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4

You are using that $\partial B^2 = S^1$, but by doing so you assumed already that your $S^1$ is homotopically trivial (it bounds a disk). So you have used the condition (simply connectedness) already. Your first answer is correct (module the fact that you can find a smooth disk $B^2$) For your second question, as $g$ is a constant map, then $g^*\omega = ...


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A differential $k$-form on a smooth manifold $M$ is a skew-symmetric $C^{\infty}(M)$-linear map $\omega :\Gamma(TM)^k \to C^{\infty}(M)$. The exterior derivative of $\omega$ is a skew-symmetric $C^{\infty}(M)$-linear map $d\omega : \Gamma(TM)^{k+1} \to C^{\infty}(M)$ where $d\omega(V_0, \dots, V_k)$ is equal to $$\sum_{i=0}^k(-1)^iV_i(\omega(V_0, \dots, ...


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I wrote a lot, due to boredom and the general hope that some of it is helpful. Much will probably be familiar to you. First: In order to work with tensors or to do calculus on manifolds at all, it's very important to start making the distinction between vectors and covectors (or "dual vectors" or whatever). Briefly, if $V$ is a (finite dimensional, real) ...


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A version of Poincare duality holds for noncompact manifolds: cap product with the fundamental class over a ring of coefficients $A$ gives an isomorphism $H^*_c(M, A) \to H_*(M, A)$, where $H^*_c$ denotes compactly-supported cohomology; see Section 3.3 of Hatcher, for example. Any manifold $M^n$ is orientable over $\mathbb{Z}_2$, so $H_n(M, \mathbb{Z}_2) = ...


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$\newcommand{\Reals}{\mathbf{R}}$In the usual sense of "depend",[*] no, the Gauss curvature, mean curvature, and shape operator of a (locally oriented) regular surface in $\Reals^{3}$ do not depend on parametrization; that's what's meant by saying these are "geometric" data. :) Depending on your definition of the shape operator (e.g., O'Neill's: If $U$ is a ...


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$\newcommand{\Reals}{\mathbf{R}}$Here's a slightly different viewpoint (even for the $2$-sphere). The point $p$ in $S^{n}$ and the normalized initial velocity $v$ of the geodesic $\gamma$ may be viewed as orthogonal unit vectors in $\Reals^{n+1}$, and they span a real $2$-plane $N$ (for "normal space") through the origin. The tangent space $T_{p} S^{n}$ is ...


1

You can always set up the coordinate as $(\sin\theta\cos\phi, \sin\theta\sin\phi, \cos\theta)$, and choose the geodesic to be $\phi=0$ and $p=(1, 0, 0)$ be a point at the equatorial. Note that $X=\frac{\partial}{\partial\theta}=(\cos\theta\cos\phi, \cos\theta\sin\phi, -sin\theta)$ and $Y=\frac{1}{\sin\theta}\frac{\partial}{\partial\phi}=(-\sin\phi, ...


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Since $\mathbb Z_2$ is a field, the functor $Ext_{\mathbb Z_2}^1(-,-)$ is identically zero.


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As I said in my comment, isometric or conformal type of a surface in $R^3$ is independent of a parameterization. This is an answer in the conformal setting (isometric setting is hopeless). You need: Uniformization Theorem: Every simply-connected Riemann surface is conformal either to the unit disk, or to the complex plane or to the 2-sphere. In view of ...


1

I know of no algebra in which 3d arrays of numbers can be manipulated with the same ease as matrices. Part of that has to do with how any linear map from one space to another space can be represented with a matrix. You can chain such maps together sensibly (and really, only in one way up to the order of how you compose those operations together), whereas ...


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The answer to the three questions is yes. If one wants to make it a little more precise, one could add the functor in play. This however isn't done in one line, which is why many people suppress it. The main ingredient of the proof is in fact Chow's theorem. Serre's GAGA is mainly concerned with the additional equivalence of coherent sheaves and cohomology ...


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It doesn't make sense to integrate $d(\gamma^*\omega)$ on $B_2$ since $\gamma^*\omega$ is not defined on $B_2$ and consequently neither is $d(\gamma^*\omega)$. Put differently an interior point of $B_2$ is not related in any way to any point of $X$. Thus your use of Stokes is indeed unwarranted.


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learner - here's one approach. If $\phi: U \rightarrow V$ is a change of coordinates, say $U$ has coordinates $x_i$ and $V$ has coordinates $y_i$, and the metric is written in $V$ as $(g_{ij})$, and in $U$ as $(h_{ij})$, then the metric transforms as $$(*) \quad \quad (h_{ij}) = d\phi^T (g_{ij}) d\phi$$. So, the top form on $U$ defined on terms of ...



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