# Tag Info

3

An orientation of a manifold can be defined in various ways. Let's suppose all throughout that $M$ is a connected smooth manifold (possibly with boundary), of dimension $m$. If $\omega, \eta \in \Omega^{m}(M)$ are two non-vanishing top forms on $M$, then one can prove that there exists a non-vanishing $f \in C^{\infty}(M)$ such that $\omega = f \eta$. Since ...

3

Your map only fails to be a submersion when that matrix is zero. So we check when that happens. This is precisely when $y = -3x^2$ and $x = -3y^2$, so $x = -27x^4$, or $x^3 = -1/27$, or $x = -1/3$. Then we must have $y = -1/3$ as well. So the map fails to be a submersion precisely at $(-1/3,-1/3)$; and this lies in $f^{-1}(0)$ for precisely one $c$: the $c$ ...

2

The only point that you get close to but not reach is $(0,0)$ which is the limit as $t \to - \infty$. So the closure is just the image plus this one point.

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Suppose $M$ is a manifold with boundary. Then there is an open set $U$ in $\mathbb{R}^2$ and a diffeomorphism $F:U\to\mathbb{R}^2$ such that $F(M\cap U)$ is either the line $\{(x,0):x\in\mathbb{R}\}$, or the half-line $\{(x,0):x\ge 0\}$ Consider the $y$ component of $F$, call it $g$. It vanishes on $M$. Consequently, $g_x$ vanishes at every point ...

2

Suppose $(X, \omega)$ is a closed symplectic manifold of dimension $2n$. If $\omega = d\alpha$, then $$\int_X\omega^n = \int_X d(\alpha\wedge \omega^{n-1}) = \int_{\partial X}\alpha\wedge \omega^{n-1} = 0$$ but this is absurd as $\omega^n$ is a volume form. Therefore, a symplectic form on a closed manifold is not exact, so it defines a non-zero element of ...

2

Compact symplectic manifolds must have nontrivial $H^2$. A symplectic manifold $M^{2n}$, by definition, possesses a closed, nondegenerate 2-form $\omega$. Because $\omega$ is closed (i.e., $d\omega = 0$), $\omega$ represents a de Rham cohomology class $[\omega] \in H^2(M)$. The nondegeneracy condition implies $\omega^n$ (the wedge product of $\omega$ with ...

2

Let us call $\pi_1 : M \times N \rightarrow M$ the projection on the first coordinate and $\pi_2 : M \times N \rightarrow N$ the projection on the second coordinate. First of all, if $p \in M$ and $q \in N$, we need to pay attention to how the natural identification $T_{(p, q)} (M \times N) \cong T_p M \times T_q N$ works. What happens is that the map $\phi ... 2 First you parametrize your ellipse :$A = (a\cos\theta, b\sin\theta) \in \mathbb{R}^2 \rvert \theta \in [0,2\pi[$. So you have : $$\left \lbrace \begin{array}{l} x = a\cos\theta\\ y = b\sin\theta \end{array} \right. \Rightarrow \left \lbrace \begin{array}{l} dx = -a\sin\theta \, d\theta\\ dy = b\cos\theta \, d\theta \end{array} \right.$$ Then rewrite your ... 2 For the vector field$X$and the associated dynamical system$\dot{x}=X(x)$we have $$x(t+h)=x(t)+\int_t^{t+h}{X(x(s))ds}\\ =x(t)+\int_t^{t+h}{[X(t)+\frac{dX}{dx}(x(t))X(t)s]ds}+0(h^2)\\ =x(t)+X(t)h+\frac{1}{2}\frac{dX}{dx}(x(t))X(x(t))h^2+o(h^2)$$ We calculate the evolution of a initial point$x_0$under the various flows. Initially under$\Phi_t^X$... 1 Consider for example one of Milnor's exotic$7-$spheres and the standard$7-$sphere: they are PL-homeomorphic since the generalized Poincaré conjecture in the PL category is true in dimensions different than 4 (i.e. any two PL-manifold which is a homotopy sphere is PL-homeomorphic to a sphere), but they are not diffeomorphic as shown by Milnor. Moreover ... 1 An explicit chart sends$GL_n(\mathbb{R}) \to \mathbb{R}^{n^2}$taking a matrix to its$n^2$coefficients strung out in a row vector. To see that multiplication is smooth, check that when you use this chart, The map is just polynomial equations in the coordinates. Since polynomials are smooth, so is matrix multiplication! 1 The great circle you ask about is parametrized by $$\sigma(t) = \cos(\|v\|t)\, p + \sin(\|v\|t)\, \frac{v}{\|v\|}.$$ However, your original strategy of picking local coordinates for the sphere and projective space looks easier; it may simply be that your coordinates for$S^{n}$(graph coordinates for a hemisphere) are not as judicious as they could be. :) ... 1 Let$M$and$N$be smooth manifolds of dimensions$m$and$n$, respectively, and let$\phi : M \to N$be a smooth map. It is sufficient to show that$\phi$is locally continuous, i.e., that every point$x \in M$has a neighborhood$U_x$such that$\phi\left|_{U_x}\right.$is continuous. Thus, let$x \in M$. Since$M$and$N$are smooth manifolds, there ... 1 Locally$\pi$is a diffeomorphism, so restricted to sufficiently small neighborhoods we have$\tilde f = f \circ \pi^{-1}$. Then$\tilde f$is differentiable with derivative given by the chain rule if such a$\tilde f$exists. In general such a$\tilde f$may not exist, i.e. if$\pi$is the map$\Bbb R \rightarrow S^1$given by$\pi(t)=e^{it}$and$f$is ... 1 That$f$is$\mathcal C^\infty$you've already done: on a ball around the origin, it is (=identity), off that ball$\|x\|$is smooth never vanishing and then$f$is the conposite of smooth functions. To check$f$is local diffeo, that is, its derivative at every point$a$is a linear iso, I suggest the following strategy. Fix a point$a\in\mathbb R^n\$, which ...

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