Why first fundamental form and second fundamental form are the same?

Surface of Revolution

$\gamma(u)=(f(u),0,g(u))$ and

$\sigma(u,v)=(f(u)\cos v, f(u)\sin v, g(u))$

Fist of all, I calculated the first fundametal form for surface of revolution.

And I obtained that $$d\theta^2+\cos^2\theta \ d\varphi^2$$

An then, I calculated the second fundametal form for surface of revalution.

I obtained that $$(\dot f(u)\ddot g(u)-\ddot f(u)\dot g(u))du^2 +f(u)\dot g(u) dv^2$$

When I take $u=\theta$ $v=\varphi$ $f(\theta)=\cos \theta$ and $g(\theta)=\sin \theta$

I get the result $$d\theta^2+\cos^2\theta \ d\varphi^2$$

That's, its 1st fundametal form and second fundametal form are the same.

Why? How does there exist a relation between them? Please explain it. Thank you.

• which surface of revolution? Dec 2 '13 at 10:10
• $\gamma(u)=(f(u),0,g(u))$ and $\sigma(u,v)=(f(u)\cos v, f(u)\sin v, g(u))$ @Avitus
– 1190
Dec 2 '13 at 10:17
• Ok, so at the end you want to select a sphere, am I right? Dec 2 '13 at 10:19
• No, I want to learn the reason why first fundamental form and second fundamental form are the same @Avitus
– 1190
Dec 2 '13 at 10:21
• Ahh, yes I take $u,v$ as it is shown above if you asked this. @Avitus
– 1190
Dec 2 '13 at 10:22

In general, the first and second fundamental forms of a surface are different objects. Let $r=r(u,v)$ be a parametrization of a given surface and let $n=n(u,v)$ denote the normal vector. We use the notation

$$I:=Edu^2 + 2Fdudv + Gdv^2$$ where $E = r_u\cdot  r_u; F = r_u\cdot  r_v;G = r_v\cdot  r_v$.

for the first fundamental form, and

$$II:=Ldu^2 + 2Mdudv + Ndv^2$$ where $L = r_{uu}\cdot  n$; $M = r_{uv}\cdot  n$; $N = r_{vv}\cdot  n$ for the second one.

If we consider the unit sphere, i.e. $r(u,v)=(\sin u\sin v, \sin u\cos v,\cos u)$, then

$$I=II$$

as the normal vector satisfies $n(u,v)=r(u,v)$ (this is evident geometrically) and

$$L=E,$$ $$M=F,$$ $$N=G;$$

this follows from the very definitions of the coefficients themselves.

Remark: if the sphere has radius $a>0$ then, in general, $$aII=I$$.

• Thank you :)) I see its reason:)
– 1190
Dec 2 '13 at 10:46
• You are welcome. Keep going with self studying and posting questions, of course. A small tip: if you describe in detail your problem in your question, explaining where you are stuck, what you did so far and the papers/books you are reading on the topic, then you will receive many more answers! Dec 2 '13 at 10:49
• Okay, I dont know this. Thanks :)
– 1190
Dec 2 '13 at 10:59