# Tag Info

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No. As an example (without being rigorous) consider a manifold which looks like an inifinite half cylinder parallel to the positive $z-$axis in Euclidean three space with a hemisphere attached at the bottom along an equator and smoothed out. (A bit like a hyperboloid or an one sided infinite cigar). If you look at the south (bottom) pole of the hemisphere ...

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Consider the "surface" defined by $$S(x, y) = (x^3, y, 1).$$ Since the image is just the plane $z = 1$, it's clearly a nice surface, even though $\partial S/\partial x$ is zero at $x = 0$. So in this case your statement "the tangent space isn't well defined when the partials are dependent is incorrect." It's a bit more subtle than that. But your ...

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Geometrically, the surface has a dent, or it degenerates to a line or curve, losing precisely the $2$-dimensionality that you want to study. For example, consider $$f(u) = \begin{cases} e^{-1/u} , &\text{if }u > 0 \\ 0, & \text{otherwise} \end{cases}$$ and plot using some program: $${\bf x}(u,v) = (f(u)\cos v, f(u)\sin v, u)$$ This is not ...

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That works, but it's working too hard... Locally you have a map $f: \Bbb R^n \to \Bbb R^m$ whose Jacobian vanishes. So in particular the partials of all of the component functions vanish. Thus it's constant on lines (because if a function $\Bbb R \to \Bbb R$ has zero derivative, then it's constant), by noting that the directional derivatives can be written ...

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Two-sided embedded surfaces which bound a smooth curve (also called a knot) $C$ are known as Seifert surfaces for the knot. I don't know a precise reference to van Kampen's result but the Wikipedia article seems to disagree he was the first to prove the existence of a Seifert surface for every knot. If you're looking for a proof, there's an algorithm for ...

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I'm a fan of Lee's Riemannian Manifolds: An Introduction to Curvature. It is definitely an introductory book; there are many deeper topics that it doesn't mention (compare to Peterson's Riemanninan Geometry). Here is an excerpt from the preface: "I have selected a set of topics that can reasonably be covered in ten to fifteen weeks, instead of making ...

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Milnor "Morse Theory" contains an extremely well written introduction to the subject.

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It is the second term that is always zero (because $d^2a = 0$), no matter what $k$ is. The first term is zero because $da$ has odd degree. In general $$x \wedge y = (-1)^{|x||y|} y \wedge x$$ where $|x|,|y|$ are the degrees of $x,y$. In particular, if $x$ has odd degree then $x \wedge x = 0$.

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By far Gallot et al is a very good choice.

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The topological space known as the square definitely does have a smooth structure, since it's homeomorphic to a disc. However, it does not have a smooth structure such that the inclusion map $i: \square \to \Bbb R^2$ is a smooth embedding. Proof: Put a smooth structure on the square. Let $(U,\varphi)$ be a boundary chart about one of the corners $p$ such ...

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@Irddo a vector field $Z$ is said to be invariant under the flow $\varphi_t$ generated by a vector field $X$ if $(\varphi_t)_*(Z) = Z$ for all values of $t$ for which the flow exists, where $(\varphi_t)_*$ is the push-forward map induced by the diffeomorphism $\varphi_t$. As a reference for such things, well, basically any book on differential geometry ...

2

Let us compute $dF(X)$ using directional derivatives: for $M\in\mathbb R^{2n\times 2n}$ we have $$dF(X)M=\lim_{t\to0}\frac{F(X+tM)-F(X)}{t}=\lim_{t\to0}\frac{(X+tM)A(X+tM)^{-1}-XAX^{-1}}{t}.$$ Now for $|t|$ small we can use the expression $$(X+tM)^{-1}=X^{-1}-tX^{-1}MX^{-1}+t^2P,$$ so that $$... 2 Not sure how the gradient of a specific function f can be interpreted as an operator. The integral of the divergence of a vector field on a compact manifold is zero. Apply this to the vector field H\nabla f (scalar times vector). But the divergence of H\nabla f is equal to the inner product of \nabla H with \nabla f plus the product (as scalar ... 2 The Laplacian is the trace of the Hessian, which is positive. However the Hessian is symmetric, and symmetric plus negative semidefinite implies nonpositive trace (this can be seen easily through diagonalization). 2 Suppose r is the distance one of the vertices of the polygon to its center. Draw the perpendicular bisector of the edge; by symmetry it will pass through the center of the polygon too. This creates a right triangle where The central angle is \pi/n. The hypotenuse is r. The opposite leg is half the edge length. Applying the appropriate rule for a ... 1 Law of cosines to the rescue: If we know that the angle at each vertex is \alpha, and the edge length to be found is x, then:$$\cos(2\pi/n)=-\cos^2(\alpha/2)+\sin^2(\alpha/2)\cosh(x/a)$$Or:$$x = a\cosh^{-1}\left(\frac{\cos(2\pi/n)+\cos^2(\alpha/2)}{\sin^2(\alpha/2)}\right)$$1 At any point P on a sphere we have a tangent plane, that is the plane orthogonal to the radius of the sphere at the point P. So there are infinetely many stright line on this plane passing thorough P. If we specify a direction on this plane than we have only one line from P with this direction . 1 If I understand your first question correctly, you ask whether you can find a \varphi \colon \Gamma(T^{*}M) \rightarrow \Gamma(T^{*}M) such that \bigwedge^{k}(\varphi) \colon \bigwedge^k\Gamma(T^{*}M) \rightarrow \bigwedge^k\Gamma(T^{*}M) coincides with the Laplacian acting on k-differentiable forms under the identification of ... 1 Let \epsilon > 0 be small enough that the exponential map is injective on B_\epsilon(0) \subset T_{p_0} M. Define a local vector field V on B_\epsilon(p_0) by parallel transporting v_0 along radial geodesics - you should be able to check that this satisfies the required conditions, with the caveat that it is not defined on all of M. Then just ... 1 Differential forms are an appropriate generalisation of derivation and integration (i.e of calculus) for arbitary (within the scope of integration) manifolds and spaces. In order to arrive at such a generalisation (e.g like E. Cartan did) one starts with the basic definitions and operations of derivation and integration, how they affect the space they are ... 1 Your statement follows from Stokes's theorem \int \limits _{\partial M} \omega = \int \limits _M \Bbb d \omega and the fact that M has no boundary (i.e. \partial M = \emptyset, so the left-hand side is 0). Note that (\Delta f) H \ \Bbb d V = (\text{div} \ \nabla f) H \ \Bbb d V = \text{div} (H \nabla f) \ \Bbb d V - (\nabla f \cdot \nabla H) \ \Bbb ... 1 Hint. (1) To show that \def\M{\mathcal M_{m,n}}\M is open, note that if M \in \M, then M has a d \times d-submatrix (where d := \min\{m,n\}) A_d(M) with nonzero determinant and the map \mathcal M \to \mathbf R, which maps M \mapsto A_d(M) is continuous. (2) By (1), \M is an open subset of \mathbf R^{mn} and hence carries naturally a ... 1 For a function f and a form \omega, we have the product rule d(f \omega) = df \wedge \omega + f d\omega. Thus$$d (\phi \iota_v \rho ) = d\phi\wedge\iota_v \rho + \phi d(\iota_v \rho)=d \phi\wedge \iota_v \rho. Now in general this is different to $d\phi(v) \rho$; but in 3 dimensions they are equal. To see this, let's choose coordinates so that at ...

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After closer perusal of Wikipedia's definitions, I think there's an omitted condition. Here are the relevant definitions: Principal bundle A principal $G$-bundle, where $G$ denotes any topological group, is a fiber bundle $\pi:P \to X$ together with a continuous right action $P × G \to P$ such that $G$ preserves the fibers of $P$ (i.e. if ...

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Look at the level sets of $r$, i.e. at the sets $\{x: r(x)=c\} =: U_c$ The condition on the Hessian allows to conclude that the curves normal to these level sets are complete geodesics and that the level sets have no focal points. The diffeomorphism $U_0\times \mathbb{R} \rightarrow M$ can be written down explicitly by looking at the exponential map ...

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Have you tried Riemannian Geometry: A Beginners Guide, by Frank Morgan?

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Imagine an ant-grid on the manifold. Each ant can determine the distance to every other ant on the manifold by means of the metric. Let each ant begin moving according to the vector field: At each moment, the vector at each ant's location determines the ant's speed and direction of motion. If the ant-to-ant distances do not change, the field is a Killing ...

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