text:
\subsection{The focal conic}

Recall that in a closed Riemannian manifold $(M,g)$, and given a submanifold $L\subset M$, we have the exponential map $\varepsilon:\nu L\to M$ from the normal bundle of $L$ to $M$, and that this is smooth and surjective. Moreover, if $d:M\times M\to{\bf R}$ denotes the distance function, then for all $x\in M$, there exists $x'\in L$ such that $$d(x,L)=\inf_{y\in L}d(x,y)=d(x,x').$$
If $\gamma:{\bf R}\to M$ is some geodesic with $\gamma(0)=x'$, $\gamma(t)=x$ and $\mathrm{length}(\gamma)=d(x,x')$, then $\dot{\gamma}(0)\in\nu_{x'}L$, with $\nu_yL\overset{\perp}{\oplus}T_yL=T_yM$ (see, for instance, \cite{BC01}). For $\overrightarrow{n}\in\nu L$, we call $\overrightarrow{n}$ a \textbf{focal vector} for $L$ if the map $\mathrm{d}_{\overrightarrow{n}}\varepsilon:T_{\overrightarrow{n}}\nu L\to T_{\varepsilon(\overrightarrow{n})}M$ is not injective, and the point $\varepsilon(\overrightarrow{n})\in M$ is called a \textbf{focal point} of $L$. From \cite{BH84}, we see that $x\in M$ is focal if and only if the distance function $d(-,x):L\to{\bf R}$ has a degenerate critical point.

The set of focal points for $L$ is called the \textbf{focal locus} of $L$.

\begin{proposition}\label{prop:focal-conic}
    The focal locus $Q$ of ${\bf RP}^2\subset{\bf CP}^2$ is the conic $$Q=\left\{[z_0:z_1:z_2]\in{\bf CP}^2\mid z_0^2+z_1^2+z_2^2=0\right\}.$$
    We call it the \textbf{focal conic}.
\end{proposition}

\begin{proof}
    We apply the very same arguments as in \cite[\S 6]{Ber97}, where the author computes the focal set of ${\bf CP}^{m+1}$ in ${\bf HP}^{m+1}$, but with the group $O(m+2)$ instead of $SU(m+2)$. Taking $m=1$, we obtain that the focal set of ${\bf RP}^2\subset{\bf CP}^2$ is the codimension-two subset $Q^2\subset{\bf CP}^2$ determined by: $$Q=\left\{\left.\tau\left(\frac{{\bf x}+i{\bf y}}{\sqrt{2}}\right)\,\right|\,{\bf x},{\bf y}\in {\bf S}^2,\;\langle{\bf x},{\bf y}\rangle_{\bf R}=0\right\},$$
    with $$\begin{matrix}\tau&:&{\bf S}^5&\to&{\bf CP}^2\\&&(z_0,z_1,z_2)&\mapsto&[z_0:z_1:z_2]\end{matrix}$$
    the Hopf map, and ${\bf S}^2\subset{\bf S}^5$ induced by the inclusion ${\bf R}^3\subset{\bf C}^3$. From the computation $$\left(\frac{a+i\alpha}{\sqrt{2}}\right)^2+\left(\frac{b+i\beta}{\sqrt{2}}\right)^2+\left(\frac{c+i\gamma}{\sqrt{2}}\right)^2=\frac{1}{2}\left\|\begin{pmatrix}a\\b\\c\end{pmatrix}\right\|^2-\frac{1}{2}\left\|\begin{pmatrix}\alpha\\\beta\\\gamma\end{pmatrix}\right\|^2+i\left\langle\!\begin{pmatrix}a\\b\\c\end{pmatrix},\begin{pmatrix}\alpha\\\beta\\\gamma\end{pmatrix}\!\right\rangle,$$
    we readily see that $Q$ is as claimed.
\end{proof}

Conversely, ${\bf RP}^2$ is the focal locus of $Q$, and if we re-normalize the function $d(-,{\bf RP}^2)$ as $f:{\bf CP}^2\to[0,1]$, we have:
\begin{enumerate}[label=(\roman*)]
    \item ${\bf RP}^2=f^{-1}(0)$ and $Q=f^{-1}(1)$.
    \item For all $0<\varepsilon<1$, $M_\varepsilon^+=f^{-1}([0,\varepsilon])$ and $M_\varepsilon^-=f^{-1}([\varepsilon,1])$ are closed regular neighborhoods of ${\bf RP}^2$ and $Q$ in ${\bf CP}^2$, respectively.
    \item $\partial M_{1/2}^+=\partial M_{1/2}^-=f^{-1}(1/2)$ is a Lens space $L(4,1)$.
\end{enumerate}
For the third claim, note that ${\bf RP}^2$ is Lagrangian in ${\bf CP}^2$, and thus $\nu{\bf RP}^2\cong-T{\bf RP}^2$. As such, the unit normal bundle of ${\bf RP}^2$ is identified with its unit tangent bundle, which is doubly-covered by $SO(3)$ the unit tangent bundle of ${\bf S}^2$. Finally, $SO(3)/\{\pm1\}\cong{\bf S}^3/\{\pm1,\pm i\}={\bf S}^3/({\bf Z}/4)\cong L(4,1)$.
\subsection{Totally flexible curves}

A \textbf{real plane algebraic curve} is a real \textit{non-singular} homogenous polynomial: $X\in{\bf R}[x_0:x_1:x_2]$. By the \textbf{real part} of the curve, we mean the set $${\bf R} X=\left\{[x_0:x_1:x_2]\in{\bf RP}^2\mid X(x_0,x_1,x_2)=0\right\},$$
and by the \textbf{complexification} of the curve, we mean $${\bf C} X=\left\{[z_0:z_1:z_2]\in{\bf CP}^2\mid X(z_0,z_1,z_2)=0\right\}.$$
Evidently, ${\bf C} X$ is invariant under complex conjugation. If $m=\deg(X)\geqslant1$, then we see that $[{\bf C} X]=m[{\bf CP}^1]\in H_2({\bf CP}^2;{\bf Z})$, and that ${\bf C} X$ is a surface of genus $g=\frac{(m-1)(m-2)}{2}$. Also, the tangent space of the complexification of the curve is the complexification of the tangent space of its real part: $T_{\bf C}({\bf C} X)={\bf C}\cdot T_{\bf R}({\bf R} X)$. Note that the focal conic itself is an algebraic curve of degree two, with empty real part ${\bf R} Q=\varnothing$. In particular, $Q$ is a 2-sphere.

We define a \textbf{flexible} curve, in the sense of Viro \cite{Vir84}, as follows:

\begin{definition}\label{def:flexible}
    Let $F$ be a closed oriented surface embedded in ${\bf CP}^2$. The surface $F$ is called a degree $m\geqslant1$ \textbf{flexible curve} if:
    \begin{enumerate}[label=(\roman*)]
        \item $\mathrm{conj}(F)=F$;
        \item $\chi(F)=-m^2+3m$;
        \item $[F]=m[{\bf CP}^1]\in H_2({\bf CP}^2;{\bf Z})$;
        \item $T_{\bf C} F={\bf C}\cdot T_{\bf R}({\bf R} F)$, where ${\bf R} F=F\cap{\bf RP}^2$.
    \end{enumerate}
\end{definition}

The following classical results are known for flexible curves (see \cite{Rok78}).
\begin{enumerate}[label=(\roman*)]
    \item If $b_0$ denotes the 0th Betti number, then $b_0({\bf R} F)\leqslant g+1=\frac{(m-1)(m-2)}{2}+1$. Curves with $b_0({\bf R} F)=g+1$ are called \textbf{$M$-curves}. On the other hand, curves with $b_0({\bf R} F)=0$ if $m$ is even and $b_0({\bf R} F)=1$ if $m$ is odd $\geqslant3$ are called \textbf{minimal} curves.
    \item If $m$ is even, then each component of ${\bf R} F$ is contractible in ${\bf RP}^2$, and if $m$ is odd, all but one components of ${\bf R} F$ are. Contractible components of ${\bf R} F$ are called \textbf{ovals}. An odd degree curve can never have ${\bf R} F=\varnothing$ (it always has the non-contractible component).
    \item A flexible curve $F$ is said to be a \textbf{type I} (\textit{resp.} \textbf{type II}) curve if $F\smallsetminus{\bf R} F$ has two connected components (\textit{resp.} is connected). An $M$-curve is always type I, and a minimal curve is always type II.
\end{enumerate}

What makes flexible curves so different from algebraic curves is the lack of rigidity, mainly seen with the Bézout theorem. If ${\bf RP}^2$ is one "end" of ${\bf CP}^2$ where we restrict the behaviour of flexible curves, then the "other end", the focal conic, should also play a role. The Bézout theorem states that a degree $m$ algebraic curve generically intersects it transversely in exactly $2m$ points.

\begin{definition}
    A flexible curve $F$ of degree $m\geqslant1$ is called \textbf{totally flexible} if $F\pitchfork Q$ consists of $2m$ points, necessarily swapped pairwise by complex conjugation.
\end{definition}

\subsection{Two 2-fold branched covers}

There is a well-known diffeomorphism between ${\bf CP}^2/\mathrm{conj}$ and ${\bf S}^4$ (see \cite{Kui74}). We denote the associated (cyclic) 2-fold branched cover as $p:({\bf CP}^2,{\bf RP}^2)\to({\bf S}^4,\mathcal{R})$, with $\mathcal{R}=p({\bf RP}^2)$ an embedded ${\bf RP}^2$ in ${\bf S}^4$. We also let $\mathcal{Q}=p(Q)$ the image of the focal conic under that branched cover. From the fact that $Q$ does not intersect the branch locus ${\bf RP}^2$, we see that the restriction $p:Q\to\mathcal{Q}$ is an unbranched 2-fold cover, with the conjugation being an orientation-reversing involution generating the group of deck transformations. As such, we see that $\mathcal{Q}$ is also an embedded ${\bf RP}^2$ in ${\bf S}^4$.

Given an embedded surface $F^2$ in a (closed oriented) 4-manifold $X^4$, we denote as $e(X,F)$ the normal Euler number of the embedding $F\subset X$; that is, the Euler class of the normal bundle $\nu F$. It is also equal to the self-intersection number $F\cdot F$, which is defined by counting signed intersection points between $F$ and a small perturbation $F'$ of $F$ in the normal direction. Moreover, if $F$ is oriented, then it corresponds to the intersection form of $X$ evaluated at $F$, with a sign depending on whether that orientation is compatible with the ambient manifold or not.

\begin{proposition}
    We have the following normal Euler numbers: $$e({\bf CP}^2,{\bf RP}^2)=-1,\; e({\bf CP}^2,Q)=+4,\;e({\bf S}^4,\mathcal{R})=-2\text{ and }e({\bf S}^4,\mathcal{Q})=+2.$$
\end{proposition}

\begin{proof}
    The focal conic $Q$ is oriented compatibly with ${\bf CP}^2$ and $[Q]=2[{\bf CP}^1]\in H_2({\bf CP}^2;{\bf Z})$, so we obtain $e({\bf CP}^2,Q)=+4$. Next, because ${\bf RP}^2$ is Lagrangian in ${\bf CP}^2$, we have $\nu{\bf RP}^2\cong-T{\bf RP}^2$, and thus, for the Euler class: $e(\nu{\bf RP}^2)=-e(T{\bf RP}^2)=-\chi({\bf RP}^2)=-1$. Finally, the computations of $e({\bf S}^4,\mathcal{R})$ and $e({\bf S}^4,\mathcal{Q})$ come from the next lemma.
\end{proof}

\begin{lemma}\label{lem:branched-cover-euler-number}
    Given a (simple) 2-fold branched cover $f:(Y^4,\tilde{B}^2)\to(X^4,B^2)$, and given $F$ an embedded surface in $X$, we denote as $\tilde{F}$ the lift $p^{-1}(F)$.
    \begin{enumerate}[label=(\roman*)]
        \item If $F\pitchfork B\neq\varnothing$, then $e(Y,\tilde{F})=2e(X,F)$.
        \item If $F\subset B$, then $e(Y,\tilde{F})=\frac{1}{2}e(X,F)$.
    \end{enumerate}
\end{lemma}

\begin{proof}
    One has to inspect what happens in each case individually. In any case, note that a perturbation of the lift is the lift of a perturbation. For the first claim, one can ensure that the self-intersection points occur away from the ramification locus $B$. As such, each of these points lifts to two intersections, and the orientations agree because $f$ is orientation-preserving.

    In the second case, note that each point of self-intersection of $F$ comes from a point of intersection of $\tilde{F}$. However, because the local model map $z\mapsto z^2$ identifies those points in pairs, one sees that $e(X,F)=2e(Y,\tilde{F})$.
\end{proof}

We now wish to consider the 2-fold branched cover of ${\bf S}^4$, ramified over $\mathcal{Q}$ this time.

\begin{proposition}
    There exists an orientation-reversing involution $\psi:{\bf S}^4\to{\bf S}^4$ with $\psi(\mathcal{R})=\mathcal{Q}$.
\end{proposition}

\begin{proof}
    We can go further in the details of the proof of Proposition \ref{prop:focal-conic}. For $\textbf{x},\textbf{y}\in{\bf S}^2$ with $\langle\textbf{x},\textbf{y}\rangle_{\bf R}=0$, set $\gamma_{\textbf{x},\textbf{y}}(t)=\tau(\cos(t)\textbf{x}+i\sin(t)\textbf{y})$, with $\tau:{\bf S}^5\to{\bf CP}^2$ the Hopf map. Note that the geodesics normal to ${\bf RP}^2$ are exactly the $\gamma_{\textbf{x},\textbf{y}}$. Then: $${\bf CP}^2=\bigsqcup_{r\in[0,\pi/4]}N_r,$$
    with $N_r=\{\gamma_{\textbf{x},\textbf{y}}(r)\mid\textbf{x},\textbf{y}\in{\bf S}^2,\,\langle\textbf{x},\textbf{y}\rangle_{\bf R}=0\}$. From the relation $$\overline{\gamma_{\textbf{x},\textbf{y}}(t)}=\gamma_{\textbf{x},-\textbf{y}}(t),$$
    we see that $\mathrm{conj}(N_r)=N_r$ for all $r$. Therefore, one has: $${\bf S}^4=\bigsqcup_{r\in[0,\pi/4]}N_r/\mathrm{conj}.$$
    Note that the $N_r/\mathrm{conj}$ are pairwise distinct, and that $N_0=\mathcal{R}$ and $N_{\pi/4}=\mathcal{Q}$. Moreover, we have already seen that $N_r\cong L(4,1)$ for all $0<r<\pi/4$. Therefore, one has a fibration $f:{\bf S}^4\to[0,\pi/4]$, with fibers $f^{-1}(0)\cong f^{-1}(\pi/4)\cong{\bf RP}^2$ and $f^{-1}(r)\cong L(4,1)/\mathrm{conj}$ for $0<r<\pi/4$. One therefore defines the involution $\psi:{\bf S}^4\to{\bf S}^4$ by reversing the orientation of $[0,\pi/4]$ in that fibration.
\end{proof}

Here, $\psi$ does \textit{not} lift to $\Psi:{\bf CP}^2\to{\bf CP}^2$, for $N_0$ and $N_{\pi/4}$ are not diffeomorphic fibers of the fibration ${\bf CP}^2\to[0,\pi/4]$. However, it is sufficient to see that $\tilde{p}=\psi\circ p$ is an orientation-reversing 2-fold branched cover ${\bf CP}^2\to{\bf S}^4$, with branch locus $\psi(\mathcal{R})=\mathcal{Q}$. This can be interpreted as a branched cover $\tilde{p}:(\overline{\bf CP}^2,\overline{\bf RP}^2)\to({\bf S}^4,\mathcal{Q})$, where $\overline{\bf RP}^2$ is $\tilde{p}^{-1}(\mathcal{Q})$ by definition.

Define $\overline{Q}=\tilde{p}^{-1}(\mathcal{R})$. We see that $\overline{\bf RP}^2$ and $\overline{Q}$ are respectively embeddings of ${\bf RP}^2$ and ${\bf S}^2$ in $\overline{\bf CP}^2$. By using Lemma \ref{lem:branched-cover-euler-number} again, we can compute the normal Euler numbers.

\begin{proposition}
    We have $e(\overline{\bf CP}^2,\overline{\bf RP}^2)=+1$ and $e(\overline{\bf CP}^2,\overline{Q})=-4$.\qed
\end{proposition}

In Figure \ref{fig:diagram} we depict a summary of the different maps in play.

\begin{figure}[ht]
    \centering
    \begin{tikzcd}
    	{{\bf RP}^2} & {{\bf CP}^2} & Q \\
    	\mathcal{R} & {{\bf S}^4} & \mathcal{Q} \\
    	\overline{Q} & {\overline{\bf CP}^2} & {\overline{\bf RP}^2}
    	\arrow["{\tilde{p}}"', from=3-2, to=2-2]
    	\arrow["p"', from=1-2, to=2-2]
    	\arrow["{||}"{description}, draw=none, from=1-1, to=2-1]
    	\arrow["{||}"{description}, draw=none, from=3-3, to=2-3]
    	\arrow[shorten <=1pt, shorten >=1pt, two heads, from=1-3, to=2-3]
    	\arrow[shorten <=2pt, shorten >=2pt, two heads, from=3-1, to=2-1]
    	\arrow["\subset"{description}, draw=none, from=1-1, to=1-2]
    	\arrow["\supset"{description}, draw=none, from=1-3, to=1-2]
    	\arrow["\supset"{description}, draw=none, from=2-3, to=2-2]
    	\arrow["\subset"{description}, draw=none, from=2-1, to=2-2]
    	\arrow["\subset"{description}, draw=none, from=3-1, to=3-2]
    	\arrow["\supset"{description}, draw=none, from=3-3, to=3-2]
    \end{tikzcd}
    \caption{The two branched coverings of interest and their associated branch loci. The arrows marked $\longrightarrow\!\!\!\!\!\to$ denote an unbranched 2-fold cover from a 2-sphere to a real projective plane.}\label{fig:diagram}
\end{figure}

\subsection{Branching over the Arnol'd surface}

Let $\Sigma\subset\overline{\bf CP}^2$ be the image of a closed surface through an immersion, with only one transverse self-intersection point $x\in\Sigma$. Take $B\subset\overline{\bf CP}^2$ to be a small ball around $x$, which meets $\Sigma$ at two disks intersecting transversely at their common center $x$. Remove these two disks from $\Sigma$, and re-glue a cylinder to the two resulting circles $\partial B\cap\Sigma$. Here, there is a choice involved in the gluing of that cylinder. Indeed, the link $\partial B\cap\Sigma\subset\partial B\cong{\bf S}^3$ is a Hopf link, and Hopf bands will be isotopic in the ball $B$. However, it is important here that this isotopy fixes the two components of the link \textit{individually}, which gives rise to both possible gluings (see Figure \ref{fig:sing-res}). We call either choice of the result of gluing a cylinder a \textbf{resolution of the singularity} of the immersed surface $\Sigma\subset\overline{\bf CP}^2$.

\begin{lemma}
    The resulting surface $\Sigma'$ is an embedded surface in $\overline{\bf CP}^2$, with the same orientability as that of $\Sigma$ and $\chi(\Sigma')=\chi(\Sigma)-1$. Moreover,  $e(\overline{\bf CP}^2,\Sigma')=e(\overline{\bf CP}^2,\Sigma)\pm2$, and the sign is determined by either choice of the two possible gluings.
\end{lemma}

\begin{proof}
    The only non-obvious claim is that about the normal Euler numbers. The fact that there are two possibilities comes from the choice in the resolution of the singularity (see Figure \ref{fig:sing-res}).

    Because everything is local, it suffices to inspect the effect of both choices on the immersion obtained by the union of two distinct projective lines $L_1$ and $L_2$. Each has $e(\overline{\bf CP}^2,L_i)=-1$ (because $e({\bf CP}^2,L_i)=+1$), so $e(\overline{\bf CP}^2,L_1\cup L_2)=-2$. Both cases yield a sphere $(L_1\cup L_2)'$. In one case, the resolution has the effect of transforming $L_1\cup L_2$ into a conic, hence $e(\overline{\bf CP}^2,(L_1\cup L_2)')=-4$. In the other case, we can fill the resulting sphere with a ball, which means it is null-homologous and as such, has $e(\overline{\bf CP}^2,(L_1\cup L_2)')=0$. In particular, we do observe that the normal Euler number is changed by $\pm2$, as claimed. The fact that $\chi(\Sigma')=\chi(\Sigma)-1$ is simply a matter of using the formula $\chi(A\cup B)=\chi(A)+\chi(B)-\chi(A\cap B)$ twice. Indeed, if $C$ denotes the cylinder that is glued to $\Sigma\smallsetminus B$, then: $$\chi(\Sigma')=\chi(\Sigma\smallsetminus B\cup_\partial C)=\chi(\Sigma\smallsetminus B)+\chi(C)-\chi({\bf S}^1\sqcup{\bf S}^1)=\chi(\Sigma\smallsetminus B),$$
    and: $$\chi(\Sigma)=\chi(\Sigma\smallsetminus B\cup_\partial B\cap\Sigma)=\chi(B\smallsetminus B)+\chi(B\cap\Sigma)-\chi({\bf S}^1\sqcup{\bf S}^1)=\chi(\Sigma\smallsetminus B)+1,$$
    by noting that $B\cap\Sigma$ is topologically a wedge of two discs.
\end{proof}

\begin{figure}[ht]
    \centering
    \begin{tikzpicture}
        \node[inner sep=0pt] at (0,0) {\includegraphics{img/singularity-resolution2.eps}};
        \node at (-2,0) {$\longleftarrow$};
        \node at (2,0) {$\longrightarrow$};
    \end{tikzpicture}
    \caption{The two possible resolutions of a singularity of an immersion. It is understood that in the middle, the two disks intersect transversely at their common center only (this 4-dimensional phenomenon cannot be depicted in 3 dimensions).}
    \label{fig:sing-res}
\end{figure}

Consider $F\subset{\bf CP}^2$ a totally flexible curve of odd degree $m$. The Arnol'd surface $\mathcal{A}(F)$ needs not be orientable, and as said before, there is no 2-fold branched cover of $(\overline{\bf CP}^2,\mathcal{A}(F))$. Note that $\mathcal{A}(F)\pitchfork\overline{\bf RP}^2$ is $m$ points, and as such, $\mathcal{A}(F)\cup\overline{\bf RP}^2$ is an immersed surface with $m$ transverse crossings only. Apply the previous resolutions of the singularities at each of those $m$ points: this yields a surface $\mathcal{X}(F)\subset\overline{\bf CP}^2$, with: $$\begin{cases}\chi(\mathcal{X}(F))=\chi(\mathcal{A}(F)\cup\overline{\bf RP}^2)-m;\\e(\overline{\bf CP}^2,\mathcal{X}(F))=e(\overline{\bf CP}^2,\mathcal{A}(F)\cup\overline{\bf RP}^2)+2r,\,r\in\{-m,\dots,m\}.\end{cases}$$
We are also free to choose any $r$ in the successive resolutions. Define $\mathcal{X}(F)$ to be the one where we pick up a $+2$ every time (that is, $r=+m$). Two applications of the topological Riemann--Hurwitz formula give $\chi(\mathcal{A}(F))=\chi(F)-m+1$. Therefore, we have: $$\chi(\mathcal{X}(F))=-m^2+2.$$

We are now interested in computing the normal Euler number of $\mathcal{X}(F)\subset\overline{\bf CP}^2$. For this, we use the following result.

\begin{proposition}
    We have $e(\overline{\bf CP}^2,\mathcal{A}(F))=m^2-2$.
\end{proposition}

\begin{proof}
    Recall that we denoted as $\overline{A^+}(F)=\tilde{p}^{-1}(A^+(F))$, and $\mathcal{A}(F)=\overline{A^+}(F)\cup\overline{Q}_+$. Viewing the normal Euler number as half the integer obstruction of extending a normal field of lines (instead of a normal field of vectors), we are allowed to make use of Lemma \ref{lem:branched-cover-euler-number} with half-integers (because one could reformulate everything in this claim with the obstruction for lines instead). As such, we can use it twice to compute that $e(\overline{\bf CP}^2,\overline{A^+}(F))=m^2$, where we see that in the middle, $e({\bf S}^4,A^+(F))=m^2/2$ is a half-integer because $m$ is odd. Now, we simply make use of the fact that $e(\overline{\bf CP}^2,\overline{Q}_+)=-2\chi(\overline{Q}_+)=-2$, because of the next lemma.
\end{proof}

\begin{lemma}\label{lem:conic-lagrangian}
    Let $X$ be a submanifold of $\overline{Q}$. Then: $e(\overline{\bf CP}^2,X)=-2\chi(X)$.
\end{lemma}

\begin{proof}
    The submanifold ${\bf RP}^2\subset{\bf CP}^2$ being Lagrangian, and the covering $p:{\bf CP}^2\to{\bf S}^4$ being branched exactly on $p({\bf RP}^2)$, we observe that $\nu\mathcal{R}\cong-T\mathcal{R}$ in ${\bf S}^4$. However, the covering $\tilde{p}:\overline{\bf CP}^2\to{\bf S}$ is \textit{unbranched} in a regular neighborhood of $\mathcal{R}$, whence $\nu\overline{Q}\cong-2T\overline{Q}$. In particular, for the Euler classes, we have: $e(\overline{\bf CP}^2,X)=e(\nu X)=-2e(TX)=-2\chi(X)$.
\end{proof}

Next, we compute: $$e(\overline{\bf CP}^2,\mathcal{X}(F))=e(\overline{\bf CP}^2,\mathcal{A}(F)\cup\overline{\bf RP}^2)+2m=m^2+2m-1.$$

Take $Y^4$ to be the 2-fold cover of $\overline{\bf CP}^2$ branched over $\mathcal{X}(F)$. This has been made possible because the surface $\mathcal{X}(F)$ has zero homology mod 2: $[\mathcal{X}(F)]=0\in H_2(\overline{\bf CP}^2;{\bf Z}/2)$. Indeed: $$H_2(\overline{\bf CP}^2;{\bf Z}/2)=\mathrm{Span}_{{\bf Z}/2}([\overline{\bf RP}^2]),$$
and $\mathcal{A}(F)$ intersects $\overline{\bf RP}^2$ in an odd number $m$ of points. Therefore, we deduce that $[\mathcal{A}(F)]=[\overline{\bf RP}^2]$ in $H_2(\overline{\bf CP}^2;{\bf Z}/2)$. Now, adding $\overline{\bf RP}^2$ and resolving the singularities means that $\mathcal{X}(F)\cap\overline{\bf RP}^2=\varnothing$, and as such $[\mathcal{X}(F)]=0$ in $H_2(\overline{\bf CP}^2;{\bf Z}/2)$. We denote as $\Theta:Y^4\to(\overline{\bf CP}^2,\mathcal{X}(F))$ the 2-fold branched cover. The previous computations of $\chi(\mathcal{X}(F))$ and $e(\overline{\bf CP}^2,\mathcal{X}(F))$ will allow us to obtain homological information about the 4-manifold $Y$.

\begin{proposition}\label{prop:torsionHomology}
    The homology groups $H_1(Y;{\bf Z})$ and $H_3(Y;{\bf Z})$ are torsion. In particular, for Betti numbers, we have: $b_1(Y)=b_3(Y)=0$.
\end{proposition}

\begin{proof}
    In order to show that $H_1(Y;{\bf Z})$ is torsion, it is sufficient to know that $H_1(Y;{\bf Z}/2)=0$, for any free part ${\bf Z}^p<H_1(Y;{\bf Z})$ would give $p$ copies of ${\bf Z}/2$ in $H_1(Y;{\bf Z}/2)$. We use a generalisation of the Gysin sequence, as stated in \cite[Theorem 1]{LW95} (with $A$ a point in $\mathcal{X}(F)$, lifting to $\tilde{A}$ a single point in $Y$): $$\cdots\longrightarrow H_1(\overline{\bf CP}^2,\mathcal{X}(F);{\bf Z}/2)\longrightarrow H_1(Y,\ast;{\bf Z}/2)\longrightarrow H_1(\overline{\bf CP}^2,\mathcal{X}(F);{\bf Z}/2)\longrightarrow\cdots$$
    Here, $H_1(Y,\ast;{\bf Z}/2)\cong\tilde{H}_1(Y;{\bf Z}/2)$ the reduced homology group, and we have $H_1(\overline{\bf CP}^2,\mathcal{X}(F);{\bf Z}/2)=0$. This provides $H_1(Y;{\bf Z}/2)=0$, as claimed. For $b_3(Y)=0$, this is a consequence of $b_1(Y)=0$ and Poincaré duality.
\end{proof}

An educated guess is that $Y$ should be simply-connected, just like the usual branched cover of ${\bf CP}^2$ branched over an \textit{algebraic} curve $\{P(x_0,x_1,x_2)=0\}$, given as the algebraic surface $\{P(x_0,x_1,x_2)=w^2\}\subset{\bf CP}^3$, is simply-connected (see \cite{Wil78}). However, this enables us to compute all the homological invariants of $Y$ that will be useful. We recall the Hirzebruch formula for the signature of 2-fold branched covers.

\begin{theorem}[{\cite[Section 3]{Hir69}}]\label{thm:H}
    Let $f:(Y,B)\to(X,A)$ be a cyclic 2-fold branched cover, with $X$ and $Y$ both closed oriented 4-manifolds, $A$ a closed surface and $f$ orientation-preserving. Then, we have the following: $$\sigma(Y)=2\sigma(X)-\frac{1}{2}e(X,A).$$
\end{theorem}

\begin{proposition}
    We have:
    $$\chi(Y)=m^2+4,\;b_2(Y)=m^2+2,\;\sigma(Y)=\frac{-m^2-2m-3}{2},$$
    $$b_2^+(Y)=\frac{(m-1)^2}{4}\text{ and }b_2^-(Y)=\frac{3m^2+2m+7}{4},$$
    where $b_2^+(Y)$ and $b_2^-(Y)$ respectively denote the ranks of the subspaces of $H_2(Y;{\bf Z})$ on which the intersection form $Q_Y$ is positive and negative definite.
\end{proposition}

\begin{proof}
    First, the topological Riemann--Hurwitz formula again yields: $$\chi(Y)=2\chi(\overline{\bf CP}^2)-\chi(\mathcal{X}(F))=m^2+4.$$
    Next, we use the Theorem \ref{thm:H} with the branched cover $\Theta$ to obtain: $$\sigma(Y)=2\sigma(\overline{\bf CP}^2)-\frac{1}{2}e(\overline{\bf CP}^2,\mathcal{X}(F))=-2-\frac{m^2+2m-1}{2}=\frac{-m^2-2m-3}{2}.$$
    Now, because of Proposition \ref{prop:torsionHomology}, we see that $\chi(Y)=2+b_2(Y)$. This finally provides $$\begin{cases}b_2^+(Y)+b_2^-(Y)=b_2(Y)\\b_2^+(Y)-b_2^-(Y)=\sigma(Y),\end{cases}$$
    which we can easily solve for $b_2^\pm(Y)$.
\end{proof}

From now on, we let $\Theta:Y\to(\overline{\bf CP}^2,\mathcal{X}(F))$ be the branched cover, and $Q_Y$ the intersection form of $Y$. That is, if $\Sigma,\Sigma'\subset Y$ are embedded closed surfaces, we denote by $Q_Y(\Sigma,\Sigma')$ the value of the intersection form on the integer homology classes they determine.
\subsection{Proving the inequality}

We will now mostly mimic the proof that Viro and Zvonilov did in \cite{VZ92}. Note that the construction of $\mathcal{X}(F)$ from $\mathcal{A}(F)\cup\overline{\bf RP}^2$ happens away from a neighborhood $\overline{Q}$. In particular, we still see $\overline{{\bf R}} F$ embedded inside $\mathcal{X}(F)$. Given an oval $o\subset{\bf R} F$, recall that $\overline{o}\subset\overline{{\bf R}} F$ is two disjoint circles. In $\overline{Q}_\pm$, there is exactly one subset $C_\pm(o)$ with two connected components that has $\partial C_\pm(o)=\overline{o}$. See Figure \ref{fig:cPMoval} for a depiction of the two subsets $C_\pm(o)$.

\begin{figure}[ht]
    \centering
    \begin{tikzpicture}
        \node[inner sep=0pt] at (0,0) {\includegraphics{img/cPMoval.eps}};
        \node at (-2.7,-2.5) {$C_-(o)\subset\overline{Q}_-$};
        \node at (2.9,-2.5) {$C_+(o)\subset\overline{Q}_+$};
    \end{tikzpicture}
    \caption{Using the same scheme $\langle J\sqcup 2\sqcup1\langle1\rangle\rangle$ as in Figure \ref{fig:arnold-surface}, we take $o$ to be the only non-empty oval. In orange we depict $\overline{o}\subset\overline{{\bf R}} F$, and in gray the two possible subsets $C_\pm(o)$.}
    \label{fig:cPMoval}
\end{figure}

We see that $C_+(o)$ is totally included in the ramification locus of the branched cover $\Theta:Y\to\overline{\bf CP}^2$, and that $C_-(o)$ intersects this ramification locus only at its boundary $\partial C_-(o)=\overline{o}$. We let: $\tilde{C}(o)=\Theta^{-1}(C_-(o))$. The restriction $\Theta:\tilde{C}(o)\to C_-(o)$ is not a branched cover, but it is close enough: it maps the boundaries $\partial\tilde{C}(o)\to\partial C_-(o)$ diffeomorphically, and is is two-to-one on the interior. Because $C_-(o)$ is planar (that is, a sphere with holes), we have that $\tilde{C}(o)$ is obtained as gluing two sphere with holes along their boundary components. Additionnally, $\Theta:Y\to\overline{\bf CP}^2$ is orientation-preserving, so that $\mathrm{Aut}(\Theta)$ is a ${\bf Z}/2$ spanned by $\tau:Y\to Y$ an orientation-reversing involution. Finally, $\tau$ swaps those two planar surfaces in $Y$ that glue to $\tilde{C}(o)$ and fixes their common boundary. As such, we have shown the next result.

\begin{proposition}
    For any oval $o\subset{\bf R} F$, $\tilde{C}(o)$ is an oriented surface in $Y$ of genus $b$ the number of ovals directly contained in $o$. The restriction $\Theta:\tilde{C}(o)\to C_-(o)$ is the result of the quotient of the surface $\Sigma_b$ by reflection along a plane of symmetry that cuts it into two planar surfaces.\qed
\end{proposition}

\begin{figure}[ht]
    \centering
    \begin{tikzpicture}
        \node[inner sep=0pt] at (0,0) {\includegraphics{img/Ctilda.eps}};
        \node at (0,0) {$\overset{\Theta}{\longleftarrow}$};
        \node at (-3,-1.1) {$\chi(o)=-1$};
        \node at (3,-1.1) {$e(Y,\tilde{C}(o))=+4$};
        \node at (-3,1.1) {$C_-(o)$};
        \node at (3,1.1) {$\tilde{C}(o)$};
    \end{tikzpicture}
    \caption{The "pseudo" branched cover $\tilde{C}(o)\to C_-(o)$.}
\end{figure}

The same construction works for $J$: there is one subset $D_\pm(J)$ of $\overline{Q}_\pm$ bounded by $\overline{J}$. Letting $\tilde{D}(J)=\Theta^{-1}(D_-(J))$, we have that $\tilde{D}(J)$ is a surface of genus $e$ the number of exterior ovals in ${\bf R} F$ (those not included in any other), and the restriction $\Theta:\tilde{D}(J)\to D_-(J)$ is again the quotient of $\Sigma_e$ by reflection along a plane in the middle.

Given an oval $o\subset{\bf R} F$, we denote as $\chi(o)=\chi(C_-(o))$ the Euler characteristic of the connected subset of ${\bf RP}^2\smallsetminus{\bf R} F$ bounded by $o$ from outside. Similarly, we let $\chi(J)=\chi(D_-(J))$. One remarks that $\chi(o)\leqslant1$, with equality if and only if $o$ is empty, and that $\chi(J)=1-e$ with $e$ the number of exterior ovals.

\begin{proposition}
    Let $o,o'\subset{\bf R} F$ be ovals, and denote again by  $J$ the non-contractible component of ${\bf R} F$.
    \begin{enumerate}
        \item We have $Q_Y(\tilde{C}(o),\tilde{C}(o))=-4\chi(o)$ and $Q_Y(\tilde{D}(J),\tilde{D}(J))=-4\chi(J)=4(e-1)$ with $e$ the number of exterior ovals.
        \item We have $Q_Y(\tilde{C}(o),\tilde{D}(J))=0$, and if $o\neq o'$, then $Q_Y(\tilde{C}(o),\tilde{C}(o'))=0$.
    \end{enumerate}
\end{proposition}

\begin{proof}
    Note that if $\Sigma\subset Y$ is an embedded non nulhomologous surface, we need not have that $Q_Y(\Sigma,\Sigma)$ is a square, for $H_2(Y;{\bf Z})$ has rank more than one. For the first statement, observe that $e(\overline{\bf CP}^2,C_-(o))=-2\chi(o)$ and $e(\overline{\bf CP}^2,D_-(J))=-2\chi(J)$, by definition of $\chi(o)$ and $\chi(J)$, and by using Lemma \ref{lem:conic-lagrangian}. Next, from Lemma \ref{lem:branched-cover-euler-number}, we can see that $e(Y,\tilde{C}(o))=2e(\overline{\bf CP}^2,C_-(o))$ and $e(Y,\tilde{D}(J))=2e(\overline{\bf CP}^2,D_-(J))$. To derive $Q_Y(\tilde{C}(o),\tilde{C}(o))$ and $Q_Y(\tilde{D}(J),\tilde{D}(J))$, remark that $\tilde{C}(o)$ and $\tilde{D}(J)$ are surfaces oriented positively in $Y$.

    For the second claim, distinct ovals $o$ and $o'$ cannot satisfy $C_-(o)\cap C_-(o')\neq\varnothing$, even if one is included inside the other (however, it is possible that $C_-(o)\cap C_+(o')\neq\varnothing$). The same goes for $C_-(o)\cap D_-(J)=\varnothing$. As such, the surfaces $\tilde{C}(o)$, $\tilde{C}(o')$ and $\tilde{D}(J)$ are non-intersecting in $Y$.
\end{proof}

The homology classes of the surfaces $\tilde{C}(o_i)$, $i\in[\![1,\ell]\!]$ and $\tilde{D}(J)$ were respectively denoted as $\beta_i$ and $\beta_0$ by Viro and Zvonilov (where $\ell$ denotes the number of ovals in ${\bf R} F$). Moreover, the set $C_-(o)$ is what they call a \textbf{membrane}, and $\tilde{C}(o)$ is its lift (and the same goes for $D_-(J)$ and $\tilde{D}(J)$). They showed the following result.

\begin{lemma}[{\cite[Lemma 1.3]{VZ92}}]\label{lem:VZ-Smith}
    Let $h=p^r$ be a prime power. Let $\nu:Y\to X$ be an $h$-sheeted cyclic covering between two $n$-manifolds, branched over a codimension two subset $A\subset X$. Let $B\subset X$ be a membrane, let $b$ be the class in $H_k(X,A)$ determined by $B$, and let $\beta$ be the class in $H_k(Y)$ determined by $\nu^{-1}(B)$, oriented coherently with $B$. Let $\tau:Y\to Y$ be a generator of $\mathrm{Aut}(\nu)$, and let $\varrho=1-\tau\in({\bf Z}/p)[\mathrm{Aut}(\nu)]$. Recall the Smith long exact sequence in homology (with coefficients in ${\bf Z}/p$):
    $$\cdots\longrightarrow H_{k+1}^\varrho(Y)\overset{\partial}{\longrightarrow}H_k(X,A)\oplus H_k(A)\overset{\alpha_k}{\longrightarrow}H_k(Y)\overset{\varrho_\ast}{\longrightarrow}H_k^\varrho(Y)\longrightarrow\cdots$$
    Then, the restriction $\tilde{\alpha}_k:H_k(X,A)\to H_k(Y)$ maps $b$ to $\beta$. Moreover:
    \begin{enumerate}
        \item $\alpha_{n-1}$ is monic if $H_n^\varrho(Y)=0$;
        \item $\tilde{\alpha}_{n-2}$ is monic if $X$ is connected and $H_{n-1}(Y)=0$;
        \item if $\lfloor(n+1)/2\rfloor\leqslant k<n-2$, then $\alpha_k$ is monic if $X$ and $A$ are connected and if $H_i(Y)=0$ for all $k+1\leqslant i\leqslant n-1$.
    \end{enumerate}
\end{lemma}

We can now prove the following result, analogue to their Corollary 1.5.C.

\begin{corollary}\label{cor:rank}
    The set $\{\tilde{C}(o_i)\mid1\leqslant i\leqslant\ell\}\cup\{\tilde{D}(J)\}$ has rank at least $\ell$ (where $\ell$ is the number of ovals $o_1,\dots,o_\ell$ of the curve). Moreover, if the family has rank $\ell+1$, then the curve is type I.
\end{corollary}

\begin{proof}
    We can apply Lemma \ref{lem:VZ-Smith} in our particular setting, where $\nu=\Theta:Y\to(\overline{\bf CP}^2,\mathcal{X}(F))$ and $h=2$. We then see that $\tilde{\alpha}_2:H_2(\overline{\bf CP}^2,\mathcal{X}(F);{\bf Z}/2)\to H_2(Y;{\bf Z}/2)$ is injective, because $\overline{\bf CP}^2$ is connected and $H_3(Y;{\bf Z}/2)=0$ (Proposition \ref{prop:torsionHomology}). Noting that $\tilde{\alpha}_2(C_-(o))=\tilde{C}(o)$ and $\tilde{\alpha}_2(D_-(J))=\tilde{D}(J)$, the claim follows from the very same arguments as in Viro and Zvonilov's paper, \S 2.4.
\end{proof}

Recall that $\ell^\pm$ and $\ell^0$ denotes the number of ovals of the curve that bound from the outside a component of ${\bf RP}^2\smallsetminus{\bf R} F$ with positive/negative or zero Euler characteristic. The previous results finally wrap up to yield the main theorem.

\begin{theorem}\label{thm:main}
    Let $F$ be a totally flexible curve of odd degree $m$. Then: $$\ell^0+\ell^-\leqslant\frac{(m-1)^2}{4}.$$
    Moreover, if equality holds, then the curve is type I.
\end{theorem}

\begin{proof}
    Take the maximal subset $\mathcal{P}$ of $\{\tilde{C}(o_i)\mid1\leqslant i\leqslant\ell\}\cup\{\tilde{D}(J)\}$ that spans a subspace of $H_2^+(Y)$, and let $r=\mathrm{rank}(\mathcal{P})$. Then, we obtain $r\leqslant b_2^+(Y)$. Moreover, because of $Q_Y(\tilde{C}(o),\tilde{C}(o))=-4\chi(o)$ and similarly for $\tilde{D}(J)$, observe that $\mathcal{P}$ has exactly $\ell^0+\ell^-+1$ elements (assuming that there is at least one oval to have $\tilde{D}(J)\in\mathcal{P}$; if there are none, the theorem is vacuous). Therefore, because of Corollary \ref{cor:rank}, we deduce $r\geqslant\#\mathcal{P}-1=\ell^0+\ell^-$. This produces: $$\ell^0+\ell^-\leqslant b_2^+(Y),$$
    which is the inequality claimed. The extremal case also follows from an almost word-for-word proof as in \cite{VZ92}.
\end{proof}
\subsection{Comparisons of our inequality}

Given a prime number $p$ and an integer $m\in{\bf N}^\star$, we denote as $\nu_p(m)=\max\{n\in{\bf N}\mid p^n\text{ divides }m\}$ the $p$-adic valuation of $m$. Define the function $h:{\bf N}^\star\to{\bf N}$ by $$h(m)=\max_{p\text{ prime}}p^{\nu_p(m)}.$$
That is: $h(m)$ is the largest prime power that divides $m$. Viro and Zvonilov's inequality, which holds for flexible curves, is: $$\ell^0+\ell^-\leqslant\frac{(m-3)^2}{4}+\frac{m^2-h(m)^2}{4h(m)^2}.$$
In infinitely many cases, ours is strictly sharper. But in infinitely many others (\textit{e.g.} when $m$ is a prime power), the converse holds. However, both are far from sharp estimates that can be obtained from considerations for \textit{algebraic} curves that come from Bézout theorem computations. For instance, Zvonilov \cite{Zvo79} has the sharper estimate, valid for algebraic curves: $$\ell^0+\ell^-\leqslant\frac{(m-1)(m-3)}{4}.$$
Yet, his proof will not adapt to flexible curves, for it requires the non-natural condition of intersecting the real line $\{z_2=0\}$ generically.

If one starts with Viro--Zvonilov's inequality in the case where $m+2$ is a prime power and the curve is totally flexible, then one can perturb its union with the focal conic into a non-singular flexible curve (which will have the same real set, for ${\bf R} Q=\varnothing$), and obtain: $$\ell^0+\ell^-\leqslant\frac{(m-1)^2}{4}+\frac{(m+2)^2-h(m+2)^2}{4h(m+2)^2}=\frac{(m-1)^2}{4}.$$
That is: one can derive our Theorem \ref{thm:main} from Viro--Zvonilov's when $m+2$ is a prime power. On a side note, if the famous twin prime conjecture happens to be true, this means that there are infinitely many degrees for which their inequality is strictly sharper than ours, and for which it is a corollary.

It turns out that by some easy number-theoretic considerations, one can show that there are infinitely many odd degrees $m$ such that neither of $m$ and $m+2$ are prime powers, and for which our inequality is strictly sharper. Indeed, the difference of the upper bounds in both inequalities is $$\frac{1}{4}\left(\left[\frac{m}{h(m)}\right]^2-4m+7\right).$$
With $m_p=1287\times429^{12p+1}$, one has $5|m_p+2$ and $7|m_p+2$, and $h(m_p)\in o(m_p^{19/40})$. In particular, the difference diverges to $+\infty$ on the degrees $m_p$.

One can mention the recent work \cite{Zvo22} from Zvonilov, which generalizes to flexible curves on almost-complex 4-manifolds. Possibly, our approach could work in this setting, and there may be a new inequality to be obtained.
\subsection{Non-orientable flexible curves}

There is a new object in itself that could be interesting to study: non-orientable flexible curve. Recall the definition of a flexible curve.

\begin{repdefinition}{def:flexible}
    Let $F$ be a closed oriented surface embedded in ${\bf CP}^2$. $F$ is called a degree $m\geqslant1$ \textbf{flexible curve} if:
    \begin{enumerate}[label=(\roman*)]
        \item $\mathrm{conj}(F)=F$;
        \item $\chi(F)=-m^2+3m$;
        \item $[F]=m[{\bf CP}^1]\in H_2({\bf CP}^2;{\bf Z})$;
        \item $T_{\bf C} F={\bf C}\cdot T_{\bf R}({\bf R} F)$, where ${\bf R} F=F\cap{\bf RP}^2$.
    \end{enumerate}
\end{repdefinition}

The first and fourth points of the definition still make sense. The third one only makes sense for orientable surfaces. However, for a positively-oriented surface $F\subset{\bf CP}^2$, we have: $$[F]=m[{\bf CP}^1]\iff e({\bf CP}^2,F)=m^2.$$
This means that we can replace the condition by $e({\bf CP}^2,F)=m^2$. Finally, the second point came from extremality in the Thom conjecture, which we briefly re-state.

\begin{theorem}[{Thom conjecture, \cite{KM94}}]
    Let $F\subset{\bf CP}^2$ be an embedded oriented surface with $[F]=m[{\bf CP}^1]$. Then $\chi(F)\leqslant-m^2+3m$.
\end{theorem}

It is also well-known that algebraic curves realize extremality in the previous bound. However, this is not yet known to generalize to non-orientable surfaces. A similar conjecture would be the following.

\begin{conjecture}\label{conj:thomNOfirst}
    Let $F\subset{\bf CP}^2$ be an embedded non-orientable surface with normal Euler number $e({\bf CP}^2,F)=m^2$. Then $\chi(F)\leqslant-m^2+3m$.
\end{conjecture}

For $m=1$ and $m=2$, the conjecture is vacuous, for $\chi(F)\leqslant2$ is always satisfied. For $m=3$, the conjecture is $\chi(F)\leqslant0$. That is, we have to prove that there is no embedding of ${\bf RP}^2\subset{\bf CP}^2$ with $e({\bf CP}^2,{\bf RP}^2)=9$. One can actually make use of the following slight generalization of the Whitney--Massey congruence.

\begin{theorem}[{\cite{Yam95}, Example 1}]\label{thm:yamada}
    Let $X=m{\bf CP}^2\# n\overline{\bf CP}^2$, and let $F\subset X$ be an embedded non-orientable surface. Identify $H_2(X;{\bf Z}/2)={\bf Z}/2\langle\xi_1,\dots,\xi_m\rangle\oplus{\bf Z}/2\langle\eta_1,\dots,\eta_n\rangle$, and assume that $$[F]=\sum_{i\in I}\xi_i+\sum_{j\in J}\eta_j.$$
    Then, we have: $$e(X,F)+2\chi(F)\equiv|I|-|J|\mod{4}.$$
\end{theorem}

This can be used in the special case where $e({\bf CP}^2,F)=m^2$.

\begin{corollary}\label{cor:evenChi}
    If $F\subset{\bf CP}^2$ is non-orientable embedded with $e({\bf CP}^2,F)=m^2$, then $\chi(F)$ is even.
\end{corollary}

\begin{proof}
    Indeed, there are two cases: $m$ even and $m$ odd. In the first, we have $[F]=0$, and in the second, we have $[F]=\xi$, which can be used to obtain $\chi(F)\equiv0\mod{2}$ always.
\end{proof}

In particular, this means that $F$ has an \textit{even} non-orientable genus. As such, there cannot be ${\bf RP}^2\subset{\bf CP}^2$ with $e({\bf CP}^2,{\bf RP}^2)=9$.

For $m=4$, we want to obtain $\chi(F)\leqslant-4$. Therefore, we only need to show that there is no embedding $F\subset{\bf CP}^2$ with $F=K$ or $K\#K$ ($K$ being the Klein bottle), and with $e({\bf CP}^2,F)=16$. So far, none of the restruictions we tried gave anything, and this becomes worse and worse in higher degrees.

Conjecture \ref{conj:thomNOfirst} means that one obtains a definition of a non-orientable flexible curve, and similarly, of a non-orientable totally flexible curve. Such objects still satisfy most of the already-known basic properties of classical flexible curves. Mainly, if $F\subset{\bf CP}^2$ is a degree $m$ non-orientable flexible curve, then for instance, we have the Harnack inequality: $b_0({\bf R} F)\leqslant\frac{(m-1)(m-2)}{2}+1$, where the Smith-theoretic proof works almost word-for-word. As such, there is a notion of a non-orientable flexible $M$-curve. Type I and type II curves are to be studied with more care, as this time the quotient $F/\mathrm{conj}$ is not orientable for a separating curve. There are also two natural questions that arise.

\begin{question}\label{qu:realSchemeNO}
    Is every real scheme of a classical flexible curve realizable as the real scheme of a non-orientable flexible curve? Conversely, is every real scheme of a non-orientable flexible curve realizable as that of a classical flexible curve?
\end{question}

Our proof of Theorem \ref{thm:main} also applies to non-orientable totally flexible curves, and still yields $$\ell^0+\ell^-\leqslant\frac{(m-1)^2}{4}.$$

However, there are a few reasons to believe that Conjecture \ref{conj:thomNOfirst} is off by 4 for flexible curves, and instead should be re-formulated as follows.

\begin{conjecture}\label{conj:thomNOsecond}
    Let $F\subset{\bf CP}^2$ be an embedded non-orientable surface with normal Euler number $e({\bf CP}^2,F)=m^2$, $T_{\bf C} F|_{{\bf R} F}={\bf C}\cdot T_{\bf R}({\bf R} F)$ and $\mathrm{conj}(F)=F$. Then $\chi(F)\leqslant-m^2+3m-4$.
\end{conjecture}

First, one can \textit{naturally} construct an embedded non-orientable surface $K(F)$ that satisfies $\chi(K(F))=-m^2+3m-4$ as well as all other points from the definition of a non-orientable flexible curve. Indeed, start with a classical flexible curve $F$, and choose an embedding of a Klein bottle $K\subset{\bf CP}^2$ with $e({\bf CP}^2,K)=0$, $K\cap\mathrm{conj}(K)=\varnothing$. and $K\cap F=\varnothing$. Now, connect $F$ to $K$ and $\mathrm{conj}(K)$ by two tubes swapped by $\mathrm{conj}$, and set $K(F)=F\#K\#\mathrm{conj}(K)$.

Second, we cannot have $\chi(F)=-m^2+3m$ for $F$ non-orientable and $m=1$ or $2$ (for this means $\chi(F)=2$, which only happens for the sphere). For $m=3$, $\chi(F)=0$ implies (if $F$ is non-orientable) that $F$ is a Klein bottle.

\begin{proposition}
    There is no embedding $K\subset{\bf CP}^2$ of the Klein bottle with $e({\bf CP}^2,K)=9$.
\end{proposition}

\begin{proof}
    Assume there is one. Then, because $e({\bf CP}^2,K)$ is odd, we see that $K$ is mod-2 characteristic (\textit{i.e.} Poincaré dual to $w_2$). Therefore, the Guillou--Marin congruence (from \cite{GM80}) applies: $$\sigma({\bf CP}^2)-e({\bf CP}^2,K)\equiv2\beta({\bf CP}^2,K)\mod{16},$$
    where $\beta({\bf CP}^2,K)$ (which is an $8$th root of $1$, but is identified with an element of ${\bf Z}/8$) is the Brown invariant of the embedding. It can be computed from the associated Guillou--Marin form $\varphi:H_1(K;{\bf Z}/2)\to{\bf Z}/4$ as follows: $$\beta({\bf CP}^2,K)=\left(\frac{1}{\sqrt{2}}\right)^{\dim H_1(K;{\bf Z}/2)}\sum_{x\in H_1(K;{\bf Z}/2)}i^{\varphi(x)}.$$
    The congurence therefore yields $-8\equiv\beta({\bf CP}^2,K)\mod{16}$, and thus $\beta({\bf CP}^2,K)=4\in{\bf Z}/8$. Now, the Guillou--Marin form can be directly computed. Recall that if $c\subset K$ is an immersed circle, then $c$ bounds an immersed disk $D\subset{\bf CP}^2$ (because $H_1({\bf CP}^2;{\bf Z})=0$). Isotope $D$ so that it is not tangent to $K$. The normal bundle $\nu D<T{\bf CP}^2$ is trivial, and as such, $\nu D|_c$ admits a unique trivialization $\nu D|_c\cong c\times{\bf R}^2$. On the other hand, $\nu c<TK$ defines a sub-bundle of $\nu D|_c$. With respect to the above trivialization, let $n(D)$ denote the number of right-handed half-twists of $\nu c$. Then, we set: $$\varphi(c)=n(D)+2D\cdot K+2e(K,c)\mod{4},$$
    with $D\cdot K$ the intersection number and $e(K,c)$ the normal Euler number of $c\subset K$. This definition ends up only depending on $c$ and not on the choice of $D$. We can use it to find the Brown invariant of $K$ by choosing a basis for the homology: $H_1(K;{\bf Z}/2)=\langle a,b\rangle$. Note that in particular, we have $H_1(K;{\bf Z}/2)=\{0,a,b,a+b\}$.

    \begin{figure}[ht]
        \centering
        \begin{tikzpicture}
            \node[inner sep=0pt] at (0,0) {\includegraphics{img/klein-bottle.eps}};
            \node at (.4,.8) {$a$};
            \node at (-.8,.2) {$b$};
        \end{tikzpicture}
        \caption{The standard basis for the first homology of the Klein bottle.}
        \label{fig:klein-bottle}
    \end{figure}

    A direct computation using Figure \ref{fig:klein-bottle} as a reference and with the previous definition of $\varphi$ gives: $$\varphi(a)\in\{1,3\}\text{ and }\varphi(b)\in\{0,2\}.$$
     Therefore, one examines the four possible values (in fact, three) for $\beta({\bf CP}^2,K)$: $$\beta({\bf CP}^2,K)\in\{0,2,6\}.$$
    Observe that this contradicts the Guillou--Marin congruence, for we never observe $\beta=4$ as a possibility.
\end{proof}

As such, Conjecture \ref{conj:thomNOfirst} cannot be optimal for $m\leqslant3$. For $m=4$, the trick of using the Guillou--Marin congruence fails, and there is no immediate contradiction. However, to explain the $-4$ in Conjecture \ref{conj:thomNOsecond}, we have the next result.

\begin{proposition}
    Let $F\subset{\bf CP}^2$ be an embedded non-orientable surface with $e({\bf CP}^2,F)=m^2$ and $\chi(F)<-m^2+3m$. Assume moreover that $m=2k$ is even, that $\mathrm{conj}(F)=F$ and that $T_{\bf C} F|_{{\bf R} F}={\bf C}\cdot T_{\bf R}({\bf R} F)$. Then $\chi(F)\leqslant-m^2+3m-4$.
\end{proposition}

\begin{proof}
    Indeed, $\chi(F)$ only takes even values, by Corollary \ref{cor:evenChi}, so we only need to prove that $\chi(F)=-m^2+3m-2$ is not possible. Because $m$ is even, the real part of the curve ${\bf R} F$ is contractible, and there is a notion of ${\bf RP}^2_+$ and ${\bf RP}^2_-$. Consider $\mathcal{A}(F)=F/\mathrm{conj}\cup{\bf RP}^2_+\subset{\bf S}^4$ the usual Arnol'd surface. Then, by looking at the map $p:(F,{\bf R} F)\to(F/\mathrm{conj},p({\bf R} F))$, we obtain, using the topological Riemann--Hurwitz formula and Lemma \ref{lem:branched-cover-euler-number}: $$\chi(\mathcal{A}(F))=-2k^2+3k-1+\chi({\bf RP}^2_+)\text{ and }e({\bf S}^4,\mathcal{A}(F))=2k^2-2\chi({\bf RP}^2_+).$$
    We can use the Whitney--Massey congruence in ${\bf S}^4$ for the non-orientable surface $\mathcal{A}(F)$: $$2\chi(\mathcal{A}(F))\equiv e({\bf S}^4,\mathcal{A}(F))\mod{4}.$$
    This implies $2k^2-2k+2\equiv4\chi({\bf RP}^2_+)\mod{4}$, \textit{i.e.} $k^2-k+1\equiv0\mod{2}$. However, this clearly never happens, for $k^2-k$ is always even.
\end{proof}

Assuming Conjecture \ref{conj:thomNOsecond} holds, one can look at the Harnack inequality again, and this time, we only obtain $$b_0({\bf R} F)\leqslant\frac{(m-1)(m-2)}{2}+3$$
from the Smith-theoretic argument. Theorem \ref{thm:main} also becomes: $$\ell^0+\ell^-\leqslant\frac{(m-1)^2}{4}+2.$$
Maybe there are ways to show that the usual Harnack bound for $b_0({\bf R} F)$ holds, by using the non-orientability of $F$, but this appears to be more difficult than at first thought.

As a last word, recall that we had a natural way to construct non-orientable flexible embeddings $K(F)\subset{\bf CP}^2$. Such embeddings realise the upper bound in Conjecture \ref{conj:thomNOsecond}. This raises the following question.

\begin{question}
    Are all non-orientable flexible curves of the form $K(F)$ for $F$ classical flexible?
\end{question}

Proving this would imply that Question \ref{qu:realSchemeNO} also has a positive answer. However, this means that the theory of non-orientable flexible is not so rich after all.
\subsection*{Acknowledgements}

The author is very grateful to T. Fiedler for all the insight he gave and for all the discussions. The author would also like to thank M. Golla for his very nice explanations, and D. Moussard for all the precious time she took. This work received support from the University Research School EUR-MINT (State support managed by the National Research Agency for Future Investments program bearing the reference ANR-18-EURE-0023).

\section{Preliminaries}
    The complex conjugation $\mathrm{conj}:{\bf CP}^2\to{\bf CP}^2$ is defined in homogeneous coordinates by $\mathrm{conj}([z_0:z_1:z_2])=[\overline{z_0}:\overline{z_1}:\overline{z_2}]$, and ${\bf RP}^2\subset{\bf CP}^2$ is the fix-point set $\mathrm{Fix}(\mathrm{conj})$. We endow ${\bf CP}^2$ with its Fubini--Study metric, which makes it a Kähler manifold.
    \input{contents/focal-conic}
    \input{contents/totally-flexible}
    \input{contents/branched-covers}

\section{The Arnol'd surface of an odd degree flexible curve}
    \input{contents/arnold-surface}

\section{Proving the inequality}
    The idea is that we would like to take the 2-fold branched cover of $\overline{\bf CP}^2$ ramified along the Arnol'd surface. This is not yet possible in this odd degree setting, for the surface $\mathcal{A}(F)$ is not null-homologous in $H_2(\overline{\bf CP}^2;{\bf Z}/2)$. In fact, this limitation is what led Viro and Zvonilov to consider $h(m)$-sheeted branched covers, where $h(m)$ denotes the highest prime power that divides $m$. However, in our favorable setting, we can perturb the Arnol'd surface in a somewhat-natural way, with the important fact that it preserves the structure of the curve $\overline{{\bf R}} F$ inside $\overline{Q}$. One last rermark is that we could \textit{not} apply the same construction to a totally flexible surface $F\subset{\bf CP}^2$ directly, for there is an \textit{even} number of points in $F\pitchfork Q$ (but $\mathcal{A}(F)\pitchfork\overline{Q}$ is an odd number of points).
    \input{contents/branching-cp2}
    \input{contents/proving-ineq}

\section{Further comments}
    \input{contents/sharpness}
    \input{contents/non-orientable}

\emergencystretch=1em
\nocite{*}
\printbibliography

\end{document}

redpajama_set_name: RedPajamaArXiv


text:
Beware the Convenient "Intent" of Software Publishers
in Blogs, IBM Audits, Microsoft Audits, Software Audits
Most software license agreements used by major publishers like Microsoft and IBM are in many ways vague with respect to license restrictions and metrics. This leaves licensees in the position of having to interpret the agreements based on whatever guidance may be available from the publisher or, often, simply based on the licensees' own experience and understanding. Unfortunately, that often lands companies in trouble in the context of an audit.
A tactic that we are seeing increasingly with many software audits is for publishers (1) to announce that audited businesses have incorrectly interpreted their software license agreements and then (2) to explain what the publishers' "intent" for the agreements really was. For example, a license agreement may prohibit using software for "commercial hosting" purposes without defining what "commercial hosting" consists of. Absent that guidance, a licensee may deploy the software in customer-facing environments but may do so in a way that, based on the licensee's understanding of the word "hosting," does not cross the line. Then, in an audit, the publisher informs the licensee that its "intent" for the prohibition was to stop just the sort of deployments used in the licensee's network and that the licensee therefore was using the software improperly for "commercial hosting" purposes. In other words, while the publisher may have had an opportunity to make its "intent" clear in the license agreement, it instead promulgated a vague license and then used that vagueness as a sword to claim a windfall during the audit.
It is important to remember in such situations that contract law typically does not favor such arguments. For instance, in interpreting contracts, courts in Washington state (the laws of which control the interpretation of Microsoft's license agreements, among others) "attempt to determine the intent of the parties by focusing on their objective manifestations as expressed in the agreement." McGuire v. Bates, 169 Wn.2d 185, 188-189 (Wash. 2010). "The subjective intent of the parties is generally irrelevant if [courts] can impute an intention corresponding to the reasonable meaning of the actual words used." Id. Moreover, to the extent that there exist ambiguities in a written agreement, those ambiguities will be construed against the party that drafted the agreement. Brinson v. Linda Rose Joint Venture, 53 F.3d 1044, 1048-1049 (9th Cir. Wash. 1995). Therefore, to the extent that a court might find that there are no ambiguities in the licensing language in question, either party's subjective intent for that language would be irrelevant. Conversely, to the extent that a court might find the language to be ambiguous, that ambiguity likely would be construed to be consistent with the licensee's interpretation, not with the publisher's interpretation.
For these and many other reasons, it is vital that a company's legal counsel be involved in any software audit as soon as the company receives an audit notice letter. Software publishers often assert legal positions that are not supported by applicable law, and a company that coordinates its response among IT, procurement, SAM and legal team members stands the best chance of mounting an effective defense.

redpajama_set_name: RedPajamaCommonCrawl


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text:
The Most Common Uses of the Internet
Mohammed Emran / August 26, 2021
A Quick History of the Uses of the Internet
The Internet in the 90s
The Internet in the 2000s
Common Uses of the Internet Today
Let's go back to the first time you used a computer to go "online."
It might have been back in 1995 or in 2015. No matter the age, it is safe to assume that the way you use the Internet today has changed since that first time your fingers touched the keyboard.
The Internet is a revolving door of technological advancement and possibilities. It only makes sense that as technology evolves and our needs progress, the way we use the Internet will also change.
For someone back in the 90s, the Internet was a way to have interactions through a screen in the form of email and chat rooms. Nobody depended on the Internet and it wasn't as essential as it is now.
In today's world, it is practically impossible to go five minutes without looking at a phone or a screen while living most of our lives online.
Think about it, we use the Internet for everything—from paying bills to binge watching shows on Netflix; the Internet is basically a family member for which we pay services every month.
As we dive deeper into the third decade of this century, we can only expect that the uses we have for the Internet continue to reshape and gain ground.
If you were not actively using the Internet in the 90s, you probably missed out on one of the funniest decades for technology.
Everything was new, and no one was really sure of what "going online" meant or why we might need to do so.
The beginning of the 90s marked the era of the Internet as we know it today. Still, it was a big change for society to understand the concept of doing things online.
For example, people did their research at the library through Encyclopedias and, you know, books. Google was not a thing yet.
Email was even more foreign because nobody ever thought they could send a letter electronically to anyone in the world. That was pretty fly.
The most common uses for the Internet in the 90s revolved around a few different things that set it apart from other decades.
Surfing the Web
Now that information was available online, people were quickly getting used to the idea of being able to do searches through the Internet from the comfort of their own home (if they owned a computer).
A ton of books came out at the time that aimed to explain how to surf the web and understand all the complexities of what was capable at the tip of their fingers. Literally.
For the society of the 90s, surfing the web mostly meant reading whatever publications were online, looking up information through forums, and having to fight your way through online ads.
During the Internet of the 90s, there was room for people to make it whatever they wanted it to be. This was a time in which many people could set up websites, rudimentary at best, for their companies and products when there was not a lot of competition online.
Floppy disks were a way to store information through a disk that you could insert into the computer. Basically, they were the first version of a USB drive.
It was during this decade that illegal downloads of music and videos were most popular. Because the Internet was so new, there were hardly any piracy laws set in place and people were taking advantage of all the "free" information they could download online.
Most users did not even have to download information onto their floppy disks, per se, because it was as easy as copying and pasting files into the storage unit.
This type of illegal activity enabled the rise in pirate copies of movies and books to be sold on the street, causing a shift in sales for the rightful owners of the material.
This was the beginning of memes. There was a big obsession over stock images back in the 90s and during the beginnings of the Internet. People got very creative with edited images anyone could view online or download into their computers as backgrounds.
HTML code was how the Internet was set up initially, which made everyone using it—especially teenagers—absolute masters at it as they were able to virtually edit any image and make it their own.
For part of the early 2000s, the uses of the Internet were still similar to the ones in the 90s. Most people now had a computer in their homes, and interactions between users were slowly improving and becoming more common.
In the late 90s, two students from Stanford came up with the idea of a search engine in which people could look up any information they wanted. What started with everyone asking "What's a Google?" slowly picked up speed and exploded in the 2000s. Nowadays, we say things like "Let me Google that," as part of our regular vocabulary. However, when Google first started, the idea of being able to look up any information through the use of a single search engine was crazy. This far-off idea paid off for the founders of Google as it became one of the main reasons for people to use the Internet in the 2000s and even today.
With the booming of the Internet in the 2000s, email became a necessary feature. In order to connect to the Internet, most users had to have an email address. This alone became a reason to use the Internet going into the millennium. Everybody was excited about email and wanted to see how this invention could help them communicate with loved ones without having to wait for days for them to receive a letter.
Of course, the entire notion of email was heavily romanticized with the film You've Got Mail which reinforced the idea that, not only could we now use the Internet to communicate, but to find love.
This was probably the main reason to use the Internet in the 2000s. Email was one thing, but the fact that people could now go online to chat with their friends in an instant was mind blowing. AOL instant messenger was mainly used by tween and teens alike. Most parents did not understand why their children would have any need to talk to their friends in a chat when they had just seen them in school, but they learned to live with this new trend. This was also where everyone mastered their typing skills along with showcasing tidbits of song lyrics on "away" messages. This was texting before texting became a thing. And if you didn't have a creative screen name, you were simply not cool.
The question we should ask ourselves is: what don't we use the Internet for today?
Some things that people used the Internet for in earlier decades still sustain today, like email and general web surfing.
However, our use of the Internet nowadays encompasses our entire lives. There's nothing we don't do online, and that includes: paying bills, making purchases, working, and even going to school.
Biggest change in the way we use the Internet from all the other decades is the addition of social media.
This started in the early 2000s with the first popular social media page of MySpace. MySpace was all the things that kids loved about the 90s into one. People could post in forum-type sections, customize their own page with basic HTML code, and interact with their peers online.
Of course, since the creation (and dissolution) of MySpace, social media pages like Facebook, Instagram, and TikTok have largely taken over our social lives and have a heavy influence on how we interact with other people.
In some aspects, social media is replacing emails and chats, by providing an all-in-one platform to communicate with one another.
Even though there is still a market for people that like to get their news through traditional mediums like TV and newspapers, a lot of the population is now shifting to the Internet for their daily news intake.
This generation of Internet users is very much accustomed to instant gratification and they seek information on a minute-by-minute basis. In fact, most news channels, if not all, run on a 24 hour basis for this very purpose. People want their news now. There is no longer a need to wait for the morning newspaper.
In today's world, the use of the Internet for entertainment plays a pivotal role in the way we relax. Since the end of Blockbuster—the retail chain where you could rent VHS movies, and later DVDs, for a small fee—streaming sites like Netflix and Hulu have completely taken over.
If you're looking for a quick way to watch something at home, or on the road, or during your lunch break, you can go on a streaming app and watch anything you want. Having a great Internet plan can aid in this by helping users enjoy their shows without any latency or freezing of the screen.
Modern use of the Internet will always change. The Internet is a tool that adapts itself to the needs of society—it helps serve the population with enough information and applications to lead more productive and satisfactory lives.
Although some of the most popular uses of the Internet today revolve around social media and entertainment, we use the Internet to work and sustain our lifestyles in the best way we can. The Internet became an essential part of day-to-day life.
We've certainly come a long way since dial-up and chat rooms.


text:
You are here: Home / Australia / Australian Wind Industry Doomed: Tony Abbott Signals the End of the Mandatory RET
Australian Wind Industry Doomed: Tony Abbott Signals the End of the Mandatory RET
August 27, 2014 by stopthesethings 8 Comments
The wind industry forecast: dark days ahead.
STT followers have been delighted with news that Tony Abbott, Joe Hockey and Mathias Cormann have teamed up to axe the mandatory RET (see our post here).
In response to the PM's mooted plan, the wind industry and its parasites have been reduced to making idle threats of "revenge" and bleating about "sovereign risk". Despite a rear-guard effort by Environment Minister, Greg Hunt to salvage something of the mandatory RET, his boss has confirmed that his mission is to kill it outright. And that pretty much means the end of the wind industry as we've known and grown to despise it. Here's the Australian Financial Review on the beginning of the end.
Coalition fails to budge on RET pruning
Phillip Coorey
Pleas by solar and wind companies to leave the Renewable Energy Target untouched have fallen on deaf ears with the government deciding to proceed with a phasing down of the scheme.
While a final position will not be announced until next month, The Australian Financial Review understands the intent is to cut the scheme harder than a compromise scenario that was being pursued by the Environment Minister, Greg Hunt.
The end result will be closer to the abolition scenario advocated by Prime Minister Tony Abbott which would end the scheme by closing it to new entrants and grandfathering existing large scale projects.
Seeking to overcome the cabinet split, Mr Hunt, Mr Abbott and Industry Minister Ian Macfarlane met on Sunday to discuss a policy position to be put to the bureaucracy for analysis and then to the cabinet for a final decision.
The government is being guided by the findings of the review into the RET conducted by businessman Dick Warburton, a person the industry has argued is ill-suited to the task because he is a climate-change sceptic.
The guiding principles of the final decision will be to balance investor risk with the impact of the RET on household and business power bills. Mr Abbott claims the RET has had a significant impact on power prices. The government's own modelling shows while the RET has added $40 a year to average household power bill, prices will fall over the medium term as more renewable energy is produced.
The industry is ramping up its warnings that any dilution of the current scheme will not only jeopardise more than $11 billion in the renewable energy investment pipeline, but create a broader sovereign risk perception for Australia.
Close watch on outcome
Philip Green, the London-based partner of the Children's Investment Master Fund (TCI), which has a 33 per cent stake in renewable energy company Infigen, said the issue was being watched closely. "Sovereign risk has already increased in Australia given the media coverage of the carbon debate and now the RET. Sovereign risk will increase more if the stories about cuts to the RET are confirmed," he said in a statement.
"This comes at a cost to the nation through higher capital costs as it seeks future investment in infrastructure. The Australian RET had strong bi-partisan political support [including from the current prime minister]. It can take a long time to restore trust and in some cases this is only achieved with a change in leadership/policy/party."
Under the RET, a policy which hitherto had bipartisan support, 20 per cent of Australian's energy production by 2020 would come from renewable sources. Based on earlier predictions of power production in 2020, this 20 per cent target was calculated at an annual production of 41,000 gigawatt hours.
But the 2020 production total has been downgraded following the decline of the manufacturing sector, including automotive and aluminium.
Consequently, 20 per cent of the revised production target is 27,000 GWh. This is the "real 20 per cent" scenario for which Mr Hunt is advocating.
Under the push by Mr Abbott, renewable energy output would be frozen at current levels of about 16,000 GWh.
Any proposed change faces a near impossible passage through the Parliament with Labor and the Greens opposed to any alteration, while Clive Palmer says he will not allow any change unless Mr Abbott goes to the next election in 2016 and wins a mandate.
Dick Warburton: perfectly suited to killing the greatest rort of all time.
STT thinks the constant reference to Dick Warburton as a "climate-change sceptic" is just churlish bitterness from the vanquished. From STT's viewpoint, Dick did precisely what he was supposed to do: standing up for Australian power consumers and helping to bring an end to the most costly and pointless piece of policy ever devised.
And, yet again, the wind industry – and those with shirts to lose when it collapses – trot out the furhpy about "sovereign risk". Not only is it utter bunkum (see our posts here and here and here and here), harping on about it won't save the wind industry from the inevitable demolition of the mandatory RET.
The AFR talks about Australia risking "$11 billion in the renewable energy investment pipeline" as if that were some kind of loss to Australian power consumers, in an already over-supplied market. As we've previously pointed out, the threatened "investment" is hardly a "no-strings attached-gift". The would be investors are after annual gross returns in the order of 20% on that figure – ie, a cool $2.2 billion, every year – which can only be recouped from power consumers through higher power bills – with a fat pile of RECs underwriting the "investment" (see our post here).
As a piece of friendly advice, we wouldn't be betting the house on Clive Palmer blocking any changes to the RET in the Senate. Horse trading is the life-blood of politics; and a week can be a very long time for anyone engaged in the political caper. As you'd expect, STT hears that Tony Abbott is already doing business with the Senate's cross-benchers, including the PUP in order to come up with a workable solution to the debacle that is the mandatory RET, which has utterly failed as a cost-effective CO2 abatement policy.
Clive Palmer wants an Emissions Trading Scheme (albeit with the price of credits set at zero). So the Coalition's Direct Action policy is being reworked by top energy market economist, Danny Price in a manner that will not only resemble something like what Clive is after, but in a way that will slash the value of the subsidies to wind power outfits (as promised by the RET) by around 90%. One of the cross-benchers, Nick Xenophon – who works closely with Danny Price – is in on the mission to kill off the wind industry, by introducing some tweaks of his own to Coalition policy, aimed at achieving least-cost CO2 abatement (see our posts here and here). Another cross-bencher, David Leyonhjelm penned a piece for The Australian today (we'll cover it shortly) setting out his eagerness to kill the mandatory RET, which he sees as "just government mandated corporate welfare" that will cost power consumers $billions "for no measurable environmental benefit". No, STT didn't write David's article.
But, in the result, whether or not changes to the mandatory RET occur during the life of this parliament is just a matter of passing academic interest. The wind industry is doomed simply because – from here on – given the very real prospect of a collapsing REC price – NO retailer in touch with their earthly senses will enter a long-term Power Purchase Agreement with a wind power outfit (why would retailers enter a 15-25 year contract that will see them end up holding a pile of worthless RECs?). Because they will never get retailers to enter PPAs, the desperados still hoping to build wind farms will never obtain the finance needed to do so (see our post here). STT hasn't heard of a retailer entering a PPA with a wind power outfit since November 2012. The result has been an almost complete collapse in new wind farm construction this year: "investment" in the construction of wind farms went from $2.69 billion in 2013 to a piddling $40 million this year (see this article).
Moreover, with the REC price bound to head south over the coming weeks and months, outfits with current wind farm operations face the prospect of imminent financial collapse.
One of those facing an early exit from the stage is our old favourite, Infigen. These boys have just announced an $8.9 million loss for 2013/14, which follows a $55 million loss in 2011/12 and an $80 million loss for 2012/13 (see our posts here and here). Those hefty losses were all racked up at a time when the mandatory RET was set in stone, such that the regulatory cards were all firmly stacked in Infigen's favour.
With the mandatory RET set for the chop, Infigen is preparing to emulate its predecessor (Babcock & Brown) with another spectacular financial collapse. Here's the Australian Financial Review setting the scene for Babcock & Brown Mk II.
Infigen at risk if RET wound up
Angela Macdonald-Smith
Wind power producer Infigen Energy has warned it could fall into breach of its debt covenants within three months should the 2020 Renewable Energy Target be wound back with no compensation for affected investors.
Managing director Miles George said either of the two outcomes apparently being favoured by the government for the overhaul of the RET would be "disastrous" for both the industry and Infigen.
He said significant write-downs would follow, with the loss of value for Infigen more than its current market cap of about $185 million.
The government is thought to be considering two potential outcomes for its RET review, one involving reining the 2020 target back to represent a "real" 20 per cent of electricity use, rather than the 26 per cent to 28 per cent it is currently expected to represent.
The other involves closing off the scheme to new entrants, while honouring existing contracts only.
"Either of these scenarios is disastrous for our industry," Mr George said, after Infigen posted an $8.9 million full-year net loss, affected by the regulatory uncertainty. "They are both death for the renewable energy industry and, to be frank, they are death for Infigen."
He said if no compensation was provided for investors, the resulting weakness in the price of large-scale renewable energy certificates would cut cash flow for debt servicing. As a result, Infigen would be at risk of breaching its covenants within three months.
This couldn't be happening to a nicer bunch of lads.
It was fun while it lasted.
Filed Under: Australia, Big wind industry, Big wind politics, power prices, RET Review Tagged With: Coalition RET review, cost of mandatory RET, cost of wind power, Dick Warburton RET review, Greg Hunt renewable target, Infigen collapse, Renewable Energy Target, scrapping renewable target, scrapping the RET, Tony Abbott renewable energy, Tony Abbott RET
« How the Public Are Deceived About the True Cost of the Mandatory RET
Senator David Leyonhjelm: "Wake Up Clive!" – It's Time to Kill the RET & Save the Poor »
Jackie Rovenksy says:
Recently I listened to John Hewson state more than a few times at a meeting that there is money to be made in renewables – I wonder where the money has gone from the wind industry as they apparently haven't enough to keep going and flourish without more hand outs from the public purse.
Where has it all gone? It certainly hasn't gone into local 'infected' communities or creating new jobs.
Good money after bad is never a good idea.
Uncle Fester says:
Still leaves the unfortunates around the ACT subject to the ACT government's imbicilic 90% RET. Jupiter is not dead yet unfortunately
Keep fighting. Get behind Angus Taylor and take it up to the ACT government. Shame them for their cynical hypocrisy: if they think giant fans are such great idea, then why not spear a few hundred across the Brindabella Ranges and "improve" their constituents' views?
Grant Winberg says:
Why on earth does the Childrens' Investment Master Fund own 33% of Infigen? Is this akin to industry super funds being overweight in wind? Both should be hastily withdrawing.
Joe Average says:
As a citizen that took up the battle locally with others against a proposed wind farm some years ago, I find it hard to believe how the tide against the wind industry is changing. Back then if you mentioned you were against wind you were looked upon as a dinosaur.
STT you have stood tall as a guiding light for all those that have fought against the lunacy of their march across the country.
You are truly the stuff LEGENDS are made of!
If you ever get a quiet moment to yourself, you can rightly say you have played a large part in making Australia a better place.
Stand against wind says:
Here's hoping RATCH will follow Infigen down the gurgler, their plans for our beautiful and wild Mount Emerald in far North Queensland are nothing short of environmental vandalism.
Fed Up says:
The wind industry is built on a house of cards which is due for collapse. Why should the industry receive compensation? They have never bothered to compensate turbine neighbours who face devaluation of properties or who suffer loss of wellbeing. It's time for these corporate conmen to get real jobs.
Martin Hayles says:
The delightful Mr. George states in regards to regulatory uncertainty "and to be frank, they are the death of Infigen"
I'm a little confused……….
Does he wish to be the equally delightful wind parasite stooge 'bestie', Frank Bestic?
Or…… was he struggling with the term 'to be honest'?
I was unaware that one can stutter attempting to pronounce a silent 'aitch'
Biden's Bonanza: Poor Will Suffer Most From Democrats Wind & Solar Obsession


Aligarh, Feb 13 (IANS) Eight students of Aligarh Muslim University (AMU) were suspended on Wednesday for their alleged involvement in beating up some students that led to police being called, an official said.
The students have been accused of violence, opening fire and vitiating the atmosphere on the campus.
Pending inquiry into the matter, the students have also been barred from entering the university campus, including hostel and classes. Safe Kidwai, public relations office in-charge, said that the two B.Tech students, Manish and Imran Khan, had fought after which complaints were lodged.
The university has suspended Manish Kumar, Ajay Kumar Singh, Aman Sharma, Pawan Jadaun, Imran Khan, Abdul Maboob, Aadil Khan and Farhan Zubairi.
There is tension on the campus as sunders are protesting outside the varsity gate. Heavy police deployment has been made to ensure that there no untoward incident, an official said.


the email account linked with your active Apple ID (the want you want to change from) a thrashy email address (one you don't use and has not been ever used for any Apple IDs) I'll explain why you'll need this later, but for now you won't be needing it.... the email account linked with your active Apple ID (the want you want to change from) a thrashy email address (one you don't use and has not been ever used for any Apple IDs) I'll explain why you'll need this later, but for now you won't be needing it.
Next under your Apple ID email, click Change Apple ID you can add an alias email address you can use as your main Apple ID email, but you can't remove the Apple domain from the account. Add dragon ball xenoverse 2 x2m how to change idle animation 1/04/2018 · Click "Change Apple ID" beneath Apple ID at the top of the Account section. A dialog box pops up prompting you to enter a new email address to use as your Apple ID.
Next under your Apple ID email, click Change Apple ID you can add an alias email address you can use as your main Apple ID email, but you can't remove the Apple domain from the account. Add how to change server in mobile legends 2018 How to Switch Your Apple ID from a Third-party to Apple Email Address. Note: Make sure to sign out of "every Apple service and device that uses your Apple ID" before going on to change the email address associated with your account.
1/04/2018 · Click "Change Apple ID" beneath Apple ID at the top of the Account section. A dialog box pops up prompting you to enter a new email address to use as your Apple ID.


An Alabama woman surprised her cancer-stricken grandmother with a gift that brought her close to her late husband, a gesture that left the beloved matriarch in tears.
Frances Jones' tight-knit family immediately rallied around the 79-year-old mom and grandma when she was diagnosed with stage 4 lung cancer several weeks before Christmas.
But one person was notably absent as Frances began her fight against cancer: Her husband of 56 years, Leroy Jones.
Under the guise that she wanted one of her grandfather's shirts for herself, Maddie sat down with her grandmother — affectionately known as MawMaw — to pick out a piece of clothing with sentimental value.
"I asked, 'Which one do you remember him wearing the most?' and she said, 'Oh, he was always in this one. He loved this one the most,'" Maddie recalled.
With the help of her aunt, Maddie sewed the shirt and fabric together to make the special blanket she later presented to her grandmother for Christmas.
Eager to see her grandmother's reaction, Maddie was filming as she opened her gift as the rest of the Oklahoma-based family looked on.
The camera continued rolling as the laughter turned to tears, and Frances broke down with gratitude and emotion.
Leroy Jones passed away in February 2014. The couple grew up together, and after college and a stint in the Army that took Leroy away for several years, the pair bumped into each other at the town library.
"He asked to take her home, and the next week called her and asked her to go to the rodeo — a great first date — and the next week they went to a drive-in movie, where he proposed," Maddie said. "Two dates and they were engaged.
Though the couple's retelling of their courtship had a comically practical tone, they remained attached at the hip until Leroy's passing.
But Frances's resolve is strong, her granddaughter said, confident that if anyone can beat cancer, it's her grandma. The family is awaiting results from a PET scan and plans to be by her side as she undergoes radiation next.


text:
title: PapljStructureViewFactory.structureOutliner - aesi-intellij
---

[aesi-intellij](../../index.html) / [org.metaborg.paplj.structureoutline](../index.html) / [PapljStructureViewFactory](index.html) / [structureOutliner](.)

# structureOutliner

`val structureOutliner: `[`PapljStructureOutliner`](../-paplj-structure-outliner/index.html)

redpajama_set_name: RedPajamaGithub


text:
Q: const and let fake hoisting in modules I've found interesting behaviour while having let and const on module level.
func.mjs
export const func = () => {
  return num1 + num2;
}

const num1 = 5;
const num2 = 6;

main.mjs
import {func} from 'func.mjs';

console.log(func())

If you run node main.mjs, you'll get 11 in console. I can explain that as all module level expressions defined and initialised on import step.
But is it falling under concepts of TDZ?

redpajama_set_name: RedPajamaStackExchange


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remote control cars that can go in water features to consider buying the best car under.
mattress zipper cover cotton zippered mattress cover.


Curl lock formula locks in a fully volumnized curl for up to 18 hours.
STEP 1: Start at the base of your lashes and slowly drag the wand upwards in a zig-zag motion.
STEP 2: Use the wand to comb out any clumps on your lash tips.


text:
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        <script src="<?=CDN?>js/vendor/bootstrap.min.js"></script>
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text:
AQSQID: 250213003
Site Name: E. Milton - Blue Hill
Location: Canton (Ponkapoag), MA
Reporting: True, Located: True
Current AQI: 28.0
Canton (Ponkapoag) MA, US: 42.2117, -71.114403
Date Observed (UTC)
NO2Y
AQI/500
Jan. 18, 2020, 3 a.m. 5.0 ppb (5) 1.0 ppb (1) N/A 4.0 ppb (4) N/A N/A 36.0 ppb (28) N/A N/A N/A 28.0
Jan. 18, 2020, midnight 6.0 ppb (6) 1.0 ppb (1) N/A 5.0 ppb (5) N/A N/A 36.0 ppb (28) N/A N/A N/A 28.0
Jan. 17, 2020, 11 p.m. 3.0 ppb (3) 0.0 ppb (0) N/A 3.0 ppb (3) N/A N/A 39.0 ppb (30) N/A N/A N/A 30.0
Jan. 17, 2020, 9 p.m. 3.0 ppb (3) 0.0 ppb (0) N/A 3.0 ppb (3) N/A N/A 41.0 ppb (32) N/A N/A N/A 32.0
Jan. 17, 2020, noon 3.0 ppb (3) 0.0 ppb (0) N/A 3.0 ppb (3) N/A N/A 38.0 ppb (29) N/A N/A N/A 29.0
Jan. 17, 2020, 11 a.m. 3.0 ppb (3) 0.0 ppb (0) N/A 2.0 ppb (2) N/A N/A 38.0 ppb (29) N/A N/A N/A 29.0
Jan. 17, 2020, 8 a.m. 2.0 ppb (2) 0.0 ppb (0) N/A 1.0 ppb (1) N/A N/A N/A N/A N/A N/A 2.0
Jan. 17, 2020, 7 a.m. N/A N/A N/A N/A N/A N/A 33.0 ppb (25) N/A N/A N/A 25.0
Jan. 16, 2020, 3 p.m. 12.0 ppb (11) 4.0 ppb (4) N/A 8.0 ppb (7) N/A N/A 15.0 ppb (12) N/A N/A N/A 12.0
Jan. 16, 2020, 2 p.m. 11.0 ppb (10) 1.0 ppb (1) N/A 10.0 ppb (9) N/A N/A 14.0 ppb (11) N/A N/A N/A 11.0
Jan. 16, 2020, noon 10.0 ppb (9) 0.0 ppb (0) N/A 9.0 ppb (8) N/A N/A 18.0 ppb (14) N/A N/A N/A 14.0
Jan. 16, 2020, 11 a.m. 17.0 ppb (16) 1.0 ppb (1) N/A 16.0 ppb (15) N/A N/A 12.0 ppb (9) N/A N/A N/A 16.0
Jan. 16, 2020, 10 a.m. 14.0 ppb (13) 1.0 ppb (1) N/A 13.0 ppb (12) N/A N/A 14.0 ppb (11) N/A N/A N/A 13.0
Jan. 15, 2020, noon 8.0 ppb (7) 1.0 ppb (1) N/A 7.0 ppb (6) N/A N/A 12.0 ppb (9) N/A N/A N/A 9.0
Jan. 15, 2020, 10 a.m. 6.0 ppb (6) 0.0 ppb (0) N/A 6.0 ppb (6) N/A N/A 12.0 ppb (9) N/A N/A N/A 9.0
Jan. 15, 2020, 9 a.m. 7.0 ppb (6) 0.0 ppb (0) N/A 7.0 ppb (6) N/A N/A 10.0 ppb (8) N/A N/A N/A 8.0
Jan. 15, 2020, 1 a.m. 10.0 ppb (9) 0.0 ppb (0) N/A 10.0 ppb (9) N/A N/A 10.0 ppb (8) N/A N/A N/A 9.0
Jan. 15, 2020, midnight 10.0 ppb (9) 0.0 ppb (0) N/A 10.0 ppb (9) N/A N/A 10.0 ppb (8) N/A N/A N/A 9.0
Jan. 14, 2020, 11 p.m. 8.0 ppb (7) 0.0 ppb (0) N/A 8.0 ppb (7) N/A N/A 12.0 ppb (9) N/A N/A N/A 9.0
Jan. 14, 2020, 7 p.m. N/A N/A N/A N/A N/A N/A 17.0 ppb (13) N/A N/A N/A 13.0
Jan. 14, 2020, 3 p.m. N/A N/A N/A N/A N/A N/A 9.0 ppb (7) N/A N/A N/A 7.0
Jan. 14, 2020, 2 p.m. 29.0 ppb (27) 16.0 ppb (15) N/A 13.0 ppb (12) N/A N/A N/A N/A N/A N/A 27.0
Jan. 14, 2020, 1 p.m. 20.0 ppb (19) 7.0 ppb (6) N/A 13.0 ppb (12) N/A N/A 4.0 ppb (3) N/A N/A N/A 19.0
Jan. 14, 2020, noon 13.0 ppb (12) 2.0 ppb (2) N/A 11.0 ppb (10) N/A N/A 5.0 ppb (4) N/A N/A N/A 12.0
Jan. 14, 2020, 11 a.m. 9.0 ppb (8) 0.0 ppb (0) N/A 8.0 ppb (7) N/A N/A 7.0 ppb (5) N/A N/A N/A 8.0
Jan. 14, 2020, 9 a.m. 7.0 ppb (6) 0.0 ppb (0) N/A 7.0 ppb (6) N/A N/A 9.0 ppb (7) N/A N/A N/A 7.0
Jan. 14, 2020, 7 a.m. N/A N/A N/A N/A N/A N/A 10.0 ppb (8) N/A N/A N/A 8.0
Jan. 14, 2020, 6 a.m. N/A N/A N/A N/A N/A N/A 9.0 ppb (7) N/A N/A N/A 7.0
Jan. 14, 2020, 4 a.m. 16.0 ppb (15) 1.0 ppb (1) N/A 16.0 ppb (15) N/A N/A 2.0 ppb (2) N/A N/A N/A 15.0
Jan. 14, 2020, midnight 34.0 ppb (31) 12.0 ppb (11) N/A 22.0 ppb (20) N/A N/A 0.0 ppb (0) N/A N/A N/A 31.0
Jan. 13, 2020, 11 p.m. 16.0 ppb (15) 3.0 ppb (3) N/A 13.0 ppb (12) N/A N/A 10.0 ppb (8) N/A N/A N/A 15.0
Data from airnow.gov


COVIL is presently collaborating with SPEC , as its Indian contractor, in multiple arenas of Project Engineering to execute a varieties of Projects, Worldwide , for various Local, National & International companies with special emphasis for Projects in India.
Our company has the French expertise and Supervisory control along with access to abundant quality Manpower resource of India. These enable us to perform jobs conforming to the high European quality at moderate Indian Price.
Complete design, drawing and programming of Automation system and accessories for all Small and Medium Furnaces and heating equipments with only basic logic and requirement input from the client. We have the experience of executing projects with Rockwel SLC-500, Siemens Step 7, Eurotherm and other systems / platforms.
PLC programming and Automation design as per the detailed logic input from the client for all types of Projects.
Design & drawing of Electrical Power and Control cabinets to suit the requirement of the Automation system and other parameters.
Design, drawing and programming for modification of old Automation system to incorporate new logic as per client's parameters.


This video doesn't feature all the latest enhancements and changes from this 3.9.2 release, but provides a great introduction to the Spring Tool Suite in general.
The default STS distribution is now based on the recently released Eclipse Oxygen.2 (4.7.2) release, please take a look at the New and Noteworthy. This includes support for JDK9 and JUnit 5 out of the box.
At this years SpringOne Platform conference, we announced the public beta access for the upcoming all-new next generation of the Spring tooling, called Spring Tools 4. If you haven't had the chance to take a look, here is the Spring Tools 4 website.
While the Spring Tools 4 Public Beta (and the Spring Tools 4 in general) come as a totally new and independent distribution of Eclipse (called the Spring Tool Suite 4), extensions for Visual Studio Code and Atom, we also offer an early access extension pack for the now-released STS 3.9.2 distribution (via an Extension Install listing on the STS dashboard). You can add this extension pack to your STS 3.9.2 install to get an early impression of the new tooling - and provide feedback, of course.
Please keep in mind that the new Spring Tools 4 distribution will replace the existing STS 3.9.x distributions in mid of 2018 and will require a fresh install of the Spring Tools 4.
In older versions of STS, dragging and dropping a project with the same name as an existing deployed application in Cloud Foundry would just open a simple dialogue warning the user that existing service bindings would be removed, but not show any differences between the manifest file in the project and the actual deployed application.
New to 3.9.2, this has been improved to show actual differences between the manifest file, if one exists in the project, and the existing application in Cloud Foundry.
Likewise, if the project does not contain a manifest file, and a user drag/drops the project into the dashboard, only the contents of the existing application in Cloud Foundry will be changed, but not any other Cloud Foundry properties, like service bindings, environment variables, etc..
STS 3.9.2 can run on top of macOS High Sierra. But due to a bug the ANSI Console feature (the one that nicely colors the output in the console view if the app uses ANSI coloring) doesn't work on macOS High Sierra.
The issue on GitHub describes various workarounds for this issue. To avoid running into this issue, STS 3.9.2 disables the ANSI color output option for Spring Boot launch configs by default when running on macOS High Sierra.
STS 3.9.2 can run on top of a JDK out-of-the-box. However, please notice that the Maven support in Eclipse/STS runs within the JVM of the IDE and therefore also uses the JDK9 runtime. While that isn't necessarily a problem, having Maven modules in your build that aren't compatible with Java 9 might cause your project build to fail in Eclipse/STS.
Workaround is open launch configuration for the Boot App. Navigate to Spring Boot tab and enter a boot property into "Override Properties" table. For example server.port and value 8080. Save the launch configuration and re-launch your Boot App.
STS 3.9.2 requires a JDK 8 or JDK9 to run on top of. If you are running it with an older JDK this may not be immediatlely apparant, but some of STS functionality will simply be 'absent' because the corresponding plugins are automatically disabled by OSGI because their requirements are not satisfied at runtime. See also below 'Setting the JDK'.
STS 3.9.2 requires a JDK 8 or JDK9 to run on top of. Nevertheless the native Eclipse launcher component might pick up a JRE or an older JDK automatically if you don't specify which JDK to run STS on top of. To avoid this, you can specify the JDK in the sts.ini file that comes with your STS installation. Add a line at the beginning "-vm" and an additional line below that which points to the "javaw" executable of the JDK on your machine. Here is a detailed explanation how to configure the JVM in the ini file.


Great masses muttering doesn't meet go on. A large inauguration is the effect of work time of researching, calligraphy and practicing, but it can inert outcome in predicament if you bury to join the desires of your viewers. Therefore, you should build all hard work to brainstorm out what your audience wants, what they look-alike and peak of all what they entail.
Believe it or not, it is much arch to run into your audience's of necessity than their requests. If you confer your viewers what they deprivation all and sundry will like you. But, give your addressees what they want and you will be their leader.
But how do you carry through such as a task? You can't publication minds and you don't even cognize these people? How can you come across their needs? Well, here are both tips to give support to you focusing on round-table the wishes of your audiences.
What is new in the organization?
What are the management's concerns?
What are the workers' concerns?
What are the members concerns?
Tip #2 Call Possible Members of Your Audience.
When you christen your introduction party to ask the preceding questions ask for the defamation and phone book of astir ten people who will be attending the thing. Then phone up these population and ask them how they knowingness around what is going on in the management. If you comprehend carefully, the facts you congregate from these mobile calls will be meriting gold bars.
While you're chitchat to forthcoming audience members ask them accurately what they are expecting from your screening. Usually they will enlighten you what they are expecting and whether they want to be in attendance or they have to be here. This is valuable rumour you can use spell you crafting your show.
Now mistreatment all the gen and investigation you have gathered gear up your inauguration to bump into your audience's necessarily and expectations. If you proudly custom-make your inauguration to the wants and wishes of your viewers you will be their hero.
Knowing your assemblage can go along way to fashioning your piece proud. So find out who they are, what they need and what they judge. Then packet it in an expected word that each person will talent from. The much you know going on for your addressees the much self-assured you will be and reliance reduces restiveness. The less troubled you are the larger you will latter-day and tennis shot your addressees.