Revisions to How to construct such a real algebraic curve

The union of $X$ and $L$ is $X\cup L$, not $X\cap L$.

edited Dec 14, 2023 at 20:21

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You can always construct an M-curve with the property you asked for.

Let $L$ be a real line in the real projective plane and $d$ an even, respectively odd, integer. Then, there exists a real M-curve of degree $d$ intersecting $L$ in only pairs of complex conjugated points, respectively in $\frac {d-1} 2$ pairs of complex conjugated points and exactly one real point.

Proof: You can just modify a little bit the Harnack method. By Harnack’s construction, there exists a real M-curve $X$ of degree $d-1$ intersecting $L(\mathbb R)$ in $d-1$ real points (the line $L$ intersects only one real connected component of $X(\mathbb R)$). We can now perturb the union $XL$$X\cup L$ to a non-singular real curve $A$ of degree $d$ with the properties that we want. In fact, in order to construct $A$, it is enough to apply a small perturbation to the union $XL$$X\cup L$ with respect to $\frac d 2$ pairs of complex conjugated lines, respectively $\frac {d-1} 2$ pairs of complex conjugated lines and one real line (which must be chosen passing through a point of an appropriate connected component of $L(\mathbb R) \setminus X(\mathbb R)$).

To be more precise, we construct $A$ as the zero set of the following real homogeneous non-singular polynomial of degree $d$:

let $f, g, h \in \mathbb R[x,y,z]$ be the homogeneous polynomials associated respectively to $X$, $L$ and the union of the lines chosen for the small perturbation;
then, there exists a small enough real non-zero number $b$ such that $$A = \{(x,y,z) \in \mathbb R \mid f(x,y,z) g(x,y,z) + b h(x,y,z) = 0 \} \ .$$

Hope it was helpful.

You can always construct an M-curve with the property you asked for.

Let $L$ be a real line in the real projective plane and $d$ an even, respectively odd, integer. Then, there exists a real M-curve of degree $d$ intersecting $L$ in only pairs of complex conjugated points, respectively in $\frac {d-1} 2$ pairs of complex conjugated points and exactly one real point.

Proof: You can just modify a little bit the Harnack method. By Harnack’s construction, there exists a real M-curve $X$ of degree $d-1$ intersecting $L(\mathbb R)$ in $d-1$ real points (the line $L$ intersects only one real connected component of $X(\mathbb R)$). We can now perturb the union $XL$ to a non-singular real curve $A$ of degree $d$ with the properties that we want. In fact, in order to construct $A$, it is enough to apply a small perturbation to the union $XL$ with respect to $\frac d 2$ pairs of complex conjugated lines, respectively $\frac {d-1} 2$ pairs of complex conjugated lines and one real line (which must be chosen passing through a point of an appropriate connected component of $L(\mathbb R) \setminus X(\mathbb R)$).

To be more precise, we construct $A$ as the zero set of the following real homogeneous non-singular polynomial of degree $d$:

let $f, g, h \in \mathbb R[x,y,z]$ be the homogeneous polynomials associated respectively to $X$, $L$ and the union of the lines chosen for the small perturbation;
then, there exists a small enough real non-zero number $b$ such that $$A = \{(x,y,z) \in \mathbb R \mid f(x,y,z) g(x,y,z) + b h(x,y,z) = 0 \} \ .$$

Hope it was helpful.

You can always construct an M-curve with the property you asked for.

Let $L$ be a real line in the real projective plane and $d$ an even, respectively odd, integer. Then, there exists a real M-curve of degree $d$ intersecting $L$ in only pairs of complex conjugated points, respectively in $\frac {d-1} 2$ pairs of complex conjugated points and exactly one real point.

Proof: You can just modify a little bit the Harnack method. By Harnack’s construction, there exists a real M-curve $X$ of degree $d-1$ intersecting $L(\mathbb R)$ in $d-1$ real points (the line $L$ intersects only one real connected component of $X(\mathbb R)$). We can now perturb the union $X\cup L$ to a non-singular real curve $A$ of degree $d$ with the properties that we want. In fact, in order to construct $A$, it is enough to apply a small perturbation to the union $X\cup L$ with respect to $\frac d 2$ pairs of complex conjugated lines, respectively $\frac {d-1} 2$ pairs of complex conjugated lines and one real line (which must be chosen passing through a point of an appropriate connected component of $L(\mathbb R) \setminus X(\mathbb R)$).

To be more precise, we construct $A$ as the zero set of the following real homogeneous non-singular polynomial of degree $d$:

let $f, g, h \in \mathbb R[x,y,z]$ be the homogeneous polynomials associated respectively to $X$, $L$ and the union of the lines chosen for the small perturbation;
then, there exists a small enough real non-zero number $b$ such that $$A = \{(x,y,z) \in \mathbb R \mid f(x,y,z) g(x,y,z) + b h(x,y,z) = 0 \} \ .$$

Hope it was helpful.

the union of $X$ and $L$ is $XL$, not $X\cap L$. $H$ is a clerical error since it is $h$ above.

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edit approved Dec 14, 2023 at 20:21

Super Sanae

85
4

You can always construct an M-curve with the property you asked for.

Let $L$ be a real line in the real projective plane and $d$ an even, respectively odd, integer. Then, there exists a real M-curve of degree $d$ intersecting $L$ in only pairs of complex conjugated points, respectively in $\frac {d-1} 2$ pairs of complex conjugated points and exactly one real point.

Proof: You can just modify a little bit the Harnack method. By Harnack’s construction, there exists a real M-curve $X$ of degree $d-1$ intersecting $L(\mathbb R)$ in $d-1$ real points (the line $L$ intersects only one real connected component of $X(\mathbb R)$). We can now perturb the union $X \cap L$$XL$ to a non-singular real curve $A$ of degree $d$ with the properties that we want. In fact, in order to construct $A$, it is enough to apply a small perturbation to the union $X \cap L$$XL$ with respect to $\frac d 2$ pairs of complex conjugated lines, respectively $\frac {d-1} 2$ pairs of complex conjugated lines and one real line (which must be chosen passing through a point of an appropriate connected component of $L(\mathbb R) \setminus X(\mathbb R)$).

To be more precise, we construct $A$ as the zero set of the following real homogeneous non-singular polynomial of degree $d$:

let $f, g, h \in \mathbb R[x,y,z]$ be the homogeneous polynomials associated respectively to $X$, $L$ and the union of the lines chosen for the small perturbation;
then, there exists a small enough real non-zero number $b$ such that $$A = \{(x,y,z) \in \mathbb R \mid f(x,y,z) g(x,y,z) + b H(x,y,z) = 0 \} \ .$$$$A = \{(x,y,z) \in \mathbb R \mid f(x,y,z) g(x,y,z) + b h(x,y,z) = 0 \} \ .$$

Hope it was helpful.

You can always construct an M-curve with the property you asked for.

Let $L$ be a real line in the real projective plane and $d$ an even, respectively odd, integer. Then, there exists a real M-curve of degree $d$ intersecting $L$ in only pairs of complex conjugated points, respectively in $\frac {d-1} 2$ pairs of complex conjugated points and exactly one real point.

Proof: You can just modify a little bit the Harnack method. By Harnack’s construction, there exists a real M-curve $X$ of degree $d-1$ intersecting $L(\mathbb R)$ in $d-1$ real points (the line $L$ intersects only one real connected component of $X(\mathbb R)$). We can now perturb the union $X \cap L$ to a non-singular real curve $A$ of degree $d$ with the properties that we want. In fact, in order to construct $A$, it is enough to apply a small perturbation to the union $X \cap L$ with respect to $\frac d 2$ pairs of complex conjugated lines, respectively $\frac {d-1} 2$ pairs of complex conjugated lines and one real line (which must be chosen passing through a point of an appropriate connected component of $L(\mathbb R) \setminus X(\mathbb R)$).

To be more precise, we construct $A$ as the zero set of the following real homogeneous non-singular polynomial of degree $d$:

let $f, g, h \in \mathbb R[x,y,z]$ be the homogeneous polynomials associated respectively to $X$, $L$ and the union of the lines chosen for the small perturbation;
then, there exists a small enough real non-zero number $b$ such that $$A = \{(x,y,z) \in \mathbb R \mid f(x,y,z) g(x,y,z) + b H(x,y,z) = 0 \} \ .$$

Hope it was helpful.

You can always construct an M-curve with the property you asked for.

Let $L$ be a real line in the real projective plane and $d$ an even, respectively odd, integer. Then, there exists a real M-curve of degree $d$ intersecting $L$ in only pairs of complex conjugated points, respectively in $\frac {d-1} 2$ pairs of complex conjugated points and exactly one real point.

Proof: You can just modify a little bit the Harnack method. By Harnack’s construction, there exists a real M-curve $X$ of degree $d-1$ intersecting $L(\mathbb R)$ in $d-1$ real points (the line $L$ intersects only one real connected component of $X(\mathbb R)$). We can now perturb the union $XL$ to a non-singular real curve $A$ of degree $d$ with the properties that we want. In fact, in order to construct $A$, it is enough to apply a small perturbation to the union $XL$ with respect to $\frac d 2$ pairs of complex conjugated lines, respectively $\frac {d-1} 2$ pairs of complex conjugated lines and one real line (which must be chosen passing through a point of an appropriate connected component of $L(\mathbb R) \setminus X(\mathbb R)$).

To be more precise, we construct $A$ as the zero set of the following real homogeneous non-singular polynomial of degree $d$:

let $f, g, h \in \mathbb R[x,y,z]$ be the homogeneous polynomials associated respectively to $X$, $L$ and the union of the lines chosen for the small perturbation;
then, there exists a small enough real non-zero number $b$ such that $$A = \{(x,y,z) \in \mathbb R \mid f(x,y,z) g(x,y,z) + b h(x,y,z) = 0 \} \ .$$

Hope it was helpful.

MathJaxed

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edited Dec 14, 2023 at 10:33

Alex M.

5.5k
11
39
54

youYou can always construct an M-curve with the property you asked for.

Let L$L$ be a real line in the real projective plane and d$d$ an even, resp.respectively odd, integer. Then, there exists a real M-curve of degree d$d$ intersecting L$L$ in only pairs of complex conjugated points, resp.respectively in (d-1)/2$\frac {d-1} 2$ pairs of complex conjugated points and exactly one real point.

Proof: You can just modify a little bit the Harnack method. By Harnack’s construction, there exists a real M-curve X$X$ of degree d-1$d-1$ intersecting L(R)$L(\mathbb R)$ in d-1$d-1$ real points (the line L$L$ intersects only one real connected component of X(R)$X(\mathbb R)$). WeWe can now perturb the union XuL$X \cap L$ to a non-singular real curve A$A$ of degree d$d$ with the properties that we want. In fact, inin order to construct A$A$, it is enough to apply a small perturbation to the union XuL$X \cap L$ with respect to d/2$\frac d 2$ pairs of complex conjugated lines, resp.respectively (d-1)/2$\frac {d-1} 2$ pairs of complex conjugated lines and one real line (which must be chosen passing through a point of an opportuneappropriate connected component of L(R) \ X(R)$L(\mathbb R) \setminus X(\mathbb R)$).

To be more precise, we construct A$A$ as the zero set of the following real homogeneous non-singular polynomial of degree d.

Let f(x,y,z), g(x,y,z) and H(x,y,z) be the homogeneous real polynomials associated respectively to X, L and the union of the lines chosen for the small perturbation. Then, there exists a small enough real non-zero number b such that {f(x,y,z)g(x,y,z)+bH(x,y,z)=0}=A.$d$:

let $f, g, h \in \mathbb R[x,y,z]$ be the homogeneous polynomials associated respectively to $X$, $L$ and the union of the lines chosen for the small perturbation;

then, there exists a small enough real non-zero number $b$ such that $$A = \{(x,y,z) \in \mathbb R \mid f(x,y,z) g(x,y,z) + b H(x,y,z) = 0 \} \ .$$

Hope it was helpful.

you can always construct M-curve with the property you asked for.

Let L be a real line in the real projective plane and d an even, resp. odd, integer. Then, there exists a real M-curve of degree d intersecting L in only pairs of complex conjugated points, resp. in (d-1)/2 pairs of complex conjugated points and exactly one real point

Proof: You can just modify a little bit the Harnack method. By Harnack’s construction, there exists a real M-curve X of degree d-1 intersecting L(R) in d-1 real points (the line L intersects only one real connected component of X(R)). We can now perturb the union XuL to a non-singular real curve A of degree d with the properties that we want. In fact, in order to construct A, it is enough to apply a small perturbation to the union XuL with respect to d/2 pairs of complex conjugated lines, resp. (d-1)/2 pairs of complex conjugated lines and one real line (which must be chosen passing through a point of an opportune connected component of L(R) \ X(R)).

To be more precise, we construct A as the zero set of the following real homogeneous non-singular polynomial of degree d.

Let f(x,y,z), g(x,y,z) and H(x,y,z) be the homogeneous real polynomials associated respectively to X, L and the union of the lines chosen for the small perturbation. Then, there exists a small enough real non-zero number b such that {f(x,y,z)g(x,y,z)+bH(x,y,z)=0}=A.

Hope it was helpful.

You can always construct an M-curve with the property you asked for.

Let $L$ be a real line in the real projective plane and $d$ an even, respectively odd, integer. Then, there exists a real M-curve of degree $d$ intersecting $L$ in only pairs of complex conjugated points, respectively in $\frac {d-1} 2$ pairs of complex conjugated points and exactly one real point.

Proof: You can just modify a little bit the Harnack method. By Harnack’s construction, there exists a real M-curve $X$ of degree $d-1$ intersecting $L(\mathbb R)$ in $d-1$ real points (the line $L$ intersects only one real connected component of $X(\mathbb R)$). We can now perturb the union $X \cap L$ to a non-singular real curve $A$ of degree $d$ with the properties that we want. In fact, in order to construct $A$, it is enough to apply a small perturbation to the union $X \cap L$ with respect to $\frac d 2$ pairs of complex conjugated lines, respectively $\frac {d-1} 2$ pairs of complex conjugated lines and one real line (which must be chosen passing through a point of an appropriate connected component of $L(\mathbb R) \setminus X(\mathbb R)$).

To be more precise, we construct $A$ as the zero set of the following real homogeneous non-singular polynomial of degree $d$:

let $f, g, h \in \mathbb R[x,y,z]$ be the homogeneous polynomials associated respectively to $X$, $L$ and the union of the lines chosen for the small perturbation;