Mixing
$df_1,…,df_k$ linearly independent $Rightarrow fracomega^nn!=df_1wedge…wedge df_kwedgesigma$
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Let $(M,omega)$ be a symplectic manifold, $H,f_1,...,f_kin C^infty(M)$ non-zero functions with $H,f_i=0$. If $cinmathbbR^k$ is a regular value of $F:=(f_1,...,f_k):Mto mathbbR^k$, consider the submanifold $M_c:=F^-1(c)$.
a) Let $U$ be a neighbourhood of $M_c$ in which $df_1,...,df_k$ are linearly independent. Show that $Lambda_omega:=fracomega^nn!$ can be written as $Lambda_omega=df_1wedge...wedge df_kwedge sigma$ for some $sigmainOmega^2n-k(M)$. [hint: find $sigma$ locally and use partitions of unity]
b) Show that $df_1wedge...wedge df_kwedge L_X_H(sigma)=0$ and use this fact to see that $L_X_H(sigma)$ can be written as $L_X_H(sigma)=sum_i=1^kdf_iwedge rho_i$. Conclude that $Lambda_c:=i^*sigma$ is invariant by the flow of $H$ (where $i:M_chookrightarrow M$ is the inclusion).
c) Show that $Lambda_c$ does not depend on the choice of $sigma$.
Here is where I'm at:
a) Taking Darboux coordinates $(x_1,...,x_n,y_1,...,y_n)$ and considering $x_n+i:=y_i$, we have $Lambda_omega=dx_1wedge...wedge dx_2n$, $df_i=sum_j=1^2nfracpartial f_ipartial x_jdx_j$. Consequently
$$df_1wedge...wedge df_k=sum_1leq j_1<...<j_kleq 2ndet(M_j_1,...,j_k),dx_j_1wedge...wedge dx_j_k
$$
where $M_j_1,...,j_k$ are $ktimes k$ minors of the matrix $left(fracpartial f_ipartial x_jright)_i=1,...,k,j=1,...,2n$ given by the columns $j_1,...,j_k$. Taking $p$ with $(df_1)_p,...,(df_k)_p$ linearly independent, we may assume $det(M_1,...,k)neq 0$, so
$$sigma:=frac1det(M_1,...,k)dx_k+1wedge...wedge dx_2n$$
is such that $df_1wedge...wedge df_kwedge sigma=dx_1wedge...wedge dx_2n=Lambda_omega$. The problem is that this works for a neighbourhood $Vsubset M$ with $Vcap Uneqemptyset$, but I don't know how to extend it for the whole $U$. I don't get the hint, because introducing a partition $U_alpha, rho_alpha$ and defining $sigma$ as a $rho_alpha$-linear combination may break the equality with $Lambda_omega$.
b) I can show that $df_1wedge...wedge df_kwedge L_X_H(sigma)=0$, but I don't see how to use this to prove $sigma$ can be written that way. Besides I don't know what it means for $Lambda_c$ to be invariant by the flow of $H$.
c) Taking another $sigma'$ with the same property, we need to prove
$$i^*(df_1)wedge...wedge i^*(df_k)wedge i^*(sigma-sigma')=0$$
but I don't know how to deal with the pullbacks.
differential-geometry differential-forms symplectic-geometry
asked Jul 19 at 23:15
rmdmc89
1,7311520
add a comment |Â
up vote
3
down vote
favorite
Let $(M,omega)$ be a symplectic manifold, $H,f_1,...,f_kin C^infty(M)$ non-zero functions with $H,f_i=0$. If $cinmathbbR^k$ is a regular value of $F:=(f_1,...,f_k):Mto mathbbR^k$, consider the submanifold $M_c:=F^-1(c)$.
a) Let $U$ be a neighbourhood of $M_c$ in which $df_1,...,df_k$ are linearly independent. Show that $Lambda_omega:=fracomega^nn!$ can be written as $Lambda_omega=df_1wedge...wedge df_kwedge sigma$ for some $sigmainOmega^2n-k(M)$. [hint: find $sigma$ locally and use partitions of unity]
b) Show that $df_1wedge...wedge df_kwedge L_X_H(sigma)=0$ and use this fact to see that $L_X_H(sigma)$ can be written as $L_X_H(sigma)=sum_i=1^kdf_iwedge rho_i$. Conclude that $Lambda_c:=i^*sigma$ is invariant by the flow of $H$ (where $i:M_chookrightarrow M$ is the inclusion).
c) Show that $Lambda_c$ does not depend on the choice of $sigma$.
Here is where I'm at:
a) Taking Darboux coordinates $(x_1,...,x_n,y_1,...,y_n)$ and considering $x_n+i:=y_i$, we have $Lambda_omega=dx_1wedge...wedge dx_2n$, $df_i=sum_j=1^2nfracpartial f_ipartial x_jdx_j$. Consequently
$$df_1wedge...wedge df_k=sum_1leq j_1<...<j_kleq 2ndet(M_j_1,...,j_k),dx_j_1wedge...wedge dx_j_k
$$
where $M_j_1,...,j_k$ are $ktimes k$ minors of the matrix $left(fracpartial f_ipartial x_jright)_i=1,...,k,j=1,...,2n$ given by the columns $j_1,...,j_k$. Taking $p$ with $(df_1)_p,...,(df_k)_p$ linearly independent, we may assume $det(M_1,...,k)neq 0$, so
$$sigma:=frac1det(M_1,...,k)dx_k+1wedge...wedge dx_2n$$
is such that $df_1wedge...wedge df_kwedge sigma=dx_1wedge...wedge dx_2n=Lambda_omega$. The problem is that this works for a neighbourhood $Vsubset M$ with $Vcap Uneqemptyset$, but I don't know how to extend it for the whole $U$. I don't get the hint, because introducing a partition $U_alpha, rho_alpha$ and defining $sigma$ as a $rho_alpha$-linear combination may break the equality with $Lambda_omega$.
b) I can show that $df_1wedge...wedge df_kwedge L_X_H(sigma)=0$, but I don't see how to use this to prove $sigma$ can be written that way. Besides I don't know what it means for $Lambda_c$ to be invariant by the flow of $H$.
c) Taking another $sigma'$ with the same property, we need to prove
$$i^*(df_1)wedge...wedge i^*(df_k)wedge i^*(sigma-sigma')=0$$
but I don't know how to deal with the pullbacks.
differential-geometry differential-forms symplectic-geometry
asked Jul 19 at 23:15
rmdmc89
1,7311520
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let $(M,omega)$ be a symplectic manifold, $H,f_1,...,f_kin C^infty(M)$ non-zero functions with $H,f_i=0$. If $cinmathbbR^k$ is a regular value of $F:=(f_1,...,f_k):Mto mathbbR^k$, consider the submanifold $M_c:=F^-1(c)$.
a) Let $U$ be a neighbourhood of $M_c$ in which $df_1,...,df_k$ are linearly independent. Show that $Lambda_omega:=fracomega^nn!$ can be written as $Lambda_omega=df_1wedge...wedge df_kwedge sigma$ for some $sigmainOmega^2n-k(M)$. [hint: find $sigma$ locally and use partitions of unity]
b) Show that $df_1wedge...wedge df_kwedge L_X_H(sigma)=0$ and use this fact to see that $L_X_H(sigma)$ can be written as $L_X_H(sigma)=sum_i=1^kdf_iwedge rho_i$. Conclude that $Lambda_c:=i^*sigma$ is invariant by the flow of $H$ (where $i:M_chookrightarrow M$ is the inclusion).
c) Show that $Lambda_c$ does not depend on the choice of $sigma$.
Here is where I'm at:
a) Taking Darboux coordinates $(x_1,...,x_n,y_1,...,y_n)$ and considering $x_n+i:=y_i$, we have $Lambda_omega=dx_1wedge...wedge dx_2n$, $df_i=sum_j=1^2nfracpartial f_ipartial x_jdx_j$. Consequently
$$df_1wedge...wedge df_k=sum_1leq j_1<...<j_kleq 2ndet(M_j_1,...,j_k),dx_j_1wedge...wedge dx_j_k
$$
where $M_j_1,...,j_k$ are $ktimes k$ minors of the matrix $left(fracpartial f_ipartial x_jright)_i=1,...,k,j=1,...,2n$ given by the columns $j_1,...,j_k$. Taking $p$ with $(df_1)_p,...,(df_k)_p$ linearly independent, we may assume $det(M_1,...,k)neq 0$, so
$$sigma:=frac1det(M_1,...,k)dx_k+1wedge...wedge dx_2n$$
is such that $df_1wedge...wedge df_kwedge sigma=dx_1wedge...wedge dx_2n=Lambda_omega$. The problem is that this works for a neighbourhood $Vsubset M$ with $Vcap Uneqemptyset$, but I don't know how to extend it for the whole $U$. I don't get the hint, because introducing a partition $U_alpha, rho_alpha$ and defining $sigma$ as a $rho_alpha$-linear combination may break the equality with $Lambda_omega$.
b) I can show that $df_1wedge...wedge df_kwedge L_X_H(sigma)=0$, but I don't see how to use this to prove $sigma$ can be written that way. Besides I don't know what it means for $Lambda_c$ to be invariant by the flow of $H$.
c) Taking another $sigma'$ with the same property, we need to prove
$$i^*(df_1)wedge...wedge i^*(df_k)wedge i^*(sigma-sigma')=0$$
but I don't know how to deal with the pullbacks.
differential-geometry differential-forms symplectic-geometry
asked Jul 19 at 23:15
rmdmc89
1,7311520
Let $(M,omega)$ be a symplectic manifold, $H,f_1,...,f_kin C^infty(M)$ non-zero functions with $H,f_i=0$. If $cinmathbbR^k$ is a regular value of $F:=(f_1,...,f_k):Mto mathbbR^k$, consider the submanifold $M_c:=F^-1(c)$.
a) Let $U$ be a neighbourhood of $M_c$ in which $df_1,...,df_k$ are linearly independent. Show that $Lambda_omega:=fracomega^nn!$ can be written as $Lambda_omega=df_1wedge...wedge df_kwedge sigma$ for some $sigmainOmega^2n-k(M)$. [hint: find $sigma$ locally and use partitions of unity]
b) Show that $df_1wedge...wedge df_kwedge L_X_H(sigma)=0$ and use this fact to see that $L_X_H(sigma)$ can be written as $L_X_H(sigma)=sum_i=1^kdf_iwedge rho_i$. Conclude that $Lambda_c:=i^*sigma$ is invariant by the flow of $H$ (where $i:M_chookrightarrow M$ is the inclusion).
c) Show that $Lambda_c$ does not depend on the choice of $sigma$.
Here is where I'm at:
a) Taking Darboux coordinates $(x_1,...,x_n,y_1,...,y_n)$ and considering $x_n+i:=y_i$, we have $Lambda_omega=dx_1wedge...wedge dx_2n$, $df_i=sum_j=1^2nfracpartial f_ipartial x_jdx_j$. Consequently
$$df_1wedge...wedge df_k=sum_1leq j_1<...<j_kleq 2ndet(M_j_1,...,j_k),dx_j_1wedge...wedge dx_j_k
$$
where $M_j_1,...,j_k$ are $ktimes k$ minors of the matrix $left(fracpartial f_ipartial x_jright)_i=1,...,k,j=1,...,2n$ given by the columns $j_1,...,j_k$. Taking $p$ with $(df_1)_p,...,(df_k)_p$ linearly independent, we may assume $det(M_1,...,k)neq 0$, so
$$sigma:=frac1det(M_1,...,k)dx_k+1wedge...wedge dx_2n$$
is such that $df_1wedge...wedge df_kwedge sigma=dx_1wedge...wedge dx_2n=Lambda_omega$. The problem is that this works for a neighbourhood $Vsubset M$ with $Vcap Uneqemptyset$, but I don't know how to extend it for the whole $U$. I don't get the hint, because introducing a partition $U_alpha, rho_alpha$ and defining $sigma$ as a $rho_alpha$-linear combination may break the equality with $Lambda_omega$.
b) I can show that $df_1wedge...wedge df_kwedge L_X_H(sigma)=0$, but I don't see how to use this to prove $sigma$ can be written that way. Besides I don't know what it means for $Lambda_c$ to be invariant by the flow of $H$.
c) Taking another $sigma'$ with the same property, we need to prove
$$i^*(df_1)wedge...wedge i^*(df_k)wedge i^*(sigma-sigma')=0$$
but I don't know how to deal with the pullbacks.
differential-geometry differential-forms symplectic-geometry
asked Jul 19 at 23:15
rmdmc89
1,7311520
edited Jul 20 at 14:54
asked Jul 19 at 23:15
rmdmc89
1,7311520
asked Jul 19 at 23:15
rmdmc89
1,7311520
asked Jul 19 at 23:15
rmdmc89
1,7311520
1,7311520
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Here are some hints.
(a) This is not a symplectic problem, as $H$ is not involved and since $Lambda_omega$ could be replaced by any volume form $Lambda$ on $M$. This suggests that the use of Darboux coordinates might be unnatural, perhaps even irrelevant. Moreover, the $f_i$'s might be very 'unrelated' to any Darboux chart. Consequently it might be best to look for another argument.
One could use the constant rank theorem in order to work in a local chart $V$ with coordinates $(f_1, dots, f_k, g_1, dots, g_2n-k)$ where the $g_i$'s are (morally speaking) coordinates on $M_c$. It shouldn't be difficult to conclude to the existence of $sigma_V in Omega^2n-k(V)$ such that $left. Lambda right|_V = df_1 wedge dots wedge df_k wedge sigma_V$. Covering $U$ by charts $V_j$ and using a partition of unity $chi_j$, it shouldn't be difficult to see that $sigma := sigma_U := sum_j chi_j sigma_V_j$ does the job.
(b) Read $L_X_Hsigma$ in any of the charts $V_j$ and express it as a linear combination of wedges of the 1-forms $df_i$ and $dg_l$. Argue that each primitive summand of $L_X_Hsigma$ has to contain at least one $df_j$.
Next, for a form $F$ to be invariant under (the flow of) a vector field $X$ means that $L_X F = 0$ where $L_X$ is the Lie derivative. By extension, $F$ is invariant under the flow of a Hamiltonian function $H$ if it is invariant under that of $X_H$.
Here it is useful to recall that pullbacks and Lie derivatives commute, for example $L_X_H circ iota^* = iota^* circ L_X_H$. Apply both sides to $sigma$; you are interested in proving that the left-hand side vanishes, so compute the right-hand side by using the fact that the $f_j$'s are constant on $M_c$.
(c) If $sigma$ and $sigma'$ both do the job in part (a), you wish to prove that $iota^*(sigma - sigma') = 0$; so study the difference $sigma - sigma'$ in a similar way to what I suggested in part (b) for $L_X_Hsigma$.
answered Jul 20 at 16:41
Jordan Payette
2,681147
I have a question about the meaning of $L_X_Hcirc iota^*$. If $etainOmega^k(M)$, then $iota^*etainOmega^k(M_c)$. In order to apply $L_X_H$ to $iota^*eta$, it would be necessary that $left.X_Hright|_M_csubset mathfrakX(M_c)$, right? Otherwise how could we guarantee $phi^*(iota^*eta)$ is a form in $M_c$? (here $phi$ is the flow of $X_H$)
– rmdmc89
Jul 21 at 12:58
1
@rmdmc89 You are absolutely right about this, and we do have here that $left. X_H right|_M_c subset mathcalX(M_c)$. Indeed, we have in general $H, f_j = pm df_j(X_H)$ (where $pm$ is a matter of convention); since we assume $H, f_j = 0$, it follows that $X_H$ is everywhere tangent to the level sets of $f_j$. As this is true for all $j=1, dots, k$, it follows that $X_H$ is tangent to $M_c$.
– Jordan Payette
Jul 21 at 14:20
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Here are some hints.
(a) This is not a symplectic problem, as $H$ is not involved and since $Lambda_omega$ could be replaced by any volume form $Lambda$ on $M$. This suggests that the use of Darboux coordinates might be unnatural, perhaps even irrelevant. Moreover, the $f_i$'s might be very 'unrelated' to any Darboux chart. Consequently it might be best to look for another argument.
One could use the constant rank theorem in order to work in a local chart $V$ with coordinates $(f_1, dots, f_k, g_1, dots, g_2n-k)$ where the $g_i$'s are (morally speaking) coordinates on $M_c$. It shouldn't be difficult to conclude to the existence of $sigma_V in Omega^2n-k(V)$ such that $left. Lambda right|_V = df_1 wedge dots wedge df_k wedge sigma_V$. Covering $U$ by charts $V_j$ and using a partition of unity $chi_j$, it shouldn't be difficult to see that $sigma := sigma_U := sum_j chi_j sigma_V_j$ does the job.
(b) Read $L_X_Hsigma$ in any of the charts $V_j$ and express it as a linear combination of wedges of the 1-forms $df_i$ and $dg_l$. Argue that each primitive summand of $L_X_Hsigma$ has to contain at least one $df_j$.
Next, for a form $F$ to be invariant under (the flow of) a vector field $X$ means that $L_X F = 0$ where $L_X$ is the Lie derivative. By extension, $F$ is invariant under the flow of a Hamiltonian function $H$ if it is invariant under that of $X_H$.
Here it is useful to recall that pullbacks and Lie derivatives commute, for example $L_X_H circ iota^* = iota^* circ L_X_H$. Apply both sides to $sigma$; you are interested in proving that the left-hand side vanishes, so compute the right-hand side by using the fact that the $f_j$'s are constant on $M_c$.
(c) If $sigma$ and $sigma'$ both do the job in part (a), you wish to prove that $iota^*(sigma - sigma') = 0$; so study the difference $sigma - sigma'$ in a similar way to what I suggested in part (b) for $L_X_Hsigma$.
answered Jul 20 at 16:41
Jordan Payette
2,681147
I have a question about the meaning of $L_X_Hcirc iota^*$. If $etainOmega^k(M)$, then $iota^*etainOmega^k(M_c)$. In order to apply $L_X_H$ to $iota^*eta$, it would be necessary that $left.X_Hright|_M_csubset mathfrakX(M_c)$, right? Otherwise how could we guarantee $phi^*(iota^*eta)$ is a form in $M_c$? (here $phi$ is the flow of $X_H$)
– rmdmc89
Jul 21 at 12:58
1
@rmdmc89 You are absolutely right about this, and we do have here that $left. X_H right|_M_c subset mathcalX(M_c)$. Indeed, we have in general $H, f_j = pm df_j(X_H)$ (where $pm$ is a matter of convention); since we assume $H, f_j = 0$, it follows that $X_H$ is everywhere tangent to the level sets of $f_j$. As this is true for all $j=1, dots, k$, it follows that $X_H$ is tangent to $M_c$.
– Jordan Payette
Jul 21 at 14:20
add a comment |Â
up vote
1
down vote
accepted
Here are some hints.
(a) This is not a symplectic problem, as $H$ is not involved and since $Lambda_omega$ could be replaced by any volume form $Lambda$ on $M$. This suggests that the use of Darboux coordinates might be unnatural, perhaps even irrelevant. Moreover, the $f_i$'s might be very 'unrelated' to any Darboux chart. Consequently it might be best to look for another argument.
One could use the constant rank theorem in order to work in a local chart $V$ with coordinates $(f_1, dots, f_k, g_1, dots, g_2n-k)$ where the $g_i$'s are (morally speaking) coordinates on $M_c$. It shouldn't be difficult to conclude to the existence of $sigma_V in Omega^2n-k(V)$ such that $left. Lambda right|_V = df_1 wedge dots wedge df_k wedge sigma_V$. Covering $U$ by charts $V_j$ and using a partition of unity $chi_j$, it shouldn't be difficult to see that $sigma := sigma_U := sum_j chi_j sigma_V_j$ does the job.
(b) Read $L_X_Hsigma$ in any of the charts $V_j$ and express it as a linear combination of wedges of the 1-forms $df_i$ and $dg_l$. Argue that each primitive summand of $L_X_Hsigma$ has to contain at least one $df_j$.
Next, for a form $F$ to be invariant under (the flow of) a vector field $X$ means that $L_X F = 0$ where $L_X$ is the Lie derivative. By extension, $F$ is invariant under the flow of a Hamiltonian function $H$ if it is invariant under that of $X_H$.
Here it is useful to recall that pullbacks and Lie derivatives commute, for example $L_X_H circ iota^* = iota^* circ L_X_H$. Apply both sides to $sigma$; you are interested in proving that the left-hand side vanishes, so compute the right-hand side by using the fact that the $f_j$'s are constant on $M_c$.
(c) If $sigma$ and $sigma'$ both do the job in part (a), you wish to prove that $iota^*(sigma - sigma') = 0$; so study the difference $sigma - sigma'$ in a similar way to what I suggested in part (b) for $L_X_Hsigma$.
answered Jul 20 at 16:41
Jordan Payette
2,681147
I have a question about the meaning of $L_X_Hcirc iota^*$. If $etainOmega^k(M)$, then $iota^*etainOmega^k(M_c)$. In order to apply $L_X_H$ to $iota^*eta$, it would be necessary that $left.X_Hright|_M_csubset mathfrakX(M_c)$, right? Otherwise how could we guarantee $phi^*(iota^*eta)$ is a form in $M_c$? (here $phi$ is the flow of $X_H$)
– rmdmc89
Jul 21 at 12:58
1
@rmdmc89 You are absolutely right about this, and we do have here that $left. X_H right|_M_c subset mathcalX(M_c)$. Indeed, we have in general $H, f_j = pm df_j(X_H)$ (where $pm$ is a matter of convention); since we assume $H, f_j = 0$, it follows that $X_H$ is everywhere tangent to the level sets of $f_j$. As this is true for all $j=1, dots, k$, it follows that $X_H$ is tangent to $M_c$.
– Jordan Payette
Jul 21 at 14:20
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Here are some hints.
(a) This is not a symplectic problem, as $H$ is not involved and since $Lambda_omega$ could be replaced by any volume form $Lambda$ on $M$. This suggests that the use of Darboux coordinates might be unnatural, perhaps even irrelevant. Moreover, the $f_i$'s might be very 'unrelated' to any Darboux chart. Consequently it might be best to look for another argument.
One could use the constant rank theorem in order to work in a local chart $V$ with coordinates $(f_1, dots, f_k, g_1, dots, g_2n-k)$ where the $g_i$'s are (morally speaking) coordinates on $M_c$. It shouldn't be difficult to conclude to the existence of $sigma_V in Omega^2n-k(V)$ such that $left. Lambda right|_V = df_1 wedge dots wedge df_k wedge sigma_V$. Covering $U$ by charts $V_j$ and using a partition of unity $chi_j$, it shouldn't be difficult to see that $sigma := sigma_U := sum_j chi_j sigma_V_j$ does the job.
(b) Read $L_X_Hsigma$ in any of the charts $V_j$ and express it as a linear combination of wedges of the 1-forms $df_i$ and $dg_l$. Argue that each primitive summand of $L_X_Hsigma$ has to contain at least one $df_j$.
Next, for a form $F$ to be invariant under (the flow of) a vector field $X$ means that $L_X F = 0$ where $L_X$ is the Lie derivative. By extension, $F$ is invariant under the flow of a Hamiltonian function $H$ if it is invariant under that of $X_H$.
Here it is useful to recall that pullbacks and Lie derivatives commute, for example $L_X_H circ iota^* = iota^* circ L_X_H$. Apply both sides to $sigma$; you are interested in proving that the left-hand side vanishes, so compute the right-hand side by using the fact that the $f_j$'s are constant on $M_c$.
(c) If $sigma$ and $sigma'$ both do the job in part (a), you wish to prove that $iota^*(sigma - sigma') = 0$; so study the difference $sigma - sigma'$ in a similar way to what I suggested in part (b) for $L_X_Hsigma$.
answered Jul 20 at 16:41
Jordan Payette
2,681147
Here are some hints.
(a) This is not a symplectic problem, as $H$ is not involved and since $Lambda_omega$ could be replaced by any volume form $Lambda$ on $M$. This suggests that the use of Darboux coordinates might be unnatural, perhaps even irrelevant. Moreover, the $f_i$'s might be very 'unrelated' to any Darboux chart. Consequently it might be best to look for another argument.
One could use the constant rank theorem in order to work in a local chart $V$ with coordinates $(f_1, dots, f_k, g_1, dots, g_2n-k)$ where the $g_i$'s are (morally speaking) coordinates on $M_c$. It shouldn't be difficult to conclude to the existence of $sigma_V in Omega^2n-k(V)$ such that $left. Lambda right|_V = df_1 wedge dots wedge df_k wedge sigma_V$. Covering $U$ by charts $V_j$ and using a partition of unity $chi_j$, it shouldn't be difficult to see that $sigma := sigma_U := sum_j chi_j sigma_V_j$ does the job.
(b) Read $L_X_Hsigma$ in any of the charts $V_j$ and express it as a linear combination of wedges of the 1-forms $df_i$ and $dg_l$. Argue that each primitive summand of $L_X_Hsigma$ has to contain at least one $df_j$.
Next, for a form $F$ to be invariant under (the flow of) a vector field $X$ means that $L_X F = 0$ where $L_X$ is the Lie derivative. By extension, $F$ is invariant under the flow of a Hamiltonian function $H$ if it is invariant under that of $X_H$.
Here it is useful to recall that pullbacks and Lie derivatives commute, for example $L_X_H circ iota^* = iota^* circ L_X_H$. Apply both sides to $sigma$; you are interested in proving that the left-hand side vanishes, so compute the right-hand side by using the fact that the $f_j$'s are constant on $M_c$.
(c) If $sigma$ and $sigma'$ both do the job in part (a), you wish to prove that $iota^*(sigma - sigma') = 0$; so study the difference $sigma - sigma'$ in a similar way to what I suggested in part (b) for $L_X_Hsigma$.
answered Jul 20 at 16:41
Jordan Payette
2,681147
answered Jul 20 at 16:41
Jordan Payette
2,681147
answered Jul 20 at 16:41
Jordan Payette
2,681147
answered Jul 20 at 16:41
Jordan Payette
2,681147
2,681147
I have a question about the meaning of $L_X_Hcirc iota^*$. If $etainOmega^k(M)$, then $iota^*etainOmega^k(M_c)$. In order to apply $L_X_H$ to $iota^*eta$, it would be necessary that $left.X_Hright|_M_csubset mathfrakX(M_c)$, right? Otherwise how could we guarantee $phi^*(iota^*eta)$ is a form in $M_c$? (here $phi$ is the flow of $X_H$)
– rmdmc89
Jul 21 at 12:58
1
@rmdmc89 You are absolutely right about this, and we do have here that $left. X_H right|_M_c subset mathcalX(M_c)$. Indeed, we have in general $H, f_j = pm df_j(X_H)$ (where $pm$ is a matter of convention); since we assume $H, f_j = 0$, it follows that $X_H$ is everywhere tangent to the level sets of $f_j$. As this is true for all $j=1, dots, k$, it follows that $X_H$ is tangent to $M_c$.
– Jordan Payette
Jul 21 at 14:20
add a comment |Â
I have a question about the meaning of $L_X_Hcirc iota^*$. If $etainOmega^k(M)$, then $iota^*etainOmega^k(M_c)$. In order to apply $L_X_H$ to $iota^*eta$, it would be necessary that $left.X_Hright|_M_csubset mathfrakX(M_c)$, right? Otherwise how could we guarantee $phi^*(iota^*eta)$ is a form in $M_c$? (here $phi$ is the flow of $X_H$)
– rmdmc89
Jul 21 at 12:58
1
@rmdmc89 You are absolutely right about this, and we do have here that $left. X_H right|_M_c subset mathcalX(M_c)$. Indeed, we have in general $H, f_j = pm df_j(X_H)$ (where $pm$ is a matter of convention); since we assume $H, f_j = 0$, it follows that $X_H$ is everywhere tangent to the level sets of $f_j$. As this is true for all $j=1, dots, k$, it follows that $X_H$ is tangent to $M_c$.
– Jordan Payette
Jul 21 at 14:20
I have a question about the meaning of $L_X_Hcirc iota^*$. If $etainOmega^k(M)$, then $iota^*etainOmega^k(M_c)$. In order to apply $L_X_H$ to $iota^*eta$, it would be necessary that $left.X_Hright|_M_csubset mathfrakX(M_c)$, right? Otherwise how could we guarantee $phi^*(iota^*eta)$ is a form in $M_c$? (here $phi$ is the flow of $X_H$)
– rmdmc89
Jul 21 at 12:58
I have a question about the meaning of $L_X_Hcirc iota^*$. If $etainOmega^k(M)$, then $iota^*etainOmega^k(M_c)$. In order to apply $L_X_H$ to $iota^*eta$, it would be necessary that $left.X_Hright|_M_csubset mathfrakX(M_c)$, right? Otherwise how could we guarantee $phi^*(iota^*eta)$ is a form in $M_c$? (here $phi$ is the flow of $X_H$)
– rmdmc89
Jul 21 at 12:58
1
1
@rmdmc89 You are absolutely right about this, and we do have here that $left. X_H right|_M_c subset mathcalX(M_c)$. Indeed, we have in general $H, f_j = pm df_j(X_H)$ (where $pm$ is a matter of convention); since we assume $H, f_j = 0$, it follows that $X_H$ is everywhere tangent to the level sets of $f_j$. As this is true for all $j=1, dots, k$, it follows that $X_H$ is tangent to $M_c$.
– Jordan Payette
Jul 21 at 14:20
@rmdmc89 You are absolutely right about this, and we do have here that $left. X_H right|_M_c subset mathcalX(M_c)$. Indeed, we have in general $H, f_j = pm df_j(X_H)$ (where $pm$ is a matter of convention); since we assume $H, f_j = 0$, it follows that $X_H$ is everywhere tangent to the level sets of $f_j$. As this is true for all $j=1, dots, k$, it follows that $X_H$ is tangent to $M_c$.
– Jordan Payette
Jul 21 at 14:20
add a comment |Â
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