Mixing
Geodesic convexity and the 2nd fundamental form
$begingroup$
Let $(M,g)$ be a Riemannian manifold, $Omegasubset M$ be a closed set with smooth boundary $partialOmega$ and $nu$ be the unit normal of $partialOmega$ pointing into $Omega$.
$Omega$ is said to be geodesically convex iff
$forall x_0, x_1inOmega$ $exists c:[0,1]stackrel{text{geodesic}}to(M,g)$ s.t. $c(0)=x_0, c(1)=x_1$, $c([0,1])subsetOmega$, $mathrm{Length}[c]=d_g(x_0,x_1)$.
Suppose $Omega$ is geodesically convex. Then...
[Q.1]
Does it hold that the 2nd fundamental form of $partialOmega$ toward $nu$ is nonnegative definite at each point on $partialOmega$?
[Q.2]
Let $psi_r(x):=mathrm{exp}^g_x [rnu(x)]in N$ $(xinpartialOmega)$. Then for small $|r|$, $psi_r$ is an embedding. Here, does it hold that the inner 2nd fundamental form of $psi_r$ is nonnegative definite at each point on $partialOmega$ when $r>0$ and is sufficinetly small?
Thank you.
geometry differential-geometry manifolds riemannian-geometry
asked Nov 4 '16 at 11:29
stb2084stb2084
444313
$endgroup$
add a comment |
$begingroup$
Let $(M,g)$ be a Riemannian manifold, $Omegasubset M$ be a closed set with smooth boundary $partialOmega$ and $nu$ be the unit normal of $partialOmega$ pointing into $Omega$.
$Omega$ is said to be geodesically convex iff
$forall x_0, x_1inOmega$ $exists c:[0,1]stackrel{text{geodesic}}to(M,g)$ s.t. $c(0)=x_0, c(1)=x_1$, $c([0,1])subsetOmega$, $mathrm{Length}[c]=d_g(x_0,x_1)$.
Suppose $Omega$ is geodesically convex. Then...
[Q.1]
Does it hold that the 2nd fundamental form of $partialOmega$ toward $nu$ is nonnegative definite at each point on $partialOmega$?
[Q.2]
Let $psi_r(x):=mathrm{exp}^g_x [rnu(x)]in N$ $(xinpartialOmega)$. Then for small $|r|$, $psi_r$ is an embedding. Here, does it hold that the inner 2nd fundamental form of $psi_r$ is nonnegative definite at each point on $partialOmega$ when $r>0$ and is sufficinetly small?
Thank you.
geometry differential-geometry manifolds riemannian-geometry
asked Nov 4 '16 at 11:29
stb2084stb2084
444313
$endgroup$
add a comment |
$begingroup$
Let $(M,g)$ be a Riemannian manifold, $Omegasubset M$ be a closed set with smooth boundary $partialOmega$ and $nu$ be the unit normal of $partialOmega$ pointing into $Omega$.
$Omega$ is said to be geodesically convex iff
$forall x_0, x_1inOmega$ $exists c:[0,1]stackrel{text{geodesic}}to(M,g)$ s.t. $c(0)=x_0, c(1)=x_1$, $c([0,1])subsetOmega$, $mathrm{Length}[c]=d_g(x_0,x_1)$.
Suppose $Omega$ is geodesically convex. Then...
[Q.1]
Does it hold that the 2nd fundamental form of $partialOmega$ toward $nu$ is nonnegative definite at each point on $partialOmega$?
[Q.2]
Let $psi_r(x):=mathrm{exp}^g_x [rnu(x)]in N$ $(xinpartialOmega)$. Then for small $|r|$, $psi_r$ is an embedding. Here, does it hold that the inner 2nd fundamental form of $psi_r$ is nonnegative definite at each point on $partialOmega$ when $r>0$ and is sufficinetly small?
Thank you.
geometry differential-geometry manifolds riemannian-geometry
asked Nov 4 '16 at 11:29
stb2084stb2084
444313
$endgroup$
Let $(M,g)$ be a Riemannian manifold, $Omegasubset M$ be a closed set with smooth boundary $partialOmega$ and $nu$ be the unit normal of $partialOmega$ pointing into $Omega$.
$Omega$ is said to be geodesically convex iff
$forall x_0, x_1inOmega$ $exists c:[0,1]stackrel{text{geodesic}}to(M,g)$ s.t. $c(0)=x_0, c(1)=x_1$, $c([0,1])subsetOmega$, $mathrm{Length}[c]=d_g(x_0,x_1)$.
Suppose $Omega$ is geodesically convex. Then...
[Q.1]
Does it hold that the 2nd fundamental form of $partialOmega$ toward $nu$ is nonnegative definite at each point on $partialOmega$?
[Q.2]
Let $psi_r(x):=mathrm{exp}^g_x [rnu(x)]in N$ $(xinpartialOmega)$. Then for small $|r|$, $psi_r$ is an embedding. Here, does it hold that the inner 2nd fundamental form of $psi_r$ is nonnegative definite at each point on $partialOmega$ when $r>0$ and is sufficinetly small?
Thank you.
geometry differential-geometry manifolds riemannian-geometry
geometry differential-geometry manifolds riemannian-geometry
asked Nov 4 '16 at 11:29
stb2084stb2084
444313
asked Nov 4 '16 at 11:29
stb2084stb2084
444313
asked Nov 4 '16 at 11:29
stb2084stb2084
444313
asked Nov 4 '16 at 11:29
stb2084stb2084
444313
asked Nov 4 '16 at 11:29
stb2084stb2084
444313
444313
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
I will address your first question. First define, for a fixed point $pinpartial Omega$, three conditions on $partial Omega$.
a) There is an open subset $Usubset M$ with $pin U$, such that any two points in $Ucap Omega$ can be joined by length minimising geodesic $c:[0,1]rightarrow M$ with $c[0,1]subset Ucap Omega$.
b) Any geodesic $c:(-epsilon,0]rightarrow M$ with $c(-epsilon,0)subset mathrm{int(Omega)}$ and $c(0)= p$ hits the boundary transversally, i.e. $g(dot c(0),nu(p))neq 0$.
c) The second fundamental form $l_nu(cdot,cdot)=g(nabla_cdot cdot, nu)$ is non-negative at $p$.
We will prove that a) $Rightarrow$ b) $Rightarrow$ c), which answers your first question.
Step 1 Take geodesic normal coordinates $x^1,dots,x^n$ , centred at $p$, with $partial_nvert_p=nu(p)$. Using the implicit function theorem, one can show that there is a smooth function $f:mathbb{R}^{n-1}rightarrow mathbb{R}$ such that
$$
Omega cap V ={x^nge f(x^1,dots,x^{n-1})},
$$
where $V$ denotes the coordinate patch. Next define smooth vector fields $X_1,dots,X_{n-1}$ on $V$ by
$$
X_j(q)=partial_jvert_q + partial_jf(x^1(q),dots,x^{n-1}(q))partial_nvert_q.
$$
If $qin V cap partial Omega$, then $X_j(q) in T_qpartial Omega$ and hence for $1le jle n-1$ we have
$$
partial_jf(0) = X_j^k(p) g_{kl}(p) nu^l(p)=g(X_j(p),nu(p)) = 0 tag{1}.
$$
Step 2 We claim that $$
text{b)} quad Leftrightarrow quad f text{ has a local minimum at 0.} tag{2}
$$
Note that in the coordinates fixed above, b) means that ${x^n=0}$ (the hypersurface spanned by geodesics through $p$ which are not transversal) does not intersect $int(Omega)cap V ={x^n> f(x^1,dots,x^{n-1})}$ near $p$, or in other words that $f(x^1,dots,x^{n-1})ge 0$ in a neighbourhood of $p$. This proves (2).
Step 3 We're now in a position to prove a) $Rightarrow$ b). Suppose b) is wrong, then $f$ does not have a local minimum at $0$. I.e. for any neighbourhood $U$ of $p$ there is a point $qin partial Omega cap U$ such that $$x^n(q)=f(x^1(q),dots,x^{n-1}(q))<0. tag{3}$$
Since we are in geodesic coordinates, the unique minimising geodesic joining $p$ and $q$ is given by $L ={x^j=tx^j(q): 0le t le 1}$. Assuming a), we must have $Lsubset Omega cap U$, which implies
$$
tx^n(q) ge f(tx^1(q),dots,tx^{n-1}(q)).
$$
Divide by $t$ and take the limit $trightarrow 0$, then the right hand side will converge to $0$, since $Df(0)=0$ (see $(1)$). We obtain $x^n(q)ge0$, which is a contradiction to $(3)$.
Step 4 We want to relate the second fundamental form to the Hessian of $f$. To this end note that, for $1le j le n-1$ and $1le k le n$ we have
$$
nabla_kX_j (p)= (Gamma_{kj}^l partial_l + partial_k partial_j f partial_n + Gamma_{kn}^r partial_r)(p) = partial_jpartial_kf(p) cdot nu(p)
$$
and since $f$ is independent of $x^n$, we further obtain
$$
nabla_{X_k}X_j(p) = partial_jpartial_kf(p) cdot nu(p),
$$
which implies $l_{nu}(X_k,X_j)vert _p = partial_jpartial_kf(p)$. Since the $X_j$ form a basis of $T_ppartial Omega$ we have
$$
text{c)}quad Leftrightarrow quad f text{ has non-negative Hessian at $0$}. tag{4}
$$
From $(2)$ and $(4)$ it is evident that b) implies c).
Remark: With the same kind of argument one obtains an interesting result in the other direction. Assuming that the second fundamental form of $partial Omega$ is strictly positive, all geodesics coming from $Omega$ hit $partial Omega$ transversally.
answered Jan 29 at 12:17
Jan BohrJan Bohr
3,3071521
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1 Answer
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1 Answer
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$begingroup$
I will address your first question. First define, for a fixed point $pinpartial Omega$, three conditions on $partial Omega$.
a) There is an open subset $Usubset M$ with $pin U$, such that any two points in $Ucap Omega$ can be joined by length minimising geodesic $c:[0,1]rightarrow M$ with $c[0,1]subset Ucap Omega$.
b) Any geodesic $c:(-epsilon,0]rightarrow M$ with $c(-epsilon,0)subset mathrm{int(Omega)}$ and $c(0)= p$ hits the boundary transversally, i.e. $g(dot c(0),nu(p))neq 0$.
c) The second fundamental form $l_nu(cdot,cdot)=g(nabla_cdot cdot, nu)$ is non-negative at $p$.
We will prove that a) $Rightarrow$ b) $Rightarrow$ c), which answers your first question.
Step 1 Take geodesic normal coordinates $x^1,dots,x^n$ , centred at $p$, with $partial_nvert_p=nu(p)$. Using the implicit function theorem, one can show that there is a smooth function $f:mathbb{R}^{n-1}rightarrow mathbb{R}$ such that
$$
Omega cap V ={x^nge f(x^1,dots,x^{n-1})},
$$
where $V$ denotes the coordinate patch. Next define smooth vector fields $X_1,dots,X_{n-1}$ on $V$ by
$$
X_j(q)=partial_jvert_q + partial_jf(x^1(q),dots,x^{n-1}(q))partial_nvert_q.
$$
If $qin V cap partial Omega$, then $X_j(q) in T_qpartial Omega$ and hence for $1le jle n-1$ we have
$$
partial_jf(0) = X_j^k(p) g_{kl}(p) nu^l(p)=g(X_j(p),nu(p)) = 0 tag{1}.
$$
Step 2 We claim that $$
text{b)} quad Leftrightarrow quad f text{ has a local minimum at 0.} tag{2}
$$
Note that in the coordinates fixed above, b) means that ${x^n=0}$ (the hypersurface spanned by geodesics through $p$ which are not transversal) does not intersect $int(Omega)cap V ={x^n> f(x^1,dots,x^{n-1})}$ near $p$, or in other words that $f(x^1,dots,x^{n-1})ge 0$ in a neighbourhood of $p$. This proves (2).
Step 3 We're now in a position to prove a) $Rightarrow$ b). Suppose b) is wrong, then $f$ does not have a local minimum at $0$. I.e. for any neighbourhood $U$ of $p$ there is a point $qin partial Omega cap U$ such that $$x^n(q)=f(x^1(q),dots,x^{n-1}(q))<0. tag{3}$$
Since we are in geodesic coordinates, the unique minimising geodesic joining $p$ and $q$ is given by $L ={x^j=tx^j(q): 0le t le 1}$. Assuming a), we must have $Lsubset Omega cap U$, which implies
$$
tx^n(q) ge f(tx^1(q),dots,tx^{n-1}(q)).
$$
Divide by $t$ and take the limit $trightarrow 0$, then the right hand side will converge to $0$, since $Df(0)=0$ (see $(1)$). We obtain $x^n(q)ge0$, which is a contradiction to $(3)$.
Step 4 We want to relate the second fundamental form to the Hessian of $f$. To this end note that, for $1le j le n-1$ and $1le k le n$ we have
$$
nabla_kX_j (p)= (Gamma_{kj}^l partial_l + partial_k partial_j f partial_n + Gamma_{kn}^r partial_r)(p) = partial_jpartial_kf(p) cdot nu(p)
$$
and since $f$ is independent of $x^n$, we further obtain
$$
nabla_{X_k}X_j(p) = partial_jpartial_kf(p) cdot nu(p),
$$
which implies $l_{nu}(X_k,X_j)vert _p = partial_jpartial_kf(p)$. Since the $X_j$ form a basis of $T_ppartial Omega$ we have
$$
text{c)}quad Leftrightarrow quad f text{ has non-negative Hessian at $0$}. tag{4}
$$
From $(2)$ and $(4)$ it is evident that b) implies c).
Remark: With the same kind of argument one obtains an interesting result in the other direction. Assuming that the second fundamental form of $partial Omega$ is strictly positive, all geodesics coming from $Omega$ hit $partial Omega$ transversally.
answered Jan 29 at 12:17
Jan BohrJan Bohr
3,3071521
$endgroup$
add a comment |
$begingroup$
I will address your first question. First define, for a fixed point $pinpartial Omega$, three conditions on $partial Omega$.
a) There is an open subset $Usubset M$ with $pin U$, such that any two points in $Ucap Omega$ can be joined by length minimising geodesic $c:[0,1]rightarrow M$ with $c[0,1]subset Ucap Omega$.
b) Any geodesic $c:(-epsilon,0]rightarrow M$ with $c(-epsilon,0)subset mathrm{int(Omega)}$ and $c(0)= p$ hits the boundary transversally, i.e. $g(dot c(0),nu(p))neq 0$.
c) The second fundamental form $l_nu(cdot,cdot)=g(nabla_cdot cdot, nu)$ is non-negative at $p$.
We will prove that a) $Rightarrow$ b) $Rightarrow$ c), which answers your first question.
Step 1 Take geodesic normal coordinates $x^1,dots,x^n$ , centred at $p$, with $partial_nvert_p=nu(p)$. Using the implicit function theorem, one can show that there is a smooth function $f:mathbb{R}^{n-1}rightarrow mathbb{R}$ such that
$$
Omega cap V ={x^nge f(x^1,dots,x^{n-1})},
$$
where $V$ denotes the coordinate patch. Next define smooth vector fields $X_1,dots,X_{n-1}$ on $V$ by
$$
X_j(q)=partial_jvert_q + partial_jf(x^1(q),dots,x^{n-1}(q))partial_nvert_q.
$$
If $qin V cap partial Omega$, then $X_j(q) in T_qpartial Omega$ and hence for $1le jle n-1$ we have
$$
partial_jf(0) = X_j^k(p) g_{kl}(p) nu^l(p)=g(X_j(p),nu(p)) = 0 tag{1}.
$$
Step 2 We claim that $$
text{b)} quad Leftrightarrow quad f text{ has a local minimum at 0.} tag{2}
$$
Note that in the coordinates fixed above, b) means that ${x^n=0}$ (the hypersurface spanned by geodesics through $p$ which are not transversal) does not intersect $int(Omega)cap V ={x^n> f(x^1,dots,x^{n-1})}$ near $p$, or in other words that $f(x^1,dots,x^{n-1})ge 0$ in a neighbourhood of $p$. This proves (2).
Step 3 We're now in a position to prove a) $Rightarrow$ b). Suppose b) is wrong, then $f$ does not have a local minimum at $0$. I.e. for any neighbourhood $U$ of $p$ there is a point $qin partial Omega cap U$ such that $$x^n(q)=f(x^1(q),dots,x^{n-1}(q))<0. tag{3}$$
Since we are in geodesic coordinates, the unique minimising geodesic joining $p$ and $q$ is given by $L ={x^j=tx^j(q): 0le t le 1}$. Assuming a), we must have $Lsubset Omega cap U$, which implies
$$
tx^n(q) ge f(tx^1(q),dots,tx^{n-1}(q)).
$$
Divide by $t$ and take the limit $trightarrow 0$, then the right hand side will converge to $0$, since $Df(0)=0$ (see $(1)$). We obtain $x^n(q)ge0$, which is a contradiction to $(3)$.
Step 4 We want to relate the second fundamental form to the Hessian of $f$. To this end note that, for $1le j le n-1$ and $1le k le n$ we have
$$
nabla_kX_j (p)= (Gamma_{kj}^l partial_l + partial_k partial_j f partial_n + Gamma_{kn}^r partial_r)(p) = partial_jpartial_kf(p) cdot nu(p)
$$
and since $f$ is independent of $x^n$, we further obtain
$$
nabla_{X_k}X_j(p) = partial_jpartial_kf(p) cdot nu(p),
$$
which implies $l_{nu}(X_k,X_j)vert _p = partial_jpartial_kf(p)$. Since the $X_j$ form a basis of $T_ppartial Omega$ we have
$$
text{c)}quad Leftrightarrow quad f text{ has non-negative Hessian at $0$}. tag{4}
$$
From $(2)$ and $(4)$ it is evident that b) implies c).
Remark: With the same kind of argument one obtains an interesting result in the other direction. Assuming that the second fundamental form of $partial Omega$ is strictly positive, all geodesics coming from $Omega$ hit $partial Omega$ transversally.
answered Jan 29 at 12:17
Jan BohrJan Bohr
3,3071521
$endgroup$
add a comment |
$begingroup$
I will address your first question. First define, for a fixed point $pinpartial Omega$, three conditions on $partial Omega$.
a) There is an open subset $Usubset M$ with $pin U$, such that any two points in $Ucap Omega$ can be joined by length minimising geodesic $c:[0,1]rightarrow M$ with $c[0,1]subset Ucap Omega$.
b) Any geodesic $c:(-epsilon,0]rightarrow M$ with $c(-epsilon,0)subset mathrm{int(Omega)}$ and $c(0)= p$ hits the boundary transversally, i.e. $g(dot c(0),nu(p))neq 0$.
c) The second fundamental form $l_nu(cdot,cdot)=g(nabla_cdot cdot, nu)$ is non-negative at $p$.
We will prove that a) $Rightarrow$ b) $Rightarrow$ c), which answers your first question.
Step 1 Take geodesic normal coordinates $x^1,dots,x^n$ , centred at $p$, with $partial_nvert_p=nu(p)$. Using the implicit function theorem, one can show that there is a smooth function $f:mathbb{R}^{n-1}rightarrow mathbb{R}$ such that
$$
Omega cap V ={x^nge f(x^1,dots,x^{n-1})},
$$
where $V$ denotes the coordinate patch. Next define smooth vector fields $X_1,dots,X_{n-1}$ on $V$ by
$$
X_j(q)=partial_jvert_q + partial_jf(x^1(q),dots,x^{n-1}(q))partial_nvert_q.
$$
If $qin V cap partial Omega$, then $X_j(q) in T_qpartial Omega$ and hence for $1le jle n-1$ we have
$$
partial_jf(0) = X_j^k(p) g_{kl}(p) nu^l(p)=g(X_j(p),nu(p)) = 0 tag{1}.
$$
Step 2 We claim that $$
text{b)} quad Leftrightarrow quad f text{ has a local minimum at 0.} tag{2}
$$
Note that in the coordinates fixed above, b) means that ${x^n=0}$ (the hypersurface spanned by geodesics through $p$ which are not transversal) does not intersect $int(Omega)cap V ={x^n> f(x^1,dots,x^{n-1})}$ near $p$, or in other words that $f(x^1,dots,x^{n-1})ge 0$ in a neighbourhood of $p$. This proves (2).
Step 3 We're now in a position to prove a) $Rightarrow$ b). Suppose b) is wrong, then $f$ does not have a local minimum at $0$. I.e. for any neighbourhood $U$ of $p$ there is a point $qin partial Omega cap U$ such that $$x^n(q)=f(x^1(q),dots,x^{n-1}(q))<0. tag{3}$$
Since we are in geodesic coordinates, the unique minimising geodesic joining $p$ and $q$ is given by $L ={x^j=tx^j(q): 0le t le 1}$. Assuming a), we must have $Lsubset Omega cap U$, which implies
$$
tx^n(q) ge f(tx^1(q),dots,tx^{n-1}(q)).
$$
Divide by $t$ and take the limit $trightarrow 0$, then the right hand side will converge to $0$, since $Df(0)=0$ (see $(1)$). We obtain $x^n(q)ge0$, which is a contradiction to $(3)$.
Step 4 We want to relate the second fundamental form to the Hessian of $f$. To this end note that, for $1le j le n-1$ and $1le k le n$ we have
$$
nabla_kX_j (p)= (Gamma_{kj}^l partial_l + partial_k partial_j f partial_n + Gamma_{kn}^r partial_r)(p) = partial_jpartial_kf(p) cdot nu(p)
$$
and since $f$ is independent of $x^n$, we further obtain
$$
nabla_{X_k}X_j(p) = partial_jpartial_kf(p) cdot nu(p),
$$
which implies $l_{nu}(X_k,X_j)vert _p = partial_jpartial_kf(p)$. Since the $X_j$ form a basis of $T_ppartial Omega$ we have
$$
text{c)}quad Leftrightarrow quad f text{ has non-negative Hessian at $0$}. tag{4}
$$
From $(2)$ and $(4)$ it is evident that b) implies c).
Remark: With the same kind of argument one obtains an interesting result in the other direction. Assuming that the second fundamental form of $partial Omega$ is strictly positive, all geodesics coming from $Omega$ hit $partial Omega$ transversally.
answered Jan 29 at 12:17
Jan BohrJan Bohr
3,3071521
$endgroup$
I will address your first question. First define, for a fixed point $pinpartial Omega$, three conditions on $partial Omega$.
a) There is an open subset $Usubset M$ with $pin U$, such that any two points in $Ucap Omega$ can be joined by length minimising geodesic $c:[0,1]rightarrow M$ with $c[0,1]subset Ucap Omega$.
b) Any geodesic $c:(-epsilon,0]rightarrow M$ with $c(-epsilon,0)subset mathrm{int(Omega)}$ and $c(0)= p$ hits the boundary transversally, i.e. $g(dot c(0),nu(p))neq 0$.
c) The second fundamental form $l_nu(cdot,cdot)=g(nabla_cdot cdot, nu)$ is non-negative at $p$.
We will prove that a) $Rightarrow$ b) $Rightarrow$ c), which answers your first question.
Step 1 Take geodesic normal coordinates $x^1,dots,x^n$ , centred at $p$, with $partial_nvert_p=nu(p)$. Using the implicit function theorem, one can show that there is a smooth function $f:mathbb{R}^{n-1}rightarrow mathbb{R}$ such that
$$
Omega cap V ={x^nge f(x^1,dots,x^{n-1})},
$$
where $V$ denotes the coordinate patch. Next define smooth vector fields $X_1,dots,X_{n-1}$ on $V$ by
$$
X_j(q)=partial_jvert_q + partial_jf(x^1(q),dots,x^{n-1}(q))partial_nvert_q.
$$
If $qin V cap partial Omega$, then $X_j(q) in T_qpartial Omega$ and hence for $1le jle n-1$ we have
$$
partial_jf(0) = X_j^k(p) g_{kl}(p) nu^l(p)=g(X_j(p),nu(p)) = 0 tag{1}.
$$
Step 2 We claim that $$
text{b)} quad Leftrightarrow quad f text{ has a local minimum at 0.} tag{2}
$$
Note that in the coordinates fixed above, b) means that ${x^n=0}$ (the hypersurface spanned by geodesics through $p$ which are not transversal) does not intersect $int(Omega)cap V ={x^n> f(x^1,dots,x^{n-1})}$ near $p$, or in other words that $f(x^1,dots,x^{n-1})ge 0$ in a neighbourhood of $p$. This proves (2).
Step 3 We're now in a position to prove a) $Rightarrow$ b). Suppose b) is wrong, then $f$ does not have a local minimum at $0$. I.e. for any neighbourhood $U$ of $p$ there is a point $qin partial Omega cap U$ such that $$x^n(q)=f(x^1(q),dots,x^{n-1}(q))<0. tag{3}$$
Since we are in geodesic coordinates, the unique minimising geodesic joining $p$ and $q$ is given by $L ={x^j=tx^j(q): 0le t le 1}$. Assuming a), we must have $Lsubset Omega cap U$, which implies
$$
tx^n(q) ge f(tx^1(q),dots,tx^{n-1}(q)).
$$
Divide by $t$ and take the limit $trightarrow 0$, then the right hand side will converge to $0$, since $Df(0)=0$ (see $(1)$). We obtain $x^n(q)ge0$, which is a contradiction to $(3)$.
Step 4 We want to relate the second fundamental form to the Hessian of $f$. To this end note that, for $1le j le n-1$ and $1le k le n$ we have
$$
nabla_kX_j (p)= (Gamma_{kj}^l partial_l + partial_k partial_j f partial_n + Gamma_{kn}^r partial_r)(p) = partial_jpartial_kf(p) cdot nu(p)
$$
and since $f$ is independent of $x^n$, we further obtain
$$
nabla_{X_k}X_j(p) = partial_jpartial_kf(p) cdot nu(p),
$$
which implies $l_{nu}(X_k,X_j)vert _p = partial_jpartial_kf(p)$. Since the $X_j$ form a basis of $T_ppartial Omega$ we have
$$
text{c)}quad Leftrightarrow quad f text{ has non-negative Hessian at $0$}. tag{4}
$$
From $(2)$ and $(4)$ it is evident that b) implies c).
Remark: With the same kind of argument one obtains an interesting result in the other direction. Assuming that the second fundamental form of $partial Omega$ is strictly positive, all geodesics coming from $Omega$ hit $partial Omega$ transversally.
answered Jan 29 at 12:17
Jan BohrJan Bohr
3,3071521
edited Jan 29 at 16:50
answered Jan 29 at 12:17
Jan BohrJan Bohr
3,3071521
answered Jan 29 at 12:17
Jan BohrJan Bohr
3,3071521
answered Jan 29 at 12:17
Jan BohrJan Bohr
3,3071521
3,3071521
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