Mixing
Projecting vector field from $mathbb{R}^n$ to $T^n$
$begingroup$
Let $T^n = S^1 times ...times S^1$. $T^n$ is Lie group and the map $pi:mathbb{R}^n to T^n$ defined by $pi(x_1,..x_n) = (e^{ix_1},..,e^{ix_n})$ gives a homomorphism between the additive group to multiplicative group $T^n$.
which vector fields on $mathbb{R}^n$ project to vector fields on
$T^n$ under $dpi$,
I'm having a hard time understanding this question.
I don't have good background in differentiable manifolds; I have the definition that a vector field $X$ on a Lie group $G$ is a linear map $X: C^{infty}(G) to C^{infty}(G)$ s.t $X(fg) = fX(g) + gX(f)$.
My question:
What is formally meant by $dpi$ and how does it send a vector field to another?
I know that if $X$ is a vector field on $mathbb{R}^n$, and $x in mathbb{R}^n$ then we have a tangent vector $X_x: C^{infty}(mathbb{R}^n) to mathbb{R}$ where $X_x(f) = (Xf)(x)$. Then $dpi_x(X_x)(f) = X_x(f circ pi)$ is a tangent vector itself.
That is, $dpi_x : T_xmathbb{R}^n to T_{pi(x)}T^n$. So how does $dpi$ operate on vector fields, and what do I need to check in this case?
Thanks!
manifolds lie-groups smooth-manifolds vector-fields
asked Jan 21 at 21:41
MariahMariah
1,5561718
$endgroup$
add a comment |
$begingroup$
Let $T^n = S^1 times ...times S^1$. $T^n$ is Lie group and the map $pi:mathbb{R}^n to T^n$ defined by $pi(x_1,..x_n) = (e^{ix_1},..,e^{ix_n})$ gives a homomorphism between the additive group to multiplicative group $T^n$.
which vector fields on $mathbb{R}^n$ project to vector fields on
$T^n$ under $dpi$,
I'm having a hard time understanding this question.
I don't have good background in differentiable manifolds; I have the definition that a vector field $X$ on a Lie group $G$ is a linear map $X: C^{infty}(G) to C^{infty}(G)$ s.t $X(fg) = fX(g) + gX(f)$.
My question:
What is formally meant by $dpi$ and how does it send a vector field to another?
I know that if $X$ is a vector field on $mathbb{R}^n$, and $x in mathbb{R}^n$ then we have a tangent vector $X_x: C^{infty}(mathbb{R}^n) to mathbb{R}$ where $X_x(f) = (Xf)(x)$. Then $dpi_x(X_x)(f) = X_x(f circ pi)$ is a tangent vector itself.
That is, $dpi_x : T_xmathbb{R}^n to T_{pi(x)}T^n$. So how does $dpi$ operate on vector fields, and what do I need to check in this case?
Thanks!
manifolds lie-groups smooth-manifolds vector-fields
asked Jan 21 at 21:41
MariahMariah
1,5561718
$endgroup$
$begingroup$
This question is more easily understood if you think of vector fields as sections of the tangent bundle; do you know this point of view ?
$endgroup$
– Max
Jan 26 at 13:25
add a comment |
$begingroup$
Let $T^n = S^1 times ...times S^1$. $T^n$ is Lie group and the map $pi:mathbb{R}^n to T^n$ defined by $pi(x_1,..x_n) = (e^{ix_1},..,e^{ix_n})$ gives a homomorphism between the additive group to multiplicative group $T^n$.
which vector fields on $mathbb{R}^n$ project to vector fields on
$T^n$ under $dpi$,
I'm having a hard time understanding this question.
I don't have good background in differentiable manifolds; I have the definition that a vector field $X$ on a Lie group $G$ is a linear map $X: C^{infty}(G) to C^{infty}(G)$ s.t $X(fg) = fX(g) + gX(f)$.
My question:
What is formally meant by $dpi$ and how does it send a vector field to another?
I know that if $X$ is a vector field on $mathbb{R}^n$, and $x in mathbb{R}^n$ then we have a tangent vector $X_x: C^{infty}(mathbb{R}^n) to mathbb{R}$ where $X_x(f) = (Xf)(x)$. Then $dpi_x(X_x)(f) = X_x(f circ pi)$ is a tangent vector itself.
That is, $dpi_x : T_xmathbb{R}^n to T_{pi(x)}T^n$. So how does $dpi$ operate on vector fields, and what do I need to check in this case?
Thanks!
manifolds lie-groups smooth-manifolds vector-fields
asked Jan 21 at 21:41
MariahMariah
1,5561718
$endgroup$
Let $T^n = S^1 times ...times S^1$. $T^n$ is Lie group and the map $pi:mathbb{R}^n to T^n$ defined by $pi(x_1,..x_n) = (e^{ix_1},..,e^{ix_n})$ gives a homomorphism between the additive group to multiplicative group $T^n$.
which vector fields on $mathbb{R}^n$ project to vector fields on
$T^n$ under $dpi$,
I'm having a hard time understanding this question.
I don't have good background in differentiable manifolds; I have the definition that a vector field $X$ on a Lie group $G$ is a linear map $X: C^{infty}(G) to C^{infty}(G)$ s.t $X(fg) = fX(g) + gX(f)$.
My question:
What is formally meant by $dpi$ and how does it send a vector field to another?
I know that if $X$ is a vector field on $mathbb{R}^n$, and $x in mathbb{R}^n$ then we have a tangent vector $X_x: C^{infty}(mathbb{R}^n) to mathbb{R}$ where $X_x(f) = (Xf)(x)$. Then $dpi_x(X_x)(f) = X_x(f circ pi)$ is a tangent vector itself.
That is, $dpi_x : T_xmathbb{R}^n to T_{pi(x)}T^n$. So how does $dpi$ operate on vector fields, and what do I need to check in this case?
Thanks!
manifolds lie-groups smooth-manifolds vector-fields
manifolds lie-groups smooth-manifolds vector-fields
asked Jan 21 at 21:41
MariahMariah
1,5561718
asked Jan 21 at 21:41
MariahMariah
1,5561718
edited Jan 26 at 12:53
Mariah
asked Jan 21 at 21:41
MariahMariah
1,5561718
asked Jan 21 at 21:41
MariahMariah
1,5561718
asked Jan 21 at 21:41
MariahMariah
1,5561718
1,5561718
$begingroup$
This question is more easily understood if you think of vector fields as sections of the tangent bundle; do you know this point of view ?
$endgroup$
– Max
Jan 26 at 13:25
add a comment |
$begingroup$
This question is more easily understood if you think of vector fields as sections of the tangent bundle; do you know this point of view ?
$endgroup$
– Max
Jan 26 at 13:25
$begingroup$
This question is more easily understood if you think of vector fields as sections of the tangent bundle; do you know this point of view ?
$endgroup$
– Max
Jan 26 at 13:25
$begingroup$
This question is more easily understood if you think of vector fields as sections of the tangent bundle; do you know this point of view ?
$endgroup$
– Max
Jan 26 at 13:25
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Functions on $mathbb{T}^n$ give functions on $mathbb{R}^n$, i.e. you have a natural map (precomposing by $pi$) $pi^*: C^infty(mathbb{T}^n)to C^infty(mathbb{R}^n)$.
The question is : for which $X: C^infty(mathbb{R}^n)to C^infty(mathbb{R}^n)$ is there $tilde{X} : C^infty(mathbb{T}^n)to C^infty(mathbb{T}^n)$ making the obvious diagram commute; that is such that $Xcirc pi^* = pi^*circ tilde{X}$ ?
Phrasing the question as "which vector fields on $mathbb{R}^n$ project to vector fields on $mathbb{T}^n$ under $mathrm{d}pi$ ?" suggests that the author was thinking of vector fields in terms of sections of the tangent bundle instead of as derivations of the function algebra.
Indeed, for a manifold $M$ you have a tangent bundle $p:TMto M$, and a smooth map $f:Mto N$ induces $mathrm{d}f : TMto TN$ defined by $mathrm{d}f (v,x) = mathrm{d}f_x(v)$. We call a vector field on $M$(this is where the terminology comes from by the way) a section of $p$, that is a map $s: Mto TM$ such that $pcirc s= id_M$. This last condition simply means that for all $x$, $s(x)in T_xM$, that is $s$ is a smooth map that associates to each point a tangent vector.
Now if you are given a vector field $X$ according to this second definition, you get a derivation of $C^infty(M)$ the following way : given $f:Mto mathbb{R}$, define $mathcal{L}_Xf : Mto mathbb{R}$ by $mathcal{L}_Xf (x) = mathrm{d}f(X(x),x)$. This is clearly a smooth function, and since you have a natural identification $Tmathbb{R}cong mathbb{Rtimes R}$, under this natural identification, you get that $mathcal{L}_Xf : Mto mathbb{R}$ is smooth, and it's a standard fact that $mathcal{L}_X$ is a derivation of $C^infty(M)$, it's called the Lie derivative along $X$.
Ok now it's a theorem that the map $mathcal{L}: Gamma(M,TM) to mathrm{Der}(C^infty(M))$ is a Lie algebra isomorphism (well actually, a vector space isomorphism, which allows you to give $Gamma(M,TM)$ the structure of Lie algebra deduced from the one on $mathrm{Der}(C^infty(M))$); where $Gamma(M,TM)$ denotes the sections of the tangent bundle, and $mathrm{Der}(A)$ is the Lie algebra of derivations of the algebra $A$.
This allows you to see a vector field as either a derivation (as you defined it) or as a section of the tangent bundle. Now with the point of view of sections, the question becomes easier to understand with this phrasing.
Indeed you have a diagram $require{AMScd} begin{CD}
Tmathbb{R}^n @>{mathrm{d}pi}>> Tmathbb{T}^n\ @V{p}VV @VV{p}V\
mathbb{R}^n @>>{pi}> mathbb{T}^n
end{CD}$ where $mathrm{d}pi$ appears explicitly. Now if you are given a section $X$ on the right side, you may be interested in finding out when there is a section $tilde{X}$ on the left side which makes this commute :
$require{AMScd} begin{CD}
mathbb{R}^n @>{pi}>> mathbb{T}^n\ @V{X}VV @VV{tilde{X}}V\
Tmathbb{R}^n @>>{mathrm{d}pi}> Tmathbb{T}^n
end{CD}$, that is "when is there $tilde{X}$ that lifts $mathrm{d}pi circ X$; that is the projection of $X$ under $mathrm{d}pi$".
So "project under $mathrm{d}pi$" now makes sense !
So you're asking for $tilde{X}$ such that $mathrm{d}picirc X = tilde{X}circ pi$. By some usual manifold nonsense, this is the same as saying that for any $f:mathbb{T}^n to mathbb{R}$, and any $xinmathbb{R}^n$, $mathrm{d}f_{pi(x)}circ mathrm{d}pi_x(X(x)) = mathrm{d}f_{pi(x)}(tilde{X}(pi(x)))$; but the LHS is just $mathrm{d}(fcircpi)_x(X(x)) = mathcal{L}_X(fcircpi)(x)$ and the RHS is just $(mathcal{L}_{tilde{X}}f)(pi(x))$.
Asking that they're equal for all $f,x$, is just asking that $mathcal{L}_Xcircpi^* = pi^*circmathcal{L}_{tilde{X}}$; which, if you use the identification between sections and derivations, is exactly the equation I gave at the very beginning.
Now I don't know if you also wanted a hint to solve the question, but I think it's way easier to understand what's happening with vector fields seen as sections rather than as derivations, it's way more geometric and the condition for a vector field to project to another one under $mathrm{d}pi$ becomes clear.
answered Jan 26 at 15:09
MaxMax
15.1k11143
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Functions on $mathbb{T}^n$ give functions on $mathbb{R}^n$, i.e. you have a natural map (precomposing by $pi$) $pi^*: C^infty(mathbb{T}^n)to C^infty(mathbb{R}^n)$.
The question is : for which $X: C^infty(mathbb{R}^n)to C^infty(mathbb{R}^n)$ is there $tilde{X} : C^infty(mathbb{T}^n)to C^infty(mathbb{T}^n)$ making the obvious diagram commute; that is such that $Xcirc pi^* = pi^*circ tilde{X}$ ?
Phrasing the question as "which vector fields on $mathbb{R}^n$ project to vector fields on $mathbb{T}^n$ under $mathrm{d}pi$ ?" suggests that the author was thinking of vector fields in terms of sections of the tangent bundle instead of as derivations of the function algebra.
Indeed, for a manifold $M$ you have a tangent bundle $p:TMto M$, and a smooth map $f:Mto N$ induces $mathrm{d}f : TMto TN$ defined by $mathrm{d}f (v,x) = mathrm{d}f_x(v)$. We call a vector field on $M$(this is where the terminology comes from by the way) a section of $p$, that is a map $s: Mto TM$ such that $pcirc s= id_M$. This last condition simply means that for all $x$, $s(x)in T_xM$, that is $s$ is a smooth map that associates to each point a tangent vector.
Now if you are given a vector field $X$ according to this second definition, you get a derivation of $C^infty(M)$ the following way : given $f:Mto mathbb{R}$, define $mathcal{L}_Xf : Mto mathbb{R}$ by $mathcal{L}_Xf (x) = mathrm{d}f(X(x),x)$. This is clearly a smooth function, and since you have a natural identification $Tmathbb{R}cong mathbb{Rtimes R}$, under this natural identification, you get that $mathcal{L}_Xf : Mto mathbb{R}$ is smooth, and it's a standard fact that $mathcal{L}_X$ is a derivation of $C^infty(M)$, it's called the Lie derivative along $X$.
Ok now it's a theorem that the map $mathcal{L}: Gamma(M,TM) to mathrm{Der}(C^infty(M))$ is a Lie algebra isomorphism (well actually, a vector space isomorphism, which allows you to give $Gamma(M,TM)$ the structure of Lie algebra deduced from the one on $mathrm{Der}(C^infty(M))$); where $Gamma(M,TM)$ denotes the sections of the tangent bundle, and $mathrm{Der}(A)$ is the Lie algebra of derivations of the algebra $A$.
This allows you to see a vector field as either a derivation (as you defined it) or as a section of the tangent bundle. Now with the point of view of sections, the question becomes easier to understand with this phrasing.
Indeed you have a diagram $require{AMScd} begin{CD}
Tmathbb{R}^n @>{mathrm{d}pi}>> Tmathbb{T}^n\ @V{p}VV @VV{p}V\
mathbb{R}^n @>>{pi}> mathbb{T}^n
end{CD}$ where $mathrm{d}pi$ appears explicitly. Now if you are given a section $X$ on the right side, you may be interested in finding out when there is a section $tilde{X}$ on the left side which makes this commute :
$require{AMScd} begin{CD}
mathbb{R}^n @>{pi}>> mathbb{T}^n\ @V{X}VV @VV{tilde{X}}V\
Tmathbb{R}^n @>>{mathrm{d}pi}> Tmathbb{T}^n
end{CD}$, that is "when is there $tilde{X}$ that lifts $mathrm{d}pi circ X$; that is the projection of $X$ under $mathrm{d}pi$".
So "project under $mathrm{d}pi$" now makes sense !
So you're asking for $tilde{X}$ such that $mathrm{d}picirc X = tilde{X}circ pi$. By some usual manifold nonsense, this is the same as saying that for any $f:mathbb{T}^n to mathbb{R}$, and any $xinmathbb{R}^n$, $mathrm{d}f_{pi(x)}circ mathrm{d}pi_x(X(x)) = mathrm{d}f_{pi(x)}(tilde{X}(pi(x)))$; but the LHS is just $mathrm{d}(fcircpi)_x(X(x)) = mathcal{L}_X(fcircpi)(x)$ and the RHS is just $(mathcal{L}_{tilde{X}}f)(pi(x))$.
Asking that they're equal for all $f,x$, is just asking that $mathcal{L}_Xcircpi^* = pi^*circmathcal{L}_{tilde{X}}$; which, if you use the identification between sections and derivations, is exactly the equation I gave at the very beginning.
Now I don't know if you also wanted a hint to solve the question, but I think it's way easier to understand what's happening with vector fields seen as sections rather than as derivations, it's way more geometric and the condition for a vector field to project to another one under $mathrm{d}pi$ becomes clear.
answered Jan 26 at 15:09
MaxMax
15.1k11143
$endgroup$
add a comment |
$begingroup$
Functions on $mathbb{T}^n$ give functions on $mathbb{R}^n$, i.e. you have a natural map (precomposing by $pi$) $pi^*: C^infty(mathbb{T}^n)to C^infty(mathbb{R}^n)$.
The question is : for which $X: C^infty(mathbb{R}^n)to C^infty(mathbb{R}^n)$ is there $tilde{X} : C^infty(mathbb{T}^n)to C^infty(mathbb{T}^n)$ making the obvious diagram commute; that is such that $Xcirc pi^* = pi^*circ tilde{X}$ ?
Phrasing the question as "which vector fields on $mathbb{R}^n$ project to vector fields on $mathbb{T}^n$ under $mathrm{d}pi$ ?" suggests that the author was thinking of vector fields in terms of sections of the tangent bundle instead of as derivations of the function algebra.
Indeed, for a manifold $M$ you have a tangent bundle $p:TMto M$, and a smooth map $f:Mto N$ induces $mathrm{d}f : TMto TN$ defined by $mathrm{d}f (v,x) = mathrm{d}f_x(v)$. We call a vector field on $M$(this is where the terminology comes from by the way) a section of $p$, that is a map $s: Mto TM$ such that $pcirc s= id_M$. This last condition simply means that for all $x$, $s(x)in T_xM$, that is $s$ is a smooth map that associates to each point a tangent vector.
Now if you are given a vector field $X$ according to this second definition, you get a derivation of $C^infty(M)$ the following way : given $f:Mto mathbb{R}$, define $mathcal{L}_Xf : Mto mathbb{R}$ by $mathcal{L}_Xf (x) = mathrm{d}f(X(x),x)$. This is clearly a smooth function, and since you have a natural identification $Tmathbb{R}cong mathbb{Rtimes R}$, under this natural identification, you get that $mathcal{L}_Xf : Mto mathbb{R}$ is smooth, and it's a standard fact that $mathcal{L}_X$ is a derivation of $C^infty(M)$, it's called the Lie derivative along $X$.
Ok now it's a theorem that the map $mathcal{L}: Gamma(M,TM) to mathrm{Der}(C^infty(M))$ is a Lie algebra isomorphism (well actually, a vector space isomorphism, which allows you to give $Gamma(M,TM)$ the structure of Lie algebra deduced from the one on $mathrm{Der}(C^infty(M))$); where $Gamma(M,TM)$ denotes the sections of the tangent bundle, and $mathrm{Der}(A)$ is the Lie algebra of derivations of the algebra $A$.
This allows you to see a vector field as either a derivation (as you defined it) or as a section of the tangent bundle. Now with the point of view of sections, the question becomes easier to understand with this phrasing.
Indeed you have a diagram $require{AMScd} begin{CD}
Tmathbb{R}^n @>{mathrm{d}pi}>> Tmathbb{T}^n\ @V{p}VV @VV{p}V\
mathbb{R}^n @>>{pi}> mathbb{T}^n
end{CD}$ where $mathrm{d}pi$ appears explicitly. Now if you are given a section $X$ on the right side, you may be interested in finding out when there is a section $tilde{X}$ on the left side which makes this commute :
$require{AMScd} begin{CD}
mathbb{R}^n @>{pi}>> mathbb{T}^n\ @V{X}VV @VV{tilde{X}}V\
Tmathbb{R}^n @>>{mathrm{d}pi}> Tmathbb{T}^n
end{CD}$, that is "when is there $tilde{X}$ that lifts $mathrm{d}pi circ X$; that is the projection of $X$ under $mathrm{d}pi$".
So "project under $mathrm{d}pi$" now makes sense !
So you're asking for $tilde{X}$ such that $mathrm{d}picirc X = tilde{X}circ pi$. By some usual manifold nonsense, this is the same as saying that for any $f:mathbb{T}^n to mathbb{R}$, and any $xinmathbb{R}^n$, $mathrm{d}f_{pi(x)}circ mathrm{d}pi_x(X(x)) = mathrm{d}f_{pi(x)}(tilde{X}(pi(x)))$; but the LHS is just $mathrm{d}(fcircpi)_x(X(x)) = mathcal{L}_X(fcircpi)(x)$ and the RHS is just $(mathcal{L}_{tilde{X}}f)(pi(x))$.
Asking that they're equal for all $f,x$, is just asking that $mathcal{L}_Xcircpi^* = pi^*circmathcal{L}_{tilde{X}}$; which, if you use the identification between sections and derivations, is exactly the equation I gave at the very beginning.
Now I don't know if you also wanted a hint to solve the question, but I think it's way easier to understand what's happening with vector fields seen as sections rather than as derivations, it's way more geometric and the condition for a vector field to project to another one under $mathrm{d}pi$ becomes clear.
answered Jan 26 at 15:09
MaxMax
15.1k11143
$endgroup$
add a comment |
$begingroup$
Functions on $mathbb{T}^n$ give functions on $mathbb{R}^n$, i.e. you have a natural map (precomposing by $pi$) $pi^*: C^infty(mathbb{T}^n)to C^infty(mathbb{R}^n)$.
The question is : for which $X: C^infty(mathbb{R}^n)to C^infty(mathbb{R}^n)$ is there $tilde{X} : C^infty(mathbb{T}^n)to C^infty(mathbb{T}^n)$ making the obvious diagram commute; that is such that $Xcirc pi^* = pi^*circ tilde{X}$ ?
Phrasing the question as "which vector fields on $mathbb{R}^n$ project to vector fields on $mathbb{T}^n$ under $mathrm{d}pi$ ?" suggests that the author was thinking of vector fields in terms of sections of the tangent bundle instead of as derivations of the function algebra.
Indeed, for a manifold $M$ you have a tangent bundle $p:TMto M$, and a smooth map $f:Mto N$ induces $mathrm{d}f : TMto TN$ defined by $mathrm{d}f (v,x) = mathrm{d}f_x(v)$. We call a vector field on $M$(this is where the terminology comes from by the way) a section of $p$, that is a map $s: Mto TM$ such that $pcirc s= id_M$. This last condition simply means that for all $x$, $s(x)in T_xM$, that is $s$ is a smooth map that associates to each point a tangent vector.
Now if you are given a vector field $X$ according to this second definition, you get a derivation of $C^infty(M)$ the following way : given $f:Mto mathbb{R}$, define $mathcal{L}_Xf : Mto mathbb{R}$ by $mathcal{L}_Xf (x) = mathrm{d}f(X(x),x)$. This is clearly a smooth function, and since you have a natural identification $Tmathbb{R}cong mathbb{Rtimes R}$, under this natural identification, you get that $mathcal{L}_Xf : Mto mathbb{R}$ is smooth, and it's a standard fact that $mathcal{L}_X$ is a derivation of $C^infty(M)$, it's called the Lie derivative along $X$.
Ok now it's a theorem that the map $mathcal{L}: Gamma(M,TM) to mathrm{Der}(C^infty(M))$ is a Lie algebra isomorphism (well actually, a vector space isomorphism, which allows you to give $Gamma(M,TM)$ the structure of Lie algebra deduced from the one on $mathrm{Der}(C^infty(M))$); where $Gamma(M,TM)$ denotes the sections of the tangent bundle, and $mathrm{Der}(A)$ is the Lie algebra of derivations of the algebra $A$.
This allows you to see a vector field as either a derivation (as you defined it) or as a section of the tangent bundle. Now with the point of view of sections, the question becomes easier to understand with this phrasing.
Indeed you have a diagram $require{AMScd} begin{CD}
Tmathbb{R}^n @>{mathrm{d}pi}>> Tmathbb{T}^n\ @V{p}VV @VV{p}V\
mathbb{R}^n @>>{pi}> mathbb{T}^n
end{CD}$ where $mathrm{d}pi$ appears explicitly. Now if you are given a section $X$ on the right side, you may be interested in finding out when there is a section $tilde{X}$ on the left side which makes this commute :
$require{AMScd} begin{CD}
mathbb{R}^n @>{pi}>> mathbb{T}^n\ @V{X}VV @VV{tilde{X}}V\
Tmathbb{R}^n @>>{mathrm{d}pi}> Tmathbb{T}^n
end{CD}$, that is "when is there $tilde{X}$ that lifts $mathrm{d}pi circ X$; that is the projection of $X$ under $mathrm{d}pi$".
So "project under $mathrm{d}pi$" now makes sense !
So you're asking for $tilde{X}$ such that $mathrm{d}picirc X = tilde{X}circ pi$. By some usual manifold nonsense, this is the same as saying that for any $f:mathbb{T}^n to mathbb{R}$, and any $xinmathbb{R}^n$, $mathrm{d}f_{pi(x)}circ mathrm{d}pi_x(X(x)) = mathrm{d}f_{pi(x)}(tilde{X}(pi(x)))$; but the LHS is just $mathrm{d}(fcircpi)_x(X(x)) = mathcal{L}_X(fcircpi)(x)$ and the RHS is just $(mathcal{L}_{tilde{X}}f)(pi(x))$.
Asking that they're equal for all $f,x$, is just asking that $mathcal{L}_Xcircpi^* = pi^*circmathcal{L}_{tilde{X}}$; which, if you use the identification between sections and derivations, is exactly the equation I gave at the very beginning.
Now I don't know if you also wanted a hint to solve the question, but I think it's way easier to understand what's happening with vector fields seen as sections rather than as derivations, it's way more geometric and the condition for a vector field to project to another one under $mathrm{d}pi$ becomes clear.
answered Jan 26 at 15:09
MaxMax
15.1k11143
$endgroup$
Functions on $mathbb{T}^n$ give functions on $mathbb{R}^n$, i.e. you have a natural map (precomposing by $pi$) $pi^*: C^infty(mathbb{T}^n)to C^infty(mathbb{R}^n)$.
The question is : for which $X: C^infty(mathbb{R}^n)to C^infty(mathbb{R}^n)$ is there $tilde{X} : C^infty(mathbb{T}^n)to C^infty(mathbb{T}^n)$ making the obvious diagram commute; that is such that $Xcirc pi^* = pi^*circ tilde{X}$ ?
Phrasing the question as "which vector fields on $mathbb{R}^n$ project to vector fields on $mathbb{T}^n$ under $mathrm{d}pi$ ?" suggests that the author was thinking of vector fields in terms of sections of the tangent bundle instead of as derivations of the function algebra.
Indeed, for a manifold $M$ you have a tangent bundle $p:TMto M$, and a smooth map $f:Mto N$ induces $mathrm{d}f : TMto TN$ defined by $mathrm{d}f (v,x) = mathrm{d}f_x(v)$. We call a vector field on $M$(this is where the terminology comes from by the way) a section of $p$, that is a map $s: Mto TM$ such that $pcirc s= id_M$. This last condition simply means that for all $x$, $s(x)in T_xM$, that is $s$ is a smooth map that associates to each point a tangent vector.
Now if you are given a vector field $X$ according to this second definition, you get a derivation of $C^infty(M)$ the following way : given $f:Mto mathbb{R}$, define $mathcal{L}_Xf : Mto mathbb{R}$ by $mathcal{L}_Xf (x) = mathrm{d}f(X(x),x)$. This is clearly a smooth function, and since you have a natural identification $Tmathbb{R}cong mathbb{Rtimes R}$, under this natural identification, you get that $mathcal{L}_Xf : Mto mathbb{R}$ is smooth, and it's a standard fact that $mathcal{L}_X$ is a derivation of $C^infty(M)$, it's called the Lie derivative along $X$.
Ok now it's a theorem that the map $mathcal{L}: Gamma(M,TM) to mathrm{Der}(C^infty(M))$ is a Lie algebra isomorphism (well actually, a vector space isomorphism, which allows you to give $Gamma(M,TM)$ the structure of Lie algebra deduced from the one on $mathrm{Der}(C^infty(M))$); where $Gamma(M,TM)$ denotes the sections of the tangent bundle, and $mathrm{Der}(A)$ is the Lie algebra of derivations of the algebra $A$.
This allows you to see a vector field as either a derivation (as you defined it) or as a section of the tangent bundle. Now with the point of view of sections, the question becomes easier to understand with this phrasing.
Indeed you have a diagram $require{AMScd} begin{CD}
Tmathbb{R}^n @>{mathrm{d}pi}>> Tmathbb{T}^n\ @V{p}VV @VV{p}V\
mathbb{R}^n @>>{pi}> mathbb{T}^n
end{CD}$ where $mathrm{d}pi$ appears explicitly. Now if you are given a section $X$ on the right side, you may be interested in finding out when there is a section $tilde{X}$ on the left side which makes this commute :
$require{AMScd} begin{CD}
mathbb{R}^n @>{pi}>> mathbb{T}^n\ @V{X}VV @VV{tilde{X}}V\
Tmathbb{R}^n @>>{mathrm{d}pi}> Tmathbb{T}^n
end{CD}$, that is "when is there $tilde{X}$ that lifts $mathrm{d}pi circ X$; that is the projection of $X$ under $mathrm{d}pi$".
So "project under $mathrm{d}pi$" now makes sense !
So you're asking for $tilde{X}$ such that $mathrm{d}picirc X = tilde{X}circ pi$. By some usual manifold nonsense, this is the same as saying that for any $f:mathbb{T}^n to mathbb{R}$, and any $xinmathbb{R}^n$, $mathrm{d}f_{pi(x)}circ mathrm{d}pi_x(X(x)) = mathrm{d}f_{pi(x)}(tilde{X}(pi(x)))$; but the LHS is just $mathrm{d}(fcircpi)_x(X(x)) = mathcal{L}_X(fcircpi)(x)$ and the RHS is just $(mathcal{L}_{tilde{X}}f)(pi(x))$.
Asking that they're equal for all $f,x$, is just asking that $mathcal{L}_Xcircpi^* = pi^*circmathcal{L}_{tilde{X}}$; which, if you use the identification between sections and derivations, is exactly the equation I gave at the very beginning.
Now I don't know if you also wanted a hint to solve the question, but I think it's way easier to understand what's happening with vector fields seen as sections rather than as derivations, it's way more geometric and the condition for a vector field to project to another one under $mathrm{d}pi$ becomes clear.
answered Jan 26 at 15:09
MaxMax
15.1k11143
answered Jan 26 at 15:09
MaxMax
15.1k11143
answered Jan 26 at 15:09
MaxMax
15.1k11143
answered Jan 26 at 15:09
MaxMax
15.1k11143
15.1k11143
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$begingroup$
This question is more easily understood if you think of vector fields as sections of the tangent bundle; do you know this point of view ?
$endgroup$
– Max
Jan 26 at 13:25