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A question about *A new proof of a theorem of Narasimhan and Seshadri*
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I am reading S. K. Donsaldson's paper A new proof of a theorem of Narasimhan and Seshadri. Now $M$ is a compact Riemann surface and $E$ is a holomorphic bundle over $M$ with a Hermitian metric. There is a statement (p.276) : for $g$ a self-adjoint complex gauge transformation $g=g^*$, $A$ a Hermitian metric
beginequation
F(A)=F(g.A)implies partialbarpartial(g^2)=-(barpartial g^2)g^-1(barpartial g^2)g^-1^*
endequation
where $F$ is the curvature form.
I try to prove this.
Here I think a problem happens.
We have
beginequation
g.A=A-(barpartial g)g^-1+((barpartial g)g^-1)^*=A-barpartial gg^-1+g^-1partial g
endequation
and for a section $sigma$
beginequation
beginsplit
F(g.A)sigma=&(d_g.A)(d_g.A)sigma\
=&(d_A-barpartial gg^-1+g^-1partial g)(d_A-barpartial gg^-1+g^-1partial g)sigma
endsplit
endequation
and just consider the second order derivative of $g$
beginequation
beginsplit
F(g.A)sigma=&F(A)sigma+d_A(-barpartial g g^-1sigma)+d_A(g^-1partial gsigma)+cdots\
=&F(A)sigma-partialbarpartial gg^-1sigma-g^-1barpartialpartial gsigma+cdots
endsplit
endequation
so from $F(A)=F(g.A)$ we will get something like
beginequation
partialbarpartial gg^-1-g^-1barpartialpartial g+cdots=0
endequation
but I think this is different from $partialbarpartial(g^2)=cdots$, since in which the second derivatives are
beginequation
partialbarpartial g g+gpartialbarpartial g
endequation
and the $partial$ operator is always ahead of $barpartial$ operator. If $partialbarpartial g+barpartialpartial g=0$, then the problem is solved but I do not think this is naturally true.
riemann-surfaces connections
asked Aug 6 at 20:27
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380212
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2
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I am reading S. K. Donsaldson's paper A new proof of a theorem of Narasimhan and Seshadri. Now $M$ is a compact Riemann surface and $E$ is a holomorphic bundle over $M$ with a Hermitian metric. There is a statement (p.276) : for $g$ a self-adjoint complex gauge transformation $g=g^*$, $A$ a Hermitian metric
beginequation
F(A)=F(g.A)implies partialbarpartial(g^2)=-(barpartial g^2)g^-1(barpartial g^2)g^-1^*
endequation
where $F$ is the curvature form.
I try to prove this.
Here I think a problem happens.
We have
beginequation
g.A=A-(barpartial g)g^-1+((barpartial g)g^-1)^*=A-barpartial gg^-1+g^-1partial g
endequation
and for a section $sigma$
beginequation
beginsplit
F(g.A)sigma=&(d_g.A)(d_g.A)sigma\
=&(d_A-barpartial gg^-1+g^-1partial g)(d_A-barpartial gg^-1+g^-1partial g)sigma
endsplit
endequation
and just consider the second order derivative of $g$
beginequation
beginsplit
F(g.A)sigma=&F(A)sigma+d_A(-barpartial g g^-1sigma)+d_A(g^-1partial gsigma)+cdots\
=&F(A)sigma-partialbarpartial gg^-1sigma-g^-1barpartialpartial gsigma+cdots
endsplit
endequation
so from $F(A)=F(g.A)$ we will get something like
beginequation
partialbarpartial gg^-1-g^-1barpartialpartial g+cdots=0
endequation
but I think this is different from $partialbarpartial(g^2)=cdots$, since in which the second derivatives are
beginequation
partialbarpartial g g+gpartialbarpartial g
endequation
and the $partial$ operator is always ahead of $barpartial$ operator. If $partialbarpartial g+barpartialpartial g=0$, then the problem is solved but I do not think this is naturally true.
riemann-surfaces connections
asked Aug 6 at 20:27
Display Name
380212
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I am reading S. K. Donsaldson's paper A new proof of a theorem of Narasimhan and Seshadri. Now $M$ is a compact Riemann surface and $E$ is a holomorphic bundle over $M$ with a Hermitian metric. There is a statement (p.276) : for $g$ a self-adjoint complex gauge transformation $g=g^*$, $A$ a Hermitian metric
beginequation
F(A)=F(g.A)implies partialbarpartial(g^2)=-(barpartial g^2)g^-1(barpartial g^2)g^-1^*
endequation
where $F$ is the curvature form.
I try to prove this.
Here I think a problem happens.
We have
beginequation
g.A=A-(barpartial g)g^-1+((barpartial g)g^-1)^*=A-barpartial gg^-1+g^-1partial g
endequation
and for a section $sigma$
beginequation
beginsplit
F(g.A)sigma=&(d_g.A)(d_g.A)sigma\
=&(d_A-barpartial gg^-1+g^-1partial g)(d_A-barpartial gg^-1+g^-1partial g)sigma
endsplit
endequation
and just consider the second order derivative of $g$
beginequation
beginsplit
F(g.A)sigma=&F(A)sigma+d_A(-barpartial g g^-1sigma)+d_A(g^-1partial gsigma)+cdots\
=&F(A)sigma-partialbarpartial gg^-1sigma-g^-1barpartialpartial gsigma+cdots
endsplit
endequation
so from $F(A)=F(g.A)$ we will get something like
beginequation
partialbarpartial gg^-1-g^-1barpartialpartial g+cdots=0
endequation
but I think this is different from $partialbarpartial(g^2)=cdots$, since in which the second derivatives are
beginequation
partialbarpartial g g+gpartialbarpartial g
endequation
and the $partial$ operator is always ahead of $barpartial$ operator. If $partialbarpartial g+barpartialpartial g=0$, then the problem is solved but I do not think this is naturally true.
riemann-surfaces connections
asked Aug 6 at 20:27
Display Name
380212
I am reading S. K. Donsaldson's paper A new proof of a theorem of Narasimhan and Seshadri. Now $M$ is a compact Riemann surface and $E$ is a holomorphic bundle over $M$ with a Hermitian metric. There is a statement (p.276) : for $g$ a self-adjoint complex gauge transformation $g=g^*$, $A$ a Hermitian metric
beginequation
F(A)=F(g.A)implies partialbarpartial(g^2)=-(barpartial g^2)g^-1(barpartial g^2)g^-1^*
endequation
where $F$ is the curvature form.
I try to prove this.
Here I think a problem happens.
We have
beginequation
g.A=A-(barpartial g)g^-1+((barpartial g)g^-1)^*=A-barpartial gg^-1+g^-1partial g
endequation
and for a section $sigma$
beginequation
beginsplit
F(g.A)sigma=&(d_g.A)(d_g.A)sigma\
=&(d_A-barpartial gg^-1+g^-1partial g)(d_A-barpartial gg^-1+g^-1partial g)sigma
endsplit
endequation
and just consider the second order derivative of $g$
beginequation
beginsplit
F(g.A)sigma=&F(A)sigma+d_A(-barpartial g g^-1sigma)+d_A(g^-1partial gsigma)+cdots\
=&F(A)sigma-partialbarpartial gg^-1sigma-g^-1barpartialpartial gsigma+cdots
endsplit
endequation
so from $F(A)=F(g.A)$ we will get something like
beginequation
partialbarpartial gg^-1-g^-1barpartialpartial g+cdots=0
endequation
but I think this is different from $partialbarpartial(g^2)=cdots$, since in which the second derivatives are
beginequation
partialbarpartial g g+gpartialbarpartial g
endequation
and the $partial$ operator is always ahead of $barpartial$ operator. If $partialbarpartial g+barpartialpartial g=0$, then the problem is solved but I do not think this is naturally true.
riemann-surfaces connections
asked Aug 6 at 20:27
Display Name
380212
asked Aug 6 at 20:27
Display Name
380212
asked Aug 6 at 20:27
Display Name
380212
asked Aug 6 at 20:27
Display Name
380212
380212
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Indeed $partial barpartial g + barpartialpartial g=0$. In fact $partial barpartial + barpartialpartial$ is the zero operator.
We have that $d= partial + barpartial$ is the induced connection on $End(E)$. Then
$$ d^2 = partial^2 + partial barpartial + barpartialpartial + barpartial^2 = partial barpartial + barpartialpartial $$
is the curvature of the induced connection on $End(E)$, since $partial^2 = barpartial^2 =0$ (by dimension). Then the result follows once we have $End(E)$ flat.
We have that the curvature of $E$ is a multiple of the identity automorphism by hypothesis, $F(A) = mu 1$ in page 276. Computing the curvature of the induced connection on $Eotimes E^ast simeq End(E)$ we see that it vanishes.
answered Aug 7 at 13:36
Alan Muniz
937518
I am not sure if it is naturally true but $d_Ad_A$ is the curvature of the connection and is not zero in general.
– Display Name
Aug 7 at 13:43
For sure. However the hypotheses imply that $End(E)$ is flat
– Alan Muniz
Aug 7 at 16:08
I just included that in my answer.
– Alan Muniz
Aug 7 at 16:09
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Indeed $partial barpartial g + barpartialpartial g=0$. In fact $partial barpartial + barpartialpartial$ is the zero operator.
We have that $d= partial + barpartial$ is the induced connection on $End(E)$. Then
$$ d^2 = partial^2 + partial barpartial + barpartialpartial + barpartial^2 = partial barpartial + barpartialpartial $$
is the curvature of the induced connection on $End(E)$, since $partial^2 = barpartial^2 =0$ (by dimension). Then the result follows once we have $End(E)$ flat.
We have that the curvature of $E$ is a multiple of the identity automorphism by hypothesis, $F(A) = mu 1$ in page 276. Computing the curvature of the induced connection on $Eotimes E^ast simeq End(E)$ we see that it vanishes.
answered Aug 7 at 13:36
Alan Muniz
937518
I am not sure if it is naturally true but $d_Ad_A$ is the curvature of the connection and is not zero in general.
– Display Name
Aug 7 at 13:43
For sure. However the hypotheses imply that $End(E)$ is flat
– Alan Muniz
Aug 7 at 16:08
I just included that in my answer.
– Alan Muniz
Aug 7 at 16:09
add a comment |Â
up vote
1
down vote
accepted
Indeed $partial barpartial g + barpartialpartial g=0$. In fact $partial barpartial + barpartialpartial$ is the zero operator.
We have that $d= partial + barpartial$ is the induced connection on $End(E)$. Then
$$ d^2 = partial^2 + partial barpartial + barpartialpartial + barpartial^2 = partial barpartial + barpartialpartial $$
is the curvature of the induced connection on $End(E)$, since $partial^2 = barpartial^2 =0$ (by dimension). Then the result follows once we have $End(E)$ flat.
We have that the curvature of $E$ is a multiple of the identity automorphism by hypothesis, $F(A) = mu 1$ in page 276. Computing the curvature of the induced connection on $Eotimes E^ast simeq End(E)$ we see that it vanishes.
answered Aug 7 at 13:36
Alan Muniz
937518
I am not sure if it is naturally true but $d_Ad_A$ is the curvature of the connection and is not zero in general.
– Display Name
Aug 7 at 13:43
For sure. However the hypotheses imply that $End(E)$ is flat
– Alan Muniz
Aug 7 at 16:08
I just included that in my answer.
– Alan Muniz
Aug 7 at 16:09
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Indeed $partial barpartial g + barpartialpartial g=0$. In fact $partial barpartial + barpartialpartial$ is the zero operator.
We have that $d= partial + barpartial$ is the induced connection on $End(E)$. Then
$$ d^2 = partial^2 + partial barpartial + barpartialpartial + barpartial^2 = partial barpartial + barpartialpartial $$
is the curvature of the induced connection on $End(E)$, since $partial^2 = barpartial^2 =0$ (by dimension). Then the result follows once we have $End(E)$ flat.
We have that the curvature of $E$ is a multiple of the identity automorphism by hypothesis, $F(A) = mu 1$ in page 276. Computing the curvature of the induced connection on $Eotimes E^ast simeq End(E)$ we see that it vanishes.
answered Aug 7 at 13:36
Alan Muniz
937518
Indeed $partial barpartial g + barpartialpartial g=0$. In fact $partial barpartial + barpartialpartial$ is the zero operator.
We have that $d= partial + barpartial$ is the induced connection on $End(E)$. Then
$$ d^2 = partial^2 + partial barpartial + barpartialpartial + barpartial^2 = partial barpartial + barpartialpartial $$
is the curvature of the induced connection on $End(E)$, since $partial^2 = barpartial^2 =0$ (by dimension). Then the result follows once we have $End(E)$ flat.
We have that the curvature of $E$ is a multiple of the identity automorphism by hypothesis, $F(A) = mu 1$ in page 276. Computing the curvature of the induced connection on $Eotimes E^ast simeq End(E)$ we see that it vanishes.
answered Aug 7 at 13:36
Alan Muniz
937518
edited Aug 7 at 16:07
answered Aug 7 at 13:36
Alan Muniz
937518
answered Aug 7 at 13:36
Alan Muniz
937518
answered Aug 7 at 13:36
Alan Muniz
937518
937518
I am not sure if it is naturally true but $d_Ad_A$ is the curvature of the connection and is not zero in general.
– Display Name
Aug 7 at 13:43
For sure. However the hypotheses imply that $End(E)$ is flat
– Alan Muniz
Aug 7 at 16:08
I just included that in my answer.
– Alan Muniz
Aug 7 at 16:09
add a comment |Â
I am not sure if it is naturally true but $d_Ad_A$ is the curvature of the connection and is not zero in general.
– Display Name
Aug 7 at 13:43
For sure. However the hypotheses imply that $End(E)$ is flat
– Alan Muniz
Aug 7 at 16:08
I just included that in my answer.
– Alan Muniz
Aug 7 at 16:09
I am not sure if it is naturally true but $d_Ad_A$ is the curvature of the connection and is not zero in general.
– Display Name
Aug 7 at 13:43
I am not sure if it is naturally true but $d_Ad_A$ is the curvature of the connection and is not zero in general.
– Display Name
Aug 7 at 13:43
For sure. However the hypotheses imply that $End(E)$ is flat
– Alan Muniz
Aug 7 at 16:08
For sure. However the hypotheses imply that $End(E)$ is flat
– Alan Muniz
Aug 7 at 16:08
I just included that in my answer.
– Alan Muniz
Aug 7 at 16:09
I just included that in my answer.
– Alan Muniz
Aug 7 at 16:09
add a comment |Â
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