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13

Ok. First of all, I'm not entirely sure what Riemann meant by that. I am going to give you a standard definition of smooth manifolds using the language that is most commonly used in mathematics today. I also include some more thoughts after the definitions. Topological Manifolds Before talking about what a smooth manifold is (one on which you can ...


10

You've stumbled across one of the interesting pitfalls of the history of science. Those words attributed to Riemann should not be understood as equivalent to the modern definition that you have learned for a manifold, i.e. as "something glued by locally homemorphic to Euclidean parts". Riemann's development of the concept of a manifold was far beyond ...


6

If you are planning to do research in modern geometry, pretty much nothing on your list is relevant. Instead: If you are planning to work in algebraic geometry, you should learn commutative algebra or/and complex differential geometry. If you are planning to work in Riemannian geometry, most likely you should spend time learning differential equations, ...


5

So, the vast majority of manifolds are not locally symmetric. The most obvious example that comes to mind is the standard torus in 3-space. This has positive curvature just to the left of the topmost circle, and negative curvature just to the right, and hence there cannot possibly be a geodesic symmetry at any of the points on that top circle. Similarly with ...


5

I think your authors misquoted Riemann. In his famous 1854 essay, Riemann mentions the possibility of passing continuously from one "mode of determination" to another, by which he means, in modern terminology, that the quadratic form determining the metric varies continuously from point to point in a manifold. Thus he did not say "continuous transition of ...


4

What's written is pretty much meaningless. If you are interested in a modern interpretation of Riemann's lecture, my suggestion is to read Spivak's "A Comprehensive Introduction to Differential Geometry," Vol. 2, where he discusses the mathematical meaning of Riemann's lecture in great detail (pages 163-180, What did Riemann say?).


4

As $dtdx =\frac{dtdx+dxdt}{2}$, then $g=\begin{pmatrix} 0 &(1+t)/2 &0 & 0\\ (1+t)/2 & 0& 0 &0 \\ 0& 0 & 1 & 0\\ 0& 0 & 0 & 1 \end{pmatrix}$, so $g$ is symmetric and it has inverse. Then, $g$ is a metric on $M$


4

Define the orthonormal basis $$ e^0 = f \, dt, \quad e^1 = dx. $$ Then $$ de^0 = \frac{f'}{f} e^1 \wedge e^0, \quad de^1 = 0. $$ Cartan's first structural equation, $$ de^i = -\omega^i{}_j \wedge e^j, $$ implies that $$ \omega^0{}_1 = \frac{f'}{f} e^0, $$ which by antisymmetry is effectively the only nonzero curvature form in two dimensions. The second ...


4

It seems you have some misprints in your formula. Start by calculating the Cristoffell symbols for the new metric $$\tilde\Gamma_{ab}^c = \Gamma_{ab}^c + g^{cd}(g_{db}\nabla_a \ln\Omega + g_{ad}\nabla_b \ln\Omega- g_{ab}\nabla_d \ln\Omega)$$ or $$\tilde\Gamma_{ab}^c = \Gamma_{ab}^c + (\delta_b^c\nabla_a \ln\Omega + \delta_a^c\nabla_b \ln\Omega- ...


4

How about this? If $\phi$ is an isometric holomorphic embedding $\Bbb C/\Bbb Z\hookrightarrow\Bbb C^n$, it lifts to a $\Bbb Z$-invariant holomorphic map $\tilde\phi\colon\Bbb C\to\Bbb C^n$ pulling back the standard Kähler form on $\Bbb C^n$ to the standard Kähler form on $\Bbb C$, i.e., $$\tilde\phi^*\left(\sum_{j=1}^n dz^j\wedge d\bar z^j\right) = dz\wedge ...


3

The condition that $k$ is a gradient implies that it is curl-free; i.e. $\nabla_i k_j = \nabla_j k_i$. Let $\gamma$ be an integral curve of $k^i$; i.e. $\dot \gamma^i(t) = k^i(\gamma(t))$. Then replacing $\dot \gamma$ with $k$ in the covariant acceleration and applying the curl-free condition we get $$\ddot \gamma^i = \dot \gamma^j \nabla_j \dot \gamma^i = ...


3

It seems you are missing some necessary background, namely, how to extend a connection on $TM$ to all tensor bundles. I will summarise this construction. Given a connection \begin{align*} \nabla : \Gamma(TM) \times \Gamma(TM) &\to \Gamma(TM)\\ (X, Y) &\mapsto \nabla_XY \end{align*} on $TM$, there is an associated connection (which I will also ...


3

Take a geodesic in the cylinder and a Jacobi field on it, and pull these back to the universal cover - this is $\Bbb R^2$, and geodesics are straight lines. Because $\Bbb R^2$ is flat, we see that the Jacobi condition is just that $J''(t) = 0$ (taking derivatives with respect to $t$), and hence that $J(t) = v+wt$. This could only vanish at both endpoints of ...


3

For each $x \in M$, let $D_x \subseteq T_x M$ be the largest set on which $\exp_x$ can be defined as a diffeomorphism (its boundary $\partial D_x$ is the cut locus of $x$ in $T_x M$). Notice that $D = \bigcup \limits _{x \in M} D_x$ is contained in $O$. For brevity, I shall denote the restriction $\exp \big| _D$ by $\exp$ again. Notice that $\exp_x D_x$ is ...


3

The torus is $$\mathbb T^n=\underbrace{S^1\times...\times S^1}$$ $$\quad\;\;\;\; n\text{ times}.$$ We parametrize the $i$th copy of $S^1$ by the angle $\theta_i$. Let $$f(\theta_1,...,\theta_n)=(\cos(\theta_1),\sin(\theta_1),...,\cos(\theta_n),\sin(\theta_n)).$$ This map should work for you.


3

I'm not sure if this is what you have in mind, but here are a couple of theorems that concern the interaction between topology and Riemannian geometry. Gauss-Bonnet theorem, and its generalization, sometimes called the Chern-Gauss-Bonnet theorem. It relates the curvature of a Riemannian metric on a manifold to its topology (Euler characteristic). ...


3

It's risky to give a blanket negative answer to so open-ended a question (particularly, one in which "canonical" is undefined), but with minor qualifications the answer appears to be "no". Here are a few easy observations, a bit too long for a comment: If for all $x$ and $y$ in $S$, there is a canonical parallel transport between $T_{x}S$ and $T_{y}S$, ...


3

The start of the computation is OK, but the partial derivative does not make sense. I think the problems come from the fact that you misinterpret $\nabla_c\nabla_dZ^a$. Evaluating the second covariant derivative $\nabla^2Z$ on $X$ and $Y$, you do not get $\nabla_X\nabla_Y Z$. What you have to do is to differentiate $\nabla Z$ as a $\binom11$-tensor field. ...


3

This seems like mainly a question about what the abstract index notation is asking you to do. Let's look at a single term from the right hand: $\nabla_X\nabla_YZ$. Recall that in abstract index notation, two tensors juxtaposed signifies their tensor product. For example, if $V^a$ and $W^b$ are vector fields, $V^aW^b$ is supposed to mean the tensor field $V ...


2

Start with the fact that the Einstein tensor (and thus $T$) is divergence-free: $$\nabla_a T^{ab} = \nabla_a(\rho u^a)u^b + \rho u^a \nabla_a u^b = 0. \tag{1}$$ The product rule expansion I've chosen is quite suggestive: in the first term we see the divergence term that appears in the continuity equation, while in the second we see the covariant ...


2

Been a long time Justin! I don't know the literature, but here is an approach towards formulating a practical relaxation of the problem. If you relax the requirements of parallel transport a bit, you might be able to get something reasonable by formulating this as a regularized optimization problem over the whole tangent bundle at once. Imagine ...


2

This is a fun question for me, since I studied precisely a connection between the two. In Riemannian geometry, a natural question is: Are there closed geodesics in my manifold? How many of them? What can I say about them? On the other hand, when you deal with Morse theory, you want to relate the topology of your manifold with the number of critical points ...


2

(1) If $\widetilde{M}={\bf R}^{n+1}$, then from Codazzi, we have $$ x(f) g(y,z) - y(f) g(x,z) =0$$ If $y=z\perp x$, then $x(f)=0$ That is $f$ is constant (2) If $\widetilde{M}$ is general, then we have an counter example : $\widetilde{M}=({\bf R}^2\times {\bf R}, G:=F(x_3)^2g_1(x_1,x_2)+ g_2(x_3)),\ M={\bf R}^2$ Here $g_i$ are canonical metric on each ...


2

You've misinterpreted the statement - you to need to show it leaves one such vector invariant, not all such vectors. (You can reasonably easily produce a counterexample to the stronger statement by constructing a flat $R^3$ bundle over $S^1$ with a twist.) Thus the problem boils down to showing that every $O \in SO(n-1)$ has a fixed point, which is true ...


2

If the map fails to be injective then $(\pi(v),\exp (v))=(\pi(w),\exp(w))$ for some $v\not=w$. But this implies in particular that $\pi(v)=\pi(w)$, i.e., both vectors are over the same point of $M$. Thus you are reduced to a local problem after all, namely that of the exponential map being a local diffeomorphism, which follows from a general fact about ...


2

For $\gamma,\ \xi >k$ and $\alpha \leq k$, assume that $$ C_\alpha^{\gamma\xi } := (v^\alpha,[v^\gamma,v^\xi ]) =0 $$ This implies that $[v^\gamma,v^\xi ]$, Lie bracket of elements in $V$ where $V$ is an orthogonal complement of the kernel, is in $V$. That is $V$ is closed under Lie bracket


2

Remember that the curvature operator is a symmetric bilinear form on the Lie algebra $$\Lambda^2T_pM\simeq \mathfrak{so}(n). $$ The structure coefficients $C^{\alpha \beta}_\gamma$ are just the components of the Lie bracket in our basis $v^\alpha$ for this algebra. By the choice of diagonalizing basis, the image of $M$ is just ...


2

The one you gave is the standard definition for a parametrized curve $\gamma:I\rightarrow M$ to be geodesic at a point $t_0\in I$. The formal meaning of the definition is that the covariant derivative $\frac{D}{dt}$ of the tangent vector field to the curve is $0$ at $t_0$. We say that a parametrized curve is a parametrized geodesic if it is geodesic at every ...


2

In my opinion this is in fact mostly a question of taste, but not completely. Different approaches have different advantages. Coordinate system based approaches (those with all the indices, this is what you are referring to when you are referring to the transformation rules. These are just the statement of coordinate invariance of the geometrical content of ...


2

Here's a counterexample on the unit disc. Let $f : \mathbb D^2 \to \mathbb R^2$ be defined by $$ f(x,y) = (x-2y^2,y). $$ This has differential $$\left(\begin{matrix}1 & -4y \\ 0 & 1\end{matrix}\right)$$ and thus is an immersion. It's also injective. The solution to the Dirichlet problem in this case is $$\omega(x,y) = (x^2 - y^2 + x - 1,y);$$ ...



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