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9

Like you say: The plane $\Bbb R^2$ can be embedded in $\Bbb R^2$ but not $\Bbb R$. The sphere $\Bbb S^2$ can be embedded in $\Bbb R^3$ but not $\Bbb R^2$. In particular, the answer depends not only on the dimension of the given curved space (which here I'll take to mean manifold), but the space itself. We also have: Real projective space $\Bbb {RP}^2$ ...


5

Given any point $p$ and $v \in T_pM$, there is a unique geodesic $\gamma$ starting at $p$ with velocity $v$, meaning $\gamma(0) = p$ and $\gamma'(0) = v$. This is a consequence of the theory of ODE's. Since every tangent vector is tangent to a point, we can use the above to conclude that the answer to your initial question is no.


5

Not a complete answer, but here's the situation for projective space. On complex projective space, the "obvious" metric is Fubini-Study. To determine the diameter, however, you have to specify a scaling, and three "interesting" scalings come to mind: An algebraic geometer might scale so that an embedded line has unit area, i.e., the Kähler form generates ...


4

Well, Singer and Thorpe do explain the notation (see p.356 in the original text) by saying that Using the usual isomorphisms defined by the inner product, a curvature tensor on $V$ may be regarded as a $2$-form on $V$ with values in the vector space of skew symmetric endomorphisms of $V$. So the meaning of $R(x,y)z$ is an endomorphism $R(x,y)$ ...


3

HINT: You can write down the expression for the Gaussian curvature of a general surface of revolution. For example, assume it's obtained by rotating the arclength parametrized curve $x=f(s)$, $z=g(s)$ about the $z$-axis. Then you should be able to show that $f''(s)+Kf(s)=0$. You will get plenty of surfaces of constant curvature (including, but not limited ...


3

For a surface $(M, g)$, there is only independent component of the curvature tensor $R$, namely, $$R_{1212} = - R_{1221} = -R_{2112} = R_{2121},$$ and this quantity (which depends on the choice of coordinates) is related to the Gaussian curvature $K$, by $$R_{1212} = -\det(g) K$$ (which is independent of coordinates). (Beware that this sign is a matter of ...


2

Parallel transport will not help, since it will not respect the endpoint values if you move from $t_i$ to $t_{i+1}$. There are several ways to do this. The most straightforward way is to cover $c([a,b])$ with coordinate neighbourhoods $U_k$ (in $M$), choose a finite subcover of $[a,b]$ of connected intervals in the sets $c^{-1}(U_k)$ and define the vector ...


2

The answer to your first question is yes. If you look at Spivak's 5-volume Differential Geometry text, this is proven several times in Volume II; he refers to it as "The Test Case." If you know the basics of differential geometry with differential forms, I can easily show you a proof for $\dim M = 2$, but higher dimensions require some version of the ...


2

To amplify slightly on the two comments by Demosthene and JHance: it means that the manifold is very locally euclidean, i.e., pointwise. At each point, you can find a basis of tangent vectors in which things are Euclidean, but this basis cannot generally be extended even to a small neighborhood. A good approximation to extending it to a small neighborhood is ...


2

Rather than convert the bivector $R(x \wedge y)$ to a 2-form, one can use clifford algebra instead. Define a product operation on arbitrary vectors $a, b, c$ such that $$aa = g(a,a), \quad (ab)c = a(bc)$$ I use $g$ here for the Riemannian metric instead of $\langle,\rangle$, for reasons that will become clear shortly. This "geometric product" produces a ...


2

You compute $q$, and $v = \alpha'(\frac{\pi\sqrt{2}}{4})$, and let $v' = v/\|v\|$, then compute a second unit vector $u'$ perpendicular to $v'$, lying in the tangent plane to the cone at $q$. Now you take $\cos \theta v' + \sin \theta u'$ (where $\theta$ is your computed angle between the transported $w$ and $v'$). In other words, you take the $t(s)$, ...


2

I'm going to skip the nonsensical first question and answer the second. Consider a nonzero meromorphic global section $f$ (i.e. a meromorphic function $f : \Bbb C^* \to \Bbb P^1(\Bbb C)$ such that $f(\alpha.w) = w^nf(w)$). We pick a fundamental domain $\Omega$ for $\Bbb C^*/A$, for example the region $|\alpha| < |w| < 1$. Then the degree of $f$ is ...


2

(1) Consider Chapter 2 exercise 2 (57p.) in the book "Riemannian Geometry" by Do Carmo: So $$ \bigg(W-\frac{J}{t}\bigg)' = \lim_t \frac{ P^{-1}_{0,t} \bigg(W-\frac{J}{t} \bigg) + \bigg(W-\frac{J}{t}\bigg)(0) }{t} = \lim_t \frac{W-\frac{J}{t} }{t} $$ where $P_{0,t} : T_{p} M \rightarrow T_{c(t)}M$ is a parallel transport. Here we used : $P_{0,t}^{-1} $ ...


1

If "between" means "$c$ lies on a minimizing geodesic from $x$ to $y$", the answer is yes. Let $\gamma:[0, \ell] \to M$ be a unit-speed geodesic with $\gamma(0) = x$ and $\gamma(\ell) = y$, and let $t_{0}$ be a point of $[0, \ell]$ at which the real-valued function $\left|g\left(\nabla f\bigl(\gamma(t)\bigr), \gamma'(t)\right)\right|$ achieves a maximum. ...


1

Hints: Let $i_{X}$ denote the identity map on $X$. The map $i_{X} \times f:X \to X \times Y$ is a holomorphic bijection to $\Gamma$, and projection $\operatorname{proj}_{X}:X \times Y \to X$ on the first factor is a holomorphic map whose restriction to $\Gamma$ is the inverse of $i_{X} \times f$.


1

Let $V(M)$ denote the space of vector fields on $M$. There are two equivalent requirements for a function $V(M)\times\ldots\times V(M)\to C^\infty(M)$ in order for it to be a tensor. One can require the function to be linear over $C^\infty(M)$ or alternatively require the function to induce a multilinear form at every point, depending only on the value of ...


1

Local invariants in a broad sense are quantities that can be used to distinguish manifolds endowed with some structure, at least locally. There are global invariants too. For instance, on a Euclidean plane, given two circles of unequal radii we can tell them one from another by a global Euclidean invariant (the perimeter), or by a local invariant (the ...


1

This will be a very sketchy argument, perhaps to be elaborated on later. First thing we may note is that a short geodesic curve $C$ joining two points $p$ and $q$ is itself a convex set. So if the two metrics are to have the same convex sets, they must also have the same geodesic curves (up to reparametrisation if by geodesic we mean a path of constant ...


1

Another way of looking at is is that what we're really interested in is not so much the metric itself as it is the norm on each tangent space we can derive from it. The norm is what actually gives structure to the manifold; representing it as a bilinear form is more of a technical convenience. Viewed in this light, the important fact is not that "the matrix ...


1

Note that if you have diagonalized the metric tensor, a further rescaling of the tangent vectors transforms the matrix of the metric tensor into the identity matrix. Clearly, then, one cannot locally diagonalize the metric tensor in general (or else every Riemann manifold would be locally isometric to some Euclidean space, which is trivially not true); one ...


1

In Lorentzian signature, the conformal Laplacian is also known as the conformal wave operator. In Riemannian signature, it is also known as the Yamabe operator. The reason for these names is mentioned on page 14 of Curry & Gover's conformal geometry notes.


1

This looks like it follows immediately from the Nash embedding theorem. Namely, let $f : M \to \mathbb{R}^N$ be a smooth isometric embedding (whose existence is guaranteed by Nash); write $f = (f^1, \dots, f^N)$. The statement that $f$ is an isometric embedding means exactly that the metric $g$ on $M$ is the pullback by $f$ of the metric $\sum_{i = 1}^N (d ...


1

There is a book by Thierry Aubin called "Some Nonlinear Problems in Riemannian Geometry" which, at least in principle, is exactly what you describe. It covers a lot of classic geometric analysis problems, with a couple of introductory chapters on geometry and PDEs. I say in principle because it might be very hard to read for somebody just starting out; it is ...



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