Does the Volume Ratio in a Complete Riemannian Manifold always tend to the Volume of the Euclidean Unit Ball?Interpreting the scalar curvature in a semi-Riemannian manifoldNon unique solution for Ricci flow equationEnergy functional(Basic) question regarding Einstein-Hilbert-functional / total scalar curvatureVolume of a complete, simply connected Riemannian manifold of constant negative curvatureComplete Riemannian manifold with fast volume growthUnderstanding the Riemannian volume density functionQuestion about a result of AndersonManifold with unit balls of unbounded volumeQuotient metric on a complete Riemannian manifold
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Does the Volume Ratio in a Complete Riemannian Manifold always tend to the Volume of the Euclidean Unit Ball?
Interpreting the scalar curvature in a semi-Riemannian manifoldNon unique solution for Ricci flow equationEnergy functional(Basic) question regarding Einstein-Hilbert-functional / total scalar curvatureVolume of a complete, simply connected Riemannian manifold of constant negative curvatureComplete Riemannian manifold with fast volume growthUnderstanding the Riemannian volume density functionQuestion about a result of AndersonManifold with unit balls of unbounded volumeQuotient metric on a complete Riemannian manifold
$begingroup$
I am reading Peter Topping's notes on Ricci flow: on page 99 a statement is made which is needed for his proof of a version of Perelman's no local volume collapse theorem, but I am not sure why it holds. If you define $omega_n$ to be the volume of the unit ball in Euclidean $n$-space and $V(p,s)$ to be the volume of the geodesic ball at point $p$ with radius $s$ given a complete Riemannian manifold $(M,g)$, then he argues that the volume ratio always tends to the volume of the Euclidean unit ball. $g(t)$ is a Ricci flow on the manifold for $t in [0,T]$ and we are working with a smooth metric $g(T)$.
$K(p,s)=fracV(p,s)s^n rightarrow omega_n$
as $s rightarrow 0$. I am not sure why this would have to hold on an arbitrary complete Riemannian manifold. If you are working on hyperbolic space or something like that with negative curvature, then now surely that relation will not hold as you will have a $sinh$ term in the denominator meaning that the ratio would tend to $0$, as opposed to the volume of the Euclidean unit ball.
differential-geometry riemannian-geometry ricci-flow
asked Mar 12 at 21:19
TomTom
316111
$endgroup$
add a comment |
$begingroup$
I am reading Peter Topping's notes on Ricci flow: on page 99 a statement is made which is needed for his proof of a version of Perelman's no local volume collapse theorem, but I am not sure why it holds. If you define $omega_n$ to be the volume of the unit ball in Euclidean $n$-space and $V(p,s)$ to be the volume of the geodesic ball at point $p$ with radius $s$ given a complete Riemannian manifold $(M,g)$, then he argues that the volume ratio always tends to the volume of the Euclidean unit ball. $g(t)$ is a Ricci flow on the manifold for $t in [0,T]$ and we are working with a smooth metric $g(T)$.
$K(p,s)=fracV(p,s)s^n rightarrow omega_n$
as $s rightarrow 0$. I am not sure why this would have to hold on an arbitrary complete Riemannian manifold. If you are working on hyperbolic space or something like that with negative curvature, then now surely that relation will not hold as you will have a $sinh$ term in the denominator meaning that the ratio would tend to $0$, as opposed to the volume of the Euclidean unit ball.
differential-geometry riemannian-geometry ricci-flow
asked Mar 12 at 21:19
TomTom
316111
$endgroup$
add a comment |
$begingroup$
I am reading Peter Topping's notes on Ricci flow: on page 99 a statement is made which is needed for his proof of a version of Perelman's no local volume collapse theorem, but I am not sure why it holds. If you define $omega_n$ to be the volume of the unit ball in Euclidean $n$-space and $V(p,s)$ to be the volume of the geodesic ball at point $p$ with radius $s$ given a complete Riemannian manifold $(M,g)$, then he argues that the volume ratio always tends to the volume of the Euclidean unit ball. $g(t)$ is a Ricci flow on the manifold for $t in [0,T]$ and we are working with a smooth metric $g(T)$.
$K(p,s)=fracV(p,s)s^n rightarrow omega_n$
as $s rightarrow 0$. I am not sure why this would have to hold on an arbitrary complete Riemannian manifold. If you are working on hyperbolic space or something like that with negative curvature, then now surely that relation will not hold as you will have a $sinh$ term in the denominator meaning that the ratio would tend to $0$, as opposed to the volume of the Euclidean unit ball.
differential-geometry riemannian-geometry ricci-flow
asked Mar 12 at 21:19
TomTom
316111
$endgroup$
I am reading Peter Topping's notes on Ricci flow: on page 99 a statement is made which is needed for his proof of a version of Perelman's no local volume collapse theorem, but I am not sure why it holds. If you define $omega_n$ to be the volume of the unit ball in Euclidean $n$-space and $V(p,s)$ to be the volume of the geodesic ball at point $p$ with radius $s$ given a complete Riemannian manifold $(M,g)$, then he argues that the volume ratio always tends to the volume of the Euclidean unit ball. $g(t)$ is a Ricci flow on the manifold for $t in [0,T]$ and we are working with a smooth metric $g(T)$.
$K(p,s)=fracV(p,s)s^n rightarrow omega_n$
as $s rightarrow 0$. I am not sure why this would have to hold on an arbitrary complete Riemannian manifold. If you are working on hyperbolic space or something like that with negative curvature, then now surely that relation will not hold as you will have a $sinh$ term in the denominator meaning that the ratio would tend to $0$, as opposed to the volume of the Euclidean unit ball.
differential-geometry riemannian-geometry ricci-flow
differential-geometry riemannian-geometry ricci-flow
asked Mar 12 at 21:19
TomTom
316111
asked Mar 12 at 21:19
TomTom
316111
asked Mar 12 at 21:19
TomTom
316111
asked Mar 12 at 21:19
TomTom
316111
asked Mar 12 at 21:19
TomTom
316111
316111
add a comment |
add a comment |
1 Answer
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Let $B^n$ denote the unit ball in $T_pM$. For $s>0$ small, let $varphi_s:B^nhookrightarrow M$ be given by $$vmapstoexp_p(sv).$$ Let $g_s$ be the pull-back metric on $B^n$ with respect to $varphi_s$, that is, $g_s=varphi_s^*g.$ Then, by definition, the volume $V(p,s)$ is equal to the volume of $B^n$ with respect to the metric $g_s$. It follows that the volume ratio, $K(p,s)$, is equal to the volume of $B^n$ with respect to the metric $h_s:=fracg_ss^2.$ A short computation shows that, as $s$ goes to $0$, the metric $h_s$ converges (uniformly) to the constant metric $$(h_0)_q(u,v)=g_p(u,v),qquad qin B^n,u,vin T_pM.$$ Hence, the corresponding Riemannian volume forms converge to the standard volume form and the claim follows.
Edit: Let us write the metric $g_s$ explicitly: for $qin B^n$ and $u,vin T_pM$ we have $$beginalign(g_s)_q(u,v)&=g_exp_p(sq)(d(exp_p)_sq(su),d(exp_p)_sq(sv))\&=s^2g_exp_p(sq)(d(exp_p)_sq(u),d(exp_p)_sq(v)).endalign$$ The metric $h_s$ is thus given by $$(h_s)_q(u,v)=g_exp_p(sq)(d(exp_p)_sq(u),d(exp_p)_sq(v)).$$By continuity of $g$, we have $$beginalignlim_sto0(h_s)_q(u,v)&=g_exp_p(0)(d(exp_p)_0(u),d(exp_p)_0(v))\&=g_p(u,v),endalign$$ where the convergence is uniform.
answered Mar 12 at 22:06
Amitai YuvalAmitai Yuval
15.5k11127
$endgroup$
$begingroup$
Could you elaborate as to why that metric $h_s$ converges to the constant metric ie. to the Kronecker delta?
$endgroup$
– Tom
Mar 13 at 23:45
$begingroup$
Also, is that meant to be $s^n$ in the denominator or is it actually $s^2$?
$endgroup$
– Tom
Mar 14 at 0:16
1
$begingroup$
@Tom Please check my edit regarding the convergence of the metric $h_s$. As for your second question, it is actually $s^2$. Whenever you multiply a Riemannian metric by a positive function $alpha$, the Riemannian volume form is multiplied by $alpha^n/2$.
$endgroup$
– Amitai Yuval
Mar 14 at 6:22
1
$begingroup$
@Tom I am not sure I understand all your questions. The first equality is just the definition of the pull-back metric and the chain rule (as the map $varphi_s$ is actually the composition of multiplication by $s$ followed by the exponential map). It is, indeed, $s^2$, as written (so this is the question I don't understand). Finally, the orientation is not an issue because your question is local, and every manifold is locally orientable. So just choose an orientation for $T_pM$ and use it to evaluate the volumes - the result is independent of the choice.
$endgroup$
– Amitai Yuval
Mar 14 at 7:21
1
$begingroup$
@Tom The derivative of a differentiable map is linear, and a Riemannian metric is bilinear. So each one of the $s$'s in the first line pops out.
$endgroup$
– Amitai Yuval
Mar 14 at 8:22
|
show 4 more comments
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1 Answer
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1 Answer
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$begingroup$
Let $B^n$ denote the unit ball in $T_pM$. For $s>0$ small, let $varphi_s:B^nhookrightarrow M$ be given by $$vmapstoexp_p(sv).$$ Let $g_s$ be the pull-back metric on $B^n$ with respect to $varphi_s$, that is, $g_s=varphi_s^*g.$ Then, by definition, the volume $V(p,s)$ is equal to the volume of $B^n$ with respect to the metric $g_s$. It follows that the volume ratio, $K(p,s)$, is equal to the volume of $B^n$ with respect to the metric $h_s:=fracg_ss^2.$ A short computation shows that, as $s$ goes to $0$, the metric $h_s$ converges (uniformly) to the constant metric $$(h_0)_q(u,v)=g_p(u,v),qquad qin B^n,u,vin T_pM.$$ Hence, the corresponding Riemannian volume forms converge to the standard volume form and the claim follows.
Edit: Let us write the metric $g_s$ explicitly: for $qin B^n$ and $u,vin T_pM$ we have $$beginalign(g_s)_q(u,v)&=g_exp_p(sq)(d(exp_p)_sq(su),d(exp_p)_sq(sv))\&=s^2g_exp_p(sq)(d(exp_p)_sq(u),d(exp_p)_sq(v)).endalign$$ The metric $h_s$ is thus given by $$(h_s)_q(u,v)=g_exp_p(sq)(d(exp_p)_sq(u),d(exp_p)_sq(v)).$$By continuity of $g$, we have $$beginalignlim_sto0(h_s)_q(u,v)&=g_exp_p(0)(d(exp_p)_0(u),d(exp_p)_0(v))\&=g_p(u,v),endalign$$ where the convergence is uniform.
answered Mar 12 at 22:06
Amitai YuvalAmitai Yuval
15.5k11127
$endgroup$
$begingroup$
Could you elaborate as to why that metric $h_s$ converges to the constant metric ie. to the Kronecker delta?
$endgroup$
– Tom
Mar 13 at 23:45
$begingroup$
Also, is that meant to be $s^n$ in the denominator or is it actually $s^2$?
$endgroup$
– Tom
Mar 14 at 0:16
1
$begingroup$
@Tom Please check my edit regarding the convergence of the metric $h_s$. As for your second question, it is actually $s^2$. Whenever you multiply a Riemannian metric by a positive function $alpha$, the Riemannian volume form is multiplied by $alpha^n/2$.
$endgroup$
– Amitai Yuval
Mar 14 at 6:22
1
$begingroup$
@Tom I am not sure I understand all your questions. The first equality is just the definition of the pull-back metric and the chain rule (as the map $varphi_s$ is actually the composition of multiplication by $s$ followed by the exponential map). It is, indeed, $s^2$, as written (so this is the question I don't understand). Finally, the orientation is not an issue because your question is local, and every manifold is locally orientable. So just choose an orientation for $T_pM$ and use it to evaluate the volumes - the result is independent of the choice.
$endgroup$
– Amitai Yuval
Mar 14 at 7:21
1
$begingroup$
@Tom The derivative of a differentiable map is linear, and a Riemannian metric is bilinear. So each one of the $s$'s in the first line pops out.
$endgroup$
– Amitai Yuval
Mar 14 at 8:22
|
show 4 more comments
$begingroup$
Let $B^n$ denote the unit ball in $T_pM$. For $s>0$ small, let $varphi_s:B^nhookrightarrow M$ be given by $$vmapstoexp_p(sv).$$ Let $g_s$ be the pull-back metric on $B^n$ with respect to $varphi_s$, that is, $g_s=varphi_s^*g.$ Then, by definition, the volume $V(p,s)$ is equal to the volume of $B^n$ with respect to the metric $g_s$. It follows that the volume ratio, $K(p,s)$, is equal to the volume of $B^n$ with respect to the metric $h_s:=fracg_ss^2.$ A short computation shows that, as $s$ goes to $0$, the metric $h_s$ converges (uniformly) to the constant metric $$(h_0)_q(u,v)=g_p(u,v),qquad qin B^n,u,vin T_pM.$$ Hence, the corresponding Riemannian volume forms converge to the standard volume form and the claim follows.
Edit: Let us write the metric $g_s$ explicitly: for $qin B^n$ and $u,vin T_pM$ we have $$beginalign(g_s)_q(u,v)&=g_exp_p(sq)(d(exp_p)_sq(su),d(exp_p)_sq(sv))\&=s^2g_exp_p(sq)(d(exp_p)_sq(u),d(exp_p)_sq(v)).endalign$$ The metric $h_s$ is thus given by $$(h_s)_q(u,v)=g_exp_p(sq)(d(exp_p)_sq(u),d(exp_p)_sq(v)).$$By continuity of $g$, we have $$beginalignlim_sto0(h_s)_q(u,v)&=g_exp_p(0)(d(exp_p)_0(u),d(exp_p)_0(v))\&=g_p(u,v),endalign$$ where the convergence is uniform.
answered Mar 12 at 22:06
Amitai YuvalAmitai Yuval
15.5k11127
$endgroup$
$begingroup$
Could you elaborate as to why that metric $h_s$ converges to the constant metric ie. to the Kronecker delta?
$endgroup$
– Tom
Mar 13 at 23:45
$begingroup$
Also, is that meant to be $s^n$ in the denominator or is it actually $s^2$?
$endgroup$
– Tom
Mar 14 at 0:16
1
$begingroup$
@Tom Please check my edit regarding the convergence of the metric $h_s$. As for your second question, it is actually $s^2$. Whenever you multiply a Riemannian metric by a positive function $alpha$, the Riemannian volume form is multiplied by $alpha^n/2$.
$endgroup$
– Amitai Yuval
Mar 14 at 6:22
1
$begingroup$
@Tom I am not sure I understand all your questions. The first equality is just the definition of the pull-back metric and the chain rule (as the map $varphi_s$ is actually the composition of multiplication by $s$ followed by the exponential map). It is, indeed, $s^2$, as written (so this is the question I don't understand). Finally, the orientation is not an issue because your question is local, and every manifold is locally orientable. So just choose an orientation for $T_pM$ and use it to evaluate the volumes - the result is independent of the choice.
$endgroup$
– Amitai Yuval
Mar 14 at 7:21
1
$begingroup$
@Tom The derivative of a differentiable map is linear, and a Riemannian metric is bilinear. So each one of the $s$'s in the first line pops out.
$endgroup$
– Amitai Yuval
Mar 14 at 8:22
|
show 4 more comments
$begingroup$
Let $B^n$ denote the unit ball in $T_pM$. For $s>0$ small, let $varphi_s:B^nhookrightarrow M$ be given by $$vmapstoexp_p(sv).$$ Let $g_s$ be the pull-back metric on $B^n$ with respect to $varphi_s$, that is, $g_s=varphi_s^*g.$ Then, by definition, the volume $V(p,s)$ is equal to the volume of $B^n$ with respect to the metric $g_s$. It follows that the volume ratio, $K(p,s)$, is equal to the volume of $B^n$ with respect to the metric $h_s:=fracg_ss^2.$ A short computation shows that, as $s$ goes to $0$, the metric $h_s$ converges (uniformly) to the constant metric $$(h_0)_q(u,v)=g_p(u,v),qquad qin B^n,u,vin T_pM.$$ Hence, the corresponding Riemannian volume forms converge to the standard volume form and the claim follows.
Edit: Let us write the metric $g_s$ explicitly: for $qin B^n$ and $u,vin T_pM$ we have $$beginalign(g_s)_q(u,v)&=g_exp_p(sq)(d(exp_p)_sq(su),d(exp_p)_sq(sv))\&=s^2g_exp_p(sq)(d(exp_p)_sq(u),d(exp_p)_sq(v)).endalign$$ The metric $h_s$ is thus given by $$(h_s)_q(u,v)=g_exp_p(sq)(d(exp_p)_sq(u),d(exp_p)_sq(v)).$$By continuity of $g$, we have $$beginalignlim_sto0(h_s)_q(u,v)&=g_exp_p(0)(d(exp_p)_0(u),d(exp_p)_0(v))\&=g_p(u,v),endalign$$ where the convergence is uniform.
answered Mar 12 at 22:06
Amitai YuvalAmitai Yuval
15.5k11127
$endgroup$
Let $B^n$ denote the unit ball in $T_pM$. For $s>0$ small, let $varphi_s:B^nhookrightarrow M$ be given by $$vmapstoexp_p(sv).$$ Let $g_s$ be the pull-back metric on $B^n$ with respect to $varphi_s$, that is, $g_s=varphi_s^*g.$ Then, by definition, the volume $V(p,s)$ is equal to the volume of $B^n$ with respect to the metric $g_s$. It follows that the volume ratio, $K(p,s)$, is equal to the volume of $B^n$ with respect to the metric $h_s:=fracg_ss^2.$ A short computation shows that, as $s$ goes to $0$, the metric $h_s$ converges (uniformly) to the constant metric $$(h_0)_q(u,v)=g_p(u,v),qquad qin B^n,u,vin T_pM.$$ Hence, the corresponding Riemannian volume forms converge to the standard volume form and the claim follows.
Edit: Let us write the metric $g_s$ explicitly: for $qin B^n$ and $u,vin T_pM$ we have $$beginalign(g_s)_q(u,v)&=g_exp_p(sq)(d(exp_p)_sq(su),d(exp_p)_sq(sv))\&=s^2g_exp_p(sq)(d(exp_p)_sq(u),d(exp_p)_sq(v)).endalign$$ The metric $h_s$ is thus given by $$(h_s)_q(u,v)=g_exp_p(sq)(d(exp_p)_sq(u),d(exp_p)_sq(v)).$$By continuity of $g$, we have $$beginalignlim_sto0(h_s)_q(u,v)&=g_exp_p(0)(d(exp_p)_0(u),d(exp_p)_0(v))\&=g_p(u,v),endalign$$ where the convergence is uniform.
answered Mar 12 at 22:06
Amitai YuvalAmitai Yuval
15.5k11127
edited Mar 14 at 6:44
answered Mar 12 at 22:06
Amitai YuvalAmitai Yuval
15.5k11127
answered Mar 12 at 22:06
Amitai YuvalAmitai Yuval
15.5k11127
answered Mar 12 at 22:06
Amitai YuvalAmitai Yuval
15.5k11127
15.5k11127
$begingroup$
Could you elaborate as to why that metric $h_s$ converges to the constant metric ie. to the Kronecker delta?
$endgroup$
– Tom
Mar 13 at 23:45
$begingroup$
Also, is that meant to be $s^n$ in the denominator or is it actually $s^2$?
$endgroup$
– Tom
Mar 14 at 0:16
1
$begingroup$
@Tom Please check my edit regarding the convergence of the metric $h_s$. As for your second question, it is actually $s^2$. Whenever you multiply a Riemannian metric by a positive function $alpha$, the Riemannian volume form is multiplied by $alpha^n/2$.
$endgroup$
– Amitai Yuval
Mar 14 at 6:22
1
$begingroup$
@Tom I am not sure I understand all your questions. The first equality is just the definition of the pull-back metric and the chain rule (as the map $varphi_s$ is actually the composition of multiplication by $s$ followed by the exponential map). It is, indeed, $s^2$, as written (so this is the question I don't understand). Finally, the orientation is not an issue because your question is local, and every manifold is locally orientable. So just choose an orientation for $T_pM$ and use it to evaluate the volumes - the result is independent of the choice.
$endgroup$
– Amitai Yuval
Mar 14 at 7:21
1
$begingroup$
@Tom The derivative of a differentiable map is linear, and a Riemannian metric is bilinear. So each one of the $s$'s in the first line pops out.
$endgroup$
– Amitai Yuval
Mar 14 at 8:22
|
show 4 more comments
$begingroup$
Could you elaborate as to why that metric $h_s$ converges to the constant metric ie. to the Kronecker delta?
$endgroup$
– Tom
Mar 13 at 23:45
$begingroup$
Also, is that meant to be $s^n$ in the denominator or is it actually $s^2$?
$endgroup$
– Tom
Mar 14 at 0:16
1
$begingroup$
@Tom Please check my edit regarding the convergence of the metric $h_s$. As for your second question, it is actually $s^2$. Whenever you multiply a Riemannian metric by a positive function $alpha$, the Riemannian volume form is multiplied by $alpha^n/2$.
$endgroup$
– Amitai Yuval
Mar 14 at 6:22
1
$begingroup$
@Tom I am not sure I understand all your questions. The first equality is just the definition of the pull-back metric and the chain rule (as the map $varphi_s$ is actually the composition of multiplication by $s$ followed by the exponential map). It is, indeed, $s^2$, as written (so this is the question I don't understand). Finally, the orientation is not an issue because your question is local, and every manifold is locally orientable. So just choose an orientation for $T_pM$ and use it to evaluate the volumes - the result is independent of the choice.
$endgroup$
– Amitai Yuval
Mar 14 at 7:21
1
$begingroup$
@Tom The derivative of a differentiable map is linear, and a Riemannian metric is bilinear. So each one of the $s$'s in the first line pops out.
$endgroup$
– Amitai Yuval
Mar 14 at 8:22
$begingroup$
Could you elaborate as to why that metric $h_s$ converges to the constant metric ie. to the Kronecker delta?
$endgroup$
– Tom
Mar 13 at 23:45
$begingroup$
Could you elaborate as to why that metric $h_s$ converges to the constant metric ie. to the Kronecker delta?
$endgroup$
– Tom
Mar 13 at 23:45
$begingroup$
Also, is that meant to be $s^n$ in the denominator or is it actually $s^2$?
$endgroup$
– Tom
Mar 14 at 0:16
$begingroup$
Also, is that meant to be $s^n$ in the denominator or is it actually $s^2$?
$endgroup$
– Tom
Mar 14 at 0:16
1
1
$begingroup$
@Tom Please check my edit regarding the convergence of the metric $h_s$. As for your second question, it is actually $s^2$. Whenever you multiply a Riemannian metric by a positive function $alpha$, the Riemannian volume form is multiplied by $alpha^n/2$.
$endgroup$
– Amitai Yuval
Mar 14 at 6:22
$begingroup$
@Tom Please check my edit regarding the convergence of the metric $h_s$. As for your second question, it is actually $s^2$. Whenever you multiply a Riemannian metric by a positive function $alpha$, the Riemannian volume form is multiplied by $alpha^n/2$.
$endgroup$
– Amitai Yuval
Mar 14 at 6:22
1
1
$begingroup$
@Tom I am not sure I understand all your questions. The first equality is just the definition of the pull-back metric and the chain rule (as the map $varphi_s$ is actually the composition of multiplication by $s$ followed by the exponential map). It is, indeed, $s^2$, as written (so this is the question I don't understand). Finally, the orientation is not an issue because your question is local, and every manifold is locally orientable. So just choose an orientation for $T_pM$ and use it to evaluate the volumes - the result is independent of the choice.
$endgroup$
– Amitai Yuval
Mar 14 at 7:21
$begingroup$
@Tom I am not sure I understand all your questions. The first equality is just the definition of the pull-back metric and the chain rule (as the map $varphi_s$ is actually the composition of multiplication by $s$ followed by the exponential map). It is, indeed, $s^2$, as written (so this is the question I don't understand). Finally, the orientation is not an issue because your question is local, and every manifold is locally orientable. So just choose an orientation for $T_pM$ and use it to evaluate the volumes - the result is independent of the choice.
$endgroup$
– Amitai Yuval
Mar 14 at 7:21
1
1
$begingroup$
@Tom The derivative of a differentiable map is linear, and a Riemannian metric is bilinear. So each one of the $s$'s in the first line pops out.
$endgroup$
– Amitai Yuval
Mar 14 at 8:22
$begingroup$
@Tom The derivative of a differentiable map is linear, and a Riemannian metric is bilinear. So each one of the $s$'s in the first line pops out.
$endgroup$
– Amitai Yuval
Mar 14 at 8:22
|
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