TITLE: Proof of Lemma 2.3 p.354 Massey
QUESTION [3 upvotes]: Lemma 2.3: Let $M$ be a an $n-$manifold and $G$ be an abelian group.
(a) For any compact $K \subset M$ and $i > n$ $$H_i(M,M-K;G) = 0$$
(b) If $u \in H_n(M,M-K;G)$ and $\rho_x(u) = 0$ for all $x\in K$, then
$u=0$.

I'd like to understand the case $4$ of the Lemma, i.e $M = \mathbb{R}^n$ and $K$ an arbitrary compact.
$\bullet$ We use the following fact which I'm unable to prove: "We assert that for any $u \in H_i(\mathbb{R}^{n},\mathbb{R}^{n}\setminus K)$ exists an open set $N$ containing $K$ and elements $u'\in H_i(\mathbb{R}^{n},\mathbb{R}^{n}\setminus N)$ such that $k_*(u') = u$ where $k$ is the inclusion map.
I do understand the "recall" part of the sentence, but I don't understand how to explicitly write the details.
Let $(X,A) \subseteq (\mathbb{R}^n,\mathbb{R}^n \setminus K)$ with $v \in H_i(X,A)\subseteq H_i(\mathbb{R}^n,A)\longmapsto u \in H_i(\mathbb{R}^n,\mathbb{R}^n \setminus K)$ under inclusion be given.
What I thought : Since $A \subseteq \mathbb{R}^n \setminus K$ and both $A,K$ are compact, we have "positive distance". Being $K$ compact, we can choose $N$ as finite union of open balls $B_j$ that cover $K\subseteq \bigcup\limits_{j=1}^n B_j$ such that $B_j \cap A =  \varnothing$ ($N \cap A = \varnothing$).
So in reality $v \in H_i(\mathbb{R}^n,\mathbb{R}^n \setminus N) \longmapsto u \in H_i(\mathbb{R}^n,\mathbb{R}^n \setminus K)$.
Is this reasoning correct? I'm not sure about last sentence since I think we could have that $v$ has image in $\text{Imm} \mathbb{R}^n \setminus N$.

REPLY [2 votes]: It is known that singular homology has compact carriers which means that for each space $X$ and each $x \in H_i(X)$ there exists a compact $C \subset X$ such that $x$ is in the image of the inclusion induced $j_* : H_i(C) \to H_i(X)$ (let me know if you want a proof). The same is of course true for reduced homology groups (which only differ in dimension $0$ from the unreduced groups).
Le $N$ be any set such that $K \subset N \subset \mathbb R^n$. Now consider the long exact sequences of reduced homology groups of the pairs $(\mathbb{R}^{n},\mathbb{R}^{n}\setminus K)$ and $(\mathbb{R}^{n},\mathbb{R}^{n}\setminus N)$.  Since all reduced homology groups of $\mathbb R^n$ are $0$, we get for $i > 0$ commutative squares
$\require{AMScd}$
$$\begin{CD}
H_i(\mathbb{R}^{n},\mathbb{R}^{n}\setminus N) @>\partial>>\tilde H_{i-1}(\mathbb{R}^{n}\setminus N)\\
@Vk_*VV@Vj_*VV@.@.\\
H_i(\mathbb{R}^{n},\mathbb{R}^{n}\setminus K) @>\partial>>\tilde H_{i-1}(\mathbb{R}^{n}\setminus K)
\end{CD}$$
where the horizontal arrows are isomorphisms and the vertical arrows are inclusion induced. Note that the long exact sequences show that $H_0(\mathbb{R}^{n},\mathbb{R}^{n}\setminus N) = H_0(\mathbb{R}^{n},\mathbb{R}^{n}\setminus K) = 0$, thus your question concerning the existence of $N$ and $u'$ is trivial for $i = 0$ (you may take $N = \mathbb{R}^{n}$).
Thus, given the element $u$, consider $x = \partial(u)$ and find compact $C \subset \mathbb{R}^{n}\setminus K$ and $x' \in H_{i-1}(C)$ such that $j_*(x') = x$. Clearly $N = \mathbb R^n \setminus C$ is an open set containing $K$. By definition $C= \mathbb{R}^{n}\setminus N$. Now let $u' = \partial^{-1}(x')$.


TITLE: The annihilator $\mathrm{Ann}_R(m)\subseteq R$ is a maximal left ideal of $R$
QUESTION [1 upvotes]: Let $R$ be a ring with $1_R$, $M$ a simple left $R$-module and $m\in M$ is a non-zero element.
We would like to prove that the annihilator $\mathrm{Ann}_R(m)\subseteq R$ is a maximal left ideal of $R$.
It may be something obvious but unfortunately I can't see it.
To prove that it is a left ideal is easy. Now I stuck at the maximality. The first thought is to assume that
$$\mathrm{Ann}_R(m) \subsetneqq J \subseteq R.$$
So, it suffices to show that in this case $J=R$.
We know that since $$\mathrm{Ann}_R(m) \subsetneqq J\implies\exists j\in J \setminus \mathrm{Ann}_R(m) \implies jm\neq 0_M \implies jm=n,$$ for some $n\neq 0_M$.
So if we take an element $x\in R$ we should show that $x\in J$. But how?
Thank you.

REPLY [3 votes]: $M$ is simple, so it's cyclic, hence $M=Rm$. But $n=jm\neq 0$ is also in $M$, so $Rn=M$. As $m\in M$, there is an $r\in R$ such that $m=rn=rjm$, meaning $rj-1\in\text{Ann}_{R}(m)$. Yet since $\text{Ann}_{R}(m)\subset J$, we have $rj-1\in J$, but $rj\in J$ so we also have $1\in J$.


TITLE: Laplace Transform of $\int_{0}^{t}u^2e^u\,du$
QUESTION [0 upvotes]: Good day,
I am having some problems with the Laplace transform of $f(t) = \int_{0}^t{u^2e^u\textrm{d}u}$.
$$\mathcal{L}\left [\int_{0}^t{u^2e^u\textrm{d}u}  \right ]$$
using the property :
$$\mathcal{L}
\left [\int_{0}^{t}f(u)\mathrm{d}u \right ] =
\frac{F(s)}{s},s>\textrm{max}\left \{ 0,a \right \}$$
This is what I have tried so far :

$$\int_{0}^{t}u^2e^u\textrm{d}u = e^tt^2-2te^t+2e^t-2 \textrm{ (Integration by parts)}$$
$$\mathcal{L}
\left [e^tt^2-2te^t+2e^t-2 \right ] = \frac{2}{(s-1)^3}-\frac{2}{(s-1)^2}+\frac{2}{s-1}-\frac{2}{s}$$
$$ \frac{\mathcal{L} \left [e^tt^2-2te^t+2e^t-2 \right ]}{s} = \frac{2}{s^2(s-1)^3} $$

The solution of the problem is : $$\frac{2}{s(s-1)^3}$$
Thanks in advance,
Emanuel Camacho

REPLY [1 votes]: You did not use the suggested property. Since:
$$ \mathcal{L}(t^2 e^t) = \frac{2}{(s-1)^3},\tag{1}$$
it follows that:
$$ \mathcal{L}\left(\int_{0}^{t}u^2 e^u\,du\right) = \frac{2}{s(s-1)^3}.\tag{2}$$
To prove $(1)$, you just have to recall that $\mathcal{L}(t^2)=\frac{2}{s^3}$.


TITLE: Prove $u(x,y,z)=f(t+r)/r+g(t-r)/r$ satisfies $u_{xx}+u_{yy}+u_{zz}=u_{tt}$.
QUESTION [3 upvotes]: I'm trying to prove that any function $u=u(x,y,z)$ of the form $$u(x,y,z)=\frac{f(t+r)}{r}+\frac{g(t-r)}{r},$$ with $r^2=x^2+y^2+z^2$, satisfies the differential equation $ u_{xx}+u_{yy}+u_{zz}=u_{tt}, $ but I'm stuck when relating the derivatives.
From what I get, $$u_{tt} = \dfrac{1}{r}(f''+g''),$$ as $r$ does not depend on $t$.
Furthermore, $$ u_x = \dfrac{r_x}{r^2}\left( r(f'-g')-(f+g) \right), $$ so, if I'm not mistaken,
$$u_{xx} = \dfrac{x^2}{r^3} (f''+g'') + \dfrac{r^2-3x^2}{r^4}(f'-g') + \dfrac{x^2-r^2}{r^5}(f+g). $$
(By symmetry, $u_{yy}$ and $u_{zz}$ have the same form).
But I don't know how to relate this expression to the other second derivatives, so I think I'm missing something...
Any ideas are greatly appreciated.
Thanks in advance!
EDIT: Corrected a $r^2$ term in the expression for $u_x$. Also, thanks to @Klaramun's and @TedShifrin's suggestions, I managed to reduce the expression for $u_{xx}$, noting that $f_x=f_y=f_z=f',\, g_x=g_y=g_z=g'$, so I fixed that a couple lines up.

REPLY [2 votes]: First,  consider the function
$$h(x,y,z)=\frac{f(t+r)}{r} \tag 1$$
and with $r^2=x^2+y^2+z^2\quad;\quad r\frac{\partial r}{\partial x}=x \quad;\quad \frac{\partial r}{\partial x}=\frac{x}{r}$
$$h_x=\left(\frac{f'(t+r)}{r}-\frac{f(t+r)}{r^2} \right)\frac{\partial r}{\partial x}=x\frac{f'}{r^2}-x\frac{f}{r^3} $$
In interest of space, we write $f$ instead of $f(t+r)$ and $f'$ instead of $f'(t+r)$ .
$$h_{xx}=\left(\frac{f'}{r^2}-\frac{f}{r^3}\right)+\left(x\frac{f''}{r^2}-x\frac{f'}{r^3} \right)\frac{x}{r}+\left(-2x\frac{f'}{r^3}+3x\frac{f}{r^4} \right)\frac{x}{r}$$
$$h_{xx}=\frac{f'}{r^2}-\frac{f}{r^3}+x^2\frac{f''}{r^3}-x^2\frac{f'}{r^4} -2x^2\frac{f'}{r^4}+3x^2\frac{f}{r^5}$$
$$h_{xx}=x^2\left(\frac{f''}{r^3}  -3\frac{f'}{r^4}+3\frac{f}{r^5}\right)+\frac{f'}{r^2}-\frac{f}{r^3}$$
On the same manner, we get :
$$h_{yy}=y^2\left(\frac{f''}{r^3}  -3\frac{f'}{r^4}+3\frac{f}{r^5}\right)+\frac{f'}{r^2}-\frac{f}{r^3}$$
$$h_{zz}=z^2\left(\frac{f''}{r^3}  -3\frac{f'}{r^4}+3\frac{f}{r^5}\right)+\frac{f'}{r^2}-\frac{f}{r^3}$$
Adding the three equations
$$h_{xx}+h_{yy}+h_{zz}=(x^2+y^2+z^2)\left(\frac{f''}{r^3}  -3\frac{f'}{r^4}+3\frac{f}{r^5}\right)+3\frac{f'}{r^2}-3\frac{f}{r^3}$$
$$h_{xx}+h_{yy}+h_{zz}=r^2\left(\frac{f''}{r^3}  -3\frac{f'}{r^4}+3\frac{f}{r^5}\right)+3\frac{f'}{r^2}-3\frac{f}{r^3}$$
$$h_{xx}+h_{yy}+h_{zz}=\frac{f''}{r}$$
From Eq.$(1)\quad:\quad h_t=\frac{f'}{r}\quad$and$\quad h_{tt}=\frac{f''}{r}\quad$thus
$$h_{xx}+h_{yy}+h_{zz}=h_{tt}$$
This proves that
$$\frac{f(t+r)}{r}\quad \text{is solution of}\quad u_{xx}+u_{yy}+u_{zz}=u_{tt}$$
Second, on the same manner one prouves that
$$\frac{g(t-r)}{r}\quad \text{is solution of}\quad u_{xx}+u_{yy}+u_{zz}=u_{tt}$$
Any linear combination of solutions of the PDE $(1)$ is solution of the PDE because the PDE is linear. Thus
$$\frac{f(t+r)}{r}+\frac{g(t-r)}{r} \quad\text{satisfies}\quad u_{xx}+u_{yy}+u_{zz}=u_{tt}$$


TITLE: From set-builder notation to limits of integration
QUESTION [1 upvotes]: I have an integral of the type:
\begin{equation}
I=\int_A f(x,y)\,dx\,dy
\end{equation}
where
\begin{equation}
A=\{(x,y) : x\leq X,y\leq Y,x+y \geq W\}
\end{equation}
Assuming that $f(x,y)$ has a closed-form anti-derivative, and that $A$ is not empty, I can handle this case by sketching the region $A$ (a triangle) and defining the limits of the double-integral to evaluate $I$.
With a 3-dimension problem maybe I can use the shadow method or the cross section method, but my problem is in dealing with higher dimensions, where I cannot sketch the region $A$. Being unable to define the limits of integration, the only way I know to evaluate $I$ is to use numerical integration methods (i.e. MonteCarlo), using an indicatory function $\mathbf{1}_A (x,y)$ so I can consider only values of $(x,y)$ actually in $A$.
So, I was wondering how to find the limits of integration given a set $A$ in the form:
\begin{equation}
A=\{(x_1,\dots,x_n) : x_i\leq X_i, \sum_{i=1}^n x_i \geq W\}
\end{equation}
Since the equations defining $A$ are all linear inequalities, is there a general way to move from the set-notation to the limits of integration? As far as I understood, this is generally a hard problem, but I was wondering whether the specific structure of the inequalities can be exploited to simplify it.
Any hint is appreciated.

REPLY [1 votes]: In your setup the $x_i$ are not apriori bounded from below, which makes things considerably simpler: When $S:=\sum_{i=1}^n X_i-W<0$ then $A$ is empty, and when  $S>0$ then $A$ is an $n$-dimensional simplex with one vertex at ${\bf X}:=(X_1,\ldots, X_n)$ and the edges  emanating from $X$  going "backwards" parallel to the axes, as in your two-dimensional example.
Maybe it helps to substitute $x_i:=X_i-y_i$, where now the $y_i$ are $\geq 0$, and $\sum_{i=1}^n y_i\leq S$. Therefore we now have to integrate the function $g(y_1,\ldots ,y_n):=f(X_1-y_1,\ldots, X_n-y_n)$ over the $n$-dimensional simplex
$$\Delta:=\left\{(y_1,\ldots,y_n)\ \biggm|y_i\geq 0\ (1\leq i\leq n),\ \sum_{i=1}^n y_i\leq S\right\}$$
in $y$-space.  As the Jacobian has absolute value $1$ we get
$$J:=\int_A f({\bf x})\ {\rm d}({\bf x})=\int_\Delta g({\bf y})\ {\rm d}({\bf y})\ .$$
The integral on the right can be reduced recursively. Note that the projection of $\Delta$ to the plane of the first $n-1$ coordinates $y_i$ is an $(n-1)$-dimensional simplex
$$\Delta':=\left\{(y_1,\ldots,y_{n-1})\ \biggm|\ y_i\geq 0\ (1\leq i\leq n-1), \ \sum_{i=1}^{n-1} y_i\leq S\right\}$$ with the same $S$. There is an "outer" integral over $\Delta'$ and an "inner" integral with respect to $y_n$: For each $(y_1,\ldots ,y_{n-1})\in \Delta'$ we have to integrate along a vertical "stalk" given by $0\leq y_n\leq S-(y_1 +\ldots+y_{n-1})$. The result of the "inner" integration is a function $g_1(y_1,\ldots,y_{n-1})$  depending only on the first $n-1$ variables $y_i$, and which then has to be integrated over $\Delta'$.
Therefore the first recursion step becomes
$$J=\int_{\Delta'}\left(\int_0^{S-(y_1+\ldots+y_{n-1})} g({\bf y})\ dy_n\right)\ {\rm d}({\bf y}')\ .$$


TITLE: Sound wave propagation through a wall
QUESTION [2 upvotes]: My question are actually two, but are somehow related.
1) Does a sound wave always (no matter what frequency) needs of the interaction (set in motion particles) with the element the wave is transmitted through?
2) Is the sound that gets transmitted to the other side of the wall only a result of the vibration transmitted through the wall, or also a combination with the sound that is radiated by the wall at its resonant frequencies after having been excited by the impinging sound wave?

REPLY [2 votes]: This could be thought of in terms of the impulse response of the wall.
If the impulse response was non oscillatory (say, a decaying exponential) then the sound on the other side would be a low pass filtered version of the impinging wave--whether it is an impulse or something else.
Otherwise the wave on the other side could include the resonant frequencies--which would be summed together with the low pass filtered impinging frequencies (assuming the impinging wave contains the some of the resonant frequencies ).

1) Does a sound wave always (no matter what frequency) needs of the
  interaction (set in motion particles) with the element the wave is
  transmitted through?

Yes--in all cases this would depend on vibrating particles since sound is a mechanical wave.

2) Is the sound that gets transmitted to the other side of the wall
  only a result of the vibration transmitted through the wall, or also a
  combination with the sound that is radiated by the wall at its
  resonant frequencies after having been excited by the impinging sound
  wave?

(As you say) It is a combination of the sound that is radiated by the wall with the wall's resonant frequencies after having been excited by the impinging sound wave? So the sound on the far side of the wall would consist of the frequency components of impinging sound wave with their amplitudes modified by the frequency response of the wall including any resonant frequencies.


TITLE: A strange answer for $\int _{-1}^1 \log x\; dx$
QUESTION [6 upvotes]: I typed $\int _{-1}^1 \log x\; dx$ on Wolfram Alpha. It is giving the answer to be $-2+i\pi$. Can someone please explain what is happening?

REPLY [0 votes]: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
With the $\ds{\ln}$-branch cut:
$$
\ln\pars{z}=\ln\pars{\verts{z}} + {\rm Arg}\pars{z}\ic\,,\quad
-\,{\pi \over 2} < {\rm Arg}\pars{z} < {3\pi \over 2}\,,\quad z \not= 0
$$

\begin{align}&\lim_{\epsilon\ \to\ 0^{+}}\braces{%
\int_{-1}^{-\epsilon}\bracks{\ln\pars{-x} + \pi\,\ic}\,\dd x
+\int_{\pi}^{0}
\ln\pars{\epsilon\expo{\ic\theta}}\,\epsilon\expo{\ic\theta}\ic\dd\theta
+\int_{\epsilon}^{1}\ln\pars{x}\,\dd x}
\\[3mm]&=\lim_{\epsilon\ \to\ 0^{+}}\braces{%
\int_{\epsilon}^{1}\bracks{\ln\pars{x} + \pi\ic}\,\dd x
+\int_{\epsilon}^{1}\ln\pars{x}\,\dd x}
=2\ \overbrace{\int_{0}^{1}\ln\pars{x}\,\dd x}
^{\ds{\color{#c00000}{\large=\ -1}}}\ +\ \pi\ic
\\[3mm]&=\color{#66f}{\Large -2 + \pi\ic}
\end{align}


TITLE: Show: $f_n \to f, a.e. \Rightarrow F_n(x) := \int_{(-\infty, x]}{f_n d\lambda^1} \to F_0$
QUESTION [1 upvotes]: Let $(f_n)_{n\in \Bbb N_0} \subset \mathcal M_+(\mathcal B^1)$, so that $\forall n \in \Bbb N_0$, $\int_{\Bbb R}{f_n d\lambda^1} = 1$. For $n \in \Bbb N_0$ we define: $$F_n : \Bbb R \to [0,1]; \:\:\: F_n(x) := \int_{(-\infty, x]}{f_n d\lambda^1}$$
Now i have to show that: $$\lim_{n\to \infty}f_n = f_0, \:\:\:  \lambda^1-a.e. \Rightarrow \forall x\in \Bbb R \lim_{n\to \infty} F_n(x) = F_0(x)$$
I also have to show whether the reverse direction is also true.
Any ideas or tips on how to show these? Thanks in advance!

REPLY [1 votes]: See the following answer for a generalisation of the dominated convergence theorem:
https://math.stackexchange.com/a/2053719/27978.
Then fix $x$ and note that with $\phi_n = f_n \cdot 1_{(-\infty,x]}$, we have $\phi_n(x) \to \phi_0(x)$ ae. $x$, $0 \le\phi_n \le f_n$,
$f_n(x) \to f_0(x)$ ae. $x$ and $\int f_n \to \int f_0$ (in fact, the integrals are all equal).
Then we can conclude from the above that $F_n(x) = \int \phi_n \to \int \phi_0 = F_0(x)$. Since $x$ was arbitrary, we are finished.


TITLE: Prove that $MN-NM$ is singular.
QUESTION [10 upvotes]: Let $M$ and $N$ be square matrices such that $M^2+N^2=MN$. Then prove that $MN-NM$ is singular.

So basically I have to prove: $\det(MN-NM)=0$. I tried to prove this by multiplying the given condition by the inverse of matrices $M$ and $N$, but could not come to the answer. Could anyone please give a hint?

REPLY [11 votes]: We work in $M_n(\mathbb{C})$.
We put $M=uX+vY,N=wX+tY$ and we seek the complex numbers $u,v,w,t$ so that $ut-vw\not= 0$ and the coefficients of $X^2,Y^2$ are $0$.
We obtain, for example, $M=X+Y,N=-jX-j^2Y$, where $j=\exp(2i\pi/3)$, and $XY=j^2YX$.
Therefore $MN-NM=(j-j^2)(XY-YX)$ and $\det(MN-NM)=0$ iff $\det(XY-YX)=0$ iff $\det(YX)=0$.
Since $\det(XY)=j^{2n}\det(YX)$, there are $2$ cases
i) $n$ is not a multiple of $3$. Then $\det(XY)=0$ and we are done.
ii) $n$ is a multiple of $3$. Then, over $\mathbb{C}$, the required result is false
Example for $n=3$: $X=\begin{pmatrix}0&0&1\\1&0&0\\0&1&0\end{pmatrix},Y=\begin{pmatrix}0&1&0\\0&0&j\\j^2
&0&0\end{pmatrix}$.


TITLE: Energy gap in Parent hamiltonian of MPS
QUESTION [1 upvotes]: Given a block injective matrix product state (MPS) with D blocks, how does the energy gap of corresponding parent hamiltonian scale with D? And is there a good reference which gives an analysis of this?
Thanks.

REPLY [3 votes]: The gap of the parent Hamiltonian does not depend on the number $D$ of blocks (at least not directly). The spectral gap of the parent Hamiltonian in the block-injective case is analyzed and a lower bound is given in B. Nachtergaele, Commun. Math. Phys. 175, 565 (1996), arXiv:cond-mat/9410110.


TITLE: Prove that for $z \in\mathbb C\setminus \{1\},\, 1 + z + z^2 +... + z^n = (1-z)^{n+1}/(1 - z)$ for $n = 1,2,3\ldots$
QUESTION [0 upvotes]: Prove that for $z \in\mathbb C\setminus \{1\}$,
$$\ 1 +z+z^2+...+z^n=\frac{1- z^{n+1}}{1-z}\quad   (n = 1,2,3\ldots)$$
This is a complex analysis problem. I have been out of an official math class for a long time and am struggling to even know where to approach a problem like this, more or less where to start. Any help is greatly appreciated.

REPLY [0 votes]: This is an example of "telescoping", since the center of the series collapses and we get $(1-z)(1+z+z^2+\cdots +z^n)=1-z+z-z^2+z^2+\cdots -z^{n-1}+z^{n-1}-z^n+z^n-z^{n+1}=1-z^{n+1}$.  Just divide by $(1-z)$ to complete the proof.


TITLE: Cryptanalyze the following cipher using modular exponentiation.
QUESTION [1 upvotes]: I am given a modulus $p=29$ and the following ciphertext: 04 19 19 11 04 24 09 15 15. I am also given that 24 corresponds to the plaintext letter U (in otherwords, 20).
I know that $C\equiv P^e$ mod $p$ in modular exponentiation. I can use the fact that 24 corresponds to the plaintext 20 so  $24 \equiv 20^e$ mod $p$. I can guess and check to find that $e=5$. However, is there a more algorithmic way of finding this to expand to cases when it isn't so easy to guess and check?

REPLY [0 votes]: This is known as the Discrete Log Problem and the site lists several algorithms to solve it.
The discrete logarithm problem is to find the exponent in the expression $$Base^{Exponent} = Power \pmod{Modulus}$$
In your case, we have
$$20^e \equiv 24  \pmod{29}$$
One of the algorithms listed (learn and try some of the others) on the site (overkill for this small problem) is Pohlig-Hellman. We will use the Discrete logarithm calculator by Dario Alpern and it immediately finds
$$e = 5$$
As an aside, you can see a more detailed and worked example here: Use Pohlig-Hellman to solve discrete log.
For the message, we have

$04 \equiv P^5 \pmod{29} \implies P = 06 = G$
$19 \equiv P^5 \pmod{29} \implies P = 14 = O$
$19 \equiv P^5 \pmod{29} \implies P = 14 = O$
$11 \equiv P^5 \pmod{29} \implies P = 03 = D$
$04 \equiv P^5 \pmod{29} \implies P = 06 = G$
$24 \equiv P^5 \pmod{29} \implies P = 20 = U$
$09 \equiv P^5 \pmod{29} \implies P = 04 = E$
$15 \equiv P^5 \pmod{29} \implies P = 18 = S$
$15 \equiv P^5 \pmod{29} \implies P = 18 = S$


TITLE: How do I divide a function into even and odd sections?
QUESTION [42 upvotes]: While working on a proof showing that all functions limited to the domain of real numbers can be expressed as a sum of their odd and even components, I stumbled into a troublesome roadblock; namely, I had no clue how one divides the function into these even and odd parts.
Looking up a solution for the proof, I found these general formulas for the even and odd parts of a function $f(n)$:
$$\begin{align*}
f_e(n)&\overset{\Delta}{=}\frac{f(n)+f(-n)}{2}\\
f_o(n)&\overset{\Delta}{=}\frac{f(n)-f(-n)}{2}
\end{align*}$$
While I understand that in an even function $f(n) = f(-n)$ and that in an odd function $f(-n) = -f(n)$, I still don't get how these general formulas for the even and odd parts were obtained. Can someone guide me through the logic?

REPLY [0 votes]: f(x)= fe(x)+fo(x).......(1)
The properties of fe(x) and fo(x) are:
fe(-x)= fe(x)
And
fo(-x)=-fo(x)
Now we have
f(-x) = fe(-x) +fo(-x)...(2)
Adding eq (1) and (2) we have
 fe(x) =( f(x)+f(-x) )/2
Subtracting (1) and (2) we have
fo(x) =( f(x)-f(-x) )/2


TITLE: On time evolution of density operator (matrix) in quantum mechanics
QUESTION [2 upvotes]: Suppose I have at time $t=0$ a statistical ensemble of quantum states
$$\{|\psi_n\rangle\langle\psi_n|\}_{n=1}^N$$
The probability of finding the result $a$ for an observable $A$ in this ensemble is given by the Born rule (assuming discrete spectrum, without degeneracy):
$$ \sum_{n=1}^N p_n \langle \psi_n | \Pi_a|\psi_n \rangle = \operatorname{Tr}\left[\rho\,\Pi_a\right]$$
where $\Pi_a$ is the projector in the $a$-eigenspace of $A$, and
$$
\rho:= \sum_{n=1}^N p_n |\psi_n\rangle\langle\psi_n|
$$
is the density operator or density matrix.
At time $t>0$, this density $\rho$ takes the form
$$
\rho(t):= \sum_{n=1}^N p_n |\psi_n(t)\rangle\langle\psi_n(t)|
$$
what puzzles me the most is that those $p_n$ stay the same. Why should I expect that? Why couldn't the probabilities change as each state $|\psi_n(t)\rangle\langle\psi_n(t)|$ evolves with time?

REPLY [4 votes]: The idea is really just that you start out with a statistical ensemble, i.e. the quantum system is in the state $\lvert \psi_n\rangle$ with probability $p_n$ at the start. The density matrix $\sum_n p_n\lvert \psi_n\rangle\langle \psi_n\rvert$ is just a way to summarize this information into a single object. The notion of density matrices does not add any new "mechanics" to quantum mechanics.
Each of these states evolves in time like quantum states normally do, i.e. $\lvert \psi_n(t)\rangle = U(t)\lvert \psi_n\rangle$ with $U(t)$ the ordinary time evolution operator. So if the system started in the state $\lvert \psi_n(0)\rangle$, of course it is in the state $\lvert \psi_n(t)\rangle$ at time $t$. Since it had probability $p_n$ to start in the state $\lvert \psi_n(0)\rangle$, it still has probability $p_n$ to be in $\lvert \psi_n(t)\rangle$ at time $t$, and the density matrix at time $t$ is $\sum_n p_n\lvert \psi_n(t)\rangle\langle \psi_n(t)\rvert$. There isn't anything more to this, the probabilities here just act like classical probabilities.


TITLE: Is the function (sum-of-squares) / sum convex on nonnegative input?
QUESTION [2 upvotes]: Let $$f \colon \mathbb{R}_{> 0}^n \to \mathbb R$$ be defined by $$f(x_1,\dotsc,x_n) = \begin{cases} 0 &\text{if }x_1 = \dotsb = x_n = 0\text{,}\\ \frac{\sum_i x_i^2}{\sum_i x_i} &\text{otherwise.} \end{cases}$$
 Is this function convex?

Motivation: The function I have described is equivalent to taking a weighted average of the $x_i$ where the weight of $x_i$ is proportional to $x_i$. If the $x_i$ represent the sizes of the various classes at a college, then this function $f$ computes the average class size experienced by students. (A large class will be experienced by more students and will consequently receive greater weight.)
This function gives a smooth approximation of the maximum function whose partial derivatives are relatively inexpensive to compute (only a single division is required for all the partial derivatives put together). As such, I imagine it might be useful as a pooling function for convolutional neural nets. But it would be helpful to know more about its optimization properties when running gradient descent and similar algorithms.

REPLY [2 votes]: Yes, it is convex.
$$
f(\lambda x+(1-\lambda) y)\le \lambda f(x) + (1-\lambda) f(y)
$$
for $x,y\in \mathbb{R}^n$, $\lambda \in [0,1]\;$.
$$
\frac{\sum(\lambda x_i+(1-\lambda)y_i)^2}{\sum(\lambda x_i+(1-\lambda)y_i)}\le\lambda \frac{\sum x_i^2}{\sum x_i}+(1-\lambda) \frac{\sum y_i^2}{\sum y_i}
$$
So
$$
\lambda^2\sum x_i^2+(1-\lambda)^2\sum y_i^2+2\lambda(1-\lambda)\sum x_iy_i\le\\
\le\lambda\sum x_i^2\Big(\lambda+(1-\lambda)\frac{\sum y_i}{\sum x_i}   \Big)+
(1-\lambda)\sum y_i^2\Big((1-\lambda)+\lambda\frac{\sum x_i}{\sum y_i}   \Big)
$$
And simplifying some addends
$$
2\lambda(1-\lambda)\sum x_iy_i\le
\lambda(1-\lambda)\sum x_i^2\frac{\sum y_i}{\sum x_i}   +\lambda(1-\lambda)\sum y_i^2\frac{\sum x_i}{\sum y_i}
$$
Here $\lambda$ disappears, and you obtain
$$
2\sum x_i\sum y_i\sum x_iy_i\le\sum x_i^2\big(\sum y_i\big)^2+\sum y_i^2\big(\sum x_i\big)^2
$$
You can prove this using first the Cauchy-Schwarz inequality, and then AM-GM (arithmetic mean-geometric mean).
$$
4\big(\sum x_i\big)^2\big(\sum y_i\big)^2\big(\sum x_iy_i\big)^2\le4\big(\sum x_i\big)^2\big(\sum y_i\big)^2\sum x_i^2\sum y_i^2=\\
=4\sum x_i^2\big(\sum y_i\big)^2\sum y_i^2\big(\sum x_i\big)^2\le\big[\sum x_i^2\big(\sum y_i\big)^2+\sum y_i^2\big(\sum x_i\big)^2\big]^2
$$
thus proving it is convex.


TITLE: Find $\cos2\theta+\cos2\phi$, given $\sin\theta + \sin\phi = a$ and $\cos\theta+\cos\phi = b$
QUESTION [1 upvotes]: If
  $$\sin\theta + \sin\phi = a \quad\text{and}\quad \cos\theta+\cos\phi = b$$
then find the value of   $$\cos2\theta+\cos2\phi$$

My attempt:
Squaring both sides of the second given equation:
$$\cos^2\theta+ \cos^2\phi + 2\cos\theta\cos\phi= b^2$$
Multiplying by 2 and subtracting 2 from both sides we obtain,
$$\cos2\theta+ \cos2\phi = 2b^2-2 - 4\cos\theta\cos\phi$$
How do I continue from here?
PS: I also found the value of $\sin(\theta+\phi)= \dfrac{2ab}{a^2+b^2}$
Edit: I had also tried to use $\cos2\theta + \cos2\phi= \cos(\theta+\phi)\cos(\theta-\phi)$ but that didn't seem to be of much use

REPLY [0 votes]: Using Prosthaphaeresis Formula,
$$\cos2\theta+\cos2\phi=2\cos(\theta+\phi)\cos(\theta-\phi),$$
Now $$a^2+b^2=2+2\cos(\theta-\phi)\implies\cos(\theta-\phi)=?$$
Again, using  Prosthaphaeresis formula,  $$a=\sin\theta+\sin\phi=2\sin\dfrac{\theta+\phi}2\cos\dfrac{\theta-\phi}2$$
and $$b=\cdots=2\cos\dfrac{\theta+\phi}2\cos\dfrac{\theta-\phi}2$$
$$\implies\dfrac ab=\tan\dfrac{\theta+\phi}2$$
Now use $$\cos2A=\dfrac{1-\tan^2A}{1+\tan^2A}$$


TITLE: Number of complex roots of the equation $x^n=a, a\ge 0$
QUESTION [0 upvotes]: I have a question about number of complex roots of the equation
$$x^n=a, a\ge 0, n\not\in \mathbb Z.$$

According to the Fundamental Theorem of Algebra if $x^n=a, n\in\mathbb N$, then we have exactly $n$ distinct complex roots. The real root is $x=\sqrt [n]{a}$, if $n$ is odd or $x_{1,2}=\pm \sqrt [n]{a}$, if $n$ is even.
But, when $n$ is not a natural number, for instance,
$$x^{\pi}=a, a\ge 0$$
What are all the complex roots and exactly how many roots are there?
I think that, $x=a^{\frac 1\pi}, a\ge 0$ is an only real root.

REPLY [1 votes]: Coincidentally I was checking this out a few months back. I'll share my findings.
The roots of said function lie along the unit circle scaled up by $a^{\frac1n}$. I'll invert your expression into $\dfrac{1}{x^n-a}$ with $a> 0$ and make all of its roots poles instead to make everything much more obvious
For an integer power $n\in\mathbb{Z}$, we will get the $n$ roots of unity. The function is entire with no branch cut.
For each noninteger $n\in\mathbb{R}$, we have the number of poles will be the nearest odd integer. There will be a branch cut since by definition $$p(z) = z^c = \exp\left(c \ln(z)\right) = \exp\left(c \left( \operatorname{Log}(z) + 2n\pi i \right)\right),\qquad n\in \mathbb Z$$ and since $c$ is not a whole multiple of the logarithm, there will be multiple branches. We can safely choose it to be along the negative real axis.
We may observe that at $n=1$, we have a singular pole at $z=1$.
As we gradually increase our $n$, the pole splits into two along the branch cut and proceeds to make its way towards the positive real axis.
More generally, at every even integer, a pole appears at $\sqrt[n]{a}$ (which we will define to be the negative real pole), and as you increase $n$ the pole splits and pushes the other poles backwards along the unit circle scaled up by $a$ until we reach the next integer, where the poles would be at some arrangement of the roots of unity again.
In your example, we have $x^\pi=a$. The nearest odd number to this is $3$ so there will be $3$ poles, one at the positive real value of $a^{\frac1{\pi}}$ and the other two roots using the complex power function definition shared earlier would be at $a^{\frac1{\pi}}\cdot e^{2i}$ and $a^{\frac1{\pi}}\cdot e^{-2i}$, pretty close to the true third roots of unity since well, $\pi$ is close to $3$.
In the most general case(you can derive this too by tinkering with the logarithm definition of the power function), the solutions can be found as follows. Given $$x^n=a$$ our real root is $$a^{\frac{1}{n}}$$ Next, the complex roots will have this form... $$a^{\frac{1}{n}}\cdot e^{\pm\frac{(2u)\pi i}{n}}$$ for some $u\in\mathbb{N}$ given that the root you get from it will not violate your chosen argument(which in this case I selected $-\pi\le\arg(z)<\pi$).
You can use the manipulate command in mathematica to watch how the roots evolve, it is pretty fun. Here, I will let $a=2$ and show the pole evolution of $n$ from $n=2$ to $n=4$


TITLE: Landau's comment on validity of Goldbach's conjecture
QUESTION [1 upvotes]: Edmund Landau had once put forth the following statement about the validity of the Goldbach conjecture.
"the goldbach conjecture is false for at most 0 % of all even integers ; this at most 0 % does not exclude, however the possibility that there are infinitely many exceptions"
I have not understood the essence of the statement, like for an ordinary novice student like me, the two statements appear contradicting, whilst they appear to make perfect sense for a number theorist. Can anyone please help me with the understanding of the essence of this statement?

REPLY [4 votes]: Consider this statement: $0\%$ of all natural numbers are powers of $2$. It is true, in this sense: if, for each natural $n$, you compute the proportion $p_n$ of powers of $2$ in $\{1,2,\ldots,n\}$, then $\lim_{n\to\infty}p_n=0$. But that does not mean that no natural number is a power of $2$.

REPLY [1 votes]: Here's an analogy.
There are infinitely many numbers of the form $10^k$ - a $1$ followed by some number of zeroes. Among the first $10$ numbers there are just $2$. Among the first $100$ numbers there are just three. Among the first $1000$, four. That pattern continues, so the fraction of all numbers of that form might reasonably be said to be $0$, even though there are infinitely many of them.
Whether there are infinitely many counterexamples to Goldbach's conjecture is unknown. We don't even know there's one.


TITLE: maximize $\tbinom{500}{x}\times\tbinom{500}{900-3x}$
QUESTION [0 upvotes]: 2 fair coins A and B, for A, tossing H gets 3 dollars; for B, tossing H gets 1 dollar. Now toss A and B 500 times each, finally we get 900 dollars. Estimate the most likely number of times of H for A and B (you are not necessary to calculate the exact value, just give the best possible bound).

It is equivalent to find the x to maximize:
$$f(x)=\tbinom{500}{x}\times\tbinom{500}{900-3x},$$
here $134\leq x \leq 500.$ Even I use the fact that the optimal $x^*$ meets:
$$f(x^*)\geq f(x^*+1)\ and\ f(x^*)\geq f(x^*-1).$$
I still can hardly solve $x^*.$

REPLY [1 votes]: Using algebra, the problem is not too bad.
Go from  binomial coefficients to the gamma function.
$$\binom{500}{x}\times\binom{500}{900-3x}=\frac{\Gamma (501)^2}{\Gamma (x+1)\,\, \Gamma (501-x)\,\, \Gamma (901-3 x) \,\,\Gamma (3x-399)}$$ which implies that you want to minimize
$$g(x)=\Gamma (x+1)\,\, \Gamma (501-x)\,\, \Gamma (901-3 x) \,\,\Gamma (3x-399)$$ or, much better, its logarithm. Using derivative
$$\frac{g'(x)}{g(x)}=3 H_{3 x-400}-H_{500-x}-3 H_{900-3 x}+H_x$$ and you want it to be equal to $0$. This is a nice and smooth function and you look for the solution between $134$ and $299$. Make an expansion to $O(x^2)$ around the midpoint and solve for $x$. You will obtain $x=220.0138$ while the "exact" solution given by Newton method is $220.0128$.
I think that we have the good approximation. Just try three points $(219,220,221)$ and you will see which is the one giving the maximum value of the initial expression.


text:
\begin{document}

\begin{frontmatter}

\titledata{On Hermitian varieties in $\PG(6,q^2)$}
{}

\authordata{Angela Aguglia}
{Dipartimento di Meccanica, Matematica e Management, Politecnico di Bari, Via Orabona 4, I-70125 Bari, Italy}
{angela.aguglia@poliba.it}
{The author was supported by  the Italian National Group for Algebraic and Geometric Structures and their Applications (GNSAGA - INdAM). }

\authordata{Luca Giuzzi}
{DICATAM, University of Brescia, Via Branze 53,
  I-25123 Brescia, Italy}
{luca.giuzzi@unibs.it}
{The author was supported by  the Italian National Group for Algebraic and Geometric Structures and their Applications (GNSAGA - INdAM). }

\authordata{Masaaki Homma}
{Department of Mathematics and Physics, Kanagawa University, Hiratsuka 259-1293, Japan}
{homma@kanagawa-u.ac.jp}
{}

\keywords{unital, Hermitian variety, algebraic hypersurface.}
\msc{51E21; 51E15; 51E20}

\begin{abstract}
 In this paper we  characterize the  non-singular Hermitian variety $\cH(6,q^2)$ of $\PG(6,q^2)$, $q\neq 2$ among the irreducible hypersurfaces of degree $q+1$ in $\PG(6,q^2)$ not containing solids
  by the number of its  points and the existence of
  a solid $S$ meeting it in $q^4+q^2+1$ points.
\end{abstract}

\end{frontmatter}

\section{Introduction}
The set of all absolute points of a non-degenerate unitary polarity in
$\PG(r,q^2)$ determines the Hermitian variety $\cH(r,q^2)$.
This is a non-singular
algebraic hypersurface of degree $q+1$ in $\PG(r,q^2)$ with a number of remarkable properties, both from
the geometrical and the combinatorial point of view; see~\cite{BC,SE2}.
In particular, $\cH(r,q^2)$ is a
$2$-character set with respect to the hyperplanes of $\PG(r,q^2)$ and $3$-character blocking set with respect to the lines of $\PG(r,q^2)$
for $r>2$.
An interesting and widely investigated problem is to provide combinatorial descriptions of $\cH(r,q^2)$ among all hypersurfaces of the same degree.

First, we observe that
a condition on the number of points and the intersection numbers
with hyperplanes is not in general sufficient to characterize Hermitian varieties; see~\cite{ACK12},\cite{A13}. On the other hand, it is
enough  to consider in addition the intersection numbers with codimension $2$ subspaces in order to get a complete description; see~\cite{DS1}.

In general, a hypersurface $\cH$ of $\PG(r,q)$ is viewed
as a hypersurface over the algebraic closure of $\GF(q)$ and a point of $\PG(r, q^i)$ in $\cH$ is called a
$\GF(q^i)$-point. A $\GF(q)$-point of $\cH$ is also said  to be a rational point of $\cH$. Throughout this paper, the
number of $\GF(q^i)$-points of $\cH$ will be denoted by $N_{q^i}(\cH)$. For simplicity, we shall also use the convention
$|\cH|=N_q(\cH)$.

In the present paper, we shall investigate a combinatorial characterization
of the Hermitian hypersurface $\cH(6,q^2)$ in $\PG(6,q^2)$ among all hypersurfaces of the same degree having also the same number of $\GF(q^2)$-rational points.

More in detail, in \cite{HK,HKf} it has been proved that if $\cX$ is a hypersurface of degree $q+1$ in $\PG(r,q^2)$, $r\geq3$ odd, with    $|\cX|=|\cH(r,q^2)|=(q^{r+1}+(-1)^r)(q^{r}-(-1)^r)/(q^2-1)$
$\GF(q^2)$-rational points,
not containing linear subspaces of dimension greater than
$\frac{r-1}{2}$, then $\cX$ is a non-singular Hermitian variety of $\PG(r,q^2)$.
This result generalizes the characterization of
\cite{HSTV} for the Hermitian curve of $\PG(2,q^2)$, $q\neq 2$.

The case where $r>4$ is even is, in general, currently open.
A starting point for a characterization in arbitrary even dimension can be found in~\cite{AAFP} where the case of a hypersurface $\cX$ of degree $q+1$
in $\PG(4,q^2)$, $q>3$ is considered. There,
it is shown that when $\cX$  has the same number of  rational points as $\cH(4,q^2)$, does not contain any subspaces of dimension greater than $1$ and meets at least
one plane $\pi$ in $q^2+1$  $\GF(q^2)$-rational points, then $\cX$ is a Hermitian variety.

In this article we deal with hypersurfaces of degree $q+1$ in $\PG(6,q^2)$
and we prove  that a characterization similar to that of~\cite{AAFP}
holds also in dimension $6$. We conjecture that this can be extended
to arbitrary even dimension.

\begin{thm}\label{main}
Let $S$ be a hypersurface of $\PG(6,q^2)$, $q>2$, defined over $\GF(q^2)$, not containing solids. If the degree of $S$ is $q+1$ and the number of its rational points is $q^{11}+q^9+q^7+q^4+q^2+1$, then every solid of $\PG(6,q^2)$ meets $S$ in at least $q^4+q^2+1$  rational points. If there is at least a solid $\Sigma_3$ such that $|\Sigma_3\cap S|=q^4+q^2+1$, then $S$ is a non-singular Hermitian variety of $\PG(6,q^2)$.
\end{thm}

Furthermore, we also extend the result obtained in \cite{AAFP} to the case $q=3$.

\section{Preliminaries and notation}

In this section we collect some useful information and results that will be crucial to our proof.

A Hermitian variety in $\PG(r,q^2)$ is the algebraic variety of
$\PG(r,q^2)$ whose points $\langle v\rangle$
satisfy the equation $\eta(v,v)=0$ where $\eta$ is a sesquilinear
form $\GF(q^2)^{r+1}\times\GF(q^2)^{r+1}\to\GF(q^2)$.
The \emph{radical} of the form $\eta$ is the vector subspace of
$\GF(q^2)^{r+1}$ given by
\[ \mathrm{Rad}(\eta):=\{ w\in\GF(q^2)^{r+1}\colon
  \forall v\in\GF(q^2)^{r+1}, \eta(v,w)=0 \}. \]
The form $\eta$ is non-degenerate if $\mathrm{Rad}(\eta)=\{\mathbf{0}\}$.
If the form $\eta$ is non-degenerate, then the corresponding
Hermitian variety is denoted by $\cH(r,q^2)$ and it is non-singular,
of degree $q+1$ and contains
$$(q^{r+1}+(-1)^r)(q^{r}-(-1)^r)/(q^2-1)$$
$\GF(q^2)$-rational points.
When $\eta$ is
degenerate we shall call \emph{vertex} $R_t$ of the degenerate
Hermitian variety associated to $\eta$ the projective
subspace $R_t:=\PG(\mathrm{Rad}(\eta)):=\{ \langle w\rangle\colon
w\in\mathrm{Rad}(\eta)\}$ of $\PG(r,q^2)$.
A degenerate Hermitian variety can always be described as
a cone of vertex $R_t$ and  basis a non-degenerate
Hermitian variety $\cH(r-t,q^2)$ disjoint from $R_t$ where $t=\dim(\mathrm{Rad}(\eta))$
is the vector dimension of the radical of $\eta$. In this case
we shall write the corresponding variety as $R_t\cH(r-t,q^2)$.
Indeed,
\[ R_t\cH(r-t,q^2):=\{ X\in\langle P,Q\rangle\colon
  P\in R_t, Q\in\cH(r-t,q^2) \}. \]

Any line of $\PG(r,q^2)$ meets a Hermitian variety (either degenerate or not)
in either $1, q+1$ or $q^2+1$ points (the latter value only for $r>2$).
The maximal dimension of projective subspaces contained in
the non-degenerate Hermitian variety $\cH(r,q^2)$ is $(r-2)/2$, if $r$ is even, or $(r-1)/2$, if $r$ is odd.
These subspaces of
maximal dimension are called {\em generators} of $\cH(r,q^2)$ and the generators of $\cH(r,q^2)$ through a point $P$ of $\cH(r, q^2)$ span a hyperplane
$P^{\perp}$ of $\PG(r,q^2)$, the {\em tangent hyperplane} at $P$.

It is well known that
this hyperplane meets $\cH(r,q^2)$ in a degenerate Hermitian variety
$P\cH(r-2,q^2)$, that is in a Hermitian
cone having as vertex the point $P$ and as base a non-singular Hermitian variety of $\Theta\cong\PG(r-2,q^2)$ contained in $P^{\perp}$ with
$P\not\in\Theta$.

Every hyperplane of $\PG(r, q^2)$, which is not tangent, meets $\cH(r, q^2)$ in a non-singular Hermitian variety $\cH(r-1, q^2)$, and is called a {\em secant hyperplane} of $\cH(r, q^2)$.
In particular, a tangent hyperplane contains
$$
1+q^2(q^{r-1}+(-1)^r)(q^{r-2}-(-1)^r)/(q^2-1)
$$
$\GF(q^2)$-rational points of $\cH(r,q^2)$, whereas a secant hyperplane contains
$$
(q^{r}+(-1)^{r-1})(q^{r-1}-(-1)^{r-1})/(q^2-1)
$$
$\GF(q^2)$-rational points of $\cH(r,q^2)$.

We now recall several results which shall be used in the course of
this paper.

\begin{lem}[\cite{SE}]\label{Se} Let $d$ be an integer with $1\leq d\leq q+1$ and let $\cC$ be a curve of degree $d$ in $\PG(2,q)$ defined over $\GF(q)$, which may have $\GF(q)$-linear components. Then the number  of its rational points is at most $dq+1$ and $N_{q}(\cC)=dq+1$ if and only if $\cC$ is a pencil of $d$ lines of $\PG(2,q)$.
\end{lem}

\begin{lem}[\cite{HK3}]\label{Szi}
Let $d$ be an integer with $2\leq d  \leq q+2$, and $\cC$
a curve of degree $d$ in $\PG(2,q)$ defined over $\GF(q)$ without any $\GF(q)$-linear components. Then $N_{q}(\cC)\leq (d-1)q+1$, except for a class of plane curves of degree $4$ over $\GF(4)$ having $14$ rational points.
\end{lem}

\begin{lem}[\cite{HK0}]\label{hk0}
Let $\cS$ be a surface of degree $d$ in $\PG(3,q)$ over $\GF(q)$. Then
\[N_{q}(\cS)\leq dq^2+q+1\]
\end{lem}

\begin{lem}[\cite{HSTV}]\label{uni}
Suppose $q\neq 2$.  Let $\cC$ be a plane curve over $\GF(q^2)$ of degree $q+1$ without $\GF(q^2)$-linear components. If $\cC$ has $q^3+1$ rational points, then $\cC$ is a Hermitian curve.
\end{lem}

\begin{lem}[\cite{DS1}]\label{ds1}
A subset of points of $\PG(r,q^2)$ having the same intersection numbers with respect to hyperplanes and spaces of codimension $2$ as non-singular Hermitian varieties, is a non-singular Hermitian variety of $\PG(r,q^2)$.
\end{lem}

From \cite[Th 23.5.1,Th 23.5.3]{HT} we have the following.
\begin{lem}\label{ch}
  If $\cW$ is a set of $q^7+q^4+q^2+1$ points of $\PG(4,q^2)$, $q > 2$, such that every line of $\PG(4,q^2)$ meets $\cW$ in $1, q+1$ or $q^2+1$ points, then $\cW$ is a
  Hermitian cone with vertex a line and base a unital.
\end{lem}
Finally, we recall that a {\em blocking set with respect to lines} of $\PG(r, q)$ is a point set which blocks all the lines, i.e., intersects each line of $\PG(r, q)$ in at least one point.

\section{Proof of Theorem~\ref{main}}
We first provide an estimate on the number of points of
a curve of degree $q+1$ in $\PG(2,q^2)$, where $q$ is any prime power.
\begin{lem}\label{Homma}
  Let $\cC$ be a plane curve over $\GF(q^2)$, without $\GF(q^2)$-lines as components and of degree $q+1$. If the number of $\GF(q^2)$-rational points
  of $\cC$ is $N<q^3+1$, then
\begin{equation} \label{num}
N\leq \left\{\begin{array}{ll}
    q^3-(q^2-2)  & {\mbox{\it if}  \ q>3}  \\
    24   & {\mbox{\it if}  \ q=3} \\
     8    & {\mbox{\it if}  \ q=2} .
   \end{array}  \right.
\end{equation}

\end{lem}
\begin{proof}
We distinguish the following three cases:
\begin{enumerate}[(a)]
\item\label{(a)} $\cC$ has two or more $\GF(q^2)$-components;
\item\label{(b)} $\cC$ is irreducible over $\GF(q^2)$, but not absolutely irreducible;
\item\label{(c)} $\cC$ is absolutely irreducible.
\end{enumerate}
Suppose first $q\neq2$.\par\noindent
{\bf{Case (\ref{(a)})}}
Suppose $\cC=\cC_1\cup\cC_2$. Let $d_i$ be the degree of $\cC_i$,  for each $i=1,2$. Hence $d_1+d_2=q+1$. By Lemma~\ref{Szi},
\[N \leq N_{q^2}(\cC_1)+N_{q^2}(\cC_2)\leq [(q+1)-2]q^2+2=q^3-(q^2-2)\]
{\bf{Case (\ref{(b)})}} Let $\cC'$ be an irreducible component of $\cC$ over the algebraic closure of $\GF(q^2)$. Let $\GF(q^{2t})$
be the minimum defining field of $\cC'$ and $\sigma$ be the Frobenius morphism of $\GF(q^{2t})$ over $\GF(q^2)$. Then
\[\cC=\cC' \cup \cC'^{\sigma} \cup \cC'^{\sigma^2}\cup \ldots \cup \cC'^{\sigma^{t-1}},    \]
and the degree of $\cC'$, say $e$, satisfies $q+1=te$ with $e>1$. Hence any $\GF(q^2)$-rational point of $\cC$ is contained in $\cap_{i=0}^{t-1} \cC'^{\sigma^i}$. In particular, $N\leq e^2\leq (\frac{q+1}{2})^2$ by Bezout's Theorem and $(\frac{q+1}{2})^2 <q^3-(q^2-2)$.\\
\noindent
{\bf{Case (\ref{(c)})}} Let $\cC$ be an absolutely irreducible curve over $\GF(q^2)$ of degree $q+1$. Either $\cC$ has a singular point or not.

In general, an absolutely irreducible plane curve $\cM$  over $\GF(q^2)$ is $q^2$-Frobenius non-classical if for a general point $P(x_0,x_1,x_2)$ of $\cM$ the point  $P^{q^2}=$ $P^{q^2}(x_0^{q^2} ,x_1^{q^2},x_2^{q^2})$ is on the tangent line to $\cM$ at the point $P$. Otherwise, the curve  $\cM$ is said to be Frobenius classical.
A lower bound of the number of $\GF(q^2)$-points for $q^2$-Frobenius non-classical curves is given by \cite[Corollary 1.4]{BH}:
for a $q^2$-Frobenius non-classical curve $\cC'$ of degree $d$,
we have $N_{q^2}(\cC') \geq d(q^2 -d +2)$.
In particular, if $d=q+1$, the lower bound is just $q^3 +1$.

Going back to our original curve $\cC$,
we know $\cC$ is Frobenius classical because $N < q^3 +1$.
Let $F(x,y,z)=0$ be an equation of $\cC$ over $\GF(q^2)$.
We consider the curve $\cD$ defined by
$
\frac{\partial F}{\partial x}x^{q^2}
+
\frac{\partial F}{\partial y}y^{q^2}
+
\frac{\partial F}{\partial z}z^{q^2} =0.
$
Then $\cC$ is not a component of $\cD$ because $\cC$ is Frobenius classical.
Furthermore, any $\GF(q^2)$-point $P$ lies on $\cC \cap \cD$
and the intersection multiplicity of $\cC$ and $\cD$ at $P$ is at least $2$
by Euler's theorem for homogeneous polynomials.
Hence by B\'{e}zout's theorem,
$
2N \leq (q+1)(q^2 +q).
$
Hence
\[
N \leq \frac{1}{2}q(q+1)^2.
\]
This argument is due to St\"{o}hr and Voloch \cite[Theorem 1.1]{SV}.
This St{\"o}hr and  Voloch's bound is lower than the estimate for $N$ in case~(\ref{(a)}) for $q>4$ and it is the same for $q=4$.
When
$q=3$ the bound in case (\ref{(a)}) is smaller than the
St{\"o}hr and  Voloch's bound.\\
\noindent
Finally, we consider the case $q=2$. Under this assumption,
$\cC$ is a cubic curve and neither case (\ref{(a)}) nor case (\ref{(b)}) might occur. For a degree $3$ curve over $\GF(q^2)$ the St{\"o}hr and  Voloch's bound is loose, thus we need to change our argument.
If $\cC$ has a singular point, then $\cC$ is a rational curve with a unique singular point. Since the degree of $\cC$ is $3$, singular points are either cusps or ordinary double points. Hence $N\in\{4,5,6\}$. If $\cC$ is nonsingular, then it is an elliptic curve and, by
the Hasse-Weil bound, see \cite{AW},  $N\in I$ where $I=\{1,2,\ldots, 9\}$ and for each number $N$ belonging to  $I$ there is an elliptic curve over $\GF(4)$ with $N$ points, from \cite[Theorem 4.2]{SH}.
This completes the proof.
\end{proof}
Henceforth, we shall always suppose $q>2$ and
we denote by $\cS$  an algebraic hypersurface of $\PG(6,q^2)$ satisfying the
following hypotheses of Theorem~\ref{main}:
\begin{enumerate}[(S1)]
\item\label{S1}
  $\cS$ is an algebraic hypersurface of degree $q+1$ defined over $\GF(q^2)$;
\item\label{S2}
  $|\cS|=q^{11}+q^9+q^7+q^4+q^2+1$;
\item\label{S3} $\cS$ does not contain projective $3$-spaces (solids);
\item\label{S4} there exists a solid $\Sigma_3$ such that
  $|\cS\cap\Sigma_3|=q^4+q^2+1$.
\end{enumerate}
We first consider the behavior of $\cS$ with respect to the lines.
\begin{lem}
  \label{block}
  An algebraic hypersurface $\cT$ of degree $q+1$
  in $\PG(r,q^2)$, $q\neq 2$, with $|\cT|=|\cH(r,q^2)|$
  is a blocking set with respect to lines of $\PG(r,q^2)$
\end{lem}
\begin{proof}
Suppose on the contrary that there is a line $\ell$
of $\PG(r,q^2)$ which is disjoint from $\cT$.
Let $\alpha$ be a plane containing  $\ell$.
The algebraic  plane curve $\cC=\alpha \cap\cT$ of degree $q+1$ cannot have $\GF(q^2)$-linear components and hence it has at most $q^3+1$ points
because of Lemma~\ref{Szi}.
If $\cC$ had $q^3+1$  rational points, then from Lemma~\ref{uni},
$\cC$ would be a Hermitian curve with an external line, a contradiction since Hermitian curves are blocking sets.
Thus $N_{q^2}(\cC)\leq q^3$.
Since $q>2$, by Lemma~\ref{Homma}, $N_{q^2}(\cC)<q^3-1$  and hence every plane through $r$ meets $\cT$ in at most $q^3-1$ rational points.
Consequently, by considering all planes through $r$, we can bound the number of rational points of $\cT$ by
$N_{q^2}(\cT)\leq (q^3-1)\frac{q^{2r-4}-1}{q^2-1}=q^{2r-3}+\dots<|\cH(r,q^2)|$,
which is a contradiction.
Therefore there are no external lines to $\cT$ and so  $\cT$ is a blocking set w.r.t. lines of $\PG(r,q^2)$.
\end{proof}
\begin{remark}
  The proof of~\cite[Lemma 3.1]{AAFP} would work
  perfectly well here under the assumption $q>3$.
  The alternative argument of Lemma~\ref{block}
  is simpler and also holds for $q=3$.
\end{remark}

By the previous Lemma and assumptions~(S\ref{S1}) and
(S\ref{S2}), $\cS$ is a blocking set for the lines of $\PG(6,q^2)$
In particular,
the intersection of $\cS$ with any $3$-dimensional subspace $\Sigma$ of $\PG(6,q^2)$ is
also a blocking set with respect to lines of $\Sigma$
and hence it contains at least $q^4+q^2+1$ $\GF(q^2)$-rational points; see~\cite{BB}.

\begin{lem}
  \label{plane}
  Let $\Sigma_3$ be a solid of $\PG(6,q^2)$
  satisfying condition~(S\ref{S4}), that is
  $\Sigma_3$ meets
  $\cS$ in exactly $q^4+q^2+1$ points. Then,
  $\Pi:=\cS\cap\Sigma_3$ is a plane.
\end{lem}
\begin{proof}
  $\cS\cap\Sigma_3$ must be a blocking set for the lines of $\PG(3,q^2)$;
  also it has
  size $q^4+q^2+1$. It follows from~\cite{BB} that $\Pi:=\cS\cap\Sigma_3$ is
  a plane.
\end{proof}
\begin{lem}
  \label{cone0}
  Let $\Sigma_3$ be a solid of satisfying condition~(S\ref{S4}).
  Then,   any $4$-dimensional projective space $\Sigma_4$ through $\Sigma_3$
  meets $\cS$ in a Hermitian cone with vertex a line
  and basis a Hermitian curve.
\end{lem}

\begin{proof}
  Consider all of the $q^6+q^4+q^2+1$
  subspaces $\overline{\Sigma}_3$ of dimension $3$  in $\PG(6,q^2)$
   containing $\Pi$.

   From Lemma~\ref{hk0} and condition~(S\ref{S3}) we have
   $|\overline{\Sigma}_3 \cap  \cS|\leq q^5+q^4+q^2+1$. Hence,
   \[|\cS|=(q^7+1)(q^4+q^2+1)\leq (q^6+q^4+q^2)q^5+q^4+q^2+1=|\cS|\]
   Consequently, $|\overline{\Sigma}_3\cap\cS|= q^5+q^4+q^2+1$ for all  $\overline{\Sigma}_3 \neq \Sigma_3$ such that $\Pi \subset \overline{\Sigma}_3$.

   Let $C:=\Sigma_4\cap\cS$. Counting the number of rational points of $C$ by considering the intersections with  the $q^2+1$ subspaces $\Sigma'_3$ of
   dimension $3$ in $\Sigma_4$ containing the plane $\Pi$  we get
  \[ |C|=q^2\cdot q^5+q^4+q^2+1=q^7+q^4+q^2+1. \]
  In particular, $C\cap\Sigma'_3$ is
  a maximal surface of degree $q+1$; so it must split in $q+1$ distinct
  planes through a line of $\Pi$; see~\cite{S92}.
  So $C$ consists of $q^3+1$ distinct planes belonging to distinct $q^2$ pencils, all  containing $\Pi$ ; denote by ${\cL}$
  the family of these planes. Also for
  each $\Sigma_3'\neq\Sigma_3$, there is a line $\ell'$
  such that all the planes
  of $\mathcal L$ in $\Sigma_3'$ pass through $\ell'$.
  It is now straightforward to see that any line contained in $C$
  must necessarily belong to one of the planes of $\mathcal L$ and
  no plane not in $\mathcal L$ is contained in $C$.

  In order to get the result it is now enough to show that
  a line of $\Sigma_4$ meets $C$ in either $1$, $q+1$ or $q^2+1$
  points. To this purpose, let $\ell$ be a line of $\Sigma_4$ and
  suppose $\ell\not\subseteq C$. Then, by Bezout's theorem,
  \[ 1\leq |\ell\cap C|\leq q+1. \]
  Assume $|\ell\cap C|>1$. Then we can distinguish two cases:
  \begin{enumerate}
  \item $\ell\cap\Pi\neq\emptyset$. If $\ell$ and $\Pi$ are incident,
    then we can consider the $3$-dimensional subspace
    $\Sigma_3':=\langle\ell,\Pi\rangle$. Then $\ell$ must meet
    each plane of $\mathcal L$ in $\Sigma_3'$ in different points (otherwise
    $\ell$ passes through the intersection of these planes and
    then $|\ell\cap C|=1$). As there are $q+1$ planes of $\mathcal L$
    in $\Sigma_3'$, we have $|\ell\cap C|=q+1$.
  \item $\ell\cap\Pi=\emptyset$. Consider the plane $\Lambda$
    generated by a point $P\in\Pi$ and $\ell$.
    Clearly $\Lambda\not\in{\mathcal L}$.
    The curve
    $\Lambda\cap S$ has degree $q+1$ by construction, does not
    contain lines (for otherwise $\Lambda\in{\mathcal L}$) and
    has $q^3+1$ $\GF(q^2)$-rational
    points (by a counting argument). So from Lemma \ref{uni} it is
    a Hermitian curve . It follows that $\ell$ is a $q+1$ secant.
  \end{enumerate}
  We can now apply   Lemma \ref{ch} to see
  that $C_1$ is a Hermitian cone with vertex a line.
\end{proof}
\begin{lem}
  \label{cone1}
  Let $\Sigma_3$ be a space satisfying condition~(S\ref{S4}) and take
  $\Sigma_5$ to be a $5$-dimensional projective space with
  $\Sigma_3\subseteq\Sigma_5$.
  Then $\cS\cap\Sigma_5$ is a Hermitian cone with vertex a point
  and basis a Hermitian hypersurface $\cH(4,q^2)$.
\end{lem}
\begin{proof}
  Let \[ \Sigma_4:=\Sigma_4^1, \Sigma_4^2,\dots, \Sigma_4^{q^2+1} \]
  be the $4$-spaces through $\Sigma_3$ contained in $\Sigma_5$.
  Put $C_i:=\Sigma_4^i\cap\cS$, for all  $i \in \{1,\ldots, q^2+1\}$ and $\Pi=\Sigma_3\cap\cC_1$.
  From Lemma \ref{cone0} $C_i$ is a Hermitian cone with vertex a line, say $\ell_i$. Furthermore
    $\Pi\subseteq\Sigma_3\subseteq\Sigma_4^i$ where $\Pi$ is a plane.
  Choose a plane $\Pi'\subseteq\Sigma_4^1$ such that
  $m:=\Pi'\cap C_1$ is a line $m$ incident with $\Pi$ but not contained
  in it. Let $P_1:=m\cap\Pi$.
  It is straightforward to see that in $\Sigma_4^1$
  there is exactly $1$ plane
  through $m$ which is a $(q^4+q^2+1)$-secant, $q^4$ planes which are
  $(q^3+q^2+1)$-secant and $q^2$ planes which are $(q^2+1)$-secant.

  Also $P_1$ belongs to the line $\ell_1$.
  There are now two cases to consider:
  \begin{enumerate}[(a)]
  \item\label{(a2)} There is a plane $\Pi''\neq\Pi'$ not contained
    in $\Sigma_4^i$ for all $i=1,\dots,q^2+1$ with
    $m\subseteq\Pi''\subseteq S\cap\Sigma_5$.

    We first show that the vertices of the cones
    $C_i$ are all concurrent.
    Consider $m_i:=\Pi''\cap\Sigma_4^i$. Then
    $\{m_i: i=1,\dots,q^2+1\}$ consists of $q^2+1$ lines
    (including $m$) all through $P_1$.
    Observe that for all $i$, the line $m_i$ meets the
    vertex $\ell_i$ of the cone $C_i$ in $P_i\in\Pi$.
    This forces $P_1=P_2=\dots=P_{q^2+1}$. So
    $P_1\in\ell_1,\dots,\ell_{q^2+1}$.

    Now let $\overline{\Sigma}_4$ be a $4$-dimensional space in
    $\Sigma_5$ with
    $P_1\not\in\overline{\Sigma}_4$; in particular
    $\Pi\not\subseteq\overline{\Sigma}_4$.
    Put also $\overline{\Sigma}_3:=\Sigma_4^1\cap\overline{\Sigma}_4$.
    Clearly, $r:=\overline{\Sigma}_3\cap\Pi$ is a line and
    $P_1\not\in r$. So $\overline{\Sigma}_3\cap\cS$ cannot be
    the union of $q+1$ planes, since if this were to be the
    case, these planes would have to pass through the vertex
    $\ell_1$. It follows that $\overline{\Sigma}_3\cap\cS$ must
    be a Hermitian cone with vertex a point and basis a
    Hermitian curve. Let
    $\cW:=\overline{\Sigma}_4\cap\cS$.
    The intersection $\cW\cap\Sigma_4^i$ as $i$ varies
    is a Hermitian cone with basis a Hermitian curve, so,
    the points of $\cW$ are
    \[ |\cW|=(q^2+1)q^5+q^2+1=(q^2+1)(q^5+1); \]
    in particular, $\cW$ is a hypersurface of $\overline{\Sigma}_4$ of
    degree $q+1$ such that there
    exists a plane of $\overline{\Sigma}_4$ meeting
    $\cW$ in just one line (such planes exist in
    $\overline{\Sigma}_3$).
    Also suppose $\cW$ to contain planes and let $\Pi'''\subseteq\cW$
    be such a plane.
    Since  $\Sigma_4^i\cap\cW$ does not contain planes,
    all  $\Sigma_4^i$ meet $\Pi'''$ in a line $t_i$. Also
    $\Pi'''$ must be contained in $\bigcup_{i=1}^{q^2+1} t_i$.
    This implies that the set $\{t_i\}_{i=1,\dots,q^2+1}$ consists
    of $q^2+1$ lines through a point $P\in \Pi \setminus \{P_1\}$.

    Furthermore each line $t_i$ passing through $P$  must meet the radical line $\ell_i$ of the Hermitian cone $\cS \cap\Sigma_4^i$  and this forces $P$ to coincide with $P_1$, a contradiction.
    It follows that $\cW$ does not contain planes.

        So by the characterization of $\cH(4,q^2)$
    of~\cite{AAFP}
    we have that $\cW$ is a Hermitian variety $\cH(4,q^2)$.

    We also have that $|\cS\cap\Sigma_5|=|P_1\cH(4,q^2)|$. Let now $r$ be any
    line of $\cH(4,q^2)=\cS\cap\overline{\Sigma}_4$ and let $\Theta$ be
    the plane $\langle r,P_1\rangle$. The plane $\Theta$ meets
    $\Sigma_4^i$ in a line $q_i\subseteq\cS$ for each $i=1,\dots,q^2+1$ and these lines
    are concurrent in $P_1$. It follows that all the points of $\Theta$ are
    in $S$. This completes the proof for the current case and shows that
    $\cS\cap\Sigma_5$ is a Hermitian cone $P_1\cH(4,q^2)$.
  \item\label{(b2)} All planes $\Pi''$ with
    $m\subseteq\Pi''\subseteq\cS\cap\Sigma_5$
    are contained in $\Sigma_4^i$ for some $i=1,\dots,q^2+1$.
    We claim that this case cannot happen.
    We can suppose without loss of generality $m\cap\ell_1=P_1$
    and $P_1\not\in\ell_i$ for all $i=2,\dots,q^2+1$.
    Since the intersection of the  subspaces $\Sigma_4^i$  is $\Sigma_3$,  there is exactly one plane through $m$  in $\Sigma_5$ which is
    $(q^4+q^2+1)$-secant,
    namely the plane $\langle\ell_1,m\rangle$.
    Furthermore, in $\Sigma_4^1$ there are $q^4$ planes
    through $m$ which are $(q^3+q^2+1)$-secant and
    $q^2$ planes which are $(q^2+1)$-secant.

    We can provide an upper bound to the points of $\cS\cap\Sigma_5$ by
    counting the number of points of $\cS\cap\Sigma_5$ on  planes in $\Sigma_5$
    through $m$ and observing that a plane through $m$ not
    in $\Sigma_5$ and not contained in $\cS$
    has at most $q^3+q^2+1$ points in common with $\cS\cap\Sigma_5$.
    So
    \[ |\cS\cap\Sigma_5|\leq q^6\cdot q^3+q^7+q^4+q^2+1. \]
    As
    $|\cS\cap\Sigma_5|=q^9+q^7+q^4+q^2+1$,
    all planes through $m$ which are neither $(q^4+q^2+1)$-secant
    nor $(q^2+1)$-secant are $(q^3+q^2+1)$-secant. That is to say
    that all of these planes meet $\cS$ in a curve of degree $q+1$
    which must split into $q+1$ lines through a point because of Lemma \ref{Se}.

    Take now $P_2\in\Sigma_4^2\cap\cS$ and consider the plane
    $\Xi:=\langle m,P_2\rangle$. The line $\langle P_1,P_2\rangle$
    is contained in $\Sigma_4^2$; so it must be a $(q+1)$-secant,
    as it does not meet the vertex line $\ell_2$ of $C_2$ in
    $\Sigma_4^2$. Now, $\Xi$ meets every of $\Sigma_4^i$ for
    $i=2,\dots,q^2+1$ in a line through $P_1$ which is either a
    $1$-secant or a $q+1$-secant; so
    \[ |\cS\cap\Xi|\leq q^2(q)+q^2+1=q^3+q^2+1. \]
    It follows $|\cS\cap\Xi|=q^3+q^2+1$ and $\cS\cap\Xi$ is
    a set of $q+1$ lines all through the point $P_1$.
    This contradicts our previous construction.
  \end{enumerate}
  \end{proof}

  \begin{lem}
    \label{sections}
       Every hyperplane of $\PG(6,q^2)$ meets $\cS$
    either in  a non-singular Hermitian variety
    $\cH(5,q^2)$ or in a cone over a Hermitian hypersurface $\cH(4,q^2)$.

  \end{lem}
  \begin{proof}
    Let $\Sigma_3$ be a solid satisfying condition~(S\ref{S4}).
    Denote by $\Lambda$  a hyperplane of $\PG(6,q^2)$. If $\Lambda$ contains $\Sigma_3$ then, from Lemma \ref{cone1} it follows that
    $\Lambda\cap\cS$ is a Hermitian cone $P\cH(4,q^2)$.

   Now assume that $\Lambda$ does not contain   $\Sigma_3$.
   Denote by $S_5^j$, with $j=1,\ldots,q^2+1$ the $q^2+1$ hyperplanes through $\Sigma_4^1$, where as before, $\Sigma_4^1$ is a $4$-space containing $\Sigma_3$. By Lemma~\ref{cone1} again we get that
   $S_5^j \cap\cS=P^j\cH(4,q^2)$.
  We count  the number of rational points of $\Lambda\cap\cS$ by
  studying the intersections  of  $S_5^j \cap\cS$ with $\Lambda$
  for all $j \in \{1,\ldots, q^2+1\}$.
  Setting  $\cW_j:=S_5^j \cap\cS \cap \Lambda$,
  $\Omega:=\Sigma_4^1 \cap\cS \cap \Lambda$ then

    \[|\cS\cap \Lambda|=\sum_{j}|\cW_j\setminus\Omega|+|\Omega|.\]
    If $\Pi$ is a plane of $\Lambda$
    then  $\Omega$ consists of $q+1$  planes of a pencil.
    Otherwise let $m$ be the line in which $\Lambda$
    meets the plane $\Pi$. Then $\Omega$
    is  either a Hermitian cone $P_0\cH(2,q^2)$, or $q+1$
    planes of a pencil, according as the vertex $P^j\in \Pi$
    is an external point with respect to $m$ or not.

    In the former case $\cW_j$ is a non singular Hermitian variety
    $\cH(4,q^2)$ and thus
    $ |\cS \cap \Lambda|= (q^2+1)(q^7) +q^5+q^2+1 = q^9+q^7+q^5+q^2+1$.

    In the case in which $\Omega$ consists of $q+1$  planes of a pencil then
    $\cW_j$  is  either a  $P_0\cH(3,q^2)$  or a Hermitian cone with vertex a line and basis a Hermitian curve $\cH(2,q^2)$.

   If there is at least one index $j$ such that $\cW_j= \ell_1\cH(2,q^2)$ then,  there must be a $3$-dimensional space $\Sigma'_3$   of $S_5^j \cap\Lambda$ meeting $\cS$ in a generator. Hence, from Lemma \ref{cone1} we get that $\cS \cap\Lambda$ is a Hermitian cone $ P'\cH(4,q^2)$.

Assume that for all $j \in \{1,\ldots, q^2+1\}$,  $\cW_j$  is  a  $P_0\cH(3,q^2)$. In this case
  \[ |\cS \cap\Lambda|=(q^2+1)q^7+ (q+1)q^4+q^2+1=q^9+q^7+q^5+q^4+q^2+1 =|\cH(5,q^2)|. \]

We are going to prove that  the intersection numbers of $\cS$ with hyperplanes are only two that is   $q^9+q^7+q^5+q^4+q^2+1$ or  $q^9+q^7+q^4+q^2+1$.

Denote by $x_i$ the number of hyperplanes meeting $\cS$ in $i$
    rational points with $i \in \{q^9+q^7+q^4+q^2+1, q^9+q^7+q^5+q^2+1, q^9+q^7+q^5+q^4+q^2+1\}$. Double counting arguments give the following equations for  the integers $x_i$:
\begin{equation} \label{tg01}
\left\{\begin{array}{l}
    \sum_{i} x_i=q^{12}+q^{10}+q^8+q^6+q^4+q^2+1 \\
    \\
    \sum_{i}ix_i=|\cS|(q^{10}+q^8+q^6+q^4+q^2+1)\\
    \\
     \sum_{i=1}i(i-1)x_i=|\cS| (|\cS|-1)(q^8+q^6+q^4+q^2+1).
   \end{array}  \right.
\end{equation}
Solving \eqref{tg01} we obtain $x_{q^9+q^7+q^5+q^2+1}=0$.
In the case in which  $|\cS \cap \Lambda|=|\cH(5,q^2)|$, since $\cS\cap\Lambda$ is an algebraic hypersurface of degree $q+1$ not containing $3$-spaces, from  \cite[Theorem 4.1]{AW} we get that  $\cS\cap\Lambda$ is a Hermitian variety $\cH(5,q^2)$ and  this completes the proof.
  \end{proof}
\begin{proof}[Proof of Theorem~\ref{main}]
The first part of  Theorem \ref{main} follows from Lemma ~\ref{plane}.
  From Lemma \ref{sections},  $\cS$ has the same intersection numbers with respect to hyperplanes and $4$-spaces as  a non-singular Hermitian variety of $\PG(6,q^2)$, hence  Lemma \ref{ds1}  applies and  $\cS$ turns out to be a $\cH(6,q^2)$.
\end{proof}

\begin{rem}
  The characterization of the non-singular Hermitian variety $\cH(4,q^2)$
  given in \cite{AAFP} is based on the property that a given hypersurface is a blocking set with respect to lines of $\PG(4,q^2)$, see \cite[Lemma 3.1]{AAFP}. This lemma holds when $q>3$.
Since Lemma~\ref{block} extends the same  property to the case $q=3$ it follows that the
result stated in \cite{AAFP} is also valid in $\PG(4,3^2)$.
\end{rem}

\section{Conjecture}

We propose a conjecture for the general $2n$-dimensional case.

\emph{
Let $\cS$ be a hypersurface of $\PG(2d,q^2)$, $q>2$, defined over $\GF(q^2)$, not containing $d$-dimensional projective subspaces. If the degree of $\cS$ is $q+1$ and the number of its rational points is $|\cH(2d,q^2)|$, then every $d$-dimensional subspace of $\PG(2d,q^2)$ meets $\cS$ in at least $\theta_{q^2}(d-1):=(q^{2d-2}-1)/(q^2-1)$  rational points. If there is at least a $d$-dimensional subspace $\Sigma_d$  such that $|\Sigma_d\cap \cS|=|\PG(d-1,q^2)|$, then $\cS$ is a non-singular Hermitian variety of $\PG(2d,q^2)$.
 }

 Lemma \ref{Homma} and Lemma \ref{block}  can be a starting point for the proof of this  conjecture  since from them we get that $\cS$ is a blocking set with respect to lines of $\PG(2d,q^2)$.

config: arxiv


text:
\section{The Contraction Hypothesis and Shrinking Curves}
\label{section: shrinking nondivergence}

	The purpose of this section is to demonstrate the link between the contraction hypothesis and the growth of orbits.
    In all the situations we consider, the height function $f$ which satisfies the contraction hypothesis
    also has the property that the ratio of $1+\log f(\cdot)$ and $1+d(\cdot,x_0)$ is uniformly bounded from above and below for any fixed base point $x_0 \in G/\G$, where $d(\cdot,\cdot)$ is the Riemannian metric  on $G/\G$.

    In fact, we establish the much stronger statement on the quantitative non-divergence of expanding translates of shrinking segments of admissible curves.
    In particular, Proposition~\ref{propn: non-divergence of shrinking curves} below implies that orbits with linear growth have measure $0$ using the Borel-Cantelli lemma along with Chebyshev's inequality.
    Throughout this section, we retain the same notation as in Section~\ref{section: abstract setup}.

	\begin{proposition} \label{propn: non-divergence of shrinking curves}
    	Let $G$ be a real Lie group and $X$ be a metric space equipped with a proper $G$-action. Suppose $g_t$ is an $\mrm{Ad}$-diagonalizable one-parameter subgroup of $G$ and $\vp$ is a $g_t$-admissible curve satisfying the $\b$-contraction hypothesis on $X$. Suppose $\d \in [0,\b)$ is fixed.
        Then, for all $x_0\in X$ with $f(x_0)<\infty$, the following holds.
    	\begin{equation*}
        	\sup\limits_{\substack{t\geqslant 0,s_0\in[-1,1]\\ J_t + s_0 \subseteq [-1,1] }}
            \frac{1}{|J_t|} \int_{J_t+s_0} f(g_t u(\vp(s)) x_0) \;ds < \infty
    	\end{equation*}
        where $J_t := [-e^{-\d \a(t)}, e^{-\d \a(t)}]$. Moreover, the supremum can be taken to be uniform over base points $x_0 \in \set{f \leqslant M}$ for any $M>0$.
	\end{proposition}

    \begin{proof}
    Let a choice of $\d \in [0,\b)$ be fixed.
    Suppose $s_0 \in [-1,1]$ and $n \geq 0$ is an integer.
    Fix $t>0$ so that~\eqref{eqn: CH} holds with constants $\tilde{c}$ and $\tilde{b}$.
   	  By Lemma~\ref{lemma: approximate curve with tangent}, we have
      \begin{equation} \label{eqn: first step in induction}
          \int_{J_{nt}+s_0} f(g_{(n+1)t} u(\vp(s))x_0)\;ds \leqslant
              C_1 \int_{J_{nt}+s_0} \int_{-1}^1 f(g_t u(r\dot{\vp}(s)) g_{nt} u(\vp(s))x_0) \;dr\;ds.
      \end{equation}
      Since $C_1$ and $\tilde{c}$ are independent of $t$, we may assume that $t>0$ is sufficiently large so that
      \begin{equation*}
      2C_1 \tilde{c} e^{-(\b-\d) \a(t)} <1.
      \end{equation*}
      Therefore, by~\eqref{eqn: first step in induction} and~\eqref{eqn: CH}, we get
      \begin{align*}
      	\int_{J_{nt}+s_0} f(g_{(n+1)t} u(\vp(s))x_0)\;ds
            &\leqslant 2C_1 \tilde{c} e^{-\b \a(t)} \int_{J_{nt}+s_0}  f(g_{nt} u(\vp(s))x_0) \;ds + 2C_1 \tilde{b} |J_{nt}|.
      \end{align*}
      Next, for all $n\geq 1$, since $J_{nt} \subseteq J_{(n-1)t}$ and $f\geq 0$, we get
      \begin{align*}
      	\int_{J_{nt}+s_0} f(g_{(n+1)t} u(\vp(s))x_0)\;ds
            &\leqslant 2C_1 \tilde{c} e^{-\b \a(t)} \int_{J_{(n-1)t}+s_0}  f(g_{nt} u(\vp(s))x_0) \;ds + 2C_1 \tilde{b} |J_{nt}|.
      \end{align*}
      Moreover, since $|J_{(n-1)t}|/|J_{nt}| = e^{\d\a(t)}$, the above inequality implies
      \begin{align} \label{eqn: inductive step}
      	\frac{1}{|J_{nt}|} \int_{J_{nt}+s_0} f(g_{(n+1)t} &u(\vp(s))x_0)\;ds \nonumber\\
            &\leqslant 2C_1 \tilde{c} e^{-(\b-\d) \a(t)}
            \frac{1}{|J_{(n-1)t}|}\int_{J_{(n-1)t}+s_0}  f(g_{nt} u(\vp(s))x_0) \;ds + 2C_1 \tilde{b}.
      \end{align}

      Define $M_0>0$ and $M$ by
      \begin{align*}
      	M_0 &=  \frac{1}{2} \int_{-1}^1 f(u(\vp(s))x_0) \;ds ,\\
        M &= \max\set{M_0,2C_1 \tilde{c} e^{-(\b-\d) \a(t)} M_0 + 2C_1 \tilde{b},
        	\frac{2C_1 \tilde{b}}{\left(1-2C_1 \tilde{c} e^{-(\b-\d) \a(t)}\right)}}.
      \end{align*}

      We claim that
      \begin{equation} \label{eqn: M as the sup}
      	\sup_{n\geqslant 0,s_0} \frac{1}{|J_{nt|}} \int_{J_{nt}+s_0} f(g_{(n+1)t}u(\vp(s)x_0)\;ds \leqslant M
      \end{equation}
      We proceed by induction on $n$.
      When $n=0$, inequality~\eqref{eqn: inductive step}, the definition of $M_0$ and the fact that $M_0\leqslant M$ show that the integrand in~\eqref{eqn: M as the sup} is bounded above by $M$.
      Inequality~\eqref{eqn: inductive step} and the definition of $M$ finish the proof of the claim by induction.

      The conclusion of the proposition follows from the log-smoothness of $f$.
      Furthermore, we note that $M$ can be chosen to be uniform over the base point $x_0$ as it varies in sublevel sets of $f$ as evident from the definition of $M_0$.

    \end{proof}


TITLE: Analytical solution for PDE-system's IBVP to validate method of lines
QUESTION [0 upvotes]: I am in search of initial-boundary value problems which are posed in the form of a system of coupled PDE's (not a single PDE) and for which an analytical  solution exists at least for some specific initial-boundary conditions.
In particular I would be interested in a PDE's system of one dynamical ($\partial_r$ on lhs) equation  coupled to one  constraint ($\partial_t$ on lhs) equation.
The generic character of the question is due to the fact that l  have not yet come acrross any PDE's system with the above characteristics. Being more specific would probably make the question overdetermined.
The ultimate purpose of my search is to use the analytical solutions as a numerical test for Mathematica Method of Lines code.
I would appreciate any recommendations.

REPLY [1 votes]: You could use the "method of manufactured solutions (MMS)", which is nicely described in
https://www.comsol.de/blogs/verify-simulations-with-the-method-of-manufactured-solutions/
Essentially, you take some system of PDE, summarized as $$F[u]=r_F,$$
where $[u]$ includes the partial derivatives of $u$ that are used, and right side $r_B$ which is to be chosen later. Denote the boundary conditions similarly as $$B[u]=r_B.$$ Now select some function $p$ that is in the class of admissible functions and perhaps contains qualities expected in a solution.
Then the test system is $$F[u]=F[p],~~ B[u]=B[p],$$ that is, the right sides are computed as $r_F=F[p]$ and $r_B=B[p]$, and $p$ is obviously the exact solution.

discussion of MMS in a very simple context, with accordingly simple example https://scicomp.stackexchange.com/q/30562/6839


TITLE: How can I find the mass of a thin wire in the shape of a helix?
QUESTION [1 upvotes]: Find the mass of a thin wire in the shape of a helix given by $\overrightarrow {r}(t)=\langle \cos t,\sin t, t\rangle$ for $0\le t\le6\pi$ with density of $z$ at the point $(x,y,z)$ on the wire (curve)?
Can you all provide me with a hint as to how I should get started with this one?
I know that:
$$M=\int^{x_1}_{x_0}\int^{y_1}_{y_0}\int^{z_1}_{z_0}\sigma(x,y,z)\;dz\;dy\;dx$$
Does this mean that:
$$M=\int^{6\pi}_{0}\int^{6\pi}_{0}\int^{6\pi}_{0}z\;dz\;dy\;dx$$

REPLY [1 votes]: In general, given the density $\rho (x,y,z)$ at each point of a curve $C$, the mass of $C$ is given by the line integral (of the first kind)
$$\int \limits _C \rho (x,y,z) \Bbb d C .$$
If your curve is parametrized by $[a,b] \ni t \mapsto \big( x(t), y(t), z(t) \big) \in \Bbb R ^3$, then the above formula means
$$\int \limits _a ^b \rho \big( x(t), y(t), z(t) \big) \sqrt{x'(t)^2 + y'(t)^2 + z'(t)^2} \Bbb d t .$$
In your concrete case $\rho \big( x(t), y(t), z(t) \big) = z(t) = t$, so the mass is
$$\int \limits _0 ^{6 \pi} t \sqrt {(-\sin t)^2 + (\cos t)^2 + 1^2} \Bbb d t = \int \limits _0 ^{6 \pi} t \sqrt 2 \Bbb d t = 18 \pi^2 \sqrt 2 .$$


TITLE: Prove using epsilon-delta that $\lim \limits_{x \to 2} 4x=8$ verification
QUESTION [0 upvotes]: The $\epsilon$-$\delta$ concept still hasn't clicked so I am doing some really easy examples to try to get it
Prove: $\lim \limits_{x \to 2} 4x=8$.
Now, I have to prove that for all $\epsilon>0$, there exists $\delta$,   such that $0 < (x-2)<\delta$  leads to $|4x-8|<\epsilon$.
Now, I know $|4x-8|=4|x-2|$, and let $x$ be not greater than $1$ distance away from $2$ (arbitrary choice), therefore, $1< x<3 \implies |x-2|<1$.
For $4|x-2| <\epsilon$ , this is the part where I get stuck.

REPLY [5 votes]: Let $\delta = \epsilon/4$. Then if $\vert x - 2 \vert < \delta$, we have
$$4\vert x - 2 \vert < 4 \cdot \epsilon/4 = \epsilon.$$
In general, if you have a linear function $y = mx + b$ with $m \neq 0$, you can show that $\lim_{x \to a}f(x) = ma + b$ as follows:
Pick $\epsilon > 0$. Let $\delta = \epsilon/m$. Then if $\vert x - a \vert < \delta$, we have
$$\vert mx + b - (ma + b)\vert = m\vert x - a \vert < m \cdot \epsilon/m = \epsilon.$$
Note that if $m = 0$, we have $m \vert x - a \vert = 0 < \epsilon$ regardless of our choice of $\delta$.