det $A^TA=1$.  show that $A$ is orthogonal if and only if the column vectors of $A$ are orthonormal?
thanks

REPLY [1 votes]: Hint: take $AA^T$ and think to matrix multiplication


TITLE: Prove that $f'(x)g(x) = 0$ for all $x \in [a,b]$
QUESTION [4 upvotes]: Let $f,g$ be continuously differentiable functions on $[a,b]$, suppose $g$ is nonnegative on $[a,b]$ and $f$ is increasing on $[a,b]$. Suppose further that $g(a) = 0$ and $g(b) = 0$, and that
$$\int_a^bf(x)g'(x)dx = 0$$
Prove that,
 $$f'(x)g(x) = 0$$ for all $x \in [a,b]$.

REPLY [3 votes]: By integration by parts, $\int f'(x) g(x) dx = f(x) g(x) - \int f(x) g'(x) dx$.
So $0 = \int_a^b f'(x) g(x) dx = f(a) g(a) - f(b)g(b) - \int_a^b f(x) g'(x) dx$
Since $g(a) = g(b) = 0$, it's clear that the first two terms are $0$.  The third term in the expression, we are given, is also $0$: $\int_a^b f(x) g'(x) dx = 0$.
We have $\int_a^b f'(x) g(x) dx = 0$, and $f'(x) g(x) \ge 0$ and is continuous, so we might conclude here that $f'(x) g(x) = 0$ for all $x \in [a,b]$.

If we cannot use the theorem that a nonnegative continuous function whose integral over some interval is identically $0$ over that integral, then we may proceed as follows:
By assumptions, $g(x) \ge 0$ and $f'(x) \ge 0$.  Suppose $f'(c) > 0$ and $g(c) > 0$ for some fixed point $c \in ( a,b)$.  By continuity, $f'(x) > 0$ and $g(x) > 0$ for some neighborhood $(c-\epsilon,c+\epsilon)$.  So $$\int_{c-\epsilon}^{c+\epsilon} f'(x) g(x) dx > 0$$
By nonnegativity of the product $f'(x) g(x)$, it follows that $$\int_a^{c-\epsilon} f'(x) g(x) dx , \int_{c+\epsilon}^b f'(x) g(x) dx \ge 0$$
So $\displaystyle \int_a^b f'(x) g(x) dx = \int_a^{c-\epsilon} f'(x) g(x) dx + \int_{c-\epsilon}^{c+\epsilon} f'(x) g(x) dx + \int_{c+\epsilon}^b f'(x) g(x) dx > 0$


TITLE: How do you extract cosmological constants like $k$ or $H_0$ from the fluctuations in the CMBR (WMAP)?
QUESTION [6 upvotes]: I am currently reading a book about Astrophysics and also tried to find some information about it on the web, but was not able to figure it out. That is why I ask you guys now. Any help is greatly appreciated :)
I understand that the CMBR of approximately $3\,\textrm{K}$ has fluctuations of $\frac{\Delta T}{T}\approx 10^{-5}$ depending in which direction of the universe you look. I also understand that his has to be some structural information (about possible pertubations) from the last scattering surface when matter an light decoupled. However, I just don't have any idea how you are able to determine the Hubble constant and the curvature constant ($k$) from this. In most sources I read it was just stated that if we assume the universe is flat ($k=0$) than we can find the Hubble constant from WMAP data and that the WMAP data agrees well with a flat universe. I don't understand that.
I am looking for a concrete explanation how the information about $H_0$ and $k$ can be extracted from the WMAP data.

REPLY [1 votes]: I’ll talk about the Planck data, just because it’s more recent, but all of this applies to WMAP as well. If you look at the Planck results (Table 1 here), you’ll see that both curvature and Hubble parameter are in the second group of quantities. This is because you do not actually measure these parameters from the data (contrary to what you may have heard). All you do is assume that the universe is flat, based off of local measurements, and from this deduce that the sum of energy densities must equal one. Hence any deviation from one we ascribe to dark energy, but it does not actually have any effect. This makes sense - how could dark energy do anything at only 300,000 years after the Big Bang. Once you have these quantities, the Hubble parameter is just given by Friedman’s equation.


TITLE: Does this turing machine exist?
QUESTION [0 upvotes]: I must test whether or not this Turing Machine exists $M(n)=0$ If $n$ is the Gödel number of a Turing Machine that stops with the empty tape $M(n)=1$ in other case.

REPLY [0 votes]: "Proof"
Reasoning for the absurd, suppose that $ M (n) $ exists, let's define $ Q (n) = 1 $ if and only if $ M (n) = 1 $ and $ Q (n) = loop $ if and only if $ n = \emptyset $. Note that $ Q $ has a Gödel number which we will denote for $ z $
If $ Q (z) = 1 $ then $ M (z) = 1 $, therefore $ z $ is the Gödel number of a Machine that stops on an empty tape, but $ Q $ does not stop on an empty tape, contradiction .
In addition $ Q (z) \neq loop$ since clearly an empty tape is not being entered.
We conclude that $ M (n) $ does not exist.
This is correct?


TITLE: Consistency strength of Turing measurability
QUESTION [16 upvotes]: This is probably well-known to recursion theorists, but as google didn't help me, I'll ask it here.
Convention: All sets of reals in the following discussion are assumed to be closed under Turing equivalence.
Consider the following regularity property: A set of reals $X$ is Turing measurable if $X$ contains a Turing cone or is disjoint to a Turing cone (recall that a Turing cone is a set of the form $\{y : x\leq_T y\}$ for some $x$).
By a well-known result of Martin, $AD$ implies that all sets of reals are Turing measurable.
My question is: What is the consistency strength of "all sets of reals are Turing measurable"? In particular, does it hold in Solovay's model?

REPLY [13 votes]: What you mean is that all Turing-invariant sets of reals (that is, if $x\in A$ and $x\equiv_T y$, then $y\in A$) are Turing measurable.
This is equiconsistent with $\mathsf{AD}$, that is, with $\mathsf{ZFC}+$ "There are infinitely many Woodin cardinals". The equiconsistency was proved by Woodin in the $80$s, but remains unpublished -- in the last few years, most of the results from that period have appeared in one way or another. This is one of the few that did not.
He showed that, in $L(\mathbb R)$, under $\mathsf{ZF}+\mathsf{DC}$, Turing determinacy (that is, "all Turing invariant sets of reals are determined", which is easily seen to be equivalent to Turing measurability) implies full determinacy.
Woodin has improved the result over the years. The strongest version I know of, from the early $2000$s, I believe, is that in $\mathsf{ZF}$, Turing determinacy implies that every Suslin set is determined. This gives the result in $L(\mathbb R)$ by reflection.
I looked at this a few years ago, when thinking about a different problem, on Galvin's multiboard determinacy. Richard Ketchersid and I noted that in $\mathsf{ZF}$, $\omega$-board determinacy is incompatible with choice, and equiconsistent with full determinacy, because it implies Turing measurability. I do not know of a direct proof that avoids the detour through Martin's result. (In the $\kappa$-board game, a set $A\subseteq\omega^\omega$ is fixed, and players I and II play the usual game on $A$ simultaneously on $\kappa$ many boards. Player I wins iff they win the usual game in at least one of the boards, else II wins.)

Update (Jun 25/22): Hugh has just published a proof that Turing determinacy implies the determinacy of Suslin sets. He indicates in the paper that the general question of whether Turing determinacy implies determinacy remains open. Coordinates:

NEW ZEALAND JOURNAL OF MATHEMATICS
Volume 52 (2022), 845–863, "Turing determinacy and Suslin sets"
https://doi.org/10.53733/140

In the comments it is mentioned that Turing determinacy is also now known to imply countable choice for sets of reals. This appears here:

ADVANCES IN MATHEMATICS
Volume 384 (2021), 107755, "TD implies CC$_{\mathbb R}$"
https://doi.org/10.1016/j.aim.2021.107755


TITLE: Matrix multiplication as composition
QUESTION [0 upvotes]: I've been trying to brush up on linear algebra, and I've encountered a problem that I haven't been able to resolve on my own.
A couple of days ago I watched a great video: "Matrix multiplication as composition | Chapter 4, Essence of linear algebra". In this video the author writes down an expression that shows how multiplication of 2 matrices (representing 2 different linear transformations) captures the overall / simultaneous effect of 2 linear transformations:
Matrix multiplication  (1)
One thing I can't wrap my head around is how to prove that the second transformation should be applied to the product of the first transformation and a vector we are transforming.
To figure out what coordinates vector $\bar v$ will have in the original basis after a linear transformation, we use the following formula:
$$
    \begin{bmatrix}
    v1 \\
    v2 \\
    \end{bmatrix}
    =
    \begin{bmatrix}
    a11 & a12 \\
    b21 & a22 \\
    \end{bmatrix}
    \begin{bmatrix}
    v'1 \\
    v'2 \\
    \end{bmatrix},\tag{*}\label{*}
$$
where $\begin{bmatrix}
    a11 & a12 \\
    b21 & a22 \\
    \end{bmatrix}$
is the matrix that shows what coordinates our new basis vectors will have (with regard to the original basis) after a linear transformation is applied, $\begin{bmatrix}
    v'1 \\
    v'2 \\
    \end{bmatrix}$ are coordinates of a transformed vector in the new basis.

I've written down a very informal proof of this formula: Matrix vector multiplication.

Turning back to the formula that is given in the video (1), I want to understand why do we multiply the matrix
$\begin{bmatrix}
    c11 & c12 \\
    d21 & d22 \\
    \end{bmatrix}$
by the vector
$\begin{bmatrix}
    v1 \\
    v2 \\
    \end{bmatrix}$ $\eqref{*}$:
$$
\begin{bmatrix}
    c11 & c12 \\
    d21 & d22 \\
    \end{bmatrix}
    \left(\begin{bmatrix}
    a11 & a12 \\
    b21 & a22 \\
    \end{bmatrix}
    \begin{bmatrix}
    v'1 \\
    v'2 \\
    \end{bmatrix}\right)
$$
I have this question because in the formula $\eqref{*}$ we use $\begin{bmatrix}
    v'1 \\
    v'2 \\
    \end{bmatrix}$ but not
$\begin{bmatrix}
    v1 \\
    v2 \\
    \end{bmatrix}$.
I would appreciate if you could point me in the right direction. Frankly speaking, I'm totally confused, despite the fact that I see that numerically everything adds up.

REPLY [1 votes]: An $m \times n$ matrix $A \in M(m \times n, \mathbb{R})$ encodes a linear transformation $L_A \colon \mathbb{R}^n \to \mathbb{R}^m$ by the following rule: $L_Ae_j = a_j$ for $1 \leq j \leq n$, where $a_j \in \mathbb{R}^m$ denotes the $j$th column of $A$ and $e_j \in \mathbb{R}^n$ is the $j$th standard basis vector.
Now given another matrix $B \in M(k \times m, \mathbb{R})$, $B$ encodes a linear transformation $L_B \colon \mathbb{R}^m \to \mathbb{R}^k$ just as $A$ does, i.e., $L_Be_i = b_i$ for $1 \leq i \leq m$. The matrix product $BA \in M(k \times n, \mathbb{R})$ is defined so that $L_{BA} = L_BL_A$, that is,
\begin{align}
(BA)_j &:= L_BL_Ae_j \\
&= L_Ba_j \\
&= L_B\sum_{i = 1}^{m}a_{ij}e_i \\
&= \sum_{i = 1}^{m}a_{ij}L_Be_i \\
&= \sum_{i = 1}^{m}a_{ij}b_i.
\end{align}
So the fact that matrix multiplication is composition is really the definition of matrix multiplication.


TITLE: continuous constant function general topology
QUESTION [6 upvotes]: I've read that if $X$ and $Y$ are topological spaces, and if the constant function $f: X \rightarrow Y$ mapping all of $X$ into the single point $y_{0}$ of $Y$, then $f$ is continuous. (Munkres page 107)
Munkres gives the following brief proof: Let $f(x) = y_{0}$ for every $x$ in $X$. Let $V$ be open in $Y$. The set $f^{-1}(V)$ equals $X$ or $\emptyset$, depending on whether $V$ contains $y_{0}$ or not. In either case, it is open.
My question is: $f$ is clearly not injective, so what does Munkres mean by the inverse function $f^{-1}$? What am i missing?

Thanks for any help/clarification!
Sincerely,
Vien

REPLY [5 votes]: $f^{-1}(V)$ is the preimage of $V$ under $f$. It is defined to be the set of all $x \in X$ such that $f(x) \in V$. So if $y_0 \in V$, then $f^{-1}(V) = X$ because all points of $X$ map into the single point $y_0$ that is now in $V$. If $y_0 \notin V$, then the entirety of $f(X)$ is disjoint from $V$, so that $f^{-1}(V) = \emptyset$.


TITLE: Is true that if $f(x)=\prod_{k=1}^nF_k(x)$ then $f'(x)=\sum_{j=1}^{n}\frac{F'_j}{F_j}\prod_{k=1}^nF_k(x)$?
QUESTION [0 upvotes]: I'm going to use this formula but I don't sure if this is true, I reached that conclusion one day but I didn't proof it.

REPLY [0 votes]: Assuming all the $F_j$ are nonzero, it is true.  You can prove it by induction on $n$.


TITLE: In what sense is the proper/effective action $\Gamma[\phi_c]$ a quantum-corrected classical action $S[\phi]$?
QUESTION [21 upvotes]: There is a difference between the classical field $\phi(x)$ (which appears in the classical action $S[\phi]$) and the quantity $\phi_c$ defined as $$\phi_c(x)\equiv\langle 0|\hat{\phi}(x)|0\rangle_J$$ which appears in the effective action. Even though $\phi_c(x)$ is referred to as the "classical field", I don't see why $\phi(x)$ and $\phi_c$ should be the same.
In what sense, therefore, is the effective action $\Gamma[\phi_c]$ a quantum-corrected classical action $S[\phi]$? How can we compare the functionals of two different objects (namely, $\phi(x)$ and $\phi_c(x)$) and claim that $\Gamma[\phi_c]$ is a correction over $S[\phi]$?
I apologize for any lack of clarity in the question and the confusion I'm hoping to clear up.

REPLY [1 votes]: Obviously, $\phi(x)$ is different from $\phi_c(x)$. The former is a classical field of a classical field theory, the latter is just a quantity that appears in the Legendre transform of the generating functional for connected Green functions. It just happens that for classical actions that an be treated as perturbations around  quadratic actions, the equations satisfied by $\phi_c(x)$ coincide with those of $\phi(x)$ in classical field theory, at the limit $\hbar\rightarrow 0$.
Except for the suggestive name, there is also no quantum-classical correspondence: $\phi_c(x)$ is not the expectation value of the field $\hat{\phi}(x)$ in presence of an external source (expressed in terms of properly defined probabilities). It does not make sense as a quasi-classical observable.
Furthermore, the effective action is non-local, and hence, it does not generate any effective quasi-classical dynamics. The effective action is only a generator for Green functions relevant to the calculation of S-matrix elements.


TITLE: torque needed to precess the angle of rotation
QUESTION [1 upvotes]: If a rigid body with the inertia matrix $I_B$, has the angular velocity $\overrightarrow{\omega_1}$ what torque is needed to rotate it around another axis say $\overrightarrow{\omega_2}$ while keeping it's original rotation ... Sort of the torque needed to change the position of the rigid body while it rotates. A know that it's angular momentum $\overrightarrow{L_1} = I_B\overrightarrow{\omega_1}$.  I don't know how to proceed ... There will be another angular momentum while rotating about $\overrightarrow{\omega_2}$ that is $\overrightarrow{L_2} = I_B\overrightarrow{\omega_2}$ so the body will feel $\overrightarrow{L} = \overrightarrow{L_1} + \overrightarrow{L_2}$ ?! The torque needed is $\overrightarrow{\tau} = \frac{d\overrightarrow{L}}{dt}$ so a sort of $\overrightarrow{\tau} \approx \frac{\overrightarrow{L_2}}{T}$ where $T$ is some time ... ?? Or the body is rotating around it's first axis of rotation hence having the angular momentum $L = I\omega$ in it's coordinate system. Now if the coordinate system is rotated around another axis $
\omega_2$, what angular momentum will the body feel? It's inertia matrix in the rotating frame will be $I_r = R^{T}IR$ and from here the angular momentum will be $L_2 = I_r\omega_2$ ?

REPLY [0 votes]: I shall present my conclusion. Given the following: the world coordinate system $(i_w, j_w, k_w)$, the body's inertia matrix in this coordinate system $I_w$ and the rotation axis $\omega$ in the world coordinate system, one can obtain the angular momentum $L$ as $L = I_w \cdot \omega$. If the rotation axis is precessed in $\omega_2 = R(t)^T\cdot \omega$, with $R(t)$ a rotation matrix then the inertia matrix of the rigid body in the world coordinate system is now $I_{w2} = R^T\cdot I_w \cdot R$ and the new angular momentum is $L_2 = I_{w2}\cdot \omega_2 = R^T\cdot I_w \cdot R \cdot R^T \cdot \omega = R^T \cdot L$. Hence, $$ \tau = \lim_{t \to 0}\frac{L2 - L}{t} = \lim_{t \to 0}\frac{(R(t)^T - I_3)}{t}\cdot L = \dot{R^T}\cdot L$$ where $I_3$ is the identity matrix.


TITLE: Function almost always periodic for every period is constant
QUESTION [3 upvotes]: Let be $(X, \mathcal{A}, \mu)$ a measure space and $f: X \to [0, +\infty]$ a measurable function such that for every $x \in X$ we have $f(t + x) = f(t)$ for almost every $t \in X$. Is true that $f$ is constant? or constant almost everywhere?
Thanks in advance.

REPLY [5 votes]: Hints: as a simple consequence of Fubini's Theorem, for almost all $t$, the equation $f(x+t)=f(t)$ holds for almost all $x$. [Use the fact that $\int \int |f(x+t)-f(t)| dx dt =\int \int |f(x+t)-f(t)| dt dx$ ].  In particular this holds for one value of $t$ and hence $f$ is constant almost everywhere. Can you see that $f$ need not be constant? [Take characteristic function of a non-empty set of measure $0$]. Will be happy to provide a detailed answer if needed.


TITLE: Find Root Solution
QUESTION [1 upvotes]: $$\frac{3 (1 - 2x)}{{\sqrt[4]{(1 - x)x}}} = \log_2\left(\frac{1 - x}{x}\right) \sqrt {2} $$
It is clear the $\frac{1}{2}$ is a solution, but there are others solutions. If we subtract this two side of equation and plot in a visualization software, we can see that are others roots. I like to find the solution that are close to $x = 0.2$.

Addendum: I need a exact solution.

REPLY [1 votes]: As you did, consider that you look for the zero's of function $$f(x)=\frac{3 (1 - 2x)}{{\sqrt {2}} {\sqrt[4]{(1 - x)x}}} - \log_2\left(\frac{1 - x}{x}\right)$$
Beside the trivial $x=0.5$, graphing, you noticed three other roots close to $x=0.2$, $x=0.8$ and another close to $x=1$; if you plot again the function between $0.95$ and $1.0$, the largest root is close to $x=0.99$.
Now, using these values a starting points, use Newton method and get the following iterates
$$\left(
\begin{array}{cc}
n & x_n \\
 0 & 0.2000000000 \\
 1 & 0.1704681790 \\
 2 & 0.1750941042 \\
 3 & 0.1752348886 \\
 4 & 0.1752350156
\end{array}
\right)$$
$$\left(
\begin{array}{cc}
n & x_n \\
 0 & 0.8000000000 \\
 1 & 0.8295318210 \\
 2 & 0.8249058958 \\
 3 & 0.8247651114 \\
 4 & 0.8247649844
\end{array}
\right)$$
$$\left(
\begin{array}{cc}
n & x_n \\
 0 & 0.9900000000 \\
 1 & 0.9912580691 \\
 2 & 0.9911109005 \\
 3 & 0.9911082777 \\
 4 & 0.9911082769
\end{array}
\right)$$


TITLE: There is only one increasing bijection from $\mathbb{N}$ to an infinite subset $X\subseteq \mathbb{N}$
QUESTION [3 upvotes]: Exercise 23, page 45, from Elon's book Curso de Análise: vol 1 (In Portuguese) I will translate it.
Let $X\subseteq \mathbb{N}$ be an infinite subset. Show that there exists only one increasing bijection $f:\mathbb{N}\rightarrow X$.
Thanks for your help!

REPLY [3 votes]: To show it's existence you have to use the well-ordering principle. Since $X\subseteq \mathbb{N}$, it admits a minimum element, say $p\in \mathbb{N}$. So you define $f:\mathbb N\to X$ recursively: put $f(0) = p$; now you define $X_1 := X-\{p\}$ and, again by the well-ordering principle, $X_1$ admits a minimum $p_1\in\mathbb{N}$; so you define $f(1) := p_1$. Assuming that you have already defined $f(n)$, in the same way as the others (taking the minimum element), you define $X_{n+1}: = X_{n}-\{f(n)\}$, and $p_{n+1}:=\min(X_{n+1})$. Since $X$ is infinite, every $X_n$ is non-empty; and  every $p_k$ is the minimum of the set $X_k$ you have $p_m < p_n$ when $m<n$. That concludes the existence of such function. Now, with the non-increasing property you can show that $f$ is injective; and, the fact that $X$ is an infinite subset of $\mathbb{N}$ guarantees that $f$ is surjective.
To ensure it's uniqueness, suppose, for the sake of the proof, that you have a bijective function $g: \mathbb{N}\to X$ such that $g(m)\leq g(n)$ when $m\leq n$. Let $A$ be any non-empty subset of $\mathbb{N}$; hence, both $g(A)$ and $f(A)$ have a minimum. If $f(x)\neq g(x), \forall x\in A$; WLOG, you have $g(x)<f(x)$. Therefore, you have $g(\min(A)) < f(\min(A))$, witch is an absurd because $f$ and $g$ are both bijective. So you must have $f(x)\geq g(x)$. By the definition of $g$, if $f(x)>g(x)$, we have an absurd. So you must have $f=g$, which concludes the demonstration. $\blacksquare$


TITLE: Is it "Valid" to prove Stokes' Theorem with Green's Theorem?
QUESTION [3 upvotes]: In my Vector Calculus course, the professor is rigorous enough that we do a decent number of proofs, but not rigorous enough to go all the way with manifolds/differential forms/etc. One proof in particular that didn't sit completely well with me was our proof of Stokes' Theorem using Green's Theorem.
Now, it's my understanding that Green's Theorem is a specific case of Stokes' Theorem (which are both specifics of the generalized Stokes' Theorem), and so it feels to me like we're sort of assuming the conclusion when we use it in the proof. Here's the outline of the proof, if it matters.
The basic idea was to show that given a parametrization of a surface in $\mathbb{R}^3$, the boundary of the surface is parameterized by the boundary of the parameterization. And so you go from a line integral around the boundary of the surface to a line integral around the boundary of the parametrization. From there you apply Green's theorem to take the double integral over the interior of the region, and then do the parametrization in reverse to get back the original surface.

REPLY [2 votes]: Good question. It is fairly common to prove a theorem by showing that the general situation can be reduced to a particular case. That's exactly what the argument you mention does.
There is no contradiction. Green is a particular case of Stokes, but Stokes (at least the version you were shown) can be proven using Green.


text:
\begin{document}

\title{On the image of the unstable Boardman map}

\author{Hadi Zare\\
        School of Mathematics, Statistics,
        and Computer Sciences\\
        University of Tehran, Tehran, Iran\\
        \textit{email:hadi.zare} at \textit{ut.ac.ir}\\
        and\\
        School of Mathematics, Institute for Research in Fundamental Sciences (IPM),\\
        P. O. Box. 19395-5746, Tehran, Iran}
\date{}

\maketitle

\begin{abstract}
We consider the `unstable Boardman map' (homomorphism if $k>0$)
$$b:\pi^{m+k}\Sigma^k\Omega^lS^{n+l}\simeq[\Omega^lS^{n+l},\Omega^kS^{m+k}]\longrightarrow \mathrm{Hom}(H_*\Omega^lS^{n+l},H_*\Omega^kS^{m+k})$$
defined by $h(f)=f_*$. We work at the prime $2$, with $k=0$, and determine the image for various in the following cases : (1) $m=n$ and $l>0$ arbitrary; (2) $m>n$ and $l=1$. We observe that in most of the cases the image is trivial with the exceptions corresponding to the cases when either there is a (commutative) $H$-space structure on $S^n$ or there is a Hopf invariant one element.
\end{abstract}

\textbf{AMS subject classification:$55Q45,55P42$}

\tableofcontents

\section{Introduction}
Suppose $E$ is a nice ring spectrum with the identity. Then there are two important homomorphisms which allow to study stable behaviour of a space $X$. The first one is a the stable Hurewicz homomorphism
$$h^s_E:\pi_n^sX\to E_nX$$
which sends $f$ to $(E_*f)x_n$ where $x^E_n\in E_n(S^n)$ is a generator provided by the unit $S^0\to E$. This homomorphism has been studies extensively for many important spectra such as $E=H\F_p,MU,BP,K,KO,tmf$ as there exists complete description of the stable Hurewicz homomorphism $h^s_E:{\pi_*^s}S^0\to E_*S^0$ for these spectra \cite{Adams-stablehomotopyandgeneralisedhomology}, \cite{Ravenel-Greenbook}, \cite{HopkinsMahowald}, \cite{Hopkins-ICM2002}. In fact computing the image of this homomorphism is so important that one tends to find a spectrum $E$ so that $h^s_E:{\pi_*^s}S^0\to E_*S^0$ is much closer to becoming an isomorphism and detects more and more elements in $\pi_*^sS^0$. From this point of view, there is an attempt to find bounds on the dimension/exponents of kernel and cokernel of this homomorphism (see \cite{Arlettaz-thestableHurewiczhomomorphism}, \cite{Mathew-exponents}).

The second homomorphism is the stable Boardman homomorphism
$$b^s_E:\pi^n_sX\to E^nX$$
which is defined by $b^s(f)=(E^*f)x^n_E$ where $x^n_E\in E^nX$ is a generator provided by the unit $S^0\to E$. There also has been detailed study on this homomorphism \cite{Hunton-Boardmanhomomorphism}, \cite{Arlettaz-Boardmanhomomorphism}.\\

These two homomorphisms are dual, possibly up to some degree shift, over $\pi_*E$ in a suitable sense (see \cite[Chapter 13]{Switzer} for a detailed discussion). Moreover, up to our knowledge, despite existence of some explicit relation among $h^s_E$ and $b^s_E$, there is no dictionary of the results about $h^s_E$ and $b^s_E$. We also note that this duality is not one that is induced by means of $S$-duality in the stable homotopy category in the sense that for a given $f:S^n\to X$ then $D(f)$ is not necessarily, up to finite number of suspensions, a map $X\to S^n$. However, knowing that these homomorphisms are dual in a suitable sense and the philosophy that sometimes solving a `dual' problem could be easier temps one to study one of these homomorphisms in order to justify some of claims about the other one. This duality could be used to obtain some information on the algebra. For instance, one may try to relate the rank of the image of $h^s_E$ to the rank of the kernel of $b^s_E$, etc.\\

The aim of this work is to apply and explore this idea in the presence of destabilisation functor $\Omega^\infty\Sigma^\infty$ when $E=H\Z/p$ for some prime $p$ with an special interest in the case of $p=2$. But, before proceeding further, there is one more piece of ingredient to bring in. We note that although given $f\in{\pi_n^s}X$ or $f\in{\pi^n_s}X$, knowing the specific images $h^s(f)$ or $b^s(f)$ could be important and useful, however, the primary aim of studying these homomorphisms is to understand and describe parts of ${\pi_n^s}X$ or ${\pi^n_s}X$ which could be detected using homology or cohomology. From this point of view, we may consider $h^s$ and $b^s$ as homomorphisms specific examples of more generalised Hurewicz, respectively Boardman homomorphisms
$$h^s:{\pi_n^s}X\longrightarrow \mathrm{Hom}(H_nS^n,H_nX),\ b^s:{\pi^n_s}X\longrightarrow\mathrm{Hom}(H^nS^n,H^nX).$$
Now, we describe the unstable Hurewicz and Boardman homomorphisms. Recall that for a spaces $X$ with base point, we have isomorphisms of groups
$$\pi_n^sX\simeq\pi_nQX\simeq [S^n,QX],\  \pi^n_sX\simeq [X,QS^n]$$
provided by the adjointness between $\Omega$ and $\Sigma$ functors where $QX=\colim\ \Omega^i\Sigma^i X$. This allows to define unstable Hurewicz and Boardman homomorphisms
$$h:{\pi_n^s}X\longrightarrow \mathrm{Hom}(H_nS^n,H_nQX),\ b:{\pi^n_s}X\longrightarrow\mathrm{Hom}(H^nQS^n,H^nX)$$
defined in the natural way $h(f)=f_*$ and $b(f)=f^*$. The evaluation map $\Sigma^\infty QX\to \Sigma^\infty X$ induces the stable homology suspension $\sigma_*^\infty:H_*QX\to H_*X$ and the stable cohomology suspension $\sigma^*_\infty:H^*X\to H^*QX$ which fit into commutative diagrams as
$$\xymatrix{
{\pi_n^s}X\ar[r]^-h\ar[rd]_-{h^s} & \mathrm{Hom}(H_nS^n,H_nQX)\ar[d]^-{\sigma_*^\infty} & {\pi^n_s}X\ar[r]^-{b^s}\ar[rd]_-{b} & \mathrm{Hom}(H^nX,H^nS^n)\ar[d]^-{\sigma^*_\infty}\\
                                                & \mathrm{Hom}(H_nS^n,H_nX)&& \mathrm{Hom}(H^nQX,H^nS^n)}$$

Next, notice that we have inclusion maps $\Omega^i\Sigma^iX\to QX$ which in the case of $h$ provides commutative diagram as
$$\xymatrix{
\pi_n\Omega^i\Sigma^i X\ar[r]^-h \ar[d] & \mathrm{Hom}(H_nS^n,H_n\Omega^i\Sigma^iX)\ar[d]\\
\pi_n^sX\simeq\pi_nQX\ar[r]^-h                        & \mathrm{Hom}(H_nS^n,H_nQX).}
$$
Also note that working at a prime $p$, the duality $H^nX\simeq \mathrm{Hom}_{\Z/p}(H_nX,\Z/p)$ between homology and cohomology provided by the Universal Coefficient Theorem, allows to consider $b^s$ and $b$ as homomorphisms
$$b^s:{\pi^n_s}X\longrightarrow\mathrm{Hom}(H_nX,H_nS^n),\ b:{\pi^n_s}X\longrightarrow\mathrm{Hom}(H_nX,H_nQS^n)$$
which send $f$ to $f_*$. For $b$, this provides a commutative diagram as
$$\xymatrix{
\pi^{n+i}\Sigma^iX\ar[r]^-{\simeq}&[X,\Omega^i\Sigma^iS^n]\ar[r]^-b\ar[d] & \mathrm{Hom}(H^n\Omega^iS^{n+i},H^nX)\ar[r]^-{\simeq} & \mathrm{Hom}(H_nX,H_n\Omega^iS^{n+i})\ar[d]\\
\pi^n_sX\ar[r]^-{\simeq}          &[X,QS^n]\ar[r]^-b          &  \mathrm{Hom}(H^nQS^n,H^nX)\ar[r]^-{\simeq}\ar[u] & \mathrm{Hom}(H_nX,H_nQS^n)
}$$
These observations motivates one to study the unstable Boardman homomorphisms
$$b:[X,\Omega^iS^{n+i}]\longrightarrow \mathrm{Hom}(H_nX,H_n\Omega^iS^{n+i})$$
where $i\geqslant 0$.

\section{Statement of results}
The main motivation for this work has been the study of spherical classes in $H_*QS^n$ and $H_*\Omega^lS^{n+l}$ which is the same as the determination of Hurewicz homomorphisms
$$h:\pi_*\Omega^lS^{n+l}\to H_*\Omega^lS^{n+l},\ h:\pi_*QS^n\to H_*QS^n$$
where $n\geqslant 0$ and $l\leqslant 0$.  The problem of determining spherical classes in $H_*X$ is not always an easy problem, e.g. in the case of $X=QS^0=\colim \Omega^iS^i$ it is an open problem (see for example \cite{Curtis}, \cite{Wellington}, \cite{Za}). The problem of determining spherical classes in finite loop spaces $\Omega^lS^{n+l}$ also is an open, although some progress for small values of $l$ has been made where we have achieved complete classification of these classes (see \cite{Zare-Els-1} and \cite{Zare-Mexicana}). For this purpose, and following the above philosophy, we are interested in looking at the dual problem and studying the image of Boardman homomorphisms
$$b:[\Omega^lS^{n+l},\Omega^kS^{m+k}]\to \mathrm{Hom}(H_*\Omega^lS^{n+l},H_*\Omega^kS^{m+k})$$
where $k\leqslant \infty$ with the convention $\Omega^{\infty}\Sigma^{\infty}X=QX$. This motivates the following definition.

\begin{defn}
A generalised cospherical class in $H_*X$ is determined by a map $f:X\to \Omega^kS^{m+k}$, $k\geqslant 0$, so that $f_*\neq 0$. A reduced generalised cospherical class is determined by a map $f:X\to\Omega^kS^{m+k}$, $k\geqslant 0$ so that $f_*:H_iX\to H_i\Omega^kS^{m+k}$ is nontrivial for some $i>m$.
\end{defn}

Here, we are speaking loosely as preimage of $f_*$ could contain more than one element. If one is hesitated by our definition, then the may call any element of $f_*^{-1}(H_*\Omega^kS^{m+k}-\{0\})$ a generalised cospherical class; this is the point of view taken in \cite{Wu-Memoire2003} when studying cospherical classes $X\to S^m$ corresponding to the case $k=0$ in our definition. Indeed, one may decide to define $z\in H^*X$ to be a cospherical class if for some $f:X\to S^m$ we have $f^*(x_m)=z$ which is the point of view considered in \cite{LamDuane} where the authors consider maps $X\to S^m$ which induce nontrivial maps in $KO$-theory; hence they have studied c-spherical classes in $KO(X)$ for specific choices of $X$.\\

Finally, note that for $k>0$ the source of the unstable Boardman map
$$b:[\Omega^lS^{n+l},\Omega^kS^{m+k}]\to \mathrm{Hom}(H_*\Omega^lS^{n+l},H_*\Omega^kS^{m+k})$$
is the `unstable' group $[\Omega^lS^{n+l},\Omega^kS^{m+k}]$ whose complete computation needs a suitable unstable Adams spectral sequence (ASS). However, we do not attempt working with any unstable ASS and instead we try to use available geometric techniques to study the image of this homomorphism. We shall state our results integrally when possible. Otherwise, we often localise at $p=2$. The following is our first result.

\begin{thm}\label{maintheorem}
Suppose all spaces are localised at the prime $2$. Suppose $k=0$ the following statement hold.\\
(i) If $m=n=0$ then for any $l>0$ the image of $b$ is isomorphic to $\Z/2\{\iota\}$ where $\iota\in H^1QS^1$ is the fundamental class.\\
(ii) If $m=n>0$ and $l=1$ then the image of $b$ is nontrivial if and only if $n\in\{1,3,7\}$ and in this case the image is isomorphic to $\Z/2\{\partial\}$ where $\partial:\Omega S^{n+1}\to S^n$ is the boundary map in the Barratt-Puppe sequence for one of the Hopf maps $\eta,\nu,\sigma$.\\
(iii) If $m=n>1$ and $l>1$ then the image of $b$ is trivial.\\
(iv) If $m=n=1$ and any $l>1$ then the image of $b$ is isomorphic to $\Z/2\{\theta_{S^1}\}$ where $\theta_{S^1}:QS^1\to S^1$ corresponds to the structure map of $S^1$ as an infinite loop space.
\end{thm}

Next we consider the cases with $f:\Omega^lS^{n+1}\to S^m$. In this case, we have only results for $l=1$. Note that by James splitting if $f:\Omega S^{n+1}\to S^m$ satisfies $f_*\neq 0$ then $m=tn$ for some $t$. We have the following.

\begin{thm}\label{maintheorem2}
Suppose $f:\Omega S^{n+1}\to S^m$ with $m=tn>n$.
(i) For $m>n$ with $m=tn$ if $n+1$ is odd or both of $n+1$ and $t$ are even then the image of $b$ is trivial.\\
(ii) For $m>n$ with $m=tn$ if $n+1$ is even, $t$ is odd, and $tn\not\in\{3,7\}$ then the image of $b$ is trivial.\\
(iii) For $tn\in\{3,7\}$ there exists a map $f:\Omega S^2\to S^t$ with $H_t(f)\neq 0$.
\end{thm}

We have excluded the case with $tn=1$ as it implies that $t=n=1$. In this case, it is impossible to have $1=m>n=1$. In this case, we have $f:\Omega S^{1+1}\to S^1$ is one of the cases studies by the previous theorem.\\

\textbf{Acknowledgements.} Some of the results in this paper were presented in Dalian workshop on algebraic topology, 2018. I am grateful to the organisers, specially Fengchun Lei, and Jie Wu for the invitation and the hospitality. I am also grateful to Mark Grant for some communications on Theorem 4.5 which resulted in a corrected version of the proof. This research was in part supported by a grant from IPM (No.98470122).

\section{Proof of Theorem \ref{maintheorem}: Case of $m=n>0$}
\subsection{Case of $l=1$}
For $l=1$ there are examples at hand which are provided by Hopf fibrations, namely maps $\Omega S^{n+1}\to S^n$ for $n=1,3,7$. The existence of
these maps also provides a decomposition
$$\Omega S^{n+1}\simeq S^n\times\Omega S^{2n+1}.$$
We show these are the only possible cases (at least modulo $2$). We have the following formulation of Adams' Hopf invariant one element result.

\begin{lmm}\label{AdamsHopf}
The followings are equivalent.\\
(i) There is a map $f:\Omega S^{n+1}\to S^n$ with $f_*\neq 0$ (modulo $2$).\\
(ii) There is a map $g:S^{2n}\to\Omega S^{n+1}$ with $g_*\neq 0$ (modulo $2$).\\
(iii) There is a map $h:S^{2n+1}\to S^{n+1}$ of unstable Hopf invariant one, and $n\in\{1,3,7\}$.\\
(iv) There is a stable splitting $P^n\simeq S^n\vee P^{n-1}$ where $P^n$ is the $n$-dimensional real projective space.
\end{lmm}

\begin{proof}
$(i)\Rightarrow (ii):$ Let $i:S^n\to \Omega S^{n+1}$ be the inclusion adjoint the to identity $S^{n+1}\to S^{n+1}$. Then $(f\circ\iota)_*\neq 0$, hence
(at $p=2$) $f\circ \iota$ is homotopic to the identity. Together with James fibration $S^n\to\Omega S^{n+1}\stackrel{H}{\to}\Omega S^{2n+1}$ it follows
that
$$(f,H):\Omega S^{n+1}\to S^n\times\Omega S^{2n+1}$$
is a homotopy equivalence. The inclusion $S^{2n}\to \Omega S^{2n+1}$ gives rise to a spherical class. Consequently, the composition
$$g:S^{2n}\to\Omega S^{2n+1}\to\Omega S^{n+1}$$
gives rise to a spherical class in $H_*(\Omega S^{2n+1};\Z/2)$, so $g_*\neq 0$.\\
$(ii)\Rightarrow (iii):$ If $g_*\neq 0$ then from James' description of $H_*\Omega S^{2n+1}$ we see that $g_*(x_{2n})=x_n^2$. It is well known that the
adjoint of $g$, say $h:S^{2n+1}\to S^{n+1}$ has unstable Hopf invariant one.\\
$(iii)\Rightarrow (i):$ As noted above, (ii) and (iii) are equivalent. It follows that the adjoint of $h$, say $g:S^{2n}\to\Omega S^{2n+1}$ maps
nontrivially under $H_\#:\pi_{2n}\Omega S^{n+1}\to\pi_{2n}\Omega S^{2n+1}$. This implies that $H\circ g$ is homotopy the identity. The claimed
decomposition follows immediately.\\
$(ii)\Longrightarrow (iv)$: First recall that the projection on the top cell $P^n\to S^n$ is an unstable map which is nontrivial in $\Z/2$-homology. For $g:S^{2n}\to\Omega S^{n+1}$ being nontrivial in homology, consider the composition
$S^{2n}\to \Omega S^{n+1}\to QS^n$ and write $g$ for this map as well. By Kahn-Priddy theorem, $g$ lifts to $Q\Sigma^n P$, and yields a map $S^{2n}\to Q\Sigma^n P$ which is nontrivial in homology. By taking adjoint, we obtain a map $\widetilde{g}:S^n\to QP$ which is nontrivial in homology. For dimensional reasons, this implies that $h(\widetilde{g})=a_n$ where $h$ is the Hurewicz homomorphism $\pi_*QP\to H_*QP$. Moreover, by cellular approximation, we may restrict to a map $\widetilde{g}:S^n\to QP^n$ which satisfies $h(\widetilde{g})=a_n$. This implies that for the stable adjoint of this map as $S^n\to P^n$ is nontrivial in homology. Therefore the composition $S^n\to P^n\to S^n$ is nontrivial in $\Z/2$-homology. Noting that the cofibre of $P^n\to S^n$ is $\Sigma P^{n-1}$ gives the other stable piece, so $P^n\simeq S^n\vee P^{n-1}$.\\
$(iv)\Longrightarrow (ii)$: Choose a splitting map $S^n\to P^{n-1}\vee S^n\simeq P^n$ given by the inclusion into the second summand. Then the adjoint $S^n\to QP^n$ is nontrivial in homology, so the composition $S^n\to QP$. For dimensional reasons, the $n$-adjoint of this map $S^{2n}\to Q\Sigma^n P$ is nontrivial in homology. After composition with the Kahn-Priddy map $Q\Sigma^n P\to QS^n$ we obtain a map $S^{2n}\to QS^n$ which is nontrivial in homology. Also note that the inclusion $\Omega S^{n+1}\to QS^n$ is a $2n$ equivalence, so we obtain a map $S^{2n}\to \Omega S^{n+1}$ which is nontrivial in homology. This completes the proof.
\end{proof}

This settles down the case with $l=1$. In this case, the case of $l>1$ reduces to the case of $l=1$ in the following sense.

\subsection{Case of $l>1$}
Unlike the case of $l=1$, for $l>1$ the existence of maps $f:\Omega^l S^{n+l}\to S^n$ with $f_*\neq 0$ is not so immediate. For $n=1$ we may choose $f:\Omega^lS^{n+l}\to S^1$ to be any representative of the identity element of $H^1(\Omega^lS^{n+l};\Z)\simeq\Z$. It is immediate that $f$ is nonzero in $H_*(-;k)$ for $k=\Z,\Z/2$. There is another way to see existence of such
maps. Since $S^1$ is an infinite loop space, let $\theta:QS^1\to S^1$ be the structure map which has the property that the composition $S^1\to QS^1\to
S^1$ is identity. In particular, $\theta_*\neq 0$. Now, the composition
$$f:\Omega^lS^{l+1}\to QS^1\to S^1$$
satisfies $f_*\neq 0$.\\

For the remaining cases, we have the following nonexistence result. Let's note that if $f:\Omega^lS^{n+l}\to S^n$ exists with $f_*\neq 0$ then the composition $\Omega S^{n+1}\to \Omega^lS^{n+l}\to S^n$ also would be nontrivial in $H_n(-;k)$, hence by Lemma \ref{AdamsHopf} we see that $n$ must be either $1$, $3$, or $7$. As we considered the case of $n=1$ above then we only need to resolve the cases $n\in\{3,7\}$.

\begin{lmm}\label{nonexistnece1}
(i) Suppose $n=3$. Then, there exists no map $f:\Omega^2S^5\to S^3$ with $f_*\neq 0$. Consequently, for $2\leqslant l\leqslant +\infty$ there exists no
map $f:\Omega^lS^{l+3}\to S^3$ with $f_*\neq0$.\\
(ii) Suppose $n=7$. Then, these exists no map $f:\Omega^2S^9\to S^7$ with $f_*\neq 0$. Consequently, for $2\leqslant l\leqslant +\infty$ there exists no
map $f:\Omega^lS^{l+7}\to S^3$ with $f_*\neq0$.\\
\end{lmm}

\begin{proof}
Proof of (i) and (ii) are similar. First note that the general case follows from our claim for double loop spaces as follows. For instance, note that
for $l\geqslant 2$, the composition
$\Omega^2S^5\to\Omega^lS^{l+3}$ is nonzero in $H_3(-;k)$ with $k=\Z,\Z/2$. Hence, existence of any map $f:\Omega^lS^{l+3}\to S^3$ with $f_*\neq 0$ would
imply that the composition $\Omega^2S^5\to\Omega^lS^{l+3}\to S^3$ is nonzero in homology, giving the desired contradiction.\\
Now we show there is no map $f:\Omega^2S^{n+2}\to S^{n}$ (with $n=3,7$) so that $f_*\neq 0$. We work at the prime $2$. Given a map
$f:\Omega^2\Sigma^2X\to X$ we may define $\mu:X\times X\to X$ by the following composition
$$\mu:X\times X \to *\times_{\Sigma_2}(X\times X)\to F(\R^2,2)\times_{\Sigma_2}(X\times X)\to\Omega^2\Sigma^2X\to X$$
where the first map on the left is projection, second and third maps are inclusion, and the last map is $f$. This map is a commutative multiplication on
$X$. Since the composition $S^n\to \Omega^2\Sigma^2S^n\to S^n$ is nonzero in homology (we may assume it is multiplication of degree $1$), hence it is homotpic to the identity. On the other hand, by construction, for a based path connected space $X$, the inclusion $X\to\Omega^2\Sigma^2 X$ can be decomposed as a composition
$$X\stackrel{\alpha}{\to}X\times X\to \Omega^2\Sigma^2 X$$
where $\alpha$ can be taken as either $(1,*)$ or $(*,1)$ with $*$ being the base point of $X$. This implies that, for $X=S^3,S^7$, $(X,\mu,*)$ is a commutative $H$ space in the sense of \cite{Gray}. But this is a contradiction as it is known that $S^3$ and $S^7$ do not admit any commutative
$H$-space structure.
\end{proof}

\section{Case of $m=n=0$}
\begin{lmm}
For any $l>0$ there exists a map $f:\Omega^lS^l\to S^0$ with $f_*\neq 0$. Moreover, any such $f$ we have $f=\Omega\iota$ where $\iota:\Omega^{l-1}S^l\to P=K(\Z/2,1)$ represents the fundamental class.
\end{lmm}

\begin{proof}
Let $l>0$, and take the fundamental class $\iota\in H^1QS^1\simeq\Z/2$ which can be realised as a map $\iota:QS^1\to P$. Define $f$ to be the composition
$$\Omega^lS^l\to QS^0\stackrel{\Omega\iota}{\to}\Z/2=S^0.$$
Clearly, $f_*\neq 0$. On the other hand, if $f:\Omega^lS^l\to S^0=\Omega P$ is given with $f_*\neq 0$, then the adjoint of $f$, say $\widetilde{f}:\Sigma\Omega^l S^l\to P$ is nontrivial in homology. Also, note we may consider the composition
$$\Sigma \Omega ^lS^l\stackrel{e}{\to}\Omega^{l-1}S^l\stackrel{\iota}{\to} P$$
which is nontrivial in homology where $e$ is the evaluation map (adjoint to the identity). Since $\widetilde{f},\iota\circ e\in [\Sigma\Omega^lS^l,P]\simeq\Z/2$ and both elements are nontrivial, therefore
$$\widetilde{f}=\iota\circ e\Rightarrow f=\Omega\iota\circ\widetilde{e}=\Omega\iota\circ 1=\Omega\iota.$$
This completes the proof.
\end{proof}

\section{Proof of Theorem \ref{maintheorem2}}

\subsection{The cases of $m=tn$ with $n+1$ odd, or $n+1$ and $t$ even}
We begin with the case $n=0$. In this case, as $\Omega S^1\simeq\Z$ it is not possible to find $m>0$ so that $g:\Omega S^1\to S^m$ is nontrivial in homology. Therefore, we consider the cases with $n>0$. If $f:\Omega S^{n+1}\to S^m$ is given with $f_*\neq 0$ then $(\Sigma f)_*\neq 0$. In the light of James splitting, $\Sigma\Omega S^{n+l}\simeq\bigvee_{t=1}S^{tn+1}$, $m=tn$ for some $t>1$ are the only possible choices for which $(\Sigma f)_*\neq0$ and consequently $f_*\neq 0$ can happen. That is

\begin{prp}
For $f:\Omega S^{n+1}\to S^m$ with $m\neq tn$ for all $t$, we have $f_*=0$.
\end{prp}

Our main result in this section is the following.

\begin{thm}\label{main-nonexistence1}
(i) Suppose $n+1$ is odd, $n>0$, and $t>1$. Then for any $f:\Omega S^{n+1}\to S^{tn}$ we have $f_*=0$ in $H_*(-;k)$ with $k=\Z,\Z/2$.\\
(ii) Suppose $n+1$, $n\geqslant1$, and $t>1$ are even. Then for any $f:\Omega S^{n+1}\to S^{tn}$ we have $f_*=0$ in $H_*(-;k)$ with $k=\Z,\Z/2$.\\
\end{thm}

\begin{proof}
The proof in both cases is by contradiction. By abuse of notation, we write $x_i$ for a generator of $H^iS^i$. Let's recall that
$$H^*(\Omega S^{n+1};\Z)\simeq\left\{\begin{array}{ll}
                                \Gamma_{\Z}(x_{n}) & \textrm{if }n+1\textrm{ is odd}\\
                                \Gamma_{\Z}(x_{2n})\otimes_{\Z}E_{\Z}(x_{n}) & \textrm{if }n+1\textrm{ is even}
                                \end{array}\right.$$
where $\Gamma_{\Z}$ and $E_{\Z}$ denote divided power algebra and exterior algebra functors over $\Z$ respectively. We first prove the integral result. \\
(i) Suppose $m=tn$ with $t>1$ and $f:\Omega S^{n+1}\to S^{tn}$ is given with $f_*\neq 0$ in $\Z$-homology. For $n>1$ with $n+1$ odd, and $t>1$, for dimensional reasons, for $X=\Omega S^{n+1},S^{tn}$ there is a duality between homology and cohomology with $H^*X\simeq\mathrm{Hom}_{\Z}(H_*X,\Z)$. It follows that $f^*\neq 0$. If $n+1$ is odd then having $f^*(x_{tn})\neq 0$ we see that $(f^*(x_{tn}))^2\neq 0$ in $\Gamma_{\Z}(x_{n})$ which implies that $x_{tn}^2\neq 0$ in $H^*S^{tn}$ which is an obvious contradiction. Hence, $f_*=0$.\\
(ii) First suppose $n>1$. Let's note that for $n>1$, $H_*\Omega S^{n+1}$ has nontrivial homology only in dimensions $kn$, with $k>0$, by James' splitting. So, if $H_i\Omega S^{n+1}\neq 0$ then $H_{i+\epsilon}\Omega S^{n+1}\simeq 0$ for $\epsilon\in\{-1,1\}$. This property is essential while appealing to the Universal Coefficient Theorem. Obviously, $H_*\Omega S^2$ does not have this property. We just note that choosing $t$ even forces $f^*(x_{tn})\in\Gamma_{\Z}(x_{2n})$. The rest of the proof goes exactly as part (i) and we leave the details to the reader.\\

Next, suppose either $n+1$ is odd with $n>0$ and $t>1$ or $n+1$ and $t$ both are even with $t,n>1$. Assume $f$ is given with $f_*\neq 0$ in $\Z/2$-homology. By the duality between homology and cohomology over $\Z/2$ we see that $f^*\neq 0$ in $\Z/2$-cohomology. For dimensional reasons, implied by our choices of $n$ and $t$, together with the naturality of the Universal Coefficient Theorem, we have a commutative diagram
$$\xymatrix{
0\ar[r] & \mathrm{Ext}(H_{tn-1}(S^{tn};\Z),\Z/2)\ar[r]\ar[d] & H^{tn}(S^{tn};\Z/2)\ar[r]\ar[d]^-{f^*} & \mathrm{Hom}(H_{tn}(S^{tn};\Z),\Z/2)\ar[r]\ar[d]^-{\mathrm{Hom}(f_*,\Z/2)} & 0\\
0\ar[r] & \mathrm{Ext}(H_{tn-1}(\Omega S^{n+1};\Z),\Z/2)\ar[r] & H^{tn}(\Omega S^{n+1};\Z/2)\ar[r] & \mathrm{Hom}(H_{tn}(\Omega S^{n+1};\Z),\Z/2)\ar[r] & 0}$$
The $\mathrm{Ext}$ terms do vanish for dimensional reasons, hence yielding a commutative diagram as
$$\xymatrix{
H^{tn}(S^{tn};\Z/2)\ar[r]^-{\simeq}\ar[d]^-{f^*} & \mathrm{Hom}(H_{tn}(S^{tn};\Z),\Z/2)\ar[d]^-{\mathrm{Hom}(f_*,\Z/2)}\\
H^{tn}(\Omega S^{n+1};\Z/2)\ar[r]^-{\simeq} & \mathrm{Hom}(H_{tn}(\Omega S^{n+1};\Z),\Z/2). }$$
We then conclude that the nonvanishing of $f^*$ in $\Z/2$-cohomology implies the nonvansihing of $\mathrm{Hom}(f_*,\Z/2)$ and consequently, nonvanishing of $f_*$ in $\Z$-homology. The result now follows by appealing to the integral case.

Finally, $n=1$ and $t$ even, notice that by the Hopf invariant one result $\Omega S^2\simeq \Omega S^3\times S^1$ and the above decomposition of cohomology is becomes a result of this splitting. The fact that $t$ is even implies that the composition $\Omega S^3\stackrel{\Omega\eta}{\to}\Omega S^2\to S^{t}$ is nonzero in homology where $t$ is even. But, this leads to a contradiction by part (i).\\

\end{proof}

\subsection{Attaching by Whitehead products}
The content of this subsection is probably well known, but we include an exposition for further reference. The idea is that given a $CW$-complex which splits after finite number of suspensions, or stable splits, certain attaching maps are obtained from Whitehead product. The main of this subsection is to make such statements more precise. In the case of splitting after finite number of suspensions, it is enough to deal with the case of splitting after one suspension. During this section and afterwards, for a CW-complex $X$, write $X^k$ for its $k$-skeleton. We write $X_k=X/X^{k-1}$ and $X^m_n=X^m/X_{n-1}$. We have the following.

\begin{lmm}\label{Whiteheadattaching1}
Suppose $X$ is a $CW$-complex with finite number of cells in each dimension. Suppose $X$ has only one cell in dimensions $n$ and $2n$. Suppose $\Sigma X^{2n}\simeq S^{2n+1}\vee \Sigma X^{2n-1}$ then the attaching map of the $2n$-cell in $X$, resulting in the $S^{2n+1}$ summand of $\Sigma X^{2n}$, is ``obtained'' by a Whitehead product on the $n$-cell. More precisely, write $\alpha$ for the attaching map $S^{2n-1}\to X^{2n-1}$, $p:X^{2n-1}\to X^{2n-1}_n$ for the pinching map, and $i:S^n\to X_n^{2n-1}$ for the inclusion of the bottom cell. If $p\circ\alpha\neq 0$ then
$$p\circ \alpha=i_\#[\iota_n,\iota_n].$$
\end{lmm}

\begin{proof}
Suppose $\alpha:S^{2n-1}\to X^{2n-1}$ is the attaching map of the desired $2n$-cell. We may compose $\alpha$ with the projection $X^{2n-1}\to X_n^{2n-1}$. The fact that $X^{2n}=X^{2n-1}\cup_{\alpha}e^{2n}$ after one suspension splits as $\Sigma X^{2n}\simeq S^{2n+1}\vee \Sigma X^{2n-1}$ is the same as saying that $\alpha$ being to the kernel of the suspension map $E:\pi_{2n-1}X^{2n-1}\to\pi_{2n-1}\Omega\Sigma X^{2n-1}$. The splitting of $\Sigma X^{2n}$ also implies that $\Sigma X^{2n}_n\simeq \Sigma X^{2n-1}_n\vee S^{2n+1}$. The space $X_n^{2n-1}$ has its bottom cell in dimension $n$, so we may use the $\mathrm{EHP}$-sequence in the meta-stable range. In particular, consider the following portion of the $\mathrm{EHP}$-sequence
$$\pi_{2n-1}\Omega^2\Sigma(X_n^{2n-1}\wedge X_n^{2n-1})\stackrel{P}{\longrightarrow} \pi_{2n-1}X_n^{2n-1}\stackrel{E}{\longrightarrow}\pi_{2n-1}\Omega\Sigma X_n^{2n-1}.$$
Since $E(p\circ\alpha)=0$ then $p\circ\alpha=P\beta$ for some $\beta\in\pi_{2n-1}\Omega^2\Sigma(X_n^{2n-1}\wedge X_n^{2n-1})$. Since $X$ has only one cell in dimension $n$ then the space $\Sigma(X_n^{2n-1}\wedge X_n^{2n-1})$ has its bottom cell in dimension $2n+1$, by Hurewicz Theorem $\Omega^2\Sigma(i\wedge i)$ is an isomorphism and there exists a unique nonzero $\beta\in\pi_{2n-1}\Omega^2\Sigma(X_n^{2n-1}\wedge X_n^{2n-1})\simeq\pi_{2n-1}\Omega^2 S^{2n+1}$ with $p\circ\alpha=P\beta$. Writing $i:S^n\to X_n^{2n-1}$ for the inclusion we have the following commutative diagram
$$\xymatrix{
\pi_{2n-1}\Omega^2 S^{2n+1}\ar[r]^-P\ar[d]_-{(\Omega^2\Sigma(i\wedge i))_\#} & \pi_{2n-1}S^n\ar[r]^-E\ar[d]_-{i_\#} & \pi_{2n-1}\Omega\Sigma S^n\ar[d]^-{(\Omega\Sigma i)_\#}\\
\pi_{2n-1}\Omega^2\Sigma(X_n^{2n-1}\wedge X_n^{2n-1})\ar[r]^-P               & \pi_{2n-1}X_n^{2n-1}\ar[r]^-E        & \pi_{2n-1}\Omega\Sigma X_n^{2n-1}.
}$$
It follows that $p\circ\alpha=i_\#P\beta$. However, it is known that $P:\pi_{2n-1}\Omega^2 S^{2n+1}\to \pi_{2n-1}S^n$ is the Whitehead product. Since, we work at the prime $2$, it allows us to think of $P\beta$ as the Whitehead product $[\iota_n,\iota_n]$ which implies that
$$p\circ\alpha=i_\#[\iota_{n},\iota_{n}].$$
This completes the proof.
\end{proof}

\subsection{The case of $n+1$ even and $t$ odd}
We localise at the prime $2$.\\

\begin{thm}\label{elimination-Whitehead1}
Suppose $f:\Omega S^{n+1}\to S^{tn}$ where $t$ is odd and $n+1$ is even. If $tn\not\in\{1,3,7\}$ then $f_*=0$.
\end{thm}

\begin{proof}
The proof is by contradiction, and we wish to apply Lemma \ref{Whiteheadattaching1} to prove the Theorem. The space $X=\Omega S^{n+1}$ satisfies conditions of Lemma \ref{Whiteheadattaching1}. Suppose $tn\not\in\{1,3,7\}$ and $f_*\neq 0$. Let $i:S^{tn}\to X^{2tn-1}_{tn}$ denote the inclusion of the bottom cell, $\alpha:S^{2tn-1}\to X^{2tn-1}$ the attaching map of the $2tn$-cell, and $p:X^{2tn-1}\to X^{2tn-1}_{tn}$ the pinching map which collapses $X^{tn-1}$. Then, by Lemma \ref{Whiteheadattaching1} we have $$i_\#[\iota_{tn},\iota_{tn}]=p\circ\alpha.$$
Moreover, for dimensional reasons, the restriction of $f$ to the $(2tn-1)$-cell of $X$, say $f|_{2tn-1}:X^{2tn-1}\to S^{tn}$ extends to a map $g:X^{2tn-1}_{tn}\to S^{tn}$ so that $gp=f|_{2tn-1}$. These data yield a commutative diagram as following.
$$\xymatrix{S^{2tn-1}\ar[r]^-{\alpha}\ar[d]_-{[\iota_{tn},\iota_{tn}]}& X^{2tn-1}\ar[d]^-p\ar[r]^-{f|_{2tn-1}} & S^{tn} \\
            S^{tn}\ar[r]^-i                                    & X^{2tn-1}_{tn}\ar[ru]_-g.}$$
On the other hand $f|_{2tn-1}$ has to extend over the $2tn$-skeleton of $X$ to a map $f|_{2tn}:X^{2tn-1}_{tn}\to S^{tn}$ where the necessity being implied by the existence of $f:X\to S^{tn}$. The existence of $f|_{2tn}$ is the same as saying that $f|_{2tn-1}\circ \alpha=0$. On the other hand $i$ and $g$ are both nonzero in $H_{tn}(-;\Z/2)$ and working at the prime $2$ this implies that we may take $g\circ i$ is the identity (over $\Z$ this implies that $g\circ i$ is an odd multiple of the identity). It follows that
$$[\iota_{tn},\iota_{tn}]=g_\#\circ i_\#[\iota_{tn},\iota_{tn}]=f|_{2tn-1}\circ \alpha=0.$$
It is known that for $m$ odd the Whitehead product $[\iota_m,\iota_m]\in\pi_{2m-1}S^m$ vanishes if and only if there is a Hopf invariant one element in $\pi_{2m-1}S^m$
\cite[Corollary 1.3, Remark 1.4]{Cohen-acourse} which together with Adams Hopf invariant one result means that $m\in\{1,3,7\}$ \cite{Adams-Hopfinv}. This shows that if $f$ extends over the $2tn$-skeleton then $tn\in\{1,3,7\}$. This gives the desired contradiction. Hence, $f_*=0$.
\end{proof}

It remains to decide about the cases with $tn\in\{1,3,7\}$ with $t$ odd and $n+1$ even with $n>0$. However, note that we have required $t>1$, so $t\geqslant 3$. Also, $n$ has to be odd. This leave us with the cases where $n=1$ and $t=3,7$. That is we have to decide about the existence of maps $\Omega S^2\to S^t$ with $t\in\{3,7\}$ such that $f_*\neq 0$. We have the following existence result and we note that its proof is also valid for the case of $\Omega S^2\to S^1$.

\begin{thm}\label{existenceHopf1}
For $t\in\{1,3,7\}$ there exists a map $f:\Omega S^2\to S^t$ with $f_*\neq 0$.
\end{thm}

\begin{proof}
For the given values of $t$, the existence of Hopf invariant one elements in $\pi_t^s$ is equivalent to existence of a decomposition $\Omega S^{t+1}\simeq \Omega S^{2t+1}\times S^t$. The examples of Hopf invariant one elements, which exist integrally and before any localisation, is provided by the Hopf fibre bundle $S^t\to S^{2t+1}\to S^{t+1}$ which could be extended to a fibre sequence
$$\Omega S^{2t+1}\to \Omega S^{t+1}\stackrel{\partial}{\to} S^t\to S^{2t+1}\to S^{t+1}.$$
We outline the proof for the case of $t=3$ and the case of $t=1,7$ are similar. In this case consider following portion of the above sequence
$$\Omega S^7\stackrel{\Omega\nu}{\to}\Omega S^4\stackrel{\partial}{\to}S^3.$$
We choose the more functorial, although less usual, notation $H_k(g)$ to denote $H_k(g;a)$ which is $g_*$ on the $k$-th homology with coefficients in a PID $a$ with an interest in $a=\Z,\Z/2$. We construct a map $g:\Omega S^2\to \Omega S^4$ with the property that $H_3(g)\neq 0$ and $H_3(\partial g)\neq 0$. It then would suffice to define $f=\partial g$. \\

By James splitting $\Sigma\Omega S^2\simeq\bigvee_{k=1}^{+\infty} \Sigma S^k$ there is an isomorphism
$$[\Omega S^2,\Omega S^4]\simeq [\Sigma\Omega S^2,S^4]\simeq [\bigvee_{k=1}^{+\infty} \Sigma S^k,S^4]\simeq\bigoplus_{k=1}^{+\infty} [S^{k+1},S^4]\simeq\bigoplus_{k=1}^{+\infty}\pi_{k+1}S^4.$$
For $g:\Omega S^2\to \Omega S^4$, write $\mathrm{ad}(g):\Sigma \Omega S^k\to S^4$ for its adjoint and $g_k:S^{k+1}\to S^4$ for the restriction of $\mathrm{ad}(g)$ to the $\Sigma S^k$ summand of $\Sigma\Omega S^2$. Consider the commutative diagram
$$\xymatrix{
[\Omega S^2,\Omega S^4]\ar[r]^-{\simeq}\ar[d]_-{H_3(-)} & [\Sigma\Omega S^2,S^4]\ar[d]^-{H_4(-)}\\
H_3\Omega S^4\ar[r]^-{\simeq}                           & H_4S^4}
$$
by which we may write
$$H_3(g)=H_4(\mathrm{ad}(g))=\sum_{k=1}^{+\infty}H_4(g_k)=H_4(g_3)$$
as for dimensional reasons $H_4(g_k)=0$ if $k\neq 3$. Now, to construct $g$ with $H_3(g)\neq 0$, choose $g_3:S^{3+1}\to S^4$ to be any nonzero element $d\in\pi_{4}S^4\simeq\Z$. Choose $g_k$ for $k\neq 3$ to be any arbitrary element. It follows that $H_3(g)=d$. If we wish to provide a proof which is valid at any prime then choose $g_3$ to be the mapping corresponding to the identity $S^4\to S^4$ which yields $H_3(g)=1$. Note that $g$ is not in the image of
$$(\Omega \nu)_\#:[\Omega S^2,\Omega S^7]\to[\Omega S^2,\Omega S^4].$$
For if $g=(\Omega\nu)_\# g'$ for some $g':\Omega S^2\to\Omega S^7$ then $H_3(g')\neq 0$. However, this is impossible for dimensional reasons as $H_3(\Omega S^7)\simeq 0$. Therefore, $g$ does not lift to $\Omega S^7$ and $f=\partial g:\Omega S^2\to S^3$ is a nontrivial element. On the other hand the mapping $\partial$ is a $5$-equivalence whose inverse is provided by the inclusion $S^3\to\Omega S^4$. Therefore, $H_3(f)\neq 0$. This completes the proof.
\end{proof}

\section{Notes on the case with $l>1$ and $m>n$}
We can use some geometry to prove -

\begin{lmm}
For any $l>1$ and $f:\Omega^l S^{n+l}\to S^{2n}$ we have $f_*=0$.
\end{lmm}

\begin{proof}
The inclusion $i:\Omega S^{n+1}\to \Omega^lS^{n+l}$ is a $2n$-equivalence. Hence, assuming $f:\Omega^2S^{n+2}\to S^{2n}$ is nontrivial in homology, $g:\Omega S^{n+1}\stackrel{i}{\to}\Omega^lS^{n+l}\stackrel{f}{\to}S^{2n}$ satisfies $g_*\neq 0$. This contradicts Theorem \ref{main-nonexistence1}.
\end{proof}

We postpone further investigation on this to a future work.

\bibliographystyle{plain}

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text:
\begin{document}

\begin{abstract}
We prove that if $\alpha\in (0,1/2]$, then the packing dimension of a set $E\subset\mathbb{R}^2$ for which there exists a set of lines of dimension $1$ intersecting $E$ in Hausdorff dimension $\ge \alpha$ is at least $1/2+\alpha+c(\alpha)$ for some $c(\alpha)>0$. In particular, this holds for $\alpha$-Furstenberg sets, that is, sets having intersection of Hausdorff dimension $\ge\alpha$ with at least one line in every direction. Together with an earlier result of T. Orponen, this provides an improvement for the packing dimension of $\alpha$-Furstenberg sets over the ``trivial'' estimate for all values of $\alpha\in (0,1)$. The proof extends to more general families of lines, and shows that the scales at which an $\alpha$-Furstenberg set resembles a set of dimension close to $1/2+\alpha$, if they exist, are rather sparse.
\end{abstract}

\maketitle

\section{Introduction and results}

\subsection{An improved lower bound on the packing dimension of Furstenberg sets}

In this paper we are concerned with Furstenberg sets and their variants. Let $\hdim$ denote Hausdorff dimension.
\begin{definition}
A set $E\subset\R^2$ is an $\alpha$-Furstenberg set if there exists a set $\Theta\subset S^1$ of positive measure, such that for all $\theta\in\Theta$ there is a line $\ell_\theta$ in direction $\theta$ such that
\[
\hdim(E\cap \ell_\theta) \ge \alpha.
\]
\end{definition}

An old problem motivated by work of Furstenberg \cite{Furstenberg70}, and first stated in print by Wolff \cite{Wolff99}, asks what is the smallest possible Hausdorff dimension of an $\alpha$-Furstenberg set. Wolff's paper contains a proof that if $E$ is an $\alpha$-Furstenberg set with $\alpha\in (0,1]$, then
\be \label{eq:easy-lower-bound}
\hdim(E) \ge \max(2\alpha,\alpha+1/2).
\ee
Note that the estimate is $\alpha+1/2$ is better for $\alpha\in (0,1/2)$ and both estimates coincide at $\alpha=1/2$. On the other hand, Wolff constructed an $\alpha$-Furstenberg set of Hausdorff dimension $\tfrac{1}{2}+\tfrac{3}{2}\al$. Thus \eqref{eq:easy-lower-bound} is sharp for $\al=1$, but it might not be otherwise.

It turned out to be quite hard to improve upon \eqref{eq:easy-lower-bound}. Katz and Tao \cite{KatzTao01} related the problem of improving the bound $\hdim(E)\ge 1$ for $1/2$-Furstenberg sets with other (at the time) outstanding problems in geometric measure theory. Shortly after, Bourgain \cite{Bourgain03} made dramatic progress on one of these related problems, the discretized sum-product problem. Combining the results of \cite{KatzTao01} and \cite{Bourgain03} it follows that $\hdim(E)\ge 1+c$ for all $1/2$-Furstenberg sets, where $c>0$ is a small universal constant. In fact, the proof shows that the same holds for $\alpha$-Furstenberg sets if $\alpha$ is close enough to $1/2$ (possibly smaller). This remains the only improvement to \eqref{eq:easy-lower-bound} to date.

In \cite{Orponen20}, however, Orponen achieved an improvement over \eqref{eq:easy-lower-bound} in the range $\alpha\in (1/2,1)$, with the caveat that he considered the \emph{packing} dimension $\pdim(E)$ of $E$, rather than Hausdorff dimension. That is, he proved that for each $\alpha\in (1/2,1)$ there is a small number $c(\alpha)>0$ such that
\[
\pdim(E) \ge 2\alpha+c
\]
for all $\alpha$-Furstenberg sets $E$. Since (upper) box-counting dimension is at least as large as packing dimension, this estimate holds for it as well. See \cite[Chapters 2 and 3]{Falconer14} for an introduction to the notions of dimension considered in this paper.

Orponen's result left open the question of whether such an improvement is also possible, at least for packing dimension, also in the range $\alpha\in (0,1/2)$.  Our main result provides such an improvement:

\begin{theorem} \label{thm:main}
For $\alpha \in (0,1/2]$ there exists a constant $c=c(\alpha)>0$ such that
\[
\pdim(E) \ge \frac{1}{2}+\alpha +c
\]
for all $\alpha$-Furstenberg sets $E\subset\R^2$.
\end{theorem}

We can in fact prove a more general result. In \cite{MolterRela12}, Molter and Rela generalized the notion of Furstenberg set by considering also fractal sets of directions. In several later papers such as \cite{HKM19, HSY20} this was expanded further to arbitrary families of lines (not necessarily pointing in different directions). Note that the Grassmanian $A(2,1)$ of affine lines in the plane is a $2$-dimensional manifold, so one can speak of the Hausdorff dimension of an arbitrary set of lines.  We say that $E\subset\R^2$ is an $(\alpha,\beta)$-Furstenberg set if
\[
\hdim\{ \ell\in A(2,1):\hdim(E\cap \ell) \ge \alpha\}\ge \beta.
\]
Clearly, a classical $\alpha$-Furstenberg set is an $(\alpha,1)$-Furstenberg set. Let $s(\al,\beta)$ be the infimum of the Hausdorff dimensions of $(\al,\beta)$-Furstenberg sets. In the range $\al\in (0,1/2]$, the following bounds are known:
\[
s(\al,\beta) \ge \left\{
\begin{array}{cccl}
  \al+\min(\al,\beta) & \text{if} & \beta \le 2\alpha & \text{(\cite{LutzStull17, HSY20})} \\
  2\al+ c(\al) & \text{if} & \beta \in (2\al-c(\al),2\al+c(\al)) & \text{(\cite{HSY20})} \\
  \al + \beta/2 & \text{if} & \beta\in (2\al, 1] & \text{(\cite{MolterRela12,Hera19})} \\
\end{array}
\right..
\]
We are able to improve upon the $\al+\beta/2$ bound in the range $\beta>2\al$, with the caveat that we consider the packing (or upper box-counting) dimension of the $(\al,\beta)$-Furstenberg set. Let $\cN_\de(A)$ denote the smallest number of balls of radius $\de$ needed to cover $A\subset\R^d$.
\begin{theorem} \label{thm:main-alpha-beta}
Given $\alpha \in (0,1/2]$ and $\beta \in (2\alpha,2]$ there is $c=c(\alpha,\beta)>0$ (depending continuously on the parameters) such that the following holds.

Let $E$ be an $(\alpha,\beta)$-Furstenberg set and fix $s\in [1/2,1)$. Then for all small enough $\de$ (depending on $E$ and $s$), either
\be  \label{eq:main-scale-dichotomy}
\cN_\de(E) \ge \de^{-\alpha-\beta/2-(1-s)c} \quad\text{or}\quad \cN_{\des}(E) \ge (\des)^{-\alpha-\beta/2-(1-s)c}.
\ee
\end{theorem}

We make a few remarks on this statement. Firstly, by taking $s=1/2$, it follows immediately that $\ubdim(E)\ge \al+\beta/2+c'(\al,\beta)$, with $c'(\al,\beta)=c(\al,\beta)/2$. By a standard argument (see \cite[Section 2]{Orponen20}), the same is true for packing dimension:
\begin{corollary} \label{cor:main-alpha-beta}
In the context of Theorem \ref{thm:main-alpha-beta},
\[
\ubdim(E) \ge \pdim(E) \ge \alpha+\frac{\beta}{2}+c'(\al,\beta).
\]
\end{corollary}
Of course, this implies Theorem \ref{thm:main} as a special case.

Secondly, although our method falls short of proving that
\be \label{eq:E-large-scale-delta}
\cN_\delta(E)\gtrsim \delta^{-\al+\beta/2+c}
\ee
for all small $\de$ (which would imply that $\hdim(E)\ge  \al+\beta/2+c$), Theorem \ref{thm:main-alpha-beta} shows that the scales $\de$ such that \eqref{eq:E-large-scale-delta} fails are quite sparse (while $\ubdim(E)\ge \al+\beta/2+c$ only requires that \eqref{eq:E-large-scale-delta} holds for arbitrarily small $\de$) . Namely, given $\e>0$, for any small $\de_0$, if \eqref{eq:E-large-scale-delta} fails for $\de_0$, then it must hold for all $\de\in [\de_0^{1/2},\de_0^2] \setminus [\de_0^{1-\e},\de_0^{1+\e}]$ (with $c=c(\al,\beta,\e)$).

\subsection{Sketch of the proof of Theorem \ref{thm:main-alpha-beta}}

The methods of \cite{KatzTao01, Orponen20, HSY20} all consist in starting with a counter-assumption that the dimension is very close to the one we want to beat, and deduce (after some substantial processing and transformation of the set) that there must be some underlying ``product structure''. The proof is concluded by using Bourgain's discretized sum-product or projection theorems \cite{Bourgain03,Bourgain10} to parlay the product structure into a small dimensional gain. We use a similar approach: under the counter-assumption that \eqref{eq:main-scale-dichotomy} fails, we show that $E$ (after refinement) must have a quite rigid structure at scales $\des$ and $\de$. This structure (which is not quite of product type) enables the application of Bourgain's projection theorem. The reduction to Bourgain's projection theorem is different from, and perhaps simpler than, those of \cite{KatzTao01, Orponen20, HSY20} .

After some standard reductions detailed in \S\ref{subsec:reductions}, at scale $\de$ we replace the $(\al,\beta)$-Furstenberg set by a union $E'=\cup_{\om\in\Om} E_\om$, where $\Om$ is a $\de$-separated set of (parameters of) lines satisfying a $\beta$-dimensional Frostman assumptions, and $E_\om$ are subsets of the corresponding lines satisfying a uniform $\alpha$-dimensional Frostman bound.

We start with a counter-assumption that $\cN_{\des}(E) \le (\des)^{-\al-\beta/2-\eta}$ for a small $\eta>0$. Using a variant of the estimate that provides the lower bound $\al+\beta/2$ for the dimension of $(\alpha,\beta)$-Furstenberg sets, given in Lemma \ref{lem:alpha+gamma bound}, we show that the sets $E_\om$ must cluster nearly as much as possible at scale $\des$. Namely, we cover $E'$ (after refinement) by $\approx$ $(\des)^{-\beta}$ sets $F_\lam$, contained in rectangles of size $\des \times 1$ whose long directions are $\des$-separated, and such that $\cN_{\des}(F_\lam)$ is not much larger than $(\des)^{-\al}$. Moreover, each $F_\lam$ contains $\approx (\dess)^{\beta}$ of the sets $E_\om$. This analysis is carried out in \S\ref{subsec:scale-des}.

In \S\ref{subsec:analysis-F-lam}, we analyze a fixed set $F_\lam$ at scale $\de$. We show that it can be essentially decomposed into $\approx (\des)^{-\al}$ squares $Q_j$ of side length $\des$ such that ``most'' of the set $E_\om\subset F_\lam$ have large intersection with $Q_j$. By the assumption $s\ge 1/2$, the lines containing the sets $E_\om$ are nearly parallel inside $Q_j$. This enables the application of Bourgain's discretized projection theorem (recalled as Theorem \ref{thm:projection-Bourgain} below) to show that for most such $Q_j$ we have a box-counting gain:
\[
\cN_{\de}(E\cap Q_j) \ge (\dess)^{-\al-\beta/2-\zeta},
\]
where $\zeta>0$ depends only on $\al$ and $\beta$. Finally, a double counting of the squares $Q_j$ presented in \S\ref{subsec:conclusion-proof} allows us to conclude the proof of Theorem \ref{cor:main-alpha-beta}.

\section{Preliminaries}

\subsection{Notation}

We use the notation $A\lesssim B$ to mean $A\le C B$ for some constant $C>0$. If $C$ depends on certain parameters $\text{par}$, we write $A\lesssim_{\text{par}} B$. We write $A\gtrsim B$ for $B\lesssim A$ and $A\sim B$ for $A\lesssim B\lesssim A$, and likewise with parameters. Sometimes we write $A=O(B)$ to mean $A\lesssim B$.

We denote (open) balls of centre $x$ and radius $r$ in $\R^d$ by $B^d(x,r)$. We often omit the superscript $d$. We write $B^d=B^d(0,1)$ for short. The notation $B_r$ denotes an arbitrary ball of radius $r$ (with the ambient dimension fixed but implicit).

Given $\om=(a,b)$, we let $L_\om(t)=at+b$. Abusing notation slightly, we will also denote the line $\{ y=at+b\}$ by $L_\om$.

Recall that $\cN_\de(A)$ denotes the $\de$-covering number of $A$.

\subsection{Discretization}

As usual in this circle of problems, in the proof of Theorem \ref{thm:main-alpha-beta}, we will work with certain discretized versions of Furstenberg sets, which we now introduce.

\begin{definition}
A probability measure $\mu$ on $B^1$ is a $(\delta,\alpha,K)$-measure if
\[
\mu(B_r)\le K r^\alpha \quad\text{for all } r\in [\delta,1].
\]
\end{definition}

Note that a $(\delta,\alpha,K)$-measure is automatically a $(\delta',\alpha',K')$-measure whenever $\delta'\in [\delta,1]$, $\alpha'\le \alpha$ and $K'\ge K$. Also, if $\mu$ is a $(\delta,\alpha,K)$ measure and $\mu(A)\ge c>0$, then $\cN_\de(A)\gtrsim_{c,K,\alpha} \delta^{-\alpha}$. We will repeatedly use these facts without further mention. For us, the crucial parameters are the scale $\delta$ and the exponent $\alpha$; the value of the constant $K$ will not concern us too much.

\begin{definition}
We say that $(\nu_\om)_{\om\in\Om}$ is a $(\delta,\alpha,K)$-Furstenberg tuple if:
\begin{itemize}
  \item $\Om\subset B^2$ is a $\delta$-separated set,
  \item For each $\om$, there is a $(\de,\al,K)$-measure $\nu'_\om$ such that $\nu_\om= L'_\om \nu'_\om$, where $L'_\om(x)=(x,L_\om(x))$.
\end{itemize}
\end{definition}

We will obtain a Furstenberg tuple by starting with a Furstenberg set $E$, taking any $\delta$-separated set $\Omega$ of the (parameters of the) lines satisfying \eqref{eq:positive-content}, and using Frostman's Lemma to obtain a suitable measure $\nu_\om$ on $E\cap L_\om$ for each $\om\in\Om$. See \S\ref{subsec:reductions} for details.

In the following two lemmas, we fix a $(\delta,\alpha,K)$-Furstenberg tuple $(\nu_\om)_{\om\in\Om}$, and write $E_\om=\supp(\nu_\om)$ and $E=\cup_{\om\in\Om} E_\om$.

Our first lemma is a small box-counting variant of the argument giving the lower bound $1/2+\alpha$ for $\alpha$-Furstenberg sets.
\begin{lemma} \label{lem:alpha+gamma bound}
Suppose $\alpha>0$ and $\cN_\delta(E_\om)\ge M$ for all $\om\in\Omega$. Then
\[
(\cN_\delta(E))^2 \gtrsim_{K,\alpha} |\Omega| M\delta^{-\alpha}.
\]
\end{lemma}
\begin{proof}
Pick $c=c_{K,\alpha}$ such that $K(3c)^\alpha\le 1/2$; hence $\nu_\om(B_{3c})\le 1/2$ for all $\om\in\Om$. By assumption, for each $\om\in\Om$ there is a segment of length $c$, say $I_\om\subset L_\om$, such that $\cN_\delta(E_\om\cap I_\om)\gtrsim_{c} M$. Since $\nu_\om(3I_\om)\le 1/2$, it follows from the Frostman condition $\nu_\om(B_r)\le K r^\alpha$ that $\cN_\delta(E_\om\setminus 3I_\om)\gtrsim_K \delta^{-\alpha}$.

Let $\ell_{x,y}$ denote the line through $x$ and $y$. If $x$ is at distance $\le \delta$ from $I_\om$ and $y$ is at distance $\le\delta$ from $L_\om \setminus 3 I_\om$, then the distance on $A(2,1)$ between $\ell_{x,y}$ and $L_\om$ is $\lesssim_K \de$.  This implies that if $d(\om_1,\om_2)\ge C\delta$, for a sufficiently large constant $C=C_K$, then the sets $(E_{\om_i}\cap I_{\om_i}) \times (E_{\om_i}\setminus 3 I_{\om_i})$, $i=1,2$, are $\delta$-separated. Thus
\[
(\cN_\delta(E))^2\sim \cN_\delta(E\times E) \ge \cN_\de\left(\bigcup_{\om\in\Om} E_\om\times E_\om\right) \gtrsim_{K,\alpha} |\Omega|M\delta^{-\alpha},
\]
giving the claim.
\end{proof}

\begin{lemma} \label{lem:fan}
If there is a point $x$ such that $\dist(x,L_\om)\le C\delta$ for all $\om\in\Om$, then
\[
\cN_\delta(E) \gtrsim_{C,K} |\Om|\delta^{-\alpha}.
\]
\end{lemma}
\begin{proof}
Pick $c=c_{K,\alpha}$ small enough that $\nu_\om(B_c)\le 1/2$ for all $\om\in\Om$. By elementary geometry, if $d(\om_1,\om_2)\ge C'\delta$ for a sufficiently large constant $C'=C'_{K,c}$, the sets $E_{\om_i}\setminus B(x,c)$ are $\delta$-separated. Since $\cN_\delta(E_\om\setminus B(x,c))\gtrsim_K \delta^{-\alpha}$, the lemma follows.
\end{proof}

\subsection{Bourgain's projection theorem}

We state Bourgain's discretized projection theorem for later reference. We borrow the version from \cite[Theorem 1]{He18}, in the special case of projections from the plane to the real line. Given $\theta\in S^1$, let $P_\theta:\R^2\to\R$, $x\mapsto \langle x,\theta\rangle$ denote projection in direction $\theta$.
\begin{theorem} \label{thm:projection-Bourgain}
Given $0<u<2$ and $0<t<1$ there exist $\e,\zeta>0$ such that the following holds for all sufficiently small $\de$. Let $X\subset B^2$, and let $\rho$ be a probability measure on $S^1$ such that the following hold:
\begin{align*}
\cN_\delta(X) &\ge \delta^{\e-u},\\
\cN_\delta(X\cap B(x,r)) &\le \delta^{-\e} r^t \cN_\delta(X) \quad\text{for all } r\in [\delta,1], x\in B^2,\\
\rho(B_r) &\le \delta^{-\e} r^t \quad\text{for all } r\in [\de,1].
\end{align*}
Then there is a set $\Theta\subset S^1$ with $\rho(\Theta)\le \delta^\e$ such that if $\theta\in S^1\setminus \Theta$ and  $X'\subset X$ satisfies $\cN_\delta(X') \ge \de^\e\cN_\delta(X)$, then
\[
\cN_\delta(P_\theta X') \ge \de^{-\frac{u}{2}-\zeta}.
\]
\end{theorem}

It follows from the robustness of the hypotheses that $\eta,\zeta$ can be taken continuous in $t,u$.

\section{Proof of Theorem \ref{thm:main-alpha-beta}}

\subsection{Initial reductions} \label{subsec:reductions}

Recall that the (spherical) $\gamma$-dimensional Hausdorff content is defined as
\[
\mathcal{H}_\infty^\gamma(A) = \inf\left\{ \sum_i r_i^\gamma: A\subset \cup_i B(x_i,r_i)\right \}.
\]
Standard, easy to verify properties are that $\mathcal{H}_\infty^\gamma$ is countably subadditive, and $\mathcal{H}_\infty^\gamma(A)>0$ if and only if $\mathcal{H}^\gamma(A)>0$, with the latter denoting standard Hausdorff measure.

Let $E$ be an $(\al,\beta)$-Furstenberg set and let $\mathcal{L}$ denote the set of lines $\ell$ such that $\hdim(E\cap\ell)\ge \al$; then $\hdim(\mathcal{L})\ge \beta$ by definition. By passing to a subset of $\mathcal{L}$, we may assume that the slopes of all lines in $\mathcal{L}$ either are $\ge 1$ or $\le 1$ in absolute value. By rotating $E$ if needed, we may assume that all slopes are $\le 1$ in absolute value. We can then identify $\mathcal{L}$ with a subset of $[-1,1]\times\R$ via $\om\leftrightarrow L_\om$; this is a smooth identification, in particular it preserves Hausdorff dimension.

Fix $\e>0$. By the countable stability of Hausdorff dimension and the countable subadditivity of Hausdorff content, we may find $b_0\in\R$ such that
\[
\mathcal{H}_\infty^{\beta-\e}(\{\om:L_\om\in\mathcal{L}\}\cap B^2((0,b_0),1)) > 0.
\]
By translating $E$, we may assume that $b_0=0$.  Now, since $\mathcal{H}_\infty^{\al-\e}(E\cap L_\om)>0$ for all $L_\om\in\mathcal{L}$, the countable subadditivity of $\mathcal{H}_\infty^{\beta-\e}$ shows that there exists $c>0$ such that
\[
\mathcal{H}_\infty^{\beta-\e}\{\om\in B^2: \mathcal{H}_\infty^{\al-\e}(E\cap L_\om)\ge c \} > 0.
\]
Since $\e>0$ is arbitrary, this shows that in order to establish Theorem \ref{thm:main-alpha-beta}, we may assume that $\mathcal{L}\subset \{ L_\om: \om\in\wt{\Om}\}$ for some $\wt{\Om}\subset B^2$, $\mathcal{H}_\infty^\beta(\wt{\Om})>0$, and
\be \label{eq:positive-content}
\mathcal{H}_\infty^\alpha(E\cap L_\om) \ge c > 0
\ee
for all $\om\in\wt{\Om}$.

Fix a small $\de>0$ and $s\in [1/2,1)$. In the rest of the paper, any implicit constants are allowed to depend on $K,c,\alpha$ and $\beta$, but they must always be independent of $\de$.

F\"{a}ssler and Orponen \cite[Proposition A.1]{FasslerOrponen14} proved a discretized version of Frostman's Lemma (they stated it in $\R^3$ but the same proof works in any dimension). Applying this result to the set $\wt{\Om}$, we obtain a $\delta$-separated set $\Om\subset \wt{\Om}$ such that
\eqref{eq:positive-content} holds for all $\om\in \Om$ and, moreover, $\Om$ satisfies the Frostman condition
\be \label{eq:Om-Frostman}
|\Om\cap B_r|\lesssim r^\beta |\Om|\quad\text{for all }r\in [\de,1].
\ee
Since $\Om$ is nonempty we get that, in particular, $|\Om|\gtrsim \de^{-\beta}$.

Let $E^{(\delta)}$ denote the $\delta$-neighborhood of $E$. Since $\cN_\de(E^{(\de)})\sim \cN_\de(E)$ and $\cN_{\des}(E^{(\de)})\sim \cN_{\des}(E)$, we may replace $E$ by $E^{(\delta)}$; in particular, this allows to assume that $E$ is Borel. Applying Frostman's Lemma to each $E\cap L_\om$, we obtain a $(\delta,\alpha,K)$-Furstenberg tuple $(\nu_\om)_{\om\in\Om}$ where $K=K(c)$. We recall that in Frostman's Lemma the constant factor depends only on the Hausdorff content on the set, and therefore \eqref{eq:positive-content} ensures that a suitable uniform $K$ can be found. Note that
\[
E\supset \bigcup_{\om\in\Om} \supp(\nu_\om) =: \bigcup_{\om\in\Om} E_\om.
\]
and hence it suffices to estimate $\cN_\de$, $\cN_{\des}$ for the union on the right-hand side.

\subsection{Analysis at scale $\des$}
\label{subsec:scale-des}

Assume that
\be  \label{eq:bound-scale-sqr-delta}
\cN_{\des}(E) \le (\des)^{-\alpha-\beta/2-\eta}
\ee
We first show that this counter-assumption forces a rather rigid structure on $E$ at scale $\des$ (at least after refinement of $E$).

Let $\Lambda$ be a maximal $\des$-separated subset of $\Om$. Note that $|\Lambda|\gtrsim (\des)^{-\beta}$ by \eqref{eq:Om-Frostman}.
\begin{lemma} \label{lem:upper-bound-Lambda}
\[
|\Lambda| \lesssim (\des)^{-\beta-2\eta}.
\]
\end{lemma}
\begin{proof}
Note that $( \nu_\om )_{\om\in\Lambda}$ is a $(\des,\alpha,K)$-Furstenberg tuple; in particular, $\cN_{\des}(E_\om)\gtrsim (\des)^{-\al}$. The claim follows from Lemma \ref{lem:alpha+gamma bound} and our assumption \eqref{eq:bound-scale-sqr-delta}.
\end{proof}

For each $\lambda\in\Lambda$, let
\[
\Om_\lam = \Om\cap B^2(\lam,\des).
\]
Note that the sets $\Om_\lam$ have bounded overlapping and cover $\Om$.

\begin{lemma} \label{lem:many-rich-deh-tubes}
There is a set $\Lam'\subset\Lam$ such that
\[
|\Lambda'| \gtrsim (\des)^{-\beta}
\]
and
\be \label{eq:Lambda'}
|\Om_\lam| \gtrsim (\des)^{\beta+2\eta}|\Om|
\ee
for all $\lam\in\Lam'$.
\end{lemma}
\begin{proof}
Let
\[
\Lambda' =\{\lam\in\Lam: |\Om_\lam| \ge C^{-1}(\des)^{\beta+2\eta}|\Om|\}.
\]
It follows from Lemma \ref{lem:upper-bound-Lambda} that if $C$ is taken large enough then
\[
\sum_{\lam\in\Lam\setminus\Lam'} |\Om_\lam| \le |\Om|/2.
\]
Therefore, using \eqref{eq:Om-Frostman},
\[
|\Om|/2\le \sum_{\lam\in\Lam'} |\Om_\lam| \lesssim |\Lam'| (\des)^{\beta}|\Om|,
\]
as claimed.
\end{proof}

Fix $\lam\in\Lam'$. Let
\begin{align*}
\rho_\lam&= \frac{1}{|\Om_\lam|} \sum_{\om\in \Om_\lam} \nu_\om,\\
F_\lam &= \bigcup_{\om\in\Om_\lam} E_\om =\supp(\rho_\lam).
\end{align*}
Let us also denote by $\wt{\rho}_\lam$, $\wt{F}_\lam$ the orthogonal projections of $\rho_\lam, F_\lam$ onto the line $L_\lam$.

\begin{lemma} \label{lem:Frostman-average}
\[
\wt{\rho}_\lam(B_r) \lesssim r^\alpha \quad \text{for } r\in [\de,1].
\]
\end{lemma}
\begin{proof}
Since each $L_\om, \om\in\Om_\lam$ makes an angle $\lesssim \des\ll 1$ with $L_\lam$, the orthogonal projection of $\nu_\om$ to $L_\lam$ satisfies the same Frostman condition as $\nu_\om$ (other than the constant), and hence so does an average of such projections.
\end{proof}

\begin{lemma} \label{lem:deh-rectangles-small-projection}
If $C=C_{K,\alpha}$ is large enough,
\begin{equation} \label{eq:deh-rectangles-small-projection}
\left|\{\lam\in\Lam': \cN_{\des}(\wt{F}_\lam) \ge C(\des)^{-\alpha-2\eta}\}\right| \le |\Lam'|/2.
\end{equation}
\end{lemma}
\begin{proof}
Since $\wt{F}_\lam$ and $F_\lambda$ are $O(\des)$-close in the Hausdorff metric, the assumption \eqref{eq:bound-scale-sqr-delta} yields
\[
\cN_{\des}\left(\bigcup_{\lam\in\Lam'}\wt{F}_\lam\right) \sim \cN_{\des}\left(\bigcup_{\lam\in\Lam'} F_\lam\right) \le \cN_{\des}(E) \le (\des)^{-\alpha-\beta/2-\eta} .
\]
Let $\Lambda''\subset\Lam'$ be the set on the left-hand side of \eqref{eq:deh-rectangles-small-projection}. By Lemma \ref{lem:Frostman-average},  $(\wt{\rho}_\lam)_{\lam\in\Lambda''}$ is a $(\des,\alpha,O_K(1))$-Furstenberg tuple. It follows from Lemma \ref{lem:alpha+gamma bound} that $|\Lambda''|\le c (\des)^{-\beta}$, where $c$ can be made arbitrarily small by taking $C$ large. The conclusion now follows from Lemma \ref{lem:many-rich-deh-tubes}.
\end{proof}
In light of this lemma, by passing to a subset of $\Lambda'$ that still satisfies the conclusions of Lemma \ref{lem:many-rich-deh-tubes}, we may assume from now on that
\begin{equation}  \label{eq:assumption-deh-rectangles-small-projection}
\cN_{\des}(\wt{F}_\lam) \lesssim (\des)^{-\alpha-2\eta} \quad\text{for all }\lam\in\Lambda'.
\end{equation}

\subsection{Analysis of a fixed $F_\lam$ at scale $\de$}
\label{subsec:analysis-F-lam}

Fix $\lam\in\Lam'$ for the rest of this section. Cover $L_\lam\cap B^2$ by intervals of length $2\des$. As above, let $\wt{F}_\lam$ be the orthogonal projection of $F_\lam$ onto the line $L_\lam$. Let us denote the subset of $(2\des)$-intervals that intersect $\wt{F}_\lam$ by $(I_j)_{j=1}^N$.  By \eqref{eq:assumption-deh-rectangles-small-projection}, we know that
\be \label{eq:N-upper-bound}
N \lesssim (\des)^{-\alpha-2\eta}.
\ee
Denote by $Q_j$ the square of side length $2\des$ whose orthogonal projection onto $L_\lam$ is $I_j$, and such that $I_j$ divides it into equal pieces.

\begin{lemma} \label{lem:J}
There exists a set $J\subset \{1,\ldots,N\}$ such that
\[
\sum_{j\in J} \wt{\rho}_\lam(I_j) \ge 1/2,
\]
and
\be \label{eq:om-j-large}
|\Om_\lam^{j}|\gtrsim (\des)^{2\eta} |\Om_\lam|
\ee
for all $j\in J$, where
\be \label{eq:om-large-int-with-square}
\Om_\lam^{j} = \{ \om\in\Om_\lam: \cN_\de(E_\om\cap Q_j)   \gtrsim \de^{2\eta s} (\dess)^{-\al} \}.
\ee
\end{lemma}
\begin{proof}
By Lemma \ref{lem:Frostman-average}, $\wt{\rho}_\lam(I_j)\lesssim (\des)^\alpha$ for all $j$. Let
\[
J = \{j\in\{1,\ldots,N\}:\wt{\rho}_\lam(I_j)\ge C^{-1}(\des)^{\alpha+2\eta}\}.
\]
Notice that
\[
\sum_{j\in\{1,\ldots,N\}\setminus J} \wt{\rho}_\lam(I_j) \lesssim  C^{-1}N(\des)^{\alpha+2\eta},
\]
and thus if $C$ is taken large enough, the first claim follows from \eqref{eq:N-upper-bound}.

We now show the second claim. It follows from the definitions that
\[
\wt{\rho}_\lam(I_j) = \frac{1}{|\Om_\lam|}\sum_{\om\in\Om_\lam} \nu_\om(Q_j).
\]
Splitting the sum depending on whether $\nu_\om(Q_j)\ge c(\des)^{\alpha+2\eta}$ or not, where $c$ is a suitably small constant, we deduce that
\[
|\{ \om\in\Om_\lam: \nu_\om(Q_j)\ge c(\des)^{\alpha+2\eta} \}| \gtrsim (\des)^{2\eta}|\Om_\lam|.
\]
The second claim now follows from the Frostman condition of $\nu_\om$.
\end{proof}

We now introduce the projections that will allow us to apply Bourgain's projection theorem in our setting, and state the key lemma that relates the box-counting numbers $\cN_\de(E\cap Q_j)$ to such projections.

Let
\[
\pi_t(\om) = L_\om(t)\quad\text{or}\quad \pi_t(a,b)=at+b.
\]
\begin{lemma} \label{lem:box-counting-from-projection}
Fix $j\in J$. Then, for any $t\in I_j$,
\[
\cN_\de(E\cap Q_j) \gtrsim   \de^{2\eta s} (\dess)^{-\al}  \, \cN_{\de}\left(\pi_t \Om_\lam^{(j)}\right).
\]
\end{lemma}
\begin{proof}
Fix $t\in I_j$. Since all $L_\om$, $\om\in\Om_\lam$ make an angle $\le 2 \des$ with each other and $Q_j$ has diameter $<4 \des$, the segments $L_\om\cap Q_j, L_{\om'}\cap Q_j$ are $(2\de)$-separated provided  $|\pi_t(\om)-\pi_t(\om')| \ge 10\de$. This is the point where we use that $s\ge 1/2$.

We can thus take a subset of $\Om_\lam^{(j)}$ of size $\sim \cN_{\de}(\pi_t \Om_\lam^{(j)})$ such that the corresponding segments $L_\om\cap Q_j$ are $(2\de)$-separated. The claim now follows from \eqref{eq:om-large-int-with-square}.
\end{proof}

We are now able to apply Bourgain's projection theorem to obtain a gain in the size of $\cN_\de(E\cap Q_j)$ for most $j$.
\begin{prop} \label{prop:application-bourgain-proj-thm}
If $\zeta,\eta$ are sufficiently small in terms of $\al,\beta$ (with the threshold depending in a continuous manner on $\al,\beta$), then for $\de$ small in terms of all the previous parameters,
\[
\cN_\de(E \cap Q_j) \gtrsim (\dess)^{-\al-\beta/2-\zeta}
\]
on a set of $j$ such that $\sum_j \wt{\rho}_\lam(I_j)\ge 1/4$.
\end{prop}
\begin{proof}
In light of Lemmas \ref{lem:J} and \ref{lem:box-counting-from-projection}, it is enough to show that there is $\zeta=\zeta(\alpha,\beta)>0$ such that, provided $\eta=\eta(\alpha,\beta)>0$ is sufficiently small, then
\be \label{proj-to-prove}
\cN_{\de}\left(\pi_t\Om_\lam^{(j)}\right) \ge (\dess)^{-\beta/2-\zeta}
\ee
for all $t$ in a set of $\wt{\rho}_\lam$-measure $\ge 3/4$. (Making $\eta$ small enough that $2\eta \le \zeta/2$, this will give the claim with $\zeta/2$ in place of $\zeta$.) We will see that this is a consequence of Theorem \ref{thm:projection-Bourgain}. To begin, recall from Lemma \ref{lem:Frostman-average} that $\wt{\rho}_\lam(B_r)\lesssim r^\alpha$ for $r\in [\delta,1]$.

We now rescale and translate $\Om_\lam$ to bring it into the setting of Theorem \ref{thm:projection-Bourgain}: let
\[
X_\lam = \de^{-s}(\Om_\lam-\lam) = \{ \de^{-s}(\om-\lam):\om\in\Om_\lam\}.
\]
Note that $X_\lam\subset B^2$ and $X_\lam$ is $(\dess)$-separated. It follows from \eqref{eq:Om-Frostman} and \eqref{eq:Lambda'} that, for $r\in [\dess,1]$,
\[
|X_\lam\cap B_r| = |\Om_\lam\cap B_{\des r}| \lesssim (\des r)^{\beta}|\Om| \lesssim \de^{-2\eta s} r^\beta |X_\lam|.
\]
In particular, taking $r=\dess$,
\[
|X_\lam| \gtrsim \de^{2\eta s} (\dess)^{-\beta}.
\]
Thus (provided $\eta$ is small enough in terms of $\al$ and $\beta$ only) $X_\lam$ and $\wt{\rho}_\lam$ satisfy the assumptions of Theorem \ref{thm:projection-Bourgain} (with $t=\alpha$, $u=\beta$ and scale $\dess$ in place of $\de$). Now the maps $\pi_t$ are, up to bounded rescaling, a smooth reparametrization of the family of orthogonal projections (given by $\theta\mapsto t=\tan(\theta)$).

Applying Theorem \ref{thm:projection-Bourgain}, at scale $\dess$, to the rescaled and translated sets $X'_j=\de^{-s}(\Om_\lam^{j}-\lam)$ (which is permissible by \eqref{eq:om-j-large}), we conclude that there are $\zeta=\zeta(\alpha,\beta)>0$, $\eta=\eta(\alpha,\beta)$ (depending continuously on the parameters) such that if $\de$ is sufficiently small in terms of all previous parameters, then the lower bound \eqref{proj-to-prove} holds for all $t$ outside of a set of $\wt{\rho}_\lam$-measure $\le (\dess)^{\eta}\le 1/4$.
\end{proof}

\subsection{Conclusion of the proof}
\label{subsec:conclusion-proof}

Fix $\zeta,\eta$ as given by Proposition \ref{prop:application-bourgain-proj-thm}. In this proposition the value of $\lam$ is fixed. We now use a double counting argument to conclude that, always under the standing assumption \eqref{eq:bound-scale-sqr-delta},  $\cN_\de(E)$ must be large.

Let $\cD$ be the grid of axes-parallel squares of side length $\des$. Call a square $D\in\cD$ \emph{rich} if
\[
\cN_\de(E\cap D) \ge c (\dess)^{-\alpha-\beta/2-\zeta},
\]
where $c>0$ is a sufficiently small constant to be chosen later. Let $\cR$ denote the collection of rich squares; our goal is to show that $|\cR|$ is large. For each $\lam\in\Lam'$, let $\cD_\lam$ be all the squares in $\cD$ intersecting the $\des$-neighborhood of $L_\lam$. It follows from Proposition \ref{prop:application-bourgain-proj-thm} (and the Frostman estimate $\wt{\rho}_\lam(B_r)\lesssim r^\alpha$ given in Lemma \ref{lem:Frostman-average}) that, provided $c$ is taken sufficiently small,
\[
|\cR\cap \cD_\lam | \gtrsim (\des)^{-\alpha}.
\]
Hence, recalling Lemma \ref{lem:many-rich-deh-tubes},
\be \label{eq:rich-squares-double-counting}
\sum_{\lam\in\Lam'} |\cR\cap \cD_\lam |\gtrsim (\des)^{-\alpha}|\Lam'| \gtrsim (\des)^{-\alpha-\beta}.
\ee
How many times is each rich square counted? If $\Lambda_D$ denotes the set of all $\lam\in\Lam'$ such that $D\in\cD_\lam$, then $(\nu_\om)_{\om\in\Lambda_D}$ is a $(\des,\alpha,K)$-Furstenberg tuple such that all lines pass at a distance $\lesssim \des$ from the center of $D$. Lemma \ref{lem:fan} and our standing assumption \eqref{eq:bound-scale-sqr-delta} then show that
\[
|\Lambda_D|\lesssim (\des)^{-\beta/2-\eta}.
\]
Hence every rich square contributes to the sum in \eqref{eq:rich-squares-double-counting} at most $C(\des)^{-\beta/2-\eta}$ times, and we conclude that
\[
|\cR|\gtrsim (\des)^{-\alpha-\beta/2+\eta}.
\]
Recalling the definition of rich square,  we conclude that
\[
\cN_\delta(E) \gtrsim \de^{\eta s-\zeta(1-s)} \de^{-\alpha-\beta/2}.
\]
So far, $\eta,\zeta$ must be small in terms of $\al,\beta$ only but are otherwise independent of each other. We can then make $\zeta$ smaller in terms of $\alpha,\beta$ only and choose $\eta$ depending also on $s$ so that $\eta = (1-s)\zeta/4$.

With this choice, and recalling \eqref{eq:bound-scale-sqr-delta} one last time,  we have concluded the proof of \eqref{eq:main-scale-dichotomy} ,with $\zeta/4$ in place of $\zeta$, and with it of Theorem \ref{thm:main-alpha-beta}.