text:
\begin{document}

\title{A Converse to a Theorem on Normal Forms of Volume Forms with Respect to a Hypersurface}
\author{Konstantinos Kourliouros}
\address{Imperial College London, Department of Mathematics,  Huxley Building 180 Queen's Gate,\\
South Kensington Campus, London SW7, United Kingdom}
\ead{k.kourliouros10@imperial.ac.uk}

\begin{abstract}
In this note we give a positive answer to a question asked by Y. Colin de Verdi\`ere concerning the converse of the following theorem, due to A. N. Varchenko: two germs of volume forms are equivalent with respect to diffeomorphisms preserving a germ of an isolated hypersurface singularity, if their difference is the differential of a form whose restriction on the smooth part of the hypersurface is exact.

\vspace{0.3cm}

\end{abstract}

\begin{keyword}
Isolated Singularities \sep De Rham Cohomology \sep Volume Forms \sep Normal Forms
\end{keyword}

\maketitle

\section{Introduction-Main Results}

In this paper we will give a positive answer to a question asked by Y. Colin de Verdi\`ere in \cite{C1} which was formulated as follows: suppose that two germs of symplectic forms at the origin of the plane are equivalent with respect to a diffeomorphism preserving a plane curve germ with an isolated singularity at the origin. Is it true that their difference is the differential of a 1-form whose restriction on the smooth part of the curve is exact? This question asks for the validity of the converse to a general normal form theorem in Lagrangian singularity theory  according to which: two germs of symplectic structures are equivalent with respect to diffeomorphisms preserving a Lagrangian variety if their difference is the differential of a 1-form whose restriction on the smooth part of the variety is exact. The proof of this theorem can be easily deduced from the reasoning in A. B. Givental's paper \cite{Gi} using Moser's homotopy method. It holds in any dimension and for arbitrary Lagrangian singularities. It's converse though is not so easy to deduce; as it turns out, the main difficulty comes from the fact that the singularities of Lagrangian varieties in dimension higher than two are non-isolated (c.f. \cite{Gi}, \cite{Sev}) and their cohomology can be rather complicated. On the other hand, for the 2-dimensional case (where the Lagrangian singularities are indeed isolated) the normal form theorem stated above can be viewed as a special case of a general theorem obtained by A. N. Varchenko in \cite{Var} concerning the normal forms of germs of (powers of) volume forms with respect to an isolated hypersurface singularity. Here we will prove a converse to Varchenko's normal form theorem, which trivially answers Verdi\`ere's question, and it can be formulated as follows:
\begin{thm}
\label{t}
Suppose that two germs of volume forms are equivalent with respect to a diffeomorphism preserving a germ of an isolated hypersurface singularity. Then their difference is the differential of a form whose restriction on the smooth part of the hypersurface is exact.
\end{thm}

The method of proof is as follows: we first prove the theorem in the formal category. For this we use a formal interpolation lemma for the elements of the isotropy group of an isolated hypersurface singularity (Lemma \ref{l1}) which is a variant of the one presented by J. -P. Fran\c{c}oise in \cite{F} and relies in a general interpolation method obtained by S. Sternberg \cite{Ste}. Then we pass to the analytic category using a comparison theorem between the corresponding de Rham cohomologies in the formal and analytic categories (Lemma \ref{l2}). This is analogous to the well known Bloom-Brieskorn theorem \cite{B} for the de Rham cohomology of an analytic space with isolated singularities. But in contrast to the ordinary Bloom-Brieskorn theorem where the cohomology of the complex of K\"ahler differentials is considered, we need to consider instead the cohomology of the so called Givental complex, i.e. the complex of germs of holomorphic forms modulo those that vanish on the smooth part of the hypersurface (which naturally appears in the statements of the theorems above).

\section{De Rham Cohomology of an Isolated Hypersurface Singularity and an Analog of the Bloom-Brieskorn Theorem}

Let $f:(\mathbb{C}^{n+1},0)\rightarrow (\mathbb{C},0)$ be a germ of a holomorphic function with an isolated singularity at the origin and let $(X,0)=\{f=0\}$ be the corresponding hypersurface germ, zero level set of $f$ (we will suppose throughout that the germ $(X,0)$ is reduced). To the germ $(X,0)$ we may associate several complexes of holomorphic forms, quotients of the complex $\Omega^{\bullet}$ of germs of holomorphic forms at the origin of $\mathbb{C}^{n+1}$, the ``largest'' one being the so called complex of K\"ahler differentials:
\[\Omega^{\bullet}_{X,0}=\frac{\Omega^{\bullet}}{df\wedge \Omega^{\bullet-1}+f\Omega^{\bullet}},\]
where the differential is induced by the differential in $\Omega^{\bullet}$ after passing to quotients. The cohomologies of this complex are finite dimensional vector spaces and they have being computed by E. Brieskorn in \cite{B}. In particular, along with the results of M. Sebastiani \cite{S} it follows that:
\begin{equation}
\label{ck}
H^p(\Omega^{\bullet}_{X,0})=\left\{\begin{array}{cl}
							 \mathbb{C}, & p=0,\\
							 0, & 0<p<n, p>n \\
							 \mathbb{C}^{d}, & p=n,\\
							 \end{array} \right..
\end{equation}
The number $d$ can be interpreted as the degree of non-quasihomogeneity of the germ $f$, i.e.
\[d=\mu-\tau,\]
where $\mu$ is the Milnor number and $\tau$ is the Tjurina number of the singularity $f$:
\[\mu=\dim_{\mathbb{C}}\frac{\Omega^{n+1}}{df\wedge \Omega^n}, \hspace{0.3cm} \tau=\dim_{\mathbb{C}}\frac{\Omega^{n+1}}{df\wedge \Omega^n+f\Omega^{n+1}},\]
\[d=\dim_{\mathbb{C}}\frac{df\wedge \Omega^n+f\Omega^{n+1}}{df\wedge \Omega^n}.\]
Indeed, it is a result of K. Saito \cite{Sa} according to which $f$ is equivalent to a quasihomogeneous germ if and only if it belongs to its gradient ideal, i.e. $f\Omega^{n+1}\subset df\wedge \Omega^n$.

Denote now by $X^*=X\setminus 0$ the smooth part of the hypersurface $X$. In \cite{Fe}, A. Ferrari introduced another important complex associated to $X$ which is the quotient complex of $\Omega^{\bullet}$ modulo the subcomplex $\Omega^{\bullet}(X^*)$ which consists of forms whose restriction on the smooth part $X^*$ of $X$ is identically zero:
\[\tilde{\Omega}^{\bullet}_{X,0}=\frac{\Omega^{\bullet}}{\Omega^{\bullet}(X^*)}.\]
This complex was also used extensively by A. B. Givental in \cite{Gi} and is called the Givental complex in \cite{He}. We adopt the same notation here as well.  As it is easy to see there is an identification of the complex of K\"ahler differentials with the Givental complex on the smooth part $X^*$ and thus there is a short exact sequence of complexes:
\begin{equation}
\label{ses0}
0\rightarrow T^{\bullet}_{X,0}\rightarrow \Omega^{\bullet}_{X,0}\rightarrow \tilde{\Omega}^{\bullet}_{X,0}\rightarrow 0,
\end{equation}
where $T^{\bullet}_{X,0}$ is the torsion subcomplex of $\Omega^{\bullet}_{X,0}$ (here is where we need $(X,0)$ to be reduced). Indeed any torsion element vanishes on the smooth part $X^*$ and thus the complex $T^{\bullet}_{X,0}$ is contained in the kernel of the natural projection $\Omega^{\bullet}_{X,0}\rightarrow \tilde{\Omega}^{\bullet}_{X,0}$.

In \cite{G}, G. M. Greuel studied the relationship of the Givental and K\"ahler complexes in the general case where $(X,0)$ defines an $n$-dimensional isolated complete intersection singularity (embedded in some $\mathbb{C}^{m}$). He proves that:
\[T^{p}_{X,0}=0, \hspace{0.3cm}  p< n, \]
\[T^p_{X,0}=\Omega^{p}_{X,0}, \hspace{0.3cm} p>n,\]
and also:
\[H^p(\Omega^{\bullet}_{X,0})=0,\hspace{0.3cm} 0<p<n,\]
\[H^p(\tilde{\Omega}^{\bullet}_{X,0})=0, \hspace{0.3cm} p\neq 0,n.\]
Thus, in the particular case where $(X,0)$ is an isolated hypersurface singularity we obtain the following analog of the Brieskorn-Sebastiani result (\ref{ck}) for the cohomology of the Givental complex:
\begin{prop}
\label{p1}
\[H^p(\tilde{\Omega}^{\bullet}_{X,0})=\left\{\begin{array}{cl}
							 \mathbb{C}, & p=0,\\
							 0, & 0<p<n, p>n \\
							 \mathbb{C}^{d}, & p=n,\\
							 \end{array} \right.,\]
where $d=\mu-\tau$ is the degree of non-quasihomogeneity of the germ $f$.
\end{prop}
\begin{proof}
It suffices only to show the following equality (the zero cohomology is trivial):
\[H^n(\tilde{\Omega}^{\bullet}_{X,0})=\mathbb{C}^d.\]
This in turn has been proved by A. N. Varchenko in \cite{Var}. Here we will give an alternative, simple proof, which is distilled from \cite{C1}.  To the germ $f$ we associate the Brieskorn module as in \cite{B}:
\[H_f''=\frac{\Omega^{n+1}}{df\wedge d\Omega^{n-1}}.\]
According to the Sebastiani theorem \cite{S} this is a free module of rank $\mu$ over $\mathbb{C}\{f\}$ and thus the quotient
\[\frac{H''_f}{fH''_f}=\frac{\Omega^{n+1}}{df\wedge d\Omega^{n-1}+f\Omega^{n+1}}\]
is a $\mu$-dimensional $\mathbb{C}$-vector space. Denote now by
\[\mathcal{Q}_{X,0}=\frac{\Omega^{n+1}}{df\wedge \Omega^n+f\Omega^{n+1}}\]
the space of deformations of the germ $(X,0)$. By the fact that $df\wedge d\Omega^{n-1}+f\Omega^{n+1}\subseteq df\wedge \Omega^n+f\Omega^{n+1}$ there is a natural projection:
\[\frac{H''_f}{fH''_f}\stackrel{\pi}{\rightarrow}\mathcal{Q}_{X,0},\]
whose kernel:
\[\ker{\pi}=\frac{df\wedge \Omega^n}{df\wedge d\Omega^{n-1}+f\Omega^{n+1}}\]
is a priori a $d=\mu-\tau$-dimensional vector space. Now, the $n$-th cohomology of the Givental complex is:
\[H^n(\tilde{\Omega}^{\bullet}_{X,0})=\frac{\tilde{\Omega}^n_{X,0}}{d\tilde{\Omega}^{n-1}_{X,0}}=\frac{\Omega^n}{\Omega^n(X^*)+d\Omega^{n-1}},\]
where:
\[\Omega^n(X^*)=\{\alpha \in \Omega^n/df\wedge \alpha \in f\Omega^{n+1}\}.\]
It follows from this that
\[\ker{\pi}=df\wedge H^n(\tilde{\Omega}^{\bullet}_{X,0})\]
and thus there is a short exact sequence:
\begin{equation}
\label{ses1}
0\rightarrow H^n(\tilde{\Omega}^{\bullet}_{X,0})\stackrel{df\wedge}{\rightarrow}\frac{H''_f}{fH''_f}\stackrel{\pi}{\rightarrow}\mathcal{Q}_{X,0}\rightarrow 0.
\end{equation}
This proves that indeed $H^n(\tilde{\Omega}^{\bullet}_{X,0})=\mathbb{C}^d$ as was asserted.
\end{proof}

It follows from the proposition above along with (\ref{ck}) that there is an isomorphism of vector spaces:
\[H^{\bullet}(\Omega^{\bullet}_{X,0})\cong H^{\bullet}(\tilde{\Omega}^{\bullet}_{X,0}).\]
Thus we may formulate  the following version of the Poincar\'e lemma for the germ $(X,0)$:
\begin{cor}[c.f. \cite{Gi} for $n=1$]
\label{c1}
The germ $(X,0)$ is quasihomogeneous if and only if its Givental (or K\"ahler) complex is acyclic (except in zero degree).
\end{cor}

Finally, we will need  the following analog of the Bloom-Brieskorn theorem \cite{B}, which is a comparison of the cohomologies of the analytic and formal Givental complexes. The proof we will give below is in fact a simple variant of the one presented in \cite{B}. Moreover, the fact that $(X,0)$ is an isolated hypersurface singularity plays no significant role; the same proof holds for any analytic space, as long as its singularities are isolated.
\begin{lem}
\label{l2}
Let $\hat{\tilde{\Omega}}^{\bullet}_{X,0}$ be the formal completion of the Givental complex. Then the natural inclusion $\tilde{\Omega}^{\bullet}_{X,0}\hookrightarrow \hat{\tilde{\Omega}}^{\bullet}_{X,0}$ induces an isomorphism of finite dimensional vector spaces:
\[H^{\bullet}(\tilde{\Omega}^{\bullet}_{X,0})\cong H^{\bullet}(\hat{\tilde{\Omega}}^{\bullet}_{X,0}).\]
\end{lem}
\begin{proof}
Following \cite{B} let $\pi:Y\rightarrow X$ be a resolution of singularities in the sense of Hironaka and denote by $A=\pi^{-1}(0)$ the exceptional set, which we may suppose it is given by some equations $y_1\cdots y_r=0$. Let $\Omega^{\bullet}_{Y}$ be the complex of holomorphic forms on $Y$ and let $\Omega^{\bullet}_Y|_{A}$ be its restriction on $A$. Let also
\[\hat{\Omega}^{\bullet}_{Y}=\lim_{\underset{k}{\leftarrow}} \frac{\Omega^{\bullet}_Y}{\frak{m}^k\Omega^{\bullet}_Y}.\]
Consider now the direct image sheaf $R^0\pi_*\Omega^{\bullet}_{Y}$ (this is also called the Noether complex). Since the map $\pi$ is proper this is a coherent sheaf (by Grauert's coherence theorem), which away from the singular point $0$ it can be identified with the Givental complex: $R^0\pi_*\Omega^{\bullet}_Y|_{X^*}\cong \tilde{\Omega}^{\bullet}_{X^*}$. In particular there is an inclusion $j:\tilde{\Omega}^{\bullet}_{X}\rightarrow R^0\pi_*\Omega^{\bullet}_Y$ whose cokernel is concentrated at the singular point $0$ and it is thus finite dimensional. Consider now the formal completion of the above complexes. It gives a commutative diagram:
\begin{equation}
\begin{CD}
\tilde{\Omega}^{\bullet}_{X,0} @>j>>H^0(A, \Omega^{\bullet}_Y|_{A}) \\
 @VVV          @VVV         \\
\hat{\tilde{\Omega}}^{\bullet}_{X,0} @>\hat{j}>>H^0(A, \hat{\Omega}^{\bullet}_Y)  \\
\end{CD}
\end{equation}
where of course $H^0(A, \Omega^{\bullet}_Y|_{A}) \cong (R^0\pi_*\Omega^{\bullet}_Y)|_0$ and $\hat{j}$ is the formal completion of the inclusion $j$. Indeed, this follows from the fact (c.f. \cite{B} and the corresponding references therein):
\[H^0(A,\hat{\Omega}^{\bullet}_Y)\cong \lim_{\underset{k}{\leftarrow}}H^0(A, \frac{\Omega^{\bullet}_Y}{\frak{m}^k\Omega^{\bullet}_Y})\cong \lim_{\underset{k}{\leftarrow}}\frac{H^0(A, \Omega^{\bullet}_Y|_{A})}{\frak{m}^kH^0(A, \Omega^{\bullet}_Y|_{A})}.\]
Now, since the completion functor is exact and by the fact that the cokernel of $j$ is already complete (by finite dimensionality), it follows that
\[\text{Coker}j\cong \text{Coker}\hat{j}.\]
Thus, in order to show the theorem starting from the commutative diagram above, it suffices to show the isomorphism:
\[H^{\bullet}(H^0(A,\Omega^{\bullet}_{Y}|_{A}))\cong H^{\bullet}(H^0(A,\hat{\Omega}^{\bullet}_Y)).\]
This is proved in turn in \cite{B} (points (b)-(d), pp. 140-142).
\end{proof}
\begin{rem}
For the hypersurface case, there is a simple alternative proof of the above lemma, only for the $n$th-cohomology of the Givental complex, without using resolution of singularities: let $\hat{H}''_f$ be the formal completion of the Brieskorn module with respect to the $\frak{m}$-adic topology. Then, by the regularity of the Gauss-Manin connection and the properties of its analytical index \cite{Mal}, there is an isomorphism of $\mathbb{C}[[f]]$-modules\footnote{or equivalently by the Bloom-Brieskorn theorem \cite{B}, but this uses again resolution of singularities.}:
\[\hat{H}''_f\cong H''_f\otimes_{\mathbb{C}\{f\}} \mathbb{C}[[f]]\]
and thus the quotient
\[\frac{\hat{H}''_f}{f\hat{H}''_f}=\frac{\hat{\Omega}^{n+1}}{df\wedge d\hat{\Omega}^{n-1}+f\hat{\Omega}^{n+1}}\]
is again a $\mu$-dimensional vector space. The space of deformations $\mathcal{Q}_{X,0}$ of the germ $(X,0)$ is finite dimensional and thus it is already complete:
\[\mathcal{Q}_{X,0}\cong \hat{\mathcal{Q}}_{X,0}.\]
Following the construction presented in the proof of Proposition \ref{p1} for the cohomology $H^n(\tilde{\Omega}^{\bullet}_{X,0})$ we obtain again a short exact sequence:
\[0\rightarrow H^n(\hat{\tilde{\Omega}}^{\bullet}_{X,0})\stackrel{df\wedge}{\rightarrow}\frac{\hat{H}''_f}{f\hat{H}''_f}\stackrel{\pi}{\rightarrow}\hat{\mathcal{Q}}_{X,0}\rightarrow 0.\]
The proof of the isomorphism
\begin{equation}
\label{iso}
H^n(\tilde{\Omega}^{\bullet}_{X,0})\cong H^n(\hat{\tilde{\Omega}}^{\bullet}_{X,0})
\end{equation}
follows then immediately by comparing the short exact sequence above with the analytic one (\ref{ses1}).
\end{rem}

\section{An Interpolation Lemma for the Isotropy Group of a Hypersurface Singularity}

Let $\mathcal{R}_{X,0}$ be the isotropy group of the germ $(X,0)$, i.e. the group of germs of diffeomorphisms at the origin tangent to the identity and preserving the hypersurface $X=\{f=0\}$. It means that for every $\Phi \in \mathcal{R}_{X,0}$ there exists an invertible function germ $g \in \mathcal{O}$ such that the following hold:
\[\Phi(x)=x \hspace{0.15cm} \text{mod} \hspace{0.15cm} \frak{m}^2, \hspace{0.3cm} g(x)=1\hspace{0.15cm} \text{mod}\hspace{0.15cm} \frak{m},\]
\[\Phi^*f=gf.\]

We will need the following interpolation lemma for the group $\mathcal{R}_{X,0}$ which is a simple variant of the one presented by J. -P. Fran\c{c}oise in \cite{F} and it relies in a general method obtained by S. Sternberg in \cite{Ste}. It can be also generalised without difficulty to any germ of an analytic subset $(X,0)$ (whose singularities can be arbitrary).

\begin{lem}
\label{l1}
Any diffeomorphism $\Phi \in \mathcal{R}_{X,0}$  can be interpolated by a 1-parameter family of formal diffeomorphisms $\Phi_t \in \hat{\mathcal{R}}_{X,0}$, i.e. there exists a family of formal function germs $g_t \in \hat{\Omega}^0$ such that:
\[\Phi_0=Id, \hspace{0.3cm} \Phi_1=\Phi,\]
\[g_0=1, \hspace{0.3cm} g_1=g,\]
\[\Phi_t^*f=g_tf.\]
\end{lem}
\begin{proof}
Denote by $(x_1,...,x_{n+1})$ the coordinates at the origin and let $x^{\beta}=x_1^{\beta_1}...x_{n+1}^{\beta_{n+1}}$, $\beta=(\beta_1,...,\beta_{n+1})\in \mathbb{N}^{n+1}$, $|\beta|=\sum_{i=1}^{n+1}\beta_i$. Let
\[\Phi_i(x)=x_i+\sum_{j}\sum_{|\beta|=j}\phi_{i,\beta}x^{\beta}, \hspace{0.3cm} i=1,...,{n+1}\]
be the components of $\Phi$. We will find the interpolation $\Phi_t$ with components in the form:
\[\Phi_{t,i}(x)=x_i+\sum_j\sum_{|\beta|=j}\phi_{i,\beta}(t)x^{\beta}, \hspace{0.3cm} i=1,...,{n+1}\]
as solution of the differential equation:
\begin{equation}
\label{fe}
\Phi'_t=\Phi'_0\circ \Phi_t,
\end{equation}
with boundary conditions $\Phi_0=Id$, $\Phi_1=\Phi$ (c.f. \cite{Ste}). We can can do this by induction on $j$ and we may assume that the $\phi_{i,\beta}$ are already known for $j\leq k-1$. Then, for $j=k$, equation (\ref{fe}) implies:
\[\phi'_{i,\beta}(t)=\phi'_{i,\beta}(0)+\psi_{i,\beta}(t),\]
where the functions $\psi_{i,\beta}(t)$ are known by induction and they vanish at zero. Integration then gives:
\[\phi_{i,\beta}(t)=\phi'_{i,\beta}(0)t+\int_0^t\psi_{i,\beta}(\tau)d\tau.\]
Obviously the initial condition $\phi_{i,\beta}(0)=0$ is satisfied, and it suffices to choose the $\phi'_{i,\beta}(0)$ such that the boundary condition $\phi_{i,\beta}(1)=\phi_{i,\beta}$ is satisfied as well. Now, by the fact that the family $\Phi_t$ is an interpolation of $\Phi$, we may choose an interpolation $g_t$ of $g$:
\[g_t(x)=g(0)+\sum_{|\beta|\geq 1}g_{\beta}(t)x^{\beta},\]
satisfying the required assumptions (recall that $g(0)=1$) and such that $\Phi_t^*f=g_tf$ for all integer values of $t$. In fact, the coefficients of $\Phi_t$ are polynomials in $t$, and choosing the interpolation $g_t$ with polynomial coefficients in $t$ as well (linear in $t$ for example), it follows that for any $k$ fixed, the homogeneous part in the Taylor expansion of $\Phi_t^*f-g_tf$ is a polynomial in $t$ which vanishes for all integer values of $t$. Thus, it vanishes for all real $t$ as well and this finishes the proof of the lemma.
\end{proof}

\section{Proof of the Theorem}
We will prove here Theorem \ref{t} which can now be restated in the following form:
\begin{thm}
\label{t2}
Let $\omega$ and $\omega'$ be two germs of volume forms which are $\mathcal{R}_{X,0}$-equivalent. Then there exists an $n$-form $\alpha$ such that $\omega-\omega'=d\alpha$ and $[\alpha]=0$ in $H^n(\tilde{\Omega}^{\bullet}_{X,0})$.
\end{thm}
\begin{proof}
Consider first the $n$-form $\alpha$ defined by $\omega-\omega'=d\alpha$ (Poincar\'e lemma) and let $\Phi \in \mathcal{R}_{X,0}$ be the diffeomorphism providing the equivalence: $\Phi^*\omega'=\omega$.
It follows that
\begin{equation}
\label{c1}
\omega-\Phi^*\omega=d\alpha
\end{equation}
holds in $\Omega^{n+1}$. Interpolate now $\Phi$ by the 1-parameter family of formal diffeomorphisms $\Phi_t\in \hat{\mathcal{R}}_{X,0}$ as in Lemma \ref{l1} above. We have that:
\[\omega-\Phi^*\omega=\int_0^1\frac{d}{dt}\Phi_t^*\omega dt=\int_0^1\Phi_t^*(L_{\hat{v}}\omega) dt=\]
\[=\int_0^1\Phi_t^*d(\hat{v}\lrcorner \omega)dt=d\int_0^1\Phi_t^*(\hat{v}\lrcorner \omega)dt,\]
holds in $\hat{\Omega}^{n+1}$, where $\hat{v}$ is the 1-parameter family of formal vector fields generating $\Phi_t$: $\exp{t\hat{v}}=\Phi_t$. Thus, in $\hat{\Omega}^{n+1}$ we may write:
\begin{equation}
\label{c2}
\omega-\Phi^*\omega=d\hat{\alpha},
\end{equation}
where the formal $n$-form $\hat{\alpha}$ is defined by:
\[\hat{\alpha}=\int_0^1\Phi_t^*(\hat{v}\lrcorner \omega)dt+d\hat{h},\]
for some formal $(n-1)$-form $\hat{h}$. Now, since $\Phi_t$ preserves the germ $(X,0)$ for all $t$ and $\hat{v}$ is tangent to its smooth part, it follows that $\hat{\alpha}|_{X^*}=d\hat{h}|_{X^*}$, i.e. that $[\hat{\alpha}]=0$ in $H^n(\hat{\tilde{\Omega}}^{\bullet}_{X,0})$. View now the relation (\ref{c1}) as a relation in $\hat{\Omega}^{n+1}$. By comparing it with the relation (\ref{c2}) we obtain $\alpha=\hat{\alpha}+d\hat{g}$ for some formal $(n-1)$-form $\hat{g}$ and thus $[\alpha]=[\hat{\alpha}]=0$ in $H^n(\hat{\tilde{\Omega}}^{\bullet}_{X,0})$ as well.  By the the Bloom-Brieskorn Lemma \ref{l2} and in particular by the isomorphism (\ref{iso}) we finally obtain that $[\alpha]=0$ in $H^n(\tilde{\Omega}^{\bullet}_{X,0})$ and this finishes the proof of the theorem.
\end{proof}

\section*{Acknowledgements}
The author is grateful to the staff of ICMC-USP (Instituto de Ci\^encias Matem\'aticas e de Computa\c{c}\~ao-Universidade de S\~ao Paulo) for their hospitality, where the paper was written.

The author is also grateful to Mauricio D. Garay for his valuable comments on a previous version of the paper, as well as to Christian Sevenheck for his interest in the work and for providing directions concerning G. M. Greuel's work.

 ``This research has been supported (in part) by EU Marie-Curie IRSES Brazilian-European partnership in Dynamical Systems (FP7-PEOPLE-2012-IRSES 318999 BREUDS)''.

\section*{Bibliography}


TITLE: Is a strictly monotone continuous function defined on $\mathbb{R}^m$ has a continuous inverse?
QUESTION [3 upvotes]: Let $F: \mathbb{R}^m \to \mathbb{R}^m$ be a strictly monotone, surjective, continuous function. Here, the strict monotonicity means that $(x - y)^T (F(x) - F(y)) > 0$ for all $x,y \in \mathbb{R}^m$ as shown at
Is every monotone map the gradient of a convex function?
Then, by strict monotonicity of $F$, it is also injective, and hence bijective. So, there is its inverse function $F^{-1}: \mathbb{R}^m \to \mathbb{R}^m$.
My question is "Is the inverse $F^{-1}$ also continuous?"
When $m = 1$, it seems true as shown at
Prove that if f is a continuous strictly monotone function defined on an interval, then its inverse is also a continuous function.
Many thanks in advance for your answer and discussion.

REPLY [0 votes]: Maybe it is helpful the local inversion theorem, that is a consequence of Dini's Theorem for systems. It states that if $f:A \subset \mathbb{R}^n \rightarrow \mathbb{R}^n$, $A$ open set, $f \in C^1(A)$ and $x_0 \in A$.
If $\det J_f(x_0) \neq 0$ then there exixts $\sigma, \delta>0$ and a unique function $\varphi$ such that
$$
\varphi: B_\sigma(y_0) \rightarrow B_\delta (x_0)
$$
is $C^1$ in $B_\sigma(y_0)$,
$$
\varphi(f(x))=x,\quad f(\varphi(y))=y,\quad \forall x \in B_\delta (x_0), \quad\forall y \in B_\sigma(y_0)
$$ where $y_0=f(x_0)$.
Here the inverse is $\varphi$ that belongs to $C^1(B_\sigma(y_0))$, so in particular $\varphi$ is continuous over $B_\sigma(y_0)$. Your function $F$ is invertible then satisfies the hypothesis $\det J_f(x_0) \neq 0$.


TITLE: Finding Singular points
QUESTION [0 upvotes]: Let $f = 3x^3 + 3x^2 − y^2 + z^2$, $g = 3x^2 + 4^x + 3y^2 + z^2$ be polynomials
in $\mathbb{C}[x, y, z]$ and let $W = V(\langle f, g\rangle) ⊂ \mathbb{A}^3(C)$. By using the Jacobian matrix, find the singular points of W or show that it has no singular points.(You may assume without proof that $\langle f, g\rangle$
is a prime ideal therefore $W$ is an irreducible variety and $I(W) = \langle f, g\rangle)$
I have $f_x=9x^2+6x$, $f_y=-2y$ and $f_z=2z$
and $g_x=6x+4$, $g_y=6y$, $g_z=2z$
so $J=\begin{bmatrix}9x^2+6x&-2y&2z\\6x+4&6y&2z\end{bmatrix}$
I don't know how to find the singular points (or show that it has none)

REPLY [0 votes]: A point on $W$ is singular if and only if the rank of the Jacobian matrix at this point is $\leq 1$, which is (by basic linear algebra) equivalent to the fact, that all $2 \times 2$-minors vanish.
Hence you should compute the three $2 \times 2$-minors and see where they will vanish simultaneously.


TITLE: Proving that if two integers have opposite parity, then their product is even
QUESTION [9 upvotes]: If two integers have opposite parity, then their product is even.
Proof Method: Direct Proof
If two integers have opposite parity, then one is even and the other is odd.
Suppose: $a$ is an even integer and $b$ is an odd integer, then by definition of even and odd integers
$$a = 2m, \quad b = 2n+1,$$ while $m$ and $n$ are integers.
$$ ab = 2m(2n+1)= 4mn+2m = 2(2mn+m)
$$
Let $c = 2mn+m$ be an integer, then $ab=2c$ is even
Therefore, the product of two opposite parity integers is even
Thank You!

REPLY [4 votes]: Euclid's Lemma asserts that if a prime number $p$ divides the product $ab$ of two integers $a$ and $b$, then $p$ must divide at least one of those integers $a$ and $b$.
Since $2$ is a prime number, we can let $p = 2$. We now want to show that for integers $a$ and $b$ that have an opposite parity, $2$ must divide the product $ab$.
When two numbers have an opposite parity, then one of those numbers is even and the remaining number is odd. The definition of an even number is such that the number is divisible by $2$. The contrary is the definition of an odd number. This is why we let $p = 2$ in the first place.
Since either $a$ or $b$ must be even, the remaining integer consequently being odd, then $2$ must divide either $a$ or $b$. Therefore, $2$ must also divide $ab$ which makes that product an even number.
Go here for a proof of Euclid's Lemma.


TITLE: Show surface integral of curl of vector field is 0
QUESTION [0 upvotes]: I need help with this question!
Let $\vec F$ = $\nabla f$ and $S$ be any surface. Show that $\int$$\int_s$ $\nabla$ $\times$ $\vec F$ $\cdot$ $d\vec S = 0$

REPLY [0 votes]: If $\vec F = \nabla f$, then $\vec F$ is conservative. Then, we must have that $\nabla \times \vec F = 0$.
See here for why conservative vector fields have zero curl.


text:
\section{Polynomial approximation results}\label{sec:polys}

In this section we state (and prove for ease of reference) two classical
results on uniform polynomial approximation
(cf.~\cite{KuczynskiWo92,Nemirovski94}) that stand at the core of the
technical development in this work.

\begin{lemma}\label{lem:cheby-first}
	Let $n \ge 1$ and $0< \alpha \le \beta$, and let $\kappa =
	\beta/\alpha$. Then
	\begin{equation*}
	\min_{p \in \Polys} \max_{x\in [\alpha,\beta]} | 1 - x p(x) | =
		\minmaxT \defeq
		2 \left(  \left(\frac{\sqrt{\kappa}+1}{\sqrt{\kappa}-1}\right)^n
		+ \left(\frac{\sqrt{\kappa}-1}{\sqrt{\kappa}+1}\right)^n \right)^{-1}
	\end{equation*}
	and
	\begin{equation*}
	2\left(e^{2n/(\sqrt{\kappa}-1)}+1\right)^{-1} \le \minmaxT \le
	2e^{-2n/\sqrt{\kappa}}.
	\end{equation*}
	Moreover, there exist $x_0, x_1, \ldots, x_{n} \in [\alpha,\beta]$ and
	probability distribution $\pi_0, \pi_1, \ldots \pi_{n}$ such that
	\begin{equation*}
	\min_{p \in \Polys} \sum_{k=0}^{n} \pi_k (1-x_k p(x_k))^2 =
	[\minmaxT] ^2.
	\end{equation*}
\end{lemma}

\begin{proof}
	Let
	\begin{equation*}
	T_n(x) = \begin{cases}
	\cos(n\arccos(x)) & |x| \le 1 \\
	\half \left(  (x + \sqrt{x^2-1})^n + (x - \sqrt{x^2-1})^n \right) & |x|
	\ge 1
	\end{cases}
	\end{equation*}
	denote the order $n$ Chebyshev polynomial of the first kind. We
	claim that $p^\star\in\Polys$ that solves the minimax problem $\min_{p
	\in
	\Polys} \max_{x\in [\alpha,\beta]} | 1 - x p(x) |$ is given by
	\begin{equation*}
	1-xp^\star(x) = \minmaxT\cdot
	 T_n \left(
	\frac{\kappa+1-2x/\alpha}
	{\kappa-1} \right),
	\end{equation*}
	where $\minmaxT = \left[ T_n \left( \frac{\kappa+1}{\kappa-1}
	\right)\right]^{-1}$ guarantees that the RHS has value 1 at $x=0$ and
	therefore $p^\star$ is well defined. Since clearly $|T_n(y)|\le 1$ for every
	$y\in[-1,1]$, we have that $\max_{x\in [\alpha,\beta]} | 1 - x p^\star(x)
	| = \minmaxT$.

	We argue that $p^\star$ is optimal using the classical alternating signs
	argument, sometimes also referred to as Chebyshev's theorem.
	First, note that $T_n(y)$ has $n+1$ extrema in $[-1,1]$ (at
	$y_k=\cos(k\pi/n)$ for $k=0,\ldots,n$) and that their values alternate
	between $-1$ and $1$ (\ie $T_n(y_k)=(-1)^k$). Therefore, there exist
	$n+1$ distinct points $x_0, x_1, \ldots, x_{n} \in [\alpha,\beta]$ for
	which $1-x_i p^\star(x_k) = (-1)^{k}\minmaxT$. Let $q\in\Polys$
	satisfy $\max_{x\in [\alpha,\beta]} | 1 - x q(x)
	| \le \minmaxT$. Then,
	\[p^\star(x_k) - q(x_k) = \frac{[1-x_k q(x_k)]
	-[1-x_k p^\star(x_k)]}{x_k} \]
	must be non-positive for
	even $k$ and non-negative for odd $k$, and therefore $p^\star- q$
	must
	have at least $n$ roots
	in $[\alpha,\beta]$. However, $p^\star- q$ is a polynomial of degree at
	most $n-1$ and can have $n$ roots only if it is identically 0, so we have
	that $q=p^\star$, proving that  $p^\star$ is the unique
	solution of the minimax problem.

	To see the upper and lower bounds on $\minmaxT$, note that
	$\minmaxT
	= 1/\cosh(n \log( 1 + \frac{2}{\sqrt{\kappa}-1}))$, that $\half e^{|y|} \le
	\cosh(y) \le \half(e^{|y|} +1)$, and that
	\begin{equation*}
	\frac{2}{z} \le \log\left( 1 + \frac{2}{z-1}\right) \le
	\frac{2}{z-1}
	\end{equation*}
	for all $z>1$,
	where the lower bound above can seen by comparing derivatives.

	To see the final part of the lemma, let $x_0, x_1, \ldots, x_{n} \in
	[\alpha,\beta]$ be the points constructed in the optimality argument
	above, and note that this argument continues to hold if the inner
	maximization is restricted to these points. Therefore,
	\begin{equation*}
	\min_{p \in \Polys}\max_{0\le k \le n} (1-x_k p(x_k))^2
	= \left[ \min_{p \in \Polys}\max_{0\le k \le n} |1-x_k p(x_k)|  \right]^2
	= [\minmaxT]^2.
	\end{equation*}
	Letting $\Delta_{n+1}$ denote the probability simplex with $n+1$
	variables, we may write
	\begin{equation*}
	\max_{0\le k \le n} (1-x_k p(x_k))^2 =
	\max_{\mu\in\Delta_{n+1}}
	\sum_{k=0}^{n} \mu_k (1-x_k p(x_k))^2.
	\end{equation*}
	Finally, noting that the objective $\sum_{k=0}^{n} \mu_k (1-x_k
	p(x_k))^2$ is linear (and hence concave) in $\mu$ and convex in (the
	coefficients of) $p$, we
	may use Von-Neumann's lemma and swap the $\min$ and $\max$
	above, writing
	\begin{equation*}
	\max_{\mu\in\Delta_{n+1}} \min_{p \in \Polys} \sum_{k=0}^{n} \mu_k
	(1-x_k p(x_k))^2
	=
	\min_{p \in \Polys}\max_{\mu\in\Delta_{n+1}}  \sum_{k=0}^{n} \mu_k
	(1-x_k p(x_k))^2
	= [\minmaxT]^2.
	\end{equation*}
	Letting $\pi$ denote the distribution attaining the outer maximum, we
	get the desired result. We remark in passing that $\pi$ may be
	constructed explicitly using the orthogonality principle of least squares
	estimation and orthogonality relations of Chebyshev polynomials.
\end{proof}

\begin{lemma}\label{lem:cheby-second}
	Let $n \ge 1$ and $0< \alpha \le \beta$, let $\kappa =
	\beta/\alpha$ and define $w(x)\defeq \sqrt{x-\alpha}$.

	Then
	\begin{equation*}
	\min_{p \in \Polys} \max_{x\in [\alpha,\beta]}
	{w(x)}| 1 - x 	p(x)
	| =
	\minmaxU \defeq
	2\sqrt{\alpha} \left(
	\left(\frac{\sqrt{\kappa}+1}{\sqrt{\kappa}-1}\right)^{n + \half}
	- \left(\frac{\sqrt{\kappa}-1}{\sqrt{\kappa}+1}\right)^{n + \half}
	\right)^{-1}
	\end{equation*}
	and
	\begin{equation*}
	2\sqrt{\alpha}
	\left(e^{2(2n+1)/(\sqrt{\kappa}-1)} - 1\right)^{-\half}
	\le \minmaxU \le
	2\sqrt{\alpha}
	\left(e^{2(2n+1)/\sqrt{\kappa}} - 2\right)^{-\half}.
	\end{equation*}
	Moreover, there exist $x_0, x_1, \ldots, x_{n} \in [\alpha,\beta]$ and
	probability distribution $\pi_0, \pi_1, \ldots \pi_{n}$ such that
	\begin{equation*}
	\min_{p \in \Polys} \sum_{k=0}^{n} \pi_k w^2(x_k)(1-x_k p(x_k))^2 =
	[\minmaxU] ^2.
	\end{equation*}
\end{lemma}

\begin{proof}
	Let
	\begin{equation*}
	U_n(x) = \begin{cases}
	\frac{1}{\sqrt{1-x^2}}\sin((n+1)\arccos(x)) & |x| \le 1 \\
	 \frac{1}{2\sqrt{x^2-1}}\left( (x + \sqrt{x^2-1})^{n+1} - (x -
	 \sqrt{x^2-1})^{n+1}\right)
	  & |x| \ge 1
	\end{cases}
	\end{equation*}
	denote the order $n$ Chebyshev polynomial of the second kind. We
	claim that $p^\star\in\Polys$ that solves the minimax problem $\min_{p
		\in
		\Polys} \max_{x\in [\alpha,\beta]} (x-\alpha)^{1/2}| 1 - x p(x) |$ is
		given by
	\begin{equation*}
	1-xp^\star(x) = \frac{\minmaxU}{w(\beta)}\cdot
	U_{2n} \left( \sqrt{
	\frac{\kappa-x/\alpha}
	{\kappa-1} }\right),
	\end{equation*}
	where $\minmaxU = w(\beta)\left[ U_{2n} \left(
	\sqrt{\frac{\kappa}{\kappa-1}}
	\right)\right]^{-1}$ guarantees that the RHS has value 1 at $x=0$ and
	therefore $p^\star$ is well defined (note that $U_{2n}(\cdot)$ is an even
	polynomial and therefore $U_{2n}(\sqrt\cdot)$ is a polynomial of degree
	$n$). For $x\in[\alpha,\beta]$, we have by the definition of $p^*$ and
	the expression for $U_{2n}$,
	\begin{equation*}
	w(x)(1-xp^\star(x)) = \minmaxU \cdot
	\sin\left( (2n+1)\arccos\left( \sqrt{\frac{\kappa-x/\alpha}{\kappa-1}
	}\right)\right).
	\end{equation*}
	Therefore, we have that $w(x)|1-xp^\star(x)| \le \minmaxU$ for every
	$x\in[\alpha,\beta]$, and moreover we have that $w(x_k)(1-x_k
	p^\star(x_k)) = (-1)^k \cdot \minmaxU$, for the points $x_0, \ldots x_n
	\in
	[\alpha, \beta]$ satisfying
	\begin{equation*}
	\sqrt{\frac{\kappa-x_k/\alpha}{\kappa-1} }
	= \cos \left(\frac{\pi}{2}\cdot \frac{2k+1}{2n+1}\right).
	\end{equation*}
	Hence, the alternating signs argument from the proof of
	Lemma~\ref{lem:cheby-first} holds here as well and we have that
	$p^\star$ is optimal and that $\min_{p \in \Polys} \max_{x\in
	[\alpha,\beta]}
	{w(x)}| 1 - x 	p(x) | = \minmaxU$.

	To see the upper and lower bounds on $\minmaxU$, note that
	$\minmaxU
	= \sqrt{\alpha}/\sinh((n+\half) \log( 1 + \frac{2}{\sqrt{\kappa}-1}))$,
	that for $y\ge 0$, $\sinh(y) =\frac{1}{\sqrt{2}}\sqrt{\cosh(2y)-1}$ gives
	$\half \sqrt{e^{2y}-2} \le
	\sinh(y) \le \half \sqrt{e^{2y}-1}$, and that  (as in
	Lemma~\ref{lem:cheby-first})
	$\frac{2}{z} \le \log\left( 1 + \frac{2}{z-1}\right) \le
	\frac{2}{z-1}$.

	The final part of the lemma follows exactly as in
	Lemma~\ref{lem:cheby-first}.
\end{proof}

config: arxiv


text:
\begin{document}
	\maketitle

\begin{abstract}
	We construct the non-linear Markov process connected with biological model of bacterial genome recombination.
	The description of invariant measures of this process gives us the solution of one problem in elementary probability theory.
\end{abstract}

The genetic recombination in bacteria can be formally described in the following way~\cite{pirogov2013recombination}.
Let $\Lambda=\{1, 2, \dots, n\}$ be a finite set and for any $i \in \Lambda$ we have a finite alphabet $K_i$.
We call by genomes elements of the set $X =\prod_{i \in \Lambda} K_i$, i.~e. words in the alphabet depending on $i$.
Suppose we have a set $\mathcal{J}$ of subsets $I \subset \Lambda$.
These subsets we call \emph{frames}.
A system of frames is called the $T_0$-system if for any $i \ne j \in \Lambda$ there is a~frame $I\in \mathcal{J}$ for which either $i \in I$, $j \notin I$ or $j \in I$, $i \notin I$.

This property we named $T_0$ by the analogue with Kolmogorov's $T_0$-axiom in the general topology.

The restriction of a word $x = \{x_i, i \in \Lambda\}$ on a subset $M \subset \Lambda$ we denote by $x_M$.

\begin{definition}
	\label{def:i-recombination}
	The transform of a word $x=(x_I,x_{\Lambda\setminus I})$ to the word $\tilde x=(y_I, x_{\Lambda\setminus I})$ is called the \emph{$I$-recombination} of the word $x$ with the word $y$.
\end{definition}

Let us suppose that for any $I \in \mathcal{J}$ we have a similarity function $\phi_I(x_I, y_I)$ which we suppose to be symmetric ($\phi_I(x_I, y_I)=\phi_I(y_I, x_I)$) and strictly positive.
 For given $I$ we consider the symmetric matrix $\Phi_I=(\phi_I(x_I, y_I))$.

The set of matrices $\mathcal{R}=\{\Phi_I, I \in \mathcal{J}\}$ we call the \emph{legend} of recombination.

Suppose we have a probability measure $\mu$ on the space $X$.
A non-linear Markov process \cite{mckean1966class} of recombination is defined by its transition rates.
By definition, for each $I \in \mathcal{J}$ the transition rate $\lambda_I(x, \tilde x, \mu)$ of the word $x=(x_I,x_{\Lambda\setminus I})$ to the word $\tilde x=(y_I, x_{\Lambda\setminus I})$ equals $\phi_I(x_I, y_I)\mu_I(y_I)$.
(Here and below we denote by $\mu_I$ the corresponding marginal distribution, i.~e. the projection of the measure $\mu$).

We suppose that we have the initial measure $\mu^0$, i.~e.~the distribution of the word~$x(0)$, and for $t>0$ transition rates $\lambda_I(x, \tilde x, \mu^t)$ are defined by the measure $\mu^t$, which is the distribution of the word $x(t)$.
So, the distribution $\mu^t$ satisfies the nonlinear differential equations:
\begin{equation}
	\label{eq:diff equation}
	\frac{d \mu(x)}{dt}=\sum_I \sum_{y_I}\left(\phi_I(y_I, x_I)\mu_I(x_I)\mu(x_{\Lambda\setminus I}, y_I) - \phi_I(x_I, y_I) \mu_I (y_I) \mu(x)\right).
\end{equation}

For a given legend $\mathcal{R}$ let us define the following properties of the measure $\mu$.
\begin{definition}
	\label{def:r-stable measure}
	 Probability measure $\mu$ is called\\
	 a) $\mathcal{R}$-stable, if it is a fixed point for equation \eqref{eq:diff equation}.\\
	 b) $\mathcal{J}$-separated, if for any $I \in \mathcal{J}$ two sets of random variables $x_I$ and $x_{\Lambda \setminus I}$ are independent with respect to measure $\mu$.
\end{definition}

\begin{theorem}
	\label{thm:r-stable measure if T_o-sistem}
	The measure $\mu$ is $\mathcal{R}$-stable if and only if it is $\mathcal{J}$-separated.
\end{theorem}

	We supposed that $\mathcal{J}$ is $T_0$-system (for Theorem~\ref{thm:r-stable measure if T_o-sistem} this condition can be omited, see Remark~\ref{rem:not T-0 system}).

The ``if'' part of Theorem \ref{thm:r-stable measure if T_o-sistem} is trivial:
any $\mathcal{J}$-separated measure $\mu$ is $\mathcal{R}$-stable.
Indeed, if the measure $\mu$ is $\mathcal{J}$-separated, then $\mu(x_{\Lambda\setminus I}, y_I)=\mu_{\Lambda \setminus I}(x_{\Lambda\setminus I})\mu(y_I)$, and $\mu(x)=\mu_{\Lambda\setminus I}(x_{\Lambda\setminus I})\mu(x_I)$.
Therefore, by the symmetry of the function $\phi_I$, all the summands in the r.~h.~s. of differential equation \eqref{eq:diff equation} are $0$.

Now we derive Theorem~\ref{thm:r-stable measure if T_o-sistem} in direction ``only if'' from the stronger theorem.

\begin{theorem}
	\label{thm:mu0 goes to bernoulli}
	Let $\mu^0$ be an arbitrary probability measure on $X$.
	Then the trajectory $\mu^t$ in the space of measure (the solution of differential equation~\eqref{eq:diff equation}with the initial condition~$\mu^0$) for $t \rightarrow \infty$ tends to the set to of the $\mathcal{J}$-separated measures $\nu$ such that $\nu_i=\mu_i^0$.
\end{theorem}

\begin{remark}
	\label{rem:measure product}
As we will see later if $\mathcal{J}$ is $T_0$-system, then this set of $\mathcal{J}$-separated measures consist of the unique point $\nu=\prod_i \mu_i^0$.
\end{remark}

\begin{proof}[Proof of Theorem~\ref{thm:mu0 goes to bernoulli}]
The proof is based on Lyapunov method.

For the Lyapunov function we take the Shannon entropy of the measure $\mu$ (with the minus sign) $H(\mu)=\sum_x \mu(x)\ln \mu(x)$.

For a given frame $I \subset \mathcal{J}$ let us consider the differential equation containing only the summands with this $I$:
\begin{equation}
	\label{eq:modification of uravnenie}
	\frac{d \mu(x)}{dt} = \sum_{y_I}\left(\phi_I(y_I, x_I) \mu_I(x_I)\mu(x_{\Lambda\setminus I}, y_I) - \phi_I (x_I, y_I) \mu_I(y_I) \mu(x)\right).
\end{equation}
Summing this equation over $x_{\Lambda \setminus I}$ we get $\frac{d \mu_I}{dt}=0$.
So, $\mu_I$ does not depend on time, and the right hand side of~\eqref{eq:modification of uravnenie} we can consider as the direct Kolomogorov equation (i.~e. the linear differential equation for the measure) for the Markov process with constant transition rates $x=(x_I, x_{\Lambda\setminus I}) \rightarrow \tilde x=(y_I, x_{\Lambda\setminus I})$ equal to $\lambda_I(x_I, y_I)=\phi_I(x_I,y_I)\mu_I(y_I)$.

This process does not change $x_{\Lambda\setminus I}$ but it changes the mutual distribution of $x_I$ and $x_{\Lambda\setminus I}$.
For fixed $x_{\Lambda\setminus I}$ it is irreducible continuous time Markov chain on the set $\supp \mu_I$.
We suppose also that $\mu_{\Lambda\setminus I}(x_{\Lambda\setminus I})>0$.

Consider the process with fixed $x_{\Lambda\setminus I}$, the previous transition rates $\lambda_I(x_I, y_I) = \phi_I(x_I,y_I)\mu_I(y_I)$ and an arbitrary ``wrong''  initial distribution $\tilde \mu_I$ on the set $\supp \mu_I$.
It is well known \cite{feller10introduction}, that due to the irreducibility the distribution of this Markov process converges to $\mu_I$ as $t \rightarrow \infty$.

Moreover, the Kullback--Leibler divergence $H(\tilde \mu|\mu)=\sum_{x_I}\tilde \mu_I(x_I)\ln \frac{\tilde \mu_I(x_I)}{\mu_I(x_I)}$ has the strictly negative time derivative via the direct Kolmogorov equation for this process \cite{batishcheva2005, malyshev2004random, pirogov2013recombination}.
Thus if we set $\tilde \mu_I = \frac{\mu(x_I, x_{\Lambda \setminus I})}{\mu_{\Lambda \setminus I}(x_{\Lambda \setminus I})}$, then for the evolution governed by equation~\eqref{eq:modification of uravnenie} the (minus-)entropy
\begin{multline}
H(\mu)=\sum_{x_I, x_{\Lambda\setminus I}} \mu(x_I, x_{\Lambda \setminus I})\ln \mu(x_I, x_{\Lambda \setminus I})=\\
=
\sum_{x_I, x_{\Lambda\setminus I}} \mu_{\Lambda \setminus I}(x_{\Lambda \setminus I})\frac{\mu(x_I, x_{\Lambda \setminus I})}{\mu_{\Lambda \setminus I}(x_{\Lambda \setminus I})} \ln \frac{\mu(x_I, x_{\Lambda \setminus I})}{\mu_{\Lambda \setminus I}(x_{\Lambda \setminus I})\mu_I(x_I)} +\\
+\sum_{x_I}\mu_I(x_I)\ln \mu_I(x_I)+\sum_{x_{\Lambda \setminus I}}\mu_{\Lambda \setminus I}(x_{\Lambda \setminus I})\ln \mu_{\Lambda \setminus I} (x_{\Lambda \setminus I})
\end{multline}
has a strictly negative time derivative if $\mu(x_I, x_{\Lambda \setminus I})$ does not coincide with $\mu_I(x_I)\mu_{\Lambda \setminus I}(x_{\Lambda \setminus I})$.
(Here $0\ln 0=0$.)

For the complete equation~\eqref{eq:diff equation} the (minus-)Shannon entropy has a strictly negative time-derivative if the equality $\mu(x_I, x_{\Lambda \setminus I})=\mu_I(x_I)\mu_{\Lambda \setminus I}(x_{\Lambda \setminus I})$ is violated for some $I \in \mathcal{J}$.

So, any trajectory of equation~\eqref{eq:diff equation} converges to the set of $\mathcal{J}$-separated measures as $t\rightarrow \infty$.

Moreover, if $i \in I$, then $\mu_i$ does not change via equation~\eqref{eq:modification of uravnenie}, since $\mu_i$ is the projection of $\mu_I$.
The same holds for $i \notin I$ because $\mu_i$ is the projection of $\mu_{\Lambda\setminus I}$.
Therefore $\mu_i$ is not time depended for any solution~\eqref{eq:diff equation}.
Thus, any limit point of a solution of this equation is a $\mathcal{J}$-separated measure $\nu$ with marginal distributions $\nu_i=\mu_i^0$.
\end{proof}

Let us describe this set of measures $\nu$ and prove the statement of Remark~\ref{rem:measure product}.
That follows from the next Theorem.

\begin{theorem}
	\label{thm:measure is bernoulli}
	Let $\xi_i, i \in \Lambda$ be the random variables.
	Suppose for any $i, j$, there is a subset $I \subset \Lambda$ containing exactly one element from $\{i, j\}$ and such that $\xi_I=\{\xi_i, i \in I\}$ and $\xi_{\Lambda \setminus I}=\{\xi_i, i \in \Lambda \setminus I\}$ are independent.
	Then all variables $\xi_i, i \in \Lambda$ are independent.
\end{theorem}

\begin{proof}
 The collection of subsets $I$ (frames) is $T_0$-system in the sense of the definition above.
 For any $\Lambda$ and $M\subset \Lambda$ by $J_M$ we denote the set of frames $I \cap M$ on the
 set $M$ for $I \in J$.
 It is evident that $J_M$ is a $T_0$-system if $J$ is a $T_0$-system.
 Now we use the induction on the cardinality of $\Lambda$.
 For $|\Lambda|=1$ the statement is trivial.
 Let $|\Lambda|>1$.
 Fix some frame $M \in J$ which is a proper subset of $\Lambda$.
 Then $J_M$ and $J_{\Lambda\setminus M}$ are $T_0$-systems.
 Cardinalities of $M$ and $\Lambda \setminus M$ are less than the cardinality of $\Lambda$.
 So we can apply the induction hypothesis and conclude that the random variables $x_i$, $i \in M$, are
 mutually independent and random variables $x_i$, $i \in \Lambda\setminus M$, are mutually independent.
 But these two sets of random variables are independent by the condition of the theorem.
 So all random variables $x_i$, $i \in \Lambda$, are mutually independent.
 By the induction the theorem follows.
\end{proof}

An example of an application of this Theorem following statement.
\begin{corollary}
	\label{cor: matrix}
	If there is a random matrix $\xi_{i,j}$ such that its columns are independent and its rows are independent.
	Then all elements $\xi_{i,j}$ are independent.
\end{corollary}

The description of limit measure $\nu$ implies the following statement.
\begin{corollary}
	\label{cor:combinatorial refolmulation}
	If $\mathcal{Y}=\{x^1, \dots, x^m\}$ is a set of words such that all letters $x_i^j$ for given $i$ take all values from $K_i$, then any word $x \in X$ can be obtained from words of the set $\mathcal{Y}$ by the finite sequence of recombinations.
\end{corollary}

Informally, from the set of genoms having all possible letters on any place we can obtain any genom by recombinations.

\begin{proof}
	Conisder the non-linear Markov recombination process with the inital measure $\mu^0$ such that $\supp \mu^0 = \mathcal{Y}$.
	Then as $t$ tends to infinity the measure $\mu^t$ tends to the product measure with strictly positive marginals.
	So the measure $\mu^t$ is strictly positive for sufficiently large $t$.
	Thus just means that any word can be obtained from the words of the set $\mathcal{Y}$ by the some finite sequence of recombination.
\end{proof}

This Corollary can also be proved without probability arguments using the induction on the cardinality as in the proof of Theorem~\ref{thm:measure is bernoulli}.

From Corollary~\ref{cor:combinatorial refolmulation} it easily follows the next statement.

\begin{corollary}

	\label{cor:R fill the space modification}
	If $\mathcal{J}$ is $T_0$-system and all the projections~$\mu_i^0$ are strictly positive then $\supp \mu^t = X$.
\end{corollary}

\begin{remark}
	\label{rem:not T-0 system}
	Let the system $\mathcal{J}$ be not a $T_0$-system.
	We say that two points $i, j \in \Lambda$ are equivalent if for any $I \in \mathcal{J}$ either $i\in I$, $j\in I$ or $i \notin I$, $j\notin I$.

Then the set $\Lambda$ is divided on equivalence classes $\Lambda=\cup \Lambda_j$, $j \in \tilde \Lambda$, where by $\tilde \Lambda$ we denote the set of equivalence classes.
The system $\mathcal{J}$ defines the system of frames $\mathcal{J}'$ on~$\tilde \Lambda$, which comes to be $T_0$-system.

Now $X=\prod_j \tilde K_j$ where $\tilde K_j = \prod_{i \in \Lambda_j}K_i$.
This construction evidently reduces the proof of Theorem~\ref{thm:r-stable measure if T_o-sistem} for arbitrary system $\mathcal{J}$ to the case of $T_0$-system.
It easy to formulate Theorem~\ref{thm:mu0 goes to bernoulli} for this case.

The invariant measure $\nu$ which is a limit of the solution of \eqref{eq:diff equation} is $\prod_{j \in \tilde \Lambda} \mu_j^0$, where~$\mu_j^0$
is marginal distribution of~$\mu^0$ on the set $\Lambda_j$.
\end{remark}

\section*{Aknowledgement}
	The authors are grateful to M.~S.~Gelfand and A.~S.~Kalinina for fruitful discussions. We also thanks G.~A.~Kabatiansky for useful advice and F.~V.~Petrov for the idea of the new proof of Theorem~\ref{thm:measure is bernoulli}.


TITLE: Is the set of vanishing $k$ derivatives of smooth functions in a null set dense in $W^{1,p}$?
QUESTION [1 upvotes]: Let $\Omega$ be an open set with compact closure denoted by $\overline{\Omega}$  and a null set $N\subset\Omega$   with respect the Lebesgue measure.
Then consider the two sets

$C^{\infty}(\overline{\Omega})$ (smooth functions in $\overline{\Omega}$).
$V=\{\psi\in C^{\infty}(\overline{\Omega}/N) \text{ with } \psi=\psi_{a_{1}}=...=\psi_{a_{1}...a_{k}}=0 \text{ in }  N  \} \cup \{\psi\in C^{\infty}(\overline{\Omega}) \text{ with } \psi=\psi_{a_{1}...a_{k}}=0 \text{ in }  N  \}$

So $V$ is the set of all  $\phi\in C^{\infty}(\overline{\Omega})$ functions where the value of $\phi$ and and its first $k$ derivatives is change to be $0$ at $N$.
Now is $V$ dense in $W^{k,p}(\Omega)$?
I think it is for the following reason:
Consider $\phi\in C^{\infty}(\overline{\Omega})$ then $\phi\in W^{k,p}$. Now this $\phi$ is in the same class as $\phi_{n}$ which is just $\phi$ everywhere except possibly at $N$.
So "according to" the $W^{k,p}(\Omega)$ space $\phi_{n}$ is $\phi$ in  $W^{k,p}(\Omega)$. And then we know that the class of $C^{\infty}(\overline{\Omega})$ is dense in $W^{k,p}(\Omega)$.
This conclude the argument as we have prove that that  $C^{\infty}(\overline{\Omega})\subset V$ as classes in $W^{k,p}(\Omega)$
Why is this argument incorrect?

REPLY [1 votes]: The argument is incorrect.
Suppose $\phi \in C^\infty(\bar{\Omega}) \subset W^{k,p}(\Omega)$.  As you say, we can define a function
$$\phi_n(x) = \begin{cases} \phi(x), & x \notin N \\ 0, & x \in N \end{cases}$$
Then $\phi_n$ is a function that vanishes on $N$ and is almost everywhere equal to $\phi$.  However, $\phi_n$ typically will not be $C^\infty$, so it is not an element of your set $V$, which you specifically defined to consist only of $C^\infty$ functions.  Generally, if you modify a continuous function on a null set, it will no longer be continuous.  (A null set has a dense complement, so if $\phi$ and $\phi_n$ are both continuous and are equal on $N^c$, they must also be equal on $N$.)
For an explicit counterexample, let $n=1$, $p=1$, $\Omega = (0,1)$ and $N = \Omega \cap \mathbb{Q}$.  If $\phi \in C^\infty$ and $\phi'$ vanishes on $N$ then $\phi$ is constant.  Likewise if $\phi$ vanishes on $N$ then $\phi = 0$.  So $V$ only contains the constants, or more properly, $V$ consists of those equivalence classes which contain a constant function.  For a function to be in one of the equivalence classes in $V$, it must be almost everywhere equal to a constant.  This is a closed subset of $W^{k,p}(\Omega)$ and is certainly not dense.


TITLE: Drawing large rectangle under concave curve
QUESTION [4 upvotes]: Let $f$ be a continuous concave  function on $[0,1]$ with $f(1)=0$ and $f(0)=1$. Does there exist a constant $k$ for which we can always draw a rectangle with area at least $k\cdot \int_0^1f(x)dx$, with sides parallel to the axes, in the area bounded by the two axes and the curve $f$?
If concavity is not required, it is possible to adapt from this example by using the curve $c/x$ to ensure that any rectangle has sufficiently small area. But with concavity, we know that $f$ lies above the line connecting the points $(0,1)$ and $(1,0)$, hence must have area at least $1/2$. If $f$ is exactly that line, then $k=1/2$ exactly. Otherwise, if $f$ is above the line, it looks like the rectangle will even get larger compared to the area under the curve.

REPLY [4 votes]: Let $t\in[0,1]$ be a value such that $f(x)\le f(t)$ for all $x\in[0,1]$. (Such a value exists since $f$ is unimodal; I don't think we need continuity for this.) The triangle formed by $(0,0)$, $(1,0)$ and $(t,f(t))$ lies under the curve. Its area is $\frac12f(t)$, and it contains an axis-parallel rectangle with half its area, $\frac14f(t)$. The area under the curve is at most $f(t)$. Thus $k=\frac14$ suffices. I'm not sure this is the best possible constant, though.


TITLE: Calculating correlation given a joint cumulative distribution function
QUESTION [0 upvotes]: I am interested in computing $Corr(X,Y)$ for $0 \leq x \leq 1$ and $0 \leq y \leq 1$, given two cases: firstly, $F_{X,Y}(x,y)=min(x,y)$ and secondly, $F_{X,Y}(x,y)=xy$.
How would I go about doing this? In the earlier part to this question, I have found that $X$ and $Y$ both follow the standard uniform distribution. I immediately thought of applying the correlation formula, as I can find all the expectations that I need, except for $E(XY)$ - how do I find that?
A worthy note is that this question is worth 6 points - 3 points for each case, so I personally do not think that my professor intended for us to use the correlation formula, as that would require calculations of $E(XY)$, $E(X)$, $E(Y)$, $E(X^2$), $E(Y^2)$, $E(X)^2$ and $E(Y)^2$. I feel like there is a more (un)intuitive approach to this question, which I clearly cannot see.

REPLY [1 votes]: The first one is simply the distribution fu nction 0f $(X,X)$ where $X\sim unif(0,1)$ and the second one is the distribution function of $(X,Y)$ where $X$ and $Y$ are i.i.d with $unif(0,1)$ distribution. So the covariance is $0$ in second case. In the first case it is $EX^{2}-(EX)^{2}=\frac 1 3 -\frac 1 4=\frac 1 {12}$.


TITLE: Invalid algebraic move in some proofs about impossibility of division by zero
QUESTION [1 upvotes]: I've come across a certain proof by contradiction at many places, textbooks and online likewise, which seems faulty to me. I'm no mathematician, but I just can't wrap my head around this issue.
The story goes like this:
Suppose dividing some integer by $0$ has some value:
$$a/0 = b$$
Now, first assume that $a$ is a non-zero integer. Then by multiplying both sides by $0$ we get:
$$a = b \cdot 0$$
but given the rules of arithmetic about multiplication with $0$ we arrive at a contradiction with our assumption that $a$ is non-zero.
Afterwards, assume that $a = 0$. Then, using the same step we get:
$$0 = b \cdot 0$$
which is unsatisfactory because it doesn't give a unique value, because for every $n$ it holds that $n  \cdot 0 = 0$.
The problem I have with this argument is that it somehow presupposes the very thing it argues against, i.e. that division by zero is defined (at least when looking at the case when zero is divided by zero). What else could justify the move from
$$a/0 = b$$
to
$$(a/0) \cdot 0 = b \cdot 0$$
to
$$a\cdot 0/0 = b \cdot 0$$
to
$$a = b \cdot 0$$
if not a presupposition that $0/0 = 1$? In no other way can we leave $a$ sitting there alone on the left-hand side.
On the other hand, the same applies for the second assumption. Namely, we can get
$$0 = b \cdot 0$$
only if we suppose that
$$0/0 \cdot 0 = b \cdot 0$$
resolves to
$$0 = b \cdot 0$$
which in this case works even if we don't set its value to $1$, because any $n$ would suffice, but for the sake of the argument suppose it's still set to that.
It seems to me that discarding this presupposition the two steps of the argument would collapse into the same
$$0/0 = b \cdot 0$$
which opens a door to an infinite regress because we're back at the starting point again, leaving us with multiplying by $0$ again and having
$$0 \cdot 0/0 = b \cdot 0 \cdot 0$$
which is again the same old left-hand
$$0/0 = b \cdot 0 \cdot 0$$
Therefore, the presented argument is invalid in its aim. Leaving
$$0/0 = b \cdot 0 \cdot 0$$
to be standardly resolved as
$$0/0 = 0$$
Now, where am I wrong and what to make of this whole business?

REPLY [1 votes]: The proper definition of $a/b$ is that it is the unique number you multiply $b$ by to get $a$.
Try to read that when $b=0$. Then you want the number that you multiply $0$ by to get $a$.
If $a \ne 0$ there is no such number.
If $a = 0$ any number will do for $0/0$. There is no unique number to define it as.
There is no need to invoke the rule that says
$$
a \times \frac{b}{c} = \frac{ab}{c}.
$$
(That rule follows from the definition of division.)


TITLE: Trouble with proof by induction
QUESTION [3 upvotes]: I am trying to prove that the following recursive function
$$f(n)=\begin{cases}1,\ n=0,\ \\ (2n-1)f(n-1),\ n>0,\end{cases}$$
can be expressed as
$$f(n)=\frac{(2n)!}{n!2^{n}}.$$
I know that this should be done by induction. It is easy to see that for $n=0$ the statement holds, i.e., $f(0)=1$. As for the induction step I have assumed that
$$f(n-1)=\frac{(2n-1)!}{(n-1)!2^{n-1}}$$
holds for some $n\in$ $\mathbb{N}$. From the recursive definition I have that $f(n)=(2n-1)f(n-1)$ in which I have tried to arrive at the desired statement but to no avail. I have tried calculating $(2n-1)f(n-1)$ in various ways but nothing seems to get me there. Furthermore, I am not sure whether this even is the right approach.
$\textbf{Side note:}$ In the problem it is not clear as to how $(2n)!$ is defined which also confuses me a bit. I am assuming it is defined as $(2n)!=2n(2n-1)!$ but it could also be $(2n)!=2n(2(n-1))!$ which obviously results in completely different answers.
Any hints would be highly appreciated. Thanks in advance.

REPLY [8 votes]: Suppose that$$f(n-1)=\frac{\bigl(2(n-1)\bigr)!}{(n-1)!2^{n-1}}=\frac{(2n-2)!}{(n-1)!2^{n-1}}.$$Then\begin{align}f(n)&=(2n-1)f(n-1)\\&=(2n-1)\frac{(2n-2)!}{(n-1)!2^{n-1}}\\&=\frac{(2n)(2n-1)(2n-2)!}{(2n)(n-1)!2^{n-1}}\\&=\frac{(2n)!}{n\times(n-1)!\times2\times2^{n-1}}\\&=\frac{(2n)!}{n!2^n}.\end{align}


TITLE: Is every odd positive integer of the form $P_{n+m}-P_n-P_m$?
QUESTION [17 upvotes]: I am looking for a comment, reference, remark, or proof of three conjectures as follows:
Conjecture 1: Let $x$ be an odd positive integer.  Then there exist two integers $n, m \ge 2$ so that  $$x=P_{n+m}-P_n-P_m,$$ where $P_n$ is the $n^{th}$ prime.

My calculations the conjecture 1 true for $x=1, 3,\dotsc, 10^7-1$ and $x = 5.4371349x10^7$, $\cdots, 5.4375349x10^7$
A general of the conjecture above as follows (but weaker):

Let every integer $r_0$ exist positive integer $x_0$ such that every odd number $x \ge x_0$  has the form $x=P_{n+m+r_0}-P_n-P_m$, where $m, n \ge 2$ (PS: inspired from the comment of Lev Borisov be low)

Or simpler:

Conjecture 2: Is every odd positive integer of the form $P_{c}-P_a-P_b$ ?

See also:

Goldbach's weak conjecture

A stronger comjecture 2

Conjecture 3 (Maillet's conjecture) Is every even positive integer of the form $P_{a}-P_b$

I have just computed the conjecture 3 is true with $x=2, 4, \cdots, 10^6$ and $3873$ numbers $x = $$9.82197492x10^8$$,$ $\cdots$$,$ $9.82226054x10^8$
Example:
$2=5-3$;
$4=7-3$;
$6=13-7$;
$8=13-5$;
$10=17-7$
See also:

Goldbach's strong conjecture

REPLY [2 votes]: This isn't an answer to conjecture 1, just an elaboration of things others mentioned.
There is every reason to think that the answer to conjecture 1 is yes and even that for fixed $m \gt 1$ we have that for each  odd integer $x \geq 3$ there are infinitely many $n$ with $p_{n+m}-p_n-p_m=x.$ We don't know that there is even one, however we can say with good precision how the number of solutions should be expected to grow as $n \rightarrow \infty.$ Computations always (seem) to conform to this with good fidelity as far as checked. I'll explain that a bit and then point out that that fixing $m=2$ would likely not be the most fruitful way to find a solution.
So with $m=3$ and $p_3=5$ solving $p_{n+3}-p_n-5=15$ amounts to finding two primes $p_N$ and $p_n$ with

$p_N-p_m=20$
$N-m=3.$

We do not know that $p_N-p_m=20$ infinitely often. For each even integer g define $\pi_g(X)$ to be the number of pairs $(p_N,p_n)$ with $p_N-p_n=g$ and $p_N \leq X.$ Then $\pi_2(X)$ is the number of twin primes up to $X$ We don't know that this grows without bound but can expect that it is assymptotic (in some sense which could be made precise) to $C_2\frac{X}{\ln^2X}$ and that the constant $C_2=\prod_p(1-\frac1{(p-1)^2})$ with the product over the odd primes.
There is a similar constant $C_g$ for each even $g.$ Namely $C_g=C_2\prod_p\frac{p-1}{p-2}$ where the product is over odd primes which  divide $g.$
So $p_n+20$ is expected to be prime more often that $p_2$  but about as often as $p+10.$ i.e. $\pi_{20}(X) \sim \pi_{10}(X)\sim \frac{4}{3}\pi_2(X).$ That is a prediction which holds up quite well as far as checked.
For an amazing exposition of this read Heuristic Reasoning In The Theory of Numbers by Polya.
But our goal had an extra condition. Want something like to have
$p_n,p_n+6,p_n+14,p_n+20$ to all be prime but $p_n+h$ be composite for $h=2,4,8,10,12,16,18.$ It is possible to similarly predict how often that happens up to $X.$ Summing over finite number of cases would give a prediction for the number of solutions of the given problem.
$m=2$ is a little easier but I wanted to use a different value.
But where is it most fruitful to look for solutions  to $p_{n+m}-p_n-p_m=x$?
Here is a graph for $p_{n+m}-p_n-p_m=101$ with $n+m\lt 200.$
I find $168$ solutions. here is a graph.

Of the $168$ solutions, $61$ of them have $m \in\{34,35,36,37,38\}$ and the ratio $\frac{n}{m}$ ranges from $3$ to $5.2$
Using $p_k \sim {k}{\ln{k}}$ one might be able to argue that for fixed $r=\frac{n}{m}$ ( really $r$ in some small range like $[3,5]$) there is a narrow range of $m$ values that would be worth searching first. Perhaps the best $r$ (given $x$) could be estimated. I would have guessed $r \sim 1$ is best, but that seems not to be the case based on this one computation. Perhaps the optimal range is past $m+n=200.$