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I'm having trouble comprehending a question from Do Carmo's Differential Forms and Applications. The question (in its entirety) is as follows:

(Exercise 5-2 in Do Carmo). Let $H^2$ be the upper half-plane, that is, $$ H^2=\{(x,y)\in\mathbb{R}^2;y>0\}. $$ Consider in $H^2$ the following inner product: If $(x,y)\in H^2$ and $u,v\in T_pH^2$, then \begin{equation} \langle u,v \rangle_p=\frac{u\cdot v}{y^2} \end{equation} where $u\cdot v$ is the canonical inner product of $\mathbb{R}^2$. Prove that this is a Riemannian metric in $H^2$ whose Gaussian curvature is $K\equiv -1$; with this Riemannian metric $H^2$ is called the hyperbolic plane. [Hint: Choose the orthonormal frame $e_1=\frac{a_1}{y}$, $e_2=\frac{a_2}{y}$, where $\{a_1,a_2\}$ is the canonical frame of $\mathbb{R}^2$.]

I'm not all that troubled with the question, except that I don't understand the hint to the point where I think there's a typo (though there probably isn't).

I get how $e_1,e_2$ are orthogonal with respect to the inner product $\langle\cdot\,,\cdot\rangle$, but $$ \langle e_1,e_1\rangle_p=\frac{e_1\cdot e_1}{y^2}=\frac{(a_1\cdot a_1)/y^2}{y^2}=\frac{1}{y^4}\neq 1. $$ So $e_1,e_2$ can't be orthonormal with the inner product. It would work perfectly with $e_1=ya_1$ and $e_2=ya_2$, but then why does the hint say something different? Am I misunderstanding what he means by "orthonormal?" Any help is appreciated. Thank you in advance.

Blake
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    Orthonormal means $e_1\cdot e_i=1, e_2\cdot e_2=1, e_1\cdot e_2=0$. Since ${a_1,a_2}$ is orthonormal, $a_1\cdot a_2=0 \Rightarrow e_1\cdot e_2=0$ – GFR Apr 18 '15 at 09:54
  • But doesn't then doesn't $e_1\cdot e_1\neq 1$? Or are you refering to the standard dot product? – Blake Apr 18 '15 at 16:54
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    You are right and I am wrong, I was reading quickly and for some reasons I thought it was the orthogonal part you did not like, I did not even check $e_i$ being of unit norm. You are right about $e_1,e_2$ not being of unit norm but $y e_i$ being so. – GFR Apr 18 '15 at 20:45
  • You're not entirely wrong either. Earlier today I noticed I had said $\langle e_1,e_2\rangle_p$ instead of $\langle e_1,e_1\rangle_p$ (what I originally intended) and I didn't leave a note. So I don't blame you for the mistake.

    I ignored my issue and went along with the problem. Everything worked out fine (even though I still don't quite understand the hint). Thank you for taking the time out to help.

     – Blake Apr 18 '15 at 20:51

1 Answers1

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Since $\{ ye_i \}$ is ON so $\{\omega_1:= \frac{dx}{y},\ \omega_2:=\frac{dy}{y}\}$ is coframe Then $$ \omega_{12}\omega_2=d\omega_1=-\frac{1}{y^2} dydx=\omega_1\omega_2 $$ $$ \omega_{21}\omega_1=d\omega_2= 0 $$

Hence $$ \omega_{12}=\omega_1 $$

And apply $ d\omega_{12}=-K\omega_1\omega_2$

HK Lee
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