Upper volume bounds for submanifolds Announcing the arrival of Valued Associate #679: Cesar...
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Upper volume bounds for submanifolds
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Upper bound on volume growthA question about an estimateGradient of a functionalwhat is the inner product appeared in front of the integral?Ricci tensor and average of a tensorCounterexample to the form of Gromov compactness theorem without a Ricci curvature boundlower bound on volume of ballsWhether Ricci flow keep the diagonal of metric and Ricci tensor?Existence of non-trivial smooth quasi-convex function on complete Riemannian manifold with finite-volume.Constant scalar curvature with positive Ricci curvature
$begingroup$
Let $f:Mto (N,g)$ be an immersion of a compact smooth manifold $M$ into a compact Riemannian manifold $(N,g)$. Let
$$
mathrm{Vol}(M,f^*g)=int_M dmathrm{vol}_{f^*g}
$$
be the volume of $M$ with respect to the pullback metric $f^*g$.
My question: Can we get an upper bound on the volume of $(M,f^*g)$ in terms of curvature properties of the metric $g$?
One possible idea I had: suppose that $(N,g)$ has Ricci curvature bounded below by $kappa$, i.e. there exists a constant $kappain mathbb{R}$ such that
$$
mathrm{Ric}^Ngeq (n-1)kappa.
$$
Does it follow that $(M,f^*g)$ also has Ricci curvature bounded below by some $kappa'in mathbb{R}$? If so, then we can bound $mathrm{Vol}(M,f^*g)$ above using the Bishop-Gromov inequality.
differential-geometry metric-spaces riemannian-geometry smooth-manifolds
$endgroup$
add a comment |
$begingroup$
Let $f:Mto (N,g)$ be an immersion of a compact smooth manifold $M$ into a compact Riemannian manifold $(N,g)$. Let
$$
mathrm{Vol}(M,f^*g)=int_M dmathrm{vol}_{f^*g}
$$
be the volume of $M$ with respect to the pullback metric $f^*g$.
My question: Can we get an upper bound on the volume of $(M,f^*g)$ in terms of curvature properties of the metric $g$?
One possible idea I had: suppose that $(N,g)$ has Ricci curvature bounded below by $kappa$, i.e. there exists a constant $kappain mathbb{R}$ such that
$$
mathrm{Ric}^Ngeq (n-1)kappa.
$$
Does it follow that $(M,f^*g)$ also has Ricci curvature bounded below by some $kappa'in mathbb{R}$? If so, then we can bound $mathrm{Vol}(M,f^*g)$ above using the Bishop-Gromov inequality.
differential-geometry metric-spaces riemannian-geometry smooth-manifolds
$endgroup$
$begingroup$
First, there is always a $kappa'$, as a consequence of compactness. Second, control of $kappa'$ is not enough - you can embed $S^1$ into $S^3$ with arbitrarily large length, but $kappa' = 0$ on $S^1$. Perhaps control of the second fundamental form is enough?
$endgroup$
– Jason DeVito
Mar 26 at 1:29
$begingroup$
You have to restrict to embedding, or you can wrap around itself as many times as you wish.
$endgroup$
– Arctic Char
Mar 26 at 3:02
add a comment |
$begingroup$
Let $f:Mto (N,g)$ be an immersion of a compact smooth manifold $M$ into a compact Riemannian manifold $(N,g)$. Let
$$
mathrm{Vol}(M,f^*g)=int_M dmathrm{vol}_{f^*g}
$$
be the volume of $M$ with respect to the pullback metric $f^*g$.
My question: Can we get an upper bound on the volume of $(M,f^*g)$ in terms of curvature properties of the metric $g$?
One possible idea I had: suppose that $(N,g)$ has Ricci curvature bounded below by $kappa$, i.e. there exists a constant $kappain mathbb{R}$ such that
$$
mathrm{Ric}^Ngeq (n-1)kappa.
$$
Does it follow that $(M,f^*g)$ also has Ricci curvature bounded below by some $kappa'in mathbb{R}$? If so, then we can bound $mathrm{Vol}(M,f^*g)$ above using the Bishop-Gromov inequality.
differential-geometry metric-spaces riemannian-geometry smooth-manifolds
$endgroup$
Let $f:Mto (N,g)$ be an immersion of a compact smooth manifold $M$ into a compact Riemannian manifold $(N,g)$. Let
$$
mathrm{Vol}(M,f^*g)=int_M dmathrm{vol}_{f^*g}
$$
be the volume of $M$ with respect to the pullback metric $f^*g$.
My question: Can we get an upper bound on the volume of $(M,f^*g)$ in terms of curvature properties of the metric $g$?
One possible idea I had: suppose that $(N,g)$ has Ricci curvature bounded below by $kappa$, i.e. there exists a constant $kappain mathbb{R}$ such that
$$
mathrm{Ric}^Ngeq (n-1)kappa.
$$
Does it follow that $(M,f^*g)$ also has Ricci curvature bounded below by some $kappa'in mathbb{R}$? If so, then we can bound $mathrm{Vol}(M,f^*g)$ above using the Bishop-Gromov inequality.
differential-geometry metric-spaces riemannian-geometry smooth-manifolds
differential-geometry metric-spaces riemannian-geometry smooth-manifolds
asked Mar 26 at 1:19
srpsrp
2828
2828
$begingroup$
First, there is always a $kappa'$, as a consequence of compactness. Second, control of $kappa'$ is not enough - you can embed $S^1$ into $S^3$ with arbitrarily large length, but $kappa' = 0$ on $S^1$. Perhaps control of the second fundamental form is enough?
$endgroup$
– Jason DeVito
Mar 26 at 1:29
$begingroup$
You have to restrict to embedding, or you can wrap around itself as many times as you wish.
$endgroup$
– Arctic Char
Mar 26 at 3:02
add a comment |
$begingroup$
First, there is always a $kappa'$, as a consequence of compactness. Second, control of $kappa'$ is not enough - you can embed $S^1$ into $S^3$ with arbitrarily large length, but $kappa' = 0$ on $S^1$. Perhaps control of the second fundamental form is enough?
$endgroup$
– Jason DeVito
Mar 26 at 1:29
$begingroup$
You have to restrict to embedding, or you can wrap around itself as many times as you wish.
$endgroup$
– Arctic Char
Mar 26 at 3:02
$begingroup$
First, there is always a $kappa'$, as a consequence of compactness. Second, control of $kappa'$ is not enough - you can embed $S^1$ into $S^3$ with arbitrarily large length, but $kappa' = 0$ on $S^1$. Perhaps control of the second fundamental form is enough?
$endgroup$
– Jason DeVito
Mar 26 at 1:29
$begingroup$
First, there is always a $kappa'$, as a consequence of compactness. Second, control of $kappa'$ is not enough - you can embed $S^1$ into $S^3$ with arbitrarily large length, but $kappa' = 0$ on $S^1$. Perhaps control of the second fundamental form is enough?
$endgroup$
– Jason DeVito
Mar 26 at 1:29
$begingroup$
You have to restrict to embedding, or you can wrap around itself as many times as you wish.
$endgroup$
– Arctic Char
Mar 26 at 3:02
$begingroup$
You have to restrict to embedding, or you can wrap around itself as many times as you wish.
$endgroup$
– Arctic Char
Mar 26 at 3:02
add a comment |
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$begingroup$
First, there is always a $kappa'$, as a consequence of compactness. Second, control of $kappa'$ is not enough - you can embed $S^1$ into $S^3$ with arbitrarily large length, but $kappa' = 0$ on $S^1$. Perhaps control of the second fundamental form is enough?
$endgroup$
– Jason DeVito
Mar 26 at 1:29
$begingroup$
You have to restrict to embedding, or you can wrap around itself as many times as you wish.
$endgroup$
– Arctic Char
Mar 26 at 3:02