Tangent space of Cotangent bundle at zero section? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Why are smooth manifolds defined to be paracompact?Riemannian metric of the tangent bundleHow to show that geodesics exist for all of time in a compact manifold?(Co)Tangent bundle of Cone manifoldCan tangent and cotangent spaces be distinguished?The topology of $T_pM$Tangent bundle of a manifold as more than a vector bundleWhen is a Divergence-Free Vector Field on the Tangent Bundle of a Riemannian Manifold Hamiltonian?Computing the flow on the cotangent bundleDependence of spinor bundle on choice of metricTangent Bundle Cotangent Bundle Isomorphism - Partitions Of Unity
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Tangent space of Cotangent bundle at zero section?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Why are smooth manifolds defined to be paracompact?Riemannian metric of the tangent bundleHow to show that geodesics exist for all of time in a compact manifold?(Co)Tangent bundle of Cone manifoldCan tangent and cotangent spaces be distinguished?The topology of $T_pM$Tangent bundle of a manifold as more than a vector bundleWhen is a Divergence-Free Vector Field on the Tangent Bundle of a Riemannian Manifold Hamiltonian?Computing the flow on the cotangent bundleDependence of spinor bundle on choice of metricTangent Bundle Cotangent Bundle Isomorphism - Partitions Of Unity
$begingroup$
Let $M$ be a differentiable manifold with cotangent bundle $T^*M$.
How can I prove that $T_(p,0)T^*M$ is naturally isomorphic to $T_pMoplus T_pM^*$?
If this true, then I think I could prove that the Hessian of $fcolon Mto mathbbR$ is well-defined (I mean, without choice of Connection or Riemannian metric) at critical point of $f$.
This is not any homework.
differential-geometry riemannian-geometry symplectic-geometry
edited Jul 25 '11 at 10:49
t.b.
63k7211289
asked Jul 25 '11 at 10:37
TopologieeeeeTopologieeeee
37539
$endgroup$
add a comment |
$begingroup$
Let $M$ be a differentiable manifold with cotangent bundle $T^*M$.
How can I prove that $T_(p,0)T^*M$ is naturally isomorphic to $T_pMoplus T_pM^*$?
If this true, then I think I could prove that the Hessian of $fcolon Mto mathbbR$ is well-defined (I mean, without choice of Connection or Riemannian metric) at critical point of $f$.
This is not any homework.
differential-geometry riemannian-geometry symplectic-geometry
edited Jul 25 '11 at 10:49
t.b.
63k7211289
asked Jul 25 '11 at 10:37
TopologieeeeeTopologieeeee
37539
$endgroup$
$begingroup$
Hint: look at the kernel of the derivative of the bundle projection. This gives you the desired decomposition.
$endgroup$
– t.b.
Jul 25 '11 at 10:49
$begingroup$
Your comment about the Hessian is also correct. This allows you to say, for instance, that a function is Morse independently of the choice of connection.
$endgroup$
– Sam Lisi
Dec 22 '11 at 13:00
add a comment |
$begingroup$
Let $M$ be a differentiable manifold with cotangent bundle $T^*M$.
How can I prove that $T_(p,0)T^*M$ is naturally isomorphic to $T_pMoplus T_pM^*$?
If this true, then I think I could prove that the Hessian of $fcolon Mto mathbbR$ is well-defined (I mean, without choice of Connection or Riemannian metric) at critical point of $f$.
This is not any homework.
differential-geometry riemannian-geometry symplectic-geometry
edited Jul 25 '11 at 10:49
t.b.
63k7211289
asked Jul 25 '11 at 10:37
TopologieeeeeTopologieeeee
37539
$endgroup$
Let $M$ be a differentiable manifold with cotangent bundle $T^*M$.
How can I prove that $T_(p,0)T^*M$ is naturally isomorphic to $T_pMoplus T_pM^*$?
If this true, then I think I could prove that the Hessian of $fcolon Mto mathbbR$ is well-defined (I mean, without choice of Connection or Riemannian metric) at critical point of $f$.
This is not any homework.
differential-geometry riemannian-geometry symplectic-geometry
differential-geometry riemannian-geometry symplectic-geometry
edited Jul 25 '11 at 10:49
t.b.
63k7211289
asked Jul 25 '11 at 10:37
TopologieeeeeTopologieeeee
37539
edited Jul 25 '11 at 10:49
t.b.
63k7211289
asked Jul 25 '11 at 10:37
TopologieeeeeTopologieeeee
37539
edited Jul 25 '11 at 10:49
t.b.
63k7211289
edited Jul 25 '11 at 10:49
t.b.
63k7211289
edited Jul 25 '11 at 10:49
t.b.
63k7211289
63k7211289
asked Jul 25 '11 at 10:37
TopologieeeeeTopologieeeee
37539
asked Jul 25 '11 at 10:37
TopologieeeeeTopologieeeee
37539
asked Jul 25 '11 at 10:37
TopologieeeeeTopologieeeee
37539
37539
$begingroup$
Hint: look at the kernel of the derivative of the bundle projection. This gives you the desired decomposition.
$endgroup$
– t.b.
Jul 25 '11 at 10:49
$begingroup$
Your comment about the Hessian is also correct. This allows you to say, for instance, that a function is Morse independently of the choice of connection.
$endgroup$
– Sam Lisi
Dec 22 '11 at 13:00
add a comment |
$begingroup$
Hint: look at the kernel of the derivative of the bundle projection. This gives you the desired decomposition.
$endgroup$
– t.b.
Jul 25 '11 at 10:49
$begingroup$
Your comment about the Hessian is also correct. This allows you to say, for instance, that a function is Morse independently of the choice of connection.
$endgroup$
– Sam Lisi
Dec 22 '11 at 13:00
$begingroup$
Hint: look at the kernel of the derivative of the bundle projection. This gives you the desired decomposition.
$endgroup$
– t.b.
Jul 25 '11 at 10:49
$begingroup$
Hint: look at the kernel of the derivative of the bundle projection. This gives you the desired decomposition.
$endgroup$
– t.b.
Jul 25 '11 at 10:49
$begingroup$
Your comment about the Hessian is also correct. This allows you to say, for instance, that a function is Morse independently of the choice of connection.
$endgroup$
– Sam Lisi
Dec 22 '11 at 13:00
$begingroup$
Your comment about the Hessian is also correct. This allows you to say, for instance, that a function is Morse independently of the choice of connection.
$endgroup$
– Sam Lisi
Dec 22 '11 at 13:00
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Things are much nicer i.e. more general and canonical than that!
a) Let $pi:Nto M$ be a submersion of manifolds. From it you obtain the exact sequence of vector bundles on $N$: $$ 0to T^vert(N) to T(N) stackrel dpito pi^-1T(M) to 0 quad (*) $$ Profound, eh? Not at all!
This is just a fancy way of looking at the differential of the map $pi: Nto M.$
In order that both the tangent bundles to $N$ and to $M$ live on $N$, you have to pull back $T(M)$ to $N$: that is why we have $pi^-1 T(M)$ on the right.
The kernel is the set of vertical tangent vectors to $N$, those that lie along the level lines $N[n]stackrel def=pi^-1(pi (n))$ of $pi$.
In other words at a point $nin N$ the fibre of $T^vert(N)$ is $$T^vert_n(N)=T_n(N[n]) quad (**)$$
b) Consider now a vector bundle $pi:Vto M$ on $M$.
You then have the canonical exact sequence of vector bundles on $V;$ (yes, vector bundles on a vector bundle!) :
$$ 0to T^vert(V) to T(V) stackrel dpito pi^-1T(M) to 0 quad (***)$$
In this new set-up you have the interesting identification $ T^vert(V)=pi^-1(V)$.
This boils down to the fact that the tangent space at any point $ein E$ of a vector space $E$ is that vector space itself: $T_e(E)=E.$
Hence $(***)$ becomes $$ 0to pi^-1(V) to T(V) stackrel dpito pi^-1T(M) to 0 quad (****) $$
c) Finally, if you restrict this last exact sequence (****) to the zero section of $V$, identified to $M$, you get
$$ 0to V to T(V)|M stackrel dpito T(M) to 0 quad quad (*****) $$
d) Although this is not very helpful, you can if the manifold $M$ is paracompact (see here) non-canonically split the exact sequence (*****) and obtain the isomorphism $$ T(V)|M cong V oplus T(M)$$
At a point $(m,0)in Msubset V$ this gives the decomposition $$ T_(m,0)(V) cong V_m oplus T_m(M)$$
and for $V=T^*(M)$ this is what you were looking for.
answered Jul 25 '11 at 11:22
Georges ElencwajgGeorges Elencwajg
120k7182335
$endgroup$
$begingroup$
dear Georges, we are ok that more generally $TT^*M$ is naturally isomorphic to $pi^-1TMoplus pi^-1T^*M$ where $pi:T^*Mto M$ ?
$endgroup$
– epsilones
Mar 26 at 22:13
1
$begingroup$
Dear @epsilones: your comment is absolutely correct . I have edited my answer in order to address your question in a completely general context.
$endgroup$
– Georges Elencwajg
Mar 27 at 7:56
add a comment |
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$begingroup$
Things are much nicer i.e. more general and canonical than that!
a) Let $pi:Nto M$ be a submersion of manifolds. From it you obtain the exact sequence of vector bundles on $N$: $$ 0to T^vert(N) to T(N) stackrel dpito pi^-1T(M) to 0 quad (*) $$ Profound, eh? Not at all!
This is just a fancy way of looking at the differential of the map $pi: Nto M.$
In order that both the tangent bundles to $N$ and to $M$ live on $N$, you have to pull back $T(M)$ to $N$: that is why we have $pi^-1 T(M)$ on the right.
The kernel is the set of vertical tangent vectors to $N$, those that lie along the level lines $N[n]stackrel def=pi^-1(pi (n))$ of $pi$.
In other words at a point $nin N$ the fibre of $T^vert(N)$ is $$T^vert_n(N)=T_n(N[n]) quad (**)$$
b) Consider now a vector bundle $pi:Vto M$ on $M$.
You then have the canonical exact sequence of vector bundles on $V;$ (yes, vector bundles on a vector bundle!) :
$$ 0to T^vert(V) to T(V) stackrel dpito pi^-1T(M) to 0 quad (***)$$
In this new set-up you have the interesting identification $ T^vert(V)=pi^-1(V)$.
This boils down to the fact that the tangent space at any point $ein E$ of a vector space $E$ is that vector space itself: $T_e(E)=E.$
Hence $(***)$ becomes $$ 0to pi^-1(V) to T(V) stackrel dpito pi^-1T(M) to 0 quad (****) $$
c) Finally, if you restrict this last exact sequence (****) to the zero section of $V$, identified to $M$, you get
$$ 0to V to T(V)|M stackrel dpito T(M) to 0 quad quad (*****) $$
d) Although this is not very helpful, you can if the manifold $M$ is paracompact (see here) non-canonically split the exact sequence (*****) and obtain the isomorphism $$ T(V)|M cong V oplus T(M)$$
At a point $(m,0)in Msubset V$ this gives the decomposition $$ T_(m,0)(V) cong V_m oplus T_m(M)$$
and for $V=T^*(M)$ this is what you were looking for.
answered Jul 25 '11 at 11:22
Georges ElencwajgGeorges Elencwajg
120k7182335
$endgroup$
$begingroup$
dear Georges, we are ok that more generally $TT^*M$ is naturally isomorphic to $pi^-1TMoplus pi^-1T^*M$ where $pi:T^*Mto M$ ?
$endgroup$
– epsilones
Mar 26 at 22:13
1
$begingroup$
Dear @epsilones: your comment is absolutely correct . I have edited my answer in order to address your question in a completely general context.
$endgroup$
– Georges Elencwajg
Mar 27 at 7:56
add a comment |
$begingroup$
Things are much nicer i.e. more general and canonical than that!
a) Let $pi:Nto M$ be a submersion of manifolds. From it you obtain the exact sequence of vector bundles on $N$: $$ 0to T^vert(N) to T(N) stackrel dpito pi^-1T(M) to 0 quad (*) $$ Profound, eh? Not at all!
This is just a fancy way of looking at the differential of the map $pi: Nto M.$
In order that both the tangent bundles to $N$ and to $M$ live on $N$, you have to pull back $T(M)$ to $N$: that is why we have $pi^-1 T(M)$ on the right.
The kernel is the set of vertical tangent vectors to $N$, those that lie along the level lines $N[n]stackrel def=pi^-1(pi (n))$ of $pi$.
In other words at a point $nin N$ the fibre of $T^vert(N)$ is $$T^vert_n(N)=T_n(N[n]) quad (**)$$
b) Consider now a vector bundle $pi:Vto M$ on $M$.
You then have the canonical exact sequence of vector bundles on $V;$ (yes, vector bundles on a vector bundle!) :
$$ 0to T^vert(V) to T(V) stackrel dpito pi^-1T(M) to 0 quad (***)$$
In this new set-up you have the interesting identification $ T^vert(V)=pi^-1(V)$.
This boils down to the fact that the tangent space at any point $ein E$ of a vector space $E$ is that vector space itself: $T_e(E)=E.$
Hence $(***)$ becomes $$ 0to pi^-1(V) to T(V) stackrel dpito pi^-1T(M) to 0 quad (****) $$
c) Finally, if you restrict this last exact sequence (****) to the zero section of $V$, identified to $M$, you get
$$ 0to V to T(V)|M stackrel dpito T(M) to 0 quad quad (*****) $$
d) Although this is not very helpful, you can if the manifold $M$ is paracompact (see here) non-canonically split the exact sequence (*****) and obtain the isomorphism $$ T(V)|M cong V oplus T(M)$$
At a point $(m,0)in Msubset V$ this gives the decomposition $$ T_(m,0)(V) cong V_m oplus T_m(M)$$
and for $V=T^*(M)$ this is what you were looking for.
answered Jul 25 '11 at 11:22
Georges ElencwajgGeorges Elencwajg
120k7182335
$endgroup$
$begingroup$
dear Georges, we are ok that more generally $TT^*M$ is naturally isomorphic to $pi^-1TMoplus pi^-1T^*M$ where $pi:T^*Mto M$ ?
$endgroup$
– epsilones
Mar 26 at 22:13
1
$begingroup$
Dear @epsilones: your comment is absolutely correct . I have edited my answer in order to address your question in a completely general context.
$endgroup$
– Georges Elencwajg
Mar 27 at 7:56
add a comment |
$begingroup$
Things are much nicer i.e. more general and canonical than that!
a) Let $pi:Nto M$ be a submersion of manifolds. From it you obtain the exact sequence of vector bundles on $N$: $$ 0to T^vert(N) to T(N) stackrel dpito pi^-1T(M) to 0 quad (*) $$ Profound, eh? Not at all!
This is just a fancy way of looking at the differential of the map $pi: Nto M.$
In order that both the tangent bundles to $N$ and to $M$ live on $N$, you have to pull back $T(M)$ to $N$: that is why we have $pi^-1 T(M)$ on the right.
The kernel is the set of vertical tangent vectors to $N$, those that lie along the level lines $N[n]stackrel def=pi^-1(pi (n))$ of $pi$.
In other words at a point $nin N$ the fibre of $T^vert(N)$ is $$T^vert_n(N)=T_n(N[n]) quad (**)$$
b) Consider now a vector bundle $pi:Vto M$ on $M$.
You then have the canonical exact sequence of vector bundles on $V;$ (yes, vector bundles on a vector bundle!) :
$$ 0to T^vert(V) to T(V) stackrel dpito pi^-1T(M) to 0 quad (***)$$
In this new set-up you have the interesting identification $ T^vert(V)=pi^-1(V)$.
This boils down to the fact that the tangent space at any point $ein E$ of a vector space $E$ is that vector space itself: $T_e(E)=E.$
Hence $(***)$ becomes $$ 0to pi^-1(V) to T(V) stackrel dpito pi^-1T(M) to 0 quad (****) $$
c) Finally, if you restrict this last exact sequence (****) to the zero section of $V$, identified to $M$, you get
$$ 0to V to T(V)|M stackrel dpito T(M) to 0 quad quad (*****) $$
d) Although this is not very helpful, you can if the manifold $M$ is paracompact (see here) non-canonically split the exact sequence (*****) and obtain the isomorphism $$ T(V)|M cong V oplus T(M)$$
At a point $(m,0)in Msubset V$ this gives the decomposition $$ T_(m,0)(V) cong V_m oplus T_m(M)$$
and for $V=T^*(M)$ this is what you were looking for.
answered Jul 25 '11 at 11:22
Georges ElencwajgGeorges Elencwajg
120k7182335
$endgroup$
Things are much nicer i.e. more general and canonical than that!
a) Let $pi:Nto M$ be a submersion of manifolds. From it you obtain the exact sequence of vector bundles on $N$: $$ 0to T^vert(N) to T(N) stackrel dpito pi^-1T(M) to 0 quad (*) $$ Profound, eh? Not at all!
This is just a fancy way of looking at the differential of the map $pi: Nto M.$
In order that both the tangent bundles to $N$ and to $M$ live on $N$, you have to pull back $T(M)$ to $N$: that is why we have $pi^-1 T(M)$ on the right.
The kernel is the set of vertical tangent vectors to $N$, those that lie along the level lines $N[n]stackrel def=pi^-1(pi (n))$ of $pi$.
In other words at a point $nin N$ the fibre of $T^vert(N)$ is $$T^vert_n(N)=T_n(N[n]) quad (**)$$
b) Consider now a vector bundle $pi:Vto M$ on $M$.
You then have the canonical exact sequence of vector bundles on $V;$ (yes, vector bundles on a vector bundle!) :
$$ 0to T^vert(V) to T(V) stackrel dpito pi^-1T(M) to 0 quad (***)$$
In this new set-up you have the interesting identification $ T^vert(V)=pi^-1(V)$.
This boils down to the fact that the tangent space at any point $ein E$ of a vector space $E$ is that vector space itself: $T_e(E)=E.$
Hence $(***)$ becomes $$ 0to pi^-1(V) to T(V) stackrel dpito pi^-1T(M) to 0 quad (****) $$
c) Finally, if you restrict this last exact sequence (****) to the zero section of $V$, identified to $M$, you get
$$ 0to V to T(V)|M stackrel dpito T(M) to 0 quad quad (*****) $$
d) Although this is not very helpful, you can if the manifold $M$ is paracompact (see here) non-canonically split the exact sequence (*****) and obtain the isomorphism $$ T(V)|M cong V oplus T(M)$$
At a point $(m,0)in Msubset V$ this gives the decomposition $$ T_(m,0)(V) cong V_m oplus T_m(M)$$
and for $V=T^*(M)$ this is what you were looking for.
answered Jul 25 '11 at 11:22
Georges ElencwajgGeorges Elencwajg
120k7182335
edited Mar 27 at 7:51
answered Jul 25 '11 at 11:22
Georges ElencwajgGeorges Elencwajg
120k7182335
answered Jul 25 '11 at 11:22
Georges ElencwajgGeorges Elencwajg
120k7182335
answered Jul 25 '11 at 11:22
Georges ElencwajgGeorges Elencwajg
120k7182335
120k7182335
$begingroup$
dear Georges, we are ok that more generally $TT^*M$ is naturally isomorphic to $pi^-1TMoplus pi^-1T^*M$ where $pi:T^*Mto M$ ?
$endgroup$
– epsilones
Mar 26 at 22:13
1
$begingroup$
Dear @epsilones: your comment is absolutely correct . I have edited my answer in order to address your question in a completely general context.
$endgroup$
– Georges Elencwajg
Mar 27 at 7:56
add a comment |
$begingroup$
dear Georges, we are ok that more generally $TT^*M$ is naturally isomorphic to $pi^-1TMoplus pi^-1T^*M$ where $pi:T^*Mto M$ ?
$endgroup$
– epsilones
Mar 26 at 22:13
1
$begingroup$
Dear @epsilones: your comment is absolutely correct . I have edited my answer in order to address your question in a completely general context.
$endgroup$
– Georges Elencwajg
Mar 27 at 7:56
$begingroup$
dear Georges, we are ok that more generally $TT^*M$ is naturally isomorphic to $pi^-1TMoplus pi^-1T^*M$ where $pi:T^*Mto M$ ?
$endgroup$
– epsilones
Mar 26 at 22:13
$begingroup$
dear Georges, we are ok that more generally $TT^*M$ is naturally isomorphic to $pi^-1TMoplus pi^-1T^*M$ where $pi:T^*Mto M$ ?
$endgroup$
– epsilones
Mar 26 at 22:13
1
1
$begingroup$
Dear @epsilones: your comment is absolutely correct . I have edited my answer in order to address your question in a completely general context.
$endgroup$
– Georges Elencwajg
Mar 27 at 7:56
$begingroup$
Dear @epsilones: your comment is absolutely correct . I have edited my answer in order to address your question in a completely general context.
$endgroup$
– Georges Elencwajg
Mar 27 at 7:56
add a comment |
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$begingroup$
Hint: look at the kernel of the derivative of the bundle projection. This gives you the desired decomposition.
$endgroup$
– t.b.
Jul 25 '11 at 10:49
$begingroup$
Your comment about the Hessian is also correct. This allows you to say, for instance, that a function is Morse independently of the choice of connection.
$endgroup$
– Sam Lisi
Dec 22 '11 at 13:00