Exact differential forms defined in path connected sets
$begingroup$
There's a theorem in mathematics that states:
Let $A$ be an open path connected set in $mathbb{R^m}$, and $omega:Atomathbb{R^{1times2}},xto[omega^1(x)space...space omega^m]$ be a 1 degree differential form, then $omega$ is exact.
But we know that:
$$omega:mathbb{R^2setminus {(0,0)}}to mathbb{R^{1times2}}\(x,y)tobigg[ -dfrac{y}{x^2 + y^2} space dfrac{x}{x^2 +y^2}bigg]$$
is not an exact differential form, however, $mathbb{R^2}setminus {(0,0)}$ is an path connected set. What am I getting wrong?
PS:I may be a bit lost in translation about some technical terms, and some rigor may also be flawed, but I hope the message gets through.
analysis differential-forms connectedness
asked Dec 27 '18 at 19:01
BidonBidon
967
$endgroup$
|
show 1 more comment
$begingroup$
There's a theorem in mathematics that states:
Let $A$ be an open path connected set in $mathbb{R^m}$, and $omega:Atomathbb{R^{1times2}},xto[omega^1(x)space...space omega^m]$ be a 1 degree differential form, then $omega$ is exact.
But we know that:
$$omega:mathbb{R^2setminus {(0,0)}}to mathbb{R^{1times2}}\(x,y)tobigg[ -dfrac{y}{x^2 + y^2} space dfrac{x}{x^2 +y^2}bigg]$$
is not an exact differential form, however, $mathbb{R^2}setminus {(0,0)}$ is an path connected set. What am I getting wrong?
PS:I may be a bit lost in translation about some technical terms, and some rigor may also be flawed, but I hope the message gets through.
analysis differential-forms connectedness
asked Dec 27 '18 at 19:01
BidonBidon
967
$endgroup$
$begingroup$
Your "theorem" is wrong and the counter-example is correct. The statement becomes correct if you assume that $A$ is simply connected.
$endgroup$
– Nicolas Hemelsoet
Dec 27 '18 at 19:11
$begingroup$
Just looked up the definitions for arc and path connected sets and I meant a path connected set (already edited it), is it still wrong?
$endgroup$
– Bidon
Dec 27 '18 at 19:17
1
$begingroup$
Yes, to my knoweldge both path and arc connected mean the same (if $A$ is open it's equivalent to ask that $A$ is connected). You need a stronger assumption which is simply connected, look here : en.wikipedia.org/wiki/Simply_connected_space
$endgroup$
– Nicolas Hemelsoet
Dec 27 '18 at 19:19
3
$begingroup$
The theorem is as follows (and it's usually called the Poincaré lemma) : let $A$ be a simply connected open subset of $Bbb R^n$. Then, any closed $1$-form on $A$ is exact.
$endgroup$
– Nicolas Hemelsoet
Dec 27 '18 at 19:23
2
$begingroup$
P.S : I didn't noticed something but in fact your version as stated is also wrong : you need to assume that $omega$ is closed, because it's a necessary condition to be exact.
$endgroup$
– Nicolas Hemelsoet
Dec 27 '18 at 19:24
|
show 1 more comment
$begingroup$
There's a theorem in mathematics that states:
Let $A$ be an open path connected set in $mathbb{R^m}$, and $omega:Atomathbb{R^{1times2}},xto[omega^1(x)space...space omega^m]$ be a 1 degree differential form, then $omega$ is exact.
But we know that:
$$omega:mathbb{R^2setminus {(0,0)}}to mathbb{R^{1times2}}\(x,y)tobigg[ -dfrac{y}{x^2 + y^2} space dfrac{x}{x^2 +y^2}bigg]$$
is not an exact differential form, however, $mathbb{R^2}setminus {(0,0)}$ is an path connected set. What am I getting wrong?
PS:I may be a bit lost in translation about some technical terms, and some rigor may also be flawed, but I hope the message gets through.
analysis differential-forms connectedness
asked Dec 27 '18 at 19:01
BidonBidon
967
$endgroup$
There's a theorem in mathematics that states:
Let $A$ be an open path connected set in $mathbb{R^m}$, and $omega:Atomathbb{R^{1times2}},xto[omega^1(x)space...space omega^m]$ be a 1 degree differential form, then $omega$ is exact.
But we know that:
$$omega:mathbb{R^2setminus {(0,0)}}to mathbb{R^{1times2}}\(x,y)tobigg[ -dfrac{y}{x^2 + y^2} space dfrac{x}{x^2 +y^2}bigg]$$
is not an exact differential form, however, $mathbb{R^2}setminus {(0,0)}$ is an path connected set. What am I getting wrong?
PS:I may be a bit lost in translation about some technical terms, and some rigor may also be flawed, but I hope the message gets through.
analysis differential-forms connectedness
analysis differential-forms connectedness
asked Dec 27 '18 at 19:01
BidonBidon
967
asked Dec 27 '18 at 19:01
BidonBidon
967
edited Dec 27 '18 at 19:15
Bidon
asked Dec 27 '18 at 19:01
BidonBidon
967
asked Dec 27 '18 at 19:01
BidonBidon
967
asked Dec 27 '18 at 19:01
BidonBidon
967
967
$begingroup$
Your "theorem" is wrong and the counter-example is correct. The statement becomes correct if you assume that $A$ is simply connected.
$endgroup$
– Nicolas Hemelsoet
Dec 27 '18 at 19:11
$begingroup$
Just looked up the definitions for arc and path connected sets and I meant a path connected set (already edited it), is it still wrong?
$endgroup$
– Bidon
Dec 27 '18 at 19:17
1
$begingroup$
Yes, to my knoweldge both path and arc connected mean the same (if $A$ is open it's equivalent to ask that $A$ is connected). You need a stronger assumption which is simply connected, look here : en.wikipedia.org/wiki/Simply_connected_space
$endgroup$
– Nicolas Hemelsoet
Dec 27 '18 at 19:19
3
$begingroup$
The theorem is as follows (and it's usually called the Poincaré lemma) : let $A$ be a simply connected open subset of $Bbb R^n$. Then, any closed $1$-form on $A$ is exact.
$endgroup$
– Nicolas Hemelsoet
Dec 27 '18 at 19:23
2
$begingroup$
P.S : I didn't noticed something but in fact your version as stated is also wrong : you need to assume that $omega$ is closed, because it's a necessary condition to be exact.
$endgroup$
– Nicolas Hemelsoet
Dec 27 '18 at 19:24
|
show 1 more comment
$begingroup$
Your "theorem" is wrong and the counter-example is correct. The statement becomes correct if you assume that $A$ is simply connected.
$endgroup$
– Nicolas Hemelsoet
Dec 27 '18 at 19:11
$begingroup$
Just looked up the definitions for arc and path connected sets and I meant a path connected set (already edited it), is it still wrong?
$endgroup$
– Bidon
Dec 27 '18 at 19:17
1
$begingroup$
Yes, to my knoweldge both path and arc connected mean the same (if $A$ is open it's equivalent to ask that $A$ is connected). You need a stronger assumption which is simply connected, look here : en.wikipedia.org/wiki/Simply_connected_space
$endgroup$
– Nicolas Hemelsoet
Dec 27 '18 at 19:19
3
$begingroup$
The theorem is as follows (and it's usually called the Poincaré lemma) : let $A$ be a simply connected open subset of $Bbb R^n$. Then, any closed $1$-form on $A$ is exact.
$endgroup$
– Nicolas Hemelsoet
Dec 27 '18 at 19:23
2
$begingroup$
P.S : I didn't noticed something but in fact your version as stated is also wrong : you need to assume that $omega$ is closed, because it's a necessary condition to be exact.
$endgroup$
– Nicolas Hemelsoet
Dec 27 '18 at 19:24
$begingroup$
Your "theorem" is wrong and the counter-example is correct. The statement becomes correct if you assume that $A$ is simply connected.
$endgroup$
– Nicolas Hemelsoet
Dec 27 '18 at 19:11
$begingroup$
Your "theorem" is wrong and the counter-example is correct. The statement becomes correct if you assume that $A$ is simply connected.
$endgroup$
– Nicolas Hemelsoet
Dec 27 '18 at 19:11
$begingroup$
Just looked up the definitions for arc and path connected sets and I meant a path connected set (already edited it), is it still wrong?
$endgroup$
– Bidon
Dec 27 '18 at 19:17
$begingroup$
Just looked up the definitions for arc and path connected sets and I meant a path connected set (already edited it), is it still wrong?
$endgroup$
– Bidon
Dec 27 '18 at 19:17
1
1
$begingroup$
Yes, to my knoweldge both path and arc connected mean the same (if $A$ is open it's equivalent to ask that $A$ is connected). You need a stronger assumption which is simply connected, look here : en.wikipedia.org/wiki/Simply_connected_space
$endgroup$
– Nicolas Hemelsoet
Dec 27 '18 at 19:19
$begingroup$
Yes, to my knoweldge both path and arc connected mean the same (if $A$ is open it's equivalent to ask that $A$ is connected). You need a stronger assumption which is simply connected, look here : en.wikipedia.org/wiki/Simply_connected_space
$endgroup$
– Nicolas Hemelsoet
Dec 27 '18 at 19:19
3
3
$begingroup$
The theorem is as follows (and it's usually called the Poincaré lemma) : let $A$ be a simply connected open subset of $Bbb R^n$. Then, any closed $1$-form on $A$ is exact.
$endgroup$
– Nicolas Hemelsoet
Dec 27 '18 at 19:23
$begingroup$
The theorem is as follows (and it's usually called the Poincaré lemma) : let $A$ be a simply connected open subset of $Bbb R^n$. Then, any closed $1$-form on $A$ is exact.
$endgroup$
– Nicolas Hemelsoet
Dec 27 '18 at 19:23
2
2
$begingroup$
P.S : I didn't noticed something but in fact your version as stated is also wrong : you need to assume that $omega$ is closed, because it's a necessary condition to be exact.
$endgroup$
– Nicolas Hemelsoet
Dec 27 '18 at 19:24
$begingroup$
P.S : I didn't noticed something but in fact your version as stated is also wrong : you need to assume that $omega$ is closed, because it's a necessary condition to be exact.
$endgroup$
– Nicolas Hemelsoet
Dec 27 '18 at 19:24
|
show 1 more comment
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$begingroup$
Your "theorem" is wrong and the counter-example is correct. The statement becomes correct if you assume that $A$ is simply connected.
$endgroup$
– Nicolas Hemelsoet
Dec 27 '18 at 19:11
$begingroup$
Just looked up the definitions for arc and path connected sets and I meant a path connected set (already edited it), is it still wrong?
$endgroup$
– Bidon
Dec 27 '18 at 19:17
1
$begingroup$
Yes, to my knoweldge both path and arc connected mean the same (if $A$ is open it's equivalent to ask that $A$ is connected). You need a stronger assumption which is simply connected, look here : en.wikipedia.org/wiki/Simply_connected_space
$endgroup$
– Nicolas Hemelsoet
Dec 27 '18 at 19:19
3
$begingroup$
The theorem is as follows (and it's usually called the Poincaré lemma) : let $A$ be a simply connected open subset of $Bbb R^n$. Then, any closed $1$-form on $A$ is exact.
$endgroup$
– Nicolas Hemelsoet
Dec 27 '18 at 19:23
2
$begingroup$
P.S : I didn't noticed something but in fact your version as stated is also wrong : you need to assume that $omega$ is closed, because it's a necessary condition to be exact.
$endgroup$
– Nicolas Hemelsoet
Dec 27 '18 at 19:24