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4

It is only a matter of notation. The parametrization $\sigma:U\to\mathbb{R}^3$ has its two directional derivatives, in the directions of the standard basis vectors. These directional derivatives are denoted by $\sigma_u$ and $\sigma_v$. Together, these directional derivatives give the differential of $\sigma$. Then we have different paths in $U$, but ...

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To find $S$ may be highly non-trivial: that implicit surface recalls some equipotential surfaces; a similar (just 2d) question of mine is still unsolved. Anyway, the section $z=0$ of that surface is a lemniscate, and by setting: $$f(x,y) = \sqrt{-(x^2+y^2)+|a|\sqrt{x^2-y^2}}$$ we have: $$S = 8\iint_{L}\sqrt{\left(\frac{\partial f}{\partial ... 1 I'm assuming you are working with the definitions given in Do Carmo's "Curves and Surfaces" book and that V \subseteq \mathbb{R}^3 is an open set. First, let us write precisely the domains and ranges of the maps involved: The map \mathbf{x}_1 \colon U_1 \rightarrow \mathbb{R}^3 is classically differentiable. Here, U_1 \subseteq \mathbb{R}^2 is an ... 1 Plugging in x+y+z=0 into the equation for the surface of the sphere.$$x^2+xy+y^2=a^2/2$$Completing the square$$(x+y/2)^2+(\sqrt{3}y/2)^2=(a/\sqrt{2})^2\sqrt{3}y/2=a/\sqrt{2}sin(t)y(t)=\sqrt{2/3}a sin(t)x+y/2=a/\sqrt{2}cos(t)x(t)=\sqrt{1/6}asin(t)+a/\sqrt{2}cos(t)z(t)=-x(t)-y(t)$$2 Your suggested answer is$$\vec{g}(t) = (a\cos(t), a\sin(t), -a(\sin(t)+\cos(t)))$$which must satisfy$$x^2+y^2+z^2=a^2$$However, here:$$x^2+y^2+z^2=a^2+a^2(1+2\sin{t}\cos{t})\neq a^2$$What I do is to use two variables for parametrizing the first surface:$$x^2+y^2+z^2=a^2$$is$$\vec g(t,u)=(a \sin{t}\sin{u}, a \sin{t}\cos{u}, a\cos{t})$$... 2 There are several things I don't understand here. What role does the abelian surface play? (None that I can see.) What is the projective bundle? Is it just \mathbf P^1 \times C? (You didn't specify any rank 2 vector bundle on C to be projectivised). Now to your questions: For a surjective projective morphism, a nontrivial bundle never pulls back to ... 1 Intuitively, this is your surface x^2 + y^2 = z^2, a lovely cone in \Bbb R^3. This is your curve r(t) = (\cos t \sin t, \cos^2 t, \cos t), also a lovely subset of \Bbb R^3. When we plot them at the same time, we see that one literally lies on the other: And another angle, just for fun: Any point on the curve automatically satisfies the ... 1 The surface S is defined as the set of all points in \mathbb{R}^3 whose coordinates satisfy x^2 + y^2 = z^2. What is the image of your curve? Well, the points of the image of your curve are precisely points of the form r(t) = (\sin t\cos t, \cos^2 t, \cos t). As t ranges over all of \mathbb{R}, r(t) ranges over all points in the image of your ... 0 I interpret this as a question of recentering a polynomial. The formula$$ \sum_{k=0}^p a_k \left(z+z'\right)^k = \sum_{\ell=0}^p \left(\sum_{k=l}^p \binom{k}{\ell} (z')^{k-\ell} a_k \right) z^\ell=\sum_{\ell=0}^p b_{\ell}\, z^\ell  can be derived using Taylor series.

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You cannot. This follows already from the fact that it scales incorrectly – if you scale the whole thing up by $\lambda$, that quantity would scale with $\lambda$ whereas it should scale with $\lambda^2$. Each curve element sweeps a surface area element of different radius. The correct way to take this into account is given in the Wikipedia article ...

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I interpret your question as "if $X$ is an open subset of a closed surface, and has only one end, is it determined by its fundamental group?" Yes. In fact, a surface $\Sigma$ with one end and fundamental group $F_{2g}$ is homeomorphic to the punctured genus $g$ surface, and this is the only possible fundamental group. To prove this use the observation of ...

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