# Problem of the Week #10 - August 6th, 2012

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#### Chris L T521

##### Well-known member
Staff member
I've posted a bunch of analysis questions as of late. I'm going to change things up a little bit and ask something that involves manifold theory. Here's this week's problem:

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Problem: (i) Let $\omega$ be a 1-form. Use the structure equations

\begin{aligned}d\theta^1 &= -\omega\wedge\theta^2\\ d\theta^2 &= \omega\wedge\theta^1\end{aligned}

to compute the curvature $K(r)$ of a surface with metric $ds^2=dr^2+f(r)^2d\theta^2$.

(ii) Verify your answer by finding $f$ in the cases where $K=constant$ and checking it with the literature (i.e. show that you end up with spherical, Euclidean, or hyperbolic geometries depending on the value of the constant).

I'll provide suggestions on how to do the problem.

(i) First write the metric in the form $ds^2 = (\theta^1)^2 + (\theta^2)^2$, where $\theta^1$ and $\theta^2$ are coframes (or 1-forms). Curvature is then defined by

$K(r) = \frac{d\omega}{\theta^1\wedge\theta^2}$

where $\omega$ is determined from the structure equations

\begin{aligned}d\theta^1 &= -\omega\wedge\theta^2\\ d\theta^2 &= \omega\wedge\theta^1.\end{aligned}

(ii) Consider the cases $C>0$, $C=0$ and $C<0$; then solve the appropriate differential equations with intial conditions $f(0)=0$ and $f^{\prime}(0)=1$.

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#### Chris L T521

##### Well-known member
Staff member
No one answered this week's question. You can find my solution below.

In terms of coframes, let $\theta^1=dr$ and $\theta^2=f(r)\,d\theta$.

(i) Proof: By definition, the curvature is defined to be
$K(r) = \frac{d\omega}{\theta^1\wedge\theta^2}$
where $\omega$ is determined from the structure equations
\begin{aligned}d\theta^1 &= -\omega\wedge\theta^2\\ d\theta^2 &= \omega\wedge\theta^1.\end{aligned}
We observe that
\begin{aligned}d\theta^2 &= f^{\prime}(r)\,dr\wedge d\theta\\ &= \omega\wedge\theta^1 \\ &= \omega\wedge dr\end{aligned}
implying that $\omega = -f^{\prime}(r)\,d\theta$. Thus, $d\theta^1 = -\omega\wedge d\theta = 0$. Next, we see that
$d\omega = -f^{\prime\prime}(r)\,dr\wedge d\theta.$
Therefore, the curvature is
$K(r) = \frac{d\omega}{\theta^1\wedge\theta^2} = \frac{-f^{\prime\prime}(r)\,dr\wedge d\theta}{f(r)\,dr\wedge d\theta} = -\frac{f^{\prime\prime}(r)}{f(r)}.$

Q.E.D.

(b) Let $K(r)=C$, where $C$ is a constant. Then $f(r)$ satisfies the homogeneous differential equation
$f^{\prime\prime}(r)+Cf(r)=0$
with characteristic equation $R^2+C=0$. If $C>0$, then the general solution is of the form
$f(r) = c_1\cos(Cr)+c_2\sin(Cr)$
for some arbitrary constants $c_1$ and $c_2$. If $C=0$, then the general solution is of the form
$f(r)=c_1+c_2r$
for some arbitrary constants $c_1$ and $c_2$. If $C<0$, then the general solution is of the form
$f(r)=c_1e^{-\sqrt{|C|}r}+c_2e^{\sqrt{|C|}r}$
for some arbitrary constants $c_1$ and $c_2$.
If the initial conditions are $f(0)=0$ and $f^{\prime}(0)=1$, we get
$f(r) = \frac{\sin(Cr)}{C}$
for $C>0$,
$f(r) = r$
for $C=0$, and
$f(r) = \frac{\sinh(\sqrt{|C|}r)}{\sqrt{|C|}}$
for $C<0$. If $C=1$, we get $f(r)=\sin(r)$ and if $C=-1$, we get $\sinh(r)$, confirming what was stated in class for the spherical and hyperbolic geometry cases.

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