When (if ever) does a connection on a Lie manifold G define that groups Lie algebra?
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Note I study physics so I apologize in advance for poor notation terminology. admittedly I'm very new to Lie algebras and groups (at least in the general sense, I've encountered specific groups many times). In studying Riemannian geometry, one ends up using covariant derivatives $nabla$ in place of ordinary derivatives. In other words one defines a connection $nabla$ on the manifold (let's call it $M$) in question.
Choosing a particular direction for the derivative (ie a directional derivative) via some vector $overrightarrow{u}$ we get $nabla_{u}=u^{i}nabla_{i}$ , where we are using the Einstein summation convention.
For some arbitrary 1-form $A$ on $M$, we can then write:
$$[nabla_{u},nabla_{v}]A_{alpha}-nabla_{[u,v]}A_{alpha}=A_{beta}R_{alphamunu}^{beta}$$
Where the brackets denote the commutator of the terms contained therein, and the tensor $R_{alphamunu}^{beta}$ is the Riemann curvature tensor (If I'm remembering correctly this is actually how Riemann first defined his curvature tensor). If we choose coordinate vectors for $u,v$ then their commutators vanish (the second term in the above expression) the expression simplifies to (in coordinate free notation):
$$R(u,v)w=[nabla_{u},nabla_{nu}]w$$
The Riemann tensor has some symmetry properties such as:
$$R(u,v)=-R(v,u)$$
$$R(u,v)w+R(v,w)u+R(w,u)v=0$$
Examining all of these terms one is reminded of a Lie algebra, with the Riemann tensor playing a role simlar to a structure constant (which makes sense since the Riemann tensor is essentially a topological obstruction measuring a deviation from euclidean space). However I get that we're in $TM$.
Entrar my question: If we're on a manifold that is also a Lie group, can we form that groups Lie algebra from this (and ensuingly the Lie group itself?). My first thought is to go straight ahead and say the Covariant derivatives themselves form components of the Lie algebra? Would that be right?
The amount of arbitrariness allowed in the manifold locally which still would adhere to a particular global topology leads me to doubt this. The second line of reasoning of mine is to take a pullback $phi$ to a manifold with the same topology but constant components of the Riemann tensor. Then we would have that:
$$[left(phi*nabla_{u}right),left(phi*nabla_{v}right)]left(phi*A_{alpha}right)=left(phi*A_{beta}right)mathbb{R}_{alphamunu}^{beta}$$
This time however, the pulled back Curvature tensor has constant entries, in essence endowing the map $phi$ with all information on the local features of our original manifold. In order for this to work, the pullback would have to be some subset of all pullbacks such that the topology of the two manifolds
are the same (the group of homeomorphisms?)
update:
It would appear that for the compact Case, The Yamabe problem would guarantee that a simple conformal map would suffice to move to a space of constant Riemann curvature.
After doing this it would appear ok to say that we can form the Lie algebra of the group manifold using the pulled back covariant derivative. In this sense we could define the Lie algebra in terms of our original covariant derivative and the pullback (map).
This might seem silly to some, but would actually be useful for a project of mine. Does anyone have any input? Have I failed to see the Geometer's forest for all of the trees contained therein?
Note, there is one more thing I'd like to mention. Considering the first equation again:
$$[nabla_{u},nabla_{v}]A_{alpha}-nabla_{[u,v]}A_{alpha}=A_{beta}R_{alphamunu}^{beta}$$
The second term on the left actually is reminiscent of the second term in the
right hand side of the exponent in the Cambell-Baker-Hausdorff formula which goes like:
$$e^{A}e^{B}=e^{A+B+[A,B]+...}$$ Now since we would have to exponentiate the Lie algebra to get the group, it's interesting that a term like that mentioned above would appear? Am I chasing ghosts or is there a way to figure this out?
differential-geometry lie-groups lie-algebras connections
asked Nov 19 at 9:18
R. Rankin
305213
|
show 4 more comments
up vote
2
down vote
favorite
Note I study physics so I apologize in advance for poor notation terminology. admittedly I'm very new to Lie algebras and groups (at least in the general sense, I've encountered specific groups many times). In studying Riemannian geometry, one ends up using covariant derivatives $nabla$ in place of ordinary derivatives. In other words one defines a connection $nabla$ on the manifold (let's call it $M$) in question.
Choosing a particular direction for the derivative (ie a directional derivative) via some vector $overrightarrow{u}$ we get $nabla_{u}=u^{i}nabla_{i}$ , where we are using the Einstein summation convention.
For some arbitrary 1-form $A$ on $M$, we can then write:
$$[nabla_{u},nabla_{v}]A_{alpha}-nabla_{[u,v]}A_{alpha}=A_{beta}R_{alphamunu}^{beta}$$
Where the brackets denote the commutator of the terms contained therein, and the tensor $R_{alphamunu}^{beta}$ is the Riemann curvature tensor (If I'm remembering correctly this is actually how Riemann first defined his curvature tensor). If we choose coordinate vectors for $u,v$ then their commutators vanish (the second term in the above expression) the expression simplifies to (in coordinate free notation):
$$R(u,v)w=[nabla_{u},nabla_{nu}]w$$
The Riemann tensor has some symmetry properties such as:
$$R(u,v)=-R(v,u)$$
$$R(u,v)w+R(v,w)u+R(w,u)v=0$$
Examining all of these terms one is reminded of a Lie algebra, with the Riemann tensor playing a role simlar to a structure constant (which makes sense since the Riemann tensor is essentially a topological obstruction measuring a deviation from euclidean space). However I get that we're in $TM$.
Entrar my question: If we're on a manifold that is also a Lie group, can we form that groups Lie algebra from this (and ensuingly the Lie group itself?). My first thought is to go straight ahead and say the Covariant derivatives themselves form components of the Lie algebra? Would that be right?
The amount of arbitrariness allowed in the manifold locally which still would adhere to a particular global topology leads me to doubt this. The second line of reasoning of mine is to take a pullback $phi$ to a manifold with the same topology but constant components of the Riemann tensor. Then we would have that:
$$[left(phi*nabla_{u}right),left(phi*nabla_{v}right)]left(phi*A_{alpha}right)=left(phi*A_{beta}right)mathbb{R}_{alphamunu}^{beta}$$
This time however, the pulled back Curvature tensor has constant entries, in essence endowing the map $phi$ with all information on the local features of our original manifold. In order for this to work, the pullback would have to be some subset of all pullbacks such that the topology of the two manifolds
are the same (the group of homeomorphisms?)
update:
It would appear that for the compact Case, The Yamabe problem would guarantee that a simple conformal map would suffice to move to a space of constant Riemann curvature.
After doing this it would appear ok to say that we can form the Lie algebra of the group manifold using the pulled back covariant derivative. In this sense we could define the Lie algebra in terms of our original covariant derivative and the pullback (map).
This might seem silly to some, but would actually be useful for a project of mine. Does anyone have any input? Have I failed to see the Geometer's forest for all of the trees contained therein?
Note, there is one more thing I'd like to mention. Considering the first equation again:
$$[nabla_{u},nabla_{v}]A_{alpha}-nabla_{[u,v]}A_{alpha}=A_{beta}R_{alphamunu}^{beta}$$
The second term on the left actually is reminiscent of the second term in the
right hand side of the exponent in the Cambell-Baker-Hausdorff formula which goes like:
$$e^{A}e^{B}=e^{A+B+[A,B]+...}$$ Now since we would have to exponentiate the Lie algebra to get the group, it's interesting that a term like that mentioned above would appear? Am I chasing ghosts or is there a way to figure this out?
differential-geometry lie-groups lie-algebras connections
asked Nov 19 at 9:18
R. Rankin
305213
It's not entirely clear to me what the question is, but as to whether you can recover the Lie algebra of a Lie group $G$ from an arbitrary choice of connection on $G$, the answer is no: One can pick Lie groups $G, G'$ that are diffeomorphic but not isomorphic.
– Travis
Nov 20 at 6:18
@Travis as per the title, i'm just trying to figure under what circumstances (if any) a connection $nabla$ on $S^3$ generates the Lie algebra of SU(2) (to give an example). I'm only dealing with the manifolds of compact Lie groups. I've been trying to learn from Milnor's lie groups and curvature.
– R. Rankin
Nov 20 at 7:06
@Travis I get that an arbitrary connection isn't enough, but perhaps positing something additional like constant Riemann curvature would suffice?
– R. Rankin
Nov 20 at 7:20
The answer probably depends on what precisely you mean by "generate". NB in this case there is only a single Lie group ($SU(2)$) with underlying smooth manifold $S^3$. Since you mention Milnor's paper and Riemann curvature, are you thinking of $S^3$ as equipped with a metric? If so is it necessarily the standard metric?
– Travis
Nov 20 at 7:25
1
One can always locally choose such vector fields, yes, but this statement is independent of the choice of metric. In general, however, one can't choose those vector fields to be Killing fields for a given metric. (In fact, being able to choose such vector fields is a sufficient condition for the metric being locally isometric to $(S^3, g)$.)
– Travis
Nov 20 at 8:02
|
show 4 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Note I study physics so I apologize in advance for poor notation terminology. admittedly I'm very new to Lie algebras and groups (at least in the general sense, I've encountered specific groups many times). In studying Riemannian geometry, one ends up using covariant derivatives $nabla$ in place of ordinary derivatives. In other words one defines a connection $nabla$ on the manifold (let's call it $M$) in question.
Choosing a particular direction for the derivative (ie a directional derivative) via some vector $overrightarrow{u}$ we get $nabla_{u}=u^{i}nabla_{i}$ , where we are using the Einstein summation convention.
For some arbitrary 1-form $A$ on $M$, we can then write:
$$[nabla_{u},nabla_{v}]A_{alpha}-nabla_{[u,v]}A_{alpha}=A_{beta}R_{alphamunu}^{beta}$$
Where the brackets denote the commutator of the terms contained therein, and the tensor $R_{alphamunu}^{beta}$ is the Riemann curvature tensor (If I'm remembering correctly this is actually how Riemann first defined his curvature tensor). If we choose coordinate vectors for $u,v$ then their commutators vanish (the second term in the above expression) the expression simplifies to (in coordinate free notation):
$$R(u,v)w=[nabla_{u},nabla_{nu}]w$$
The Riemann tensor has some symmetry properties such as:
$$R(u,v)=-R(v,u)$$
$$R(u,v)w+R(v,w)u+R(w,u)v=0$$
Examining all of these terms one is reminded of a Lie algebra, with the Riemann tensor playing a role simlar to a structure constant (which makes sense since the Riemann tensor is essentially a topological obstruction measuring a deviation from euclidean space). However I get that we're in $TM$.
Entrar my question: If we're on a manifold that is also a Lie group, can we form that groups Lie algebra from this (and ensuingly the Lie group itself?). My first thought is to go straight ahead and say the Covariant derivatives themselves form components of the Lie algebra? Would that be right?
The amount of arbitrariness allowed in the manifold locally which still would adhere to a particular global topology leads me to doubt this. The second line of reasoning of mine is to take a pullback $phi$ to a manifold with the same topology but constant components of the Riemann tensor. Then we would have that:
$$[left(phi*nabla_{u}right),left(phi*nabla_{v}right)]left(phi*A_{alpha}right)=left(phi*A_{beta}right)mathbb{R}_{alphamunu}^{beta}$$
This time however, the pulled back Curvature tensor has constant entries, in essence endowing the map $phi$ with all information on the local features of our original manifold. In order for this to work, the pullback would have to be some subset of all pullbacks such that the topology of the two manifolds
are the same (the group of homeomorphisms?)
update:
It would appear that for the compact Case, The Yamabe problem would guarantee that a simple conformal map would suffice to move to a space of constant Riemann curvature.
After doing this it would appear ok to say that we can form the Lie algebra of the group manifold using the pulled back covariant derivative. In this sense we could define the Lie algebra in terms of our original covariant derivative and the pullback (map).
This might seem silly to some, but would actually be useful for a project of mine. Does anyone have any input? Have I failed to see the Geometer's forest for all of the trees contained therein?
Note, there is one more thing I'd like to mention. Considering the first equation again:
$$[nabla_{u},nabla_{v}]A_{alpha}-nabla_{[u,v]}A_{alpha}=A_{beta}R_{alphamunu}^{beta}$$
The second term on the left actually is reminiscent of the second term in the
right hand side of the exponent in the Cambell-Baker-Hausdorff formula which goes like:
$$e^{A}e^{B}=e^{A+B+[A,B]+...}$$ Now since we would have to exponentiate the Lie algebra to get the group, it's interesting that a term like that mentioned above would appear? Am I chasing ghosts or is there a way to figure this out?
differential-geometry lie-groups lie-algebras connections
asked Nov 19 at 9:18
R. Rankin
305213
Note I study physics so I apologize in advance for poor notation terminology. admittedly I'm very new to Lie algebras and groups (at least in the general sense, I've encountered specific groups many times). In studying Riemannian geometry, one ends up using covariant derivatives $nabla$ in place of ordinary derivatives. In other words one defines a connection $nabla$ on the manifold (let's call it $M$) in question.
Choosing a particular direction for the derivative (ie a directional derivative) via some vector $overrightarrow{u}$ we get $nabla_{u}=u^{i}nabla_{i}$ , where we are using the Einstein summation convention.
For some arbitrary 1-form $A$ on $M$, we can then write:
$$[nabla_{u},nabla_{v}]A_{alpha}-nabla_{[u,v]}A_{alpha}=A_{beta}R_{alphamunu}^{beta}$$
Where the brackets denote the commutator of the terms contained therein, and the tensor $R_{alphamunu}^{beta}$ is the Riemann curvature tensor (If I'm remembering correctly this is actually how Riemann first defined his curvature tensor). If we choose coordinate vectors for $u,v$ then their commutators vanish (the second term in the above expression) the expression simplifies to (in coordinate free notation):
$$R(u,v)w=[nabla_{u},nabla_{nu}]w$$
The Riemann tensor has some symmetry properties such as:
$$R(u,v)=-R(v,u)$$
$$R(u,v)w+R(v,w)u+R(w,u)v=0$$
Examining all of these terms one is reminded of a Lie algebra, with the Riemann tensor playing a role simlar to a structure constant (which makes sense since the Riemann tensor is essentially a topological obstruction measuring a deviation from euclidean space). However I get that we're in $TM$.
Entrar my question: If we're on a manifold that is also a Lie group, can we form that groups Lie algebra from this (and ensuingly the Lie group itself?). My first thought is to go straight ahead and say the Covariant derivatives themselves form components of the Lie algebra? Would that be right?
The amount of arbitrariness allowed in the manifold locally which still would adhere to a particular global topology leads me to doubt this. The second line of reasoning of mine is to take a pullback $phi$ to a manifold with the same topology but constant components of the Riemann tensor. Then we would have that:
$$[left(phi*nabla_{u}right),left(phi*nabla_{v}right)]left(phi*A_{alpha}right)=left(phi*A_{beta}right)mathbb{R}_{alphamunu}^{beta}$$
This time however, the pulled back Curvature tensor has constant entries, in essence endowing the map $phi$ with all information on the local features of our original manifold. In order for this to work, the pullback would have to be some subset of all pullbacks such that the topology of the two manifolds
are the same (the group of homeomorphisms?)
update:
It would appear that for the compact Case, The Yamabe problem would guarantee that a simple conformal map would suffice to move to a space of constant Riemann curvature.
After doing this it would appear ok to say that we can form the Lie algebra of the group manifold using the pulled back covariant derivative. In this sense we could define the Lie algebra in terms of our original covariant derivative and the pullback (map).
This might seem silly to some, but would actually be useful for a project of mine. Does anyone have any input? Have I failed to see the Geometer's forest for all of the trees contained therein?
Note, there is one more thing I'd like to mention. Considering the first equation again:
$$[nabla_{u},nabla_{v}]A_{alpha}-nabla_{[u,v]}A_{alpha}=A_{beta}R_{alphamunu}^{beta}$$
The second term on the left actually is reminiscent of the second term in the
right hand side of the exponent in the Cambell-Baker-Hausdorff formula which goes like:
$$e^{A}e^{B}=e^{A+B+[A,B]+...}$$ Now since we would have to exponentiate the Lie algebra to get the group, it's interesting that a term like that mentioned above would appear? Am I chasing ghosts or is there a way to figure this out?
differential-geometry lie-groups lie-algebras connections
differential-geometry lie-groups lie-algebras connections
asked Nov 19 at 9:18
R. Rankin
305213
asked Nov 19 at 9:18
R. Rankin
305213
edited Nov 20 at 3:10
asked Nov 19 at 9:18
R. Rankin
305213
asked Nov 19 at 9:18
R. Rankin
305213
asked Nov 19 at 9:18
R. Rankin
305213
305213
It's not entirely clear to me what the question is, but as to whether you can recover the Lie algebra of a Lie group $G$ from an arbitrary choice of connection on $G$, the answer is no: One can pick Lie groups $G, G'$ that are diffeomorphic but not isomorphic.
– Travis
Nov 20 at 6:18
@Travis as per the title, i'm just trying to figure under what circumstances (if any) a connection $nabla$ on $S^3$ generates the Lie algebra of SU(2) (to give an example). I'm only dealing with the manifolds of compact Lie groups. I've been trying to learn from Milnor's lie groups and curvature.
– R. Rankin
Nov 20 at 7:06
@Travis I get that an arbitrary connection isn't enough, but perhaps positing something additional like constant Riemann curvature would suffice?
– R. Rankin
Nov 20 at 7:20
The answer probably depends on what precisely you mean by "generate". NB in this case there is only a single Lie group ($SU(2)$) with underlying smooth manifold $S^3$. Since you mention Milnor's paper and Riemann curvature, are you thinking of $S^3$ as equipped with a metric? If so is it necessarily the standard metric?
– Travis
Nov 20 at 7:25
1
One can always locally choose such vector fields, yes, but this statement is independent of the choice of metric. In general, however, one can't choose those vector fields to be Killing fields for a given metric. (In fact, being able to choose such vector fields is a sufficient condition for the metric being locally isometric to $(S^3, g)$.)
– Travis
Nov 20 at 8:02
|
show 4 more comments
It's not entirely clear to me what the question is, but as to whether you can recover the Lie algebra of a Lie group $G$ from an arbitrary choice of connection on $G$, the answer is no: One can pick Lie groups $G, G'$ that are diffeomorphic but not isomorphic.
– Travis
Nov 20 at 6:18
@Travis as per the title, i'm just trying to figure under what circumstances (if any) a connection $nabla$ on $S^3$ generates the Lie algebra of SU(2) (to give an example). I'm only dealing with the manifolds of compact Lie groups. I've been trying to learn from Milnor's lie groups and curvature.
– R. Rankin
Nov 20 at 7:06
@Travis I get that an arbitrary connection isn't enough, but perhaps positing something additional like constant Riemann curvature would suffice?
– R. Rankin
Nov 20 at 7:20
The answer probably depends on what precisely you mean by "generate". NB in this case there is only a single Lie group ($SU(2)$) with underlying smooth manifold $S^3$. Since you mention Milnor's paper and Riemann curvature, are you thinking of $S^3$ as equipped with a metric? If so is it necessarily the standard metric?
– Travis
Nov 20 at 7:25
1
One can always locally choose such vector fields, yes, but this statement is independent of the choice of metric. In general, however, one can't choose those vector fields to be Killing fields for a given metric. (In fact, being able to choose such vector fields is a sufficient condition for the metric being locally isometric to $(S^3, g)$.)
– Travis
Nov 20 at 8:02
It's not entirely clear to me what the question is, but as to whether you can recover the Lie algebra of a Lie group $G$ from an arbitrary choice of connection on $G$, the answer is no: One can pick Lie groups $G, G'$ that are diffeomorphic but not isomorphic.
– Travis
Nov 20 at 6:18
It's not entirely clear to me what the question is, but as to whether you can recover the Lie algebra of a Lie group $G$ from an arbitrary choice of connection on $G$, the answer is no: One can pick Lie groups $G, G'$ that are diffeomorphic but not isomorphic.
– Travis
Nov 20 at 6:18
@Travis as per the title, i'm just trying to figure under what circumstances (if any) a connection $nabla$ on $S^3$ generates the Lie algebra of SU(2) (to give an example). I'm only dealing with the manifolds of compact Lie groups. I've been trying to learn from Milnor's lie groups and curvature.
– R. Rankin
Nov 20 at 7:06
@Travis as per the title, i'm just trying to figure under what circumstances (if any) a connection $nabla$ on $S^3$ generates the Lie algebra of SU(2) (to give an example). I'm only dealing with the manifolds of compact Lie groups. I've been trying to learn from Milnor's lie groups and curvature.
– R. Rankin
Nov 20 at 7:06
@Travis I get that an arbitrary connection isn't enough, but perhaps positing something additional like constant Riemann curvature would suffice?
– R. Rankin
Nov 20 at 7:20
@Travis I get that an arbitrary connection isn't enough, but perhaps positing something additional like constant Riemann curvature would suffice?
– R. Rankin
Nov 20 at 7:20
The answer probably depends on what precisely you mean by "generate". NB in this case there is only a single Lie group ($SU(2)$) with underlying smooth manifold $S^3$. Since you mention Milnor's paper and Riemann curvature, are you thinking of $S^3$ as equipped with a metric? If so is it necessarily the standard metric?
– Travis
Nov 20 at 7:25
The answer probably depends on what precisely you mean by "generate". NB in this case there is only a single Lie group ($SU(2)$) with underlying smooth manifold $S^3$. Since you mention Milnor's paper and Riemann curvature, are you thinking of $S^3$ as equipped with a metric? If so is it necessarily the standard metric?
– Travis
Nov 20 at 7:25
1
1
One can always locally choose such vector fields, yes, but this statement is independent of the choice of metric. In general, however, one can't choose those vector fields to be Killing fields for a given metric. (In fact, being able to choose such vector fields is a sufficient condition for the metric being locally isometric to $(S^3, g)$.)
– Travis
Nov 20 at 8:02
One can always locally choose such vector fields, yes, but this statement is independent of the choice of metric. In general, however, one can't choose those vector fields to be Killing fields for a given metric. (In fact, being able to choose such vector fields is a sufficient condition for the metric being locally isometric to $(S^3, g)$.)
– Travis
Nov 20 at 8:02
|
show 4 more comments
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It's not entirely clear to me what the question is, but as to whether you can recover the Lie algebra of a Lie group $G$ from an arbitrary choice of connection on $G$, the answer is no: One can pick Lie groups $G, G'$ that are diffeomorphic but not isomorphic.
– Travis
Nov 20 at 6:18
@Travis as per the title, i'm just trying to figure under what circumstances (if any) a connection $nabla$ on $S^3$ generates the Lie algebra of SU(2) (to give an example). I'm only dealing with the manifolds of compact Lie groups. I've been trying to learn from Milnor's lie groups and curvature.
– R. Rankin
Nov 20 at 7:06
@Travis I get that an arbitrary connection isn't enough, but perhaps positing something additional like constant Riemann curvature would suffice?
– R. Rankin
Nov 20 at 7:20
The answer probably depends on what precisely you mean by "generate". NB in this case there is only a single Lie group ($SU(2)$) with underlying smooth manifold $S^3$. Since you mention Milnor's paper and Riemann curvature, are you thinking of $S^3$ as equipped with a metric? If so is it necessarily the standard metric?
– Travis
Nov 20 at 7:25
1
One can always locally choose such vector fields, yes, but this statement is independent of the choice of metric. In general, however, one can't choose those vector fields to be Killing fields for a given metric. (In fact, being able to choose such vector fields is a sufficient condition for the metric being locally isometric to $(S^3, g)$.)
– Travis
Nov 20 at 8:02