Almost volume preserving charts for a riemannian manifold
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Let $M$ be riemannian manifold. Is it true that for all $pin M$ and $varepsilon>0$ there is a chart $phi:Urightarrowmathbb{R^n}$ arround $p$ such that for all nonempty open $Vsubseteq U$ $frac{vol(V)}{vol(phi (V))}in (1-varepsilon,1+varepsilon)$ ?
integration analysis differential-geometry smooth-manifolds
asked Dec 11 '18 at 13:50
triitrii
715
$endgroup$
add a comment |
$begingroup$
Let $M$ be riemannian manifold. Is it true that for all $pin M$ and $varepsilon>0$ there is a chart $phi:Urightarrowmathbb{R^n}$ arround $p$ such that for all nonempty open $Vsubseteq U$ $frac{vol(V)}{vol(phi (V))}in (1-varepsilon,1+varepsilon)$ ?
integration analysis differential-geometry smooth-manifolds
asked Dec 11 '18 at 13:50
triitrii
715
$endgroup$
add a comment |
$begingroup$
Let $M$ be riemannian manifold. Is it true that for all $pin M$ and $varepsilon>0$ there is a chart $phi:Urightarrowmathbb{R^n}$ arround $p$ such that for all nonempty open $Vsubseteq U$ $frac{vol(V)}{vol(phi (V))}in (1-varepsilon,1+varepsilon)$ ?
integration analysis differential-geometry smooth-manifolds
asked Dec 11 '18 at 13:50
triitrii
715
$endgroup$
Let $M$ be riemannian manifold. Is it true that for all $pin M$ and $varepsilon>0$ there is a chart $phi:Urightarrowmathbb{R^n}$ arround $p$ such that for all nonempty open $Vsubseteq U$ $frac{vol(V)}{vol(phi (V))}in (1-varepsilon,1+varepsilon)$ ?
integration analysis differential-geometry smooth-manifolds
integration analysis differential-geometry smooth-manifolds
asked Dec 11 '18 at 13:50
triitrii
715
asked Dec 11 '18 at 13:50
triitrii
715
asked Dec 11 '18 at 13:50
triitrii
715
asked Dec 11 '18 at 13:50
triitrii
715
asked Dec 11 '18 at 13:50
triitrii
715
715
add a comment |
add a comment |
1 Answer
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Yes, this is true (in fact, such an estimate holds for all Borel measurable $V$).
Fix $p$ and let $phi: U to mathbb{R}^n$ give Riemannian normal coordinates at $p$.
For $V subseteq U$ Borel measurable, the volume $text{vol}_M(V)$ is given by
$$
text{vol}_M(V) = int_{phi(V)} sqrt{det(g(x))} d mu(x),
$$
where $dmu$ is Lebesgue measure, and $g(x)$ is the matrix representing the metric tensor in $phi$-coordnates (i.e., $g_{ij}(x) = langle psi^*(partial_i), psi^*(partial_j) rangle_{psi^{-1}(x)}$).
At $0 = phi(p)$, the matrix $g_{ij}$ is just the identity matrix (because of our choice of $phi$), and the functions $g_{ij}$ vary smoothly on $phi(V)$, so, by shrinking $U$ if necessary, we can assume that, for all $x in U$, we have
$$
1 - epsilon leq sqrt{det(g(x))} leq 1 + epsilon.
$$
Your estimate follows immediately.
In this context it seems reasonable also to mention the formula given on this Wikipedia page relating the Riemannian and Lebesgue volume forms in normal coordinates -- the error is quadratic in the coordinate $x$ and determined by the Ricci tensor.
answered Dec 11 '18 at 22:14
mollyerinmollyerin
2,69576
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
Yes, this is true (in fact, such an estimate holds for all Borel measurable $V$).
Fix $p$ and let $phi: U to mathbb{R}^n$ give Riemannian normal coordinates at $p$.
For $V subseteq U$ Borel measurable, the volume $text{vol}_M(V)$ is given by
$$
text{vol}_M(V) = int_{phi(V)} sqrt{det(g(x))} d mu(x),
$$
where $dmu$ is Lebesgue measure, and $g(x)$ is the matrix representing the metric tensor in $phi$-coordnates (i.e., $g_{ij}(x) = langle psi^*(partial_i), psi^*(partial_j) rangle_{psi^{-1}(x)}$).
At $0 = phi(p)$, the matrix $g_{ij}$ is just the identity matrix (because of our choice of $phi$), and the functions $g_{ij}$ vary smoothly on $phi(V)$, so, by shrinking $U$ if necessary, we can assume that, for all $x in U$, we have
$$
1 - epsilon leq sqrt{det(g(x))} leq 1 + epsilon.
$$
Your estimate follows immediately.
In this context it seems reasonable also to mention the formula given on this Wikipedia page relating the Riemannian and Lebesgue volume forms in normal coordinates -- the error is quadratic in the coordinate $x$ and determined by the Ricci tensor.
answered Dec 11 '18 at 22:14
mollyerinmollyerin
2,69576
$endgroup$
add a comment |
$begingroup$
Yes, this is true (in fact, such an estimate holds for all Borel measurable $V$).
Fix $p$ and let $phi: U to mathbb{R}^n$ give Riemannian normal coordinates at $p$.
For $V subseteq U$ Borel measurable, the volume $text{vol}_M(V)$ is given by
$$
text{vol}_M(V) = int_{phi(V)} sqrt{det(g(x))} d mu(x),
$$
where $dmu$ is Lebesgue measure, and $g(x)$ is the matrix representing the metric tensor in $phi$-coordnates (i.e., $g_{ij}(x) = langle psi^*(partial_i), psi^*(partial_j) rangle_{psi^{-1}(x)}$).
At $0 = phi(p)$, the matrix $g_{ij}$ is just the identity matrix (because of our choice of $phi$), and the functions $g_{ij}$ vary smoothly on $phi(V)$, so, by shrinking $U$ if necessary, we can assume that, for all $x in U$, we have
$$
1 - epsilon leq sqrt{det(g(x))} leq 1 + epsilon.
$$
Your estimate follows immediately.
In this context it seems reasonable also to mention the formula given on this Wikipedia page relating the Riemannian and Lebesgue volume forms in normal coordinates -- the error is quadratic in the coordinate $x$ and determined by the Ricci tensor.
answered Dec 11 '18 at 22:14
mollyerinmollyerin
2,69576
$endgroup$
add a comment |
$begingroup$
Yes, this is true (in fact, such an estimate holds for all Borel measurable $V$).
Fix $p$ and let $phi: U to mathbb{R}^n$ give Riemannian normal coordinates at $p$.
For $V subseteq U$ Borel measurable, the volume $text{vol}_M(V)$ is given by
$$
text{vol}_M(V) = int_{phi(V)} sqrt{det(g(x))} d mu(x),
$$
where $dmu$ is Lebesgue measure, and $g(x)$ is the matrix representing the metric tensor in $phi$-coordnates (i.e., $g_{ij}(x) = langle psi^*(partial_i), psi^*(partial_j) rangle_{psi^{-1}(x)}$).
At $0 = phi(p)$, the matrix $g_{ij}$ is just the identity matrix (because of our choice of $phi$), and the functions $g_{ij}$ vary smoothly on $phi(V)$, so, by shrinking $U$ if necessary, we can assume that, for all $x in U$, we have
$$
1 - epsilon leq sqrt{det(g(x))} leq 1 + epsilon.
$$
Your estimate follows immediately.
In this context it seems reasonable also to mention the formula given on this Wikipedia page relating the Riemannian and Lebesgue volume forms in normal coordinates -- the error is quadratic in the coordinate $x$ and determined by the Ricci tensor.
answered Dec 11 '18 at 22:14
mollyerinmollyerin
2,69576
$endgroup$
Yes, this is true (in fact, such an estimate holds for all Borel measurable $V$).
Fix $p$ and let $phi: U to mathbb{R}^n$ give Riemannian normal coordinates at $p$.
For $V subseteq U$ Borel measurable, the volume $text{vol}_M(V)$ is given by
$$
text{vol}_M(V) = int_{phi(V)} sqrt{det(g(x))} d mu(x),
$$
where $dmu$ is Lebesgue measure, and $g(x)$ is the matrix representing the metric tensor in $phi$-coordnates (i.e., $g_{ij}(x) = langle psi^*(partial_i), psi^*(partial_j) rangle_{psi^{-1}(x)}$).
At $0 = phi(p)$, the matrix $g_{ij}$ is just the identity matrix (because of our choice of $phi$), and the functions $g_{ij}$ vary smoothly on $phi(V)$, so, by shrinking $U$ if necessary, we can assume that, for all $x in U$, we have
$$
1 - epsilon leq sqrt{det(g(x))} leq 1 + epsilon.
$$
Your estimate follows immediately.
In this context it seems reasonable also to mention the formula given on this Wikipedia page relating the Riemannian and Lebesgue volume forms in normal coordinates -- the error is quadratic in the coordinate $x$ and determined by the Ricci tensor.
answered Dec 11 '18 at 22:14
mollyerinmollyerin
2,69576
answered Dec 11 '18 at 22:14
mollyerinmollyerin
2,69576
answered Dec 11 '18 at 22:14
mollyerinmollyerin
2,69576
answered Dec 11 '18 at 22:14
mollyerinmollyerin
2,69576
2,69576
add a comment |
add a comment |
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