text:
\section{Varieties with ample  cotangent bundle}\label{SectionAmplitude}
\subsection{Statements}
In this section we prove our main application of the results of Section  \ref{MainSection}. Our statement  is the following partial result towards Debarre's conjecture (denoted by Theorem \ref{theoD} in the introduction).

\begin{theorem}\label{Amplitude}
Let $N,c,e\in\bN$ such that $N\geqslant 2$, $c\geqslant \frac{3N-2}{4}$ and $e\geqslant 2N+3$. If $X\subseteq \bP^N$ be a general complete intersection of codimension $c$ and multidegree $(e,\dots, e)$, then $\Om_X$ is ample.
\end{theorem}
\begin{remark}Observe that the condition $c\geqslant \frac{3N-2}{4}$ can be rephrased by $\codim_{\bP^N}(X)\geqslant 3\dim(X)-2$. \end{remark}
Since ampleness is an open condition (see for instance Theorem 1.2.17 in \cite{Laz04I} and Proposition 6.1.9 in \cite{Laz04II}), it suffices to construct one example of a smooth complete intersection variety  with ample cotangent bundle satisfying the hypothesis of the theorem to prove that the result holds for a general complete intersection variety. We will construct such an example by considering deformations of Fermat type complete intersection varieties. To do so, let us introduce some notation.

Fix $N\geqslant 2$, $\ep \in \bN$ and $e\in \bN^*$. Set $\bA_\ep:=\bC[Z_0,\dots,Z_N]_\ep.$ For any $s=(s_0,\dots,s_N)\in \bA_\ep^{\oplus N+1}$ we set:
\begin{eqnarray}
F_s(Z)=F(s,Z):=\sum_{i=0}^Ns_iZ_i^e.
\end{eqnarray}
This is a homogeneous equation of degree $e_0:=\ep+e$ and for a general choice of $s,$ it defines a smooth hypersurface in $\bP^N$ which we will denote by $X_s$.
For any $m,a\in \bN$, and any smooth variety $X\subseteq \bP^N$, we will write $\bL_X:=\cO_{\bP(\Om_X)}(1)$ and $\bL_X^m(-a):=\bL_X^{\otimes m}\otimes \pi_X^*\cO_X(-a).$ With those notation we have the following.
\begin{theorem}\label{AmplitudeExemple} Let $N,c,e,\ep,a\in\bN$ such that $N\geqslant 2$, $c\geqslant \frac{3N-2}{4}$ and set $n:=N-c$. Suppose that  $\ep\geqslant 1$  and that $e\geqslant N+1+a+N\ep$. Then for any $0\leqslant j\leqslant N$ there exists $s^j\in \bA_\ep^{\oplus N+1}$ such that $X:=X_{s^1}\cap \cdots \cap X_{s^c}$ is a smooth complete intersection variety such that $\bL_X^n(-a)$ is nef.
\end{theorem}
Let us first prove that this result implies Theorem \ref{Amplitude}.
\begin{proof}[Proof of Theorem \ref{AmplitudeExemple} $\Rightarrow$ Theorem \ref{Amplitude}] Take $a=\ep=1$, $e\geqslant N(1+\ep)+a+1=2N+2$ (this is equivalent to $e_0\gs 2N+3$) and $c\geqslant \frac{3N-2}{4}$. Then Theorem \ref{AmplitudeExemple} implies that there exists a smooth complete intersection variety $X\subset\bP^N$ of codimension $c$ and multidegree $(e,\dots,e)$ such that $\bL^n_X(-1)$ is nef. But this implies that $\bL^n_X$ is ample and therefore that $\bL_X$ is ample, and thus, $\Om_X$ is ample. But since ampleness is an open condition in families, we deduce that a general complete intersection variety in $\bP^N$ of codimension $c$ and multidegree $(e,\dots,e)$ has ample cotangent bundle.
\end{proof}
The proof of Theorem \ref{AmplitudeExemple} is an induction based on the following technical lemma.
\begin{lemma}\label{EliminationLemma}  Let $N,c,e,\ep,a\in\bN$ such that $N\geqslant 2$, $c\geqslant \frac{3N-2}{4}$, $\ep\geqslant 1$ and $e\geqslant N+1+a+N\ep$. Set $n:=N-c$. For any $0\leqslant j\leqslant N$ take $s^j\in \bA_\ep^{\oplus N+1}$ such that $X:=X_{s^1}\cap \cdots \cap X_{s^c}$ is a smooth complete intersection variety.
 For any $i\in \{0,\dots,N\}$, set $H_i:=(Z_i=0)$ and $W_i:=X\cap H_i$. We look at $\bP(\Om_{W_i})$ as a subvariety of $\bP(\Om_X)$. Then, for a general choice of  $s^1,\dots,s^c$, there exists $E\subseteq \bP(\Om_X)$, such that $\dim E=0$ and such that   $$\bB(\bL_X^n(-a))\subseteq\bigcup_{i=0}^N \bP(\Om_{W_i})\cup E$$
\end{lemma}
The proof of Lemma \ref{EliminationLemma} is the content of Section \ref{ExpliciteDescription} and Section \ref{EliminationSection}. But for now let us explain how to obtain  Theorem \ref{AmplitudeExemple} from Lemma \ref{EliminationLemma}.
\begin{proof}[Proof of Theorem \ref{AmplitudeExemple}] Let $H$ be a hyperplan section of $X$. Fixing $c\geqslant \frac{3N-2}{4}$, the proof is an induction on $n=\dim X$ (or equivalently on $N$). If $n=1$, then $X$ is a curve. By the adjunction formula, $K_X=\cO_{X}(c(e+\ep)-N-1)$. Therefore $\Om_X(-a)=K_X(-a)$ is nef. \\
Now suppose that $n\geqslant 2$. We have to prove that for any irreducible curve $C\subseteq \bP(\Om_X)$, $C \cdot \bL_X^n(-a)\geqslant 0$. Certainly, if $C\not\subset \bB(\bL_X^n(-a))$, $C\cdot \bL_X^n(-a)\geqslant 0$. Now suppose that $C\subseteq \bB(\bL_X^n(-a))$, from Lemma \ref{EliminationLemma} we know that $\bB(\bL_X^n(-a))\subseteq \bigcup_{i=0}^N \bP(\Om_{W_i})\cup E$ for some zero-dimensional set $E$. Therefore,  there exists $i\in \{0,\dots, N\}$ such that $C\subseteq \bP(\Om_{W_i})$. But on can view $W_i$ as a codimension $c$ complete intersection variety in $H_i\cong \bP^{N-1}$ defined by equations of the same type than $X$ (with one less variable). Observe moreover that $\bL_X(-a)|_{\bP(W_i)}= \bL_{W_i}(-a)$ and that if $c\geqslant \frac{3N-2}{4}$, one has $c\geqslant\frac{3(N-1)-2}{4}$. From our  induction hypothesis we therefore  know that $\bL_{W_i}^{n-1}(-a)$ is nef.
Thus, $\bL_{W_i}^{n-1}(-a)\cdot C\gs 0$ and in particular $(n-1)\bL_{W_i}\cdot C\gs a\pi_X^*H\cdot C.$
Hence,
$$n\bL_X\cdot C=\frac{n}{n-1}(n-1)\bL_{W_i}\cdot C\gs \frac{n}{n-1}a\pi_X^*H\cdot C\gs a\pi^*H\cdot C.$$
And finally $\bL_X^n(-a)\cdot C\gs 0.$
\end{proof}

\subsection{Constructing symmetric differential forms}\label{ExpliciteDescription} To prove Lemma \ref{EliminationLemma} we first need to construct sufficiently many symmetric differential forms   on complete intersection varieties of the above type, this is the purpose of this section. The setting is the following. We fix $N,c,\ep,e,a,r,n\in \bN$ such that $N\geqslant 2$, $N\geqslant c\geqslant\frac{N}{2}$, $a\geqslant 0$, $\ep\geqslant 0$, $e\geqslant a+N(\ep+1)+1$, $r=e-1$ and $n=N-c.$ For any $1\leqslant j\leqslant c$, we take $s^{j}\in \bA_\ep^{N+1}$ such that for any $1\leqslant p\leqslant c$ and for any $I:=(i_1,\dots,i_p)\in \{1,\dots,c\}_{\neq}^p$, the set $X^I:=X_{s^{i_1}}\cap\cdots\cap X_{s^{i_p}}$ is a smooth complete intersection variety. Of course, this last condition holds if the $s^{i_j}$'s are general. We also denote $X:=X^{(1,\dots,c)}$. For any $0\leqslant i\leqslant N$, any  $1\leqslant j\leqslant c$ and any $v\in \bA_\ep$, we set $a_i(v):=Z_iv$ and $\al_i(v):=Z_idv+evdZ_i.$ So that :
\begin{eqnarray}F({s^{j}},Z)=\sum_{i=0}^N a_i(s_i^{j})Z_i^r\ \ \ \text{and}\ \ \ dF(s^j,Z):=dF_{s^{j}}(Z)=\sum_{i=0}^N \al_i(s_i^{j})Z_i^r\label{EquationFermat}\end{eqnarray}
For any $i\in \{0,\dots,n\}$ set $U_i:=(Z_i\neq 0)$, $\mathfrak{U}=(U_i)_{0\leqslant i\leqslant N}$ and $\mathfrak{U}_X:=(U_i\cap X)_{0\leqslant i\leqslant N}$, $\mathfrak{U}_X$ is an open covering of $X$. Our first result is the following.
\begin{lemma}\label{TildeDiffForms}
With the above notations, for any $I=(i_1,\dots,i_n)\in \{1,\dots,c\}^n_{\neq}$, and for any degree $e-a-N\ep-N-1$ homogeneous polynomial $P\in\bC[Z_0,\dots,Z_N],$ there is a non-zero element $\tom^{I,P}\in H^0(X,S^n\tOm_X(-a))$ such that, when computed in \v{C}ech cohomology one has  $\tom^{I,P}=(\tom_0^{I,P},\dots, \tom_N^{I,P})\in \check{H}^0(\fU_X,S^n\tOm_X(-a))$ with
$$\tom_j^{I,P}=\frac{(-1)^j P}{Z_i^r}\left|\begin{array}{cccccc}a_0(s_0^1)&\cdots&a_{j-1}(s_{j-1}^1)&a_{j+1}(s_{j+1}^1)&\cdots&a_N(s_N^1)\\ \vdots&&\vdots&\vdots&&\vdots\\ a_0(s_0^c)&\cdots&a_{j-1}(s_{j-1}^c)&a_{j+1}(s_{j+1}^{c})&\cdots&a_N(s_N^{c})\\ \al_0(s_0^{i_1})&\cdots&\al_{j-1}(s_{j-1}^{i_1})&\al_{j+1}(s_{j+1}^{i_1})&\cdots&\al_N(s_N^{i_1})\\ \vdots&&\vdots&\vdots&&\vdots\\ \al_0(s_0^{i_n})&\cdots&\al_{j-1}(s_{j-1}^{i_n})&\al_{j+1}(s_{j+1}^{i_n})&\cdots&\al_N(s_N^{i_n})\end{array}\right|.$$
\end{lemma}
\begin{proof} For simplicity, for any $i\in \{1,\dots,N\}$ and any $j\in \{1,\dots,c\}$, we write $a_i^j:=a_i(s_i^j)$, $\al_i^j:=\al_i(s_i^j)$ and $F_j=F_{s^j}.$ As in the statement, take $I=(i_1,\dots,i_n)\in \{1,\dots,c\}^n_{\neq}$,  and take $(i_{n+1},\dots,i_c)\in \{1,\dots,c\}^{c-n}_{\neq}$ such that $\{i_1,\dots,i_c\}=\{1,\dots,c\}.$  For any $j\in \{0,\dots, c\}$ write $I_j=(i_1,\dots, i_j).$

We can apply Theorem  \ref{LimiteSetting} and Theorem \ref{LimitePair} to the simple $\lambda$-setting $\Si=(X,\lambda^0,\dots,\lambda^c)$ where $\lambda^i=\varnothing$ if $i\neq n$ and $\lambda^n=n$ (note that $q(\Si)=0$). Here the order in which we intersect the different hypersurfaces to obtain $X$ is crucial. We obtain an injection
$$H^0\left(X,S^n\tOm_{X^{I_n}}|_X\otimes \cO_{X}(-a)\right)\stackrel{\tilde{\varphi}}{\rightarrow}H^N\left(\bP^N,\cO_{\bP^N}(-a-Ne_0)\right)$$
such that \begin{eqnarray}\im(\tphi)=\left(\bigcap_{j=n+1}^c\ker(\cdot F_{{i_j}})\right)\bigcap\left(\bigcap_{i=1}^n\ker(\cdot F_{i_j})\cap\ker(\cdot dF_{i_j})\right).\label{description}\end{eqnarray}
Observe that our hypothesis on the degree ensures that $e-a-N\ep-N-1\geqslant 0$. Fix any degree $e-a-N\ep-N-1$ polynomial $P$. Then we get a well defined element $$\tom^{I,P}_{0,\dots, N}:=\frac{P}{Z_0^r\cdots Z_N^r}\in \check{H}^N\left(\fU,\cO_{\bP^N}(-a-Ne_0)\right).$$
From \eqref{EquationFermat} and \eqref{description} we see that $\tom^{I,P}_{0,\dots, N}\in \im(\tphi)$. To obtain the explicit description of $\tphi^{-1}(\tom^{I,P}_{0\cdots N})\in  \check{H}^0\left(\fU_X,S^n\tOm_{X^{I_n}}|_X(-a)\right)$ we have to describe explicitly the inclusion $\tphi$, to do so, we have to unravel the proof of Theorem \ref{LimiteSetting} in our situation. This inclusion is described in Proposition \ref{ChainInclusion}, applying this proposition in our situation, we see that $\tphi$ is obtained as the composition of the following chain of inclusions:
\begin{eqnarray*}
H^0(X,S^n\tOm_{X^{I_n}}|_X(-a))&\hookrightarrow& H^1\left(X,S^{n-1}\tOm_{X^{I_{n-1}}}|_{X}(-a-e_0)\right)\\
&\hookrightarrow & H^2\left(X,S^{n-2}\tOm_{X^{I_{n-2}}}|_{X}(-a-2e_0)\right)\\
&\vdots&\ \ \ \ \ \ \ \ \ \ \vdots\\
&\hookrightarrow & H^{n-1}(X,\tOm_{X^{I_1}}|_{X}(-a-(n-1)e_0))\\
&\hookrightarrow & H^{n}(X,\cO_X(-a-ne_0))\\
&\hookrightarrow & H^{n+1}(X^{I_{c-1}},\cO_{X^{I_{c-1}}}(-a-(n+1)e_0))\\
&\vdots&\ \ \ \ \ \ \ \ \ \ \vdots\\
&\hookrightarrow & H^{N}(\bP^N,\cO_{\bP^N}(-a-Ne_0))\\
\end{eqnarray*}
For any $0\leqslant \ell\leqslant N$, we denote by $\tom^{\ell}$ element in the $H^\ell$ group in the above chain of inclusion whose image in $H^N(\bP^N,\cO_{\bP^N}(-a-N(e+\ep)))$ under the above inclusions is $\tom^{I,P}.$ Moreover, for any $0\leqslant\ell\leqslant N$ and for any $K=(k_0,\dots,k_\ell)\in \{0,\dots,N\}_{<}^{\ell+1}$ if we let $ K^\circ=(k_{\ell+1},\dots,k_N)\in \{0,\dots,N\}^{N-\ell}_<$ such that $\{k_0,\dots,k_N\}=\{0,\dots,N\}$, we set $\ep(K)$ to be the signature of the following permutation
$$\si_K:=\left(\begin{array}{cccc}0&1&\cdots & N\\ k_0&k_1&\cdots&k_N\end{array}\right),$$
and for any $m\in \{1,\dots,c\}$, we  set $L^{m}_K:=(a^{m}_{k_{\ell+1}},\dots,a^{m}_{k_{N}})$ and $\La^{m}_K:=(\al^{m}_{k_{\ell+1}},\dots,\al^{m}_{k_{N}}).$ We also set $\bX_K:=\frac{1}{Z_{k_0}^r\cdots Z_{k_\ell}^r}$.\\

Let $\ell\in \{0,\dots,N\},$ we are going to prove by induction that $\tom^{\ell}$ can be represented in \v{C}ech cohomology by the following cocycle $(\tom^\ell_K)_{K\in \{0,\dots, N\}^{\ell+1}_{<}}:$
\begin{enumerate}
\item If $n\leqslant \ell\leqslant N$, write $m:=N-\ell$, then $\tom^\ell_K=(-1)^{N(N-\ell+1)}\ep(K)P\left|\begin{array}{c}L_K^{i_1}\\ \vdots\\ L_K^{i_m}\end{array}\right|\bX_K.$
\item If $0\leqslant \ell\leqslant n$, write $m:=n-\ell$, then $\tom^\ell_K=(-1)^{N(N-\ell+1)}\ep(K)P\left|\begin{array}{c}L_K^{i_1}\\ \vdots\\ L_K^{i_c}\\ \La_K^{i_1}\\ \vdots \\ \La_K^{i_{m}}\end{array}\right|\bX_K.$
\end{enumerate}
Of course, if we look at $\ell=0$ in the last case, up to a sign, we get the description announced in the statement of the lemma, and it suffices to look at $\tom^{I,P}$ as an element in $H^0(X,S^n\tOm_X(-a))$ by considering its image under the natural restriction map $H^0(X,S^n\tOm_{X^{I_n}}|_X(-a))\to H^0(X,S^n\tOm_X(-a)).$ \\

We start by computing a \v{C}ech cocycle for $\tom^{N-1}$. Recall that one has the twisted restriction short exact sequence
$$0\to \cO_{\bP^N}(-a-Ne_0)\stackrel{\cdot F_{i_1}}{\rightarrow} \cO_{\bP^N}(-a-(N-1)e_0)\to \cO_{X_{i_1}}(-a-(N-1)e_0)\to 0, $$
which yields the following in cohomology:
$$0\to H^{N-1}(X_{i_1},\cO_{X_{i_1}}(-a-(N-1)e_0))\stackrel{\delta_{\cdot F_{i_1}}}{\rightarrow} H^N(\bP^N, \cO_{\bP^N}(-a-Ne_0))\stackrel{\cdot F_{i_1}}{\rightarrow} H^N(\bP^N,\cO_{\bP^N}(-a-(N-1)e_0)). $$
By the very definition of $\tom^{N-1}$, we have $\delta_{\cdot F_{i_1}}(\tom^{N-1})=\tom^N=\tom^{I,P}$.
As explained in Section \ref{Cech}, to compute a cocycle for $\tom^{N-1}$, we compute $\tom^{N}_{0,\dots, N}\cdot F_{i_1}$ and write it as the \v{C}ech differential of the desired cocycle. We have
\begin{eqnarray*}
\tom^N_{0,\dots, N}\cdot F_{i_1}&=&P\cdot \bX_{0,\dots,N}\cdot \sum_{j=0}^Na_j^{i_j}Z_j^r\\
&=& P\cdot \sum_{j=0}^Na_j^{i_j}\bX_{0,\dots,\hat{j},\dots,N}= P\cdot \sum_{j=0}^N(-1)^j\left((-1)^j(-1)^{N(N-1+1)}a_j^{i_j}\bX_{0,\dots,\hat{j},\dots,N}\right)
\end{eqnarray*}
If we write $K=(0,\dots,\hat{j},\dots,N)$, we have $\ep(K)=(-1)^j$ and $L^{i_1}_{K}=(a^{i_1}_j)$ and $\bX_K=\bX_{0,\dots,\hat{j},\dots,N}$. Therefore, we see that $\tom^N_{0,\dots, N}\cdot F_{i_1}=\check{d}\left((\tom_K^{N-1})_{K\in\{0,\dots,N\}^{N}_<}\right)$ where as announced $$\tom_K^{N-1}=(-1)^{N(N-1+1)}\ep(K)P|L_K^{i_1}|\bX_K.$$
We will only treat one case of the induction, the other case is treated the exact same way. Let $n< \ell\ls N$, let $m=N-\ell$ and suppose that we know that $\tom^\ell$ can be represented by a cocycle $(\tom_K^\ell)_{K\in \{0,\dots,N\}^{\ell+1}_<}$ as in the case $1$. We have to prove that $\tom^{\ell-1}$ can be represented by a cocycle of the form described in case $1.$ We have $$\tom^\ell\in H^\ell\left(X^{I_{m}},\cO_{X^{I_m}}(-a-\ell e_0)\right),$$
and we have the following short exact sequence
\begin{eqnarray*}
0\to \cO_{X^{I_m}}(-a-\ell e_0)& \stackrel{\cdot F_{i_{m+1}}}{\to}& \cO_{X^{I_m}}(-a-(\ell-1)e_0) \to \cO_{X^{I_{m+1}}}(-a-(\ell-1)e_0)\to 0.
\end{eqnarray*}
In cohomology, it yields an injection (which is precisely the one appearing in the above chain of inclusions)
$$H^{\ell-1}\left(X^{I_{m+1}},\cO_{X^{I_{m+1}}}(-a-(\ell-1)e_0)\right)\stackrel{\delta_{\cdot F_{i_{m+1}}}}{\hookrightarrow} H^{\ell}\left(X^{I_{m}},\cO_{X^{I_{m}}}(-a-\ell e_0)\right)$$
To compute a \v{C}ech cohomology cocycle of $\om^{\ell-1}$, we proceed as before. For any $K\in \{0,\dots, N\}_<^{\ell+1}$ we are going to compute $\om_K^\ell\cdot F_{i_{m+1}}$ modulo $F_{i_1},\dots,F_{i_m}.$
Fix $K=(k_0,\dots,k_\ell)\in\{0,\dots,N\}_<^{\ell+1}$ and as usual, take $K^\circ=(k_{\ell+1},\dots,k_N)\in\{0,\dots,N\}^{N-\ell}_<$ such that $K\cup K^\circ=\{0,\dots,N\}.$ Because we are working modulo $F_{i_1},\dots,F_{i_m},$ we have
$$\left\{\begin{array}{ccc}F_{i_1}&=&0\\ \vdots &&\vdots \\ F_{i_m}&=&0 \\ F_{i_{m+1}}-F_{i_{m+1}}&=&0\end{array}\right.\Leftrightarrow \left\{\begin{array}{ccc}\sum_{j=\ell+1}^Na_{k_j}^{i_1}Z_{k_j}^r&=&-\sum_{j=0}^\ell a_{k_j}^{i_1}Z_{k_j}^r\\ \vdots &&\vdots \\ \sum_{j=\ell+1}^Na_{k_j}^{i_m}Z_{k_j}^r&=&-\sum_{j=0}^\ell a_{k_j}^{i_m}Z_{k_j}^r \\ \sum_{j=\ell+1}^N a_{k_j}^{i_{m+1}}Z_{k_j}^r-F_{i_{m+1}}&=&-\sum_{j=0}^\ell a_{k_j}^{i_{m+1}}Z_{k_j}^r\end{array}\right.$$
We may write this as follows:
$$\left(\begin{array}{cccc}a^{i_1}_{k_{\ell+1}}&\cdots&a^{i_1}_{k_N}&0\\  \vdots & & \vdots & \vdots \\ a^{i_m}_{k_{\ell+1}}&\cdots&a^{i_m}_{k_N}&0\\ a^{i_{m+1}}_{k_{\ell+1}}&\cdots&a^{i_{m+1}}_{k_N}&-1\end{array}\right)\left(\begin{array}{c}Z^r_{k_{\ell+1}}\\ \vdots \\ Z^r_{k_N}\\ F_{i_{m+1}}\end{array}\right)=\left(\begin{array}{c}-\sum_{j=0}^\ell a_{k_j}^{i_1}Z_{k_j}^r\\ \vdots \\ -\sum_{j=0}^\ell a_{k_j}^{i_m}Z_{k_j}^r \\ -\sum_{j=0}^\ell a_{k_j}^{i_{m+1}}Z_{k_j}^r\end{array}\right).$$
By a simple application of Cramer's rule we obtain
$$F_{i_m}\left|\begin{array}{cccc}a^{i_1}_{k_{\ell+1}}&\cdots&a^{i_1}_{k_N}&0\\ \vdots & & \vdots & \vdots \\ a^{i_m}_{k_{\ell+1}}&\cdots&a^{i_m}_{k_N}&0\\ a^{i_{m+1}}_{k_{\ell+1}}&\cdots&a^{i_m}_{k_N}&-1\end{array}\right|=\left|\begin{array}{cccc}a^{i_1}_{k_{\ell+1}}&\cdots&a^{i_1}_{k_N}&-\sum_{j=0}^\ell a_{k_j}^{i_1}Z_{k_j}^r\\ \vdots & & \vdots & \vdots \\ a^{i_m}_{k_{\ell+1}}&\cdots&a^{i_m}_{k_N}&-\sum_{j=0}^\ell a_{k_j}^{i_m}Z_{k_j}^r\\ a^{i_{m+1}}_{k_{\ell+1}}&\cdots&a^{i_{m+1}}_{k_N}&-\sum_{j=0}^\ell a_{k_j}^{i_{m+1}}Z_{k_j}^r\end{array}\right|$$
which yields
$$\left|\begin{array}{c}L_K^{i_1}\\  \vdots \\ L^{i_m}_K\end{array}\right|F_{i_{m+1}}=\sum_{j=0}^\ell \left|\begin{array}{cccc}a^{i_1}_{k_{\ell+1}}&\cdots&a^{i_1}_{k_N}&a_{k_j}^{i_1}\\  \vdots & & \vdots & \vdots \\ a^{i_{m+1}}_{k_{\ell+1}}&\cdots&a^{i_{m+1}}_{k_N}&a_{k_j}^{i_{m+1}}\end{array}\right|Z_{k_j}^r.$$
We just have to be a bit careful with the signs. For any $j\in \{0,\dots, \ell\}$ we set $K_j=K\setminus \{k_j\}$. We also take $\nu_j\in \{\ell+1,\dots, N-1\}$ such that $k_{\nu_j}<k_j<k_{\nu_j+1}$ (if $k_j< k_{\ell+1} $, we just take $\nu_j=\ell$). With this notation, we have $K^\circ_j=(k_{\ell+1},\dots, k_{\nu_j},k_j,k_{\nu_j+1},\dots, k_N)$ (or $K^\circ_j=(k_j,k_{\nu_j+1},\dots, k_N) $ if $k_j<k_{\ell+1}$). A straightforward computation shows that $\si_{K_j}=(k_j,k_{j+1},\dots, k_{\nu_j})^{-1}\circ \si_K$, hence $\ep(K_j)=(-1)^{\nu_j-j}\ep(K).$ On the other hand, observe that
$$\left|\begin{array}{cccc}a^{i_1}_{k_{\ell+1}}&\cdots&a^{i_1}_{k_N}&a_{k_j}^{i_1}\\  \vdots & & \vdots & \vdots \\ a^{i_{m+1}}_{k_{\ell+1}}&\cdots&a^{i_{m+1}}_{k_N}&a_{k_j}^{i_{m+1}}\end{array}\right|

=(-1)^{N-\nu_j}\left|\begin{array}{c}L_{K_j}^{i_1}\\  \vdots \\ L_{K_j}^{i_{m+1}} \end{array}\right|.$$
Now we are ready for our computation:
\begin{eqnarray*}
\om^\ell_{K}\cdot F_{i_{m+1}}&=&(-1)^{N(N-\ell)}\ep(K)P\bX_K\left|\begin{array}{c}L_K^{i_1}\\  \vdots \\ L^{i_{m}}_K\end{array}\right|F_{i_{m+1}}= (-1)^{N(N-\ell)}\ep(K)P\bX_K\sum_{j=0}^\ell (-1)^{N-\nu_j}\left|\begin{array}{c}L_{K_j}^{i_1}\\ \vdots \\ L^{i_{m+1}}_{K_j}\end{array}\right|Z_{k_j}^r\\
&=&(-1)^{N(N-\ell)}P\sum_{j=0}^\ell (-1)^{N}\ep(K)(-1)^j(-1)^{j-\nu_j}\left|\begin{array}{c}L_{K_j}^{i_1}\\ \vdots \\ L^{i_{m+1}}_{K_j}\end{array}\right| \bX_{K_j}\\
&=&\sum_{j=0}^\ell(-1)^{N(N-\ell+1)}\ep(K_j)(-1)^jP\left|\begin{array}{c}L_{K_j}^{i_1}\\  \vdots \\ L^{i_{m+1}}_{K_j}\end{array}\right| \bX_{K_j}.
\end{eqnarray*}
Hence $\left(\tom^\ell\cdot F_{i_{m+1}}\right)_K=\check{d}\left(\left(\tom^{\ell-1}_{K'}\right)_{K'\in\{0,\dots,N\}^\ell_<}\right),$ where for any $K'\in\{0,\dots,N\}_<^{\ell}$, $$\om^{\ell-1}_{K'}=(-1)^{N(N-\ell+1)}\ep(K')P\left|\begin{array}{c}L_{K'}^{i_1}\\  \vdots \\ L^{i_{m+1}}_{K'}\end{array}\right| \bX_{K'}$$ as announced.
\end{proof}
The last step is to use the tilde-symmetric differential forms constructed in Lemma \ref{TildeDiffForms} to get usual symmetric differential forms. Recall that in our situation, from the exact sequence
$$0\to S^n\Om_X(-a)\to S^n\tOm_X(-a)\to S^{n-1}\tOm_X(-a)\to 0$$
and the vanishing $H^0\left(X,S^{n-1}\tOm_X(-a)\right)$ (deduced from Lemma \ref{VanishingLemma}), we get an isomorphism
$$H^0\left(X,S^n\Om_X(-a)\right)\cong  H^0\left(X,S^n\tOm_X(-a)\right).$$
Therefore we just have to compute the image $\om^{I,P}$ of $\tom^{I,P}$ under this isomorphism. To do this we just have to dehomogenize $\tom^{I,P}.$ For simplicity, we only work in the $U_0=(Z_0\neq 0)$ chart. We use the following notation: for any $i\in \{1,\dots,N\}$, $z_i=\frac{Z_1}{Z_0}$, so that $dZ_i=Z_0dz_i+z_idZ_0.$ Let us make an elementary observation. Take $\ell\gs 1$, and let $G\in \bC[Z_0,\dots,Z_N]$ be a homogeneous degree $\ell$ polynomial. Let $g$ be the dehomogenized of $G$ with respect to $Z_0$. Then we have $$dG=\ell Z_0^{\ell-1}gdZ_0+Z_0^\ell dg.$$ Indeed,
\begin{eqnarray*}
Z_0dG&=&\sum_{i=0}^NZ_0\frac{\partial G}{\partial Z_i}dZ_i = Z_0\frac{\partial G}{\partial Z_0}dZ_0+\sum_{i=1}^N\frac{\partial G}{\partial Z_i}(Z_0^2dz_i+Z_idZ_0)\\
&=& \sum_{i=0}^N\frac{\partial G}{\partial Z_i}Z_idZ_0+Z_0^2\sum_{i=1}^N\frac{\partial G}{\partial Z_i}dz_i=\ell GdZ_0+Z_0^2\sum_{i=1}^NZ_0^{\ell-1}\frac{\partial g}{\partial z_i}dz_i= \ell Z_0^{\ell}gdZ_0+Z_0^{\ell+1} dg.
\end{eqnarray*}
We introduce some more notation. For any $u\in \bA_\ep\cong\bC[z_1,\dots,z_N]_{\ls\ep}$ and any $q\in \{1,\dots,N\}$  we set
 \begin{eqnarray*}
 b_q(u):=z_qu \ \ \ \text{and} \ \ \ \be_q(u):=z_qdu+eudz_q.
 \end{eqnarray*}
For any $1\ls j\ls c$ and for any $0\ls i\ls N$ we let $t_i^{j}$ (resp. $f_{j}$) to be the dehomogenized of $s_{i}^{j}$ (resp. $F_j$) with respect to $Z_0$.
Therefore we have $a_{i}(s_i^j)=Z_is_i^{j}=Z_0^{\ep+1}z_it_i^{j}=Z_0^{\ep+1}b_i(t_i^j)$ and
\begin{eqnarray*}
\al_i(s_i^j)&=& Z_ids_i^{j}+es_i^{j}dZ_i=Z_0z_i(\ep Z_0^{\ep-1}t_i^{j}dZ_0+Z_0^\ep dt_i^{j})+eZ_0^\ep t_i^{j}(z_idZ_0+Z_0dz_i)\\
&=& e_0Z_0^\ep z_it_i^{j}dZ_0+Z_0^{\ep+1}(z_idt_i^{j}+et_i^{j}dz_i)=e_0Z_0^\ep b_i(t_i^j)dZ_0+Z_0^{\ep+1}\be_i(t_i^j).
\end{eqnarray*}
Therefore, by the elementary properties of the determinant we get
$$\tom^{I,P}_0=\frac{P}{Z_0^r}\left|\begin{array}{ccc}a_1(s_{1}^{1})&\cdots&a_N(s_N^{1})\\ \vdots&&\vdots\\ a_1(s_{1}^{c})&\cdots&a_N(s_N^{c})\\ \al_1(s_{1}^{i_1})&\cdots&\al_N(s_N^{i_1})\\ \vdots&&\vdots\\ \al_1(s_{1}^{i_n})&\cdots&\al_N(s_N^{i_n})\end{array}\right|=PZ_0^{N(\ep+1)-r}\left|\begin{array}{ccc}b_1(t_{1}^{1})&\cdots&b_N(t_N^{1})\\ \vdots&&\vdots\\ b_1(t_{1}^{c})&\cdots&b_N(t_N^{c})\\ \be_1(t_{1}^{i_1})&\cdots&\be_N(t_N^{i_1})\\ \vdots&&\vdots\\ \be_1(t_{1}^{i_n})&\cdots&\be_N(t_N^{i_n})\end{array}\right|.$$
This completes the proof of the following.
\begin{proposition}\label{DiffForms} With the above notation. Take a homogenous polynomial $P\in \bC[Z_0,\dots,Z_N]$ of degree $e-a-N\ep-N-1$ and let $Q\in \bC[z_1,\dots,z_N]$ be the dehomogenization of $P$ with respect to $Z_0.$ Then, for any $I=(i_1,\dots,i_n)\in\{1,\dots,c\}_{\neq}^{n}$, the symmetric differential form
$$\om_0^{I,Q}(t):=Q\left|\begin{array}{ccc}b_1(t_{1}^{1})&\cdots&b_N(t_N^{1})\\ \vdots&&\vdots\\ b_1(t_{1}^{c})&\cdots&b_N(t_N^{c})\\ \be_1(t_{1}^{i_1})&\cdots&\be_N(t_N^{i_1})\\ \vdots&&\vdots\\ \be_1(t_{1}^{i_n})&\cdots&\be_N(t_N^{i_n})\end{array}\right|\in H^0(U_0\cap X,S^n\Om_{X})$$
extends to a twisted symmetric differential form $\om^{I,P}\in H^0(X,S^n\Om_X(-a))$.
\end{proposition}
\subsection{Estimating the base locus}\label{EliminationSection}
\subsubsection{Dimension count lemmas}
We will need some elementary results. The following is classical, we provide a proof because we will use the idea of it again afterwards.
\begin{lemma} Let $p,q\in \bN$ such that $n\ls p$. Fix a rank $n$ matrix $A\in \Mat_{n,p}(\bC).$ For any $r\in \bN^*$ with $r<n$, let
$$\Si^{p,q}_r:=\left\{B\in \Mat_{p,q}(\bC)\ \text{such that}\ \ \rk{AB}\ls r\right\}.$$
Then, $\dim \Si_r^{p,q}\ls pq-(q-r)(n-r).$
\end{lemma}
\begin{proof} Set $$\De:=\left\{(B,\Ga)\in \Mat_{p,q}(\bC)\times \Grass(r,\bC^n)\ \text{such that}\ \im(AB)\subseteq \Ga\right\}$$
Let $\pr_1:\De\to \Mat_{p,q}(\bC)$ and $\pr_2:\De\to \Grass(r,\bC^n) $ be the natural projections. For any $\Ga\in \Grass(r,\bC^n)$ we consider $\De_\Ga=\pr_1(\pr_2^{-1}(\Ga)).$ Of course,
\begin{eqnarray*}
\De_\Ga=\left\{B\in \Mat_{p,q}(\bC)\ \text{such that}\ \im(AB)\subseteq \Ga\right\}=\left\{B\in \Mat_{p,q}(\bC)\ \text{such that}\ \im(B)\subseteq A^{-1}(\Ga)\right\}\end{eqnarray*}
But since $\rk A=n,$ $A$ is surjective and therefore $\dim(A^{-1}(\Ga))=r+p-n$. Therefore, $\De_\Ga\cong \Hom(\bC^q,\bC^{r+p-n})$ and thus  $\dim (\De_\Ga)=q(r+p-n).$ In particular, $$\dim \De= q(r+p-n)+\dim\Grass(r,\bC^n)=q(r+p-n)+r(n-r)=qp-(q-r)(n-r).$$
Since $\Si_r^{p,q}=\pr_1(\De),$ the result follows.
\end{proof}

From the previous lemma one can easily deduce the following.
\begin{corollary}\label{determinental} Let $M,N,c\in \bN^*$ such that $c\ls N$. Let $\ell_1,\dots, \ell_N\in (\bC^M)^\vee$ be non-zero linear forms.  For any $r\in \bN^*$ such that $r<c,$ let
$$\Si_r=\left\{(x_i^j)^{1\leqslant j\ls c}_{1\ls i \ls N}\in (\bC^M)^{Nc} \ \text{such that } \rk\left(\begin{array}{ccc}\ell_1(x_1^1)&\cdots & \ell_N(x_N^1)\\ \vdots & &\vdots \\ \ell_1(x_1^c)&\cdots & \ell_N(x_N^c)\end{array}\right)\ls r\right\}.$$
Then, $\dim(\Si_r)\ls MNc-(N-r)(c-r).$
\end{corollary}
The following lemma will be crucial to us.
\begin{lemma}\label{FormesIndep}
Take $c,n,M\in \bN^*$ such that $n\ls c$ and let $N:=n+c$. Take linear forms $\ell_1,\dots, \ell_n,\la_1,\dots ,\la_n\in (\bC^M)^\vee$ such that for any $i\in \{1,\dots,c\}$, $\la_i\neq 0$ and $\ker\ell_i\neq \ker \la_i$. Fix $A\in \Gl_c(\bC)$ and  $B\in \Mat_{c}(\bC)$. For any $x=(x_i^j)^{1\ls j\ls c}_{1\ls i\ls n}\in (\bC^M)^{cn}$ consider
$$S\left(x\right)=\left(\begin{array}{ccc|ccc}&&&\ell_1(x_1^1)&\cdots&\ell_n(x_n^1)\\&A&&\vdots&&\vdots \\ &&&\ell_1(x_1^c)&\cdots&\ell_n(x_n^c)\\ \hline &&&\la_1(x_1^1)&\cdots&\la_n(x_n^1)\\&B&&\vdots&&\vdots \\ &&&\la_1(x_1^c)&\cdots&\la_n(x_n^c) \end{array}\right).$$
For any $r\in \{c,\dots, N-1\}$, let $$\Si_r:=\left\{x\in (\bC^M)^{cn}\ \text{such that }\ \rk S\left(x\right)\ls r\right\}\subseteq (\bC^M)^{cn}.$$
Then, $\dim(\Si_r)\leqslant Mnc-(N-r)(2c-r).$
\end{lemma}
\begin{proof}
Using elementary operations on the $c$ first  columns, we see that we may suppose that $A=I_c$ is the identity matrix of size $c$. We will write everything in matricial notation. For any $i\in \{1,\dots, M\}$ any $j\in \{1,\dots, c\}$ and any $x=(x_i^j)^{1\ls j\ls c}_{1\ls i\ls n}\in (\bC^M)^{cn}$ we let $$L_i=(\ell_{i,1},\dots, \ell_{i,M})\in (\bC^M)^\vee,\ \ \La_i=(\la_{i,1},\dots, \la_{i,M})\in (\bC^M)^\vee \ \text{and} \ X_{i}^j=\left(\begin{array}{c}x_{i,1}^j\\ \vdots\\ x_{i,M}^j\end{array}\right)\in \bC^M$$
such that $\ell_i(x_i^j)=L_iX_i^j$ and $\la_i(x_i^j)=\La_i X_i^j.$ For any matrix $Q$ we denote the transposed of $Q$ by ${}^tQ.$ Set moreover
$$L:=\left(\begin{array}{ccc}{}^tL_1&\cdots &0\\ \vdots &\ddots&\vdots \\ 0&\cdots &{}^tL_n\end{array}\right),\ \ \La:=\left(\begin{array}{ccc}{}^t\La_1&\cdots &0\\ \vdots &\ddots&\vdots \\ 0&\cdots &{}^t\La_n\end{array}\right)\ \text{and}\ X:=\left(\begin{array}{ccc}{}^tX_1^1&\cdots &{}^tX_n^1\\ \vdots &&\vdots \\ {}^tX_1^c&\cdots &{}^tX_n^c\end{array}\right).$$
 With those notations, $S(x)=\left(\begin{array}{cc}I_c&XL\\ B& X\La\end{array}\right),$ and by elementary operations on the lines we get that
 $$\rk(S(x))=\rk\left(\begin{array}{cc}I_c&XL\\ 0& X\La-BXL\end{array}\right)=c+\rk(X\La-BXL).$$
 Write $B=(b_{i,j})_{1\ls i,j\ls c}.$ By a straightforward computation we obtain that for any $j\in \{1,\dots, n\}$ the $j$-th column of the matrix $X\La-BXL$ is exactly $K_jX_j$ where $$X_j=\left(\begin{array}{c}X_j^1\\ \vdots\\ X_j^c\end{array}\right)\ \ \ \text{and} \ \ \
 K_j=\left(\begin{array}{ccc}\La_j&\cdots&0\\ \vdots&\ddots&\vdots\\ 0&\cdots & \La_j\end{array}\right)-\left(\begin{array}{ccc}b_{11}L_j&\cdots&b_{1c}L_j\\ \vdots&&\vdots\\ b_{c1}L_j&\cdots & b_{cc}L_j\end{array}\right)$$

In particular, we have $X\La-BXL=\left(\begin{array}{ccc}K_1X_1&\cdots&K_nX_n\end{array}\right).$ An easy computation shows that under our hypothesis on the $\ell_i'$s and the $\la_i'$s, $\rk K_j=c$ for all $j\in \{1,\dots,n\}.$ The end of the proof goes as in Lemma~\ref{determinental}. Let
$$\De:=\left\{(X,\Ga)\in (\bC^M)^{nc}\times \Grass(r-c,\bC^c)\ \text{such that} \ \im(X\La-BXL)\subseteq \Ga\right\}$$
 Let $\pr_1:\De\to (\bC^M)^{cn}$ and $\pr_2:\De\to \Grass(r-c,\bC^c) $ be the natural projections. For any $\Ga\in \Grass(r-c,\bC^c)$ let $\De_\Ga=\pr_1(\pr_2^{-1}(\Ga)).$ Then
 \begin{eqnarray*}
 \De_{\Ga}&=&\left\{X\in (\bC^M)^{nc}\ \text{such that} \ \im(X\La-BXL)\subseteq \Ga\right\}\\
 &=&\left\{X\in (\bC^M)^{nc}\ \text{such that} \ \im\left(\begin{array}{ccc}K_1X_1&\cdots&K_nX_n\end{array}\right)\subseteq \Ga\right\}\\
 &=&\left\{X\in (\bC^M)^{nc}\ \text{such that} \ \im(X_1)\subseteq K_1^{-1}(\Ga),\dots, \im(X_n)\subseteq K_n^{-1}(\Ga)\right\}
 \end{eqnarray*}
 Since for each $j\in \{1,\dots,n\},$ $K_j$ is surjective, we obtain that $\dim(K_j^{-1}(\Ga))=r-c+Mc-c=Mc+r-2c.$
 In particular
$ \De_\Ga\cong \Hom(\bC,\bC^{Mc+r-2c})^{\oplus n},$
which implies that $\dim(\De_\Ga)=n(Mc+r-2c)$. And therefore,
$$\dim(\De)=n(Mc+r-2c)+\dim\Grass(r-c,\bC^c)=n(Mc+r-2c)+(r-c)(2c-r)=Mnc-(n+c-r)(2c-r).$$
But since $\Si_r=\pr_1(\De)$, this concludes the proof of the lemma.
 \end{proof}
We are also going to need the following elementary algebraic geometry lemma.
\begin{lemma}\label{stratification} Let $X$ and $Y$ be two algebraic varieties (not necessarily irreducible). Let $f:X\to Y$ be a regular map. Suppose that $Y=Y_0\cup \cdots\cup Y_r$ such that $Y_i\cap Y_j=\varnothing$  if $i\neq j$. For each $i\in \{0,\dots,r\}$ take $n_i\in \bN$. Suppose that for each $y\in Y_i$ and for each $i\in \{0,\dots,r\}$ , $\dim X_y\leqslant n_i$. Then
$$\dim X\leqslant \max_{0\ls i\ls r}(\dim(Y_i)+n_i).$$
\end{lemma}
\subsubsection{Proof of Lemma \ref{EliminationLemma}}
We are now in position to prove Lemma \ref{EliminationLemma}. It will be a immediate consequence of the following more precise statement.
\begin{lemma}\label{RefinedElimitationLemma}Let $N,c,e,\ep,a\in\bN$ such that $N\geqslant 2$, $c\geqslant \frac{3N-2}{4}$, $\ep\geqslant 1$ and $e\geqslant N+1+a+N\ep$, set $q:=e- (N+1+a+N\ep)$. Set $n:=N-c$. For any $0\leqslant j\leqslant N$ take $s^j\in \bA_\ep^{\oplus N+1}$ such that $X:=X_{s^1}\cap \cdots \cap X_{s^c}$ is a smooth complete intersection variety.
 For any $i\in \{0,\dots,N\}$, set $H_i:=(Z_i=0)$ and $W_i:=X\cap H_i$. We look at $\bP(\Om_{W_i})$ as a subvariety of $\bP(\Om_X)$. Then, for a general choice of  $s^1,\dots,s^c$, there exists $E\subseteq \bP(\Om_X)$ such that $\dim E \ls 0$ and  $$\bB(\bL_X(-a))\subseteq \bigcap_{\substack{I\in \{1,\dots, c\}^n_{\neq}\\ P\in\bC[Z_0,\dots, Z_N]_q}}\!\!\!\!\!\!\!Ò (\om^{I,P}=0)=\bigcup_{i=0}^N \bP(\Om_{W_i})\cup E.$$
 Where $\om^{I,P}\in H^0(\bP(\Om_X),\bL_X^n(-a))\cong H^0(X,S^n\Om_X(-a))$ is the symmetric differential form constructed in Proposition \ref{DiffForms} viewed as a global section of $\bL_X^n(-a).$
\end{lemma}

\begin{proof}
For simplicity, we only treat the chart $U_0$, the other charts are dealt with in the exact same way. Let us precise the notation of Proposition \ref{DiffForms}: for any $(u,z,\xi)\in \bA_\ep\times \bC^N\times \bP^{N-1}$ we set
 \begin{eqnarray*}
 b_q(u,z):=z_qu(z),\ \text{and} \ \  \be_q(u,z,\xi):=z_qdu_z(\xi)+eu(z)\xi_q.
 \end{eqnarray*}   Moreover, for any $t=(t^1,\dots,t^c)=(t^j_i)^{1\ls j\ls c}_{1\ls i\ls N}\in (\bA_\ep^N)^c$ and for any $(z,\xi)\in \bC^N\times \bP^{N-1}$ we set
$$B(t,z):=\left(\begin{array}{ccc}b_1(t_1^1,z)&\cdots & b_N(t_N^1,z)\\ \vdots & & \vdots \\ b_1(t_1^c,z)& \cdots &b_N(t_N^c,z)\end{array}\right)\ \ \text{and}\ \
B'(t,z,\xi):=\left(\begin{array}{ccc}\be_1(t_1^1,z,\xi)&\cdots & \be_N(t_N^1,z,\xi)\\ \vdots & & \vdots \\ \be_1(t_1^c,z,\xi)& \cdots &\be_N(t_N^c,z,\xi)\end{array}\right).$$
From now on we make the identification $\bP({\Om_{\bP^N}}|_{U_0})\cong \bC^{N}\times \bP^{N-1}.$ So that if for any $j\in \{1,\dots, c\}$ we take $s^j\in \bA_\ep^{N+1}$ such that $X_s=(F_{s^1}=0)\cap \cdots \cap (F_{s^c}=0)$ is a smooth complete intersection in $\bP^N$ and if for any $j\in \{1,\dots,c\}$, we let $t^j$ be the dehomogeneization on $s^j$, $f_{t^j}$ be the dehomogeneization of $F_{s^j}$ and $X_t=X_s$, then, one naturally has
$$\bP(\Om_{X_t\cap U_0})=\left\{(x,\xi)\in \bC^N\times \bP^{N-1}\ \text{such that }\ \ \left\{\begin{array}{c}f_{t^{1}}(z)=\cdots=f_{t^{c}}(z)=0\\ df_{t^{1},z}(\xi)=\cdots=df_{t^{c},z}(\xi)=0\end{array}\right.\right\}\subseteq \bC^{N}\times \bP^{N-1}\cong \bP(\Om_{U_0})$$
Moreover, set $W:=\bigcup_{i=1}^N\left((z_i=0)\cap(\xi_i=0)\right)\cong \bigcup_{i=1}^N \bP(\Om_{H_i\cap U_0}),$ where $H_i=(Z_i=0).$
Our aim is to prove that, for a general $t\in \bA_{\ep}^{(N+1)c},$ there exists $E\subseteq \bP(\Om_{X_t\cap U_0})$ such that $\dim E\ls 0$ and $$ \bP(\Om_{X_t\cap U_0})\cap\bigcap_{I,Q}\left(\om^{I,Q}_0(t,z,\xi)=0\right)=\left(\bP(\Om_{X_t\cap U_0})\cap W\right)\cup E.$$
The first thing to observe is that if $(x,\xi)\in W$ then for any $I\in \{1,\dots,c\}^n_{\neq}$ and any $Q\in\bC[z_1,\dots,z_N]_{\ls q}$ one has $\om^{I,Q}(t,z,\xi)=0$, and therefore $W\subseteq \bigcap_{I,Q}\left(\om^{I,Q}_0(t,z,\xi)=0\right).$ The more difficult part is to prove that, generically, this is actually an equality up to a finite number of points.

Observe that if $\rk(B(t,z))<c$, then $\om_0^{I,Q}(t,z,\xi)=0$ for all $I$ and all $Q$. On the other hand, if $ \rk(B(t,z))=c$ then
$$\om_0^{I,Q}(t,z,\xi)=0 \ \forall I, \ \forall Q \ \ \iff \ \ \rk\left(\begin{array}{c}B(t,z)\\ B'(t,z,\xi)\end{array}\right)<N.$$
Therefore, for any $t\in \bA_\ep^{(N+1)c},$ the locus $\bigcap_{I,Q}\left(\om^{I,Q}_0(t,z,\xi)=0\right)$ is precisely
$$\left\{(z,\xi)\in \bC^N\times \bP^{N-1} \ \ \text{such that}\ \ \rk B(t,z)<c \ \ \text{or}\ \ \rk B(t,z)=c \ \text{and}\ \rk\left(\begin{array}{c}B(t,z)\\ B'(t,z,\xi)\end{array}\right)<N \right\}$$

 To understand this locus for general $t\in \bA_\ep^{(N+1)c}$ we will consider the problem in family, and study the corresponding incidence varieties.
 Let us introduce some more notation. Consider the natural projections
\begin{eqnarray*}
\rho_1:\bA_\ep^{(N+1)c}\times \bC^N \to \bA_\ep^{(N+1)c}\ \ \ \ \ \rho_2:\bA_\ep^{(N+1)c}\times \bC^N \to \bC^N\ \ \ \ \ p_{1}:\bA_\ep^{(N+1)c}\times \bC^N \times \bP^{N-1}\to \bA_\ep^{(N+1)c}\\
p_{12}:\bA_\ep^{(N+1)c}\times \bC^N \times \bP^{N-1}\to \bA_\ep^{(N+1)c}\times \bC^N\ \ \ \ \ p_{23}:\bA_\ep^{(N+1)c}\times \bC^N\times \bP^{N-1} \to \bC^N\times \bP^{N-1}.
\end{eqnarray*}
For any $k\in \{0,\dots, N\},$ set $Y_k=\left\{(z_1,\dots,z_N)\in\bC^N \ \ /\ \ \exists I\in \{1,\dots,N\}^k_{\neq} \  \text{such that} \ \ z_i=0\iff i\in I \right\}.$
Observe that $\dim Y_k=N-k.$ Moreover, set
\begin{eqnarray*}
\ccX_c&:=&\left\{\left((t^{1},\dots,t^{c}),z\right)\in (\bA_\ep^{(N+1)})^c\times \bC^N\ \text{such that}\ \ f_{t^{1}}(z)=\cdots=f_{t^{c}}(z)=0\right\}\\
\ccX_c'&:=&\left\{\left((t^{1},\dots,t^{c}),z,\xi\right)\in (\bA_\ep^{(N+1)})^c\times \bC^N\times \bP^{N-1}\ \text{such that}\ \ \left\{\begin{array}{c}f_{t^{1}}(z)=\cdots=f_{t^{c}}(z)=0\\ df_{t^{1},z}(\xi)=\cdots=df_{t^{c},z}(\xi)=0\end{array}\right.\right\}\\
\De_c&:=&\left\{\left(t,z\right)\in \ccX_c \ \text{such that}\ \ \rk B(t,z)<c\right\}\\
E_c^0&:=&\left\{\left(t,z,\xi\right)\in \ccX_c' \ \text{such that}\ \ \rk B(t,z)=c \ \text{and}\ \ \rk\left(\begin{array}{c}B(t,z)\\ B'(t,z,\xi)\end{array}\right)<N\right\}\\
\De_c'&:=&p_{12}^{-1}(\De_c)\cap\ccX_c',\ \ \ E_c:=E^0_c\cup \De_c', \ \ \ \Ga_c:=E_c\setminus p_{23}^{-1}(W) \ \text{and} \ \ \ \Ga_c^{0}:=E_c^0\setminus p_{23}^{-1}(W).
\end{eqnarray*}
Of course, $\ccX_c\stackrel{\rho_1}{\to}\bA_\ep^{(N+1)c}$ is just the universal family of complete intersection of Fermat type we are considering (or to be precise, the dehomogeneization of it), and $\ccX_c'$ is just the projectivization of the relative cotangent bundle of $\ccX_c\stackrel{\rho_1}{\to}\bA_\ep^{(N+1)c}.$ For any $t\in \bA_\ep^{(N+1)c},$ we write $\ccX_{c,t}=\rho_1^{-1}(t)$ (and similarly for $\ccX'_{c,t},$ $\De_{c,t}$, etc.). By the above discussion, our aim is to prove that for a general $t\in \bA_\ep^{(N+1)c},$ $\Ga_{c,t}$ is finite. To do so, we will prove that for a general $t\in \bA_{\ep}^{(N+1)c}$, $\De'_{c,t}$ is empty and $\dim \Ga_{c,t}^0\ls 0.$ This will follow at once from the two following claims.
\begin{claim}\label{claim2}$\dim \De_c<\dim\bA_\ep^{(N+1)c}$\end{claim}
\begin{claim}\label{claim3}$\dim \Ga_c^{0}\ls\dim\bA_\ep^{(N+1)c}$\end{claim}
\begin{proof}[Proof of Claim \ref{claim2}] Take $k\in \{0,\dots, N\}$. Fix $(z_1,\dots, z_N)\in Y_k.$ Without loss of generality, we may assume that $z_1=\cdots =z_k=0$ and that $z_{k+1}\cdots z_N\neq 0.$ In particular,
$$\rk B(t,z)=\rk\left(\begin{array}{ccc}t_{k+1}^1(z)&\cdots & t_N^1(z)\\ \vdots & &\vdots \\ t_{k+1}^c(z)&\cdots & t_N^c(z)\end{array}\right).$$
Set $\De_{c}^z:=\rho_1\left(\rho_2^{-1}(\{z\})\cap \De_c\right).$ Of course,
$$\De_{c}^z=\left\{\left((t_i^j)^{1\ls j\ls c}_{1\ls i\ls N}\right)\in \bA_\ep^{(N+1)c}\ \text{such that}\ \ \rk\left(\begin{array}{ccc}t_{k+1}^1(z)&\cdots & t_N^1(z)\\ \vdots & &\vdots \\ t_{k+1}^c(z)&\cdots & t_N^c(z)\end{array}\right)<c,\ \text{and}\ f_{t^1}(z)=\cdots=f_{t^c}(z)=0\right\}.$$
By Lemma \ref{determinental} the set defined by $\rk\left(\begin{array}{ccc}t_{k+1}^1(z)&\cdots & t_N^1(z)\\ \vdots & &\vdots \\ t_{k+1}^c(z)&\cdots & t_N^c(z)\end{array}\right)<c$ is of dimension less than  $\dim \bA_\ep^{(N+1)c}-\max\{0,N-c-k+1\}$. Therefore, $$\dim \De_{c}^z\ls\dim\bA_\ep^{(N+1)c}-\max\{0,N-k-c+1\}-c.$$
Indeed, for any $j\in \{1,\dots, c\},$ the equation $f_{t^j}(z)=t_0^j(z)+t_1^j(z)z_1^e+\cdots +t_N^j(z)z_N^e=0$ is affine (in $t_i^j$) and involves the term $t_0^j(z)$ which appears in none of the other equations defining $\De_{c}^z$. So that each of the equations $f_{t^j}(z)=0$ increases the codimension by one, and we get the announced dimension. Therefore by Lemma \ref{stratification} we obtain that
\begin{eqnarray*}
\dim \De_c &\ls& \max_{0\leqslant k\ls N}\left(\dim Y_k+\dim\bA_\ep^{(N+1)c}-\max\{0,N-k-c+1\}-c\right)\\
&=& \max_{0\leqslant k\ls N}\left(\dim\bA_\ep^{(N+1)c}-\max\{0,N-k-c+1\}+N-k-c\right)\ls \dim\bA_\ep^{(N+1)c}-1<\dim \bA_\ep^{(N+1)c}.
\end{eqnarray*}
\end{proof}
To prove Claim \ref{claim3} we will need the following observation.
\begin{claim}\label{claim0}Let $(x,\xi)\in\bC^N\times\bP^{N-1}\setminus W$. For any $q\in \{1,\dots, N\}$, $\be_q(\cdot,z,\xi)\neq 0$ and $\ker b_q(\cdot, z)\neq \ker\be_q(\cdot,z,\xi).$\end{claim}
\begin{proof}[Proof of Claim \ref{claim0}]
For $k\in\{1,\dots,N\}$, we set $\de_k=(0,\dots,0,1,0,\dots, 0)\in \bN^N$ be the multi-index whose only non-zero term is in the $k$-th slot. For any $u\in \bA_\ep$ any $(z,\xi)\in \bC^N\times \bP^{N-1}$ and any $q\in \{1,\dots,N\}$ we have by definition,
\begin{eqnarray*}
b_q(u,z)&=&z_q\left(\sum_{|I|\ls \ep}u_Iz^I\right)=z_qu_0+z_1z_qu_{\de_1}+\cdots +z_Nz_qu_{\de_N}+\sum_{2\ls |I|\ls \ep}z^{I+\de_q}u_I\\
\be_q(u,z,\xi)&=& z_q\left(\sum_{|I|\ls \ep}\sum_{k=1}^Ni_ku_Iz^{I-\de_k}\xi_k\right)+e\xi_q\left(\sum_{|I|\ls \ep}u_Iz^I\right)\\
&=&e\xi_qu_0+(z_q\xi_1+ez_1\xi_q)u_{\de_1}+\cdots+(z_q\xi_N+ez_N\xi_q)u_{\de_N}+\sum_{2\ls |I|\ls \ep}\left(z_q\sum_{k=1}^Ni_kz^{I-\de_k}\xi_k+ez^I\xi_q\right)u_I.
\end{eqnarray*}
Take $(x,\xi)\in \bC^{N-1}\times \bP^{N-1}\setminus W.$ Let $q\in \{1,\dots, N\}$. In view of the above expression, if $\be_q=0$ then in particular $$e\xi_q=z_q\xi_1+ez_1\xi_q=\cdots=z_q\xi_N+ez_N\xi_q=0,$$
and because $(z,\xi)\notin W$, we get $z_q\neq 0$, so that $\xi_q=\xi_1=\cdots=\xi_N=0$, which is not possible because $\xi\in\bP^{N-1}.$ It remains to prove that $\ker b_q(\cdot,z)\neq \ker \be_q(\cdot,z,\xi).$ If $b_q(\cdot,z)=0$, this is obvious. Suppose $b_q(\cdot,z)\neq 0$ and therefore $z_q\neq 0$. If $\ker b_q(\cdot,z)=\ker \be_q(\cdot,z,\xi),$ then  from the above expression we get in particular that
$$\left|\begin{array}{cc}1& z_1\\ e\xi_q& z_q\xi_1+ez_1\xi_q\end{array}\right|=\cdots=\left|\begin{array}{cc}1& z_N\\ e\xi_q& z_q\xi_N+ez_N\xi_q\end{array}\right|=0,$$
From which we deduce once again that $\xi_1=\cdots=\xi_N=0.$
\end{proof}

\begin{proof}[Proof of Claim \ref{claim3}] For any $K=(k_1,\dots, k_c)\in \{1,\dots, N\}^c_{\neq}$ let
$$U^K=\left\{(t,z,\xi)\in \bA_\ep^{(N+1)c}\times \bC^N\times \bP^{N-1}\ \text{such that }\ \det\left(\begin{array}{ccc}b_{k_1}(t_{k_1}^1,z)&\cdots &b_{k_c}(t_{k_c}^1,z)\\ \vdots&&\vdots\\ b_{k_1}(t_{k_1}^c,z)&\cdots &b_{k_c}(t_{k_c}^c,z)\end{array}\right)\neq 0\right\}.$$
For any $K$, $U^K_\Ga:=U^K\cap\Ga_c^0$ is an open subset of $\Ga_c^{0}$ and $\Ga_c^{0}=\bigcup_KU^K_\Ga,$ therefore for our dimension estimation, we might as well restrict ourselves to $U:=U^{(1,\dots,c)}$ and $U_\Ga:=U_\Ga^{(1,\dots,c)}.$ Fix $(z,\xi)\in \bC^N\times\bP^{N-1}$ and set $U^{z,\xi}_\Ga:=p_1(p_{23}^{-1}(z,\xi)\cap U_\Ga)$. Using Claim \ref{claim0} and Lemma \ref{FormesIndep} we see that the set defined in $\bA_\ep^{(N+1)c}$ by
$$ \rk\left(\begin{array}{c}B(t,z)\\ B'(t,z,\xi)\end{array}\right)<N\ \ \text{and}\ \ \ (t,z,\xi)\in U$$
is of dimension less than $\dim\bA_\ep^{(N+1)c}-2c+N-1$, and by the same argument as in the proof of Claim \ref{claim2}, we obtain
$$\dim U_\Ga^{z,\xi}\ls\dim\bA_{\ep}^{(N+1)c}-2c+N-1-2c.$$
Hence $\dim U_\Ga\ls\dim\bA_\ep^{(N+1)c}-4c+N-1+\dim(\bC^N\times\bP^{N-1})=\dim\bA_\ep^{(N+1)c}-4c+N-1+2N-1.$ And therefore (because the same argument holds for all the $U^K_\Ga$)
$$\dim \Ga_c^{0}=\dim\bA_\ep^{(N+1)c}-4c+3N-2.$$
In particular, $\dim \Ga_c^{0}\ls\dim\bA_\ep^{(N+1)c}$ as soon as $-4c+3N-2\ls 0$. This last condition is equivalent to $c\gs \frac{3N-2}{4},$ which is exactly our hypothesis.
\end{proof}
\end{proof}

config: arxiv


TITLE: For which $m,n$ is the graph $K_{m,n}$ integral?
QUESTION [0 upvotes]: A graph is integral if the eigenvalues of its adjacency matrix are all integers. For which integers $m$ and $n$ is $K_{m,n}$, the complete bipartite graph on $m$ and $n$ vertices, integral?

REPLY [2 votes]: Let $A$ denote the adjacency matrix of $K_{m,n}$. Now up to reordering of rows and columns $A$ must look like
\begin{equation*}
    A = \begin{bmatrix}
           0&\dots&0&1&\dots&1 \\
           \vdots&\ddots&\vdots&\vdots&\ddots&\vdots\\
           0&\dots&0&1&\dots&1 \\
           1&\dots&1&0&\dots&0 \\
           \vdots&\ddots&\vdots&\vdots&\ddots&\vdots\\
           1&\dots&1&0&\dots&0 \\
        \end{bmatrix}
\end{equation*}
where the blocks of ones have dimension $m \times n$. Now let's calculate some successive powers of $A$. This can done quickly by recalling that the $(i,j)$ entry of $A^n$ represents the number of distinct paths of length $n$ from vertex $i$ to vertex $j$.
\begin{equation*}
    A^2 = \begin{bmatrix}
           m&\dots&m&0&\dots&0 \\
           \vdots&\ddots&\vdots&\vdots&\ddots&\vdots\\
           m&\dots&m&0&\dots&0 \\
           0&\dots&0&n&\dots&n \\
           \vdots&\ddots&\vdots&\vdots&\ddots&\vdots\\
           0&\dots&0&n&\dots&n \\
        \end{bmatrix}
        \quad
    A^3 = \begin{bmatrix}
           0&\dots&0&mn&\dots&mn \\
           \vdots&\ddots&\vdots&\vdots&\ddots&\vdots\\
           0&\dots&0&mn&\dots&mn \\
           mn&\dots&mn&0&\dots&0 \\
           \vdots&\ddots&\vdots&\vdots&\ddots&\vdots\\
           mn&\dots&mn&0&\dots&0 \\
        \end{bmatrix}
\end{equation*}
So from this we see that $A$ has minimal polynomial $A^3-mnA$. The minimal polynomial divides the characteristic polynomial, and all the roots of the characteristic polynomial (eigenvalues) are roots of this minimum polynomial too (Cayley-Hamilton theorem), so $A$ has integer eigenvalues exactly when the roots of $A^3-mnA$ are integers. The roots of $A^3-mnA$ are $0$ and $\pm\sqrt{mn}$, so we can conclude that $K_{m,n}$ is integral exactly when the product $mn$ is a square.


TITLE: Why does the chemical potential show up in the Gross-Pitaevskii equation?
QUESTION [2 upvotes]: In the case of a weakly interacting Bose-Einstein condensate the Gross-Pitaevskii equation is the motion equation that governs the evolution of the N-particle wavefunction. Starting from the Time-Dependent Gross-Pitaevskii equation:
$$i\hbar\dot{\Psi}(x,t) = \left(-\frac{\hbar^2}{2m}\nabla^2_x + V(x,t) + g|\Psi(x,t)|^2 \right) \Psi(x,t)$$
Assuming the system to be time-independent the solution can be expressed as: $\Psi(x,t) = \psi(x) \phi(t)$, where, both fields satisfy:
$$i\hbar\dot{\phi(t)} = -\mu \phi(t)$$
$$\mu\psi(x) = \left(-\frac{\hbar^2}{2m}\nabla^2_x + V(x,t) + g|\psi(x)|^2 \right)\psi(x)$$
My question is very simple but I don't see the point: how do we know that the constant of the separation of variables method corresponds to the chemical potential? Is related with the fact that $N=\int dx |\psi(x)|^2$ ?

REPLY [3 votes]: Energy eigenstates are stationary states, so their time dependence is only contained in a phase factor.
The GPE described a BEC at $T=0$, so all atoms in one state with one energy. That energy is the ground state energy $E_0$ which is equal to the chemical potential $E_0 = \mu$. This ensures that the ground state has infinite occupancy because:
$$f(E) = \frac{1}{\mathrm{e}^{\frac{E-\mu}{k T}}-1}, \quad\text{so} \quad f(E_0) \rightarrow \infty \quad \text{as}\quad  E_0 \rightarrow \mu. $$
Hence, the phase factor of the energy eigenstate of the GPE will just be:
$$\phi(t) = \exp \left ( \frac{-\mathrm{i}E_0t}{\hbar} \right ) =   \exp \left ( \frac{-\mathrm{i}\mu t}{\hbar} \right ).$$
Substituting the latter equation into the time-dependent GPE gives you the time-independent equation.


TITLE: Weierstrass to general form
QUESTION [0 upvotes]: Can we go from  short Weierstrass equation equation $y^2=x^3+Ax^2+Bx+C$ to general
$$y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6$$?

REPLY [5 votes]: What you call the "general equation" is the Weierstrass equation, and what you call the "Weierstrass equation" is the so-called "short" Weierstrass equation, which is just the general Weierstrass equation with some coefficients $0$, so your question doesn't really make sense.
But maybe you meant to ask it the other way around: how to get a short Weierstrass equation from a general one? We perform a change of variables which preserves the behavior of the curve, called an admissible change of variables. If we are over a field of characteristic not $2$ or $3$, it is always possible to get a short equation from a general one through an admissible change of variables. Otherwise, not always.


text:
\begin{document}

\title[Material interpenetration]{Elliptic systems and material interpenetration}
\author{Giovanni Alessandrini}

\address{Dipartimento di Matematica e Informatica, Universit\`a degli Studi di Trieste, Italy}

\email{alessang@units.it}

\thanks{The first author was supported  in part by MiUR, PRIN no. 2006014115}

\author{Vincenzo Nesi}
\address{Dipartimento di Matematica,
La Sapienza, Universit\`a di Roma, Italy}
\email{nesi@mat.uniroma1.it}
\thanks{The second author was supported in part by MiUR, PRIN
no. 2006017833.}

\begin{abstract}
We classify the second order, linear, two by two systems for which the two fundamental theorems for planar harmonic mappings, the Rad\'{o}--Kneser--Choquet Theorem and the H. Lewy Theorem, hold. They are those which, up to a linear change of variable, can be written in diagonal form with \emph{the same} operator on both diagonal blocks.
In particular, we prove that the aforementioned Theorems  cannot be extended to solutions of either the Lam\`{e} system of elasticity, or of elliptic systems in diagonal form, even with just slightly different operators for the two components.
\end{abstract}

\maketitle

\section{Introduction}
\label{intro}

``A basic requirement of continuum mechanics is that interpenetration of matter
does not occur, i.e. that in any deformed configuration the mapping giving the position $u(x)$ of a particle in terms of its position $x$ in the reference configuration be invertible''. J. M. Ball \cite{jmball}.

It is then a natural question to ask what systems of equations, among those used as models for elastostatics, give rise to invertible solutions when reasonable boundary conditions are prescribed.

In this note we shall prove by an example that the Lam\`e system of isotropic, linearized elasticity in the plane, with \emph{constant} Lam\`e coefficients, may lead to physically unacceptable solutions, because interpenetration of matter occurs. Let us recall here that the same phenomenon was previously found by Fosdick and Royer--Carfagni \cite{frc} for a more involved anisotropic linear system, by elaborating on an example due to Lekhnitskii \cite{lek}. In higher dimensions similar phenomena occur. From the mathematical point of view a basic example is due to De Giorgi \cite{degiorgi}. In all these examples, however, a basic common feature is the presence of some sort of point singularity in the solution itself (in dimension greater than two) or at least in the gradient in any dimension.
Such a singularity can only be present when the coefficients of the elliptic system are irregular, for instance discontinuous. Our examples are different from those previously known in several ways, but the most crucial difference is that we choose smooth and in fact \emph{constant} coefficients in our systems. Therefore our examples, besides bringing a new argument to the many already well known, see for instance Ciarlet \cite[p. 286]{ciarlet} and the references therein, about the limitations of linearized elasticity, shed some new light on the tightness of certain classical properties enjoyed by harmonic mapping showing that they cannot be easily extended even within the class of constant coefficient systems.

We now recall some fundamental properties of planar harmonic mappings, see the book of Duren \cite{duren} for a very broad treatment of this subject.

 We begin with the classical theorem of  Rad\'{o}, Kneser and Choquet. This theorem which was first stated by  Rad\'{o} \cite{rado1}, proved by Kneser \cite{kneser} immediately after   and then independently rediscovered by Choquet \cite{Choquet}, remains the basic unequalled result of invertibility for mappings solving an elliptic system of equations.

Let $B$ be an open disk in the plane, let $\Phi:\partial B\to \gamma \subset \mathbb R^2$ be a homeomorphism of $\partial B$ onto a simple closed curve $\gamma$. Let $u\in C^2(B,\mathbb R^2)\cup C(\overline{B},\mathbb R^2)$ be the solution to the Dirichlet problem
\begin{equation}\label{rado}
\left\{
\begin{array}{ccc}
\Delta u=0&,&\hbox{in } B,\\
u=\Phi&,& \hbox{on } \partial B.
\end{array}
\right.
\end{equation}
Let $D$ be the bounded region such that $\partial D=\gamma$. The Rad\'{o}--Kneser--Choquet Theorem states the following.
\begin{theorem}
If $D$ is convex, then $u$ is a homeomorphism of $\overline{B}$ onto $\overline{D}$.
\end{theorem}
The main reasons why this theorem remains substantially unequalled are

\noindent
(i) no analogue of this theorem holds true in dimension three or higher, as it was shown by a striking example by Laugesen \cite{laugesen}, see also Melas \cite{melas},

\noindent
(ii) the convexity assumption on the target domain $D$ is optimal. In fact, it is known since Choquet \cite{Choquet}, that if $D$ is not convex then there exists homeomorphisms $\Phi:\partial B\to \partial D$, for which the injectivity of $u$, the solution to (\ref{rado}), fails. See also \cite{ankneser} for a thorough investigation of this issue.

Nevertheless various kinds of generalizations of the Rad\'{o}--Kneser--Choquet Theorem have been obtained. Regarding harmonic mappings between manifolds, see Schoen and Yau \cite{sy} and Jost \cite{jost}. For mappings $u$ whose components solve a linear elliptic equation, let us mention Bauman, Marini and Nesi \cite{bmn} and also \cite{an:arch}, \cite{an:beltrami}. It is worth pointing out that, in these last two papers the Rad\'{o}--Kneser--Choquet Theorem has been extended to linear elliptic equations in divergence form with merely bounded measurable coefficients.  For quasilinear equations of the $p$-Laplacian type see \cite{as}.

    As a remarkable special case of our  Theorem \ref{generale.th}, which will be stated here below, we prove that no analogue of the Rad\'{o}--Kneser--Choquet Theorem holds when the diagonal Laplacian system is replaced by a Lam\`e system with constant moduli of the following form
    \begin{equation*}
        \mu ~{\rm div} ((\nabla u)^T+ \nabla u) + \lambda ~\nabla ({\rm div} ~u)=0.
    \end{equation*}
More precisely,  we have the following.
\begin{theorem}\label{main.th}
Let  $\mu,\lambda \in \mathbb R$ with $\mu>0$ and $\mu+\lambda>0$. There exist a disk $B\subset \mathbb R^2$, a bounded convex domain $D\subset \mathbb R^2$ and a smooth diffeomorphism $\Phi:\partial B \to \partial D$, so that the unique solution $u\in W^{1,2}(B,\mathbb R^2)$ to
\begin{equation*}
\left\{
\begin{array}{ccc}
    \mu ~{\rm div} ((\nabla u)^T+ \nabla u) + \lambda ~\nabla ({\rm div} ~u)=0&,&\hbox{in }B,\\
    u=\Phi&,&\hbox{on } \partial B,
\end{array}
    \right.
    \end{equation*}
    is \emph{not} a homeomorphism of $B$ onto $D$.
    \end{theorem}

Note that the natural unknown for the Lam\`e system, should be the \emph{displacement} field $\delta$, rather than the \emph{deformation} field $u$. However,  due to the fact the the identity mapping $I$ is also a solution of the Lam\`e system, we have trivially that $\delta$ solves the Lam\`e system if and only $u =I + \delta$ solves the same system. This is the reason why in Theorem \ref{main.th} it is understood that the solution $u$ is representing a deformation field.

    In Remark \ref{more},  we shall see that the solution fails to be an homeomorphism in a very strong way, in fact $u$ maps $B$ onto a domain larger than $D$, moreover it folds itself along a curve, across which the orientation is reversed.

    Our next Theorem is a far reaching generalization of the previous one. It shows that the Rad\'{o}--Kneser--Choquet Theorem holds only if one deals with elliptic systems  \emph{of diagonal form with the same scalar elliptic operator on both components}.
We need some definitions. Consider a constant coefficients second order elliptic system of the form
\begin{equation}\label{gen.syst}
\left\{
\begin{array}{ccc}
  {\rm div}( A \nabla u^1+ B \nabla u^2)=0,      \\
 {\rm div}( C \nabla u^1+ D \nabla u^2)=0.
\end{array}
\right.
\end{equation}
where $A, B, C$ and $D$ are $2\times 2$ real constant matrices,  and the unknowns $u^1$ and $u^2$ are real valued functions.
we say that the system (\ref{gen.syst}) is elliptic if it satisfies the Legendre--Hadamard condition
\begin{equation}\label{ell-1}
\eta_1^2 A \xi\cdot \xi+\eta_1 \eta_2  (B+C) \xi\cdot \xi+ \eta_2^2 D\xi\cdot \xi>0,\quad \hbox{for every\quad} \xi, \eta\in \mathbb R^2\backslash\{0\}.
\end{equation}
This condition is weaker than the strong convexity condition, namely the positivity of the $4\times 4$ matrix given in block form as
\begin{equation*}
M= \left(
\begin{array}{ccc}
A& B      \\
C&D
\end{array}
\right)
\end{equation*}
and, as it is well known, ellipticity \eqref{ell-1} is the same as rank one convexity of the quadratic form associated to $M$, see \cite[Theorem 5.3]{dacorogna}.
We note that there is no loss of generality in assuming that the matrices $A,B,C$ and $D$ are symmetric.
\begin{definition}
We shall say that the system (\ref{gen.syst}) is \emph{equivalent} to the system
\begin{equation*}
\left\{
\begin{array}{ccc}
  {\rm div}( A^{\prime} \nabla u^1+ B^{\prime} \nabla u^2)=0,      \\
 {\rm div}( C^{\prime} \nabla u^1+ D^{\prime} \nabla u^2)=0
\end{array}
\right.
\end{equation*}
if there exists a non-singular $2\times 2$ matrix
$\left(
\begin{array}{cc}
  \alpha&  \beta   \\
  \gamma&   \delta\end{array}
\right)$
such that
\begin{equation}\label{equiv}
\left(
\begin{array}{ccc}
A& B      \\
C&D
\end{array}
\right)=
\left(
\begin{array}{cc}
  \alpha {\rm Id}&  \beta {\rm Id}  \\
  \gamma {\rm Id}&   \delta {\rm Id}
\end{array}
\right)
\left(
\begin{array}{cc}
A^{\prime}& B^{\prime}      \\
C^{\prime}&D^{\prime}
\end{array}
\right).
\end{equation}
\end{definition}
\begin{theorem}\label{generale.th}
Let the ellipticity condition (\ref{ell-1}) be satisfied.
The following alternative holds.

\noindent
Either

\noindent
i) the system (\ref{gen.syst}) is equivalent to the system
\begin{equation}\label{diag}
\left\{
\begin{array}{ccc}
  {\rm div}( A \nabla u^1)=0,      \\
 {\rm div}(A\nabla u^2)=0,
\end{array}
\right.
\end{equation}

\noindent
or

\noindent
ii)
     there exist a disk $B\subset \mathbb R^2$, a bounded convex domain $D\subset \mathbb R^2$  and a smooth diffeomorphism $\Phi:\partial B \to \partial D$, so that the unique solution $u=(u^1,u^2)\in W^{1,2}(B,\mathbb R^2)$ to
     \begin{equation*}
\left\{
\begin{array}{rcl}
  {\rm div}( A \nabla u^1+ B \nabla u^2)=0&,&\hbox{in $B$},      \\
 {\rm div}( C \nabla u^1+ D \nabla u^2)=0&,&\hbox{on $B$} ,\\
u=\Phi&,& \hbox{on } \partial B.
\end{array}
\right.
\end{equation*}
is \emph{not} a homeomorphism of $B$ onto $D$.
    \end{theorem}
    The above results show that one of the most basic properties enjoyed by planar harmonic mappings cannot be extended to other elliptic systems in the plane. It is then natural to ask similar questions for another fundamental property of injective harmonic mapping. A benchmark of the theory is a result of H. Lewy \cite{lewy} proving that harmonic homeomorphisms are, in fact, diffeomorphisms. More precisely we have the following result.
    \begin{theorem}\label{lewy.th}(H. Lewy.)
    Lut $u=(u^1,u^2) :B\to \mathbb R^2$ be a harmonic mapping. If
    $u$ is invertible, then
    \begin{equation}
    \det Du\neq 0\quad\hbox{for every \quad} (x,y)\in B.
    \end{equation}
    \end{theorem}
    Also in this case the validity is limited to two dimensions. J. C. Wood \cite{wood2} found a  third degree polynomial harmonic mapping which provides a counterexample in dimension three. On the positive side, Hans Lewy \cite{lewy2} recognized that, in three dimensions, if $u$ is the \emph{gradient} of an harmonic function and it is a homeomorphism, then it is a diffeomorphism. This result was extended to any dimension in the remarkable paper by Gleason and Wolff \cite{gleasonwolff}.
In a different direction, several generalizations of Lewy's
Theorem have been achieved in dimension two when the components of
$u$ satisfy the \emph{same} linear elliptic equation of the form
${\rm div} (\sigma \nabla u^i)=0$. For the case of sufficiently
smooth $\sigma$ see \cite{bmn}. When $\sigma$ is allowed to be
discontinuous, weak forms of Lewy's Theorem have been obtained in
\cite{an:arch} and \cite{an:beltrami}. A version for $p$-Laplacian
type equations can be found in \cite{as}.

In the next theorem, we show, by means of examples, that Theorem
\ref{lewy.th} cannot  be extended to an arbitrary elliptic system
with constant coefficients, unless, again,  the systems has the
special  form \eqref{diag}.

\begin{theorem}\label{no-HL.th}
Let the ellipticity condition (\ref{ell-1}) be satisfied.
The following alternative holds.

\noindent
Either

\noindent
i) the system (\ref{gen.syst}) is equivalent to the system \eqref{diag}

\noindent
or

\noindent
ii)
     there exists a polynomial solution  to
     \begin{equation*}
\left\{
\begin{array}{rcl}
  {\rm div}( A \nabla u^1+ B \nabla u^2)=0&,&\hbox{in $B$},      \\
 {\rm div}( C \nabla u^1+ D \nabla u^2)=0&,&\hbox{on $B$} ,\end{array}
\right.
\end{equation*}
which is a homeomorphism of a closed disk $\overline{B}$ onto $u(\overline{B})$ and such that in the center of the disk, denoted by $O$, we have
\begin{equation*}
\det Du(O)=0.
\end{equation*}
    \end{theorem}
     \begin{remark}
    It may be evident that the Rad\'{o}--Kneser--Choquet and the H. Lewy Theorems continue to hold for any system of the form (\ref{diag}),
    since it can be elementarily reduced to a Laplacian diagonal system via a linear change of the independent coordinates.
    It is although rather remarkable that Theorems \ref{generale.th}, \ref{no-HL.th} show that the Rad\'{o}--Kneser--Choquet and the H. Lewy Theorems do
    not extend to very slight perturbations of the Laplacian diagonal system such as, for instance, the following one
\begin{equation*}
\left\{
\begin{array}{ccc}
    u^1_{xx}+ u^1_{yy}=0,\\
    (1+\varepsilon)u^2_{xx}+ u^2_{yy}=0,\
\end{array}
    \right.
    \end{equation*}
where $\varepsilon$ is any positive number.
\end{remark}

    \section{Proofs}
In what follows, when no ambiguity occurs, we shall identify points $(x,y)\in \mathbb R^2$ with column vectors
$\left(
    \begin{array}{c}
    x\\
    y
    \end{array}
    \right)$.
    Also, for $\theta \in \mathbb R$, we shall denote  $c_{\theta} = \cos\theta$ ,  $s_{\theta} = \sin \theta$.
For the proofs of Theorems  \ref{generale.th} and \ref{no-HL.th} we shall make use of the following two propositions, which we will prove at the end of this Section.
\begin{proposition}\label{A.prop}
Let the ellipticity condition (\ref{ell-1}) be satisfied.
If the system (\ref{gen.syst}) is \emph{not} equivalent to
(\ref{diag}), then there exists $\theta \in [0,2 \pi]$ and a
quadratic polynomial $p(x,y)= \frac{1}{2}(a x^2+ 2 b x y + c y^2)$
such that
\begin{equation*}
\left(
\begin{array}{c}
  u^1   \\
 u^2
\end{array}
\right)
=
\left(
\begin{array}{cc}
  c_{\theta}  & -s_{\theta}   \\
  s_{\theta}&   c_{\theta}
\end{array}
\right)
\left(
\begin{array}{ccc}
x^2+y^2  \\
p(x,y)
\end{array}
\right)
\end{equation*}
is a solution to (\ref{gen.syst}).
\end{proposition}
\begin{proposition}\label{B.prop}
Let the ellipticity condition (\ref{ell-1}) be satisfied.
If the system (\ref{gen.syst}) is \emph{not} equivalent to
(\ref{diag}), then there exists $\theta \in [0,2 \pi]$ and a cubic
polynomial $$q(x,y)= \frac{1}{2}\left(a \frac{x^3}{3}+  b x^2 y +
c x y^2 +d \frac{y^3}{3}\right)$$ such that
\begin{equation*}
\left(
\begin{array}{c}
  u^1   \\
 u^2
\end{array}
\right)
=
\left(
\begin{array}{cc}
  c_{\theta}  & -s_{\theta}   \\
  s_{\theta}&   c_{\theta}
\end{array}
\right)
\left(
\begin{array}{ccc}
x(x^2+y^2)  \\
q(x,y)
\end{array}
\right)
\end{equation*}
is a solution to (\ref{gen.syst}).
\end{proposition}

    \begin{proof}[Proof of Theorem \ref{generale.th}.]
    We assume that (\ref{gen.syst}) is \emph{not} equivalent to (\ref{diag}). We choose $\theta$ according to Proposition \ref{A.prop} and we write
    \begin{equation*}
    R_{\theta}=
\left(
\begin{array}{cc}
  c_{\theta}  & -s_{\theta}   \\
  s_{\theta}&   c_{\theta}
\end{array}
\right).
\end{equation*}
    Being linear mappings solutions to  (\ref{gen.syst}), we have that also the following is solution to  (\ref{gen.syst})
    \begin{equation*}
    \left(
\begin{array}{c}
  u^1   \\
 u^2
\end{array}
\right)
=
R_{\theta}
\left(
\begin{array}{ccc}
x^2+y^2 -1 \\
k y +p(x,y)
\end{array}
\right)
\end{equation*}
where $k\neq 0$ is a constant to be determined later on.

    We choose $B=\left\{(x,y)\in \mathbb R^2 : \left(x-\frac 1 2\right)^2 + y^2<\frac 5 4\right\}$. We have
\begin{equation*}
    x=-1 + x^2+y^2\qquad\hbox{on}\quad\partial B.
    \end{equation*}
    We now select $\Phi$. We set
    \begin{equation*}
    \begin{array}{ccc}
    M=  \left(
    \begin{array}{cc}
    1&0\\
    0&k
    \end{array}\right),
&\Psi(x,y)=
    \left(
    \begin{array}{c}
    x\\
    y + \frac{1}{k} p(x, y)
    \end{array}
    \right),& \Phi(x,y)=R_{\theta}
 M\Psi(x,y).
    \end{array}
    \end{equation*}
    A straightforward calculation shows that when
    \begin{equation}\label{k}
    k\geq |b|
    \end{equation}
    $\Psi$ is a homeomorphism of $\partial B$ onto a closed convex curve $\Gamma$. Consequently $\Phi$ is also a homeomorphism of $\partial B$ onto the closed convex curve $\gamma=R_{\theta}
M \Gamma$.

    Let $D$ be the bounded convex domain such that $\partial D =\gamma$. It is easy to check that $\overline{D}\subset R_{\theta} M
S$ where
    \begin{equation*}
    S=\left\{(v^1,v^2)\in \mathbb R^2: \frac{1-\sqrt{5}}{2}\leq v^1\leq \frac{1+\sqrt{5}}{2}\right\}.
    \end{equation*}
    However
    \begin{equation*}
     u(0,0)=-R_{\theta}M\left(
    \begin{array}{c}    1\\0
    \end{array}
    \right)\notin R_{\theta} M S.
    \end{equation*}
\end{proof}
    \begin{remark}
    \label{more}
    Note that we can compute the the Jacobian determinant of $u$ in terms of the coefficients of $p$ and obtain $\det Du = 2 k x + 2(b(x^2-y^2)+(c-a) x y)$. Hence the Jacobian determinant vanishes at $(0,0)$ and, in fact, it changes sign across its \emph{nodal line}
    \begin{equation*}
    H=\left\{(x,y)\in \mathbb R^2:  k x + (b(x^2-y^2)+(c-a) x y)=0\right\}.
    \end{equation*}
    which is always an hyperbola (unless $a=b=c=0$ when it degenerates in a straight line). Note that $k x + (b(x^2-y^2)+(c-a) x y)$ is positive in $(1,0)$ zero in $(0,0)$ and negative in $(-1,0)$ because of (\ref{k}). See Figures $1$ and $2$
where the behaviour of $u$ and its Jacobian determinant are depicted in the specific case of the Lam\`e system.
    \end{remark}

\begin{proof}[Proof of Theorem \ref{no-HL.th}.]
We assume again that (\ref{gen.syst}) is \emph{not} equivalent to (\ref{diag}). We choose $\theta$ and $q$ according to Proposition \ref{B.prop}. Being linear mappings solutions to  (\ref{gen.syst}), we have that also the following is solution to  (\ref{gen.syst})
    \begin{equation*}
        \left(
\begin{array}{c}
  u^1   \\
 u^2
\end{array}
\right)
=
\left(
\begin{array}{cc}
  c_{\theta}  & -s_{\theta}   \\
  s_{\theta}&   c_{\theta}
\end{array}
\right)
\left(
\begin{array}{ccc}
x(x^2+y^2) \\
y +q(x,y)
\end{array}
\right).
\end{equation*}
It is now easy to check that, once we have chosen the polynomial $q$ according to Proposition \ref{B.prop}, the following two properties hold. First there exists a positive radius $r$ such that
one has $\det Du(x,y)>0$ if $(x,y)\in B_r(O)\backslash \{O\}$, where we have set $O=(0,0)$, and second  $\det Du(O)=0$.
Choose $0<\rho<r$, and denote by
\begin{equation*}
\Phi=u \big|_{\partial B_{\rho}(O)}.
\end{equation*}
A very simple calculation shows that $\Phi$ maps $\partial B$ in a one to one way onto a closed curve $\gamma$ provided one has
\begin{equation*}
2+b \rho^2>0\quad\hbox{and \quad} 3+d \rho^2>0.
\end{equation*}
Let $D$ be the bounded domain such that $\partial D=\gamma$. We can now apply a topological result of Meisters and Olech \cite{meistol}. Indeed we have a smooth mapping $u$ defined on a closed disk $\overline{B}$ and which is a local homeomorphism at each point of $\overline{B}$, with the possible exception of the point $O$ only.  Moreover, the restriction of $u$ to $\partial B$ is a homeomorphism. Therefore the hypotheses of  Theorem 1 in \cite{meistol}  (see also Corollary 2) are satisfied and we can conclude  that $u$ is a homeomorphism of $\overline{B_{\rho}(O)}$ onto $\overline{D}$.
\end{proof}
\begin{proof}[Proof of Theorem \ref{main.th}.]
It suffices to verify that the Lam\`e system  is not equivalent to any elliptic system of the form (\ref{diag}). In fact it can be rewritten in the form (\ref{gen.syst}) with the
following choices
\begin{equation*}
\begin{array}{ccccc}
A=\left(
\begin{array}{cc}
 2 \mu +\lambda&0\\
 0&\mu
  \end{array}
\right),
&
B=C\left(
\begin{array}{cc}
 0 &\frac{ \mu +\lambda}{2}\\
 \frac{ \mu +\lambda}{2}&0     \end{array}
\right),
&
D=\left(
\begin{array}{cc}
\mu&0\\
 0&     2 \mu +\lambda
 \end{array}
\right).
\end{array}
\end{equation*}
Now $B, C$ and $D$ can be scalar multiples of $A$ only if $\mu+\lambda=0$, which contradicts the assumption $\mu+\lambda>0$.
\end{proof}
\begin{remark}
Note that our assumptions $\mu>0$, $\mu + \lambda > 0$ correspond the the strong convexity assumption, which, as is well known, is stronger that the ellipticity condition \eqref{ell-1}. \end{remark}

    In the following picture we illustrate our results for the case of the Lam\`e system. We choose
    $\lambda=\mu
=1$. With this choice, condition \eqref{k} takes the form $k\geq(1+\sqrt{10})\frac{\lambda+ 3 \mu}{\lambda+\mu}$ and, for the picture we have chosen the limiting value $k=2(1+\sqrt{10})$.
Note that, in order to facilitate visibility, the coordinates in the $u^1$ and the $u^2$ directions are scaled differently.

\begin{figure}[h]
\label{gamma}
\centerline{
\includegraphics[scale=.55,angle=0]{gamma.pdf}
}
\caption{$\partial B$ and its image $\Phi(\partial B)$.}
\end{figure}

\begin{figure}[h] \label{nodal}
\centerline{
\includegraphics[scale=1,angle=0]{nodale-noassi.pdf}
}
\caption{Left: circles $C_r$ of varying radii and the nodal  line of the Jacobian (an hyperbola) drawn within $B$. Right:   the images $U(C_r)$.}
\end{figure}

\begin{proof}[Proof of Proposition \ref{A.prop}]
We look for $\theta\in [0,2 \pi]$
and a quadratic polynomial $p(x,y)= \frac{1}{2}(a x^2+ 2 b x y + c y^2)$ such that
\begin{equation}\label{quad.sol}
\left(
\begin{array}{c}
  u^1   \\
 u^2
\end{array}
\right)
=
\left(
\begin{array}{cc}
  c_{\theta}  & -s_{\theta}   \\
  s_{\theta}&   c_{\theta}
\end{array}
\right)
\left(
\begin{array}{ccc}
x^2+y^2  \\
p(x,y)
\end{array}
\right)
\end{equation}
is a solution to (\ref{gen.syst}).

Since the Hessian matrices of $u^i$ are constant, the system (\ref{gen.syst}) is equivalent to the following two equations
\begin{equation*}
\left(-s_{\theta} F+c_{\theta} G\right) \left(\begin{array}{c}
 a\\b\\c
  \end{array}
 \right)= Y
\end{equation*}
where
\begin{equation*}
\begin{array}{cccc}
F=\left(\begin{array}{ccc}
 a_{11}&2a_{12}& a_{22}\\
  c_{11}&2c_{12}& c_{22}
 \end{array}
 \right),
 &
G=\left(\begin{array}{ccc}
 b_{11}&2b_{12}& b_{22}\\
  d_{11}&2d_{12}& d_{22}
 \end{array}
 \right),
 \end{array}
\end{equation*}
and $ Y=\left(\begin{array}{c}
 Y^1\\
 Y^2
 \end{array}
 \right)$ is a vector of known data, possibly depending on $\theta$. Given $\theta$,
the above system has at least one solution if  the rank of  $-s_{\theta} F+c_{\theta} G$ is two and consequently we find a solution of the form \eqref{quad.sol} to \eqref{gen.syst}. If this were not the case, then for every $\theta$ there exists $\phi$ such that
\begin{equation}\label{linear.dip}
c_{\phi} (-s_{\theta} A +c_{\theta} B)+ s_{\phi}(-s_{\theta} C +c_{\theta} D)=0.
\end{equation}
Choosing $\theta=0$ yields that $B$ and $D$ are linearly
dependent.  Similarly, by choosing $\theta=\frac{\pi}{2}$, we get
that $A$ and $C$ are linearly dependent. Recalling that, by
ellipticity \eqref{ell-1}, $A$ and $D$ are positive definite, and
hence nontrivial, we obtain  that $B= \sigma D$ and $C= \gamma A$
for suitable constants $\sigma, \gamma \in \mathbb{R}$. Plugging
these linear dependencies into \eqref{linear.dip} and using once
more ellipticity, we obtain that also
 $D$ is a scalar multiple of $A$. In conclusion $B, C$ and $D$ are scalar multiples of $A$.

\end{proof}
\begin{proof}[Proof of Proposition \ref{B.prop}]
We look for  $\theta \in [0,2 \pi]$ and a cubic polynomial $$q(x,y)= \frac{1}{2}\left(a \frac{x^3}{3}+  b x^2 y + c x y^2 +d \frac{y^3}{3}\right)$$ such that
\begin{equation*}
\left(
\begin{array}{c}
  u^1   \\
 u^2
\end{array}
\right)
=
\left(
\begin{array}{cc}
  c_{\theta}  & -s_{\theta}   \\
  s_{\theta}&   c_{\theta}
\end{array}
\right)
\left(
\begin{array}{ccc}
x(x^2+y^2)  \\
q(x,y)
\end{array}
\right)
\end{equation*}
is a solution to (\ref{gen.syst}).
This time the Hessian matrices of $u^i$ are of the form $x H_1+y H_2$ for suitable constant matrices $H_1$ and $H_2$ and thus (\ref{gen.syst}) imposes the following four conditions.
\begin{equation*}
\left(-s_{\theta} F+c_{\theta} G\right) \left(\begin{array}{c}
 a\\b\\c\\d
  \end{array}
 \right) = Y
\end{equation*}
where
\begin{equation*}
\begin{array}{cccc}
F=\left(\begin{array}{cccc}
 a_{11}&2a_{12}& a_{22}&0\\
 0& a_{11}&2a_{12}& a_{22}\\
  c_{11}&2c_{12}& c_{22}&0\\
0& c_{11}&2c_{12}& c_{22}
 \end{array}
 \right),
 &
G=\left(\begin{array}{cccc}
 b_{11}&2b_{12}& b_{22}&0\\
  0&b_{11}&2b_{12}& b_{22}\\
   d_{11}&2d_{12}& d_{22}&0\\
  0&d_{11}&2d_{12}& d_{22}
 \end{array}
 \right),
 \end{array}
\end{equation*}
and $ Y\in \mathbb R^4 $ is a data vector.
We make use of the following linear algebra fact.
 Given any two $2\times 2$ symmetric
matrices
\begin{equation*}
M=\left(\begin{array}{cccc}
 m_{11}&m_{12}\\
m_{12}& m_{22}
 \end{array}
 \right)\hbox{and\quad}
S= \left(\begin{array}{cccc}
 s_{11}&s_{12}\\
s_{12}& s_{22}
 \end{array}
 \right),
 \end{equation*}
if  we have
 \begin{equation*}
 \det \left(\begin{array}{cccc}
 m_{11}&2m_{12}& m_{22}&0\\
 0& m_{11}&2m_{12}& m_{22}\\
  s_{11}&2s_{12}& s_{22}&0\\
0& s_{11}&2s_{12}& s_{22}\\
 \end{array}
 \right)=0,
\end{equation*}
and either $M$ or $S$ is positive definite, then $M$ and $S$ are
linearly dependent. This fact may be verified in many ways, for
instance with the aid of Gaussian elimination.
Assume that $(-s_{\theta} F+ c_{\theta} G)$ is singular for every $\theta$, then we must have that the matrix $(-s_{\theta} C+ c_{\theta} D)$
 is a scalar multiple of  $
(-s_{\theta} A+ c_{\theta} B) $ at least for all those   $\theta$
for which $ (-s_{\theta} A+ c_{\theta} B) $ is positive definite.
Equivalently, switching the roles of $A, B$ and $D, C$, if
$(-s_{\theta} C+ c_{\theta} D)$ is positive definite, then  $
(-s_{\theta} A+ c_{\theta} B) $  is a scalar multiple of
$(-s_{\theta} C+ c_{\theta} D)$. Recalling that the matrices $A,
D$ are positive definite by ellipticity, and choosing $\theta=0,
\frac{3\pi}{2}$, we deduce $B= \sigma D$ and $C= \gamma A$ for
suitable constants $\sigma, \gamma \in \mathbb{R}$. Moreover, it
is evident that there exists an open interval $I$ containing
$\frac{3\pi}{2}$ for which $(-s_{\theta} A+ c_{\theta} B) $
remains positive definite as long as $\theta \in I$. Thus, for all
$\theta\in I$, there exists $\phi$ such that \eqref{linear.dip}
holds true, and from now
 we  can argue similarly as in the proof of Proposition \ref{A.prop}.
\end{proof}

\providecommand{\bysame}{\leavevmode\hbox to3em{\hrulefill}\thinspace}
\providecommand{\MR}{\relax\ifhmode\unskip\space\fi MR }

\providecommand{\MRhref}[2]{
  \href{http://www.ams.org/mathscinet-getitem?mr=#1}{#2}
}
\providecommand{\href}[2]{#2}


TITLE: Functional equation for ray class zeta functions
QUESTION [2 upvotes]: I understand that there is a functional equation for ideal class zeta functions of number fields.  That is, if $k$ is a number field and $C$ is a class of fractional ideals of $k$, then
$\zeta_C(s) = \sum_{\substack{\mathfrak{a} \in C\\ \mathfrak{a} \text{ integral}}} \frac{1}{N(\mathfrak{a})^s}$
has a functional equation relating $\zeta_C(s)$ to $\zeta_C(s-1)$ (I guess this is in Neukirch, Algebraic Number Theory, Chapter VII).  Does such a functional equation exist for general ray class zeta functions, where $C$ is replaced by a ray class modulo some modulus?  I am especially interested in when the modulus includes some infinite places, i.e., for "narrow ideal class zeta functions".
(I guess that maybe the answer is yes, since Hurwitz zeta functions satisfy a functional equation, at least for a limited range of $s$.)

REPLY [3 votes]: As @user1952009 commented, yes, with characters with the same conductor (and maybe same sign/epsilon factor in functional equation), linear combinations obviously inherit the functional equation.
For example, for not-narrow ideal classes, all the Hecke L-functions have the same functional equation, so arbitrary linear combinations inherit it, and the "ideal class zeta" you wrote is such a linear combination (by the usual Fourier inversion on the finite abelian group that is the class group).
A superficial thought experiment makes me think that the narrow-class-group characters all give L-functions with the same functional equation, too.
For general ray-class things not all characters will give L-functions with the same functional equation, so $\zeta_C(1-s)$ will be a messier linear combination of $\Lambda(1-s,\chi)$'s.


TITLE: Recurrence relations for a substring that contains 0000 and for partioning an integer
QUESTION [0 upvotes]: Having trouble with these two questions:

Let $S_n$ be the number of bitstrings of length $n$ that contain the substring $0000$. Which of
  the following is true?
(a) $S_n = S_{n-1} + S_{n-2} + S_{n-3} + S_{n-4}$
(b) $S_n = S_{n-1} + S_{n-2} + S_{n-3} + S_{n-4} + 2^{n-4}$
  (c) $S_n = S_{n-1} + S_{n-2} + S_{n-3}$
  (d) $S_n = S_{n-1} + S_{n-2} + S_{n-3} + S_{n-4} + 2^{n-4}$

The answer is (b).
I accounted for the bitstrings that begin with:
$1 \rightarrow (S_{n-1})$,
$01 \rightarrow (S_{n-2})$,
$001 \rightarrow (S_{n-3})$,
$0001 \rightarrow (S_{n-4}),$ and,
$0000  \rightarrow (2^{n-4})$
This gives $S_n = S_{n-1} + S_{n-2} + S_{n-3} + S_{n-4} + 2^{n-4}$
Is my logic correct?
How can I do this if I wanted to count bitstrings that do not contain $0000$ and then subtract those from all $2^n$ bitstrings to arrive at the same recurrence relation?

Next question (this one I am not sure how to do):

Let $n \ge 1$ be an integer and let $S_n$ be the number of ways in which $n$ can be written as a sum of 1s and 2s, such that:

the order in which the 1s and 2s occur in the sum matters, and
it is not allowed to have two consecutive 2s.

For example, if $n = 7$, then both,
  $7 = 1 + 2 + 1 + 2 + 1$ and $7 = 2 + 1 + 1 + 2 + 1$ are allowed, whereas $7 = 1 + 2 + 2 + 1 + 1$ is not allowed.
Which of the following is true?
(a) $S_n = S_{n-1} + S_{n-2}$
(b) $S_n = S_{n-1} + S_{n-3}$
  (c) $S_n = S_{n-2} + S_{n-3}$
  (d) $S_n = S_{n-1} + S_{n-2} + S_{n-3}$

The answer is (b).
Why is this true? I can't wrap my head around it. If all possibile sequences must start with $1$ and $21$ this should be (a) shouldn't it?

REPLY [1 votes]: For (2), you have 2 cases. a) the "partition" starts with a 1, and b) the partition starts with a 2.
a) This is just 1+(partition of n-1), and since here, the partition of n-1 used uniquely determines your partition of n, there are $S_{n-1}$ of these.
b) Since you start with 2, the next thing to be summed must be a 1. So your partition is 2+1+(partition of n-$\underbrace{3}_{=2+1}$). This is uniquely determined by the partition of n-3 that you use, so there are $S_{n-3}$ of these.
Combining cases a) and b) gives you $S_n=S_{n-1}+S_{n-3}$.


TITLE: Does the significance of frequency domain imply the resonance?
QUESTION [0 upvotes]: Two mobile phone, A and B, are placed on the top of a wooden desk without direct contact, the gap between these two phones is about 5 centimeters.
When A is vibrating, e.g., a call comes in, a time series data is captured by B's accelerator sensor, which represents the acceleration in x, y, and z axis.
After transforming this time series data into frequency domain, I notice there is a significant power around 22 Hz.
Besides, from some paper, I have read a statement：

“If an object is struck by a rod, it will vibrate at its natural
  frequency no matter how hard it is struck. The magnitude of the strike
  will increase the amplitude of vibration, but not its frequency...if a
  periodic force is applied at the same natural frequency of the object,
  the object exhibits amplified vibration – resonance.”

If this is true, does it suggestion the 22 Hz is the natural frequency of wooden desk? or it is just the vibration frequency of phone A? or even the resonance frequency of phone A?

REPLY [0 votes]: My bet is it's the vibration frequency of phone A. The vibrator in phone A causes pressure waves through the material to end up driving a 22 Hz oscillation on phone B's accelerometer.