Calibrations vs. Riemannian holonomyHolonomy group of quotient manifoldBerger's theorem on holonomyAbout two notions of holonomyHolonomy of Lie groupsIntegrability of the holonomy invariant distributionRecovering Curvature Endomorphism out of HolonomyWhat is the holonomy of homogenous spaces?Translation HolonomyWhat is the Riemannian holonomy of a quotient of Lie groups?Riemannian holonomy of a covering
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Calibrations vs. Riemannian holonomy
Holonomy group of quotient manifoldBerger's theorem on holonomyAbout two notions of holonomyHolonomy of Lie groupsIntegrability of the holonomy invariant distributionRecovering Curvature Endomorphism out of HolonomyWhat is the holonomy of homogenous spaces?Translation HolonomyWhat is the Riemannian holonomy of a quotient of Lie groups?Riemannian holonomy of a covering
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
I've began to study the relationship between calibrations and holonomy, mainly through D.D. Joyce's Riemannian Holonomy Groups and Calibrated Geometry and partly through internet material.
Pretty much everyone explains this relationship by the holonomy principle: if $H=textHol_x$ and $varphi_0$ is an $H$-invariant $k$-form in $T_pM$, then there is a parallel $k$-form $varphi$ in $M$ with $nablavarphi=0$. In particular, this means $dvarphi=0$. Rescaling $varphi_0$ if necessary, we get that $varphi$ is a calibration.
So far, so good.
After this, people start saying something about special holonomy and invariably mention Berger's classification.
1) What does special mean in this context? I thought this was an informal adjective used by Joyce, but apparently everyone uses it and I haven't found a definition.
2) I understand Berger's list is interesting since it deals with irreducible manifolds. But why don't they mention symmetric manifolds, which are not on the list, like $mathbbR^n$, $mathbbS^n,mathbbRH^n$, compact Lie groups etc. They seem pretty interesting (and numerous) to me, so why not consider them?
riemannian-geometry holonomy calibrations
asked 9 hours ago
rmdmc89rmdmc89
2,5371924
$endgroup$
add a comment |
$begingroup$
I've began to study the relationship between calibrations and holonomy, mainly through D.D. Joyce's Riemannian Holonomy Groups and Calibrated Geometry and partly through internet material.
Pretty much everyone explains this relationship by the holonomy principle: if $H=textHol_x$ and $varphi_0$ is an $H$-invariant $k$-form in $T_pM$, then there is a parallel $k$-form $varphi$ in $M$ with $nablavarphi=0$. In particular, this means $dvarphi=0$. Rescaling $varphi_0$ if necessary, we get that $varphi$ is a calibration.
So far, so good.
After this, people start saying something about special holonomy and invariably mention Berger's classification.
1) What does special mean in this context? I thought this was an informal adjective used by Joyce, but apparently everyone uses it and I haven't found a definition.
2) I understand Berger's list is interesting since it deals with irreducible manifolds. But why don't they mention symmetric manifolds, which are not on the list, like $mathbbR^n$, $mathbbS^n,mathbbRH^n$, compact Lie groups etc. They seem pretty interesting (and numerous) to me, so why not consider them?
riemannian-geometry holonomy calibrations
asked 9 hours ago
rmdmc89rmdmc89
2,5371924
$endgroup$
add a comment |
$begingroup$
I've began to study the relationship between calibrations and holonomy, mainly through D.D. Joyce's Riemannian Holonomy Groups and Calibrated Geometry and partly through internet material.
Pretty much everyone explains this relationship by the holonomy principle: if $H=textHol_x$ and $varphi_0$ is an $H$-invariant $k$-form in $T_pM$, then there is a parallel $k$-form $varphi$ in $M$ with $nablavarphi=0$. In particular, this means $dvarphi=0$. Rescaling $varphi_0$ if necessary, we get that $varphi$ is a calibration.
So far, so good.
After this, people start saying something about special holonomy and invariably mention Berger's classification.
1) What does special mean in this context? I thought this was an informal adjective used by Joyce, but apparently everyone uses it and I haven't found a definition.
2) I understand Berger's list is interesting since it deals with irreducible manifolds. But why don't they mention symmetric manifolds, which are not on the list, like $mathbbR^n$, $mathbbS^n,mathbbRH^n$, compact Lie groups etc. They seem pretty interesting (and numerous) to me, so why not consider them?
riemannian-geometry holonomy calibrations
asked 9 hours ago
rmdmc89rmdmc89
2,5371924
$endgroup$
I've began to study the relationship between calibrations and holonomy, mainly through D.D. Joyce's Riemannian Holonomy Groups and Calibrated Geometry and partly through internet material.
Pretty much everyone explains this relationship by the holonomy principle: if $H=textHol_x$ and $varphi_0$ is an $H$-invariant $k$-form in $T_pM$, then there is a parallel $k$-form $varphi$ in $M$ with $nablavarphi=0$. In particular, this means $dvarphi=0$. Rescaling $varphi_0$ if necessary, we get that $varphi$ is a calibration.
So far, so good.
After this, people start saying something about special holonomy and invariably mention Berger's classification.
1) What does special mean in this context? I thought this was an informal adjective used by Joyce, but apparently everyone uses it and I haven't found a definition.
2) I understand Berger's list is interesting since it deals with irreducible manifolds. But why don't they mention symmetric manifolds, which are not on the list, like $mathbbR^n$, $mathbbS^n,mathbbRH^n$, compact Lie groups etc. They seem pretty interesting (and numerous) to me, so why not consider them?
riemannian-geometry holonomy calibrations
riemannian-geometry holonomy calibrations
asked 9 hours ago
rmdmc89rmdmc89
2,5371924
asked 9 hours ago
rmdmc89rmdmc89
2,5371924
asked 9 hours ago
rmdmc89rmdmc89
2,5371924
asked 9 hours ago
rmdmc89rmdmc89
2,5371924
asked 9 hours ago
rmdmc89rmdmc89
2,5371924
2,5371924
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
(1) A generic metric has (restricted) holonomy group $SO(n)$ (More proper statement: the set of holonomy $SO(n)$ metrics is comeagre in the space of all Riemannian metrics). Hence the adjective special is coined (as in the opposite of "generic") when we can reduce it to smaller subgroups. It definitely predates Joyce (certainly Harvey and Lawson used that in their seminal paper introducing calibrations in the early 1980s).
(2) Elie Cartan proved that for Riemannian symmetric spaces $G/H$, the restricted holonomy group is the identity component of the isotropy group $H$. So this is just a pure algebra problem as to which Lie group is a subgroup of another Lie group (or equivalently which Lie algebra is a subalgebra of another), hence not interesting (as in having little if not no geometry content).
answered 8 hours ago
user10354138user10354138
16.8k21130
$endgroup$
$begingroup$
If I got it right, you're saying that determining $textHol_p$ of a symmetric manifold is essentially an algebraic problem, not a geometric one. But in the context of calibrations, our goal is to find calibrated submanifolds, right? By the method I described, we try that by looking for $textHol_p$-invariant forms.
$endgroup$
– rmdmc89
7 hours ago
$begingroup$
My point is: if we already know $textHol_p$ (like we do in $mathbbS^n$, for example), we have everything we need to find new calibrations, so being symmetric or not is irrelevant, right?
$endgroup$
– rmdmc89
6 hours ago
1
$begingroup$
Yes, you have that. But finding calibration alone isn't the end-goal --- we want to study the submanifolds they calibrate in order to learn more about the manifold. The questions we ask in symmetric spaces (knowing it is symmetric) is, therefore, in a sense the opposite of what we do elsewhere. For nonsymmetric spaces we don't have this luxury of turning to algebra (well, unless you count algebraic geometry for things like Kahler or Calabi-Yau) so a lot remains in analysis of PDEs and suchlike.
$endgroup$
– user10354138
5 hours ago
$begingroup$
By the way, you mentioned a symmetric space $G/H$. Does every symmetric space have that form?
$endgroup$
– rmdmc89
5 hours ago
$begingroup$
A symmetric space $M$ is the quotient of two Lie groups $operatornameIsom(M)/operatornameIsom_p(M)$.
$endgroup$
– user10354138
4 hours ago
add a comment |
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1 Answer
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votes
1 Answer
1
active
oldest
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active
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active
oldest
votes
$begingroup$
(1) A generic metric has (restricted) holonomy group $SO(n)$ (More proper statement: the set of holonomy $SO(n)$ metrics is comeagre in the space of all Riemannian metrics). Hence the adjective special is coined (as in the opposite of "generic") when we can reduce it to smaller subgroups. It definitely predates Joyce (certainly Harvey and Lawson used that in their seminal paper introducing calibrations in the early 1980s).
(2) Elie Cartan proved that for Riemannian symmetric spaces $G/H$, the restricted holonomy group is the identity component of the isotropy group $H$. So this is just a pure algebra problem as to which Lie group is a subgroup of another Lie group (or equivalently which Lie algebra is a subalgebra of another), hence not interesting (as in having little if not no geometry content).
answered 8 hours ago
user10354138user10354138
16.8k21130
$endgroup$
$begingroup$
If I got it right, you're saying that determining $textHol_p$ of a symmetric manifold is essentially an algebraic problem, not a geometric one. But in the context of calibrations, our goal is to find calibrated submanifolds, right? By the method I described, we try that by looking for $textHol_p$-invariant forms.
$endgroup$
– rmdmc89
7 hours ago
$begingroup$
My point is: if we already know $textHol_p$ (like we do in $mathbbS^n$, for example), we have everything we need to find new calibrations, so being symmetric or not is irrelevant, right?
$endgroup$
– rmdmc89
6 hours ago
1
$begingroup$
Yes, you have that. But finding calibration alone isn't the end-goal --- we want to study the submanifolds they calibrate in order to learn more about the manifold. The questions we ask in symmetric spaces (knowing it is symmetric) is, therefore, in a sense the opposite of what we do elsewhere. For nonsymmetric spaces we don't have this luxury of turning to algebra (well, unless you count algebraic geometry for things like Kahler or Calabi-Yau) so a lot remains in analysis of PDEs and suchlike.
$endgroup$
– user10354138
5 hours ago
$begingroup$
By the way, you mentioned a symmetric space $G/H$. Does every symmetric space have that form?
$endgroup$
– rmdmc89
5 hours ago
$begingroup$
A symmetric space $M$ is the quotient of two Lie groups $operatornameIsom(M)/operatornameIsom_p(M)$.
$endgroup$
– user10354138
4 hours ago
add a comment |
$begingroup$
(1) A generic metric has (restricted) holonomy group $SO(n)$ (More proper statement: the set of holonomy $SO(n)$ metrics is comeagre in the space of all Riemannian metrics). Hence the adjective special is coined (as in the opposite of "generic") when we can reduce it to smaller subgroups. It definitely predates Joyce (certainly Harvey and Lawson used that in their seminal paper introducing calibrations in the early 1980s).
(2) Elie Cartan proved that for Riemannian symmetric spaces $G/H$, the restricted holonomy group is the identity component of the isotropy group $H$. So this is just a pure algebra problem as to which Lie group is a subgroup of another Lie group (or equivalently which Lie algebra is a subalgebra of another), hence not interesting (as in having little if not no geometry content).
answered 8 hours ago
user10354138user10354138
16.8k21130
$endgroup$
$begingroup$
If I got it right, you're saying that determining $textHol_p$ of a symmetric manifold is essentially an algebraic problem, not a geometric one. But in the context of calibrations, our goal is to find calibrated submanifolds, right? By the method I described, we try that by looking for $textHol_p$-invariant forms.
$endgroup$
– rmdmc89
7 hours ago
$begingroup$
My point is: if we already know $textHol_p$ (like we do in $mathbbS^n$, for example), we have everything we need to find new calibrations, so being symmetric or not is irrelevant, right?
$endgroup$
– rmdmc89
6 hours ago
1
$begingroup$
Yes, you have that. But finding calibration alone isn't the end-goal --- we want to study the submanifolds they calibrate in order to learn more about the manifold. The questions we ask in symmetric spaces (knowing it is symmetric) is, therefore, in a sense the opposite of what we do elsewhere. For nonsymmetric spaces we don't have this luxury of turning to algebra (well, unless you count algebraic geometry for things like Kahler or Calabi-Yau) so a lot remains in analysis of PDEs and suchlike.
$endgroup$
– user10354138
5 hours ago
$begingroup$
By the way, you mentioned a symmetric space $G/H$. Does every symmetric space have that form?
$endgroup$
– rmdmc89
5 hours ago
$begingroup$
A symmetric space $M$ is the quotient of two Lie groups $operatornameIsom(M)/operatornameIsom_p(M)$.
$endgroup$
– user10354138
4 hours ago
add a comment |
$begingroup$
(1) A generic metric has (restricted) holonomy group $SO(n)$ (More proper statement: the set of holonomy $SO(n)$ metrics is comeagre in the space of all Riemannian metrics). Hence the adjective special is coined (as in the opposite of "generic") when we can reduce it to smaller subgroups. It definitely predates Joyce (certainly Harvey and Lawson used that in their seminal paper introducing calibrations in the early 1980s).
(2) Elie Cartan proved that for Riemannian symmetric spaces $G/H$, the restricted holonomy group is the identity component of the isotropy group $H$. So this is just a pure algebra problem as to which Lie group is a subgroup of another Lie group (or equivalently which Lie algebra is a subalgebra of another), hence not interesting (as in having little if not no geometry content).
answered 8 hours ago
user10354138user10354138
16.8k21130
$endgroup$
(1) A generic metric has (restricted) holonomy group $SO(n)$ (More proper statement: the set of holonomy $SO(n)$ metrics is comeagre in the space of all Riemannian metrics). Hence the adjective special is coined (as in the opposite of "generic") when we can reduce it to smaller subgroups. It definitely predates Joyce (certainly Harvey and Lawson used that in their seminal paper introducing calibrations in the early 1980s).
(2) Elie Cartan proved that for Riemannian symmetric spaces $G/H$, the restricted holonomy group is the identity component of the isotropy group $H$. So this is just a pure algebra problem as to which Lie group is a subgroup of another Lie group (or equivalently which Lie algebra is a subalgebra of another), hence not interesting (as in having little if not no geometry content).
answered 8 hours ago
user10354138user10354138
16.8k21130
edited 8 hours ago
answered 8 hours ago
user10354138user10354138
16.8k21130
answered 8 hours ago
user10354138user10354138
16.8k21130
answered 8 hours ago
user10354138user10354138
16.8k21130
16.8k21130
$begingroup$
If I got it right, you're saying that determining $textHol_p$ of a symmetric manifold is essentially an algebraic problem, not a geometric one. But in the context of calibrations, our goal is to find calibrated submanifolds, right? By the method I described, we try that by looking for $textHol_p$-invariant forms.
$endgroup$
– rmdmc89
7 hours ago
$begingroup$
My point is: if we already know $textHol_p$ (like we do in $mathbbS^n$, for example), we have everything we need to find new calibrations, so being symmetric or not is irrelevant, right?
$endgroup$
– rmdmc89
6 hours ago
1
$begingroup$
Yes, you have that. But finding calibration alone isn't the end-goal --- we want to study the submanifolds they calibrate in order to learn more about the manifold. The questions we ask in symmetric spaces (knowing it is symmetric) is, therefore, in a sense the opposite of what we do elsewhere. For nonsymmetric spaces we don't have this luxury of turning to algebra (well, unless you count algebraic geometry for things like Kahler or Calabi-Yau) so a lot remains in analysis of PDEs and suchlike.
$endgroup$
– user10354138
5 hours ago
$begingroup$
By the way, you mentioned a symmetric space $G/H$. Does every symmetric space have that form?
$endgroup$
– rmdmc89
5 hours ago
$begingroup$
A symmetric space $M$ is the quotient of two Lie groups $operatornameIsom(M)/operatornameIsom_p(M)$.
$endgroup$
– user10354138
4 hours ago
add a comment |
$begingroup$
If I got it right, you're saying that determining $textHol_p$ of a symmetric manifold is essentially an algebraic problem, not a geometric one. But in the context of calibrations, our goal is to find calibrated submanifolds, right? By the method I described, we try that by looking for $textHol_p$-invariant forms.
$endgroup$
– rmdmc89
7 hours ago
$begingroup$
My point is: if we already know $textHol_p$ (like we do in $mathbbS^n$, for example), we have everything we need to find new calibrations, so being symmetric or not is irrelevant, right?
$endgroup$
– rmdmc89
6 hours ago
1
$begingroup$
Yes, you have that. But finding calibration alone isn't the end-goal --- we want to study the submanifolds they calibrate in order to learn more about the manifold. The questions we ask in symmetric spaces (knowing it is symmetric) is, therefore, in a sense the opposite of what we do elsewhere. For nonsymmetric spaces we don't have this luxury of turning to algebra (well, unless you count algebraic geometry for things like Kahler or Calabi-Yau) so a lot remains in analysis of PDEs and suchlike.
$endgroup$
– user10354138
5 hours ago
$begingroup$
By the way, you mentioned a symmetric space $G/H$. Does every symmetric space have that form?
$endgroup$
– rmdmc89
5 hours ago
$begingroup$
A symmetric space $M$ is the quotient of two Lie groups $operatornameIsom(M)/operatornameIsom_p(M)$.
$endgroup$
– user10354138
4 hours ago
$begingroup$
If I got it right, you're saying that determining $textHol_p$ of a symmetric manifold is essentially an algebraic problem, not a geometric one. But in the context of calibrations, our goal is to find calibrated submanifolds, right? By the method I described, we try that by looking for $textHol_p$-invariant forms.
$endgroup$
– rmdmc89
7 hours ago
$begingroup$
If I got it right, you're saying that determining $textHol_p$ of a symmetric manifold is essentially an algebraic problem, not a geometric one. But in the context of calibrations, our goal is to find calibrated submanifolds, right? By the method I described, we try that by looking for $textHol_p$-invariant forms.
$endgroup$
– rmdmc89
7 hours ago
$begingroup$
My point is: if we already know $textHol_p$ (like we do in $mathbbS^n$, for example), we have everything we need to find new calibrations, so being symmetric or not is irrelevant, right?
$endgroup$
– rmdmc89
6 hours ago
$begingroup$
My point is: if we already know $textHol_p$ (like we do in $mathbbS^n$, for example), we have everything we need to find new calibrations, so being symmetric or not is irrelevant, right?
$endgroup$
– rmdmc89
6 hours ago
1
1
$begingroup$
Yes, you have that. But finding calibration alone isn't the end-goal --- we want to study the submanifolds they calibrate in order to learn more about the manifold. The questions we ask in symmetric spaces (knowing it is symmetric) is, therefore, in a sense the opposite of what we do elsewhere. For nonsymmetric spaces we don't have this luxury of turning to algebra (well, unless you count algebraic geometry for things like Kahler or Calabi-Yau) so a lot remains in analysis of PDEs and suchlike.
$endgroup$
– user10354138
5 hours ago
$begingroup$
Yes, you have that. But finding calibration alone isn't the end-goal --- we want to study the submanifolds they calibrate in order to learn more about the manifold. The questions we ask in symmetric spaces (knowing it is symmetric) is, therefore, in a sense the opposite of what we do elsewhere. For nonsymmetric spaces we don't have this luxury of turning to algebra (well, unless you count algebraic geometry for things like Kahler or Calabi-Yau) so a lot remains in analysis of PDEs and suchlike.
$endgroup$
– user10354138
5 hours ago
$begingroup$
By the way, you mentioned a symmetric space $G/H$. Does every symmetric space have that form?
$endgroup$
– rmdmc89
5 hours ago
$begingroup$
By the way, you mentioned a symmetric space $G/H$. Does every symmetric space have that form?
$endgroup$
– rmdmc89
5 hours ago
$begingroup$
A symmetric space $M$ is the quotient of two Lie groups $operatornameIsom(M)/operatornameIsom_p(M)$.
$endgroup$
– user10354138
4 hours ago
$begingroup$
A symmetric space $M$ is the quotient of two Lie groups $operatornameIsom(M)/operatornameIsom_p(M)$.
$endgroup$
– user10354138
4 hours ago
add a comment |
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