Tag Info

Hot answers tagged differential-geometry

6

Since the tangent bundle of a manifold $M$ is the disjoint union of tangent spaces $$TM=\bigcup_{p \in M}T_pM$$ usually a tangent vector is denoted by $(p,v) \in TM$, where $p \in M$ is a point and $v \in T_pM$ is a vector. This notation has the purpose to remember that the vector $v$ is tangent to $M$ in the point $p$. Now, it seems that the correct ...

6

i think it is unlikely that this is true. Take a dumbbell type shape with rotational symmetry. Make the bar bit pinch in. The smallest circle is a totally geodesic submanifold and the invariant under rotations. Now deform the rest of the barbell whilst leaving a small neighbourhood of the smallest circle unchanged. If you make it lumpy enough there won't be ...

5

Denote by $E$ your ellipsoid, and by $S^2$ the sphere $\{x \in \mathbf R^3 \mid x_1^2 + x_2^2 + x_3^2 = 1\}$. Define a map $f \colon E \to S^2$ by $$f(x) = \left(\frac{x_1}a, \frac{x_2}b, \frac{x_3}c \right)$$ Then $f$ is a diffeomorphism.

4

When constructing or defining a mathematical object, you have to strike a certain balance: You want your object to be general enough to have lots of applications but structured enough to let you answer important questions. Fiber bundles are very general (in the sense that many spaces can be viewed as fiber bundles) and very structured (in the sense that you ...

4

If a group $G$ acts on itself by left multiplication and if $p$, $q$ are arbitrary points of $G$, there is exactly one element $g$ such that $gp = L_{g}(p) = q$. Consequently, a tangent vector at $p$ corresponds to a unique tangent vector at $q$ via the tangent map $dL_{g}$. In general, however, the left $G$-action on $G/H$ has non-trivial stabilizers, and ...

3

Actually, the Möbius strip $M$ is not trivial as an $(\mathbf{R}, +)$ principal bundle over the circle $S^{1}$, because $M$ is not the total space of a principal bundle with structure group $G = (\mathbf{R}, +)$ at all: There's no continuous action of $G$ on $M$ reducing to addition in the fibres. It's likely the intuition you're seeking is to view $S^{1}$ ...

3

What follows is a topologist's solution, but I believe quite revealing. A torus minus one point is a torus minus a disc. Expanding the disc one gets two cilinders (without border, but for the picture it's easier to draw it) attached as seen. These are just a couple of annuli with a common rectangle, which can be projected onto the plane. It's an open ...

3

Here is what was probably intended. Observe that $\overline{\phi}_{\alpha}(q)=g_{\alpha}(p)g_{\alpha}(q)^{-1}$. Thus $g_{\alpha}(p)=\overline{\phi}_{\alpha}(q)g_{\alpha}(q)$.

3

The sheaf condition in the definition of a diffeological space should imply that the inclusion functor $\mathsf{Man} \hookrightarrow \mathsf{DiffLo}$ preserves finite coproducts and Hausdorff pushouts along open embeddings. It also preserves infinite coproducts, if you replace "second countable" by "paracompact" in the definition of a manifold, which is ...

3

This simply cannot be true. A generic Riemannian manifold does not have any isometries, but still the image of any geodesic is a totally geodesic submanifold.

2

$dF(u)\cdot dF(v) = u\cdot v \implies |dF(u)|^2 = |u|^2.$ So $dF$ is an isometry at each point. Suppose $\gamma : [0,1] \to \mathbb {R}^3$ is a nice curve from $a$ to $b.$ Then $F\circ \gamma$ is a nice curve from $F(a)$ to $F(b).$ The length of $F\circ \gamma$ is $\int_0^1|(F\circ \gamma)'(t)|\, dt = \int_0^1|dF(\gamma (t))[\gamma '(t)]|\, dt = ... 2 For the vector field$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$\Phi_t^X$... 2 $$t:= \alpha'(s) \in \mathbb{R}^2,\ t'=\kappa n\in \mathbb{R}^2$$ And$|t|=1$implies that $$t\cdot n=0$$ Then $$\cos\ \theta = t\cdot e_1,\ -\sin\ \theta =n\cdot e_1$$ so that $$-\sin\ \theta \theta ' = t'\cdot e_1=\kappa n\cdot e_1$$ 2 If$\theta(s)$is the angle ´twixt the$x$-axis and the tangent line at the point$\alpha(s)$, the unit tangent vector$\mathbf T(s) = \alpha'(s)$is given by$\mathbf T(s) = (\cos \theta(s), \sin \theta (s)); \tag{1}$then we have$\mathbf T'(s) = (-\sin \theta(s), \cos \theta (s)) \theta'(s); \tag{2}$setting$\mathbf n(s) = (-\sin \theta(s), \cos ...

2

Here are some things which make sense on complex manifolds, but not almost complex manifolds: Complex coordinates $(z_1, \ldots, z_n)$. In particular, vectors like $\frac{\partial}{\partial z_k}$ only make sense on a complex manifold. On complex manifolds, we have $\overline{\partial}^2 = 0$. This gives rise to the Dolbeault complex $$\Omega^{p,0}(M) ... 2 The Plücker embedding is an isometry of G(k,\Bbb C^n) to its image in \Bbb P(\Lambda^k\Bbb C^n) with the standard Fubini-Study metric. In the moving frames notation, for example, the Kähler form on \Bbb P^N is given by$$\frac i2\sum_{j=1}^N \omega_{0\bar j}\wedge\overline\omega_{0\bar j},$$where \{f_0;f_1,\dots,f_N\} is a unitary frame at the ... 2 A 2-form is a smooth choice of a skew-symmetric bilinear form on each of the tangent spaces of your manifold. A 2-form is non-degenerate if it is non-degenerate (as a bilinear form) when restricted to each tangent space. For a general introduction to differential forms and smooth manifolds see Lee's Introduction to Smooth Manifolds. For a general ... 1 HINTS: This is going to be a common, everyday surface, so you need to do some algebraic manipulations and more multivariable analysis. You should first prove that the surface is compact. Then you will need to recognize that this surface is a branched double cover of the 2-sphere and use basic facts about Euler characteristic. 1 Using the definition from Wu-Sachs: By definition (M,g) obeys the chronology condition iff no points in M chronologically precedes itself (Rougthly: ancestorcide is impossible) Now M is diffeomorphic to a strip, say (t,x) with 0\le t <1 and x\in \mathbf{R}. The lines t=0 and t=1 are identified by (t=0,x) \sim (t=1,x+1). You can ... 1 First, you need to understand what a derivative at x is in this context. Df(x) is the linear function such that$$ \begin{array} .Df(x) : & \mathbb{R}^m & \to & \mathbb{R}^m \\ & u & \mapsto & Df(x)(u) \end{array} Personnally, I find it more clear to write it as Df_x, to show that x is a ... 1 To answer your first question: Df(x) is a linear map. So you can think of Df(x) as a matrix and u the vector it's acting on. Ie, just matrix multiplication. Perhaps more formally: \begin{align*} DF:\mathbb{R}^m&\rightarrow \mathcal{L}(\mathbb{R}^n)\\ x&\mapsto DF(x) \end{align*} where \mathcal{L}(\mathbb{R}^n) is the set of linear maps on ... 1 There is no need to proceed via the Stiefel-Manifolds. You can directly realize the Grassmannians as homogeneous spaces. In the real case, this takes the form G(k,n)=SO(n)/S(O(k)\times O(n-k)). This corresponds to a so-called symmetric decomposition of the Lie algebra \mathfrak{so}(n), thus making G(k,n) into a compact symmetric space. (The Riemannian ... 1 Notice that there is no Riemann tensor for an inner product space. Moreover, at one chosen point p \in M the metric inner product in T_p M can always be made Euclidean. In other words, the condition for a metric to be flat is local, not pointwise. To see what is going on, one may look at the Taylor expansion of the metric. In special ("geodesic" or ... 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 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 You should read more carefully the page before 203 (namely 202). J in his example (or \beta in yours) is not a differential 3-form per se. It is a differential 3-form in the distributional sense. You can think of it as \delta_{f(C)} \iota_{f'} \omega  by which I mean a Dirac delta function supported on the curve $f(C)$ times the three form ...

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

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 ...

1

You have the right idea to look at the preimage and compose the charts with the diffeomorphism. To proceed, remember that diffeomorphism is, in particular, a homeomorphism, and so identifies open sets in the domain and the range. In considering the ellipsoid, is there something similar to an ellipsoid you already know is a regular surface? Can you relate ...

1

Question i: From the definition of Reeb vector field we know that $\omega(R_{\omega})=1$ and being positively-tangent to the binding implies that $T_\theta B = f(\theta) R_{\omega}$ for some positive function $f$. Hence $\omega(T_\theta B) = f(\theta) > 0$. Question ii: Again from the definition of Reeb vector field we know that \$\iota_{R_\omega} d ...

Only top voted, non community-wiki answers of a minimum length are eligible