# Tag Info

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Yes. Deleting a point from a manifold of dimension $n \geq 3$ doesn't change the fundamental group, so the result is still simply connected. (Either use van Kampen or transversality.) Mayer-Vietoris shows that the homology of the resulting space is the same as $S^2$, whence $\pi_2(U \setminus p)$ is isomorphic to $H_2(U \setminus p) = \Bbb Z$, generated by a ...

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A sphere is a (2D) manifold that can admit a smooth structure. A cube, I interpret to mean as $[0,1]^3$ is not even a topological manifold. A line is a (1D) manifold that can admit a smooth structure. A point is a (0D) manifold that can admit a smooth structure. If you are looking for a manifold without a smooth structure, you will have a difficult ...

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Hint Use Cartan's Magic Formula, which says that the Lie derivative $\mathcal L_X$ of a differential form $\alpha$ satisfies $$\mathcal L_X \alpha = \iota_X d \alpha + d (\iota_X \alpha) .$$ From the statement of the original problem in the comments, $L_X(\omega \wedge \mu)$ can be integrated, so $L_X(\omega \wedge \mu)$, and hence $\alpha := \omega \wedge ... 2 I'm confused.$H_1(\mathbb{R}P^3)=\mathbb{Z}/{2\mathbb{Z}}$, but is orientable (on edit, this answers your question 2). EDIT: Maybe it refers to the fact that$H^{m-1}(M)$can only be torsion if the manifold$M$is non-orientable. It is Corollary 3.28 in Hatcher. 2 To clear up some confusion, recall that a$k$-manifold$M$is a second-countable, Hausdorff space which is locally homeomorphic to$\mathbb{R}^k$. The last requirement is the most important, at least it's the one that can be understood better since it says something about the intrinsic geometry. To make things a bit easier, lets look at all of your examples ... 2 No, having a CW complex structure is absolutely not enough to be a manifold. Being a CW complex is very easy, being a manifold is hard. Take the wedge sum of two circles$S^1 \vee S^1$for example: it's a CW complex, but not a manifold. If a space$M$is a compact manifold without boundary, then you can read its dimension using singular homology: it is the ... 1 The Frobenius theorem implies that you can find a chart which has$V_1=\frac{\partial}{\partial x^1}$and$V_2=\frac{\partial}{\partial x^2}$, so locally your$f$, which is a function of$(x^1,x^2,x^3)$, is such that $$\frac{\partial f}{\partial x^1}=\frac{\partial f}{\partial x^2}=0.$$ Of course, this means that (in an appropriate neighborhood of each ... 1 In general this is not true. Recall that $$L_X(\omega) = i_x d\omega + d i_x\omega$$ where you see that the right part is exact and the left part mustn't be. As an example for your case take$N$a manifold with a non exact form$\mu$and let$\omega$be a 0-form (function) on$\mathbb{R}$and define$M=N\times\mathbb{R}$and note that the extension of ... 1 The tangent space at$I_n$is the space of antisymmetric matrices defined by$A+A^T=0$. Given$g\in SO(n)$, and let$AS(n)$the space of antisymmetric matrices, the tangent space at$g$is$gAS(n)$. That is the image of the tangent space$T_{I_n}SO(n)$by the left translation defined by$g$. Suppose that$g\in SO(n)$,$A\in AS(n)\$, you have ...

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