Laurent series and tensorProblem with integration of $1$-form on surfaceResidue of $mathrmexp (z+1/z)/((z-z_1)(z-z_2))$ at z=0Contour Integration - my solution for real integral is complex?Residue Theorem for trigonometric integrals.Suppose that $F$ is analytic in a convex domain $G$ and $Re(F')>0$. Prove or disprove: $F$ is univalent in $G$Multivariate/multidimensional residuesProbably incorrect method of integration gives the correct result. Why?How to view this pseudocode in math symbols?compute residual of $A/(overlinez-overlinelambda_1)$Show that $iint_gamma Arg(z)dz=fracpi2+ln2$
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Laurent series and tensor
Problem with integration of $1$-form on surfaceResidue of $mathrmexp (z+1/z)/((z-z_1)(z-z_2))$ at z=0Contour Integration - my solution for real integral is complex?Residue Theorem for trigonometric integrals.Suppose that $F$ is analytic in a convex domain $G$ and $Re(F')>0$. Prove or disprove: $F$ is univalent in $G$Multivariate/multidimensional residuesProbably incorrect method of integration gives the correct result. Why?How to view this pseudocode in math symbols?compute residual of $A/(overlinez-overlinelambda_1)$Show that $iint_gamma Arg(z)dz=fracpi2+ln2$
$begingroup$
Let us begin with the complex vector space
beginequation
V_z=Bigomegain mathbbC[[z,z^-1]]dz mid operatornameRes_z=0 omega (z)Big
endequation
We could define the tensor product of $V_z_1 otimes V_z_2$.
My question is why $$omega_0,2:= fracdz_1 dz_2(z_1 - z_2)^2$$ does not belong to $V_z_1 otimes V_z_2$.
Residue of $omega_0,2$ at $z_1 =0 , z_2 =0 $ is zero. What is going here?
complex-analysis meromorphic-functions
edited Mar 24 at 1:39
Andrews
1,2812422
asked Mar 22 at 3:43
GGTGGT
485311
$endgroup$
add a comment |
$begingroup$
Let us begin with the complex vector space
beginequation
V_z=Bigomegain mathbbC[[z,z^-1]]dz mid operatornameRes_z=0 omega (z)Big
endequation
We could define the tensor product of $V_z_1 otimes V_z_2$.
My question is why $$omega_0,2:= fracdz_1 dz_2(z_1 - z_2)^2$$ does not belong to $V_z_1 otimes V_z_2$.
Residue of $omega_0,2$ at $z_1 =0 , z_2 =0 $ is zero. What is going here?
complex-analysis meromorphic-functions
edited Mar 24 at 1:39
Andrews
1,2812422
asked Mar 22 at 3:43
GGTGGT
485311
$endgroup$
add a comment |
$begingroup$
Let us begin with the complex vector space
beginequation
V_z=Bigomegain mathbbC[[z,z^-1]]dz mid operatornameRes_z=0 omega (z)Big
endequation
We could define the tensor product of $V_z_1 otimes V_z_2$.
My question is why $$omega_0,2:= fracdz_1 dz_2(z_1 - z_2)^2$$ does not belong to $V_z_1 otimes V_z_2$.
Residue of $omega_0,2$ at $z_1 =0 , z_2 =0 $ is zero. What is going here?
complex-analysis meromorphic-functions
edited Mar 24 at 1:39
Andrews
1,2812422
asked Mar 22 at 3:43
GGTGGT
485311
$endgroup$
Let us begin with the complex vector space
beginequation
V_z=Bigomegain mathbbC[[z,z^-1]]dz mid operatornameRes_z=0 omega (z)Big
endequation
We could define the tensor product of $V_z_1 otimes V_z_2$.
My question is why $$omega_0,2:= fracdz_1 dz_2(z_1 - z_2)^2$$ does not belong to $V_z_1 otimes V_z_2$.
Residue of $omega_0,2$ at $z_1 =0 , z_2 =0 $ is zero. What is going here?
complex-analysis meromorphic-functions
complex-analysis meromorphic-functions
edited Mar 24 at 1:39
Andrews
1,2812422
asked Mar 22 at 3:43
GGTGGT
485311
edited Mar 24 at 1:39
Andrews
1,2812422
asked Mar 22 at 3:43
GGTGGT
485311
edited Mar 24 at 1:39
Andrews
1,2812422
edited Mar 24 at 1:39
Andrews
1,2812422
edited Mar 24 at 1:39
Andrews
1,2812422
1,2812422
asked Mar 22 at 3:43
GGTGGT
485311
asked Mar 22 at 3:43
GGTGGT
485311
asked Mar 22 at 3:43
GGTGGT
485311
485311
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The tensor product consists of elements
$$f(z_1)g(z_2),dz_1,dz_2$$
where $f$ and $g$ are Laurent series satisfying your residue condition.
But
$$h(z_1,z_2)=frac1(z_1-z_2)^2$$
cannot be written as a product of a Laurent series in $z_1$ and
a Laurent series in $z_2$. Nothing mysterious about that; most functions
of two variables are not products of two functions of one variable.
answered Mar 24 at 5:57
Lord Shark the UnknownLord Shark the Unknown
108k1162135
$endgroup$
$begingroup$
Is this obvious in this case it can't be written? Is there some proof?
$endgroup$
– GGT
Mar 24 at 11:58
$begingroup$
For an $h$ to have a such a factorisation one needs $f(z_1,z_2)f(z_1',z_2')=f(z_1,z_2')f(z_1',z_2)$ etc.
$endgroup$
– Lord Shark the Unknown
Mar 24 at 12:03
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The tensor product consists of elements
$$f(z_1)g(z_2),dz_1,dz_2$$
where $f$ and $g$ are Laurent series satisfying your residue condition.
But
$$h(z_1,z_2)=frac1(z_1-z_2)^2$$
cannot be written as a product of a Laurent series in $z_1$ and
a Laurent series in $z_2$. Nothing mysterious about that; most functions
of two variables are not products of two functions of one variable.
answered Mar 24 at 5:57
Lord Shark the UnknownLord Shark the Unknown
108k1162135
$endgroup$
$begingroup$
Is this obvious in this case it can't be written? Is there some proof?
$endgroup$
– GGT
Mar 24 at 11:58
$begingroup$
For an $h$ to have a such a factorisation one needs $f(z_1,z_2)f(z_1',z_2')=f(z_1,z_2')f(z_1',z_2)$ etc.
$endgroup$
– Lord Shark the Unknown
Mar 24 at 12:03
add a comment |
$begingroup$
The tensor product consists of elements
$$f(z_1)g(z_2),dz_1,dz_2$$
where $f$ and $g$ are Laurent series satisfying your residue condition.
But
$$h(z_1,z_2)=frac1(z_1-z_2)^2$$
cannot be written as a product of a Laurent series in $z_1$ and
a Laurent series in $z_2$. Nothing mysterious about that; most functions
of two variables are not products of two functions of one variable.
answered Mar 24 at 5:57
Lord Shark the UnknownLord Shark the Unknown
108k1162135
$endgroup$
$begingroup$
Is this obvious in this case it can't be written? Is there some proof?
$endgroup$
– GGT
Mar 24 at 11:58
$begingroup$
For an $h$ to have a such a factorisation one needs $f(z_1,z_2)f(z_1',z_2')=f(z_1,z_2')f(z_1',z_2)$ etc.
$endgroup$
– Lord Shark the Unknown
Mar 24 at 12:03
add a comment |
$begingroup$
The tensor product consists of elements
$$f(z_1)g(z_2),dz_1,dz_2$$
where $f$ and $g$ are Laurent series satisfying your residue condition.
But
$$h(z_1,z_2)=frac1(z_1-z_2)^2$$
cannot be written as a product of a Laurent series in $z_1$ and
a Laurent series in $z_2$. Nothing mysterious about that; most functions
of two variables are not products of two functions of one variable.
answered Mar 24 at 5:57
Lord Shark the UnknownLord Shark the Unknown
108k1162135
$endgroup$
The tensor product consists of elements
$$f(z_1)g(z_2),dz_1,dz_2$$
where $f$ and $g$ are Laurent series satisfying your residue condition.
But
$$h(z_1,z_2)=frac1(z_1-z_2)^2$$
cannot be written as a product of a Laurent series in $z_1$ and
a Laurent series in $z_2$. Nothing mysterious about that; most functions
of two variables are not products of two functions of one variable.
answered Mar 24 at 5:57
Lord Shark the UnknownLord Shark the Unknown
108k1162135
answered Mar 24 at 5:57
Lord Shark the UnknownLord Shark the Unknown
108k1162135
answered Mar 24 at 5:57
Lord Shark the UnknownLord Shark the Unknown
108k1162135
answered Mar 24 at 5:57
Lord Shark the UnknownLord Shark the Unknown
108k1162135
108k1162135
$begingroup$
Is this obvious in this case it can't be written? Is there some proof?
$endgroup$
– GGT
Mar 24 at 11:58
$begingroup$
For an $h$ to have a such a factorisation one needs $f(z_1,z_2)f(z_1',z_2')=f(z_1,z_2')f(z_1',z_2)$ etc.
$endgroup$
– Lord Shark the Unknown
Mar 24 at 12:03
add a comment |
$begingroup$
Is this obvious in this case it can't be written? Is there some proof?
$endgroup$
– GGT
Mar 24 at 11:58
$begingroup$
For an $h$ to have a such a factorisation one needs $f(z_1,z_2)f(z_1',z_2')=f(z_1,z_2')f(z_1',z_2)$ etc.
$endgroup$
– Lord Shark the Unknown
Mar 24 at 12:03
$begingroup$
Is this obvious in this case it can't be written? Is there some proof?
$endgroup$
– GGT
Mar 24 at 11:58
$begingroup$
Is this obvious in this case it can't be written? Is there some proof?
$endgroup$
– GGT
Mar 24 at 11:58
$begingroup$
For an $h$ to have a such a factorisation one needs $f(z_1,z_2)f(z_1',z_2')=f(z_1,z_2')f(z_1',z_2)$ etc.
$endgroup$
– Lord Shark the Unknown
Mar 24 at 12:03
$begingroup$
For an $h$ to have a such a factorisation one needs $f(z_1,z_2)f(z_1',z_2')=f(z_1,z_2')f(z_1',z_2)$ etc.
$endgroup$
– Lord Shark the Unknown
Mar 24 at 12:03
add a comment |
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