Welcome to mathimatikoi.org forum; Enjoy your visit here.

Parallel

Differential Geometry
Post Reply
PJPu17
Articles: 0
Posts: 10
Joined: Mon Oct 17, 2016 9:33 pm

Parallel

#1

Post by PJPu17 » Mon Oct 24, 2016 12:20 am

Hi, i´ve working hard on this problem but i don´t get the solution. It is the exercise 2.12 of this notes
http://www.maths.ed.ac.uk/~aar/papers/dupontnotes.pdf" onclick="window.open(this.href);return false;

I´ve computed Christoffel´ symbols of the induced conection, and the metric´s matrix, but i don´t know how to prove the final result, and why $V$ and $W$ are orthogonal ( using the induced metric, in cartesians is trivial).

Thank you for your time :roll: :roll: :roll:
User avatar
Grigorios Kostakos
Founder
Founder
Articles: 0
Posts: 460
Joined: Mon Nov 09, 2015 1:36 am
Location: Ioannina, Greece

Re: Parallel

#2

Post by Grigorios Kostakos » Mon Oct 24, 2016 4:43 am

Assuming that you are referring to part a) of the exercise 2.12, did you try this?
\begin{align*}
\frac{D Z}{dt}(t)&=\frac{D}{dt}\big(\cos(\theta_0-\beta t)\,X(t)+\sin(\cos(\theta_0-\beta t)\,Y(t)\big)\\
&=\frac{D}{dt}\big(\cos(\theta_0-\beta t)\,X(t)\big)+\frac{D}{dt}\big(\sin(\cos(\theta_0-\beta t)\,Y(t)\big)\\
&=\frac{d}{dt}\cos(\theta_0-\beta t)\,X(t)+\cos(\theta_0-\beta t)\,\frac{DX}{dt}(t)+\frac{d}{dt}\sin(\theta_0-\beta t)\,Y(t)+\cos(\theta_0-\beta t)\,\frac{DY}{dt}(t)\\
\end{align*}
and then calculate $\frac{DX}{dt}(t)$, $\frac{DY}{dt}(t)$ separately.
Grigorios Kostakos
PJPu17
Articles: 0
Posts: 10
Joined: Mon Oct 17, 2016 9:33 pm

Re: Parallel

#3

Post by PJPu17 » Mon Oct 24, 2016 1:36 pm

I came up to these point but I don´t know how derivate these fields because they´re in cartesians. On the other hand, Is there any geometric argument to prove the statement without doing any operation? Using only the compatibility of the connection with the metric, and the orthogonality of both fields in cartesians?
User avatar
Grigorios Kostakos
Founder
Founder
Articles: 0
Posts: 460
Joined: Mon Nov 09, 2015 1:36 am
Location: Ioannina, Greece

Re: Parallel

#4

Post by Grigorios Kostakos » Mon Oct 24, 2016 6:08 pm

PJPu17 wrote:...Is there any geometric argument to prove the statement without doing any operation? Using only the compatibility of the connection with the metric, and the orthogonality of both fields in cartesians?
Maybe you're right! But I have no answer about this. My suggestion is: write $X(t)=\mathop{\sum}\limits_{k=1}^3v^k(t)\,\partial_k\big|_{\gamma(t)}$ and then apply the formula \[\frac{dX}{dt}(t)=\mathop{\sum}\limits_{k=1}^3\bigg(\frac{dv^i}{dt}+\mathop{\sum}\limits_{ij=1}^3\frac{d\gamma^j}{dt}\,\Gamma_{ji}^{k}\,v^i\bigg)\,\partial_k\big|_{\gamma(t)}\,.\]
Do the same for $Y(t)$. And don't forget the restrictions for $\alpha$ and $\beta$.
Grigorios Kostakos
PJPu17
Articles: 0
Posts: 10
Joined: Mon Oct 17, 2016 9:33 pm

Re: Parallel

#5

Post by PJPu17 » Mon Oct 24, 2016 11:04 pm

I can´t do this, because i have the metric induce by the euclidean ( the first fundamental form on the sphere), and these fields are in cartesians. :S
User avatar
Grigorios Kostakos
Founder
Founder
Articles: 0
Posts: 460
Joined: Mon Nov 09, 2015 1:36 am
Location: Ioannina, Greece

Re: Parallel

#6

Post by Grigorios Kostakos » Tue Oct 25, 2016 3:51 am

PJPu17 wrote:I can´t do this, because i have the metric induce by the euclidean ( the first fundamental form on the sphere), and these fields are in cartesians. :S
Which are the charts for the sphere that you are considering?
Grigorios Kostakos
Post Reply