Is there an irreducible projective hypersurface such that its complement has zero Euler characteristic?
$begingroup$
We know that, if $f=X_0X_1...X_n in mathbb{C}[X_0,...,X_n]$ and $Z(f)subset mathbb{CP}^n$, then the Euler characteristic of its complement is zero, i.e.
$$
chi(mathbb{CP}^nsetminus Z(f))=0.
$$
But $f$ is not irreducible.
Let $Zsubset mathbb{CP}^n$ be a smooth, irreducible hypersurface. Then, we know that
$$
chi(Z)=frac{1}{d}((1-d)^{n+1}-1)+n+1,
$$
where $d$ degree of $Z$.
In particular, if $g=X_0^2+...+X_3^2 in mathbb{C}[X_0,...,X_3]$, we have
$$
chi(Z(g))=frac{1}{2}((1-2)^4-1)+4=4,
$$
then $chi(mathbb{CP}^3setminus Z(g))=0$, since $chi(mathbb{CP}^3)=4$.
So I ask: is there an irreducible homogeneous polynomial $h in mathbb{C}[X_0,...,X_n]$ such that deg$h>2$ and $chi(mathbb{CP}^nsetminus Z(h))=0$?
Remark: this is not possible if $Z(h)$ is smooth (with deg$h>2$).
algebraic-geometry algebraic-topology complex-geometry
asked Dec 29 '15 at 17:05
user73454user73454
501314
$endgroup$
add a comment |
$begingroup$
We know that, if $f=X_0X_1...X_n in mathbb{C}[X_0,...,X_n]$ and $Z(f)subset mathbb{CP}^n$, then the Euler characteristic of its complement is zero, i.e.
$$
chi(mathbb{CP}^nsetminus Z(f))=0.
$$
But $f$ is not irreducible.
Let $Zsubset mathbb{CP}^n$ be a smooth, irreducible hypersurface. Then, we know that
$$
chi(Z)=frac{1}{d}((1-d)^{n+1}-1)+n+1,
$$
where $d$ degree of $Z$.
In particular, if $g=X_0^2+...+X_3^2 in mathbb{C}[X_0,...,X_3]$, we have
$$
chi(Z(g))=frac{1}{2}((1-2)^4-1)+4=4,
$$
then $chi(mathbb{CP}^3setminus Z(g))=0$, since $chi(mathbb{CP}^3)=4$.
So I ask: is there an irreducible homogeneous polynomial $h in mathbb{C}[X_0,...,X_n]$ such that deg$h>2$ and $chi(mathbb{CP}^nsetminus Z(h))=0$?
Remark: this is not possible if $Z(h)$ is smooth (with deg$h>2$).
algebraic-geometry algebraic-topology complex-geometry
asked Dec 29 '15 at 17:05
user73454user73454
501314
$endgroup$
2
$begingroup$
As a note it is not generally true for CW complexes that $chi(X setminus Y) + chi(Y)=chi(X)$. It's true here by a trick of dimension.
$endgroup$
– user98602
Dec 29 '15 at 19:32
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Have you tried to calculate the top Chern class of the tangent buundle of a smooth hypersurface?
$endgroup$
– Alan Muniz
Jan 21 '16 at 0:43
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@AlanMuniz, sorry, but how I will use top Chern class? I'm asking this, because $Z(h)$ is not smooth, since I know that there is no such $h$ if $Z(h)$ is smooth.
$endgroup$
– user73454
May 9 '18 at 14:42
add a comment |
$begingroup$
We know that, if $f=X_0X_1...X_n in mathbb{C}[X_0,...,X_n]$ and $Z(f)subset mathbb{CP}^n$, then the Euler characteristic of its complement is zero, i.e.
$$
chi(mathbb{CP}^nsetminus Z(f))=0.
$$
But $f$ is not irreducible.
Let $Zsubset mathbb{CP}^n$ be a smooth, irreducible hypersurface. Then, we know that
$$
chi(Z)=frac{1}{d}((1-d)^{n+1}-1)+n+1,
$$
where $d$ degree of $Z$.
In particular, if $g=X_0^2+...+X_3^2 in mathbb{C}[X_0,...,X_3]$, we have
$$
chi(Z(g))=frac{1}{2}((1-2)^4-1)+4=4,
$$
then $chi(mathbb{CP}^3setminus Z(g))=0$, since $chi(mathbb{CP}^3)=4$.
So I ask: is there an irreducible homogeneous polynomial $h in mathbb{C}[X_0,...,X_n]$ such that deg$h>2$ and $chi(mathbb{CP}^nsetminus Z(h))=0$?
Remark: this is not possible if $Z(h)$ is smooth (with deg$h>2$).
algebraic-geometry algebraic-topology complex-geometry
asked Dec 29 '15 at 17:05
user73454user73454
501314
$endgroup$
We know that, if $f=X_0X_1...X_n in mathbb{C}[X_0,...,X_n]$ and $Z(f)subset mathbb{CP}^n$, then the Euler characteristic of its complement is zero, i.e.
$$
chi(mathbb{CP}^nsetminus Z(f))=0.
$$
But $f$ is not irreducible.
Let $Zsubset mathbb{CP}^n$ be a smooth, irreducible hypersurface. Then, we know that
$$
chi(Z)=frac{1}{d}((1-d)^{n+1}-1)+n+1,
$$
where $d$ degree of $Z$.
In particular, if $g=X_0^2+...+X_3^2 in mathbb{C}[X_0,...,X_3]$, we have
$$
chi(Z(g))=frac{1}{2}((1-2)^4-1)+4=4,
$$
then $chi(mathbb{CP}^3setminus Z(g))=0$, since $chi(mathbb{CP}^3)=4$.
So I ask: is there an irreducible homogeneous polynomial $h in mathbb{C}[X_0,...,X_n]$ such that deg$h>2$ and $chi(mathbb{CP}^nsetminus Z(h))=0$?
Remark: this is not possible if $Z(h)$ is smooth (with deg$h>2$).
algebraic-geometry algebraic-topology complex-geometry
algebraic-geometry algebraic-topology complex-geometry
asked Dec 29 '15 at 17:05
user73454user73454
501314
asked Dec 29 '15 at 17:05
user73454user73454
501314
edited Jan 4 '16 at 2:49
user73454
asked Dec 29 '15 at 17:05
user73454user73454
501314
asked Dec 29 '15 at 17:05
user73454user73454
501314
asked Dec 29 '15 at 17:05
user73454user73454
501314
501314
2
$begingroup$
As a note it is not generally true for CW complexes that $chi(X setminus Y) + chi(Y)=chi(X)$. It's true here by a trick of dimension.
$endgroup$
– user98602
Dec 29 '15 at 19:32
$begingroup$
Have you tried to calculate the top Chern class of the tangent buundle of a smooth hypersurface?
$endgroup$
– Alan Muniz
Jan 21 '16 at 0:43
$begingroup$
@AlanMuniz, sorry, but how I will use top Chern class? I'm asking this, because $Z(h)$ is not smooth, since I know that there is no such $h$ if $Z(h)$ is smooth.
$endgroup$
– user73454
May 9 '18 at 14:42
add a comment |
2
$begingroup$
As a note it is not generally true for CW complexes that $chi(X setminus Y) + chi(Y)=chi(X)$. It's true here by a trick of dimension.
$endgroup$
– user98602
Dec 29 '15 at 19:32
$begingroup$
Have you tried to calculate the top Chern class of the tangent buundle of a smooth hypersurface?
$endgroup$
– Alan Muniz
Jan 21 '16 at 0:43
$begingroup$
@AlanMuniz, sorry, but how I will use top Chern class? I'm asking this, because $Z(h)$ is not smooth, since I know that there is no such $h$ if $Z(h)$ is smooth.
$endgroup$
– user73454
May 9 '18 at 14:42
2
2
$begingroup$
As a note it is not generally true for CW complexes that $chi(X setminus Y) + chi(Y)=chi(X)$. It's true here by a trick of dimension.
$endgroup$
– user98602
Dec 29 '15 at 19:32
$begingroup$
As a note it is not generally true for CW complexes that $chi(X setminus Y) + chi(Y)=chi(X)$. It's true here by a trick of dimension.
$endgroup$
– user98602
Dec 29 '15 at 19:32
$begingroup$
Have you tried to calculate the top Chern class of the tangent buundle of a smooth hypersurface?
$endgroup$
– Alan Muniz
Jan 21 '16 at 0:43
$begingroup$
Have you tried to calculate the top Chern class of the tangent buundle of a smooth hypersurface?
$endgroup$
– Alan Muniz
Jan 21 '16 at 0:43
$begingroup$
@AlanMuniz, sorry, but how I will use top Chern class? I'm asking this, because $Z(h)$ is not smooth, since I know that there is no such $h$ if $Z(h)$ is smooth.
$endgroup$
– user73454
May 9 '18 at 14:42
$begingroup$
@AlanMuniz, sorry, but how I will use top Chern class? I'm asking this, because $Z(h)$ is not smooth, since I know that there is no such $h$ if $Z(h)$ is smooth.
$endgroup$
– user73454
May 9 '18 at 14:42
add a comment |
1 Answer
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Let $S = { x_1^j x_2^{d-j} = x_3^k x_4^{d-k} }$ in $mathbb{P}^3$, with $0 < j,k < d$ and $GCD(j,k,d)=1$. I claim that $S$ is irreducible and $chi(S)=4$ and hence $chi(mathbb{P}^3 setminus S)=0$.
Let $T ={ x_1 x_2 x_3 x_4 neq 0 } subset mathbb{P}^3$, this is isomorphic to $(mathbb{C}^{ast})^3$. Since $GCD(j,k,d)=GCD(j,k,d-j,d-k)=1$, the locus in $T$ where $x_1^j x_2^{d-j} x_3^{-k} x_4^{-(d-k)}=1$ is isomorphic to $(mathbb{C}^{ast})^2$, with Euler characteristic $0$. So $chi(S cap T) =0$ and thus $chi(S) = chi(S setminus T)$.
But (using the assumptions $0 < j,k < d$) the complement $S setminus T$ is just four $mathbb{P}^1$'s: ${x_1=x_3=0}$, ${x_1=x_4=0}$, ${x_2=x_3=0}$ and ${x_2=x_4=0}$, and this has Euler characteristic $4$. Also, $T cap S$ is obviously irreducible and is dense in $S$, so $S$ is irreducible.
answered Jan 7 at 19:22
David E SpeyerDavid E Speyer
46.3k4127211
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add a comment |
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$begingroup$
Let $S = { x_1^j x_2^{d-j} = x_3^k x_4^{d-k} }$ in $mathbb{P}^3$, with $0 < j,k < d$ and $GCD(j,k,d)=1$. I claim that $S$ is irreducible and $chi(S)=4$ and hence $chi(mathbb{P}^3 setminus S)=0$.
Let $T ={ x_1 x_2 x_3 x_4 neq 0 } subset mathbb{P}^3$, this is isomorphic to $(mathbb{C}^{ast})^3$. Since $GCD(j,k,d)=GCD(j,k,d-j,d-k)=1$, the locus in $T$ where $x_1^j x_2^{d-j} x_3^{-k} x_4^{-(d-k)}=1$ is isomorphic to $(mathbb{C}^{ast})^2$, with Euler characteristic $0$. So $chi(S cap T) =0$ and thus $chi(S) = chi(S setminus T)$.
But (using the assumptions $0 < j,k < d$) the complement $S setminus T$ is just four $mathbb{P}^1$'s: ${x_1=x_3=0}$, ${x_1=x_4=0}$, ${x_2=x_3=0}$ and ${x_2=x_4=0}$, and this has Euler characteristic $4$. Also, $T cap S$ is obviously irreducible and is dense in $S$, so $S$ is irreducible.
answered Jan 7 at 19:22
David E SpeyerDavid E Speyer
46.3k4127211
$endgroup$
add a comment |
$begingroup$
Let $S = { x_1^j x_2^{d-j} = x_3^k x_4^{d-k} }$ in $mathbb{P}^3$, with $0 < j,k < d$ and $GCD(j,k,d)=1$. I claim that $S$ is irreducible and $chi(S)=4$ and hence $chi(mathbb{P}^3 setminus S)=0$.
Let $T ={ x_1 x_2 x_3 x_4 neq 0 } subset mathbb{P}^3$, this is isomorphic to $(mathbb{C}^{ast})^3$. Since $GCD(j,k,d)=GCD(j,k,d-j,d-k)=1$, the locus in $T$ where $x_1^j x_2^{d-j} x_3^{-k} x_4^{-(d-k)}=1$ is isomorphic to $(mathbb{C}^{ast})^2$, with Euler characteristic $0$. So $chi(S cap T) =0$ and thus $chi(S) = chi(S setminus T)$.
But (using the assumptions $0 < j,k < d$) the complement $S setminus T$ is just four $mathbb{P}^1$'s: ${x_1=x_3=0}$, ${x_1=x_4=0}$, ${x_2=x_3=0}$ and ${x_2=x_4=0}$, and this has Euler characteristic $4$. Also, $T cap S$ is obviously irreducible and is dense in $S$, so $S$ is irreducible.
answered Jan 7 at 19:22
David E SpeyerDavid E Speyer
46.3k4127211
$endgroup$
add a comment |
$begingroup$
Let $S = { x_1^j x_2^{d-j} = x_3^k x_4^{d-k} }$ in $mathbb{P}^3$, with $0 < j,k < d$ and $GCD(j,k,d)=1$. I claim that $S$ is irreducible and $chi(S)=4$ and hence $chi(mathbb{P}^3 setminus S)=0$.
Let $T ={ x_1 x_2 x_3 x_4 neq 0 } subset mathbb{P}^3$, this is isomorphic to $(mathbb{C}^{ast})^3$. Since $GCD(j,k,d)=GCD(j,k,d-j,d-k)=1$, the locus in $T$ where $x_1^j x_2^{d-j} x_3^{-k} x_4^{-(d-k)}=1$ is isomorphic to $(mathbb{C}^{ast})^2$, with Euler characteristic $0$. So $chi(S cap T) =0$ and thus $chi(S) = chi(S setminus T)$.
But (using the assumptions $0 < j,k < d$) the complement $S setminus T$ is just four $mathbb{P}^1$'s: ${x_1=x_3=0}$, ${x_1=x_4=0}$, ${x_2=x_3=0}$ and ${x_2=x_4=0}$, and this has Euler characteristic $4$. Also, $T cap S$ is obviously irreducible and is dense in $S$, so $S$ is irreducible.
answered Jan 7 at 19:22
David E SpeyerDavid E Speyer
46.3k4127211
$endgroup$
Let $S = { x_1^j x_2^{d-j} = x_3^k x_4^{d-k} }$ in $mathbb{P}^3$, with $0 < j,k < d$ and $GCD(j,k,d)=1$. I claim that $S$ is irreducible and $chi(S)=4$ and hence $chi(mathbb{P}^3 setminus S)=0$.
Let $T ={ x_1 x_2 x_3 x_4 neq 0 } subset mathbb{P}^3$, this is isomorphic to $(mathbb{C}^{ast})^3$. Since $GCD(j,k,d)=GCD(j,k,d-j,d-k)=1$, the locus in $T$ where $x_1^j x_2^{d-j} x_3^{-k} x_4^{-(d-k)}=1$ is isomorphic to $(mathbb{C}^{ast})^2$, with Euler characteristic $0$. So $chi(S cap T) =0$ and thus $chi(S) = chi(S setminus T)$.
But (using the assumptions $0 < j,k < d$) the complement $S setminus T$ is just four $mathbb{P}^1$'s: ${x_1=x_3=0}$, ${x_1=x_4=0}$, ${x_2=x_3=0}$ and ${x_2=x_4=0}$, and this has Euler characteristic $4$. Also, $T cap S$ is obviously irreducible and is dense in $S$, so $S$ is irreducible.
answered Jan 7 at 19:22
David E SpeyerDavid E Speyer
46.3k4127211
edited Jan 9 at 15:58
answered Jan 7 at 19:22
David E SpeyerDavid E Speyer
46.3k4127211
answered Jan 7 at 19:22
David E SpeyerDavid E Speyer
46.3k4127211
answered Jan 7 at 19:22
David E SpeyerDavid E Speyer
46.3k4127211
46.3k4127211
add a comment |
add a comment |
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$begingroup$
As a note it is not generally true for CW complexes that $chi(X setminus Y) + chi(Y)=chi(X)$. It's true here by a trick of dimension.
$endgroup$
– user98602
Dec 29 '15 at 19:32
$begingroup$
Have you tried to calculate the top Chern class of the tangent buundle of a smooth hypersurface?
$endgroup$
– Alan Muniz
Jan 21 '16 at 0:43
$begingroup$
@AlanMuniz, sorry, but how I will use top Chern class? I'm asking this, because $Z(h)$ is not smooth, since I know that there is no such $h$ if $Z(h)$ is smooth.
$endgroup$
– user73454
May 9 '18 at 14:42