Isotropy group of connection is isomorphic to centraliser of holonomy group
$begingroup$
I am asking for a proof of Lemma (4.2.8) of Donaldson, Kronheimer: The Geometry of Four-Manifolds.
Let $P rightarrow X$ be a principal bundle with structure group $G$.
Denote by $mathcal{G}$ the gauge group.
Let $A$ be a connection on $P$ and $H_A subset G$ its holonomy.
Denote by
$$
Gamma_A=
{u in mathcal{G} | u(A)=A }
$$
the isotropy group of $A$ in $mathcal{G}$.
Then the claim of the lemma is:
For any connection $A$ over a connected base $X$, $Gamma_A$ is isomorphic to the centraliser of $H_A$ in $G$.
My attempt to prove it:
Now if the bundle was trivial, i.e. $P=X times G$, we could take $g in G$ in the centraliser of $H_A$ and get a unique element $u_g in mathcal{G}$ satisfying $u_g(x,e)=(x,g)$.
If the connection $A$ is the trivial connection then it is easy to check that this satisfies $u_g in Gamma_A$.
In matrix notation we have $u(A)=u^{-1}Au+u^{-1} du$.
Plugging in an $A$-horizontal vector field to this formula shows that any $u in Gamma_A$ needs to be constant on $X times {e}$, i.e. $u=u_g$ for some $g in G$.
Plugging in a vertical vector field shows that $g$ must be in the centraliser of $H_A$ in $G$, because $u^{-1} du$ vanishes on vertical vector fields.
I believe I can make this argument work for non-trivial connections $A$.
But no matter if I can or cannot: all of this assumed that the bundle was trivial.
If the bundle isn't trivial I don't even know how to write down a map from the centraliser of $H_A$ to $Gamma_A$.
differential-geometry principal-bundles gauge-theory
asked Dec 19 '18 at 12:01
user505117user505117
233
$endgroup$
add a comment |
$begingroup$
I am asking for a proof of Lemma (4.2.8) of Donaldson, Kronheimer: The Geometry of Four-Manifolds.
Let $P rightarrow X$ be a principal bundle with structure group $G$.
Denote by $mathcal{G}$ the gauge group.
Let $A$ be a connection on $P$ and $H_A subset G$ its holonomy.
Denote by
$$
Gamma_A=
{u in mathcal{G} | u(A)=A }
$$
the isotropy group of $A$ in $mathcal{G}$.
Then the claim of the lemma is:
For any connection $A$ over a connected base $X$, $Gamma_A$ is isomorphic to the centraliser of $H_A$ in $G$.
My attempt to prove it:
Now if the bundle was trivial, i.e. $P=X times G$, we could take $g in G$ in the centraliser of $H_A$ and get a unique element $u_g in mathcal{G}$ satisfying $u_g(x,e)=(x,g)$.
If the connection $A$ is the trivial connection then it is easy to check that this satisfies $u_g in Gamma_A$.
In matrix notation we have $u(A)=u^{-1}Au+u^{-1} du$.
Plugging in an $A$-horizontal vector field to this formula shows that any $u in Gamma_A$ needs to be constant on $X times {e}$, i.e. $u=u_g$ for some $g in G$.
Plugging in a vertical vector field shows that $g$ must be in the centraliser of $H_A$ in $G$, because $u^{-1} du$ vanishes on vertical vector fields.
I believe I can make this argument work for non-trivial connections $A$.
But no matter if I can or cannot: all of this assumed that the bundle was trivial.
If the bundle isn't trivial I don't even know how to write down a map from the centraliser of $H_A$ to $Gamma_A$.
differential-geometry principal-bundles gauge-theory
asked Dec 19 '18 at 12:01
user505117user505117
233
$endgroup$
$begingroup$
I think you want $A$ to be a $G$-connection not just any connection.
$endgroup$
– user10354138
Dec 19 '18 at 14:22
$begingroup$
That's right. I copied the wording from the book. In the book it is clear from the context that I am asking for $A$ to be what I'd call a "principal bundle connection", which should be the same as your $G$-connection. But I didn't provide that context here.
$endgroup$
– user505117
Dec 20 '18 at 14:06
add a comment |
$begingroup$
I am asking for a proof of Lemma (4.2.8) of Donaldson, Kronheimer: The Geometry of Four-Manifolds.
Let $P rightarrow X$ be a principal bundle with structure group $G$.
Denote by $mathcal{G}$ the gauge group.
Let $A$ be a connection on $P$ and $H_A subset G$ its holonomy.
Denote by
$$
Gamma_A=
{u in mathcal{G} | u(A)=A }
$$
the isotropy group of $A$ in $mathcal{G}$.
Then the claim of the lemma is:
For any connection $A$ over a connected base $X$, $Gamma_A$ is isomorphic to the centraliser of $H_A$ in $G$.
My attempt to prove it:
Now if the bundle was trivial, i.e. $P=X times G$, we could take $g in G$ in the centraliser of $H_A$ and get a unique element $u_g in mathcal{G}$ satisfying $u_g(x,e)=(x,g)$.
If the connection $A$ is the trivial connection then it is easy to check that this satisfies $u_g in Gamma_A$.
In matrix notation we have $u(A)=u^{-1}Au+u^{-1} du$.
Plugging in an $A$-horizontal vector field to this formula shows that any $u in Gamma_A$ needs to be constant on $X times {e}$, i.e. $u=u_g$ for some $g in G$.
Plugging in a vertical vector field shows that $g$ must be in the centraliser of $H_A$ in $G$, because $u^{-1} du$ vanishes on vertical vector fields.
I believe I can make this argument work for non-trivial connections $A$.
But no matter if I can or cannot: all of this assumed that the bundle was trivial.
If the bundle isn't trivial I don't even know how to write down a map from the centraliser of $H_A$ to $Gamma_A$.
differential-geometry principal-bundles gauge-theory
asked Dec 19 '18 at 12:01
user505117user505117
233
$endgroup$
I am asking for a proof of Lemma (4.2.8) of Donaldson, Kronheimer: The Geometry of Four-Manifolds.
Let $P rightarrow X$ be a principal bundle with structure group $G$.
Denote by $mathcal{G}$ the gauge group.
Let $A$ be a connection on $P$ and $H_A subset G$ its holonomy.
Denote by
$$
Gamma_A=
{u in mathcal{G} | u(A)=A }
$$
the isotropy group of $A$ in $mathcal{G}$.
Then the claim of the lemma is:
For any connection $A$ over a connected base $X$, $Gamma_A$ is isomorphic to the centraliser of $H_A$ in $G$.
My attempt to prove it:
Now if the bundle was trivial, i.e. $P=X times G$, we could take $g in G$ in the centraliser of $H_A$ and get a unique element $u_g in mathcal{G}$ satisfying $u_g(x,e)=(x,g)$.
If the connection $A$ is the trivial connection then it is easy to check that this satisfies $u_g in Gamma_A$.
In matrix notation we have $u(A)=u^{-1}Au+u^{-1} du$.
Plugging in an $A$-horizontal vector field to this formula shows that any $u in Gamma_A$ needs to be constant on $X times {e}$, i.e. $u=u_g$ for some $g in G$.
Plugging in a vertical vector field shows that $g$ must be in the centraliser of $H_A$ in $G$, because $u^{-1} du$ vanishes on vertical vector fields.
I believe I can make this argument work for non-trivial connections $A$.
But no matter if I can or cannot: all of this assumed that the bundle was trivial.
If the bundle isn't trivial I don't even know how to write down a map from the centraliser of $H_A$ to $Gamma_A$.
differential-geometry principal-bundles gauge-theory
differential-geometry principal-bundles gauge-theory
asked Dec 19 '18 at 12:01
user505117user505117
233
asked Dec 19 '18 at 12:01
user505117user505117
233
asked Dec 19 '18 at 12:01
user505117user505117
233
asked Dec 19 '18 at 12:01
user505117user505117
233
asked Dec 19 '18 at 12:01
user505117user505117
233
233
$begingroup$
I think you want $A$ to be a $G$-connection not just any connection.
$endgroup$
– user10354138
Dec 19 '18 at 14:22
$begingroup$
That's right. I copied the wording from the book. In the book it is clear from the context that I am asking for $A$ to be what I'd call a "principal bundle connection", which should be the same as your $G$-connection. But I didn't provide that context here.
$endgroup$
– user505117
Dec 20 '18 at 14:06
add a comment |
$begingroup$
I think you want $A$ to be a $G$-connection not just any connection.
$endgroup$
– user10354138
Dec 19 '18 at 14:22
$begingroup$
That's right. I copied the wording from the book. In the book it is clear from the context that I am asking for $A$ to be what I'd call a "principal bundle connection", which should be the same as your $G$-connection. But I didn't provide that context here.
$endgroup$
– user505117
Dec 20 '18 at 14:06
$begingroup$
I think you want $A$ to be a $G$-connection not just any connection.
$endgroup$
– user10354138
Dec 19 '18 at 14:22
$begingroup$
I think you want $A$ to be a $G$-connection not just any connection.
$endgroup$
– user10354138
Dec 19 '18 at 14:22
$begingroup$
That's right. I copied the wording from the book. In the book it is clear from the context that I am asking for $A$ to be what I'd call a "principal bundle connection", which should be the same as your $G$-connection. But I didn't provide that context here.
$endgroup$
– user505117
Dec 20 '18 at 14:06
$begingroup$
That's right. I copied the wording from the book. In the book it is clear from the context that I am asking for $A$ to be what I'd call a "principal bundle connection", which should be the same as your $G$-connection. But I didn't provide that context here.
$endgroup$
– user505117
Dec 20 '18 at 14:06
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Fix some $xin X$ (and assume $X$ is connected).
There is an evaluation map $$operatorname{ev}_xcolonGamma_Atooperatorname{Aut}P_x=G.$$
The crucial property that we will exploit is:
Any path $gamma$ joining $x$ to $x'$ in $X$ gives an isomorphism between the $operatorname{ev}_x(Gamma_A)$ and $operatorname{ev}_{x'}(Gamma_A)$ (conjugation by the parallel transport along $gamma$)
This gives us:
$operatorname{ev}_x$ is injective.- When $gamma$ runs through loops at $x$, we get $operatorname{ev}_x(Gamma_A)subseteq C_G(H_A)$.
Conversely, given any $gin C_G(H_A)$, we can define $uinmathcal{G}$ by:
- $u(x):=g$
- for any $x'neq x$, let $gamma$ be a path joining $x$ to $x'$, and
$$
u(x'):=(Pi_gamma)g(Pi_gamma)^{-1}
$$
where $Pi_gamma$ is the map "parallel transport along $gamma$". The condition $gin C_G(H_A)$ ensures $(Pi_gamma)g(Pi_gamma)^{-1}$ does not depend on the choice of $gamma$. It is easy to check the horizontal subspace at $x$ (and hence at all other $x'$) is preserved by $u$, so $uinGamma_A$.
It is clear the constructions are inverse of each other, so $Gamma_Acong C_G(H_A)$.
answered Dec 19 '18 at 14:58
user10354138user10354138
7,5472925
$endgroup$
$begingroup$
You write $text{Aut} P_x=G$. What is that identification? I don't see a canonical one. You can fix an identification for one $x$ which explains how to go from $Gamma_A$ to $C_G(H_A)$. But for the converse direction you seem to use that you have such an identification for all $x' neq x$.
$endgroup$
– user505117
Dec 20 '18 at 14:27
$begingroup$
We have a (non-canonical) identification $G$ with $operatorname{Aut}P_x$ for every $x$. Yes technically everything should have $G$ replaced by $operatorname{Aut} P_x$ (or $operatorname{Aut}P_{x'}$ as appropriate), including the $H_Asubset G$ and the $G$ in $C_G$, so what we have is $Gamma_Acong C_{operatorname{Aut}P_x}(H_{A,x})$.
$endgroup$
– user10354138
Dec 20 '18 at 14:52
$begingroup$
I understand now and your solution makes perfect sense. Thank you!
$endgroup$
– user505117
Dec 20 '18 at 16:31
add a comment |
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$begingroup$
Fix some $xin X$ (and assume $X$ is connected).
There is an evaluation map $$operatorname{ev}_xcolonGamma_Atooperatorname{Aut}P_x=G.$$
The crucial property that we will exploit is:
Any path $gamma$ joining $x$ to $x'$ in $X$ gives an isomorphism between the $operatorname{ev}_x(Gamma_A)$ and $operatorname{ev}_{x'}(Gamma_A)$ (conjugation by the parallel transport along $gamma$)
This gives us:
$operatorname{ev}_x$ is injective.- When $gamma$ runs through loops at $x$, we get $operatorname{ev}_x(Gamma_A)subseteq C_G(H_A)$.
Conversely, given any $gin C_G(H_A)$, we can define $uinmathcal{G}$ by:
- $u(x):=g$
- for any $x'neq x$, let $gamma$ be a path joining $x$ to $x'$, and
$$
u(x'):=(Pi_gamma)g(Pi_gamma)^{-1}
$$
where $Pi_gamma$ is the map "parallel transport along $gamma$". The condition $gin C_G(H_A)$ ensures $(Pi_gamma)g(Pi_gamma)^{-1}$ does not depend on the choice of $gamma$. It is easy to check the horizontal subspace at $x$ (and hence at all other $x'$) is preserved by $u$, so $uinGamma_A$.
It is clear the constructions are inverse of each other, so $Gamma_Acong C_G(H_A)$.
answered Dec 19 '18 at 14:58
user10354138user10354138
7,5472925
$endgroup$
$begingroup$
You write $text{Aut} P_x=G$. What is that identification? I don't see a canonical one. You can fix an identification for one $x$ which explains how to go from $Gamma_A$ to $C_G(H_A)$. But for the converse direction you seem to use that you have such an identification for all $x' neq x$.
$endgroup$
– user505117
Dec 20 '18 at 14:27
$begingroup$
We have a (non-canonical) identification $G$ with $operatorname{Aut}P_x$ for every $x$. Yes technically everything should have $G$ replaced by $operatorname{Aut} P_x$ (or $operatorname{Aut}P_{x'}$ as appropriate), including the $H_Asubset G$ and the $G$ in $C_G$, so what we have is $Gamma_Acong C_{operatorname{Aut}P_x}(H_{A,x})$.
$endgroup$
– user10354138
Dec 20 '18 at 14:52
$begingroup$
I understand now and your solution makes perfect sense. Thank you!
$endgroup$
– user505117
Dec 20 '18 at 16:31
add a comment |
$begingroup$
Fix some $xin X$ (and assume $X$ is connected).
There is an evaluation map $$operatorname{ev}_xcolonGamma_Atooperatorname{Aut}P_x=G.$$
The crucial property that we will exploit is:
Any path $gamma$ joining $x$ to $x'$ in $X$ gives an isomorphism between the $operatorname{ev}_x(Gamma_A)$ and $operatorname{ev}_{x'}(Gamma_A)$ (conjugation by the parallel transport along $gamma$)
This gives us:
$operatorname{ev}_x$ is injective.- When $gamma$ runs through loops at $x$, we get $operatorname{ev}_x(Gamma_A)subseteq C_G(H_A)$.
Conversely, given any $gin C_G(H_A)$, we can define $uinmathcal{G}$ by:
- $u(x):=g$
- for any $x'neq x$, let $gamma$ be a path joining $x$ to $x'$, and
$$
u(x'):=(Pi_gamma)g(Pi_gamma)^{-1}
$$
where $Pi_gamma$ is the map "parallel transport along $gamma$". The condition $gin C_G(H_A)$ ensures $(Pi_gamma)g(Pi_gamma)^{-1}$ does not depend on the choice of $gamma$. It is easy to check the horizontal subspace at $x$ (and hence at all other $x'$) is preserved by $u$, so $uinGamma_A$.
It is clear the constructions are inverse of each other, so $Gamma_Acong C_G(H_A)$.
answered Dec 19 '18 at 14:58
user10354138user10354138
7,5472925
$endgroup$
$begingroup$
You write $text{Aut} P_x=G$. What is that identification? I don't see a canonical one. You can fix an identification for one $x$ which explains how to go from $Gamma_A$ to $C_G(H_A)$. But for the converse direction you seem to use that you have such an identification for all $x' neq x$.
$endgroup$
– user505117
Dec 20 '18 at 14:27
$begingroup$
We have a (non-canonical) identification $G$ with $operatorname{Aut}P_x$ for every $x$. Yes technically everything should have $G$ replaced by $operatorname{Aut} P_x$ (or $operatorname{Aut}P_{x'}$ as appropriate), including the $H_Asubset G$ and the $G$ in $C_G$, so what we have is $Gamma_Acong C_{operatorname{Aut}P_x}(H_{A,x})$.
$endgroup$
– user10354138
Dec 20 '18 at 14:52
$begingroup$
I understand now and your solution makes perfect sense. Thank you!
$endgroup$
– user505117
Dec 20 '18 at 16:31
add a comment |
$begingroup$
Fix some $xin X$ (and assume $X$ is connected).
There is an evaluation map $$operatorname{ev}_xcolonGamma_Atooperatorname{Aut}P_x=G.$$
The crucial property that we will exploit is:
Any path $gamma$ joining $x$ to $x'$ in $X$ gives an isomorphism between the $operatorname{ev}_x(Gamma_A)$ and $operatorname{ev}_{x'}(Gamma_A)$ (conjugation by the parallel transport along $gamma$)
This gives us:
$operatorname{ev}_x$ is injective.- When $gamma$ runs through loops at $x$, we get $operatorname{ev}_x(Gamma_A)subseteq C_G(H_A)$.
Conversely, given any $gin C_G(H_A)$, we can define $uinmathcal{G}$ by:
- $u(x):=g$
- for any $x'neq x$, let $gamma$ be a path joining $x$ to $x'$, and
$$
u(x'):=(Pi_gamma)g(Pi_gamma)^{-1}
$$
where $Pi_gamma$ is the map "parallel transport along $gamma$". The condition $gin C_G(H_A)$ ensures $(Pi_gamma)g(Pi_gamma)^{-1}$ does not depend on the choice of $gamma$. It is easy to check the horizontal subspace at $x$ (and hence at all other $x'$) is preserved by $u$, so $uinGamma_A$.
It is clear the constructions are inverse of each other, so $Gamma_Acong C_G(H_A)$.
answered Dec 19 '18 at 14:58
user10354138user10354138
7,5472925
$endgroup$
Fix some $xin X$ (and assume $X$ is connected).
There is an evaluation map $$operatorname{ev}_xcolonGamma_Atooperatorname{Aut}P_x=G.$$
The crucial property that we will exploit is:
Any path $gamma$ joining $x$ to $x'$ in $X$ gives an isomorphism between the $operatorname{ev}_x(Gamma_A)$ and $operatorname{ev}_{x'}(Gamma_A)$ (conjugation by the parallel transport along $gamma$)
This gives us:
$operatorname{ev}_x$ is injective.- When $gamma$ runs through loops at $x$, we get $operatorname{ev}_x(Gamma_A)subseteq C_G(H_A)$.
Conversely, given any $gin C_G(H_A)$, we can define $uinmathcal{G}$ by:
- $u(x):=g$
- for any $x'neq x$, let $gamma$ be a path joining $x$ to $x'$, and
$$
u(x'):=(Pi_gamma)g(Pi_gamma)^{-1}
$$
where $Pi_gamma$ is the map "parallel transport along $gamma$". The condition $gin C_G(H_A)$ ensures $(Pi_gamma)g(Pi_gamma)^{-1}$ does not depend on the choice of $gamma$. It is easy to check the horizontal subspace at $x$ (and hence at all other $x'$) is preserved by $u$, so $uinGamma_A$.
It is clear the constructions are inverse of each other, so $Gamma_Acong C_G(H_A)$.
answered Dec 19 '18 at 14:58
user10354138user10354138
7,5472925
answered Dec 19 '18 at 14:58
user10354138user10354138
7,5472925
answered Dec 19 '18 at 14:58
user10354138user10354138
7,5472925
answered Dec 19 '18 at 14:58
user10354138user10354138
7,5472925
7,5472925
$begingroup$
You write $text{Aut} P_x=G$. What is that identification? I don't see a canonical one. You can fix an identification for one $x$ which explains how to go from $Gamma_A$ to $C_G(H_A)$. But for the converse direction you seem to use that you have such an identification for all $x' neq x$.
$endgroup$
– user505117
Dec 20 '18 at 14:27
$begingroup$
We have a (non-canonical) identification $G$ with $operatorname{Aut}P_x$ for every $x$. Yes technically everything should have $G$ replaced by $operatorname{Aut} P_x$ (or $operatorname{Aut}P_{x'}$ as appropriate), including the $H_Asubset G$ and the $G$ in $C_G$, so what we have is $Gamma_Acong C_{operatorname{Aut}P_x}(H_{A,x})$.
$endgroup$
– user10354138
Dec 20 '18 at 14:52
$begingroup$
I understand now and your solution makes perfect sense. Thank you!
$endgroup$
– user505117
Dec 20 '18 at 16:31
add a comment |
$begingroup$
You write $text{Aut} P_x=G$. What is that identification? I don't see a canonical one. You can fix an identification for one $x$ which explains how to go from $Gamma_A$ to $C_G(H_A)$. But for the converse direction you seem to use that you have such an identification for all $x' neq x$.
$endgroup$
– user505117
Dec 20 '18 at 14:27
$begingroup$
We have a (non-canonical) identification $G$ with $operatorname{Aut}P_x$ for every $x$. Yes technically everything should have $G$ replaced by $operatorname{Aut} P_x$ (or $operatorname{Aut}P_{x'}$ as appropriate), including the $H_Asubset G$ and the $G$ in $C_G$, so what we have is $Gamma_Acong C_{operatorname{Aut}P_x}(H_{A,x})$.
$endgroup$
– user10354138
Dec 20 '18 at 14:52
$begingroup$
I understand now and your solution makes perfect sense. Thank you!
$endgroup$
– user505117
Dec 20 '18 at 16:31
$begingroup$
You write $text{Aut} P_x=G$. What is that identification? I don't see a canonical one. You can fix an identification for one $x$ which explains how to go from $Gamma_A$ to $C_G(H_A)$. But for the converse direction you seem to use that you have such an identification for all $x' neq x$.
$endgroup$
– user505117
Dec 20 '18 at 14:27
$begingroup$
You write $text{Aut} P_x=G$. What is that identification? I don't see a canonical one. You can fix an identification for one $x$ which explains how to go from $Gamma_A$ to $C_G(H_A)$. But for the converse direction you seem to use that you have such an identification for all $x' neq x$.
$endgroup$
– user505117
Dec 20 '18 at 14:27
$begingroup$
We have a (non-canonical) identification $G$ with $operatorname{Aut}P_x$ for every $x$. Yes technically everything should have $G$ replaced by $operatorname{Aut} P_x$ (or $operatorname{Aut}P_{x'}$ as appropriate), including the $H_Asubset G$ and the $G$ in $C_G$, so what we have is $Gamma_Acong C_{operatorname{Aut}P_x}(H_{A,x})$.
$endgroup$
– user10354138
Dec 20 '18 at 14:52
$begingroup$
We have a (non-canonical) identification $G$ with $operatorname{Aut}P_x$ for every $x$. Yes technically everything should have $G$ replaced by $operatorname{Aut} P_x$ (or $operatorname{Aut}P_{x'}$ as appropriate), including the $H_Asubset G$ and the $G$ in $C_G$, so what we have is $Gamma_Acong C_{operatorname{Aut}P_x}(H_{A,x})$.
$endgroup$
– user10354138
Dec 20 '18 at 14:52
$begingroup$
I understand now and your solution makes perfect sense. Thank you!
$endgroup$
– user505117
Dec 20 '18 at 16:31
$begingroup$
I understand now and your solution makes perfect sense. Thank you!
$endgroup$
– user505117
Dec 20 '18 at 16:31
add a comment |
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$begingroup$
I think you want $A$ to be a $G$-connection not just any connection.
$endgroup$
– user10354138
Dec 19 '18 at 14:22
$begingroup$
That's right. I copied the wording from the book. In the book it is clear from the context that I am asking for $A$ to be what I'd call a "principal bundle connection", which should be the same as your $G$-connection. But I didn't provide that context here.
$endgroup$
– user505117
Dec 20 '18 at 14:06