# Tag Info

6

If $G$ is compact and connected, one can prove (by constructing a bi-invariant metric on $G$ and relating the metric and Lie group exponential maps) that the exponential map $\exp : {\frak g} \to G$ is surjective. This justifies the claim for, e.g., $SU(2)$. Using the Baker-Campbell-Hausdorff formula, one can show that $\exp$ is also surjective for Lie ...

5

In fact they are not even homotopy equivalent. $\mathbb{R}^3$ is homotopy equivalent to a point, whereas $\mathbb{R}^3 \setminus \{ 0 \}$ is homotopy equivalent to $S^2$. These can be distinguished by e.g. their second homology $H_2$. For a more differential argument you can try comparing second de Rham cohomology. On $\mathbb{R}^3 \setminus \{ 0 \}$ you ...

4

I strongly agree with Karl's and Björn's comments regarding the Latin orgin: spatium. See: Leonhard Euler, Mechanica sive motus scientia analytice exposita, Tomus I, Petropoli, 1736 : Propositio 4 [ page 13 ] Sit spatium $AM$, sive sit linea recta sive curva, $=s$, et celeritas, quam corpus habet in M sit $c$, quae erit functio quaedam ipsius $s$. Ab ...

4

For a simple counterexample, take $X=S^1$ and $Z$ to be a point. Then $\chi(X\setminus Z)=1\neq \chi(X)-\chi(Z)=-1$. Morally, what's going wrong here is that you should consider $\chi(X\setminus Z)$ to be $-1$ rather than $1$, since it is made up of only a single (open) $1$-cell and no other cells. But if you take the standard homological definition of ...

3

Hint the determinant is continuous so the inverse image of the real line -0 by the determinant is an open subset so it is a submanifold.

3

Before saying what I think are good introductions to mathematical gauge theory, I should say what I think gauge theory is. What gauge theory means to me is the application of certain PDEs, relevant in physics, to the topology and geometry of manifolds. I note here that one can be ignorant of the actual physics (as I am). Here is one example. Given a ...

2

To show that the Jacobian really is of rank $1$ (i.e., not equal to the zero vector), I suggest you expand $\det A$ by minors along the first column, say: $$\det A = \sum_{i=1}^n (-1)^{i+1}a_{i,1}M_{i,1}.$$ Of course, $(-1)^{i+1}M_{i,1}$ is in fact an entry of the Jacobian. So the Jacobian can only be zero, if also $\det A=0$. But we are looking at the ...

2

The Lie bracket $[\dot c,X]$ is not well defined. A Lie bracket requires two vector fields defined on an open set, so as you noted, to make sense of this bracket you need to extend $\dot c$ to a vector field. But different extensions will give different results. For example, in $\mathbb R^2$, suppose $c(t) = (t,0)$ and $X(x,y) = \partial/\partial y$. Then ...

2

The surface is contained inside the sphere, but the tangent plane is outside the sphere, except at $p$. Therefore as you leave $p$ in any direction, the surface will curve away from the tangent plane at least as fast as the sphere does.

2

Write $p_1=\gamma(t_1)$ and $p_2=\gamma(t_2)$. Consider $f(t)=<\gamma(t)-\gamma(t_1),\gamma(t)-\gamma(t_1)>$. The distance between $p_1$ and $p_2$ is maximal implies that $f'(t_2)=0$. This is equivalent to $<\gamma'(t_2),\gamma(t_2)-\gamma(t_1)>=0$. To show that $<\gamma'(t_1),\gamma(t_1)-\gamma(t_2)>=0$, use ...

1

Let $\gamma:\ t\mapsto {\bf x}(t)$ be the given curve, whereby it is assumed that ${\bf x}(\cdot)$ is $C^1$ and periodic with some period $T>0$, and that ${\bf x}'(t)\ne{\bf 0}$ for all $t$. We want to find the two most distant points on $\gamma$. To this end we consider the doubly periodic function $$f(s,t):=|{\bf x}(t)-{\bf x}(s)|^2=\langle{\bf ... 1 Let's think about how this works for a curve in \Bbb R^3. We show that a curve \gamma has torsion identically 0 if and only if \gamma lies in an affine plane. Namely: If there is a constant unit vector \mathbf C so that \mathbf C\cdot(\gamma-\gamma_0) = 0, then we differentiate a few times to find that \mathbf e_1\cdot\mathbf C = \mathbf ... 1 TX= \{(p,v) : p \in X, v \in T_{p}X \} since X and T_{p}(X) are connected Hausdorff spaces ( T_{p}(X) is a subspace of manifold X ) Hence TX is a connected Hausdorff space 1 First of all, "equivariant version of \mathcal{O}(1)" sounds a bit confusing to me. What we actually do, we endow \mathcal{O}(1) with equivariant structure. U(1) acts as follows. Point of total space is a pair (x, \xi), where x \in \mathbb{P}^1 i.e. x is line in \mathbb{C}^2. \xi is a linear function on this line l. U(1) acts on hole ... 1 In other words, you are asking if there are "exotic" coequalisers in the category of manifolds. In fact, there are. Let S^1 = \{ z \in \mathbb{C} : \left| z \right| = 1 \} be the circle, let R = \mathbb{Z} \times S^1, and let d_0, d_1 : R \to S^1 be defined as follows:$$d_0 (n, z) = \exp (i n) zd_1 (n, z) = z$$It is not hard to see that the ... 1 In the meantime I have come up with a similar answer to Zhen Lin's. Consider \mathbb{Q} with the discrete topology, then the colimit of the coproduct map \mathbb{Q} \rightrightarrows \mathbb{R} \coprod \mathbb{R}, where the two arrows are the two obvious inclusions, has the colimit$$ \begin{array}{rcl} \mathbb{R} \coprod \mathbb{R} & \to & ...

1

Notice that $\int_A\omega$ is equal to the integral of $\omega$ over the lower half of the sphere and so $\int_A\omega=\frac12\int_Bd\omega$, where $B=\partial A$ is the unit ball. Since $d\omega=0$ we get $\int_A\omega=0$.

1

I hope this answer satisfies the sadistic requirement of the problem. The correct pullback $\phi^*\omega$ is \begin{align} \phi^*\omega & = \left( xy \begin{vmatrix} \frac{\partial\phi_1}{\partial u} & \frac{\partial\phi_1}{\partial v}\\ \frac{\partial\phi_2}{\partial u} & \frac{\partial\phi_2}{\partial v} \end{vmatrix} + 2x \begin{vmatrix} ...

1

According to "Hamiltonian manifolds and moment maps" by Nicole Berline and Michèle Vergne (chapter 2.3, page 19), or to the lectures notes by Yael Karshon (page 1), Darboux coordinate patches can be obtained as follows: let $U = \{ [z_0 : \dots : z_n] \in \Bbb P ^n \Bbb C \mid z_0 \ne 0 \}$, as usual let $B = \{ (x_1, \dots, x_n, y_1, \dots, y_n) \in \Bbb ... 1 HINT:$\gamma_1$and$\gamma_2$are asymptotic curves (their tangent vectors are directions of normal curvature$0$). Write down Euler's formula for the normal curvature in direction$\theta$(relative to the first principal direction, say). 1 There are at least two ways to calculate the Gaussian Curvature, when$M$is a regular surfaces in$\mathbb R^3$. By the second fundamental form: $$K = \frac{\det II}{\det I} = \frac{\cos^2 u}{\cos^2 u} = 1.$$ By the first fundamental form alone (this equation holds when$F = 0$, there's another formula when$F\neq 0): $$K = -\frac{1}{2\sqrt{EG}} ... 1 Firstly, note that two non-zero vectors are orthogonal if and only if their dot product is zero. Begin by differentiating the dot product:$$\begin{align*} \frac{d}{dt}\alpha(t)\cdot v & = \alpha'(t)\cdot v = 0 \end{align*}$$This means that the dot product of \alpha and v is constant. Since you know it is 0 at t_0, it is 0 everywhere. Thus ... 1 I think you're right,$${x^1}^2+\dots+{x^n}^2$$is hard to understand (although not really ambiguous once you've looked at it for long enough) but$${(x^1)}^2+\dots+{(x^n)}^2$$is totally clear. (You could also write x^1x^1+\dots+x^nx^n or \sum_i x^ix^i or \sum_i(x^i)^2.) 1 In general, no, absolutely not! Consider M = \Bbb R^2/\Bbb Z^2 and X= \partial/\partial x + a\partial/\partial y, where a is irrational. Then an integral curve of this is a line with irrational slope, which is dense in M. About as far from embedded you can get! You should expect that the "generic" vector field has non-embedded integral curves. 1 Examining the Gauss map, we see that \overline{g} is covered by the bundle map$$g \oplus g^\perp: \tau \oplus \nu \to \gamma_n \oplus \gamma^\perp$$and the differential$$D\overline{g}: \tau M \to \tau G_n(\mathbb{R}^{n+k}) \simeq \text{Hom}(\gamma^n, \gamma^\perp).$$This thus yields a fiber-linear morphism$$\sigma: \tau M \to \text{Hom}(\tau M, ... 1 Or does everyone really just refer to points and vectors interchangeably in the Euclidean space\Bbb R^n$? Yes. Everyone does. Okay, hyperbolic statements like that one are almost guaranteed to not be true. But still, it is close to true. Practically no one balks at the concept. However, it is also not completely true that they are used ... 1 They do cancel, if you keep track of the signs correctly. For example, the second term produces$(Yf) \omega(Z,X)$, while the fourth produces$-\omega( -(Yf)X,Z) = (Yf)\omega(X,Z)$. These are negatives of each other because of the antisymmetry of$\omega$. 1 We are given the vector$\vec F=\frac{\vec r}{r^2}=\hat r$, where$\hat r$is the radial unit vector. The divergence of$\vec F$for$0<a\le r\le b$is$\begin{align}\nabla \cdot \vec F&=\frac{1}{r^2}\frac{\partial (r^2F_r)}{\partial r}+\frac{1}{r\sin \theta}\frac{\partial (\sin \theta F_{\theta})}{\partial \theta}+\frac{1}{r\sin ... 1 Your question is rather psychological. But nevertheless I will write some remarks. Remark 1 Maybe formulation of your question is not the best. Non compact space may have finite volume (I mean just classical volume, not equivariant). For instance you may consider\mathbb{C} = \mathbb{CP}^1 \ \backslash \ \{ \infty \}$. And consider symplectic volume ... 1 Though you've got an excellent answer, here's a lower-tech argument for posterity: If the equation of the surface in cylindrical coordinates is$r = f(z)$for some positive,$C^{2}$function$f$, it's well-known that: A latitude$z = c$is a geodesic if and only if$f'(c) = 0$. The Gaussian curvature and$f''$have opposite sign. Particularly, if$K < ...

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