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Hot answers tagged differential-geometry

8

That $f$ is a local diffeomorphism implies that it is an open map. Hence $f(M)$ is an open subset of $S^n$. But $M$ is compact, so $f(M)$ is also closed. Since $S^n$ is connected, it follows that $f$ is surjective. Further, since $df_x$ has full rank everywhere, the fibres $f^{-1}(p)$ are discrete. Hence the fibres are finite. Use that to find that $f$ is a ...

6

I like $(x,y)\mapsto(x,y,x^2+y^2)$. $\ddot\smile$

5

Yes. One of many ways to see this is to fit $\mathbb{CP}^n$ into a fiber sequence $$S^1 \to S^{2n+1} \to \mathbb{CP}^n$$ (since $S^1 \cong \text{U}(1)$ acts by scalars on the unit sphere $S^{2n+1} \subset \mathbb{C}^{n+1}$ with quotient $\mathbb{CP}^n$) and apply the long exact sequence in homotopy. The long exact sequence also shows that ...

4

No. In fact compact Riemannian manifolds can have zero sectional curvature; take flat tori $\mathbb{R}^n/\Gamma$ where $\mathbb{R}^n$ has the usual Euclidean metric and $\Gamma$ is a lattice. Apparently every compact Lie group is the isometry group of some compact Riemannian manifold, but I don't know how cavalier one can be about specifying its curvature. ...

2

When you ask that $G \times G \to G$ is a smooth map, it means "with respect to the smooth structure you put on $G$ and the product differentiable structure on $G \times G$", so that asking this map to be smooth makes sense since it becomes a map between smooth manifolds and you can ask yourself this question. So somewhere you already assumed there was a ...

2

Yes. Since the product of a form and a smooth function is again a form, extend the form to a small tubular neighborhood of your submanifold and multiply by a bump function. Georges Elencwajg has suggested I explain further. To extend a $p$-form $\omega\in\Omega^p(M)$ to a $p$-form on $\mathbb{R}^n$, we need to first define a $p$-form on ...

2

It is a great mystery as to why there are so many excellent Russian born geometers/topologists (S. Novikov, M. Kontsevich, V. Voevodsky, G. Perelman, I. Shafarevich, D. Fuchs, M. Postnikov) and so few good textbooks in Russian on those subjects. So let me list the textbooks that are (in my opinion) still on par with the best modern textbooks: 1) M. M. ...

2

In Lee's 'Intro to Smooth Manifolds', $\Lambda^k(V)$ refers to the space of alternating $k$-tensors on a vector space $V$, as you mentioned. However, the space $\Omega^k(M)$ is the space of smooth $k$-forms on a smooth manifold $M$. That is, an element of $\omega \in \Omega^k(M)$ is a smooth map $M \to \Lambda^k(T^* M)$ (called a smooth section of ...

2

If you take the case where $a$ is negative, just say $-1$ for ease, you get, $$z_0-z_3+z_1=0$$ or $$z_0-z_3-z_1=0$$ the union of two planes, so their points of intersection are not smooth. If $a$ is positive, say $1$, then you have $$z_0=z_3$$and $$z_1=0$$ this seems to be the intersection and its a line, so it seems to me to be smooth ? But its dimension ...

2

Since $B = T \times N$, differentiate with respect to $s$, obtaining $$\frac{dB}{ds} = -\tau N = \frac{dT}{ds} \times N + T \times \frac{dN}{ds} = T \times \frac{dN}{ds}$$ since $dT/ds= \kappa N$ and $N \times N = 0$. There are some conclusions to be drawn from this: $dN/ds$ is not in the direction of $T$, else this cross product would be null. $dN/ds$ is ...

2

A geodesical ball $B$ centered in $p\in M$ of radius $r>0$, in a connected Riemannian manifold $(M,g)$, is a set of the form $$B:= \{x\in M \:|\: d(x,p) < r\}\:,$$ where $d(x,y) = \inf\left\{ L(\gamma)\:|\: \gamma: [0,1] \to M\:, \gamma \in C^1([0,1])\:, \gamma(0)=x\:, \gamma(1)=y \right\}$ and $L(\gamma)$ is the arch length of $\gamma$, $$L(\gamma) ... 2 Levap gave a great answer and I would like to summarize and put it in my own words. We want to prove the following claim: Claim: let B, C, and D smooth manifolds and$$ h:B\times C\to D $$a smooth map. Then, for any (X,Y)\in\mathfrak X(B\times C) we have$$ \boxed{D_{b,c}h.(X,Y) = D_bh(\bullet,c).X + D_ch(b,\bullet).Y} $$Proof: ... 2 This is indeed legitimate. To justify it rigorously without getting too boggled down with notation, I suggest that you do first the case where A = (-\varepsilon, \varepsilon), D = B \times C, h = \mathrm{id}, p = 0 and X_p = \left. 1 \cdot \frac{d}{dt} \right|_{t=0}. That is, handle first the case of curves. Rename f = \beta and g = \gamma. ... 2 Let \theta be the angle that the slanted red (?) line on the right makes with the horizontal. Then the height of the rectangle is r\sin\theta and the base is 2r\cos\theta, for an area of r^2\sin\theta\cos\theta. This is \frac{r^2}{2}\sin 2\theta. But \sin 2\theta has a maximum value of 1, at \theta=\frac{\pi}{4}. 1 You have dropped an x in calculating your derivative. By applying the product rule:$$\begin{align}A'(x) &= 2x\left(\frac{1}{2}(r^2-x^2)^{-1/2}(-2\color{red}{x})\right) + 2\sqrt{r^2-x^2}\\ &= \frac{-2x^{\color{red}{2}}}{\sqrt{r^2-x^2}} + 2\sqrt{r^2-x^2}\end{align}$$1 hint :x\sqrt{r^2-x^2}=\sqrt{x^2(r^2-x^2)}\le \dfrac{x^2+(r^2-x^2)}{2}=\dfrac{r^2}{2} 1 An example of a solenoid field is the vector field V(x,y)= (y,-x). This vector field is ''swirly" in that when you plot a bunch of its vectors, it looks like a vortex. It is solenoid since$$ \text{div} V = \frac{\partial}{\partial x}(y) + \frac{\partial}{\partial y}(-x) = 0. $$The divergence being zero means that locally no field is being "created" at ... 1 can we directly compare g_X and g_Y, since they both are defined on T\mathbb R^2 Generally, one should think of parameter spaces as different copies of \mathbb R^2 floating somewhere in Platonic universe and not interacting with each other at all. But yes, you could check if g_X is a scalar multiple of g_Y; if they are, the surfaces are ... 1 The g=dx^2 + dy^2 notation really means g = dx \otimes dx + dy \otimes dy, using the tensor product. Tensor products build up linear maps. That way, you get a map that takes in two vectors to return a scalar--exactly what you would want the metric to do. In other coordinates (like polar), you must remember to keep a distinction between positions and ... 1 Unless I'm missing something, your "strengthened" version is correct as well. If you don't mind, I'm going to use notation which is more familiar to me. So, I'll let X denote a vector field. Since you already have shown that if X = Y on an open subset U\subseteq M, that F(X)(p) = F(Y)(p) for all p\in U, we may reduce your question to a local ... 1 F can be written as$$ F = h \circ (f \times g) \circ \Delta_A $$where f \times g \colon A \times A \to B \times C is the map (f\times g)(a,a') = \bigl(f(a), g(a')\bigr), with differential$$ D_{(a,a')}(f\times g) = D_af \times D_{a'}g $$(using the identification T(B \times C) \cong TB \times TC of the tangent bundles, induced by the ... 1 \frac{dN}{ds} is orthogonal to N by differentiating the identity N\cdot N=1. So, \frac{dN}{ds} is a combination of T and B, i.e. \frac{dN}{ds}=x\,T+y\,B. Comparing the "known identity" T\cdot \frac{dN}{ds}=-\kappa with$$ T\cdot \frac{dN}{ds}=x\,(T\cdot T)+y\,(T\cdot B)=x $$(T is perpendicular to B by the definition of B) you get ... 1 We are given a curve in {\mathbb R}^3:$$\gamma: \quad s\mapsto{\bf x}(s)\ ,$$parametrized with respect to arc length s. Assume that \ddot {\bf x}(s)\ne{\bf 0}. Along this curve the so-called Frenet frame, a moving orthonormal frame, is defined as follows: Begin with$$T=T(s):=\dot{\bf x}(s)\ .$$This is a unit vector for all s. Therefore \dot ... 1 With the representation of 2, we have$$f(x) - f(a) - \phi_a(a)(x-a) = \left(\phi_a(x) - \phi_a(a)\right)(x-a).$$Thus$$\lVert f(x) - f(a) - \phi_a(a)(x-a)\rvert \leqslant \lVert \phi_a(x) - \phi_a(a)\rVert\cdot \lVert x-a\rVert.$$Now the continuity of \phi_a in a says that for every \varepsilon > 0 there is a \delta > 0 with$$\lVert ...

1

Schouten's "Tensor Analysis for Physicists" is a kind of gentle introduction, and he does not use the word "concomitant" there. The reference in Defrise-Carter's paper is to the English translation (1954) of Schouten's "Ricci Kalkül" (1924). The term "concomitant" comes from the ancient (XIX century) invariant theory. It has many uses and varying senses. ...

1

Sure. The tangent bundle to $S^1 \times S^1$ is simply the product of the tangent bundles of the two factors. Since each of these is a cylinder, a point of the object you're interested in can be thought of as a pair of points, one on each of two cylinders. I'm not certain that this helps, but... ...to be honest, if I wanted to visualize the tangent bundle ...

1

Because of the striking fact that (a) The area of [a slice of the spherical surface between two parallel planes (within the poles)] is proportional to its width. . . . here's a rarely (if ever) mentioned way to integrate over a spherical surface. We assume the radius = 1. (b) Note that every point on the sphere is uniquely determined by its z-coordinate ...

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