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Laurent series and tensor

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0

$begingroup$

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

share|cite|improve this question

edited Mar 24 at 1:39

Andrews

1,2812422

asked Mar 22 at 3:43

GGTGGT

485311

$endgroup$

add a comment | 

0

$begingroup$

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

share|cite|improve this question

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

share|cite|improve this question

edited Mar 24 at 1:39

Andrews

1,2812422

asked Mar 22 at 3:43

GGTGGT

485311

$endgroup$

share|cite|improve this question

edited Mar 24 at 1:39

Andrews

1,2812422

asked Mar 22 at 3:43

GGTGGT

485311

share|cite|improve this question

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

asked Mar 22 at 3:43

GGTGGT

485311

asked Mar 22 at 3:43

GGTGGT

485311

1 Answer
1

active

oldest

votes

1

$begingroup$

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.

share|cite|improve this answer

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 | 

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1

$begingroup$

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.

share|cite|improve this answer

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 | 

1

$begingroup$

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.

share|cite|improve this answer

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.

share|cite|improve this answer

answered Mar 24 at 5:57

Lord Shark the UnknownLord Shark the Unknown

108k1162135

$endgroup$

share|cite|improve this answer

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