A closed set in the plane that is pluripolar in $mathbb{C}^{2}$
$begingroup$
Let $ E $ be a compact set in $mathbb{R}^{2}$. We know that if $E$ is polar, then it is a set of isolated points. Now, suppose $E$ is pluripolar when regarded as $Esubsetmathbb{R}^{2}subset mathbb{C}^{2}$, i.e., there is a plurisubharmonic function on an open neighborhood of $E$ in $mathbb{C}^{2}$ that is not identically $-infty$, but on $E$ it is equal to $-infty$. Can we still say that $E$ is a set of isolated points?
complex-analysis
edited Jan 22 at 20:31
Alon Amit
10.6k3768
asked Jan 20 at 1:14
M. RahmatM. Rahmat
291212
$endgroup$
add a comment |
$begingroup$
Let $ E $ be a compact set in $mathbb{R}^{2}$. We know that if $E$ is polar, then it is a set of isolated points. Now, suppose $E$ is pluripolar when regarded as $Esubsetmathbb{R}^{2}subset mathbb{C}^{2}$, i.e., there is a plurisubharmonic function on an open neighborhood of $E$ in $mathbb{C}^{2}$ that is not identically $-infty$, but on $E$ it is equal to $-infty$. Can we still say that $E$ is a set of isolated points?
complex-analysis
edited Jan 22 at 20:31
Alon Amit
10.6k3768
asked Jan 20 at 1:14
M. RahmatM. Rahmat
291212
$endgroup$
add a comment |
$begingroup$
Let $ E $ be a compact set in $mathbb{R}^{2}$. We know that if $E$ is polar, then it is a set of isolated points. Now, suppose $E$ is pluripolar when regarded as $Esubsetmathbb{R}^{2}subset mathbb{C}^{2}$, i.e., there is a plurisubharmonic function on an open neighborhood of $E$ in $mathbb{C}^{2}$ that is not identically $-infty$, but on $E$ it is equal to $-infty$. Can we still say that $E$ is a set of isolated points?
complex-analysis
edited Jan 22 at 20:31
Alon Amit
10.6k3768
asked Jan 20 at 1:14
M. RahmatM. Rahmat
291212
$endgroup$
Let $ E $ be a compact set in $mathbb{R}^{2}$. We know that if $E$ is polar, then it is a set of isolated points. Now, suppose $E$ is pluripolar when regarded as $Esubsetmathbb{R}^{2}subset mathbb{C}^{2}$, i.e., there is a plurisubharmonic function on an open neighborhood of $E$ in $mathbb{C}^{2}$ that is not identically $-infty$, but on $E$ it is equal to $-infty$. Can we still say that $E$ is a set of isolated points?
complex-analysis
complex-analysis
edited Jan 22 at 20:31
Alon Amit
10.6k3768
asked Jan 20 at 1:14
M. RahmatM. Rahmat
291212
edited Jan 22 at 20:31
Alon Amit
10.6k3768
asked Jan 20 at 1:14
M. RahmatM. Rahmat
291212
edited Jan 22 at 20:31
Alon Amit
10.6k3768
edited Jan 22 at 20:31
Alon Amit
10.6k3768
edited Jan 22 at 20:31
Alon Amit
10.6k3768
10.6k3768
asked Jan 20 at 1:14
M. RahmatM. Rahmat
291212
asked Jan 20 at 1:14
M. RahmatM. Rahmat
291212
asked Jan 20 at 1:14
M. RahmatM. Rahmat
291212
291212
add a comment |
add a comment |
1 Answer
1
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oldest
votes
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Zero sets of non-constant holomorphic functions $f$ are pluripolar (because $log |f|$ is plurisubharmonic), so the example $f(z,w) = w$ shows that the (complex one-dimensional) $z$-plane is a pluripolar set in $mathbb{C}^2$ (viewed as the complex two-dimensional $(z,w)$-space.)
answered Jan 22 at 21:53
Lukas GeyerLukas Geyer
13.8k1555
$endgroup$
$begingroup$
Sorry! I didn't get it. But the $z-$plane does not turn $log|f|$ into $-infty$...
$endgroup$
– M. Rahmat
Jan 24 at 2:35
1
$begingroup$
@M.Rahmat: $log |f|$ is $-infty$ exactly at the zeros of $f$, which in this case is given by the equation $w=0$, i.e., the $z$-plane.
$endgroup$
– Lukas Geyer
Jan 24 at 4:03
$begingroup$
Now i got it! Thanks
$endgroup$
– M. Rahmat
Jan 24 at 19:45
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Zero sets of non-constant holomorphic functions $f$ are pluripolar (because $log |f|$ is plurisubharmonic), so the example $f(z,w) = w$ shows that the (complex one-dimensional) $z$-plane is a pluripolar set in $mathbb{C}^2$ (viewed as the complex two-dimensional $(z,w)$-space.)
answered Jan 22 at 21:53
Lukas GeyerLukas Geyer
13.8k1555
$endgroup$
$begingroup$
Sorry! I didn't get it. But the $z-$plane does not turn $log|f|$ into $-infty$...
$endgroup$
– M. Rahmat
Jan 24 at 2:35
1
$begingroup$
@M.Rahmat: $log |f|$ is $-infty$ exactly at the zeros of $f$, which in this case is given by the equation $w=0$, i.e., the $z$-plane.
$endgroup$
– Lukas Geyer
Jan 24 at 4:03
$begingroup$
Now i got it! Thanks
$endgroup$
– M. Rahmat
Jan 24 at 19:45
add a comment |
$begingroup$
Zero sets of non-constant holomorphic functions $f$ are pluripolar (because $log |f|$ is plurisubharmonic), so the example $f(z,w) = w$ shows that the (complex one-dimensional) $z$-plane is a pluripolar set in $mathbb{C}^2$ (viewed as the complex two-dimensional $(z,w)$-space.)
answered Jan 22 at 21:53
Lukas GeyerLukas Geyer
13.8k1555
$endgroup$
$begingroup$
Sorry! I didn't get it. But the $z-$plane does not turn $log|f|$ into $-infty$...
$endgroup$
– M. Rahmat
Jan 24 at 2:35
1
$begingroup$
@M.Rahmat: $log |f|$ is $-infty$ exactly at the zeros of $f$, which in this case is given by the equation $w=0$, i.e., the $z$-plane.
$endgroup$
– Lukas Geyer
Jan 24 at 4:03
$begingroup$
Now i got it! Thanks
$endgroup$
– M. Rahmat
Jan 24 at 19:45
add a comment |
$begingroup$
Zero sets of non-constant holomorphic functions $f$ are pluripolar (because $log |f|$ is plurisubharmonic), so the example $f(z,w) = w$ shows that the (complex one-dimensional) $z$-plane is a pluripolar set in $mathbb{C}^2$ (viewed as the complex two-dimensional $(z,w)$-space.)
answered Jan 22 at 21:53
Lukas GeyerLukas Geyer
13.8k1555
$endgroup$
Zero sets of non-constant holomorphic functions $f$ are pluripolar (because $log |f|$ is plurisubharmonic), so the example $f(z,w) = w$ shows that the (complex one-dimensional) $z$-plane is a pluripolar set in $mathbb{C}^2$ (viewed as the complex two-dimensional $(z,w)$-space.)
answered Jan 22 at 21:53
Lukas GeyerLukas Geyer
13.8k1555
answered Jan 22 at 21:53
Lukas GeyerLukas Geyer
13.8k1555
answered Jan 22 at 21:53
Lukas GeyerLukas Geyer
13.8k1555
answered Jan 22 at 21:53
Lukas GeyerLukas Geyer
13.8k1555
13.8k1555
$begingroup$
Sorry! I didn't get it. But the $z-$plane does not turn $log|f|$ into $-infty$...
$endgroup$
– M. Rahmat
Jan 24 at 2:35
1
$begingroup$
@M.Rahmat: $log |f|$ is $-infty$ exactly at the zeros of $f$, which in this case is given by the equation $w=0$, i.e., the $z$-plane.
$endgroup$
– Lukas Geyer
Jan 24 at 4:03
$begingroup$
Now i got it! Thanks
$endgroup$
– M. Rahmat
Jan 24 at 19:45
add a comment |
$begingroup$
Sorry! I didn't get it. But the $z-$plane does not turn $log|f|$ into $-infty$...
$endgroup$
– M. Rahmat
Jan 24 at 2:35
1
$begingroup$
@M.Rahmat: $log |f|$ is $-infty$ exactly at the zeros of $f$, which in this case is given by the equation $w=0$, i.e., the $z$-plane.
$endgroup$
– Lukas Geyer
Jan 24 at 4:03
$begingroup$
Now i got it! Thanks
$endgroup$
– M. Rahmat
Jan 24 at 19:45
$begingroup$
Sorry! I didn't get it. But the $z-$plane does not turn $log|f|$ into $-infty$...
$endgroup$
– M. Rahmat
Jan 24 at 2:35
$begingroup$
Sorry! I didn't get it. But the $z-$plane does not turn $log|f|$ into $-infty$...
$endgroup$
– M. Rahmat
Jan 24 at 2:35
1
1
$begingroup$
@M.Rahmat: $log |f|$ is $-infty$ exactly at the zeros of $f$, which in this case is given by the equation $w=0$, i.e., the $z$-plane.
$endgroup$
– Lukas Geyer
Jan 24 at 4:03
$begingroup$
@M.Rahmat: $log |f|$ is $-infty$ exactly at the zeros of $f$, which in this case is given by the equation $w=0$, i.e., the $z$-plane.
$endgroup$
– Lukas Geyer
Jan 24 at 4:03
$begingroup$
Now i got it! Thanks
$endgroup$
– M. Rahmat
Jan 24 at 19:45
$begingroup$
Now i got it! Thanks
$endgroup$
– M. Rahmat
Jan 24 at 19:45
add a comment |
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