Gaussian curvature

• Dec 16th 2011, 03:00 PM
Sogan
Gaussian curvature
Hello everybody,

when we consider some regular surface $S\subset \mathbb{R}^3$ with constant gaussian curvature.
Then my question is: Why and how the curvature does changes, when i multiply the givin scalar product by a constant $a\neq 0$.

I'm cannot see, why a different 1.fundamental form changes the determinant of the shape-map? For instance, what is wrong with the following ideas:

Let (S,I) be a geometric surface with curvature K and a given 1.Fundamental form I.

Now i change my quadratic form by $I'(x):=a\cdot I(x)$. Then i get the same curvature, because i have:

$K=k_1 \cdot k_2=II(e_1)\cdot II(e_2)=a\cdot \cdot a=II'(\frac{e_1}{\sqrt{a}})\cdot II'(\frac{e_2}{\sqrt{a}})=k'_1 \cdot k'_2 =K'$

here the vectors e_1 and e_2 are the corresponding eigenvectors, which form an orthonormal basis.

I hope, someone can help me...
• Dec 16th 2011, 08:52 PM
xxp9
Re: Gaussian curvature
why k1=I(e1)?
• Dec 16th 2011, 09:10 PM
xxp9
Re: Gaussian curvature
deleted
• Dec 17th 2011, 12:18 AM
Sogan
Re: Gaussian curvature

So the idea is to fix II and just change I. Because if i change I and II by I'=a*I and II'=a*II, then the curvature keeps the same! Is this statement correct?

My second question is, why is it possible to change just I, but let II fixed?
I was thinking, that the definition of II on a given Surface is in terms of I. So if i change I, then we have $II'(x,y)='=a*=a*II(x,y)$??

Thank you a lot!!
• Dec 17th 2011, 03:27 AM
Sogan
Re: Gaussian curvature
Ok, i meant not I(e_1) but II(e_1). So i have changed it in the first thread.
• Dec 17th 2011, 08:56 AM
xxp9
Re: Gaussian curvature
Let's clearly define what does "changing the first fundamental form" means here.

If we consider only the intrinsic geometry here, since the Gaussian curvature depends only on intrinsic geometry, the meaning of "changing the first fundamental form" is clear: let g be the metric tensor, then $a^2g$ defines a new metric tensor, where $a$ is a positive real number. The vector length $\|v\|=\sqrt{g(v,v)}$ is changed by a ratio $a$. In this case the Gaussian curvature $K$ changes to $\frac{K}{a^2}$. You can easily see this since the area element $\omega=\sqrt{det{g}}du\wedge dv$ becomes $a^2\omega$, and areas are changed by a ratio of $a^2$, then use Gauss-Bonnet theorem on a small geodesic triangle you'll find the Gaussian curvature turns to $\frac{K}{a^2}$.
If we instead consider an embedded surface in $R^3$, the meaning of "changing the first fundamental form" could be defined as follows:
If $r(u,v)=(x(u,v),y(u,v),z(u,v))$ is the parameterization of the surface, then $r_a(u,v)=a\cdot r(u,v)=(a\cdot x(u,v),a\cdot y(u,v),a\cdot z(u,v))$ defines a new surface, with a corresponding first fundamental form of $I_a=a^2 \cdot I$ and the corresponding second fundamental form is $II_a=a \cdot II$, so the normal curvature is changed to $\frac{II_a}{I_a}=\frac{II}{a\cdot I}$