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4

A large class of structures on manifolds are defined using "reduction of the structure group" applied to the tangent bundle: these include, but are not limited to, orientations, spin structures, almost complex structures, almost symplectic structures, etc. A parallelizable manifold admits all of these structures. Is that the sort of thing you were looking ...


4

Since you did not show your work, I can only conjecture that you forgot to differentiate $a$ in $$\Delta f = \frac{1}{a^2 \sin\phi} \frac{\partial}{\partial \phi} \left( \sin\phi \frac{\partial f}{\partial \phi} \right) + \frac{1}{a^2\sin^2\phi} \frac{\partial^2 f}{\partial \theta^2} \, , $$ In the next step, if you want to take $\Delta$ again, note then ...


3

The first thing is purely pedagogical: A parallelizable manifold is the one which has trivial tangent bundle. When one is first introduced to (locally trivial) fiber bundles, trivial bundles are easy comparing to the general bundles; computations are sometimes easier for trivial bundles. (Later on you just get used to the concept.) Second, in the case of ...


3

This is nowhere near an answer. I'm just throwing some thoughts out there. Hopefully this inspires others to write down better results. I assume the domain is closed. We can always take $k \leq n-2$. One path at attempting to get lower bounds at $k$ is to find manifolds that immerse into small-dimensional spaces but embed only in large-dimensional spaces. ...


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 ...


3

Find the normal vector in terms of $x$ and $y$. Then show $\vec{n} \cdot (x, y, \sqrt{xy}) = 0$. If the dot product is 0, then the two vectors are orthogonal, "perpendicular".


3

Use the Corollary of this paper: If a compact Riemannian manifold $M$ splits as $M = M_1 \times M_2$, then the identity component of the isometry group splits as $I_0 (M) = I_0(M_1) \times I_0(M_2)$. (This is authors' stated Corollary, but from their main theorem it looks like we can weaken compactness to your requirement that $M$ has no Euclidean ...


3

Suppose there are infinitely many $t$ such that $\gamma(t)=x$. Since $[a,b]$ is compact there is actually a sequence $t_n$ that converges to some $t_0$ such that $\gamma(t_n)=x$. In particular $\gamma(t_0)=x$. Show this implies that $\gamma'(t_0)=0$.


2

I spent a few more hours on the problem, and eventually found the solution. No major breakthroughs, but a lot of algebra. First, I reversed the problem, instead maximizing the volume subject to a surface area constraint. Not strictly necessary, but it simplifies the computation. I also gave the surface area a magnitude $k$, per user7530's comment. The new ...


2

There are a few problems with the formulation: The volume constraint: right now you are constraining the total volume to be zero (check by differentiating $S$ with respect to $\lambda$). If you want the total volume to be equal to some fixed value $C$, you need instead the action $$S = \int_{-1}^1 \left[2\pi y \sqrt{1+y'^2} + \lambda(\pi ...


2

Noting that by the equation$$p^2 + q^2 + 1 = z^{-2}$$the equations for $\dot{p}$ and $\dot{q}$ can be simplified. Furthermore, the equation for $z$ is independent from the rest, which means that we can solve for $z$ first, then for $p$ and $q$. Geometrically, our equation looks like an eikonal equation since it can be written as$$\left|\nabla u\right|^2 = ...


2

It's always risky to answer "no" to open-ended "is it possible"-type questions. That said, in the case of using the volume formula for a family of regions to deduce surface area (the way the area of a sphere of radius $r$ is the derivative with respect to $r$ of the volume of a ball of radius $r$), the answer is probably "no": Think, for example, of a ...


2

I am not sure whether I understand the question correctly, but for the diffeomorphism $\theta_{t_0}$, the derivative in $0$ will be a linear isomorphism. By the flow property this will map $V(0)$ to $V(p)$ and the tangent vector to $S$ in $0$ to the tangent vector of $\theta_{t_0}(S\cap O)$ in $p$, so these two vectors have to be linearly independent, and $ ...


2

I am afraid you are trying to get information in a situation where not a lot of information is available. The only "general" statement that you can make about the derivative is that it is always a linear isomorphism, everything else depends on the choice of coordinates. For the choice of local charts (or coordinates) you made, the computation you did ...


2

I pulled up my sleeves and did the computation for $M^2 \subseteq \Bbb R^4$. I'll post it here, but I'll leave the question open for a while in case someone sees a non-medieval way to do this. First of all, take $\{e_1,e_2\}$ an orthonormal positive basis of $T_xM$, and pick any orthonormal positive basis $\{n,\nu\}$ of $T_xM^\perp$ with $\{e_1,e_2,n,\nu\}$ ...


2

Some remarks on infinite-dimensional manifolds. There are two approaches which to me still feel very manifold-ish (there are others yet: Frolicher spaces and diffeological spaces, which feel a bit less so); the approach of Banach manifolds and the much more general approach of Frechet manifolds. The former is almost precisely the same theory as the theory ...


2

The confusion here seems to be one of linear algebra, and not of differential geometry. The elements of a tensor product $\Bbb V \otimes \Bbb W$ of two vector spaces can be written as sums $\sum_a v_a \otimes w_a$, and this is enough to describe the manifold structure on the tensor product vector bundle $V \times W \to X$. Choosing bases $(E_1, \ldots, ...


2

Here are hints for parts of the questions: 1) Note that if $\nabla f(x_0) \neq 0$ then $\dot{\gamma}(t) \neq 0 $ for all $t \ge 0$. To see this, if $\dot{\gamma}(t_0) = 0$ for some $t_0 >0$, then $t \mapsto \gamma(t_0)$ and $t \mapsto \gamma(t)$ would be two solutions passing through $\gamma(t_0)$ and by uniqueness they must be the same hence a ...


2

I think the issue is that you are conflating the Biharmonic operator $\Delta^2$ of the sphere (the first one in your question) with the one $\bar \Delta^2$ of $\mathbb R^3;$ i.e. you are assuming that when $f : \mathbb R^3 \to \mathbb R$ depends only on $\phi, \theta$ that $\Delta^2 f = \bar \Delta^2 f$. The equivalent statement for the Laplacian is true ...


2

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 ...


2

I can see a problem right off the bat: $$L=\dot x^2+x^2\dot y^2$$ $$p_x=\frac{\partial L}{\partial\dot x}=2\dot x$$ $$\dot p_x=2\ddot x=\frac{\partial L}{\partial x}=2x\dot y^2$$ Since your equation for $\dot p_x$ was not just a typo in my reading of your solution, you were hosed at this point. $$p_y=\frac{\partial L}{\partial\dot y}=2x^2\dot y$$ $$\dot ...


2

As John Ma correctly points out, your space is not smooth along the intersection of the hemisphere and the disk, hence the smooth notion of orientation does not make sense. However, there is a topological notion of orientability, using homology, which applies to your space. It might be intuitively acceptable to you to think of "smoothing" your space along ...


2

One can do without smoothing: An orientation of the hemisphere $H$ induces an orientation (i.e., a sense of direction) of the boundary circle $\partial H$, say counterclockwise when seen from above. Similarly for the disk $D$. The question is whether you can orient the disk in such a way that the induced orientation of the boundary circle $\partial D$ is now ...


2

That formula is often used as the definition! Since you're asking this question, I'll assume you're using the other common definition of $d$ for one-forms, which is the coordinate formula $d\omega_{ij} = \partial_i \omega_j - \partial_j \omega_i.$ Contracting with $X^i Y^j$ we get $$d \omega(X,Y) = Y^j X(\omega_j) - X^j Y(\omega_j).$$ From the Leibniz rule ...


2

Expanding my comment: \begin{align*}(g^*\omega)(x)(v_x^1,\dots,v_x^{2n+1}) & =\omega(g(x))(d_xg(v_x^1),\dots,d_xg(v_x^{2n+1})) \\ & =\omega(\color{red}{-x})(-v_x^1,\dots,-v_x^{2n+1})\\ & =\sum_{j=1}^{n+1} (-x_j) (-v_x^{n+j}) - (-y_j) (-v_x^j)\\ &=\sum_{j=1}^{n+1} x_jv_x^{n+j} - y_jv_x^j\\ &=\omega(x)(v_x^1,\dots,v_x^{2n+1}).\end{align*} ...


1

Since the question is tagged differential-geometry, I'm assuming that the "isometry" you're talking about is not an isometry between metric spaces, with the global metrics each surface inherits from $\mathbb R^3$. (Every such isometry between arbitrary subsets of $\mathbb R^3$ can be extended to a rigid motion). Instead, I suppose that an isomorphy merely ...


1

Take $x,y\in \mathbb R^3$ and $\gamma$ any curve connecting them. You want to show that $\ell(f(\gamma))=\ell(\gamma)$ then since the metric is given by the following: $d(x,y)=\inf \ell(\gamma)$ or $\gamma$ connecting $x$ to $y$ you get the result (you need to do it for the inverse as well, but that is just noting that if $f$ is a diffeomorphism all such ...


1

Note that the right side is periodic in $x$, with period $2\pi$, and varies between $B - |A|$ and $B + |A|$. The left side, on the other hand, takes any value $\ge 1$. There are three main cases (I won't bother with the cases of equality): $B + |A| < 1$: no real solutions. $B - |A| < 1 < B + |A|$: infinitely many closed curves. $1 < B ...


1

Since you know the tangent plane, the vector normal to plane will also be normal to the surface. We know normal vector to plane of equation $ax+by+cz+d=0$ is $n=<a,b,c>$. We also know a point on the line. Hence we can form a equation, $x = 2 -16t $ $y = 1 -2t$ $z = -61 + t$ This is same as your equation. (For forming line equation with a given ...


1

Sometimes, when doing calculations, it is convenient to view tangent vectors as velocities on smooth trajectories. Formally, if $x \in S^n$ and $v \in T_x S^n$, then there exist a smooth curve $\gamma : (-\varepsilon, \varepsilon) \to S^n$ with $\gamma (0) = x$ and $\dot \gamma (0) = v$. The correct consequence of $q(x) = q(-x)$ is that $$(\Bbb d _x q) (v) ...



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