���������������� ��������� Input the parametrization: �[�_� �_] �= {�� �� � / � ���[���[� �] / ���[� �]]} Here’s a graph: ����������������[�[�� �] /� � → �� {�� - � π / �� � π / �}� {�� - � π / �� � π / �}] This surface is called Scherk’s Surface. Notice that it is doubly-periodic and, as we’ll see momentarily, it is a minimal surface. Now, we compute Xu , Xv , Xuu , Xuv , Xvv , and N (which I’ll call “Gauss”, since the capital N is a reserved symbol in Mathematica): �� = �[�[�� �]� �] {�� �� ���[� �]} �� = �[�[�� �]� �] {�� �� - ���[� �]} 2 ��� Sols.nb ��� = �[��� �] �� �� � ���[� �]� ��� = �[��� �] {�� �� �} ��� = �[��� �] �� �� - � ���[� �]� �����[��� ��] {- ���[� �]� ���[� �]� �} ����� = % ����[�����[# � � � /@ %]] ���[� �] - � + ���[� �]� � + ���[� �]� ���[� �] � + ���[� �]� + ���[� � � �]� � + ���[� �]� + ���[� �]� From these it’s straightforward to compute E, F, G, e, f, and g (again, I can’t use a capital E, as this is Mathematica’s symbol for the number 2.71828...): �� = ����� � + ���[� �]� � = ����� - ���[� �] ���[� �] � = ����� � + ���[� �]� � = ��������� � ���[� �]� � + ���[� �]� + ���[� �]� � = ��������� � � = ��������� � ���[� �]� - � + ���[� �]� + ���[� �]� And now, of course, we can just use the formulas K= e g - f2 E G - F2 and H= 1 eG-2fF+gE 2 E G - F2 to get the Gaussian curvature: Sols.nb ��� 3 ������������� � - � � � �� � - � � � - �� ���[� �]� ���[� �]� ���[� �]� + ���[� �]� � and the mean curvature: � / � � � - � � � + � �� �� � - � � � - � ���[� �]� � + ���[� �]� � + ���[� �]� + ���[� �]� + � ���[� �]� � + ���[� �]� � + ���[� �]� + ���[� �]� � - ���[� �]� ���[� �]� + � + ���[� �]� � + ���[� �]� ������������[%] � Since the mean curvature is zero, this really is a minimal surface. ��������� Enter the parametrization: �[�_� �_] �= {� ���[�]� � ���[�]� �}� The goal is to compute the Gaussian curvature using the Gauss equation -E K = Γ112 Γ211 + Γ212 u + Γ212 Γ221 - Γ111 Γ212 - Γ211 v - Γ211 Γ222 , so we need to compute the various Christoffel symbols. In turn, we can compute the Christoffel symbols from the equations Γ111 Γ211 Γ112 Γ212 Γ122 Γ222 E F = F G Eu Fu - E F F G -1 E F = F G -1 = 1 2 -1 1 2 1 2 1 2 Ev 1 2 Gu Ev Gu Fv 1 2 Gv So now we get computing: �� = �[�[�� �]� �] {���[�]� ���[�]� �} �� = �[�[�� �]� �] {- � ���[�]� � ���[�]� �} 4 ��� Sols.nb �� = ������������[�����] � � = ����� � � = ������������[�����] �� �� = �[��� �] � �� = �[��� �] � �� = �[�� �] � �� = �[�� �] � �� = �[�� �] �� �� = �[�� �] � And then solve for the Christoffel symbols: {Γ���� Γ���} = �������[{{��� �}� {�� �}}]�{� / � ��� �� - � / � ��} {�� �} {Γ���� Γ���} = �������[{{��� �}� {�� �}}]�{� / � ��� � / � ��} �� � � {Γ���� Γ���} = �������[{{��� �}� {�� �}}]�{�� - � / � ��� � / � ��} {- �� �} And so the Gaussian curvature is: - (Γ��� * Γ��� + �[Γ���� �] + Γ��� * Γ��� - Γ��� * Γ��� - �[Γ���� �] - Γ��� * Γ���) / �� � (More precisely, Γ212 u = -1 u2 and Γ212 Γ221 = 1 , u2 so those two terms cancel out. Since the rest of the terms in the numerator are 0, the whole expression is 0.) Sols.nb ��� 5 ��������� Again, first enter the parametrization: �[�_� �_] �= � α[�] The surface is regular ⇔ X has a continuous inverse and Xu ×Xv ≠ 0. To be invertible, clearly we need that α is injective, meaning that the curve is simple. As for Xu ×Xv ≠ 0, we compute Xu , Xv , and their cross product: �� = �[�[�� �]� �] α[�] �� = �[�[�� �]� �] � α′ [�] �����[��� ��] α[�] ⨯ (� α′ [�]) Now, this will be a nonzero vector provided: (i) α(v) ≠ 0, meaning that the curve can’t pass through the origin; (ii) α' (v) ≠ 0, meaning that the curve is a regular curve (which was part of the assumption of the problem; and (iii) α(v)×α' (v) ≠ 0, which means that α(v) and α' (v) can’t be parallel, or equivalently that the curve α never travels radially outward or inward. So the surface will be regular provided the regular curve α is simple, doesn’t pass through the origin, and never travels radially. Now, for the second part of the problem, we’ll want to compute Xuu , Xuv , Xvv , and N; since we’ve already got Xu ×Xv , we can do N first: ����� = �����[α[�]� � α �[�]] (� ����[α[�]]) α[�] ⨯ (� α′ [�]) � ����[α[�]] In fact, looking at this makes it obvious we should have just cancelled the u’s: ����� = �����[α[�]� α �[�]] ( ����[α[�]]) α[�] ⨯ α′ [�] ����[α[�]] And then the second derivatives: ��� = �[��� �] � ��� = �[��� �] α′ [�] 6 ��� Sols.nb ��� = �[��� �] � α′′ [�] So now we can straightforwardly compute E,F,G,e,f,g: �� = ����� α[�]�α[�] � = ����� α[�]�(� α′ [�]) which is zero since α is parametrizedy by arclength �=� � � = ����� (� α′ [�])�(� α′ [�]) � = ��������� α[�] ⨯ α′ [�] ����[α[�]] �� which is just zero: �=� � � = ��������� α[�] ⨯ α′ [�] ����[α[�]] �α′ [�] Also zero since α[�] ⨯ α′ [�] has to be perpendicular to α’(v). �=� � � = ��������� α[�] ⨯ α′ [�] ����[α[�]] �(� α′′ [�]) This gives us the Gaussian curvature: � � - � � � �� � - � � � � It’s not so surprising that the Gaussian curvature is 0; after all, the line from the origin to any point on the curve lies in the surface. Also the mean curvature: Sols.nb ��� 7 � / � � � - � � � + � �� �� � - � � � // ������������ α[�]⨯α′ [�] �(� ����[α[�]] α′′ [�]) � (� α′ [�])�(� α′ [�]) and now we do some fancy simplifications: ������% //� �_ α �[�]��_ α �[�] ⧴ � �� �����[�_� � �_] ⧴ � �����[�� �]� ���[�_� � �_] ⧴ � ���[�� �]� ���[� / � �_� �_] ⧴ � / � (���) α[�]⨯α′ [�] �α′′ [�] ����[α[�]] �� which is just (α(v)×α' (v))·α'' (v) 2 u α(v) ��������� Consider the function g : ℝ3 → ℝ given by g(p) = p·p. Then the restriction of g to Σ is a smooth map. Since Σ is compact, g has a global max at some point p ∈ Σ. (Another way to think of p is as follows: since Σ is compact, it is bounded and hence contained in some very large sphere centered at the origin. Now, let this sphere shrink until it first touches the surface: the point p is the point where this first contact occurs [really, the first contact may occur at several points simultaneously, but p is guaranteed to be one of them]). v . Then Now, let v ∈ Tp Σ and let α be a curve in Σ passing through p so that α(0) = p and α' (0) = f (s) := g(α(s)) has a maximum at s = 0, so f ' (0) = 0 and f '' (0) ≤ 0. But, of course, 0 = f ' (0) = d [(f (s))]s=0 ds = d [g(α(s))]s=0 ds = d [α(s)· α(s)]s=0 ds = 2 α(0)·α' (0) = 2 p · v. Since v was arbitrary, this means that p is perpendicular to Tp Σ, and hence α(0) = p = a N, where N is the Gauss map and a ≠ 0 is a real number (if a were 0, then p would be the origin; since p was the global max of g, that would mean Σ consisted of only a single point at the origin, which is clearly not a regular surface). Moreover, since f '' (0) ≤ 0, we have 0 ≥ f '' (0) = d [f ds ' (s)]s=0 = d [2 α(s)·α' (s)]s=0 ds In other words, a N ·α'' (0) ≤ -2 v 2 = 2 (α ' (0)·α' (0) + α(0) ·α'' (0)) = 2 v 2 +a N ·α'' (0) . . Since the right hand side is negative, we see that N ·α'' (0) is either always negative or always positive (depending on the sign of a). But then this means all normal curvatures at p have the same sign, and hence that the principal curvatures k1 and k2 both have the same sign. Therefore, the Gaussian curvature is 8 ��� Sols.nb K = k1 k2 > 0, as desired.