Mixing
Does this condition on the curvature implies existence of a parallel section?
$begingroup$
Let $E$ be a smooth vector bundle over a manifold $M$ ($dim M>1$), equipped with a metric. Let $nabla$ be a metric connection on $E$.
Suppose there exist locally a non-zero section $sigma in Gamma(E)$ which lies in $ker R(X,Y)$ for all $X,Y in Gamma(TM)$. Does $nabla$ admit a parallel section (locally)?
Note that even if $| sigma|=1$, it is not necessarily true that $sigma$ parallel. (e.g. if $nabla$ is flat).
(We have to normalize: the point is that if $sigma in ker R(X,Y)$ so is $fsigma$ for any function $f$. A parallel section has a constant norm though.)
Clearly, this is a necessary condition:
If $sigma$ is parallel, then
$R(X,Y)sigma=d_{nabla}^2sigma(X,Y)=0$
differential-geometry riemannian-geometry vector-bundles curvature connections
asked May 30 '18 at 5:31
Asaf ShacharAsaf Shachar
5,79031145
$endgroup$
add a comment |
$begingroup$
Let $E$ be a smooth vector bundle over a manifold $M$ ($dim M>1$), equipped with a metric. Let $nabla$ be a metric connection on $E$.
Suppose there exist locally a non-zero section $sigma in Gamma(E)$ which lies in $ker R(X,Y)$ for all $X,Y in Gamma(TM)$. Does $nabla$ admit a parallel section (locally)?
Note that even if $| sigma|=1$, it is not necessarily true that $sigma$ parallel. (e.g. if $nabla$ is flat).
(We have to normalize: the point is that if $sigma in ker R(X,Y)$ so is $fsigma$ for any function $f$. A parallel section has a constant norm though.)
Clearly, this is a necessary condition:
If $sigma$ is parallel, then
$R(X,Y)sigma=d_{nabla}^2sigma(X,Y)=0$
differential-geometry riemannian-geometry vector-bundles curvature connections
asked May 30 '18 at 5:31
Asaf ShacharAsaf Shachar
5,79031145
$endgroup$
1
$begingroup$
Notice that (Ambrose-Singer theorem, see e.g. Kobayashi-Nomizu's book) endomorphisms of the form $Zmapsto R(X,Y)Z$ span the Lie algebra of the holonomy group of the connection. Hence, the reduced holonomy group $H_p$ of your connection has a fixed nonzero vector $vin E_p$. Now, parallel-translate this vector in a small neighborhood of $p$. The result is the required parallel section.
$endgroup$
– Moishe Kohan
Jan 29 at 23:19
$begingroup$
Thank you. By the reduced holonomy group do you refer to restriction to contractible loops? (sometimes this is called the reduced holonomy group $text{Hol}^0$, right?). I don't understand why the exponential map should be surjective on $text{Hol}^0$: If I understood correctly, your argument is the following: Let $G le text{GL}(n)$ be a Lie subgroup and $mathfrak{g} le M_n$ be its Lie algebra. We assume that $v in M_n$ satisfies $Av=0$ for every $A in mathfrak{g} $. Then $e^{tA} v=v$ for every $A in mathfrak{g} $...
$endgroup$
– Asaf Shachar
Feb 11 at 9:11
$begingroup$
Now, we need to know that the exponential map of $exp:mathfrak{g} to G$ is surjective, in order to conclude that $Gv=v$. In our case, $text{Hol}$ is contained in $O(n)$ since our connection is metric, but I don't see why this implies that $exp$ is surjective, as $text{Hol}^0$ might not be closed inside $text{Hol}$, right? (I know surjectivity of $exp$ is automatic only when the group is compact...)
$endgroup$
– Asaf Shachar
Feb 11 at 9:11
add a comment |
$begingroup$
Let $E$ be a smooth vector bundle over a manifold $M$ ($dim M>1$), equipped with a metric. Let $nabla$ be a metric connection on $E$.
Suppose there exist locally a non-zero section $sigma in Gamma(E)$ which lies in $ker R(X,Y)$ for all $X,Y in Gamma(TM)$. Does $nabla$ admit a parallel section (locally)?
Note that even if $| sigma|=1$, it is not necessarily true that $sigma$ parallel. (e.g. if $nabla$ is flat).
(We have to normalize: the point is that if $sigma in ker R(X,Y)$ so is $fsigma$ for any function $f$. A parallel section has a constant norm though.)
Clearly, this is a necessary condition:
If $sigma$ is parallel, then
$R(X,Y)sigma=d_{nabla}^2sigma(X,Y)=0$
differential-geometry riemannian-geometry vector-bundles curvature connections
asked May 30 '18 at 5:31
Asaf ShacharAsaf Shachar
5,79031145
$endgroup$
Let $E$ be a smooth vector bundle over a manifold $M$ ($dim M>1$), equipped with a metric. Let $nabla$ be a metric connection on $E$.
Suppose there exist locally a non-zero section $sigma in Gamma(E)$ which lies in $ker R(X,Y)$ for all $X,Y in Gamma(TM)$. Does $nabla$ admit a parallel section (locally)?
Note that even if $| sigma|=1$, it is not necessarily true that $sigma$ parallel. (e.g. if $nabla$ is flat).
(We have to normalize: the point is that if $sigma in ker R(X,Y)$ so is $fsigma$ for any function $f$. A parallel section has a constant norm though.)
Clearly, this is a necessary condition:
If $sigma$ is parallel, then
$R(X,Y)sigma=d_{nabla}^2sigma(X,Y)=0$
differential-geometry riemannian-geometry vector-bundles curvature connections
differential-geometry riemannian-geometry vector-bundles curvature connections
asked May 30 '18 at 5:31
Asaf ShacharAsaf Shachar
5,79031145
asked May 30 '18 at 5:31
Asaf ShacharAsaf Shachar
5,79031145
edited Jul 6 '18 at 14:57
Asaf Shachar
asked May 30 '18 at 5:31
Asaf ShacharAsaf Shachar
5,79031145
asked May 30 '18 at 5:31
Asaf ShacharAsaf Shachar
5,79031145
asked May 30 '18 at 5:31
Asaf ShacharAsaf Shachar
5,79031145
5,79031145
1
$begingroup$
Notice that (Ambrose-Singer theorem, see e.g. Kobayashi-Nomizu's book) endomorphisms of the form $Zmapsto R(X,Y)Z$ span the Lie algebra of the holonomy group of the connection. Hence, the reduced holonomy group $H_p$ of your connection has a fixed nonzero vector $vin E_p$. Now, parallel-translate this vector in a small neighborhood of $p$. The result is the required parallel section.
$endgroup$
– Moishe Kohan
Jan 29 at 23:19
$begingroup$
Thank you. By the reduced holonomy group do you refer to restriction to contractible loops? (sometimes this is called the reduced holonomy group $text{Hol}^0$, right?). I don't understand why the exponential map should be surjective on $text{Hol}^0$: If I understood correctly, your argument is the following: Let $G le text{GL}(n)$ be a Lie subgroup and $mathfrak{g} le M_n$ be its Lie algebra. We assume that $v in M_n$ satisfies $Av=0$ for every $A in mathfrak{g} $. Then $e^{tA} v=v$ for every $A in mathfrak{g} $...
$endgroup$
– Asaf Shachar
Feb 11 at 9:11
$begingroup$
Now, we need to know that the exponential map of $exp:mathfrak{g} to G$ is surjective, in order to conclude that $Gv=v$. In our case, $text{Hol}$ is contained in $O(n)$ since our connection is metric, but I don't see why this implies that $exp$ is surjective, as $text{Hol}^0$ might not be closed inside $text{Hol}$, right? (I know surjectivity of $exp$ is automatic only when the group is compact...)
$endgroup$
– Asaf Shachar
Feb 11 at 9:11
add a comment |
1
$begingroup$
Notice that (Ambrose-Singer theorem, see e.g. Kobayashi-Nomizu's book) endomorphisms of the form $Zmapsto R(X,Y)Z$ span the Lie algebra of the holonomy group of the connection. Hence, the reduced holonomy group $H_p$ of your connection has a fixed nonzero vector $vin E_p$. Now, parallel-translate this vector in a small neighborhood of $p$. The result is the required parallel section.
$endgroup$
– Moishe Kohan
Jan 29 at 23:19
$begingroup$
Thank you. By the reduced holonomy group do you refer to restriction to contractible loops? (sometimes this is called the reduced holonomy group $text{Hol}^0$, right?). I don't understand why the exponential map should be surjective on $text{Hol}^0$: If I understood correctly, your argument is the following: Let $G le text{GL}(n)$ be a Lie subgroup and $mathfrak{g} le M_n$ be its Lie algebra. We assume that $v in M_n$ satisfies $Av=0$ for every $A in mathfrak{g} $. Then $e^{tA} v=v$ for every $A in mathfrak{g} $...
$endgroup$
– Asaf Shachar
Feb 11 at 9:11
$begingroup$
Now, we need to know that the exponential map of $exp:mathfrak{g} to G$ is surjective, in order to conclude that $Gv=v$. In our case, $text{Hol}$ is contained in $O(n)$ since our connection is metric, but I don't see why this implies that $exp$ is surjective, as $text{Hol}^0$ might not be closed inside $text{Hol}$, right? (I know surjectivity of $exp$ is automatic only when the group is compact...)
$endgroup$
– Asaf Shachar
Feb 11 at 9:11
1
1
$begingroup$
Notice that (Ambrose-Singer theorem, see e.g. Kobayashi-Nomizu's book) endomorphisms of the form $Zmapsto R(X,Y)Z$ span the Lie algebra of the holonomy group of the connection. Hence, the reduced holonomy group $H_p$ of your connection has a fixed nonzero vector $vin E_p$. Now, parallel-translate this vector in a small neighborhood of $p$. The result is the required parallel section.
$endgroup$
– Moishe Kohan
Jan 29 at 23:19
$begingroup$
Notice that (Ambrose-Singer theorem, see e.g. Kobayashi-Nomizu's book) endomorphisms of the form $Zmapsto R(X,Y)Z$ span the Lie algebra of the holonomy group of the connection. Hence, the reduced holonomy group $H_p$ of your connection has a fixed nonzero vector $vin E_p$. Now, parallel-translate this vector in a small neighborhood of $p$. The result is the required parallel section.
$endgroup$
– Moishe Kohan
Jan 29 at 23:19
$begingroup$
Thank you. By the reduced holonomy group do you refer to restriction to contractible loops? (sometimes this is called the reduced holonomy group $text{Hol}^0$, right?). I don't understand why the exponential map should be surjective on $text{Hol}^0$: If I understood correctly, your argument is the following: Let $G le text{GL}(n)$ be a Lie subgroup and $mathfrak{g} le M_n$ be its Lie algebra. We assume that $v in M_n$ satisfies $Av=0$ for every $A in mathfrak{g} $. Then $e^{tA} v=v$ for every $A in mathfrak{g} $...
$endgroup$
– Asaf Shachar
Feb 11 at 9:11
$begingroup$
Thank you. By the reduced holonomy group do you refer to restriction to contractible loops? (sometimes this is called the reduced holonomy group $text{Hol}^0$, right?). I don't understand why the exponential map should be surjective on $text{Hol}^0$: If I understood correctly, your argument is the following: Let $G le text{GL}(n)$ be a Lie subgroup and $mathfrak{g} le M_n$ be its Lie algebra. We assume that $v in M_n$ satisfies $Av=0$ for every $A in mathfrak{g} $. Then $e^{tA} v=v$ for every $A in mathfrak{g} $...
$endgroup$
– Asaf Shachar
Feb 11 at 9:11
$begingroup$
Now, we need to know that the exponential map of $exp:mathfrak{g} to G$ is surjective, in order to conclude that $Gv=v$. In our case, $text{Hol}$ is contained in $O(n)$ since our connection is metric, but I don't see why this implies that $exp$ is surjective, as $text{Hol}^0$ might not be closed inside $text{Hol}$, right? (I know surjectivity of $exp$ is automatic only when the group is compact...)
$endgroup$
– Asaf Shachar
Feb 11 at 9:11
$begingroup$
Now, we need to know that the exponential map of $exp:mathfrak{g} to G$ is surjective, in order to conclude that $Gv=v$. In our case, $text{Hol}$ is contained in $O(n)$ since our connection is metric, but I don't see why this implies that $exp$ is surjective, as $text{Hol}^0$ might not be closed inside $text{Hol}$, right? (I know surjectivity of $exp$ is automatic only when the group is compact...)
$endgroup$
– Asaf Shachar
Feb 11 at 9:11
add a comment |
1 Answer
1
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$begingroup$
Consider the reduced holonomy group $Hol^0_p=G$ of $nabla$ ($p$ is a point in $M$), i.e. the subgroup of the full holonomy group whose elements are defined via parallel transport along null-homotopic loops based at $p$. This is a connected subgroup of the holonomy group. According to Ambrose-Singer theorem (see e.g. Kobayashi-Nomizu's book) endomorphisms of the fiber $E_p$ of the form $Zmapsto R(X,Y)Z$ span the Lie algebra ${mathfrak g}$ of $G$. Let $Z$ be a nonzero local section of $Eto M$ near $p$ such that $R(X,Y)Z=0$ for all local sections $X, Y$ of $Eto M$ near $p$. Define the (nonzero) vector $vin E_p$, $v=Z(p)$. Exponentiating the Lie algebra, we obtain an open neighborhood $U$ of $ein G$ whose elements fix $v$ (since, by the assumption, $R(X,Y)Z=0$ for all local sections $X, Y$). Let $H$ denote the subgroup of $G$ generated by $U$. Then $H$ is an open subgroup of $G$ which fixes $v$. However, it is also a closed subgroup of $G$ (every open subgroup of a topological group is also closed, this is a nice exercise). Since $G$ is connected, $H=G$. Thus, the entire group $G$ fixes $v$. Now, parallel-translate $v$ in a small (simply-connected) neighborhood $V$ of $𝑝$: Since $v$ is fixed by $G$, the parallel transport is independent of the path in $V$. The result is a nonzero local parallel section of $Eto M$.
answered Feb 11 at 16:17
Moishe KohanMoishe Kohan
48.4k344110
$endgroup$
$begingroup$
Thank you, I understand now. Just to be sure then: By the same argument, can we conclude that the dimension of the space of local parallel sections of $(E,nabla)$ equals to the dimension of the kernel of the Lie algebra of $Hol^0$? i.e. there exist $k$ independent local parallel sections $sigma_i$ of $E$ if and only if there exist $k$ independent local sections $sigma_i$ satisfying $R(cdot,cdot)sigma_i=0$. Am I right?
$endgroup$
– Asaf Shachar
Feb 11 at 16:26
$begingroup$
@AsafShachar: Right.
$endgroup$
– Moishe Kohan
Feb 11 at 16:28
add a comment |
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$begingroup$
Consider the reduced holonomy group $Hol^0_p=G$ of $nabla$ ($p$ is a point in $M$), i.e. the subgroup of the full holonomy group whose elements are defined via parallel transport along null-homotopic loops based at $p$. This is a connected subgroup of the holonomy group. According to Ambrose-Singer theorem (see e.g. Kobayashi-Nomizu's book) endomorphisms of the fiber $E_p$ of the form $Zmapsto R(X,Y)Z$ span the Lie algebra ${mathfrak g}$ of $G$. Let $Z$ be a nonzero local section of $Eto M$ near $p$ such that $R(X,Y)Z=0$ for all local sections $X, Y$ of $Eto M$ near $p$. Define the (nonzero) vector $vin E_p$, $v=Z(p)$. Exponentiating the Lie algebra, we obtain an open neighborhood $U$ of $ein G$ whose elements fix $v$ (since, by the assumption, $R(X,Y)Z=0$ for all local sections $X, Y$). Let $H$ denote the subgroup of $G$ generated by $U$. Then $H$ is an open subgroup of $G$ which fixes $v$. However, it is also a closed subgroup of $G$ (every open subgroup of a topological group is also closed, this is a nice exercise). Since $G$ is connected, $H=G$. Thus, the entire group $G$ fixes $v$. Now, parallel-translate $v$ in a small (simply-connected) neighborhood $V$ of $𝑝$: Since $v$ is fixed by $G$, the parallel transport is independent of the path in $V$. The result is a nonzero local parallel section of $Eto M$.
answered Feb 11 at 16:17
Moishe KohanMoishe Kohan
48.4k344110
$endgroup$
$begingroup$
Thank you, I understand now. Just to be sure then: By the same argument, can we conclude that the dimension of the space of local parallel sections of $(E,nabla)$ equals to the dimension of the kernel of the Lie algebra of $Hol^0$? i.e. there exist $k$ independent local parallel sections $sigma_i$ of $E$ if and only if there exist $k$ independent local sections $sigma_i$ satisfying $R(cdot,cdot)sigma_i=0$. Am I right?
$endgroup$
– Asaf Shachar
Feb 11 at 16:26
$begingroup$
@AsafShachar: Right.
$endgroup$
– Moishe Kohan
Feb 11 at 16:28
add a comment |
$begingroup$
Consider the reduced holonomy group $Hol^0_p=G$ of $nabla$ ($p$ is a point in $M$), i.e. the subgroup of the full holonomy group whose elements are defined via parallel transport along null-homotopic loops based at $p$. This is a connected subgroup of the holonomy group. According to Ambrose-Singer theorem (see e.g. Kobayashi-Nomizu's book) endomorphisms of the fiber $E_p$ of the form $Zmapsto R(X,Y)Z$ span the Lie algebra ${mathfrak g}$ of $G$. Let $Z$ be a nonzero local section of $Eto M$ near $p$ such that $R(X,Y)Z=0$ for all local sections $X, Y$ of $Eto M$ near $p$. Define the (nonzero) vector $vin E_p$, $v=Z(p)$. Exponentiating the Lie algebra, we obtain an open neighborhood $U$ of $ein G$ whose elements fix $v$ (since, by the assumption, $R(X,Y)Z=0$ for all local sections $X, Y$). Let $H$ denote the subgroup of $G$ generated by $U$. Then $H$ is an open subgroup of $G$ which fixes $v$. However, it is also a closed subgroup of $G$ (every open subgroup of a topological group is also closed, this is a nice exercise). Since $G$ is connected, $H=G$. Thus, the entire group $G$ fixes $v$. Now, parallel-translate $v$ in a small (simply-connected) neighborhood $V$ of $𝑝$: Since $v$ is fixed by $G$, the parallel transport is independent of the path in $V$. The result is a nonzero local parallel section of $Eto M$.
answered Feb 11 at 16:17
Moishe KohanMoishe Kohan
48.4k344110
$endgroup$
$begingroup$
Thank you, I understand now. Just to be sure then: By the same argument, can we conclude that the dimension of the space of local parallel sections of $(E,nabla)$ equals to the dimension of the kernel of the Lie algebra of $Hol^0$? i.e. there exist $k$ independent local parallel sections $sigma_i$ of $E$ if and only if there exist $k$ independent local sections $sigma_i$ satisfying $R(cdot,cdot)sigma_i=0$. Am I right?
$endgroup$
– Asaf Shachar
Feb 11 at 16:26
$begingroup$
@AsafShachar: Right.
$endgroup$
– Moishe Kohan
Feb 11 at 16:28
add a comment |
$begingroup$
Consider the reduced holonomy group $Hol^0_p=G$ of $nabla$ ($p$ is a point in $M$), i.e. the subgroup of the full holonomy group whose elements are defined via parallel transport along null-homotopic loops based at $p$. This is a connected subgroup of the holonomy group. According to Ambrose-Singer theorem (see e.g. Kobayashi-Nomizu's book) endomorphisms of the fiber $E_p$ of the form $Zmapsto R(X,Y)Z$ span the Lie algebra ${mathfrak g}$ of $G$. Let $Z$ be a nonzero local section of $Eto M$ near $p$ such that $R(X,Y)Z=0$ for all local sections $X, Y$ of $Eto M$ near $p$. Define the (nonzero) vector $vin E_p$, $v=Z(p)$. Exponentiating the Lie algebra, we obtain an open neighborhood $U$ of $ein G$ whose elements fix $v$ (since, by the assumption, $R(X,Y)Z=0$ for all local sections $X, Y$). Let $H$ denote the subgroup of $G$ generated by $U$. Then $H$ is an open subgroup of $G$ which fixes $v$. However, it is also a closed subgroup of $G$ (every open subgroup of a topological group is also closed, this is a nice exercise). Since $G$ is connected, $H=G$. Thus, the entire group $G$ fixes $v$. Now, parallel-translate $v$ in a small (simply-connected) neighborhood $V$ of $𝑝$: Since $v$ is fixed by $G$, the parallel transport is independent of the path in $V$. The result is a nonzero local parallel section of $Eto M$.
answered Feb 11 at 16:17
Moishe KohanMoishe Kohan
48.4k344110
$endgroup$
Consider the reduced holonomy group $Hol^0_p=G$ of $nabla$ ($p$ is a point in $M$), i.e. the subgroup of the full holonomy group whose elements are defined via parallel transport along null-homotopic loops based at $p$. This is a connected subgroup of the holonomy group. According to Ambrose-Singer theorem (see e.g. Kobayashi-Nomizu's book) endomorphisms of the fiber $E_p$ of the form $Zmapsto R(X,Y)Z$ span the Lie algebra ${mathfrak g}$ of $G$. Let $Z$ be a nonzero local section of $Eto M$ near $p$ such that $R(X,Y)Z=0$ for all local sections $X, Y$ of $Eto M$ near $p$. Define the (nonzero) vector $vin E_p$, $v=Z(p)$. Exponentiating the Lie algebra, we obtain an open neighborhood $U$ of $ein G$ whose elements fix $v$ (since, by the assumption, $R(X,Y)Z=0$ for all local sections $X, Y$). Let $H$ denote the subgroup of $G$ generated by $U$. Then $H$ is an open subgroup of $G$ which fixes $v$. However, it is also a closed subgroup of $G$ (every open subgroup of a topological group is also closed, this is a nice exercise). Since $G$ is connected, $H=G$. Thus, the entire group $G$ fixes $v$. Now, parallel-translate $v$ in a small (simply-connected) neighborhood $V$ of $𝑝$: Since $v$ is fixed by $G$, the parallel transport is independent of the path in $V$. The result is a nonzero local parallel section of $Eto M$.
answered Feb 11 at 16:17
Moishe KohanMoishe Kohan
48.4k344110
answered Feb 11 at 16:17
Moishe KohanMoishe Kohan
48.4k344110
answered Feb 11 at 16:17
Moishe KohanMoishe Kohan
48.4k344110
answered Feb 11 at 16:17
Moishe KohanMoishe Kohan
48.4k344110
48.4k344110
$begingroup$
Thank you, I understand now. Just to be sure then: By the same argument, can we conclude that the dimension of the space of local parallel sections of $(E,nabla)$ equals to the dimension of the kernel of the Lie algebra of $Hol^0$? i.e. there exist $k$ independent local parallel sections $sigma_i$ of $E$ if and only if there exist $k$ independent local sections $sigma_i$ satisfying $R(cdot,cdot)sigma_i=0$. Am I right?
$endgroup$
– Asaf Shachar
Feb 11 at 16:26
$begingroup$
@AsafShachar: Right.
$endgroup$
– Moishe Kohan
Feb 11 at 16:28
add a comment |
$begingroup$
Thank you, I understand now. Just to be sure then: By the same argument, can we conclude that the dimension of the space of local parallel sections of $(E,nabla)$ equals to the dimension of the kernel of the Lie algebra of $Hol^0$? i.e. there exist $k$ independent local parallel sections $sigma_i$ of $E$ if and only if there exist $k$ independent local sections $sigma_i$ satisfying $R(cdot,cdot)sigma_i=0$. Am I right?
$endgroup$
– Asaf Shachar
Feb 11 at 16:26
$begingroup$
@AsafShachar: Right.
$endgroup$
– Moishe Kohan
Feb 11 at 16:28
$begingroup$
Thank you, I understand now. Just to be sure then: By the same argument, can we conclude that the dimension of the space of local parallel sections of $(E,nabla)$ equals to the dimension of the kernel of the Lie algebra of $Hol^0$? i.e. there exist $k$ independent local parallel sections $sigma_i$ of $E$ if and only if there exist $k$ independent local sections $sigma_i$ satisfying $R(cdot,cdot)sigma_i=0$. Am I right?
$endgroup$
– Asaf Shachar
Feb 11 at 16:26
$begingroup$
Thank you, I understand now. Just to be sure then: By the same argument, can we conclude that the dimension of the space of local parallel sections of $(E,nabla)$ equals to the dimension of the kernel of the Lie algebra of $Hol^0$? i.e. there exist $k$ independent local parallel sections $sigma_i$ of $E$ if and only if there exist $k$ independent local sections $sigma_i$ satisfying $R(cdot,cdot)sigma_i=0$. Am I right?
$endgroup$
– Asaf Shachar
Feb 11 at 16:26
$begingroup$
@AsafShachar: Right.
$endgroup$
– Moishe Kohan
Feb 11 at 16:28
$begingroup$
@AsafShachar: Right.
$endgroup$
– Moishe Kohan
Feb 11 at 16:28
add a comment |
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1
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Notice that (Ambrose-Singer theorem, see e.g. Kobayashi-Nomizu's book) endomorphisms of the form $Zmapsto R(X,Y)Z$ span the Lie algebra of the holonomy group of the connection. Hence, the reduced holonomy group $H_p$ of your connection has a fixed nonzero vector $vin E_p$. Now, parallel-translate this vector in a small neighborhood of $p$. The result is the required parallel section.
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– Moishe Kohan
Jan 29 at 23:19
$begingroup$
Thank you. By the reduced holonomy group do you refer to restriction to contractible loops? (sometimes this is called the reduced holonomy group $text{Hol}^0$, right?). I don't understand why the exponential map should be surjective on $text{Hol}^0$: If I understood correctly, your argument is the following: Let $G le text{GL}(n)$ be a Lie subgroup and $mathfrak{g} le M_n$ be its Lie algebra. We assume that $v in M_n$ satisfies $Av=0$ for every $A in mathfrak{g} $. Then $e^{tA} v=v$ for every $A in mathfrak{g} $...
$endgroup$
– Asaf Shachar
Feb 11 at 9:11
$begingroup$
Now, we need to know that the exponential map of $exp:mathfrak{g} to G$ is surjective, in order to conclude that $Gv=v$. In our case, $text{Hol}$ is contained in $O(n)$ since our connection is metric, but I don't see why this implies that $exp$ is surjective, as $text{Hol}^0$ might not be closed inside $text{Hol}$, right? (I know surjectivity of $exp$ is automatic only when the group is compact...)
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– Asaf Shachar
Feb 11 at 9:11