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9

Nice question! If you fix $x\in S$ and find the point $y\in S$ farthest from $x$, you can check easily that the chord joining $x$ and $y$ must be normal to $S$ at $y$. By symmetry, then, if we take any pair of points $x,y\in S$ with maximal distance (which exists by compactness), the chord joining $x$ and $y$ must be normal to $S$ at both points.


8

If you want to have a general idea of what a subject is about, then casual reading will be sufficient. It will help you understand what the problems are in the field, but not a lot more. Also, in any mathematics book, after you get through the first couple of chapters, it will get hard to read if you haven't been working along the way. The exercises in a ...


6

What you propose is perfectly fine... although all too often textbooks (intentionally or not) play a sort of passive-aggressive game wherein the main text does not "let on" certain subtleties, on which the "exercises" specifically prank the reader. Or major results are posed as "exercises", for which there's no methodological prototype in the text. This is ...


6

I've checked five of your claimed errors, none of which you're right about. There are other cases in which you've found obvious typos, which are admittedly annoying but which shouldn't cause any real trouble. It may be you've found some significant errors in the other points which are more complicated for me to check, but based on the following evidence it ...


6

The differential is an isometry of vector spaces. The local map between Manifolds is a radial isometry, but in general it will fail to map geodesic distance spheres from the base point to isometric distance spheres. Simply because these need not admit an isometry. So a path orthogonal to geodesic rays starting in the base points will be mapped onto a path ...


4

This somehow 'depends'. If, as usual, in Riemannian Geometry, you are looking at a smooth ($C^\infty$) manifold with smooth metric, then the answer is, for sure, yes. Only if you are reducing the differentiability assumptions on the manifold and the metric you may run into trouble at a certain point. What you are looking at (e.g. in local coordinates) is an ...


4

Using the nice formula at wikipedia. I calculate: $$ K = \frac{9x^2y^2z^2}{x^6+y^6+z^6}. $$


3

You can use the shape operator. You know that $U = \frac{1}{4\sqrt{x^6 + y^6 + z^6}}(4x^3,4y^3,4z^3)$ is the unit normal vector to this object. Now the shape operator is defined by $$S_p(v) = - \nabla_vU.$$ The gaussian curvature is defined by $$k_g := \det S_p.$$


3

$$ \sum_j g^{ij} g_{kj} = \delta^i_k $$ Now, $\delta^i_i = 1$, with no summation, so $$ \sum_i \sum_j g^{ij} g_{ij} = \sum_i \delta_i^i = \sum_i 1 = n. $$ You may be confusing whether or not you are using summation convention.


3

Note: $$\sum_{i,j=1}^n g^{ij} g_{ij} = \text{tr} (g^{-1}g) = \text{tr} (I_n) = n.$$


3

NB: in the following, it is assumed that $X, Y, Z$ and $R$ are continuous. If $C \subset TM$ is the compact subset from which $X$, $Y$ and $Z$ are taken, then $C \times C \times C$ is compact (Tychonoff's theorem), and the map $R(X, Y)Z: C \times C \times C \to TM \tag{1}$ is continuous; this follows by virtue of the fact that in any local coordinate ...


3

In general, points have no length. But if you mean "are there line segments whose projections have the same length", the answer is "yes, but they're not all that easy to identify and don't seem to have a lot of geometric meaning." (I'm assuming that by a "segment" on a sphere you're willing to consider arcs-of-shortest-distance as segments, and their ...


3

One (of the possible two ones) unit normal vector of $\mathbb{S}^1 \times \mathbb{S}^1 \subset \mathbb{S}^3$ at the point $(e^{i \theta},e^{i\psi})$ is the vector $$\eta(\theta,\psi) := \frac{(e^{i \theta},-e^{i\psi})}{\sqrt{2}} \, .$$ Indeed, the scalar dot product of two vectors $\mathbb{v} = (v_1,v_2) ,\mathbb{w} = (w_1,w_2) $ in $\mathbb{C}^2$ is given ...


3

Exponential maps are in general not isometries (otherwise every Riemannian manifold would be flat), and the term "linear" does not make sense since general Riemannian manifolds do not have a linear structure.


3

Yes, can you can - it's just that the sum of two "points", i.e., the sum of the pure states associated to two points, is not necessarily itself a pure state. As your excerpt says: ... whereas a general state $\rho:A\to \mathbb{R}$ is a distribution of such specific states. So there is certainly not, in general, an addition operation on the points of ...


2

Recall that a compact manifold has empty boundary. You can apply the divergence theorem, $$ \int_M \operatorname{div}{X} \, dV_g = \int_{\partial M} \langle X, N \rangle \, dV_{\bar{g}}, $$ where $g$ is the metric, $\bar{g}$ the metric induced on the boundary, and $N$ a unit normal vector field (this is basically the Hodge dual of the general fundamental ...


2

I think that for finite $G$, $M/G$ is a smooth manifold, regardless of any compactness assumptions. For a general discrete group $G$ acting on a manifold $M$, you need the action to be free and properly discontinuous in order to make $M/G$ into a smooth manifold (and $M\to M/G$ into a covering). The second condition means that for each $x\in M$ there is a ...


2

Proffering a counterexample: If the set of non-fixed points of $f$ is non-empty and finite, then we cannot find such a differentiable structure. The set of fixed points of a diffeomorphism is always closed, but the complement of a finite non-empty set on a manifold is never closed. I don't know what extra requirements would make this true.


2

The implication is a consequence of Ricci identity. The notations $\nabla_k$ is standard: Let, in general, that $A$ is a $(p, q)$-tensor, one can define a $(p, q+1)$ tensor $\nabla A$. In your specific example, there are two tensors: The function, $\omega$, which is a $(0,0)$ tensor, and It's exterior derivative $d\omega$, which is a one form (that is, a ...


2

Here's one way to view this: For any manifold, the Ricci curvature is an invariant of the metric, and so it must be invariant under the isometries (symmetries) of the metric. Now, the group of isometries of the $n$-sphere $\Bbb S^n$ (endowed with the round metric $g$ is $O(n + 1, \Bbb R)$, which acts transtively, and the isotropy group preserving any point ...


2

If "stereographic projection" refers to "projecting the unit sphere centered at the origin away from the north pole $(0, 0, 1)$", then the induced metric on the plane with Cartesian coordinates $(u, v)$ is $$ \left[\frac{2}{1 + u^{2} + v^{2}}\right]^{2} (du^{2} + dv^{2}). $$ Since the term in parentheses is the Euclidean metric, the question of area or ...


2

I think one of the reasons why textbooks encourage you to do a lot of exercises is so that you can see things for yourself. Sometimes even the derivation of important results will be left as an exercise to the reader. One of the best teachers I've ever had almost never presented us with a statement of fact or lectured us while we took notes. Instead, she ...


2

At this point you have expressed $F : \mathbb{R} \to S^1$ as $$F(\varphi^{-1}(y))=(\frac{5-y}{\sqrt{2y^2-10y+25}},\frac{y}{\sqrt{2y^2-10y+25}})$$ You also have the mapping $\psi : U \to ]-1, 1[$ given by $$\psi(x, y) = y$$ and have demonstrated you would like the differential at the point $0$ in the domain of $F \circ \varphi^{-1}$. So we can compose ...


2

Time ago I studied this concepts and I remember I was convinced that the situation is the same as in Riemannian geometry with the curvature $R$ of a Riemannian manifold $(M,g)$. Namely, you have the definition of $(M,g)$ having positive sectional curvature which means that for all orthogonal unit vectors $\mathbf{u},\mathbf{v} \in T_pM$: $$\langle ...


1

Let $\mathbf{y}:E\to M\subseteq\mathbb{R^3}$, $M$ a surface, be a proper patch. Then $\mathbf{y}^{-1}:\mathbf{y}(E)\to E$ is continuous. If $\mathbf{y}(E)$ contains any boundary points (i.e. it is not open), the function $\mathbf{y}^{-1}$ cannot be continuous, as a boundary point has no equivalent in an open set (every point in an open set has a ...


1

The answer is that for an open set $M$ in $\mathbb R^n$, and $p\in M$, the tangent space $T_pM=\mathbb R^n$, and $N_pM=0$. Here's the reason: If you know some differential topology, you will be clear that each tangent vector $v\in T_pM$ is a equivalent class of beam of osculating curves. To show that $T_pM=\mathbb R^n$, you need only to show that each ...


1

It depends on the books that you read. Some books are very dense, or leave out a ton of steps. If by "reading" you mean actually filling in all the gaps and rewriting the incomplete proofs to make sure you understand, then you are actively learning the subject. Some of Serre's books are dense (e.g. cours d'arithmetique), but when I read through the whole ...


1

The reason this stuff is hard is because there are a lot of identifications going on behind the scenes when we write down equations like they one you're meant to prove. It can be tricky to keep track of them all until you gain some intuition. In the meantime, make everything explicit. Let's start with $v_p \in T_p U$. It's a derivation of $C^\infty(U)$ at ...


1

$\mathfrak{h}$-invariant complements to $\mathfrak{h}\subseteq\mathfrak{g}$ correspond to splittings of the canonical short exact sequence of $\mathfrak{h}$-modules: $$0\to \mathfrak{h}\to\mathfrak{g}\to\mathfrak{g}/\mathfrak{h}\to 0.$$ Let $s_0:\mathfrak{g}/\mathfrak{h}\to \mathfrak{g}$ denote the splitting whose image is $\mathfrak{h}^\perp$. Given a ...


1

What you write before "Correct me if I'm wrong" is indeed wrong. In $\mathbb{R}^n$ we know that for any smooth function $f$ we have$$\frac{\partial^2f}{\partial x_1\partial x_2}=\frac{\partial^2f}{\partial x_2\partial x_1}.$$In other words:$$\left[\frac{\partial}{\partial x_1},\frac{\partial}{\partial x_2}\right]=0.$$Hence, whenever $X_1,\ldots,X_n$, are ...



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