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5

The Bonnet theorem (see also here) states, roughly speaking, that if we are given the data $\tilde{I}$ and $\tilde{II}$ that are proposed to be the first and the second fundamental forms of the surface, then provided they satisfy the equations of Gauss and Codazzi (and Ricci in a more general setting) there exist a surface in the Euclidean space such that ...


4

Your vector fields really are derivation, so think that they will be applied to functions! Take a test function $f:\mathbb{R}^2\longrightarrow\mathbb{R}$ of class $C^\infty$. Then apply your vector field to this function (and you'll have to use the product rule and Schwarz' theorem at some point): $$\begin{align*} &\left(\left(-y\frac{\partial}{\partial ...


3

The idea is to operate on a (sufficiently nice) function. In your case, "sufficiently nice" means at least twice continuously differentiable (since you will have mixed partials). Here's an example that can guide you through the computation. Suppose we want to evaluate $\left[x,\frac{d}{dx}\right]$. Derivatives don't make sense on their own. It isn't until ...


3

I did the case for a $(0, 1)$ tensor $A = A_i$, it's $$ (\Delta A)_i = \Delta (A_i) - g^{jk}\partial _k \Gamma_{ij}^l A_l - 2 g^{jk}\Gamma_{ij}^l A_{l,k} +g^{jk}\Gamma_{ik}^l \Gamma_{jl}^m A_m + g^{jk} \Gamma_{jk}^l \Gamma_{il}^m A_m$$ In general there is no hope that $(\Delta A)_i = \Delta (A_i)$, since the equations should be all coupled and the term ...


3

I think your questions will be answered most clearly if we try to identify the assumptions under which the definitions you wrote for differentiability make sense. The first thing to note is, like you wrote, that the first definition tells you when a map is differentiable and what is its differential while the second definition only tells you when a map is ...


3

Every torus are Kahler: the standard form $\omega = \frac{\sqrt{-1}}{2}\sum_{i=1}^n dz^i \wedge d\bar z ^i$ descends to a Kahler form on the torus. Let $\alpha$ be a 1 form. Write $$\alpha = \alpha^{1,0} + \alpha^{0,1}$$ so $$d \alpha = d\alpha^{1,0} + d\alpha^{0,1}$$ If $d\alpha$ has no $(0,2)$ component, then $d\alpha^{0,1}$ is a $(1,1)$ form and the ...


3

I'm not sure why you were asked to "compute" those forms in all their lengthy glory. The key thing to realize is this: As you suggested, $\partial S$ is the union of those two $3$-spheres (appropriately oriented—I'll get to that in a moment). Since $dt=0$ on $\partial S$ (as each component is contained in a slice with $t=\text{constant}$), any term in ...


2

The principle curvatures are extrinsic quantities, the depend on the embedding of the manifold into some exterior manifold, and describe how the embedded manifold curves in that space. The curvature tensor of a Riemannian (or Lorentzian) manifold is an intrinsic quantity which measures to which extend covariant derivatives commute.


2

If $G$ is a compact Lie group and $H$ is a closed subgroup, then $G\rightarrow G/H$ is a principal $H$-bundle. In particular, it is a fibre bundle with fibre $H$. So, $SO(n)\rightarrow SO(n)/SO(n-1)$ is a fibre bundle with fibre $SO(n-1)$. We then compose this quotient map with the diffeomorphism $SO(n)/SO(n-1)\rightarrow S^{n-1}$ to get a fibre bundle ...


2

$z$ is just the complex function on the complex plane and $\bar z$ is its conjugate. Writing in usual Euclidean coordinate, $z =x+ \sqrt{-1} y$ and $\bar z = x - \sqrt{-1} y$, so $$dz =dx+ \sqrt{-1} dy \text{ and } d\bar z = dx - \sqrt{-1} dy\ .$$ If you put this into the metric, $$\lambda dzd\bar z = \lambda (dx dx + dydy)\ ,$$ which means that ...


2

There is always some disagreement on such notation, but you should probably think of $dz$ as acting on tangent vectors (assuming we're in one dimension) by $dz(v) = v\in\Bbb C$ and $d\bar z$ as acting by $d\bar z(v) = \bar v$. (In general, in working with complex manifolds, one works with the complexified tangent bundle, which is spanned by the ...


2

No, you're misunderstanding the notation. The author is evaluating the derivative of $n$ at $p$ (that was omitted) on the tangent vector $\phi_v$. The chain rule tells us that $$Df_p(\xi) = \frac d{dt}\Big|_{t=0} f(g(t)) \quad \text{for any curve } g \text{ in $M$ with } g(0)=p \text{ and } g'(0)=\xi\,.$$ In this case, he's just taking the path $g(t) = ...


2

Searching through the literature suggests (see e.g. M. Audin, Torus actions on symplectic manifolds, 2004, p. 89) that a Lagrangian torus is a submanifold which is (diffeomorphic to) the torus $\mathbb{T}^n \equiv \mathbb{S}^1 \times \dots \times \mathbb{S}^1$ and is a Lagrangian submanifold at the same time.


1

Since the boundary is the level set $r = 2m$, the (unnormalized) normal can be found as the gradient of $r$: $$ \nu^a = \nabla^a r = g^{ab} \nabla_b r = g^{ab} \delta_b^r = g^{ar}.$$ To get the unit normal, just divide by the length: $$ n^a = (g(\nu,\nu))^{-1/2} \nu^a = \left(\left( 1-\frac{2GM}{r}\right)^{-1}\right)^{-1/2} g^{ar}. $$ The minus sign shows ...


1

$\newcommand{\R}{\mathbf{R}}$Your cover of $S^{n-1}$ is suitable. Let $x = (x_{1}, \dots, x_{n-1})$ denote an element of $\R^{n-1}$, let $\Pi:U^{+} \to \R^{n-1}$ be stereographic projection from $e_{n}$, i.e., $$ \Pi(x, x_{n}) = \frac{x}{1 - x_{n}},\quad \|x\|^{2} + x_{n}^{2} = 1,\ x_{n} \neq 1, $$ and let $\Sigma:\R^{n-1} \to U^{+}$ be the inverse map $$ ...


1

Yes, you are correct: a 1-form is $C^\infty(M)$ linear so if $\theta(X) = 0$ then $\theta(fX) = f\theta(X) = 0$ for all smooth functions $f$. This means that $\ker\theta$ is a module over smooth functions on $M$. If $\theta$ is nowhere vanishing, then $\ker \theta$ will be a vector subbundle (i.e. distribution) of $TM$. If $\theta$ has some zeros then ...


1

The cylinder (of unit radius, for simplicity) is obtained from the infinite strip $S = \{z\in \mathbb C: |\operatorname{Im}z|<\pi \}$ by identifying its sides. What you need is Neumann Green's function for $S$ (i.e., with zero normal derivative on $\partial S$), because it will remain harmonic after gluing the sides. And to find this function, you can ...


1

You're correct that $[\tilde{v}, \tilde{w}]$ is the Lie bracket of vector fields. $[\tilde{v},\tilde{w}]_p(f) = 0$ since $$[\tilde{v},\tilde{w}]_p(f) = df_p([\tilde{v},\tilde{w}]) = 0,$$ as $p$ is a critical point of $f$. This definition of the Hessian only makes sense at critical points. One way to get a coordinate-free definition of the Hessian everywhere ...


1

Here's the idea: define a covariant $5$-tensor $T$ by $$T_{ijklm} = R_{ijkl;m} + R_{ijlm;k} + R_{ijmk;l}.$$ The differential Bianchi identity says that $T$ is the zero tensor. First contract $T$ on its first and fourth indices, yielding a $3$-tensor $U$ which is also zero: $$0 = U_{jkm}= T_{ijk}{}^i{}_m = R_{jk;m} - R_{jm;k} + R_{ijmk;}{}^{i}.$$ Then ...


1

We may have a good notion of arc length even for some nonregular (or not even differentiable) curves; all we need (by definition) is that the curve $a\colon[0,b]\to \mathbb R^n$ is rectifiable, that is: $$ l(a):=\sup\{\,d(a(0),a(t_1))+d(a(t_1),a(t_2))+\ldots+d(a(t_{n}),a(b))\mid 0<t_1<\ldots<t_n<b\,\}$$ exists (is finite). For the case you ...


1

Simply if $$\alpha: I\to R^3$$ is a regular curve and the arc length is $$s(t)=\int_{t_0}^t |\alpha'|dt$$ Now we solve for $t$ as $t=t(s)$ to get the function $$t:J\to I$$So we define a new curve $\beta(s)=\alpha\circ t=\alpha(t(s))$ where $$\beta: J\to I \to R^3 \\ |\beta'(s)|=|\frac{d\beta}{ds}|=|\frac{d\beta}{dt}\times ...


1

(a) is a simple calculation. (b) is the difference of volumes of two 4-balls of given radii. Let $H=p_1^2+p_2^2+q_1^2+q_2^2.$ Let $D$ be the subset of $\mathbb R^4$ given by $1\leq H\leq 2$. Then $S$ is the graph of the function $t=H$ over $D$. Now change variables in the integral to integrate over $D$ instead of $S$ and you get the volume of $D$ (since ...


1

As for the first question, since a permutation of the coordinates is a diffeomorphism, any set of coordinate functions is admitted. It's just easier to write it down that way. As for the second question, note that the $y-axis$ is divided in two parts which meet a the orign (and the origin is excluded from the set, note the tip of the arrows).


1

You have to look at G as a function of 2 variables, no $a_2$ fixed. As you figured out earlier, this would result in a (special) curve (a coordinate line). You need to totally differentiate G, not only consider partial derivatives. That is, you have to look at the linear map $DG: \mathbb{R}^2 \rightarrow \mathbb{R}^3$ If that map has maximal rank ($2$ in ...


1

When you see a computer-drawn picture of a surface $S = G(D)$, there's often a "curvilinear grid" on the surface. These curves are precisely the "coordinate curves" obtained by holding one variable in the parametrization $G$ constant and varying the other. Two coordinate curves pass through each point $G(a_{1}, a_{2})$ of $S$, each obtained by holding one ...



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