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Show that the map $F(q)=\frac{q}{|q|^2}$ is conformal.

One way to show that a map is conformal is by looking at the first fundamental form. But since we don't have matrix here, we can recall that $\cos\theta=\frac{X\cdot Y}{|X||Y|}$. So if we can show that $\frac{X\cdot Y}{|X||Y|}=\frac{DF(X)\cdot DF(Y)}{|DF(X)DF(Y)|}$, we win.

Then I computed what $DF(v)$ is, assuming that $v \in T_p M$:

Consider a curve parametrized curve $\alpha(t)$ on the surface, such that $\alpha(0)=p$ and $\alpha'(0)=v$, then we have:

$DF(v)=\frac{dF\circ\alpha}{dt}=\frac{|q|^2v-q\frac{d}{dt}(|q|^2)}{|q|^4}=\frac{|q|^2v-2q(q\cdot v)}{|q|^4}$

Once we have this, I computed $DF(X)\cdot DF(Y)$, which yields $|q|^4(X\cdot Y)$. But I don't know how to calculate $|DF(X)DF(Y)|$. Any idea? Thanks!

3x89g2
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1 Answers1

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To calculate $|DF(X)||DF(Y)|$ use the definition for the norm. We have

$$|DF(X)|^2 \equiv DF(X)\cdot DF(X) = \frac{(|q|^2X - 2q(q\cdot X))\cdot (|q|^2X - 2q(q\cdot X)) }{|q^8|}\\= \frac{|q|^2|X|^2 + 4|q|^2(q\cdot X)^2 - 4|q|^2(q\cdot X)^2}{|q^8|} = \frac{|X|^2}{|q|^4}$$

so $|DF(X)| = \frac{|X|}{|q|^2}$. It therefore follows that $|DF(X)||DF(Y)| = \frac{|X||Y|}{|q|^4}$ and with $DF(X)\cdot DF(Y) = \frac{X\cdot Y}{|q|^4}$ it follows that $\frac{X\cdot Y}{|X||Y|} = \frac{DF(X)\cdot DF(Y)}{|DF(X)||DF(Y)|}$ and the map is conformal.

Winther
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