Mixing
$int_{T_pM}phi dx = int_M lambda^{-d}phi(lambda^{-1}exp_p^{-1}(cdot)) dV_g$?
$begingroup$
Let $(M,g)$ be a Riemannian manifold, $r>0$ be its injecitivity radius, $p$ be a point on $M$. Let $phi:T_p(M)to mathbf R$ be a function supported in
$$B_{T_p M}(0_p,r):={Xin T_p M:|X|_g<r }.$$
It is easy to see that the exponential map $exp_p:B_{T_p M}(0_p,r)to M$ is a diffeomorphism to its image.
Then the following family of scaling functions of $phi$ is well-defined on $M$,
begin{align}
phi_p^lambda(q):=
begin{cases}
lambda^{-d}phi(lambda^{-1}exp_p^{-1}(q)), & qin exp(B_{T_p M}(0_p,lambda r)), \
0, & text{otherwise}.
end{cases} quad lambdain (0,1]
end{align}
My question is: does the following equality holds
$$int_M phi_p^lambda dV_g = int_{T_pM}phi dx?$$
Here $dV_g$ is the volume form on $M$, $dx$ is the lebesgue measure on $T_pM$.
differential-geometry manifolds differential-topology riemannian-geometry smooth-manifolds
asked Jan 28 at 5:47
Q. HuangQ. Huang
603213
$endgroup$
add a comment |
$begingroup$
Let $(M,g)$ be a Riemannian manifold, $r>0$ be its injecitivity radius, $p$ be a point on $M$. Let $phi:T_p(M)to mathbf R$ be a function supported in
$$B_{T_p M}(0_p,r):={Xin T_p M:|X|_g<r }.$$
It is easy to see that the exponential map $exp_p:B_{T_p M}(0_p,r)to M$ is a diffeomorphism to its image.
Then the following family of scaling functions of $phi$ is well-defined on $M$,
begin{align}
phi_p^lambda(q):=
begin{cases}
lambda^{-d}phi(lambda^{-1}exp_p^{-1}(q)), & qin exp(B_{T_p M}(0_p,lambda r)), \
0, & text{otherwise}.
end{cases} quad lambdain (0,1]
end{align}
My question is: does the following equality holds
$$int_M phi_p^lambda dV_g = int_{T_pM}phi dx?$$
Here $dV_g$ is the volume form on $M$, $dx$ is the lebesgue measure on $T_pM$.
differential-geometry manifolds differential-topology riemannian-geometry smooth-manifolds
asked Jan 28 at 5:47
Q. HuangQ. Huang
603213
$endgroup$
add a comment |
$begingroup$
Let $(M,g)$ be a Riemannian manifold, $r>0$ be its injecitivity radius, $p$ be a point on $M$. Let $phi:T_p(M)to mathbf R$ be a function supported in
$$B_{T_p M}(0_p,r):={Xin T_p M:|X|_g<r }.$$
It is easy to see that the exponential map $exp_p:B_{T_p M}(0_p,r)to M$ is a diffeomorphism to its image.
Then the following family of scaling functions of $phi$ is well-defined on $M$,
begin{align}
phi_p^lambda(q):=
begin{cases}
lambda^{-d}phi(lambda^{-1}exp_p^{-1}(q)), & qin exp(B_{T_p M}(0_p,lambda r)), \
0, & text{otherwise}.
end{cases} quad lambdain (0,1]
end{align}
My question is: does the following equality holds
$$int_M phi_p^lambda dV_g = int_{T_pM}phi dx?$$
Here $dV_g$ is the volume form on $M$, $dx$ is the lebesgue measure on $T_pM$.
differential-geometry manifolds differential-topology riemannian-geometry smooth-manifolds
asked Jan 28 at 5:47
Q. HuangQ. Huang
603213
$endgroup$
Let $(M,g)$ be a Riemannian manifold, $r>0$ be its injecitivity radius, $p$ be a point on $M$. Let $phi:T_p(M)to mathbf R$ be a function supported in
$$B_{T_p M}(0_p,r):={Xin T_p M:|X|_g<r }.$$
It is easy to see that the exponential map $exp_p:B_{T_p M}(0_p,r)to M$ is a diffeomorphism to its image.
Then the following family of scaling functions of $phi$ is well-defined on $M$,
begin{align}
phi_p^lambda(q):=
begin{cases}
lambda^{-d}phi(lambda^{-1}exp_p^{-1}(q)), & qin exp(B_{T_p M}(0_p,lambda r)), \
0, & text{otherwise}.
end{cases} quad lambdain (0,1]
end{align}
My question is: does the following equality holds
$$int_M phi_p^lambda dV_g = int_{T_pM}phi dx?$$
Here $dV_g$ is the volume form on $M$, $dx$ is the lebesgue measure on $T_pM$.
differential-geometry manifolds differential-topology riemannian-geometry smooth-manifolds
differential-geometry manifolds differential-topology riemannian-geometry smooth-manifolds
asked Jan 28 at 5:47
Q. HuangQ. Huang
603213
asked Jan 28 at 5:47
Q. HuangQ. Huang
603213
edited Feb 24 at 22:37
Q. Huang
asked Jan 28 at 5:47
Q. HuangQ. Huang
603213
asked Jan 28 at 5:47
Q. HuangQ. Huang
603213
asked Jan 28 at 5:47
Q. HuangQ. Huang
603213
603213
add a comment |
add a comment |
1 Answer
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$begingroup$
Yes. It is simply derived as follows:
begin{equation}
begin{split}
int_M phi_p^lambda dV_g &= int_{exp(B_{T_p M}(0_p,lambda r))} lambda^{-d}phi(lambda^{-1}exp_p^{-1}(q)) dV_g(q) \
(x=exp_p^{-1} q) &= int_{B_{T_p M}(0_p,lambda r)}lambda^{-d}phi(lambda^{-1}x) dx^1wedgecdotswedge dx^d \
(v=lambda^{-1}x) &= int_{B_{T_p M}(0_p,r)}phi(v) dv \
&= int_{T_pM} phi(v) dv.
end{split}
end{equation}
Note that in the second equality, we used the normal coordinates, then $g_{ij}(x) = delta_{ij}$.
answered Feb 24 at 22:52
Q. HuangQ. Huang
603213
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Yes. It is simply derived as follows:
begin{equation}
begin{split}
int_M phi_p^lambda dV_g &= int_{exp(B_{T_p M}(0_p,lambda r))} lambda^{-d}phi(lambda^{-1}exp_p^{-1}(q)) dV_g(q) \
(x=exp_p^{-1} q) &= int_{B_{T_p M}(0_p,lambda r)}lambda^{-d}phi(lambda^{-1}x) dx^1wedgecdotswedge dx^d \
(v=lambda^{-1}x) &= int_{B_{T_p M}(0_p,r)}phi(v) dv \
&= int_{T_pM} phi(v) dv.
end{split}
end{equation}
Note that in the second equality, we used the normal coordinates, then $g_{ij}(x) = delta_{ij}$.
answered Feb 24 at 22:52
Q. HuangQ. Huang
603213
$endgroup$
add a comment |
$begingroup$
Yes. It is simply derived as follows:
begin{equation}
begin{split}
int_M phi_p^lambda dV_g &= int_{exp(B_{T_p M}(0_p,lambda r))} lambda^{-d}phi(lambda^{-1}exp_p^{-1}(q)) dV_g(q) \
(x=exp_p^{-1} q) &= int_{B_{T_p M}(0_p,lambda r)}lambda^{-d}phi(lambda^{-1}x) dx^1wedgecdotswedge dx^d \
(v=lambda^{-1}x) &= int_{B_{T_p M}(0_p,r)}phi(v) dv \
&= int_{T_pM} phi(v) dv.
end{split}
end{equation}
Note that in the second equality, we used the normal coordinates, then $g_{ij}(x) = delta_{ij}$.
answered Feb 24 at 22:52
Q. HuangQ. Huang
603213
$endgroup$
add a comment |
$begingroup$
Yes. It is simply derived as follows:
begin{equation}
begin{split}
int_M phi_p^lambda dV_g &= int_{exp(B_{T_p M}(0_p,lambda r))} lambda^{-d}phi(lambda^{-1}exp_p^{-1}(q)) dV_g(q) \
(x=exp_p^{-1} q) &= int_{B_{T_p M}(0_p,lambda r)}lambda^{-d}phi(lambda^{-1}x) dx^1wedgecdotswedge dx^d \
(v=lambda^{-1}x) &= int_{B_{T_p M}(0_p,r)}phi(v) dv \
&= int_{T_pM} phi(v) dv.
end{split}
end{equation}
Note that in the second equality, we used the normal coordinates, then $g_{ij}(x) = delta_{ij}$.
answered Feb 24 at 22:52
Q. HuangQ. Huang
603213
$endgroup$
Yes. It is simply derived as follows:
begin{equation}
begin{split}
int_M phi_p^lambda dV_g &= int_{exp(B_{T_p M}(0_p,lambda r))} lambda^{-d}phi(lambda^{-1}exp_p^{-1}(q)) dV_g(q) \
(x=exp_p^{-1} q) &= int_{B_{T_p M}(0_p,lambda r)}lambda^{-d}phi(lambda^{-1}x) dx^1wedgecdotswedge dx^d \
(v=lambda^{-1}x) &= int_{B_{T_p M}(0_p,r)}phi(v) dv \
&= int_{T_pM} phi(v) dv.
end{split}
end{equation}
Note that in the second equality, we used the normal coordinates, then $g_{ij}(x) = delta_{ij}$.
answered Feb 24 at 22:52
Q. HuangQ. Huang
603213
answered Feb 24 at 22:52
Q. HuangQ. Huang
603213
answered Feb 24 at 22:52
Q. HuangQ. Huang
603213
answered Feb 24 at 22:52
Q. HuangQ. Huang
603213
603213
add a comment |
add a comment |
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