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












2












$begingroup$


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.










share|cite|improve this question









$endgroup$












  • $begingroup$
    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
















2












$begingroup$


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.










share|cite|improve this question









$endgroup$












  • $begingroup$
    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














2












2








2





$begingroup$


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.










share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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


















  • $begingroup$
    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










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