Question

Differential Geometry

Prove that for a coordinate patch x(u,v), where U is the unit normal defined as , and K is the Gaussian Curvature.

L, V 1,0) (0,1 1,0)

$U_u \times U_v = K(x_u\times x_v)$

Answer 1

A coordinate patch on manifold M is an open subset U
A coordinate patch on manifold M is an open subset U subset M together with a map phi: U rightarrow R^n that is homeomorphism between U and its message phi (U). With this a manifold is a topological space where every point can be contained in some coordinate patch. Here a coordinate patch is x(u, v). Given U is unit normal given by x^(1, 0) (u, v) * x^(0, 1) (u, v)/squareroot x^(1, 0) (u, v) * x^(0, 1) (u, v) middot x^(1, 0) (u, v) * x^(0, 1) (u, v) = U_u times V_v and K is Gaussian Curvature, where K = K1K2 Therefore U_u times V_v = K (X_u * X_v)
M together with a map $\phi :U \rightarrow R^{n}$ that is homeomorphism between U and its message . With this a manifold is a topological space where every point can be contained in some coordinate patch.

Here a coordinate patch is x(u,v). Given U is unit normal given by

= U_u x U_v

and K is Gaussian Curvature, where K=K₁K₂

Therefore U_u x U_v = K( x_u * x_v )