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9

The Riemann sphere is an essential object, and is certainly no mere didactic tool. For one thing the Riemann sphere can be given the structure of a complex manifold by using the maps $z \to z, z \to 1/z$ as charts from $\mathbb{C} \to S^2$. Then meromorphic functions on a domain $\Omega$ are simply the functions $f: \Omega \to S^2$ that are complex ...


8

The Riemann sphere is not just a visual aid or an intuitive heuristic, it is a bona fide mathematical object that exists in its own right. For example, the uniformization theorem states the three simply connected Riemann surfaces are the open unit disk, the complex plane and the Riemann sphere. The definition of Riemann surfaces mimics the definition of ...


6

One common way to visualize the "intrinsic appearance" of a simply-connected universe of constant curvature $\pm 1$ is to give the angular size of an object modeled as a geodesic arc of (sufficiently small) length $\ell$ placed at distance $d$ from one's eye. (I didn't try to run the linked applets, and am not sure if either implements this strategy.) On a ...


5

Connections are not tensors, but that does not mean they are not coordinate-independent objects! A linear connection is a map sending two vector fields $X,Y$ to another vector field $\nabla_X Y$ which satisfies the rules $\nabla_{fX+Z} Y = f \nabla_X Y + \nabla_Z Y$ and $\nabla_X (fY + Z) = f\nabla_X Y + (\nabla_X f)Y + \nabla_X Z$. This is an abstract ...


5

When you look at a circle, you are seeing its extrinsic curvature, which is also what your link is calculating. That is a property of how the circle is imbedded into another manifold, not a property of the circle as a manifold itself. The curvature being referred to here is the intrinsic curvature, which is defined strictly in terms of the manifold itself, ...


4

Initially, the sphere $\mathbb{S}^2$ is just a set of points. Sets are okay, but kind of boring. But we can go further: we can endow the sphere with a topology to make it into a topological space. The standard way of doing this makes the sphere into a compact surface, which is nice. We can go further still: having given the sphere its usual topology, we ...


4

As mentioned in a previous answer, connections are quite intrinsic. I will take the slightly more pedestrian view that a connection is a collection of maps from type $(k,l)$ tensor fields to type $(k,l+1)$ fields that is linear, satisfies a Leibniz rule, commutes with contraction, agrees with the differential for smooth functions and is usually assumed to be ...


3

In contrast to what is written in the question and some of the comments, a connection is independent of coordinates. The intrinsic way to define connections is the following. For simplicity, we treat only connections on the tangent bundle, even though a similar definition can be applied for any vector bundle. Let $M$ be a smooth manifold. Let $TM$ denote ...


3

Hint: Use Lemma $4.6$ (ii) (i.e. the formula displayed here) to deduce that $$(\nabla_X g)(Y, Z) = \nabla_Xg(Y, Z) - g(\nabla_XY, Z) - g(Y, \nabla_XZ).$$


3

Here are two suggestions: Morse theory is a central tool in analyzing the topological structure of manifolds. A key step in the theory is studying the gradient flow of a Morse function, which requires a Riemannian metric to define. Donaldson's theorem says that if $M$ is a smooth, simply connected, compact $4$-manifold with definite intersection form, then ...


3

The definition of isotropic (which I found here) ensures only the existence of some isometries satisfying the conditions. Since $x$ and $y$ are isometries, so is $xyx^{-1}$. There is no requirement that the set of isometries that you "use" to satisfy these conditions (which is not necessarily unique) have any specific topology or even be closed under ...


3

You shouldn't assume that the naive parametrization $(\text{cos}(t), \text{sin}(t))$ will solve the equation. It doesn't, as it clearly doesn't satisfy even the first differential equation. However, if you assume a more general form $x(t) = \text{cos}(f(t))$, $y(t)= \text{sin}(f(t))$, which still traces out a semicircle (but with a different speed), then it ...


3

First, we should note that a very similar question has already been asked here, and several interesting answers were given. But because the question keeps coming up, I'm going to go out on a limb and suggest that there still might be room for a more complete list of reasons why hyperbolic geometry is important in its own right. It's hard to know where to ...


3

No. Let $(X,d')$ be a metric space and let $f \colon X \rightarrow X$ be a homeomorphism. You can define a new metric on $X$ (which you might call the pullback metric) by $d(x,y) := d'(f(x),f(y))$. The topology induced by $d$ is the same as the topology induced by $d'$ and $\varphi \colon X \rightarrow X$ is an isometry for $(X,d')$ if and only if $f^{-1} ...


2

Attempt #2: I think the issue here is that we need to transform from the $(1, 3)$ curvature tensor to the $(0, 4)$ one. The formula for the components of the $(1, 3)$ version are given by: $$R^l_{ijk} = \partial_i\Gamma^l_{jk} - \partial_j\Gamma^l_{ik} + \sum_r(\Gamma^r_{jk}\Gamma^l_{ir} - \Gamma^r_{ik}\Gamma^l_{jr})$$ This means that the $(1, 3)$ curvature ...


2

The induced metric is just the Euclidean metric on $\mathbb R^{n+1}$, i.e. $g_{ij} = \delta_{ij}$, but restricted to act on vectors tangent to $S^n$. You need to choose a system of $n$ coordinates parametrizing $S^n$ alone (or an $n$-frame tangent to $S^n$) if you want to get an $n \times n$ matrix of components for $g$.


2

You're absolutely right. Without using the fact that the full isometry group is a Lie group, I think it would be very difficult to show that there's a Lie group containing all the elements of the form $x^{-1}yx$ for $x\in G$ and $y\in H$. I think the Wikipedia definition pointed out by @MattSamuel has it right: There's no good reason to include the ...


2

If you expand the first two brackets in terms of the covariant derivative you get $$\begin{align} [\Omega ,\nabla _UV]-\nabla _U([\Omega, V]) - \nabla _{[\Omega ,U]}V &= \nabla_\Omega \nabla_U V - \nabla_U \nabla_\Omega V - \nabla _{[\Omega ,U]}V + \nabla_U \nabla_V \Omega - \nabla_{\nabla_U V} \Omega \\ &= R(\Omega, U)V+\nabla^2_{U,V} \Omega, ...


2

For (1), depending on whether you are a geometer or an analyst you may have a different sign on your Laplacian. However, if you locally write $$ \Delta f = \sum \frac{\partial^{2} f}{\partial x_{j}^{2}} $$ then this is negative definite. This being the classic example of the sort of operator $L$ you are interested in, the author might be assuming $L$ to ...


2

If you already reduced this problem to the $\mathbb{R}^n$ case, then we should be able to tackle it with the usual analytical methods. The following is probably a bit of technical overkill but should work. As far as I can see, the only problem is to smoothly connect two pieces with an arbitrarily small loss of length. Assume we have two smooth paths $$p_1: ...


2

The answer of user1551 is perfectly fine, but I found a highschool level proof that I want to share here: Since $v$ is time-like, we follow $$v_1^2> v_2^2+v_3^2+v_4^2.$$ Assume that $\langle w,w \rangle\leq 0$. Then $$w_1^2\geq w_2^2+w_3^2+w_4^2.$$ Now by assumption it is $\langle v,w \rangle=0$ and therefore $$v_1w_1=v_2w_2+v_3w_3+v_4w_4.$$ Taking the ...


2

It is known that There is no isometric immersion of $\mathbb{H}^n$ into $\mathbb{R}^{2n-2}$, even a local one. There is a local immersion into $\mathbb{R}^{2n-1}$. This is due to E. Cartan (1919-20) and T. Otsuki (1955). There is a complete isometric immersion of $\mathbb{H}^n$ into $\mathbb{R}^{4n-3}$. This was proved recently (1980s) by W. Henke. There ...


2

As expected in the comment of @MikeMiller, I think the answer is "You need a connection on $E$", although I don't think that it has to be a linear connetion. As you pointed out in the question, for each point $u\in E$, there is an exact sequence $0\to V_uE\to T_uE\to T_{\pi(u)}B\to 0$. Since $E$ is a vector bundle $V_uE$ is canonically isomorphic to $E_u$, ...


1

If $f$ is only Lipschitz continuous, then $\Delta f$ is not in general an $L^2$ function. Instead, it has to be interpreted as a distribution. (For example, on $\mathbb R$, the Laplacian of $|x|$ is a delta distribution concentrated at the origin.) It follows easily from the definition of distributional derivatives that $\int_M\Delta f=0$ for any $L^2$ ...


1

Hint : Parallel transport along $\gamma (t)=(t,\phi_0)$ : Consider a cone $C$ tangent to a sphere $S$ s.t. $$ C\cap S = {\rm Im}\ \gamma $$ So we consider a parallel transport on a cone.


1

I think I see another argument, but I'm not sure it is valid, because everyone is using different terminology and I'm a bit confused by the profusion of terms. I know "geodesic coordinates" as coordinates in which the Christoffel symbols of the second kind vanish at a point. Their derivatives do not vanish here. Various people talk about "normal ...


1

I think it's true. If $G$ is your metric tensor, then, using the formula for the derivative of the inverse, your expression is zero if $tr(G^{-1}G_{x_j}G^{-1}G_{x_k})=tr(G^{-1}G_{x_k}G^{-1}G_{x_j})$, where $G_{x_i}$ is the derivative of $G$ wrt $x_i$. Let $A=G^{-1}G_{x_k}$ and $B=G^{-1}G_{x_j}$. Since $G$ is symmetric, so are $A$ and $B$, and your expression ...


1

An additional remark to the answer. On a Riemannian manifold (without "missing" points, e.g., complete) the minimum length in any homotopy class of path exists and is attained by a geodesic path, which is necessarily smooth. If the manifold is of some reasonably finite topological type (compact is much more than enough), the infimum of the geodesic lengths ...


1

I'm guessing the equality you need is $$h^i_j \frac{\partial P}{\partial h^i_j} = \alpha P.$$ To see that this is true, note that the LHS is the derivative of $P$ in the outwards radial direction; i.e. $$ h^i_j \frac{\partial P}{\partial h^i_j} = \frac{d}{dt}\Big|_{t=0} P((1+t)W).$$ If we apply the $\alpha$-homogeneity of $P$ to the RHS we get ...


1

For 1, positive mass theorem can be one of the examples. Here is some information about positive mass theorem. Schoen and Yau used the minimal surface to prove the positive mass theorem, which is originally a problem in general relativity.



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