CLASSIFICATIONS OF HELICOIDAL SURFACES WITH L₁-POINTWISE 1-TYPE GAUSS MAP

Abstract

Abstract. In this paper, we study rotational and helicoidal surfaces in Euclidean 3-space in terms of their Gauss map. We obtain a complete classification of these type of surfaces whose Gauss maps G satisfy L

On the Gauss map of a class of hypersurfaces in ℍ4

“…for a constant λ ∈ R and a constant vector C, where M is a submanifold of a Euclidean space. When the codimension of M is 1, by replacing the condition of satisfying (6) with a weaker one, one can define pointwise 1-type Gauss map (See, for example [12,13,14]).…”

“…where K and H are Gaussian and mean curvature of M. In [13], the following theorems obtained. • In [14], helicoidal surfaces of E 3 is studied in terms of having -pointwise 1-type Gauss map. It is proved that a helicoidal surface with -pointwise 1-type Gauss map of the second must necessarily be a rotational surface with a specifically chosen profile curve.…”

The Third Laplace-Beltrami Operator of the Rotational Hypersurface in 4-Space

“…Choi et al [8] study on helicoidal surfaces and their Gauss map in Minkowski 3-space. Kim and Turgay [15] classify the helicoidal surfaces with L 1 -pointwise 1-type Gauss map.…”

Rotational hypersurfaces with $L_r$-pointwise 1-type Gauss map

“…Based on this definition rotational, helicoidal and canal surfaces in E 3 with L 1 -pointwise 1-type Gauss map were discussed in [16,18]. Motivated by such an idea, the following definition was given by the author in [17].…”

“…Then (1.1) holds for some function f and some vector C. When the Gauss map is not L r -harmonic (i.e. L r G = 0), (1.1), (3.1), (3.11) and (3.12) imply that the first n components of C must be zero and Suppose that M is a hypersurface of revolution of polynomial kind, that is, g(t) is a polynomial in t. Suppose that deg g(t) = m. Eliminating f in (3.15) and, using (3.13) and (3.14), we get following relation 16) where A(t) and B(t) are polynomials in t of degrees (m−1)(r+9) and (m−1)(r+7) respectively. So the polynomial g(t) that satisfies (3.16) has degree m = 1.…”