# Tag Info

## New answers tagged differential-geometry

0

In fact $[g]$ is path-connected with respect to the $C^k$-, $C^\infty$-, or $C^{k,\alpha}$-topologies. $C^\infty(M,\mathbb{R})$ is path-connected: if $f,h\in C^\infty(M,\mathbb{R})$, then $\gamma(t)=th+(1-t)f$ is a continuous path connecting them. The map $$C^\infty(M,\mathbb{R}) \ni f \mapsto e^{2f}g \in \mathcal{G}$$ is continuous. Hence the image of ...

1

First recall theorem: Let $\gamma= [0, a] \to M$ be a geodesic and put $\gamma(0) =p$. The point $q = \gamma(t_0 )$, $t_0 \in (0, a]$, is conjugate to $p$ along $\gamma$ if and only if $v_0 = t_0 \gamma'(0)$ is a critical point of $exp_p$. Now assume by contradiction that there are infinitely many geodesic of length $\le a$ from $p$ to $q$. Each geodesic $... 0 If$L$is an embedded submanifold, then$f:L\to N$is smooth embedding since the inclusion$i:L\to M$is smooth embedding and composition of smooth embeddings is smooth embedding. But if$L$is only an immersion submanifold, it could happen that$f_{|N}$is not smooth embedding. Take the classical example: let$\beta:(-\pi,\pi)\to\mathbb{R}, \beta(t)=(\...

1

The idea is to write $cosh$ in function of $tanh$ Let be $tanh(x)=y$. $cosh(2x)=1+sinh(2x)=$ $=1+2cosh(x)sinh(x)=1+2\frac{cosh(x)sinh(x)}{1}=$ $=1+2 \frac{cosh(x)sinh(x)}{cosh(x)^2-sinh(x)^2}=1+2 \frac{\frac{cosh(x)sinh(x)}{cosh^2(x)}}{\frac{cosh(x)^2-sinh(x)^2}{cosh^2(x)}}=$ $=1+\frac{2y}{1-y^2}$ Then in your case if you choose $x=tanh^{-1}(a)$ you ...

1

In fact, we can show more: That $$\theta \wedge d\theta \wedge \cdots \wedge d\theta$$ vanishes nowhere. As you've pointed out, this fact will imply that $\ker \theta$ is nonintegrable and moreover by definition that $\theta$ is a contact form---so, maximally nonintegrable---which is a strictly stronger condition than nonintegrability for $n \geq 3$. First, ...

1

I find the wording of your question sufficiently ambiguous that it could lead to confusion already among mathematicians, without even going so far as to interfacing with physicists! It is clear to me that any connection is determined by its parallel transport: in other words, for every function which assigns to each (smooth) path a (smoothly varying) ...

0

In fact, consider the smooth map: $$f: \mathbb R^{n+1}-\{0\}\longrightarrow S^n, x\longmapsto \frac{x}{|x|}.$$ Let $\pi_2: S^n\longrightarrow S^n/(v\sim -v)$ be the canonical projection. This is also smooth. In particular, the composite is smooth $$\pi_2\circ f: \mathbb R^{n+1}-\{0\}\longrightarrow S^n/(v\sim -v).$$ This induces a smooth map at the level ...

4

Here's a sketch for the case $n=2$, but it will generalize without difficulty. Note first of all that on $S^3$ we have the equation $$\sum_{i=1}^4 x^i\,dx^i = 0.$$ In particular, at the point $P=(0,0,0,1)$, we have $dx^4=0$ (as a form on $S^3$), and so (again, as forms on $S^3$) $$\theta(P) = dx^3 \quad\text{and}\quad d\theta(P) = -2dx^1\wedge dx^2,$$ so ...

0

(1) The key is that $u$ is positive (as the exponential of something). As you already noted, $u$ is an eigenfunction of the Laplacian. But the strong maximum principle implies that (on a compact manifold) the only positive eigenfunctions of the Laplacian are the positive constants. (2) The literature now contains far more sufficient conditions for the ...

6

Here is a longer derivation that requires mostly basic calculus. The Laplacian is invariant under rotations. Specifically, if $f:\mathbb R^n\rightarrow \mathbb R^n$ and $\nabla^2 f(x)\;=\alpha$, then $\nabla^2 g\; (R^{-1} x)=\alpha$ where $R$ is any rotation matrix ($R^TR=I$ and $\det(R)=1$) and $g(x) := f(Rx)$. Now suppose that $f:\mathbb R^n\rightarrow ... 3 Since you're a physicist, I thought a variational derivation might be appealing. Consider the functional $$F[\phi]=\int \left[\frac12(\nabla \phi)^2+U(\phi)\right] dV.$$ Then the functional derivative is $$\frac{\delta}{\delta \phi}F[\rho]=\frac{\partial U}{\partial \phi}-\nabla^2 \phi =U'(\phi)-\Delta_D \phi,$$ so your first equation is equivalent to$\...

1

There's a bunch more things to say. First of all, the opposite inequality $\text{ord}(A) \le \text{ord}(A_T)$ is also true, for the simple reason that the function $A \mapsto A_T$ is an injective homomorphism from the group $SL(n,\mathbb Z)$ to the group of automorphisms of $T$. It follows that you can convert your problem entirely into algebra, by ...

0

There are (too) many points of view on connections. A connection is not a pointwise object so that there is no hope of representing it as a section of a tensor bundle over $M$. However, both $G$-connections on principal bundles and linear connections on vector bundles are instances of Ehresmann connections which make sense for general smooth fiber bundles ...

1

Here is something that may help you. Consider $G=GL_n(\Bbb R)$, that is the group of real invertible $n\times n$ matrices. Now show the following statements: Show that $GL_n(\Bbb R)$ is open in the vector space $M_{n\times n}(\Bbb R)$ of real $n\times n$ matrices. As such you may identify every tangent space with $M_{n\times n}(\Bbb R)$. Let $g, h\in GL_n(\... 1 It's taking in a vector in$\mathfrak{g}$and outputting what exactly? It is taking a vector in$\mathfrak{g}$and giving a vector in$\mathfrak{g}$. Given$g\in G$, we have$Ad(g):G\rightarrow G$defined as$h\mapsto ghg^{-1}$. This map sends$e\in G$to$e\in G$. So, if we consider differential of$Ad(g):G\rightarrow G$at$e\in G$, we get$Ad(g)_{*,e}...

0

To get equispaced point you should use path length $s$ along the curve as parameter instead of $t$. A straightforward computation gives: $$s=\int_0^t\sqrt{dx^2+dy^2}=\int_0^tr\tau\,d\tau={1\over2}rt^2,$$ hence make the substitution: $$t=\sqrt{2s\over r}$$ into your equations, compute $s_\max=(1/2)rt_\max^2$ and use $s$ instead of $t$ to get your points.

1

Why do we write $T_h$ as the partial derivative w.r.t $h$ and what does it actually mean to take the derivative of the left action as we are doing in this definition? $T_h$ is not at all a partial derivative with respect to $h$; note that in the setting of general manifolds, you cannot use the naive difference-quotient definition of derivatives, because in ...

2

It is not really a partial derivative, it is first of all the differential or tangent map of a smooth map on a manifold, which is evaluated at a point. In this case, it is $L_g$, and its differential gets evaluated at a point $h$, resulting in a linear map between the corresponding tangent spaces $T_h G$ and $T_{L_g(h)}G = T_{gh}G$. Wether you write $(d L_g)... 2 Let$g, h \in G$. Both$T_h$and$(d \ \cdot)_h$denote taking the tangent map at the point$h \in G$. Notably we do have any vector space structure on the Lie-Group$G$. Therefore naively defining derivatives via differential quotients is not possible, because differences such as$g - h$are not defined without a vector space being involved. The ... 0 Hint: Let$x_0\in M$, define for every$x\in M$, consider a path$c_x$such that$c_x(0)=x_0, c_x(1)=x$and define$g(x)=\int_0^1c_x^*\omega dt$.$g$is well defined since$\int_S^1f^*\omega=0$for every$f:S^1\rightarrow M$. Show that$dg=\omega$. To show that look the situation locally, that is in a chart, to evaluate$dg$at$v\in T_xM$, consider ... 0 HINT: Suppose the geodesic has$\kappa(P) = 0$for some$P$. Then$P$must be a planar point, a parabolic point, or a hyperbolic point of the surface. In the latter two cases, you get$\kappa_n\ne 0$for all but one or two directions, and you're done. Now what if$P$is a planar point? Now it's a set-theoretic argument. You need to assume the surface is ... 1 One sufficient condition is that the distance of the surface to the origin is attained at at most one point. 1 Consider all points in$[0,1]^n$with rational coordinates. We can enumerate them with natural numbers, so they produce a sequence$P_i$. We can also enumerate them in such a way so$|P_{i+1}-P_i|>\epsilon$for every$i$and fixed$\epsilon$. Now you can assign a spline that goes through all points in a succession. By construction, the closure of the ... 5 We can extend the example you mention, sometimes called an irrational winding of a torus, to higher-dimensional tori$\Bbb T^n$: Pick a constants$\alpha_1, \ldots, \alpha_n \in \Bbb R$that are linearly independent over$\Bbb Q$and denote$\alpha := (\alpha_1, \ldots, \alpha_n) \in \Bbb R^n$. Then, the image of the curve $$\gamma : \Bbb R \to \Bbb T^n \... 1 It's not really clear what kind of relation OP expects, but here's something ... Let the generators of a cone (with vertex at the origin and z-axis as axis) make angle \alpha with the xy-plane. Let a cutting plane parallel to the y-axis make angle \beta with the xy-plane, and let it meet the cone at V := (-v \cos\alpha,0,v \sin\alpha) (which ... 0 WARNING: this solution refers only to the case when the plane is perpendicular to the base of the cone. See Blue's answer for the general case. Let G' be the projection of G (vertex of the hyperbola) on the axis of the cone, H the projection of P on the axis of the hyperbola and H' be the projection of H on the axis of the cone. If we set:$$ x=... 1 Partial derivative differs from full derivative in the way that the result of$\partial/\partial x_1$is dependent on the choice of basis$x_1,\ldots,x_n$. When you change the basis, the derivative also changes (it forms covector space dual to vector space of coordinates, so it is no surprise). In Lagrange representation, the independent variables are is$...

3

The prerequisites for a course I took in Spring 1980 (at a U.S. university) that used Spivak’s Volume I were the following three courses: 1. Graduate linear algebra, which roughly covered chapters III, XIII, XIV, XVI in Lang’s Algebra, but I think knowing the first 5 chapters of Hoffman/Kunze’s Linear Algebra would easily be sufficient for the course I took,...

2

The main components that you need to be comfortable with to start studying differential geometry -- other than multivariable calculus and linear algebra that you can find in Apostol's II -- are basics from point-set topology and multivariate analysis/metric spaces. Respectively, you need to have a firm grasp on 1) Abstract topological spaces, separability (...

0

By Holder's inequality with $p=n$ and $q=\frac n {n-1}$ you can see that RHS of Brunn-Minkowski inequality is less than or equal to RHS of Bonnesen inequlity. Let $x=\frac {\mu(A)} a,y=\frac {\mu(B)} b$. Then $\mu(A)^{1/n}+\mu(B) ^{1/n} =a^{1/n}x^{1/n}+b^{1/n}y ^{1/n}\leq (a^{q/n}+(b^{q/n})^{1/q} (x+y)^{1/n}$. Hence $(\mu(A)^{1/n}+\mu(B) ^{1/n} )^{1/n} \... 0 Since the sign of$\kappa(s)$depends only on the choice of orientation of$\alpha$(and in particular has identical magnitude to$k(s)$) I can, without loss of generality, choose$\alpha$to be oriented such that $$\begin{equation*} \kappa(0) = k(0) \end{equation*}$$ And so$\textrm{sgn}({\kappa(0)})=1$, showing the result. 0 The identification$X:=X^i\partial_i$gives$XYf=X^i\partial_i(Y^j\partial_jf)=X^iY^j\partial_i\partial_jf+(X^i\partial_iY^j)\partial_jf$so $$[X,\,Y]f=XYf-YXf=X^iY^j\partial_i\partial_jf+(X^i\partial_iY^j)\partial_jf-Y^iX^j\partial_i\partial_jf-(Y^i\partial_iX^j)\partial_jf.$$The$\partial_i\partial_jf$terms cancel, since$$Y^iX^j\partial_i\partial_jf=Y^jX^... 1 Since you know Einstein field equations, then you must have heard of metric tensor and curvature tensor and their relationship to the field equations. In local coordinates (normal coordinates) we can make the first derivative of the metric tensor to vanish at a given point but not the second derivative simultaneously. But curvature tensor is a function of ... 2 Put your system in a more systematic form$$ \pmatrix{g&f\\-fg&1+f^2\\-f&g\\1+g^2&-fg} \pmatrix{f''\\g''} = \pmatrix{6f'g'\\-3gf'^2+3ff'g'\\−3f'^2+3g'^2\\3gf'g'−3fg'^2} $$The first and third equations can be combined to$$ \pmatrix{f&-g\\g&f}\pmatrix{f''\\g''}=3\pmatrix{f'^2-g'^2\\2f'g'} \\\text{or}\\ (f+ig)(f''+ig'')=3(f'+ig')^2 $$... 2 Hint Use the fact that a circle is a one dimensional manifold based on the mapping t \mapsto (\cos t, \sin t) and that (u,v) \mapsto (\cos u, \sin u, v) is a parameterization of a cylinder. 3 Ten is simply the number of distinct components of a second order symmetric tensor in a space of dimension four. Writing these equations in tensor form enables us to write EFE as a single equation instead of ten, just as \vec F=m\ddot{\vec x} is a single vector equation written in place of three scalar equations. 1 Hint: Denote the canonical projection by p:\land^2M\to M, then for each U, take p^{-1}(U)\cong U\times A^2(T_pM). 0 Your approach does work well with one modification, although it is probably not what do Carmo intended. As you note, the objective is to minimize \theta-\phi or to maximize \cos(\theta-\phi) = \cos\theta\cos\phi + \sin\theta\sin\phi subject to the constraint k_1 \cos\theta\cos\phi + k_2 \sin\theta\sin\phi = 0. While it may work if you think of \phi ... 1 Let \alpha:I\to\mathbb{R}^3 be a parametrized curve, suppose that [a,b]\subset I and define \alpha(a)=p, \alpha(b)=q. Then, for every vector v\in\mathbb{R}^3 with unit length we have$$(q-p)\cdot v=(\alpha(b)-\alpha(a))\cdot v=\int_a^b\alpha'(t)\cdot v\,dt\leq\int_a^b|\alpha'(t)||v|\,dt=\int_a^b|\alpha'(t)|\,dt.$$Thus if v=\dfrac{q-p}{|q-p|} ... 0 To construct geodesic parallel coordinates locally start with any regular curve \gamma with \gamma(0)=p and let the u-curves v\equiv t be the unit speed geodesics crossing \gamma orthogonally at \gamma(t). For example starting with the curve \gamma(t)=(\cos t,\sin t,0) on the unit sphere S^2 you get spherical coordinates$$x(u,v)=(\cos u\... 0 To make Somos's comments more explicit: it is not very hard to use the classical results of the "lemniscatic case" of the Weierstrass$\wp$function to derive the required invariants$g_2,g_3$(the proper term for what OP termed the "parameters", as they enter in the defining cubic$4u^3-g_2 u -g_3). I gave a derivation of the parametric equations for the ... 1 Your question is confuse because on one hand you ask to express a (3D) parametric curve in spherical coordinates, but on the other hand you give the example of a (2D) parametric curve turned to an implicit equation and not using polar coordinates. 2D case: $$\begin{cases}x=\cos t,\\y=\sin t\end{cases}\to\begin{cases}\rho\cos\theta=\cos t,\\\rho\sin\theta=\... 1 I cannot improve on Ted Shifrin's answer describing why the bracket term is necessary. So, I'll only attempt in this answer to elaborate the sense in which the exterior derivative and bracket are dual. Fix a local frame (E_a) and let (\theta^a) denote its dual coframe, so that \theta^a (E_b) = \delta^a{}_b; in particular each such contraction is ... 1 There is a minor issue with your problem. You need two equations to describe the helix. To explain why, think about how would you write the equation of a circle in 3D. x^2+y^2=1 is the equation of the surface of the cylinder. You would need to specify another equation (for example z=0) to get a single circle. As you can see, the equations for x and y... 1 The same way that you convert any coordinate to spherical$$\begin{align} r(\lambda) &= \sqrt{x^2+y^2+z^2} = \sqrt{1 + \lambda^2} \\ \theta(\lambda) &= \arctan(\frac{y}{x}) = \lambda \\ \phi(\lambda) &= \cos^{-1}(\frac{z}{\sqrt{x^2 + y^2 + z^2}}) = \cos^{-1}(\frac{\lambda}{\sqrt{1+\lambda^2}}) \end{align}$$-1 Given a 1-form \omega:M\rightarrow \Lambda^1T^*M, we want to associate a 2-form d\omega:M\rightarrow \Lambda^2T^*M. Forget about smoothness/alternating and all that for some time. Given a point m\in M and two vectors v,w\in T_mM, we are looking for a real number, which we want to declare as (d\omega)(m)(v,w). For fixed m\in M, the 1-... 2 One can do this without directly thinking about or computing flows. Hint Taking a general 1-form$$\omega_0 := p(x, y, z) \,dx + q(x, y, z) \,dy + r(x, y, z) \,dz$$on \Bbb R^3, imposing \omega(X) = \omega(Y) = 0, and solving for the coefficient functions p, q, r, shows that the annihilator of X and Y at each point is spanned by$$\omega_0 := dx ... 0 Ok, I finally figured out what I was doing wrong: I messed up the computations for the flows of the vector fields. The flows\theta_t$of$X$and$\phi_t$of$Yshould be \begin{align*} \theta_t(x,y,z) &= (x + zt, y + t, z)\\ \phi_s(x,y,z) &= (x + ys, y, z + s). \end{align*} This would then give us that$\theta_t\circ \phi_s = \phi_s \circ \... 5 One can prove that the Chern class is well-defined, independent of the choice of section. More precisely,$\text{Tr} F \wedge F / 8 \pi^2$is not the Chern class itself, instead it is a closed form. One must prove that the cohomology class of that form is well-defined independent of the choice of$A\$. That makes it a property of the vector bundle alone.

4

Scherk Surface How the coefficients of FFF and SFF are calculated is mentioned.

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