Hyperbolic Geometry: Representation of the metric in a chart












1














Let $mathbb{R}^{1,n}$ denote Lorentzian $n$-space, i.e., $mathbb{R}^n$ equipped with the Lorentzian inner product $$langle x,y rangle = - x_0 y_0 + x_1 y_1 + cdots + x_n y_n.$$



Define $mathbb{H}^n : = left { xi in mathbb{R}^{1,n} : langle xi, xi rangle = -1, xi_0 > 0 right }$ to be hyperbolic $n$-space. I have been battling with the following seemingly standard/trivial exercises in Jöst's Riemannian Geometry and Geometric Analysis and have had little success:



5) Show that $langle cdot, cdot rangle$ induces a Riemannian metric on the tangent spaces $T_p mathbb{H}^n subset T_p mathbb{R}^{n+1}$ for $p in mathbb{H}^n$.



6) Let $s = (-1, 0, ..., 0) in mathbb{R}^{n+1}$, and define $$f(x) = s - frac{2(x-s)}{langle x-s, x-s rangle}.$$ Show that $f : mathbb{H}^n longrightarrow { xi in mathbb{R}^n : | xi | < 1 }$ is a diffeomorphism, and that in this chart, the metric assumes the form $$frac{4}{(1 - | xi |^2)^2} dxi_i otimes dxi_i.$$



Thoughts/Progress:



The fact that $langle cdot, cdot rangle$ defines an inner product is clear, but I cannot convince myself that $langle cdot, cdot rangle$ is positive definite. For Q6., I am aware that a similar question has been posted on here, but the solution is not clear to me.



Note that I do not want answers to these problems, just guidance.



Thanks in advance.










share|cite|improve this question






















  • For $5)$ I have a trick so I don't know how to give a hint about this solution.
    – Dante Grevino
    Nov 25 at 1:29






  • 1




    Are you familiar with differential forms and their use in differential geometry? If so, I can give some hints.
    – Ted Shifrin
    Nov 25 at 2:41










  • @TedShifrin I am familiar with differential forms. I am just new to the hyperbolic material.
    – AmorFati
    Nov 25 at 20:21










  • Actually, I take it back. The check that the induced metric on the orthogonal complement of $xi$ is positive-definite is just basic linear algebra. If you want to see an easy moving-frames computation that this space has constant curvature $-1$, I'll show you that ... :)
    – Ted Shifrin
    Nov 25 at 22:34






  • 1




    Sure thing. If my answer here makes sense to you, I'll show you how things get modified in the $O(n,1)$ case. Just post a separate question.
    – Ted Shifrin
    Nov 26 at 0:13


















1














Let $mathbb{R}^{1,n}$ denote Lorentzian $n$-space, i.e., $mathbb{R}^n$ equipped with the Lorentzian inner product $$langle x,y rangle = - x_0 y_0 + x_1 y_1 + cdots + x_n y_n.$$



Define $mathbb{H}^n : = left { xi in mathbb{R}^{1,n} : langle xi, xi rangle = -1, xi_0 > 0 right }$ to be hyperbolic $n$-space. I have been battling with the following seemingly standard/trivial exercises in Jöst's Riemannian Geometry and Geometric Analysis and have had little success:



5) Show that $langle cdot, cdot rangle$ induces a Riemannian metric on the tangent spaces $T_p mathbb{H}^n subset T_p mathbb{R}^{n+1}$ for $p in mathbb{H}^n$.



6) Let $s = (-1, 0, ..., 0) in mathbb{R}^{n+1}$, and define $$f(x) = s - frac{2(x-s)}{langle x-s, x-s rangle}.$$ Show that $f : mathbb{H}^n longrightarrow { xi in mathbb{R}^n : | xi | < 1 }$ is a diffeomorphism, and that in this chart, the metric assumes the form $$frac{4}{(1 - | xi |^2)^2} dxi_i otimes dxi_i.$$



Thoughts/Progress:



The fact that $langle cdot, cdot rangle$ defines an inner product is clear, but I cannot convince myself that $langle cdot, cdot rangle$ is positive definite. For Q6., I am aware that a similar question has been posted on here, but the solution is not clear to me.



Note that I do not want answers to these problems, just guidance.



Thanks in advance.










share|cite|improve this question






















  • For $5)$ I have a trick so I don't know how to give a hint about this solution.
    – Dante Grevino
    Nov 25 at 1:29






  • 1




    Are you familiar with differential forms and their use in differential geometry? If so, I can give some hints.
    – Ted Shifrin
    Nov 25 at 2:41










  • @TedShifrin I am familiar with differential forms. I am just new to the hyperbolic material.
    – AmorFati
    Nov 25 at 20:21










  • Actually, I take it back. The check that the induced metric on the orthogonal complement of $xi$ is positive-definite is just basic linear algebra. If you want to see an easy moving-frames computation that this space has constant curvature $-1$, I'll show you that ... :)
    – Ted Shifrin
    Nov 25 at 22:34






  • 1




    Sure thing. If my answer here makes sense to you, I'll show you how things get modified in the $O(n,1)$ case. Just post a separate question.
    – Ted Shifrin
    Nov 26 at 0:13
















1












1








1







Let $mathbb{R}^{1,n}$ denote Lorentzian $n$-space, i.e., $mathbb{R}^n$ equipped with the Lorentzian inner product $$langle x,y rangle = - x_0 y_0 + x_1 y_1 + cdots + x_n y_n.$$



Define $mathbb{H}^n : = left { xi in mathbb{R}^{1,n} : langle xi, xi rangle = -1, xi_0 > 0 right }$ to be hyperbolic $n$-space. I have been battling with the following seemingly standard/trivial exercises in Jöst's Riemannian Geometry and Geometric Analysis and have had little success:



5) Show that $langle cdot, cdot rangle$ induces a Riemannian metric on the tangent spaces $T_p mathbb{H}^n subset T_p mathbb{R}^{n+1}$ for $p in mathbb{H}^n$.



6) Let $s = (-1, 0, ..., 0) in mathbb{R}^{n+1}$, and define $$f(x) = s - frac{2(x-s)}{langle x-s, x-s rangle}.$$ Show that $f : mathbb{H}^n longrightarrow { xi in mathbb{R}^n : | xi | < 1 }$ is a diffeomorphism, and that in this chart, the metric assumes the form $$frac{4}{(1 - | xi |^2)^2} dxi_i otimes dxi_i.$$



Thoughts/Progress:



The fact that $langle cdot, cdot rangle$ defines an inner product is clear, but I cannot convince myself that $langle cdot, cdot rangle$ is positive definite. For Q6., I am aware that a similar question has been posted on here, but the solution is not clear to me.



Note that I do not want answers to these problems, just guidance.



Thanks in advance.










share|cite|improve this question













Let $mathbb{R}^{1,n}$ denote Lorentzian $n$-space, i.e., $mathbb{R}^n$ equipped with the Lorentzian inner product $$langle x,y rangle = - x_0 y_0 + x_1 y_1 + cdots + x_n y_n.$$



Define $mathbb{H}^n : = left { xi in mathbb{R}^{1,n} : langle xi, xi rangle = -1, xi_0 > 0 right }$ to be hyperbolic $n$-space. I have been battling with the following seemingly standard/trivial exercises in Jöst's Riemannian Geometry and Geometric Analysis and have had little success:



5) Show that $langle cdot, cdot rangle$ induces a Riemannian metric on the tangent spaces $T_p mathbb{H}^n subset T_p mathbb{R}^{n+1}$ for $p in mathbb{H}^n$.



6) Let $s = (-1, 0, ..., 0) in mathbb{R}^{n+1}$, and define $$f(x) = s - frac{2(x-s)}{langle x-s, x-s rangle}.$$ Show that $f : mathbb{H}^n longrightarrow { xi in mathbb{R}^n : | xi | < 1 }$ is a diffeomorphism, and that in this chart, the metric assumes the form $$frac{4}{(1 - | xi |^2)^2} dxi_i otimes dxi_i.$$



Thoughts/Progress:



The fact that $langle cdot, cdot rangle$ defines an inner product is clear, but I cannot convince myself that $langle cdot, cdot rangle$ is positive definite. For Q6., I am aware that a similar question has been posted on here, but the solution is not clear to me.



Note that I do not want answers to these problems, just guidance.



Thanks in advance.







differential-geometry riemannian-geometry hyperbolic-geometry






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asked Nov 24 at 21:49









AmorFati

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391529












  • For $5)$ I have a trick so I don't know how to give a hint about this solution.
    – Dante Grevino
    Nov 25 at 1:29






  • 1




    Are you familiar with differential forms and their use in differential geometry? If so, I can give some hints.
    – Ted Shifrin
    Nov 25 at 2:41










  • @TedShifrin I am familiar with differential forms. I am just new to the hyperbolic material.
    – AmorFati
    Nov 25 at 20:21










  • Actually, I take it back. The check that the induced metric on the orthogonal complement of $xi$ is positive-definite is just basic linear algebra. If you want to see an easy moving-frames computation that this space has constant curvature $-1$, I'll show you that ... :)
    – Ted Shifrin
    Nov 25 at 22:34






  • 1




    Sure thing. If my answer here makes sense to you, I'll show you how things get modified in the $O(n,1)$ case. Just post a separate question.
    – Ted Shifrin
    Nov 26 at 0:13




















  • For $5)$ I have a trick so I don't know how to give a hint about this solution.
    – Dante Grevino
    Nov 25 at 1:29






  • 1




    Are you familiar with differential forms and their use in differential geometry? If so, I can give some hints.
    – Ted Shifrin
    Nov 25 at 2:41










  • @TedShifrin I am familiar with differential forms. I am just new to the hyperbolic material.
    – AmorFati
    Nov 25 at 20:21










  • Actually, I take it back. The check that the induced metric on the orthogonal complement of $xi$ is positive-definite is just basic linear algebra. If you want to see an easy moving-frames computation that this space has constant curvature $-1$, I'll show you that ... :)
    – Ted Shifrin
    Nov 25 at 22:34






  • 1




    Sure thing. If my answer here makes sense to you, I'll show you how things get modified in the $O(n,1)$ case. Just post a separate question.
    – Ted Shifrin
    Nov 26 at 0:13


















For $5)$ I have a trick so I don't know how to give a hint about this solution.
– Dante Grevino
Nov 25 at 1:29




For $5)$ I have a trick so I don't know how to give a hint about this solution.
– Dante Grevino
Nov 25 at 1:29




1




1




Are you familiar with differential forms and their use in differential geometry? If so, I can give some hints.
– Ted Shifrin
Nov 25 at 2:41




Are you familiar with differential forms and their use in differential geometry? If so, I can give some hints.
– Ted Shifrin
Nov 25 at 2:41












@TedShifrin I am familiar with differential forms. I am just new to the hyperbolic material.
– AmorFati
Nov 25 at 20:21




@TedShifrin I am familiar with differential forms. I am just new to the hyperbolic material.
– AmorFati
Nov 25 at 20:21












Actually, I take it back. The check that the induced metric on the orthogonal complement of $xi$ is positive-definite is just basic linear algebra. If you want to see an easy moving-frames computation that this space has constant curvature $-1$, I'll show you that ... :)
– Ted Shifrin
Nov 25 at 22:34




Actually, I take it back. The check that the induced metric on the orthogonal complement of $xi$ is positive-definite is just basic linear algebra. If you want to see an easy moving-frames computation that this space has constant curvature $-1$, I'll show you that ... :)
– Ted Shifrin
Nov 25 at 22:34




1




1




Sure thing. If my answer here makes sense to you, I'll show you how things get modified in the $O(n,1)$ case. Just post a separate question.
– Ted Shifrin
Nov 26 at 0:13






Sure thing. If my answer here makes sense to you, I'll show you how things get modified in the $O(n,1)$ case. Just post a separate question.
– Ted Shifrin
Nov 26 at 0:13












1 Answer
1






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oldest

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3














We will prove $5)$. As differential manifold, $mathbb{H}^n$ is a regular submanifold of $mathbb{R}^{n+1}$ so we can identify tangent vectors in $mathbb{H}^n$ as velocities of differentiable curves in $mathbb{H}^n$. Take a point $p=(p_0,ldots,p_n)$ in $mathbb{H}^n$ and a differentiable curve $sigma : (-varepsilon,varepsilon)to mathbb{H}^n$ such that $sigma(0)=p$. Write the coordinates of $sigma$ as $sigma_0,ldots,sigma_n$ and let $v_i=frac{dsigma_i}{dt}(0)$ for every $i$. So we want to prove that $sum_{i=1}^nv_i^2>v_0^2$.



We have that $langle sigma(t),sigma(t) rangle =-1$ for every $t$. So in particular at $0$ we have that



$$
sum_{i=1}^np_i^2 = p_0^2-1
$$



On the other hand, by the product rule for derivatives we have that



$$
sum_{i=1}^nv_ip_i = v_0p_0
$$



So, by the Cauchy-Schwarz inequality and the first equation, it follows that



$$
v_0^2p_0^2=bigg{(}sum_{i=1}^nv_ip_ibigg{)}^2leq bigg{(}sum_{i=1}^nv_i^2bigg{)} bigg{(}sum_{i=1}^np_i^2bigg{)} = bigg{(}sum_{i=1}^nv_i^2bigg{)} bigg{(}p_0^2-1bigg{)} < bigg{(}sum_{i=1}^nv_i^2bigg{)} p_0^2
$$



We know that $p_0>0$ by definition of $mathbb{H}^n$ so the proposition follows.



EDIT: Sketch of $6)$. Let $mathbb{D}^n={xiinmathbb{R}^n:|xi|<1}$. Note that $f:mathbb{H}^ntomathbb{D}^n$ is defined by the formula
$$
f(x_0,ldots,x_n)=bigg{(}frac{x_1}{1+x_0},ldots,frac{x_n}{1+x_0}bigg{)}
$$



and it is differentiable. The inverse $f^{-1}:mathbb{D}^ntomathbb{H}^n$, defined by the formula



$$
f^{-1}(xi_1,ldots,xi_n)=bigg{(}frac{1+|xi|^2}{1-|xi|^2},frac{2xi_1}{1-|xi|^2},ldots,frac{2xi_n}{1-|xi|^2}bigg{)},
$$



is differentiable so $f$ is a diffeomorphism. Prove that the pullback by $f^{-1}$ of the metric defined in $mathbb{H}^n$ is equal to the metric given on $mathbb{D}^n$, that is



$$
bigg{langle} frac{partial f^{-1}}{partial xi_i}(xi),frac{partial f^{-1}}{partialxi_j}(xi) bigg{rangle} = frac{4delta_{ij}}{(1-|xi|^2)^2}
$$



for every $i,jin {1,ldots,n}$ and every $xiinmathbb{D}^n$.



Note that $6)$ implies $5)$ because the pullback by a diffeomorphism of a riemannian metric is a riemannian metric.






share|cite|improve this answer























  • Thank you for your time. I very much appreciate this.
    – AmorFati
    Nov 25 at 20:20










  • You are welcome! Please select my answer as the "accepted" answer if it works for you. See: meta.stackexchange.com/questions/5234/…
    – Dante Grevino
    Nov 25 at 20:42










  • Very pleased with this answer, thanks again.
    – AmorFati
    Nov 25 at 23:59











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1 Answer
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3














We will prove $5)$. As differential manifold, $mathbb{H}^n$ is a regular submanifold of $mathbb{R}^{n+1}$ so we can identify tangent vectors in $mathbb{H}^n$ as velocities of differentiable curves in $mathbb{H}^n$. Take a point $p=(p_0,ldots,p_n)$ in $mathbb{H}^n$ and a differentiable curve $sigma : (-varepsilon,varepsilon)to mathbb{H}^n$ such that $sigma(0)=p$. Write the coordinates of $sigma$ as $sigma_0,ldots,sigma_n$ and let $v_i=frac{dsigma_i}{dt}(0)$ for every $i$. So we want to prove that $sum_{i=1}^nv_i^2>v_0^2$.



We have that $langle sigma(t),sigma(t) rangle =-1$ for every $t$. So in particular at $0$ we have that



$$
sum_{i=1}^np_i^2 = p_0^2-1
$$



On the other hand, by the product rule for derivatives we have that



$$
sum_{i=1}^nv_ip_i = v_0p_0
$$



So, by the Cauchy-Schwarz inequality and the first equation, it follows that



$$
v_0^2p_0^2=bigg{(}sum_{i=1}^nv_ip_ibigg{)}^2leq bigg{(}sum_{i=1}^nv_i^2bigg{)} bigg{(}sum_{i=1}^np_i^2bigg{)} = bigg{(}sum_{i=1}^nv_i^2bigg{)} bigg{(}p_0^2-1bigg{)} < bigg{(}sum_{i=1}^nv_i^2bigg{)} p_0^2
$$



We know that $p_0>0$ by definition of $mathbb{H}^n$ so the proposition follows.



EDIT: Sketch of $6)$. Let $mathbb{D}^n={xiinmathbb{R}^n:|xi|<1}$. Note that $f:mathbb{H}^ntomathbb{D}^n$ is defined by the formula
$$
f(x_0,ldots,x_n)=bigg{(}frac{x_1}{1+x_0},ldots,frac{x_n}{1+x_0}bigg{)}
$$



and it is differentiable. The inverse $f^{-1}:mathbb{D}^ntomathbb{H}^n$, defined by the formula



$$
f^{-1}(xi_1,ldots,xi_n)=bigg{(}frac{1+|xi|^2}{1-|xi|^2},frac{2xi_1}{1-|xi|^2},ldots,frac{2xi_n}{1-|xi|^2}bigg{)},
$$



is differentiable so $f$ is a diffeomorphism. Prove that the pullback by $f^{-1}$ of the metric defined in $mathbb{H}^n$ is equal to the metric given on $mathbb{D}^n$, that is



$$
bigg{langle} frac{partial f^{-1}}{partial xi_i}(xi),frac{partial f^{-1}}{partialxi_j}(xi) bigg{rangle} = frac{4delta_{ij}}{(1-|xi|^2)^2}
$$



for every $i,jin {1,ldots,n}$ and every $xiinmathbb{D}^n$.



Note that $6)$ implies $5)$ because the pullback by a diffeomorphism of a riemannian metric is a riemannian metric.






share|cite|improve this answer























  • Thank you for your time. I very much appreciate this.
    – AmorFati
    Nov 25 at 20:20










  • You are welcome! Please select my answer as the "accepted" answer if it works for you. See: meta.stackexchange.com/questions/5234/…
    – Dante Grevino
    Nov 25 at 20:42










  • Very pleased with this answer, thanks again.
    – AmorFati
    Nov 25 at 23:59
















3














We will prove $5)$. As differential manifold, $mathbb{H}^n$ is a regular submanifold of $mathbb{R}^{n+1}$ so we can identify tangent vectors in $mathbb{H}^n$ as velocities of differentiable curves in $mathbb{H}^n$. Take a point $p=(p_0,ldots,p_n)$ in $mathbb{H}^n$ and a differentiable curve $sigma : (-varepsilon,varepsilon)to mathbb{H}^n$ such that $sigma(0)=p$. Write the coordinates of $sigma$ as $sigma_0,ldots,sigma_n$ and let $v_i=frac{dsigma_i}{dt}(0)$ for every $i$. So we want to prove that $sum_{i=1}^nv_i^2>v_0^2$.



We have that $langle sigma(t),sigma(t) rangle =-1$ for every $t$. So in particular at $0$ we have that



$$
sum_{i=1}^np_i^2 = p_0^2-1
$$



On the other hand, by the product rule for derivatives we have that



$$
sum_{i=1}^nv_ip_i = v_0p_0
$$



So, by the Cauchy-Schwarz inequality and the first equation, it follows that



$$
v_0^2p_0^2=bigg{(}sum_{i=1}^nv_ip_ibigg{)}^2leq bigg{(}sum_{i=1}^nv_i^2bigg{)} bigg{(}sum_{i=1}^np_i^2bigg{)} = bigg{(}sum_{i=1}^nv_i^2bigg{)} bigg{(}p_0^2-1bigg{)} < bigg{(}sum_{i=1}^nv_i^2bigg{)} p_0^2
$$



We know that $p_0>0$ by definition of $mathbb{H}^n$ so the proposition follows.



EDIT: Sketch of $6)$. Let $mathbb{D}^n={xiinmathbb{R}^n:|xi|<1}$. Note that $f:mathbb{H}^ntomathbb{D}^n$ is defined by the formula
$$
f(x_0,ldots,x_n)=bigg{(}frac{x_1}{1+x_0},ldots,frac{x_n}{1+x_0}bigg{)}
$$



and it is differentiable. The inverse $f^{-1}:mathbb{D}^ntomathbb{H}^n$, defined by the formula



$$
f^{-1}(xi_1,ldots,xi_n)=bigg{(}frac{1+|xi|^2}{1-|xi|^2},frac{2xi_1}{1-|xi|^2},ldots,frac{2xi_n}{1-|xi|^2}bigg{)},
$$



is differentiable so $f$ is a diffeomorphism. Prove that the pullback by $f^{-1}$ of the metric defined in $mathbb{H}^n$ is equal to the metric given on $mathbb{D}^n$, that is



$$
bigg{langle} frac{partial f^{-1}}{partial xi_i}(xi),frac{partial f^{-1}}{partialxi_j}(xi) bigg{rangle} = frac{4delta_{ij}}{(1-|xi|^2)^2}
$$



for every $i,jin {1,ldots,n}$ and every $xiinmathbb{D}^n$.



Note that $6)$ implies $5)$ because the pullback by a diffeomorphism of a riemannian metric is a riemannian metric.






share|cite|improve this answer























  • Thank you for your time. I very much appreciate this.
    – AmorFati
    Nov 25 at 20:20










  • You are welcome! Please select my answer as the "accepted" answer if it works for you. See: meta.stackexchange.com/questions/5234/…
    – Dante Grevino
    Nov 25 at 20:42










  • Very pleased with this answer, thanks again.
    – AmorFati
    Nov 25 at 23:59














3












3








3






We will prove $5)$. As differential manifold, $mathbb{H}^n$ is a regular submanifold of $mathbb{R}^{n+1}$ so we can identify tangent vectors in $mathbb{H}^n$ as velocities of differentiable curves in $mathbb{H}^n$. Take a point $p=(p_0,ldots,p_n)$ in $mathbb{H}^n$ and a differentiable curve $sigma : (-varepsilon,varepsilon)to mathbb{H}^n$ such that $sigma(0)=p$. Write the coordinates of $sigma$ as $sigma_0,ldots,sigma_n$ and let $v_i=frac{dsigma_i}{dt}(0)$ for every $i$. So we want to prove that $sum_{i=1}^nv_i^2>v_0^2$.



We have that $langle sigma(t),sigma(t) rangle =-1$ for every $t$. So in particular at $0$ we have that



$$
sum_{i=1}^np_i^2 = p_0^2-1
$$



On the other hand, by the product rule for derivatives we have that



$$
sum_{i=1}^nv_ip_i = v_0p_0
$$



So, by the Cauchy-Schwarz inequality and the first equation, it follows that



$$
v_0^2p_0^2=bigg{(}sum_{i=1}^nv_ip_ibigg{)}^2leq bigg{(}sum_{i=1}^nv_i^2bigg{)} bigg{(}sum_{i=1}^np_i^2bigg{)} = bigg{(}sum_{i=1}^nv_i^2bigg{)} bigg{(}p_0^2-1bigg{)} < bigg{(}sum_{i=1}^nv_i^2bigg{)} p_0^2
$$



We know that $p_0>0$ by definition of $mathbb{H}^n$ so the proposition follows.



EDIT: Sketch of $6)$. Let $mathbb{D}^n={xiinmathbb{R}^n:|xi|<1}$. Note that $f:mathbb{H}^ntomathbb{D}^n$ is defined by the formula
$$
f(x_0,ldots,x_n)=bigg{(}frac{x_1}{1+x_0},ldots,frac{x_n}{1+x_0}bigg{)}
$$



and it is differentiable. The inverse $f^{-1}:mathbb{D}^ntomathbb{H}^n$, defined by the formula



$$
f^{-1}(xi_1,ldots,xi_n)=bigg{(}frac{1+|xi|^2}{1-|xi|^2},frac{2xi_1}{1-|xi|^2},ldots,frac{2xi_n}{1-|xi|^2}bigg{)},
$$



is differentiable so $f$ is a diffeomorphism. Prove that the pullback by $f^{-1}$ of the metric defined in $mathbb{H}^n$ is equal to the metric given on $mathbb{D}^n$, that is



$$
bigg{langle} frac{partial f^{-1}}{partial xi_i}(xi),frac{partial f^{-1}}{partialxi_j}(xi) bigg{rangle} = frac{4delta_{ij}}{(1-|xi|^2)^2}
$$



for every $i,jin {1,ldots,n}$ and every $xiinmathbb{D}^n$.



Note that $6)$ implies $5)$ because the pullback by a diffeomorphism of a riemannian metric is a riemannian metric.






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We will prove $5)$. As differential manifold, $mathbb{H}^n$ is a regular submanifold of $mathbb{R}^{n+1}$ so we can identify tangent vectors in $mathbb{H}^n$ as velocities of differentiable curves in $mathbb{H}^n$. Take a point $p=(p_0,ldots,p_n)$ in $mathbb{H}^n$ and a differentiable curve $sigma : (-varepsilon,varepsilon)to mathbb{H}^n$ such that $sigma(0)=p$. Write the coordinates of $sigma$ as $sigma_0,ldots,sigma_n$ and let $v_i=frac{dsigma_i}{dt}(0)$ for every $i$. So we want to prove that $sum_{i=1}^nv_i^2>v_0^2$.



We have that $langle sigma(t),sigma(t) rangle =-1$ for every $t$. So in particular at $0$ we have that



$$
sum_{i=1}^np_i^2 = p_0^2-1
$$



On the other hand, by the product rule for derivatives we have that



$$
sum_{i=1}^nv_ip_i = v_0p_0
$$



So, by the Cauchy-Schwarz inequality and the first equation, it follows that



$$
v_0^2p_0^2=bigg{(}sum_{i=1}^nv_ip_ibigg{)}^2leq bigg{(}sum_{i=1}^nv_i^2bigg{)} bigg{(}sum_{i=1}^np_i^2bigg{)} = bigg{(}sum_{i=1}^nv_i^2bigg{)} bigg{(}p_0^2-1bigg{)} < bigg{(}sum_{i=1}^nv_i^2bigg{)} p_0^2
$$



We know that $p_0>0$ by definition of $mathbb{H}^n$ so the proposition follows.



EDIT: Sketch of $6)$. Let $mathbb{D}^n={xiinmathbb{R}^n:|xi|<1}$. Note that $f:mathbb{H}^ntomathbb{D}^n$ is defined by the formula
$$
f(x_0,ldots,x_n)=bigg{(}frac{x_1}{1+x_0},ldots,frac{x_n}{1+x_0}bigg{)}
$$



and it is differentiable. The inverse $f^{-1}:mathbb{D}^ntomathbb{H}^n$, defined by the formula



$$
f^{-1}(xi_1,ldots,xi_n)=bigg{(}frac{1+|xi|^2}{1-|xi|^2},frac{2xi_1}{1-|xi|^2},ldots,frac{2xi_n}{1-|xi|^2}bigg{)},
$$



is differentiable so $f$ is a diffeomorphism. Prove that the pullback by $f^{-1}$ of the metric defined in $mathbb{H}^n$ is equal to the metric given on $mathbb{D}^n$, that is



$$
bigg{langle} frac{partial f^{-1}}{partial xi_i}(xi),frac{partial f^{-1}}{partialxi_j}(xi) bigg{rangle} = frac{4delta_{ij}}{(1-|xi|^2)^2}
$$



for every $i,jin {1,ldots,n}$ and every $xiinmathbb{D}^n$.



Note that $6)$ implies $5)$ because the pullback by a diffeomorphism of a riemannian metric is a riemannian metric.







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edited Nov 25 at 3:57

























answered Nov 25 at 2:00









Dante Grevino

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  • Thank you for your time. I very much appreciate this.
    – AmorFati
    Nov 25 at 20:20










  • You are welcome! Please select my answer as the "accepted" answer if it works for you. See: meta.stackexchange.com/questions/5234/…
    – Dante Grevino
    Nov 25 at 20:42










  • Very pleased with this answer, thanks again.
    – AmorFati
    Nov 25 at 23:59


















  • Thank you for your time. I very much appreciate this.
    – AmorFati
    Nov 25 at 20:20










  • You are welcome! Please select my answer as the "accepted" answer if it works for you. See: meta.stackexchange.com/questions/5234/…
    – Dante Grevino
    Nov 25 at 20:42










  • Very pleased with this answer, thanks again.
    – AmorFati
    Nov 25 at 23:59
















Thank you for your time. I very much appreciate this.
– AmorFati
Nov 25 at 20:20




Thank you for your time. I very much appreciate this.
– AmorFati
Nov 25 at 20:20












You are welcome! Please select my answer as the "accepted" answer if it works for you. See: meta.stackexchange.com/questions/5234/…
– Dante Grevino
Nov 25 at 20:42




You are welcome! Please select my answer as the "accepted" answer if it works for you. See: meta.stackexchange.com/questions/5234/…
– Dante Grevino
Nov 25 at 20:42












Very pleased with this answer, thanks again.
– AmorFati
Nov 25 at 23:59




Very pleased with this answer, thanks again.
– AmorFati
Nov 25 at 23:59


















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