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7

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 ...


3

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 ...


3

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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