text:
\begin{document}
\maketitle
\begin{abstract}
We suggest an elementary Harmonic Analysis approach to canceling and weakly canceling differential operators, which allows to extend these notions to anisotropic setting and also replace differential operators with Fourier multiplies with mild smoothness regularity. In this more general setting of anisotropic Fourier multipliers, we prove the inequality~$\|f\|_{L_{\infty}} \lesssim \|Af\|_{L_1}$ if~$A$ is a weakly canceling operator of order~$d$ and the inequality~$\|f\|_{L_2} \lesssim \|Af\|_{L_1}$ if~$A$ is a canceling operator  of order~$\frac{d}{2}$, provided~$f$ is a function in~$d$ variables.
\end{abstract}

\section{Introduction}
Consider two similar inequalities for smooth compactly supported functions in two variables:
\eq{\label{Basic}
\|f\|_{L_2} \lesssim \|\nabla f\|_{L_1(\R^2)};\qquad \|\nabla f\|_{L_2} \lesssim \|\Delta f\|_{L_1(\R^2)}.
}
Both inequalities are dilation invariant. The first one was proved by Gagliardo and Nirenberg in late 50s and the second one is false. The reason for this phenomenon is that~$\Delta f$ may be almost an arbitrary~$L_1$ function whereas the vectorial function~$\nabla f$ possesses hidden cancellations that distinguish it from an arbitrary~$L_1$ function. In~\cite{VanSchaftingen2013}, Van Schaftingen gave a simple necessary and sufficient condition on a differential operator on the right hand side of such type inequalities for the inequality to be true (we refer the reader to the early papers~\cite{BourgainBrezis2002}, \cite{BourgainBrezis2007}, \cite{VanSchaftingen2004}, \cite{VanSchaftingen2008}, and the survey~\cite{VanSchaftingen2014} for historical information about the preceding development of the subject). Let~$A$ be a differential operator that maps~$V$-valued functions into~$E$-valued functions, where~$V$ and~$E$ are finite dimensional spaces over~$\Co$. We will denote the symbol of~$A$ (which is a polynomial in~$d$ variables with values in the space of linear operators from~$V$ to~$E$) by the same letter~$A$. Assume~$A$ is elliptic, which in this setting means that the kernel of the linear operator~$A(\xi)$ is trivial for any~$\xi \in \R^d\setminus \{0\}$, and homogeneous of order~$m$. The operator~$A$ is called canceling provided
\eq{\label{Cancel}
\bigcap\limits_{\zeta \in S^{d-1}} \Image (A(\zeta)) = \{0\};
}
here~$S^{d-1}$ denotes the unit sphere. Note that, in particular, this is possible when~$\dim E > \dim V$ only. In other words, the corresponding system of partial differential equations~$A f = u$ is overdetermined. In~\cite{VanSchaftingen2013}, it was proved that an elliptic homogeneous of order~$m$ operator~$A$ is canceling if and only if
\eq{\label{canceling1}
\|\nabla^{m-1} f\|_{L_{\frac{d}{d-1}}} \lesssim \|A f\|_{L_1(\R^d,E)}.
}
Here and in what follows, the notation~$A \lesssim B$ means~$A \leq CB$, where the constant~$C$ may be chosen uniformly with respect to parameters that are clear from the context. For example, the constant in the inequality above  should be independent of~$f$. The space~$L_1(\R^d, E)$ is the space of summable~$E$-valued functions on~$\R^d$ equipped with the norm
\eq{
\|g\|_{L_1(\R^d,E)} = \int\limits_{\R^d}\|g(x)\|_E\,dx;
}
the particular choice of the norm on~$E$ is not important. By the classical Sobolev embedding, inequality~\eqref{canceling1} implies
\eq{\label{canceling2}
\|\nabla^{m-n} f\|_{L_{\frac{d}{d-n}}} \lesssim \|A f\|_{L_1(\R^d,E)}
}
whenever~$1 \leq n \leq m$ and~$n < d$. The case~$n=d$ (with additional assumption~$m \geq d$) is special; we postpone the discussion of this case.

Van Schaftingen's theorem may be restated in another way. Consider the space
\alg{\label{SpacesPolynom}
W_1^{A} = \clos_{L_1(\R^d,E)}\set{Af}{f \in C_0^{\infty}(\R^d,V)},
}
which may be thought of as  a generalization of the classical homogeneous Sobolev space~$\dot{W}_1^1$: the space~$W_1^A$ becomes~$\dot{W}^1_1$ in the case~$A = \nabla$.  The space~\eqref{SpacesPolynom} inherits the norm from~$L_1(\R^d,E)$. In the case~$n < d$, inequality~\eqref{canceling2} is equivalent via the Calder\'on--Zygmund theory to
\eq{\label{canceling2}
\|\I_{n}[g]\|_{L_{\frac{d}{d-n}}} \lesssim \|g\|_{W_1^A};
}
where~$\I_\alpha$ is the Riesz potential of order~$\alpha \in (0,d)$, i.e. the Fourier multiplier with the symbol~$|\fdot|^{-\alpha}$. The space~\eqref{SpacesPolynom} has a good description in terms of the Fourier transform:
\eq{\label{Intermediatespace}
W_1^{A} = \Set{g\in L_1(\R^d,E)}{\forall \xi \in \R^d\setminus \{0\}\quad \hat{g}(\xi) \subset \Image(A(\xi))}.
}
We note that, by ellipticity of~$A$, the linear spaces~$A(\xi)$ have dimension~$\dim V$ that does not depend on~$\xi$. The paper~\cite{Raita2018} extends Van Schaftingen's theory to the setting of differential operators of constant rank; this class of operators is wider than the class of elliptic operators, in this case the inequality~\eqref{canceling2} is also true provided one modifies the function on the left hand side in an appropriate way. The representation~\eqref{Intermediatespace} is suitable for generalizations.  We borrow the notation from~\cite{ASW2018}.

Let us identify~$E$ with~$\Co^\ell$, where $\dim E = \ell$. By~$G(\ell,k)$ we denote the complex Grassmannian of~$k$-dimensional subspaces of~$\Co^\ell$, i.e. the collection of all~$k$-dimensional linear subspaces of~$\Co^\ell$. This set has natural metric space structure. We may identify~$L\in G(\ell,k)$ with the orthogonal projection~$\pi_L$ of~$\Co^\ell$ onto~$L$, thus embedding~$G(\ell,k)$ into the space of~$\ell\times\ell$ matrices (which equals~$\Co^{\ell^2}$). The set~$G(\ell,k)$ inherits Euclidean metric from the ambient space: for any~$L_1, L_2 \in G(\ell,k)$, the quantity~$\|\pi_{L_1} - \pi_{L_2}\|_{\mathrm{HS}}$ plays the role of the distance between~$L_1$ and~$L_2$; the norm~$\|\cdot\|_{\mathrm{HS}}$ is the Hilbert--Schmidt norm on the space of~$\ell \times \ell$ complex matrices. Fix~$\ell$ and~$k$ and let~$\Omega \colon S^{d-1}\to G(\ell,k)$ be a continuous function, define
\eq{\label{IsotropicOmega}
W_1^\Omega = \Set{g\in L_1(\R^d,\Co^\ell)}{\forall \xi \in \R^d\setminus \{0\}\quad \hat{g}(\xi) \subset \Omega\Big(\frac{\xi}{|\xi|}\Big)}.
}
By a H\"older continuous mapping between metric spaces, we mean a function~$F\colon M \to N$ such that
\eq{
\dist_N (F(x),F(y)) \lesssim \dist_M (x,y)^{\alpha}
}
from some uniform~$\alpha > 0$ and all~$x,y\in M$ such that~$ \dist_M (x,y) \leq 1$.
\begin{Th}\label{L2Theorem}
Let~$\Omega$ be H\"older continuous and canceling in the sense that
\eq{\label{CancelBundle}
\bigcap_{\zeta \in S^{d-1}}\Omega(\zeta) = \{0\}.
}
Then\textup,~$\I_{\frac{d}{2}}$ maps~$W_1^\Omega$ to~$L_2$ continuously.
\end{Th}
It was conjectured in~\cite{ASW2018} that if~$\Omega$ is smooth and canceling in the same sense as above, then~$\I_{\alpha}$ maps~$W_1^\Omega$ to~$L_{\frac{d}{d-\alpha}}$ continuously. Theorem~\ref{L2Theorem} confirms the conjecture for~$\alpha  \geq \frac{d}{2}$. In particular, Theorem~\ref{L2Theorem} provides an alternative proof of Van Schaftingen's result~\eqref{canceling1} in dimension~$d=2$. Let us now return to the case~$\alpha = d$ and the language of differential operators for a moment. It appears that the cancellation condition~\eqref{Cancel} is only sufficient for the inequality
\eq{\label{canceling3}
\|\nabla^{m-d} f\|_{L_{\infty}} \lesssim \|A f\|_{L_1(\R^d,E)}.
}
In~\cite{Raita2019}, Raita gave necessary and sufficient conditions on~$A$ for this inequality to be true, the corresponding (elliptic) operators are called weakly canceling (similar effect is also present in the setting of constant rank operators, see~\cite{Raita2018}). Note that in this case the Calder\'on--Zygmund theory is not applicable directly, so the result depends not only on the collection~$\{\Image(A(\zeta))\}_{\zeta \in S^{d-1}}$, but on the structure of~$A$ as well. Now let us return to the Fourier transform notation.

Consider the set~$\omega$ of pairs~$(\zeta,v)$, where~$\zeta \in S^{d-1}$ and~$v\in \Omega(\zeta)$ and a function~$B \colon \omega \to \mathbb{C}$. The function~$B$ is assumed to be linear with respect to the second variable and H\"older continuous as a mapping of~$\omega \subset S^{d-1}\times \Co^\ell$ to~$\Co$. In other words,~$B(\zeta,\cdot)$ is a linear functional on~$\Omega(\zeta)$ that depends in a H\"older continuous way on~$\zeta$. We prefer to write~$B(\zeta)[v]$ instead of~$B(\zeta,v)$ since the two arguments have different nature. Consider the vectorial Fourier multiplier~$\B$ acting on functions~$g \in W_1^\Omega$ by the rule
\eq{\label{Functional}
\B[g] = \mathcal{F}^{-1}\Big[|\xi|^{-d}B\Big(\frac{\xi}{|\xi|}\Big)\big[\hat{g}(\xi)\big]\Big],\qquad g\in W_1^\Omega.
}
This definition requires explanations. First, we apply the functional~$B(\zeta)$,~$\zeta = \xi/|\xi|$, to a vector~$\hat{g}(\xi)$, which, by the assumption~$g\in W_1^\Omega$, belongs to the domain of~$B$, that is~$\Omega(\zeta)$. Second, even if the formula is logically correct, it is not clear why the function on the right hand side is even a distribution since its Fourier transform might have a singularity~$|\xi|^{-d}$ at the origin. There are two ways to get around this peculiarity: either regularize the symbol of our multiplier or consider only good functions~$g$ that form a dense set in~$W_1^\Omega$.  While we will follow the former way in Section~\ref{S2}, we provide a short comment about the latter. It is not clear whether the Schwartz functions are dense in~$W_1^\Omega$ when~$\Omega$ is not smooth; for smooth~$\Omega$ this is true, see~\cite{AyoushWojciechowski2017} (though this density statement is not formulated in that paper, the notation and definitions there make the statement almost obvious). However, the set of functions~$g\in W_1^\Omega$ whose spectrum is compact and does not contain the origin, is dense in~$W_1^\Omega$; one may use the formula~\eqref{Functional} for such functions~$g$ only and then pass to the limit whenever an appropriate uniform estimate is proved.
\begin{Th}\label{LinftyTheorem}
Let~$\Omega$ and~$B$ be H\"older continuous and let~$B$ be weakly canceling\textup, that is
\eq{\label{Weaklycanceling}
\forall v\in V\quad \int\limits_{S^{d-1}} B(\zeta)[v]\,d\sigma(\zeta) = 0,\qquad V = \bigcap_{\zeta \in S^{d-1}} \Omega(\zeta).
}
Then\textup, the operator~$\B$ given by~\eqref{Functional} maps~$W_1^\Omega$ to~$L_{\infty}$ continuously.
\end{Th}
Here and in what follows,~$\sigma$ stands for the Lebesgue measure on the unit sphere. We will comment in Section~\ref{S4} how this theorem generalizes Raita's result. Now we only mention that if~$\Omega$ is canceling (satisfies~\eqref{CancelBundle}), then~$V$ is trivial, and any H\"older continuous functional~$B$ is admissible.
\begin{Rem}
By a simple limit argument\textup, the space~$L_{\infty}$ in Theorem~\ref{LinftyTheorem} may be replaced with~$C(\R^d)$. Note that, in the classical setting of differential operators, a similar effect holds for a related~$\BV$ space \textup(defined similarly to~\eqref{BV} below\textup) if and only if~$\Omega$ is canceling\textup, i. e. satisfies~{\eqref{CancelBundle}}\textup, see~\textup{\cite{RaitaSkorobogatova2020}} for details.
\end{Rem}

Let us provide a short comment about the proofs. In fact, the proofs are not longer than either introduction, or statement of the results in higher generality in Section~\ref{S2}, or the applications at the end of the paper. Theorem~\ref{L2Theorem} follows from Theorem~\ref{LinftyTheorem} by simple bilinearization; such a trick was used in~\cite{KMS2015} in a similar context. The proof of Theorem~\ref{LinftyTheorem} is not more difficult: we extend the functional~$B$ to the whole space~$\Co^\ell$ preserving the weak cancellation condition~\eqref{Weaklycanceling} and H\"older continuity and use Mikhlin's principle that a singular integral has bounded Fourier transform. In fact, the proof of Theorem~\ref{LinftyTheorem} mimics the proof of the corresponding theorem in discrete setting in~\cite{Stolyarov2019} (see~\cite{ASW2018} for discrete probabilistic interpretation of canceling operators).

Classical embedding theorems allow a useful generalization that replaces classical homogeneity with respect to isotropic dilations with the so-called anisotropic homogeneity (see, e.g. the book~\cite{BIN1978}). The progress in this direction related to canceling and weakly canceling operators is desirable and seemingly, not straightforward. However, the methods of the present paper are so elementary that we are able to obtain anisotropic versions for Theorems~\ref{L2Theorem} and~\ref{LinftyTheorem}; we present them in Section~\ref{S2}.

\section{Anisotropic homogeneity and anisotropic singular integrals}\label{S2}
Our notation is close to that of~\cite{BesovLizorkin1967}. Let~$a\in \mathbb{R}^d$ be a vector with positive coordinates such that~$\sum_1^d a_j = d$. We call such~$a$ a pattern of homogeneity. Define the~$a$-dilations by the formula
\eq{
\Dil_t(\xi) = \big(t^{-a_1}\xi_1, t^{-a_2}\xi_2,\ldots, t^{-a_d}\xi_d\big),\quad \xi = (\xi_1,\xi_2,\ldots, \xi_d)\in \mathbb{R}^d,\ t > 0.
}
A function~$\Phi\colon\mathbb{R}^d \to \mathbb{C}$ is called~$a$-homogeneous of order~$m$ provided
\eq{
\Phi(\Dil_t(\xi)) = t^{-m}\Phi(\xi)
}
for any~$\xi\in\mathbb{R}^d$ and any~$t > 0$. This definition also applies to vector-valued functions. Let us define the function~$\eta \colon \R^d\setminus \{0\} \to\R$ implicitly:
\eq{
\sum\limits_{j=1}^d \frac{\xi_j^2}{\eta(\xi)^{2a_j}} = 1.
}
In other words, for any~$\xi$ the trajectory~$t\mapsto \Dil_t(\xi)$ intersects the unit sphere~$S^{d-1}$ at the moment~$t = \eta(\xi)$; the quantity~$\eta(\xi)$ may be regarded as the "norm" of~$\xi$. One may replace~$\eta$ with an equivalent expression that is more convenient for computations:
\eq{\label{AsymptoticEta}
\eta(\xi) \lesssim  \Big(\sum\limits_{j=1}^d|\xi_j|^{\frac{2}{a_j}}\Big)^{\frac12} \lesssim \eta(\xi).
}
The function~$\eta$ gives rise to a "polar change of variables"~$\xi = (\eta, \zeta)$, here~$\zeta = \zeta(\xi) \in S^{d-1}$ is given by
\eq{
\zeta_j = \eta(\xi)^{-a_j}\xi_j,\quad j = 1,2,\ldots,d.
}
In other words,~$\zeta = \Dil_{\eta(\xi)}(\xi)$.

The polar change of variables reads as follows:
\eq{
\int\limits_{\mathbb{R}^d}\Psi(\xi)\,d\xi = \int\limits_{\mathbb{R}_+} \eta^{d-1}\int\limits_{S^{d-1}} \Big(\sum\limits_{j=1}^d a_j \zeta_j^2\Big) \Psi(\Dil_{\eta^{-1}}(\zeta))\,d\sigma(\zeta)\,d\eta.
}
The proof of the formula relies upon homogeneity considerations and the fact that the scalar product of the unit normal to the sphere~$S^{d-1}$ and the tangent to the trajectory~$t\mapsto \Dil_t(\zeta)$ at their intersection point~$\zeta$ equals exactly~$\sum_{j=1}^d a_j \zeta_j^2$. In particular,
\eq{\label{SubstitutionHom}
\int\limits_{\mathbb{R}^d}\Phi(\xi)\,d\xi = \int\limits_{\mathbb{R}_+} \eta^{m+d-1}\int\limits_{S^{d-1}} \Big(\sum\limits_{j=1}^d a_j \zeta_j^2\Big) \Phi(\zeta)\,d\sigma(\zeta)\,d\eta,
}
provided~$\Phi$ is~$a$-homogeneous of order~$m$. We will need an anisotropic version of Mikhlin's principle that a singular integral has bounded Fourier transform. We will use a version from~\cite{BesovLizorkin1967} (see~\cite{FabesRiviere1966} for a similar theory; we prefer a version from~\cite{BesovLizorkin1967} since it indicates smoothness assumptions on the kernel in a more explicit way; see the short report~\cite{BIL1966} as well).

\begin{Le}\label{SingularIntegral}
Let~$K\colon\mathbb{R}^d\setminus \{0\}\to \Co$ be a continuous function that is $a$-homogeneous of order~$-d$. Assume it satisfies the cancellation condition
\eq{\label{MikhlinCanc}
\int\limits_{S^{d-1}}\Big(\sum\limits_{j=1}^d a_j \zeta_j^2\Big) K(\zeta)\,d\sigma(\zeta) = 0
}
and smoothness condition
\eq{\label{Dini}
\int\limits_{0}^1\frac{w(t)}{t}\,dt < \infty; \quad w(t) = \sup\Set{|K(\zeta_1) - K(\zeta_2)|}{\zeta_1,\zeta_2\in S^{d-1},\ |\zeta_1-\zeta_2|\leq t}.
}
Then\textup, the truncated functions~$K_{\eps,R}$\textup, $0 < \eps< R$\textup, defined by the formula
\eq{
K_{\eps,R}(\xi) = \begin{cases}K(\xi),\quad &\eta(\xi) \in [\eps,R];\\ 0,\quad &\hbox{otherwise},\end{cases}\qquad \xi= (\eta,\zeta),
}
have uniformly bounded Fourier transforms. In particular\textup, the singular integral
\eq{
\varphi \mapsto \vp\int\limits_{\R^d}K(\xi)\varphi(\xi)\,d\xi = \lim\limits_{\genfrac{}{}{0pt}{-2}{\eps \to 0}{R\to \infty}}\int\limits_{\R^d} K_{\eps,R}(\xi)\varphi(\xi)\,d\xi, \quad \varphi \in C_0^{\infty}(\R^d),
}
defines a distribution with bounded Fourier transform.
\end{Le}
\begin{Rem}
There is a little mess in the terminology. As we will see\textup, the classical cancellation condition~\eqref{MikhlinCanc} is more related to weakly cancelling condition~\eqref{Weaklycanceling} rather than Van Schaftingen's cancellation condition~\eqref{Cancel}.
\end{Rem}
\begin{Rem}
The smoothness condition~\eqref{Dini} holds true provided~$K|_{S^{d-1}}$ is H\"older continuous.
\end{Rem}

We transfer the notions from introduction to the anisotropic setting. Let us start with~\eqref{IsotropicOmega}:
\eq{\label{GeneralOmega}
W_1^{\Omega,a} = \Set{f \in L_1(\R^d,\Co^\ell)}{\forall \xi \in \R^d\setminus \{0\}\quad \hat{f}(\xi)\in \Omega(\Dil_{\eta(\xi)}(\xi))}.
}
Note that this space is translation invariant and dilation invariant with respect to~$a$-dilations. We equip it with the norm inherited from of~$L_1(\R^d,\Co^\ell)$. We will suppress the dependence on the pattern of homogeneity in our notation and write simply~$W_1^\Omega$ instead of~$W_1^{\Omega,a}$; the same principle applies to functionals introduced below.

Consider the vectorial Fourier multipliers~$\B_{\eps,R}^a$, where~$0 < \eps < R$, acting on functions~$f\in W_\Omega^1$ by the rule
\eq{\label{AeR}
\B_{\eps,R}^a[f] = \mathcal{F}^{-1}\Big[\chi_{\eta\in [\eps,R]}(\xi) \eta^{-d} B(\Dil_{\eta}(\xi))[\hat{f}(\xi)]\Big],\quad \xi = (\eta,\zeta).
}
Note that the formula is correct since~$\hat{f}(\xi) \in \Omega(\Dil_{\eta}(\xi))$ for any~$\xi$ and thus, the functional~$B(\Dil_{\eta}(\xi))$ may be applied to~$\hat{f}(\xi)\in \Omega(\Dil_{\eta(\xi)}(\xi))$. The version of the operator~$\B^a$ without the cutoff function is~$a$-homogeneous. The cutoff function is needed to make the formula correct (note that~$\B_{\eps,R} [f]$ is a real analytic function).

\begin{Th}\label{RaitaTheorem}
The operator~$\B_{\eps,R}$ is uniformly with respect to~$\eps$ and~$R$ bounded as a mapping of~$W_1^\Omega$ to~$L_{\infty}$ if and only if the symbol~$B$ satisfies the weak cancellation condition
\eq{\label{WeakCancellation}
\int\limits_{S^{d-1}} B(\zeta)[v] \Big(\sum\limits_{j=1}^d a_j \zeta_j^2\Big)\,d\zeta = 0
}
for any~$v \in V$\textup, where
\eq{\label{DefV}
V =  \bigcap\limits_{\zeta \in S^{d-1}} \Omega(\zeta).
}
\end{Th}
The theorem above implies Theorem~\ref{LinftyTheorem}.
Let us write
\eq{\label{Jacobian}
\J(\zeta) = \sum\limits_{j=1}^d a_j \zeta_j^2
}
for brevity. The lemma below is the core of the matter.
\begin{Le}\label{Extension}
Let~$B$ and~$\Omega$ be the same as defined above. There exists a function~$\tilde{B}\colon S^{d-1}\times \Co^\ell \to \Co$ that is linear with respect to the second variable\textup, H\"older continuous\textup, extends~$B$ in the sense that
\eq{
\forall \zeta \in S^{d-1}, \ v\in \Omega(\zeta)\qquad \tilde{B}(\zeta)[v] = B(\zeta)[v],
}
and satisfies the cancellation condition
\eq{\label{CancellationTilde}
\int\limits_{S^{d-1}} \tilde{B}(\zeta)[v] \J(\zeta)\,d\zeta = 0
}
for any~$v\in \Co^\ell$.
\end{Le}

By~$\I_\beta$ we denote the $a$-homogeneous version of the classical Riesz potential:
\eq{\label{AnisotropicRiesz}
\I_\beta[f]=\mathcal{F}^{-1}\Big[\hat{f}(\xi)\eta^{-\beta}(\xi)\Big],\quad f\in L_1(\R^d),\ \beta \in (0,d).
}
Note that this definition applies to the case~$f$ is a vectorial measure of bounded variation as well (however, the result of application of~$\I_\beta$ might be a distribution).

The theorem below is an anisotropic generalization of Theorem~\ref{L2Theorem}.
\begin{Th}\label{AnisotropicL2}
Assume the function~$\Omega \colon S^{d-1}\to G(\ell,k)$ is H\"older continuous and canceling in the sense that the corresponding space~$V$ \textup(defined in~\eqref{DefV}\textup) is trivial. Then\textup, the $a$-homogeneous Riesz potential~$\I_{\frac{d}{2}}$ maps the~$a$-homogeneous space~$W_1^\Omega$ to~$L_2$.
\end{Th}

\section{Proofs}
\begin{proof}[Proof of Lemma~\ref{Extension}]
Let~$\Omega^\perp(\zeta)$ be the orthogonal complement of~$\Omega(\zeta)$ in~$\Co^\ell$. We will search for~$\tilde{B}$ of the form
\eq{
\tilde{B}(\zeta)[v] = B(\zeta)[\pi_{\Omega(\zeta)}[v]] + \scalprod{b(\zeta)}{\pi_{\Omega^\perp(\zeta)}[v]},
}
where~$b\colon S^{d-1} \to \Co^\ell$ is a vectorial function to be chosen later. Note that this~$\tilde{B}$ extends~$B$ since the second term vanishes for~$v\in \Omega(\zeta)$ and the first equals~$B(\zeta)[v]$ in this case. Moreover, if~$b$ is H\"older continuous, then~$\tilde{B}$ is H\"older continuous as well.

Note that the linear functional
\eq{
v\mapsto \int\limits_{S^{d-1}} B(\zeta)[\pi_{\Omega(\zeta)}[v]] \J(\zeta)\,d\sigma(\zeta),\qquad v\in \Co^\ell,
}
vanishes on~$V$ (recall~\eqref{WeakCancellation} and~\eqref{DefV}). Therefore, there exists~$a\in V^\perp$ such that
\eq{
\scalprod{a}{u} = -\int\limits_{S^{d-1}} B(\zeta)[\pi_{\Omega(\zeta)}[u]] \J(\zeta)\,d\sigma(\zeta),\qquad u \in\Co^\ell.
}
Thus, in order to fulfill~\eqref{CancellationTilde}, the function~$b$ suffices to satisfy
\eq{\label{SearchForb}
\int\limits_{S^{d-1}}\pi_{\Omega^\perp(\zeta)}[b(\zeta)] \J(\zeta)\,d\sigma(\zeta) = a.
}

Let~$T\colon C^{\infty}(S^{d-1},\Co^\ell)\to \Co^\ell$ be the linear operator that maps~$b$ to
\eq{
\int\limits_{S^{d-1}}\pi_{\Omega^\perp(\zeta)}[b(\zeta)] \J(\zeta)\,d\sigma(\zeta).
}
It suffices to show that the image of~$T$ is dense in~$V^{\perp}$ (in such a case this image coincides with~$V^{\perp}$ and there exists a smooth solution~$b$ to the equation~\eqref{SearchForb}).

First, it is clear that~$\Image[T] \subset V^\perp$ since for any~$v \in V$ we have~$\pi_{\Omega^\perp(\zeta)}[v] = 0$ for any~$\zeta$ and thus,
\eq{
\scalprod{T[b]}{v} = \int\limits_{S^{d-1}}\scalprod{b(\zeta)}{\pi_{\Omega^\perp(\zeta)}[v]} \J(\zeta)\,d\sigma(\zeta) = 0,
}
which means~$T[b]$ is orthogonal to~$V$ for any choice of~$b$.

Second, if the image of~$T$ were not dense in~$V^\perp$, there would have existed~$w\in V^{\perp}$ such that
\eq{
\Scalprod{\int\limits_{S^{d-1}}\pi_{\Omega^\perp(\zeta)}[b(\zeta)] \J(\zeta)\,d\sigma(\zeta)}{w} = 0
}
for any~$b$. This is the same as
\eq{
\int\limits_{S^{d-1}} \scalprod{b(\zeta)}{\pi_{\Omega^\perp(\zeta)}[w]}\J(\zeta)\,d\sigma(\zeta) = 0.
}
Note that the function~$\zeta \mapsto \J(\zeta)\pi_{\Omega^\perp(\zeta)}[w]$ is H\"older continuous. Consequently, by the standard limit argument, the same identity also holds true for any continuous function~$b$, and therefore,~$\pi_{\Omega^\perp(\zeta)}[w]$ vanishes identically. This means~$w = 0$ since~$w\in V^\perp$.
\end{proof}
\begin{proof}[Proof of Theorem~\ref{RaitaTheorem}]
The "only if part". Note that if~$v\in V$, then~$\delta_0\otimes v \in \BV^\Omega$, where~$\delta_0$ is the Dirac delta situated at the origin and the definition of the corresponding~$\BV$ space is similar to~\eqref{GeneralOmega}:
\eq{\label{BV}
\BV^{\Omega,a} = \Set{\mu \in \Me(\R^d,\Co^\ell)}{\forall \xi \in \R^d\setminus \{0\}\quad \hat{\mu}(\xi)\in \Omega(\Dil_{\eta(\xi)}(\xi))};
}
by~$\Me(\R^d,\Co^\ell)$ we denote the space of~$\Co^\ell$-valued measures of bounded variation. The norm of~$\B_{\eps, R}$ as an operator between~$\BV^\Omega$ and~$L_\infty$ equals the norm of the same multiplier treated as an~$W_1^\Omega \to L_{\infty}$ operator. It remains to notice that
\eq{
\B_{\eps,R}[\delta_0\otimes v](0) = \int\limits_{\eps}^R\eta^{-1}\,d\eta \int\limits_{S^{d-1}} B(\zeta)[v] \Big(\sum\limits_{j=1}^d a_j \zeta_j^2\Big)\,d\zeta
}
by virtue of~\eqref{SubstitutionHom}.

To prove the "if" part, apply Lemma~\ref{Extension} and construct the operators~$\tilde{\B}_{\eps,R}$ substituting~$\tilde{B}$ for~$B$ in~\eqref{AeR}. Since~$\tilde{B}$ is an extension of~$B$, we have
\eq{
\B_{\eps,R}[f] = \tilde{\B}_{\eps,R}[f],\qquad f\in W_1^{\Omega}.
}
Therefore, it suffices to show that~$\tilde{\B}_{\eps,R}$ maps~$L_1(\R^d,\Co^\ell)$ to~$L_{\infty}$ (with uniform norm bounds w.r.t.~$\eps$ and~$R$). This is the same as to show that the Fourier transform of the function
\eq{
\xi \mapsto \chi_{\eta\in [\eps,R]}(\xi) \eta^{-d} \tilde{B}(\Dil_{\eta}(\xi))
}
has uniformly w.r.t $\eps$ and~$R$ bounded Fourier transform. This function is vector-valued and~$a$-homogeneous of order~$-d$. So it remains to pair it with an arbitrary vector~$v\in \Co^\ell$ and apply Lemma~\ref{SingularIntegral}. The application of Lemma~\ref{SingularIntegral} is legal by~\eqref{CancellationTilde}.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{AnisotropicL2}]
Pick an arbitrary~$f\in W_1^\Omega$. By the Plancherel theorem, it suffices to prove a uniform with respect to~$f$,~$\eps$, and~$R$ bound
\eq{
\int\limits_{\eta \in [\eps,R]}\frac{|\hat{f}(\xi)|^2}{\eta^d(\xi)}\,d\xi \lesssim \|f\|_{W_1^\Omega}^2.
}
Let us bilinearize this quadratic inequality:
\eq{
\Big|\int\limits_{\eta \in [\eps,R]}\eta^{-d}(\xi) \scalprod{\hat{f}(\xi)}{\hat{g}(\xi)}_{\Co^\ell}\,d\xi\Big| \lesssim \|f\|_{W_1^\Omega}\|g\|_{L_1(\R^d,\Co^\ell)}.
}
The expression on the left hand side might be represented as
\eq{
\Big|\sum\limits_{j=1}^\ell \int\limits_{\R^d}  \B_{j,\eps,R}[f](x)\overline{g_j(x)}\,dx \Big|,
}
where~$\B_{j,\eps,R}$ is the Fourier multiplier with the symbol~\eqref{AeR}, where the corresponding functional~$B$ maps~$v\in\Omega(\zeta)$ to its~$j$-th coordinate~$v_j$ (in particular,~$B$ does not depend on~$\zeta$). Since~$V = \{0\}$, the operators~$\B_j$ fulfill the requirements of Theorem~\ref{RaitaTheorem}, and thus,
\eq{
\|\B_{j,\eps,R}[f]\|_{L_{\infty}} \lesssim \|f\|_{W_1^\Omega}
}
uniformly with respect to~$j, \eps, R$, and~$f$, and the theorem is proved.
\end{proof}

\section{Examples}\label{S4}
\paragraph{Theorem~\ref{LinftyTheorem} and differential operators.} Let us rewrite condition~\eqref{Weaklycanceling} in terms of differential operators. Let~$A$ be an elliptic differential operator that maps~$V$-valued functions to~$E$-valued functions and is (isotropic) homogeneous of order~$m \geq d$. We are interested in the inequality
\eq{\label{Interest}
\|P f\|_{L_{\infty}} \lesssim \|A f\|_{L_1},\quad f\in C_0^{\infty}(\R^d,V),
}
here~$P$ is a homogeneous differential operator of order~$m-d$. We choose
\eq{
\Omega(\zeta) = \Image (A(\zeta));\quad B(\zeta) = P(\zeta)(A^*(\zeta)A(\zeta))^{-1}A^*(\zeta),
}
where~$P$ is identified with its symbol. Note that
\eq{
\B[A f] = Pf,\quad f\in C_0^{\infty}(\R^d,V),
}
so~\eqref{Interest} is equivalent to
\eq{
\|\B g\|_{L_{\infty}} \lesssim \|g\|_{W_1^\Omega}.
}
By Theorem~\ref{LinftyTheorem}, the latter inequality holds true if and only if
\eq{
\forall v\in \bigcap_{\zeta \in S^{d-1}}\Image(A(\zeta))\qquad \int\limits_{S^{d-1}} P(\zeta)(A^*(\zeta)A(\zeta))^{-1}A^*(\zeta)[v]\,d\sigma(\zeta) = 0;
}
this coincides with the weak cancellation condition in~\cite{Raita2019}.

\paragraph{Embedding theorems from~\cite{KMS2015}.} The paper~\cite{KMS2015} concerned the isomorphism problem for the Banach spaces~$\BV^A$ and their relatives. Embedding theorems serve as useful tools for such results (this goes back to~\cite{Kislyakov1975}).  We will show how the embedding theorems from~\cite{KMS2015} may be derived from Theorem~\ref{AnisotropicL2} on two examples (we do not dwell on an interesting effect of transference of Fourier coefficients present in the most general embedding theorem in~\cite{KMS2015}).

Consider the inequality
\eq{\label{PreGN}
\|f\|_{L_2(\R^3)} \lesssim \Big\|\frac{\partial f}{\partial x_1}\Big\|_{L_1(\R^3)}\|\Delta_{x_2,x_3} f\|_{L_1(\R^3)}
}
for smooth compactly supported functions on~$\R^3$; this inequality was proved in~\cite{KMS2015}. The operator~$\Delta_{x_2, x_3}$ is the Laplacian with respect to the variables~$x_2$ and~$x_3$. By homogeneity considerations,~\eqref{PreGN} is equivalent to
\eq{\label{GN}
\|f\|_{L_2(\R^3)} \lesssim \Big\|\frac{\partial f}{\partial x_1}\Big\|_{L_1(\R^3)} + \|\Delta_{x_2,x_3} f\|_{L_1(\R^3)}.
}
Consider the pattern of homogeneity~$a = (\frac32,\frac34,\frac34)$, let~$\ell = 2$,  and define the function~$\Omega \colon S^2\to G(2,1)$ by the rule
\eq{
S^2\ni \zeta = (\zeta_1,\zeta_2,\zeta_3)\ \longmapsto\ \Omega(\zeta) = \Co\cdot(\zeta_1,\zeta_2^2+\zeta_3^2) \subset \Co^2.
}
This function is smooth and satisfies the cancellation condition~$V = \{0\}$ (the space~$V$ is defined in~\eqref{DefV}). Therefore, we may apply Theorem~\ref{AnisotropicL2}. Using formula~\eqref{AsymptoticEta}, we obtain
\eq{
\eta^{-\frac32}(\xi) \asymp \Big(|\xi_1|^{\frac43} + |\xi_2|^{\frac83} + |\xi_3|^{\frac83}\Big)^{-\frac34},
}
and Theorem~\ref{AnisotropicL2} leads to
\mlt{
\Big\|\Big(|\xi_1|^{\frac43} + |\xi_2|^{\frac83} + |\xi_3|^{\frac83}\Big)^{-\frac34} \xi_1\hat{f}(\xi)\Big\|_{L_2(\R^3)} + \Big\|\Big(|\xi_1|^{\frac43} + |\xi_2|^{\frac83} + |\xi_3|^{\frac83}\Big)^{-\frac34} (\xi_2^2 + \xi_3^2) \hat{f}(\xi)\Big\|_{L_2(\R^3)} \lesssim\\ \Big\|\frac{\partial f}{\partial x_1}\Big\|_{L_1(\R^3)} + \|\Delta_{x_2,x_3} f\|_{L_1(\R^3)},
}
which implies~\eqref{GN}.

Let us cite Theorem $0.2$ in~\cite{KMS2015}. In that theorem,~$\kappa$ and~$\lambda$ were integer numbers,~$\alpha$ and~$\beta$ were non-negative reals, and
\begin{equation}\label{sobolevspace}
W_2^{\alpha,\beta}(\mathbb{R}^2)=
\{f \in\mathcal{S}'(\mathbb{R}^2) \mid
|\xi|^{\alpha} |\eta|^{\beta}\hat{f}(\xi,\eta) \in
L^2(\mathbb{R}^2)\};
\end{equation}
there are small inaccuracies in this definition, see~\cite{KMS2015}; by~$\mathcal{S}'$ we denote the class of Schwartz distributions on~$\R^2$.

\begin{Th}[Theorem~$0.2$ in~\cite{KMS2015}]\label{KMSMain}
Let~$\varphi_j$\textup,~$j=1,\dots,N$\textup, be compactly supported distributions
on the plane~$\mathbb{R}^2$. Suppose that they satisfy the equations
\begin{equation}\label{syst}
\left\{
\begin{aligned}
-\partial_1^\kappa \varphi_1 &&&=  &\mu_0& ;\\
-\partial_1^\kappa \varphi_2 &+\partial_2^\lambda \varphi_1&& =  &\mu_1& ;\\
&\vdots&&   = &\,\vdots\,& ;\\
-\partial_1^\kappa\varphi_j &+\partial_2^\lambda \varphi_{j-1}&& = &\mu_{j-1}& ;\\
&\vdots&&   = &\,\vdots\,& ;\\
-\partial_1^\kappa \varphi_N &+\partial_2^\lambda \varphi_{N-1}&& = &\mu_{N-1}& ;\\
 &\phantom{+}\;\,\partial_2^\lambda \varphi_N&& = &\mu_N&,
\end{aligned}
\right.
\end{equation}
where~$\mu_0,\dots,\mu_N$ are finite \textup{(}compactly
supported\textup{)} measures on the plane. The inequality
\begin{equation}\label{est1}
\sum\limits_{j = 1}^{N}\|\varphi_j\|_{W_2^{\alpha,\beta}(\mathbb{R}^2)}
\lesssim \sum\limits_{j = 0}^{N}\mathrm{var}\mu_j
\end{equation}
holds true whenever~$\alpha$ and~$\beta$ are nonnegative and satisfy
\begin{equation}\label{line}
\frac{\alpha + \frac{1}{2}}{\kappa} + \frac{\beta + \frac{1}{2}}{\lambda} = 1.
\end{equation}
\end{Th}
Let us deduce this theorem from Theorem~\ref{AnisotropicL2}. For that, set~$a=\big(\frac{2\kappa}{\kappa+\lambda},\frac{2\lambda}{\kappa+\lambda}\big)$,~$\ell = N+1$,~$k = N$, and
\eq{
\Omega(\zeta) = \Set{v\in \Co^{N+1}}{\sum\limits_{j=0}^N \zeta_1^{j \kappa}\zeta_2^{(N+1-j)\lambda}v_j = 0},\qquad \zeta \in S^1.
}
This~$\Omega$ satisfies the cancellation condition. Note that by~\eqref{AsymptoticEta},
\eq{
\eta^{-1}(\xi) \asymp \Big(|\xi_1|^{\frac{\kappa + \lambda}{\kappa}} + |\xi_2|^{\frac{\kappa + \lambda}{\lambda}}\Big)^{-\frac12},
}
and Theorem~\ref{AnisotropicL2} (modulo the equivalence of embeddings of~$W_1^\Omega$ and~$\BV_1^\Omega$, which is a routine approximation argument) implies Theorem~\ref{KMSMain} similar to the previous case.

Embedding theorems in the style of Theorem~\ref{KMSMain} for more general systems of equations were later obtained in~\cite{KislyakovMaximov2018}. These theorems also follow from Theorem~\ref{AnisotropicL2} (however, this provides no simplification against the original reasoning in~\cite{KislyakovMaximov2018}).

\paragraph{Comment on bilinear estimates in~\cite{Stolyarov2015}.} The paper~\cite{Stolyarov2015} concerned bilinear estimates similar to that used in~\cite{KMS2015}. Namely, the inequalities
\begin{equation}\label{BilinearEmbeddings}
\left|\langle f,g\rangle_{W_2^{\alpha,\beta}}\right| \lesssim \left\|(\partial_1^\kappa -\tau\partial_2^\lambda)f\right\|_{L_1}\left\|(\partial_1^\kappa -\sigma\partial_2^\lambda)g\right\|_{L_1};\quad \partial_1 = \frac{\partial}{\partial x}, \partial_2 = \frac{\partial}{\partial y},\ f,g\in C_0^{\infty}(\R^2),
\end{equation}
were investigated, the Hilbert space on the left hand side is the same as in~\eqref{sobolevspace}. Under the assumption that the operators on the right are elliptic, that paper provided simple if and only if conditions on the parameters~$\sigma,\tau \in \Co$,~$\alpha,\beta \geq 0$, when~\eqref{BilinearEmbeddings} holds true. It appears this result is closely related to Lemma~\ref{SingularIntegral}. What is more, Lemma~\ref{SingularIntegral} allows to work in a more general setting.

\begin{St}\label{BilinearProp}
Let~$P_1$ and~$P_2$ be elliptic~$a$-homogeneous scalar differential operators in~$d$ variables of orders~$d_1$ and~$d_2$ correspondingly. Let~$Q$ be a scalar Fourier multiplier with H\"older continuous symbol that is~$a$-homogeneous of order~$d_1 + d_2 - d$. The inequality
\eq{
\Big|\int\limits_{\R^d}Qf(x) \overline{g(x)}\,dx\Big|\lesssim \|P_1 f\|_{L_1(\R^d)} \|P_2 g\|_{L_1(\R^d)}
}
holds true for any smooth compactly supported functions~$f$ and~$g$ with uniform constant if and only if
\eq{
\int\limits_{\zeta \in S^{d-1}} \frac{Q(\zeta) \J(\zeta)}{P_1(\zeta)\overline{P_2(\zeta)}}\,d\sigma(\zeta) = 0,\qquad \J\hbox{is defined in~\eqref{Jacobian}}.
}
\end{St}
In the formula above, we identify the differential operators~$P_i$ with their symbols; the same principle applies to~$Q$. Proposition~\ref{BilinearProp} is an immediate consequence of Lemma~\ref{SingularIntegral}. The integral that appears if apply this proposition to inequality~\eqref{BilinearEmbeddings} looks like this:
\eq{
\int\limits_{0}^{2\pi}\frac{|\cos \varphi|^{2\alpha}|\sin\varphi|^{2\beta}(\frac{\cos^2\varphi}{\kappa} + \frac{\sin^2\varphi}{\lambda})\,d\varphi}{(\cos^\kappa \varphi -\tau_1\sin^\lambda\varphi)(\cos^\kappa \varphi -\sigma_1\sin^\lambda\varphi)},
}
where~$\tau_1 = (2\pi i)^{\lambda - \kappa}\tau$,~$\sigma_1 = (-1)^{\lambda - \kappa}(2\pi i)^{\lambda - \kappa}\overline{\sigma}$. The latter integral is proportional to the integral
\eq{
\int\limits_{-\infty}^{\infty}\frac{|\rho|^{2\alpha}\,d\rho}{\big(\rho^k - \tau_1\big)\big(\rho^k - \sigma_1\big)} = 0.
}
via the change of variables~$\rho = \cos\varphi \sin^{-\frac{\kappa}{\lambda}}\varphi$ and~\eqref{line}, which is, by homogeneity, necessary for~\eqref{BilinearEmbeddings}. This integral is~$(2.6)$ in~\cite{Stolyarov2015}; see~\cite{Stolyarov2015} why the vanishing of this integral is equivalent to the condition that one of the numbers~$\kappa$ and~$\lambda$ is odd,~$\alpha = \frac{\kappa-1}{2}$\textup,~$\beta=\frac{\lambda-1}{2}$\textup, and the numbers~$\sigma_1$ and~$\tau_1$ have non-zero imaginary parts of the same sign; these reasonings are based upon elementary complex variable techniques: the residue formula and the uniqueness theorem for analytic functions.

Inequality~\eqref{BilinearEmbeddings} remains true in some cases where the operators on the right are non-elliptic, i.e. when~$\sigma_1$ and~$\tau_1$ are real; see~\cite{Stolyarov2015}. This gives hope that Proposition~\ref{BilinearProp} and Lemma~\ref{SingularIntegral} may be generalized to the non-elliptic setting (by ellipticity in Lemma~\ref{SingularIntegral} we mean the smoothness condition~\eqref{Dini}, it seems that in some cases~$K|_{S^{d-1}}$ might be a distribution).

\bibliography{mybib}{}
\bibliographystyle{amsplain}

St. Petersburg State University, Department of Mathematics and Computer Science;

d.m.stolyarov at spbu dot ru.
\end{document}

config: arxiv


TITLE: Where can I find good resources explaining conic sections?
QUESTION [1 upvotes]: I'm looking for something that explains them simply or something that has a lot of examples. The online course I'm taking doesn't explain them in much detail and I haven't managed to find something that explains it better. Mainly parabolas and hyperbolas. For parabolas, my course uses the formula $(x-h)^2 = 4p(y-k)$, so while I did find a useful video on KhanAcademy, on the subject, he uses a different formula and that just left me more confused.

REPLY [0 votes]: A good source which is a little more advanced is Companion to Concrete Mathematics by Melzak, first chapter.


text:
\begin{document}

\begin{abstract}
We prove that an analogue of Jordan's theorem on finite subgroups of general linear groups does not hold for the groups of birational automorphisms of products of an elliptic curve and the projective line. This gives a negative answer to a question of V. L. Popov.
\end{abstract}

\subjclass{14E07, 14K05}

\maketitle

\section{Introduction}

Throughout this paper, $k$ is an algebraically closed field of characteristic zero, $\A^1$ and $\P^1$ the affine line and projective line respectively (both over $k$).
If $U$ is an irreducible algebraic variety over $k$ then we write $k[U], k(U)$ and $\Bir(U)$ for its ring ($k$-algebra) of regular functions, the field of rational functions and the group of birational $k$-automorphisms respectively.

The following definition was inspired by the classical theorem of Jordan about finite subgroups of general linear groups.

\begin{defn}[Definition 2.1 of \cite{Popov}]
A group $B$ is called a {\sl Jordan group} if there exists a positive integer $J_B$ such that every finite subgroup $B_1$ of $B$ contains a (normal) commutative subgroup, whose index in $B_1$ is at most $J_B$.
\end{defn}

 V. L. Popov \cite[Sect. 2]{Popov} posed a question whether $\Bir(Y)$ is a Jordan group when $Y$ is an irreducible surface. He obtained a positive answer to his question for almost all smooth projective minimal surfaces. One of the few remaining cases is a product $E\times\P^1$ of an elliptic curve $E$ and the projective line.

 Our main result is the following statement, which gives a negative answer to Popov's question.

 \begin{thm}
 \label{elliptic}
 If $E$ is an elliptic curve over $k$ then $\Bir(E\times\P^1)$  is not a Jordan group.
 \end{thm}

 Since, $U\times \A^1$ is birationally isomorphic to $U\times \P^1$, the groups
$\Bir(U\times \A^1)$ and $\Bir(U\times \P^1)$ are isomorphic and
 Theorem \ref{elliptic} becomes equivalent to the assertion that $\Bir(E\times\A^1)$  is not a Jordan group, which, in turn, is a special case of the following statement.

\begin{thm}
\label{main} Let $X$ be an abelian variety of positive dimension  over $k$. Then  $\Bir(X\times\A^1)$  is not a Jordan group.
\end{thm}

\begin{cor}
\label{rational}

 Let $X$ be an abelian variety of positive dimension  over $k$ and $Z$ is a rational variety of positive dimension  over $k$. Then $\Bir(X\times Z)$  is not a Jordan group.
\end{cor}

\begin{proof}[Proof of Corollary \ref{rational} (modulo Theorem \ref{main})]
Since $Z$ is birationally isomorphic to the $d$-dimensional affine space $\A^d$ with $d=\dim(Z)\ge 1$, the groups $\Bir(X\times Z)$ and $\Bir(X\times\A^d)$ are isomorphic. So, it suffices to check that $\Bir(X\times\A^d)$ is {\sl not} a Jordan group. If $d=1$ the result follows from Theorem \ref{main}. If $d>1$ then
$X\times\A^d=(X\times\A^1) \times \A^{d-1}$ and one may view $\Bir(X\times\A^1)$ as the certain subgroup of $\Bir(X\times\A^d)$ and again Theorem \ref{main} gives us the desired result.
\end{proof}

The paper is organized as follows. Section \ref{Bir1} deals with the certain subgroup $\Bir_1(X\times\A^1)$ of $\Bir(E\times\A^1)$ that is generated by translations on $X$ and multiplications of the global coordinate $t$ on $\A^1$ by nonzero rational functions on $X$. We assert that  $\Bir_1(X\times\A^1)$  is not a Jordan group; obviously, this assertion implies that $\Bir(X\times\A^1)$  is also  not a Jordan group. In Section \ref{group} we discuss a {\sl symplectic geometry} related to certain analogues of Heisenberg groups that were introduced by Mumford \cite[Sect. 1]{Mumford66}. In Sect. \ref{theta}, using results of  Mumford \cite[Sect. 1]{Mumford66},  we realize these analogues as subgroups of  $\Bir_1(X\times\A^1)$, which allows us to prove that
$\Bir_1(A\times\A^1)$  is not a Jordan group.

I am grateful to Volodya Popov for a stimulating question, his interest in this paper and useful discussions.

\section{Birational automorphisms of products of an abelian variety and the affine line}
\label{Bir1}

Let $X$ be an abelian variety of positive dimension over $k$. If $y\in X(k)$ then we write $T_y$ for the translation map
$$T_y: X\to X, \ x \mapsto x+y.$$
As usual, we write $\div(f)$ for the divisor of a rational function $f \in k(X)^{*}$. Clearly,  $T_y^{*}f$ is the
rational function $x \mapsto f(x+y)$, whose divisor coincides with $T_y^{*}(\div(f))$. Let $t$ be the global coordinate on $\A^1$.

We write $\Bir_1(X\times\A^1)\subset \Bir(X\times\A^1)$
for the set of birational automorphisms of the form
$$A(y,f): X \times \A^1 \dashrightarrow  X \times \A^1, (x,t)\mapsto (x+y, f(x)\cdot t)=(T_y(x), f(x)\cdot t)$$
where $y$ runs through $X(k)$ and $f$ through $k(X)^{*}$. Actually, $\Bir_1(X\times\A^1)$ is a subgroup of  $\Bir(X\times\A^1)$.
Indeed, one may easily check that
$$A(y_2,f_2) A(y_1,f_1)=A(y_1+y_2, T_{y_1}^{*}(f_2) \cdot f_1) \in \Bir_1(X\times\A^1) $$
and the inverse of $A(y,f)$ in  $\Bir(X\times\A^1)$ coincides with $A(-y, T_{-y}^{*}(1/f)) \in \Bir_1(X\times\A^1)$.

Now Theorem \ref{main} becomes an immediate corollary of the following statement.

\begin{thm}
\label{main1} Let $X$ be an abelian variety of positive dimension  over $k$. Then  $\Bir_1(X\times\A^1)$  is not a Jordan group.
\end{thm}

We prove Theorem \ref{main1} in Section \ref{theta}.

\section{Group theory}
\label{group}
 Let ${\K}$ be a finite commutative group.

 Let $\hat{\K}:=\Hom(\K,k^{*})$ be the group of
{\sl characters} of $\K$. We write the group law on $\K$ additively and on $\hat{\K}$ multiplicatively. In particular,
we write $\l1$ for the trivial character of $\K$. Clearly, the groups ${\K}$ and $\hat{\K}$ are isomorphic
(noncanonically); in particular, they have the same order, which we denote by $N=N_{\K}$.

Let $\mu_N\subset k^{*}$ be the (sub)group of $N$th roots of unity.
Clearly, for every nonzero
$x\in \K$ there exists $\ell\in \hat{\K}$ with $\ell(x)\ne 1$. On the other hand,
$$N x=0, \ \ell(x)\in \mu_N \ \forall x\in \K, \ell \in \hat{\K}.$$
Let us consider the commutative finite group $\H_{\K}= \K\times \hat{\K}$ and the nondegenerate alternating
bi-additive form
$$e_{\K}: \H_{\K}\times \H_{\K} \to k^{*}, \ ((x,\ell),(x^{\prime},\ell^{\prime})) \mapsto
\ell^{\prime}(x)/\ell(x^{\prime}).$$
Clearly, all the values of $e_{\K}$ lie in $\mu_N$.

Let $E$ be an {\sl isotropic} subgroup of $\H_{\K}$ with respect to $e_{\K}$.
Let $E^{\bot}$ be the orthogonal complement of $E$ in $\H_{\K}$ with respect to $e_{\K}$. Then $E\subset E^{\bot}$ and the
 nondegeneracy of
$e_{\K}$  gives rise to a group isomorphism
$$\H_{\K}/E^{\bot}\cong \Hom(E,k^{*})=\hat{E}.$$
In particular, $E$ and $\H_{\K}/E^{\bot}$ have the same order.
 The inclusions $E \subset E^{\bot}\subset \H_{\K}$ imply that
$$\#(E)^2=\#(E)\cdot \#(\H_{\K}/E^{\bot})$$
divides $\#(\H_{\K})=N^2$ and therefore $\#(E)$ divides $N$. Since
$$N^2=\#(\H_{\K})=\#(E)\cdot \#(\H_{\K}/E),$$
the index of $E$ in $\H_{\K}$ is divisible by $N$.
This means that {\sl the index of every isotropic subgroup in $\H_{\K}$ is divisible by $N$ and therefore is greater or equal than $N$}.

 Following \cite[Sect. 1]{Mumford66}, let us consider the set
$$\G_{\K}=k^{*}\times \H_{\K}= k^{*}\times \K\times \hat{\K}$$ and introduce
on it the group structure, by defining the product
$$(a,x,\ell)\ (a^{\prime},x^{\prime},\ell^{\prime}):=(a a^{\prime}\ell^{\prime}(x),x+x^{\prime},\ell \ell^{\prime}).$$
One may naturally identify $k^{*}$ with the central subgroup $\{(a,0,\l1) \mid a \in k^{*}\}$. In fact, $\G_{\K}$ sits in
the short exact sequence
$$0 \to k^{*} \to \G_{\K}\stackrel{\pi}{\to} \H_{\K}\to 0$$
where $\pi: \G_{\K} \twoheadrightarrow \H_{\K}$ sends $(a,x,\ell)$ to $(x,\ell)$. One may easily check that if $g,
g^{\prime} \in \G_{\K}$ then

$$g g^{\prime} g^{-1} {g^{\prime}}^{-1}=e_{\K}(\pi(g),\pi(g^{\prime}))\in k^{*} \subset \G_{\K}.$$
It follows that a subgroup $\tilde{E}\subset \G_{\K}$ is {\sl commutative} if and only if its image $\pi(\tilde{E})$ is an
{\sl isotropic} subgroup in $\H_{\K}$; if this is the case then the index of  $\pi(\tilde{E})$ in $\H_{\K}$ is greater or equal than $N=N_{\K}$.

Clearly, the subset
$$\G_{\K}^{1}=\mu_N\times \H_{\K}= \mu_N\times \K\times \hat{\K}\subset \G_{\K}$$
is actually a subgroup of $\G_{\K}$. We have
 $\pi(\G_{\K}^{1})=\H_{\K}$. Therefore if $\tilde{E}$ is a commutative subgroup in $\G_{\K}^{1}$ then
the index of $\pi(\tilde{E})$ in $\H_K=\pi(\G_{\K}^{1})$ is greater or equal than $N=N_{\K}$. This implies that index of
 $\tilde{E}$ in $\G_{\K}^{1}$ is also greater or equal than $N=N_{\K}$.

\section{Mumford's theta groups}
\label{theta}

We keep all the notation and assumptions of Section \ref{Bir1}.

We denote by $\M_X$ the  constant sheaf (of rational functions) on the abelian variety  $X$ with respect to Zariski topology,
which assigns to every non-empty open subset $U$ of $X$ its field of rational functions $k(U)=k(X)$. For every $f \in
k(X)^{*}$ let us consider the sheaf (auto)morphism
$$[f]: \M_X \to \M_X$$
that is induced by multiplication by $f$ in $k(X)$. If $y \in X(k)$ then $T_y^{*}\M_X=\M_X$ and the induced (by functoriality)  sheaf
(auto)morphism $[f]: T_y^{*}[f]: T_y^{*}\M_X \to T_y^{*}\M_X$ coincides (after the identification of $T_y^{*}\M_X$ and $\M_X$) with
$$[T_y^{*}f]: \M_X \to \M_X.$$
 If $D$ is a divisor on $X$ then we view the invertible sheaf $\OC_X(D)$ as a certain subsheaf of $\M_X$ (see
\cite[Vol. II, Ch. 6, Sect. 1]{Sh}). Notice that for all $y\in X(k)$
$$T_y^{*}\OC_X(D)=\OC_X(T_y^{*}D).$$
If $D_1$ and $D_2$ are  linearly equivalent divisors on $X$  then  isomorphisms of invertible sheaves
$\OC_X(D_1)\cong \OC_X(D_2)$ are exactly (all the) morphisms of the form
$$[f]:\OC_X(D_1)\cong\OC_X(D_2)$$
with $\div(f)=D_1-D_2$. In particular, this set of isomorphisms is a $k^{*}$-torsor, since $\div(f)$ determines the
rational function $f$ up to multiplication by a nonzero constant.

If $[f]:\OC_X(D_1)\cong\OC_X(D_2)$ is an isomorphism of invertible sheaves and $y\in X(k)$ then the induced (by
functoriality) the isomorphism of invertible sheaves $T_y^{*}[f]:T_y^{*}\OC_X(D_1)\cong T_y^{*}\OC_X(D_2)$ coincides with
$$[T_y^{*}f]:\OC_X(T_y^{*}D_1)\cong \OC_X(T_y^{*}D_2).$$

Now let us choose an ample divisor on $X$ (e.g., a hyperplane section) and put $L=\OC_X(D)$. Then $L$ is an ample
invertible sheaf. Let us consider the (finite) commutative group
$$H(L)=\{x \in X(k)\mid L \cong T_x^{*}L \}.$$

\begin{rem}
\label{bign} Let $n$ be a positive integer. Then $nD$ remains ample, $\OC_X(nD)=L^n$ and
$$H(L^n)=\{x \in X(k)\mid nx \in H(L)\}$$
(see \cite[Sect. 1, Prop. 4]{Mumford66}). In particular, $H(L^n)$ contains the group $X_n$ of all points of order $n$ on
$X$. Since the order of $X_n$ is $n^{2\dim(X)}$ \cite[Ch. 2, Sect. 6]{MumfordAV}, the order of $H(L^n)$ is divisible by  $n^{2\dim(X)}$.

\end{rem}

 Following Mumford \cite[Sect. 1]{Mumford66}, let us consider the {\sl theta group} $\G(L)$ that consists of all pairs $(x,\phi)$ where $x \in H(L)$ and $\phi$ is an isomorphism
of invertible sheaves $L \cong T_x^{*}L$. The group law on $\G(L)$ is defined as follows. If $(x, \phi:L \cong
T_x^{*}L)\in \G(L)$ and $(y, \psi:L \cong T_y^{*}L)\in \G(L)$ then its composition $(y,\psi) (x,\phi)$ is defined as $$(x+y, T_{x}^{*}\phi \
\psi:L \cong T_y^{*}L \cong T_x^{*}(T_y^{*}L)=T_{x+y}^{*}L).$$ Taking into account our considerations in the beginning
of this Section and the equality $L=\OC_X(D)$, we conclude that $H(L)$ coincides with the set of $x\in X(k)$ such that
$D$ is linearly equivalent to $T_x^{*}D$, the theta group $\G(L)$ is the set of all pairs $(x,[f])$ where $x \in H(L)$
and $f$ is a nonzero rational function on $X$ such that

 $\div(f)=D-T_x^{*}D$.

  In addition, if $(y,[h]) \in \G(L)$ then
$$(x,[f]) (y, [h])=(x+y, [T_x^{*}h \cdot f]) \in \G(L).$$

\begin{rem}
\label{bigindex} It is known \cite[Sect. 1, Cor. of Th. 1]{Mumford66} that there exists a finite sequence of positive integers (elementary divisors) $\delta=(d_1, \dots , d_r)$ such that $d_{i+1}\mid d_i$ and the finite commutative group
$K(\delta)=\oplus_{i=1}^r \Z/d_i\Z$ enjoys the following properties:

\begin{itemize}
\item
 $H(L)$ is isomorphic to $\H_{K(\delta)}$;
 \item
  the groups $\G_{K(\delta)}$ and $\G(L)$ are
isomorphic.
\end{itemize}

 Applying the results of Section
\ref{group}, we conclude that $\G(L))$ contains a finite subgroup $G$ that enjoys the following property: every
commutative subgroup in  $G$ has index that is greater or equal than $\#(K(\delta))=\sqrt{\#(H(L))}$.
\end{rem}

\begin{proof}[Proof of Theorem \ref{main1}]
Comparing the multiplication formulas for $(x,[f])$'s and $A(y,f)$'s (Sect. \ref{Bir1}), we conclude that the embedding
$$\G(L) \hookrightarrow \Bir_1(X\times \AA^1), \ (y,[h])\mapsto A(y,h)$$
is actually a group homomorphism. So $\G(L)$
is isomorphic to a subgroup of $\Bir_1(X\times \AA^1)$. Applying this assertion to all ample divisors  $nD$ and invertible sheaves $L^n=\OC_X(nD)$ (where $n$ is a
positive integer) and combining it with Remarks \ref{bign}  and \ref{bigindex}, we conclude that for every positive
integer $n$ there exists a finite subgroup
$$G \subset \G(L^n)\hookrightarrow \Bir_1(X\times \AA^1)$$
that enjoys the following property: every commutative subgroup in $G$ has index that is greater or equal than
${(n^{2\dim(X)})}^{1/2}=n^{\dim(X)}$; in particular, this index is greater or equal than $n$. This proves that
$\Bir_1(X\times \AA^1)$ is {\sl not} a Jordan group.
\end{proof}


TITLE: Calculating the fixed points of a model
QUESTION [0 upvotes]: Does anyone know how I would go about calculating the fixed points of this model?

REPLY [2 votes]: A fixed point of a function $f$ means that $f(x) = x$. In this instance, you want $f(x_{n-1}, y_{n-1}) = (x_n, y_n)$, so if $(x,y)$ is a fixed point, this means that you want $f(x,y) = (x,y)$, i.e.
$x = x + 0.004x + 0.002xy$
$y = y + 0.005y + 0.010xy - 0.001y^2$
So all you have to do is to solve for $x$ and $y$, the most trivial solution being $x = 0$ and $y = 0$, but there are more than one solution.

To find all solutions, we solve the equations:
$0 = 0.004x + 0.002xy = x(0.004 + 0.002y)$
$0 = 0.005y + 0.010xy - 0.001y^2 = y(0.005 + 0.010x - 0.001y)$
For the first equation, there are two solutions, either $x = 0$ or $0.004 + 0.002y = 0$
For the second equation, there are two solutions, either $y = 0$ or $0.005 + 0.010x - 0.001y = 0$.
All you have to do now is to go through all the possibilities of what $x$ and $y$ can be. $x = 0$ and $y = 0$ is the one most easily spotted, try to find the others.


TITLE: "Orientation" of a surface boundary
QUESTION [1 upvotes]: Given is a surface patch embedded in $R^3$ with disk topology, i.e. it has a single boundary. I would like to know if it is possible to determine if the boundary is "interior" or "exterior" - an intuitive definition is given below, but can it be made mathematically rigorous?
For example, take a sphere that is cut in two by a plane (the two parts need not have equal size). The boundary of the larger part I would classify as "interior", since it could be seen as a hole in the surface (just think of an extreme case in which only a small cap is removed). The boundary of the smaller part would be "exterior". If the sphere is cut exactly along the equator, we get the indeterminate case.
Now, imagine you're walking on the surface along the boundary, so that the surface always lies to your right. Intuitively, the boundary should be classified as "interior" if you make more left turns than right turns on a complete loop, and vice versa.
Is there a way to rigorously define this notion?
EDIT:
I would actually like to have a computational criterion that distinguishes the cases. One idea, which I do not know how to formulate explicitly, would be to integrate along the boundary, summing up the deviations from the straight line in the tangent plane, e.g. a curve to the left counting positively, a curve to the right negatively. The sign of the integral over the entire contour would decide on the "orientation" (sorry I don't have a better term to describe this).

REPLY [2 votes]: One possibility would be to look at the curvature of the boundary curve, "as seen from the surface":
On a smooth curve in $\mathbb{R}^3$, every point has its associated "osculating circle", the circle which closest approximates the curve at that point among all circles in $\mathbb{R}^3$.  Associated to this is the curvature vector, which points towards the center of the circle with magnitude $(radius)^{-1}$.  (If the curve is locally straight, this osculating circle degenerates to the linear approximation, which is a "circle" with radius $\infty$ and hence curvature $0$.)  You can take the component of this curvature vector in the tangent plane (this is where you get the "as seen by the surface" from).  This will always be perpendicular to the boundary curve, so it'll be some scalar multiple of the inward-pointing normal curve.  The obvious thing to do, then, would be to take the integral of this inward-tangential-curvature function; if it's positive then you're on an exterior curve, and if it's negative then you're on an interior curve.  If you think about it, you'll see that this matches up with your intuitive definition on the sphere.

REPLY [1 votes]: One possibility that could at least distinguish the two cases you've given as examples, is to demand that an exterior boundary have a kind of convexity: any two points $x$ and $y$ on the boundary should be able to be connected by a minimizing geodesic on the patch which does not intersect the boundary except at $x$ and $y$.
If this is a more restrictive condition for an exterior boundary than you were hoping for, we could try to weaken it by adding a parameter $\epsilon$. Then we could consider a closed curve $\gamma$ which separates the patch $D$ into two pieces $D_1$, and $D_2$, and is such that the boundary is contained in $D_2$, and that the area of $D_2$ is less than $\epsilon$ times the area of $D_1$. We would say that the boundary is exterior if such a $\gamma$ could be found that has the above mentioned convexity property with respect to $D_1$. This idea could be adapted to handle boundaries with multiple components.


TITLE: Deduce the Hölder Inequality
QUESTION [0 upvotes]: If $f, g$ are two functions $\geq 0$ in an interval $I$ such that the
integrals $\int_I f(t) d t$ and $\int_I g(t) d t$ are convergent, the
integral $\int_I(f(t))^\alpha(g(t))^{1-\alpha} d t$ is convergent and
we have $$ \int_I(f(t))^\alpha(g(t))^{1-\alpha} d t \leq a \int_I f(t)
d t+b \int_I g(t) d t . $$ Deduce the Hölder inequality $$
 \int_I(f(t))^\alpha(g(t))^{1-\alpha} d t \leq\left(\int_I f(t) d
 t\right)^\alpha\left(\int_I g(t) d t\right)^{1-\alpha} $$

My attempt: By Young Inequality we know that for $0 \leq \alpha \leq 1$, we have
$$x^{\alpha} y^{1- \alpha} \leq ax + by,
$$where $a + b = 1$.
Thus if we set $x = f(t)$ and $y=g(t)$, then we get
$$ f(t)^{\alpha}g(t)^{1-\alpha} \leq af(t) + bg(t).
$$
If we integrate over $I$, we get the wanted inequality.
I don't see how to proceed to get the Hölder inequality now.

REPLY [0 votes]: This is somewhat using a sledgehammer to crack a nut, but you can use something called the "weighted AM-GM inequality" and the result drops out.
$$
{1\over 1-\alpha+\alpha}
\left(
\alpha\left(\int f\right)^\alpha+
(1-\alpha)
\left(\int g\right)^{1-\alpha}\right)\geq
\sqrt[1-\alpha+\alpha]{
\left(\int f\right)^\alpha\left(\int g\right)^{1-\alpha}
}
$$
The $1-\alpha+\alpha$ in the above is just $1$, and both the division and the root are redundant, so
$$
\alpha
\left(
\int f\right)^\alpha+
(1-\alpha)
\left(\int g\right)^{1-\alpha}\geq
\left(\int f\right)^\alpha\left(\int g\right)^{1-\alpha}
$$


TITLE: Why $f$ is invariant for $G_f$.
QUESTION [1 upvotes]: Let $F$ a field and $f\in F[X]$ a separable polynomial. Let $K_f$ the splitting field of $f$ and $G_f=Gal(K_f/F)$ its Galois group. We suppose it act transitively on the roots of $f$.$\alpha _1,...,\alpha _n$ the roots of $f$. Then $$f=\prod_{i=1}^n(X-\alpha _i)\in K_f[X].$$
In my solution, it's written that $f$ is invariant by $G_f$ (i.e. $\sigma (f)=f$ for all $\sigma \in G_f$), But to me $$\sigma (f(X))=\sigma \left(\prod_{i=1}^n (X-\alpha_i) \right)=\prod_{i=1}^n(\sigma (X)-\alpha _i)=f(\sigma (X))$$
since $G_f$ act transitively on the roots of $f$. So finally, we don't have $\sigma (f(X))=f(X)$, so what's wrong in my argument ? May be my teacher is wrong and $f$ is not a fix point of $\sigma $, what do you think ?

REPLY [3 votes]: Note that the standard way for the automorphisms $\sigma$ of the field $K_f$ being extended to the polynomial ring $K_f[X]$ is precisely by imposing that $\sigma(X)= X$. It is then usually also denoted by $\sigma$. Thus $\sigma(X)= X$ by definition.


TITLE: Inversion of n x n matrix
QUESTION [1 upvotes]: A matrix F is given:
$$
F = [e^{i\frac{2\pi kl}{n}}]_{k,l=0}^{n-1}
$$
Find $$ F^{-1} $$
I know Gaussian method for inverting matrices but I suppose it doesn't apply to  matrices with not given exact n value. Could you tell me what are the methods for inverting matrices like this?

REPLY [1 votes]: Hint: It's a Vandermonde matrix (link1 and link2).


TITLE: 4-coloring of 4-cycle
QUESTION [1 upvotes]: I am working on a problem related to counting the number of different colorings on the vertices of a square assuming that the colorings resulting from rotations and reflections are considered the same.
The problem is as follows:
How many ways are there to 4-color the corners of a square with rotations and reﬂections allowed if adjacent corners must have diﬀerent colors.
If we label the corners as $\{1,2,3,4\}$, then under rotation and reflection, we have the following cycles:
$(1)(2)(3)(4)$, $(1234)$, $(13)(24)$, $(1432)$, $(14)(23)$, $(12)(43)$, $(1)(24)(3)$, $(13)(2)(4)$.
I tried to use the Cauchy-Frobenius-Burnside (CFB) Theorem to count the different possibilities but could not get anything. Any help will be appreciated.

REPLY [3 votes]: Let's do the Burnside. There are $8$ symmetries (I like to think of them geometrically rather than in abstract cycle notation). The number of squares which are invariant for each of these symmetries are:

The trivial symmetry: This is just all possible colourings, without taking symmetries into account. Corner 1 can have $4$ different colours, and corner 2 can then have $3$ different colours. If the corner diagonally opposite corner 2 has the same colour as corner $2$, then the final corner has $3$ options. Otherwise it has $2$. That's a total of $4\cdot 3(3 + 2\cdot 2)= 84$
The two $90^\circ$ rotations: Adjacent corners must have different colours, so no invariants here
The $180^\circ$ rotation: Corner 1 can be any of the $4$ colours, corner $2$ can have any of the remaining $3$, and the two remaining corners are then fixed. $12$ invariants
The two mirrorings with axis a diagonal: The two corners not on the axis must have the same colour, and can have any of $4$ colours, and the two fixed corners can choose freely and independently from the $3$ remaining colours, giving $4\cdot 3^2 = 36$ invariants
The two mirrorings with axis perpendicular to a side: Adjacent corners are sent to one another, so no invariants here

Burnside's lemma then tells us that the number of distinct squares, taking the symmetries into account, is
$$
\frac{84 + 0 + 0 + 12 + 36 + 36 + 0 + 0}8 = 21
$$


TITLE: Prove that the solutions to an ODE are non-periodic oscillating.
QUESTION [1 upvotes]: I have the ODE $$\ddot{x}+ \dot{x}+k^2x=0$$
where the dot means derivative with respect to $t$ ($x$ is a function of $t$) and $k$ is a constant such that $k>\frac{1}{2}$
prove that the solutions to the ODE are non-periodic and oscillating.
What I tried is the following.
I solved the characteristic polynomial equation $$\lambda^2+\lambda+k^2=0$$
which gave me $$ \lambda_{1,2} = \frac{-1 \pm 2 \sqrt{k^2-\frac{1}{4}}}{2} $$
Now since $k>\frac{1}{2}$
the solutions are real so the ODE has no periodic solutions, but I can't understand how to prove that the solutions oscillate, what does it even mean for the solutions to oscillate?

REPLY [3 votes]: Given the equation
$\ddot x + \dot x + k^2x = 0, \tag 1$
the characteristic equation is indeed
$\lambda^2 + \lambda + k^2 = 0, \tag 2$
but the "quadratic formula" used to find $\lambda_{1,2}$ is misstated; it should read
$\lambda_{1,2} = \dfrac{-1 \pm \sqrt{1 - 4k^2}}{2}; \tag 3$
from this we see that when $k > 1/2$ the values of $\lambda_{1,2}$ are complex conjugate numbers with real part $-1/2$, hence the solutions exponentially decay to $0$; however the presence of a non-vanishing imaginary component in the $\lambda_{1,2}$ forces the solutions to venture
back and forth across $0$ whilst they decay; thus the solutions are non-periodic and oscillating.
We can in fact write down explicit solutions to (1); they serve to reify and illustrate the words of the preceding paragraph.  Writing (3) in the form
$\lambda_{1, 2} = -\dfrac{1}{2} \pm i \dfrac{1}{2}\sqrt{4k^2 - 1}, \tag 4$
we may express two relatively simple solution as
$x_\pm(t) = \exp \left (\lambda_{1,2} t\right ) = \exp \left (\left (-\dfrac{1}{2} \pm i \dfrac{1}{2}\sqrt{4k^2 - 1} \right)  t\right )$ =
$\exp \left (-\dfrac{1}{2}  t\right )\exp \left (\pm i \dfrac{1}{2}\sqrt{4k^2 - 1} t\right )$
$= \exp \left (-\dfrac{1}{2}  t\right )\left (\cos \left (\dfrac{1}{2}\sqrt{4k^2 - 1} t \right ) \pm i \sin \left (\dfrac{1}{2}\sqrt{4k^2 - 1} t \right )\right). \tag 5$
Since (1) is a real differential equation, the real and imaginary parts of $x_\pm(t)$ are individually solutions, thus for example
$x(t) =  \exp \left (-\dfrac{1}{2}  t\right )\left (\cos \left (\dfrac{1}{2}\sqrt{4k^2 - 1} t \right ) \right) \tag 6$
is a solution which exhibits both exponential decay and oscillation.


TITLE: Kinetic energy always time independent?! Where is my mistake?
QUESTION [2 upvotes]: I have some problems understanding the Lagrangian and the Hamiltonian formalism. Those can be condensed in the following "derivation" of $\frac{\partial T}{\partial t} = 0$ from the equation $\frac{\partial H}{\partial t} = - \frac{\partial L}{\partial t}$. Since the kinetic energy might be time dependent (for example when our frame of reference accelerates), it seems that I missed something really important.
Question:  . Where is my mistake or where did I missunderstood the Lagrangian / Hamiltonian formalism?

Derivation: One of the Hamiltonian equations is $\frac{\partial H}{\partial t} = - \frac{\partial L}{\partial t}$ (see section "Deriving Hamilton's equations" of the Wikipedia article "Hamiltonian mechanics"). With $H=T+V$ and $L=T-V$ we get ($T$ stands for kinetic energy and $V$ for potential energy):
$$\begin{array}{rrl}
& \frac{\partial H}{\partial t} & = - \frac{\partial L}{\partial t} \\
\iff & \frac{\partial (T+V)}{\partial t} & = - \frac{\partial (T-V)}{\partial t} \\
\iff & \frac{\partial T}{\partial t} + \frac{\partial V}{\partial t} & = - \frac{\partial T}{\partial t} + \frac{\partial V}{\partial t} \\
\iff & 2\frac{\partial T}{\partial t} &  = 0 \\
\iff & \frac{\partial T}{\partial t} &  = 0
\end{array}$$

I want to elaborate the given answer by Qmechanic, since it took me some time to understand it.
Extended Answer: We have to keep in mind, that $H = H(p,q,t)$ and $L(\dot q, q, t)$ have different signatures. While the Hamiltonian depends on the generalized impuls $p$, the Lagrangian depends on the velocity $\dot q$.
Therefore $\frac{\partial H}{\partial t}$ and $\frac{\partial L}{\partial t}$ are different partial derivations. In $\frac{\partial H}{\partial t}$ the variables $p$ and $q$ are held constant while in $\frac{\partial L}{\partial t}$ the variables $\dot q$ and $q$ are held constant. When we notate the first derivation with $\partial_t^{p,q}$ and the second with $\partial_t^{\dot q,q}$ we see were I made a mistake in the above derivation:
$$\begin{array}{rrl}
& \partial_t^{p,q} H & = - \partial_t^{\dot q,q} L \\
\iff & \partial_t^{p,q} (T+V) & = - \partial_t^{\dot q,q} (T-V) \\
\iff & \partial_t^{p,q} T + \partial_t^{p,q} V & = - \partial_t^{\dot q,q}  T + \partial_t^{\dot q,q} V \\
\iff & \partial_t^{p,q} T + \partial_t^{\dot q,q} T & = \partial_t^{\dot q,q} V - \partial_t^{p,q} V\\
\end{array}$$
Since $V$ doesn't depend on $p$ nor $\dot q$ we have $\partial_t^{\dot q,q} V - \partial_t^{p,q} V = 0$:
$$\partial_t^{p,q} T + \partial_t^{\dot q,q} T = 0$$
However, $\partial_t^{p,q} T$ do not have to be the same as $\partial_t^{\dot q,q} T$. This is not the case, when the impuls-velocity-connection is time dependent, i.e. $p = p(\dot q, t)$. An example is a launching rocket whose mass decreases with time. Here we have $T=\frac 12 m(t)\dot q^2$ and thus $p=m(t)\dot q$ (see example in the answer by Qmechanic).
Both partial derivations are only the same, when the following property is fulfilled:
$$p \text{ is constant over time} \iff \dot q \text{ is constant over time}$$

REPLY [3 votes]: In a nutshell, even if we assume the non-generic relations $$L(q,v,t)~=~T(v,t)~-~V(q,t)\quad\text{and}\quad H(q,p,t)~=~T(p,t)~+~V(q,t),$$ then OP's mistake is to be cavalier about functional dependence of $T$, and in particular, its explicit time dependence.
Perhaps a simple example is in order, cf. above comment by jacob1729:
$$\begin{align} L(q,v,t)~=~\frac{m(t)}{2}v^2 \quad &\Rightarrow\quad \frac{\partial L(q,v,t)}{\partial t} ~=~ \color{red}{+}\frac{m^{\prime}(t)}{m(t)}L(q,v,t)\cr\cr
\updownarrow\text{identify}\qquad & \qquad\qquad\text{sum up to zero }\updownarrow\cr\cr
 H(q,p,t)~=~\frac{p^2}{2m(t)}\quad &\Rightarrow\quad \frac{\partial H(q,p,t)}{\partial t} ~=~ \color{red}{-}\frac{m^{\prime}(t)}{m(t)}H(q,p,t). \end{align}$$


TITLE: Question from Nov 2006 ( cambridge A-level)
QUESTION [0 upvotes]: I've got stuck with this question (Cambridge A-level question Nov 2006 paper 3)
(first part) Simplify $$ \left( \sqrt {1+x} + \sqrt {1-x}\right)  \left( \sqrt {1+x} - \sqrt {1-x} \right) $$
(second part) showing your working, and deduce that
$$ \frac{1}{\sqrt {1+x} +\sqrt {1-x}} = \frac {\sqrt {1+x}- \sqrt {1-x}}{2x} $$
Actually I succeeded in simplifying the first part and got $(1+x)-(1-x)$ but I'm stuck with the second part. We should use the result of the first part to deduce for the second part.

REPLY [0 votes]: You can think of this like any question where you have surds, and you need to rationalise the denominator of a fraction. If you had:
$$\frac{1}{a-\sqrt{b}}$$
You would simply do:
$$\frac{1}{a-\sqrt{b}} \times \frac{a+\sqrt{b}}{a+\sqrt{b}}$$
This would simply rationalise the denominator, and you have to do a similar thing in this particular case. Firstly, you say that you got to:
$$(1+x)−(1−x)$$
Simplify that to $$2x$$
Then, start to do the second part, and rationalise its denominator:
$$\frac{1}{\sqrt{1+x}+\sqrt{1-x}} \times \frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}$$
The denominator has already been worked out (always look back at previous parts of the question to see what you have already done). The denominator is $$2x$$ and the numerator is just $$\sqrt{1+x}-\sqrt{1-x}$$
That is your answer.


text:
\section{Summation over k of Ceiling of k over 2}
Tags: Ceiling Function, Summations

\begin{theorem}
:$\displaystyle \sum_{k \mathop = 1}^n \left \lceil{\dfrac k 2}\right \rceil = \left \lceil{\dfrac {n \left({n + 2}\right)} 4}\right \rceil$
\end{theorem}

\begin{proof}
By [[Permutation of Indices of Summation]]:
:$\displaystyle \sum_{k \mathop = 1}^n \left \lceil{\dfrac k 2}\right \rceil = \sum_{k \mathop = 1}^n \left \lceil{\dfrac {n + 1 - k} 2}\right \rceil$
and so:
:$\displaystyle \sum_{k \mathop = 1}^n \left \lceil{\dfrac k 2}\right \rceil = \dfrac 1 2 \sum_{k \mathop = 1}^n \left({\left \lceil{\dfrac k 2}\right \rceil + \left \lceil{\dfrac {n + 1 - k} 2}\right \rceil}\right)$
First take the case where $n$ is [[Definition:Even Integer|even]].
For $k$ [[Definition:Odd Integer|odd]]:
:$\left \lceil{\dfrac k 2}\right \rceil = \dfrac k 2 + \dfrac 1 2$
and:
:$\left \lceil{\dfrac {n + 1 - k} 2}\right \rceil = \dfrac {n + 1 - k} 2$
Hence:
{{begin-eqn}}
{{eqn | l = \left \lceil{\dfrac k 2}\right \rceil + \left \lceil{\dfrac {n + 1 - k} 2}\right \rceil
      | r = \dfrac k 2 + \dfrac 1 2 + \dfrac {n + 1 - k} 2
      | c =
}}
{{eqn | r = \dfrac {k + 1 + n + 1 - k} 2
      | c =
}}
{{eqn | r = \dfrac {n + 2} 2
      | c =
}}
{{end-eqn}}
For $k$ [[Definition:Even Integer|even]]:
:$\left \lceil{\dfrac k 2}\right \rceil = \dfrac k 2$
and:
:$\left \lceil{\dfrac {n + 1 - k} 2}\right \rceil = \dfrac {n + 1 - k} 2 + \dfrac 1 2 = \dfrac {n - k + 2} 2$
Hence:
{{begin-eqn}}
{{eqn | l = \left \lceil{\dfrac k 2}\right \rceil + \left \lceil{\dfrac {n + 1 - k} 2}\right \rceil
      | r = \dfrac k 2 + \dfrac {n - k + 2} 2
      | c =
}}
{{eqn | r = \dfrac {k + n - k + 2} 2
      | c =
}}
{{eqn | r = \dfrac {n + 2} 2
      | c =
}}
{{end-eqn}}
So:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n \left \lceil{\dfrac k 2}\right \rceil
      | r = \dfrac 1 2 \sum_{k \mathop = 1}^n \left({\left \lceil{\dfrac k 2}\right \rceil + \left \lceil{\dfrac {n + 1 - k} 2}\right \rceil}\right)
      | c =
}}
{{eqn | r = \dfrac 1 2 \sum_{k \mathop = 1}^n \left({\dfrac {n + 2} 2}\right)
      | c =
}}
{{eqn | r = \dfrac 1 2 n \left({\dfrac {n + 2} 2}\right)
      | c =
}}
{{eqn | r = \dfrac {n \left({n + 2}\right)} 4
      | c =
}}
{{eqn | r = \left \lceil{\dfrac {n \left({n + 2}\right)} 4}\right \rceil
      | c = as $\dfrac {n \left({n + 2}\right)} 4$ is an [[Definition:Integer|integer]]
}}
{{end-eqn}}
{{qed|lemma}}
Next take the case where $n$ is [[Definition:Odd Integer|odd]].
For $k$ [[Definition:Odd Integer|odd]]:
:$\left \lceil{\dfrac k 2}\right \rceil = \dfrac k 2 + \dfrac 1 2$
and:
:$\left \lceil{\dfrac {n + 1 - k} 2}\right \rceil = \dfrac {n + 1 - k} 2 + \dfrac 1 2$
Hence:
{{begin-eqn}}
{{eqn | l = \left \lceil{\dfrac k 2}\right \rceil + \left \lceil{\dfrac {n + 1 - k} 2}\right \rceil
      | r = \dfrac k 2 + \dfrac 1 2 + \dfrac {n + 1 - k} 2 + \dfrac 1 2
      | c =
}}
{{eqn | r = \dfrac {k + 1 + n + 1 - k + 1} 2
      | c =
}}
{{eqn | r = \dfrac {n + 3} 2
      | c =
}}
{{end-eqn}}
For $k$ [[Definition:Even Integer|even]]:
:$\left \lceil{\dfrac k 2}\right \rceil = \dfrac k 2$
and:
:$\left \lceil{\dfrac {n + 1 - k} 2}\right \rceil = \dfrac {n + 1 - k} 2$
Hence:
{{begin-eqn}}
{{eqn | l = \left \lceil{\dfrac k 2}\right \rceil + \left \lceil{\dfrac {n + 1 - k} 2}\right \rceil
      | r = \dfrac k 2 + \dfrac {n - k + 1} 2
      | c =
}}
{{eqn | r = \dfrac {k + n - k + 1} 2
      | c =
}}
{{eqn | r = \dfrac {n + 1} 2
      | c =
}}
{{end-eqn}}
Let $n = 2 t + 1$.
Then:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n \left \lceil{\dfrac k 2}\right \rceil
      | r = \dfrac 1 2 \sum_{k \mathop = 1}^n \left({\left \lceil{\dfrac k 2}\right \rceil + \left \lceil{\dfrac {n + 1 - k} 2}\right \rceil}\right)
      | c =
}}
{{eqn | r = \dfrac 1 2 \sum_{k \mathop = 1}^{2 t + 1} \left({\left \lceil{\dfrac k 2}\right \rceil + \left \lceil{\dfrac {2 t + 2 - k} 2}\right \rceil}\right)
      | c =
}}
{{eqn | r = \dfrac t 2 \dfrac {\left({2 t + 1}\right) + 1} 2 + \dfrac {t + 1} 2 \dfrac {\left({2 t + 1}\right) + 3} 2
      | c = there are $t$ [[Definition:Even Integer|even]] terms and $t + 1$ [[Definition:Odd Integer|odd]] terms
}}
{{eqn | r = \dfrac {2 t^2 + 2 t} 4 + \dfrac {2 t^2 + 6 t + 4} 4
      | c = multiplying out
}}
{{eqn | r = \dfrac {4 t^2 + 8 t + 3} 4 + \dfrac 1 4
      | c =
}}
{{eqn | r = \dfrac {\left({2 t + 1}\right) \left({2 t + 3}\right)} + \dfrac 1 4
      | c =
}}
{{eqn | r = \dfrac {n \left({n + 2}\right)} 4 + \dfrac 1 4
      | c =
}}
{{eqn | r = \left \lceil{\dfrac {n \left({n + 2}\right)} 4}\right \rceil
      | c =
}}
{{end-eqn}}
{{qed}}
\end{proof}

config: wiki


TITLE: $34!=295232799cd96041408476186096435ab000000$, find $a$ ,$b$, $c$, and $d$
QUESTION [4 upvotes]: There was a number theory question that I have to do for homework.

$34!=295232799cd96041408476186096435ab000000$, find $a$ ,$b$, $c$, and $d$

I know $b=0$ because $10^7\big|34!$ only. But how can I find other variables?
Remark:
No electronic calculation device can be used for calculation. (I need to write the full solution.)
Can someone help me?

REPLY [3 votes]: Knowing $b=0$ and $\frac{34!}{10^7}$ being a multiple of $8$, the last three non-zero digits $35a$ must be divisible by $8$, so $a=2$ for sure. Now set up linear equations modulo $9$ and $11$ (this is just a digit sum and alternating digit sum, and I learned this from studying the Trachtenberg system a long time ago) and we get
$$d+c\equiv3\bmod9$$
$$d-c\equiv3\bmod11$$
since $9,11\mid34!$ By trial and error we see that $d=3$ and $c=0$, so the solution is $\overline{abcd}=2003$.

REPLY [1 votes]: Sum of digits is divisible by $9$, which gives
$$a+c+d\equiv5(\mod9).$$
$$2-9+5-2+3-2+7-9+9-c+d-9+6-0+4-1+4-0+8-4+7-6+1-8+6-0+9-6+4-3+5-a\equiv0(\mod11),$$
which gives
$$a+c-d\equiv-1(\mod11).$$
From here we can get $a+c=2$ and $d=3$.
Now, since $a\in\{2,6\}$, we obtain $a=2$, $c=0$ and we are done!