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4

Things scale in a reasonable way: The metric $$ds^2 = \frac{4}{(1-x^2-y^2)^2}(dx^2 + dy^2)$$ has $K=-1$. The metric $$ds_c^2 = \frac{c^2}{(1-x^2-y^2)^2}(dx^2 + dy^2)$$ has $K=-4/c^2$.


4

A Riemannian metric does indeed give an isomorphism between $TX$ and $T^*X$ as real vector bundles. However, if $X$ is a complex manifold, then $TX$ can also be viewed as a complex vector bundle, but the isomorphism $TX\cong T^*X$ provided by a Riemannian metric is not complex linear. If the metric is Hermitian, then the isomorphism complex anti-linear. ...


4

I think there are two slightly different questions here. The first question is the one you specifically asked: If we are given the manifold $M$ and a metric defined on a dense coordinate chart $U\subseteq M$, what is a sufficient condition for the metric to have a smooth extension to all of $M$? This is basically pretty easy to answer: just find, for each ...


4

It looks like here on functions $\nabla^k \nabla_k f$ means $g^{jk}\nabla_j\nabla_k f$ where $\nabla_j\nabla_k f = \nabla_{\partial_j}\nabla_{\partial_k}f = (\nabla \nabla f)_{jk}$. So for an $r$-tensor $\alpha_{i_1 \cdots i_r}$, we could lift an index on the $(r+1)$-tensor $\nabla_k \alpha_{i_1 \cdots i_r} = (\nabla \alpha)_{ki_1 \cdots i_r}$ and write ...


2

As you observed, the ellipticity of the Ricci operator (or any partial differential operator) does not depend on a choice of coordinates. The Ricci operator is never elliptic. If you assume that the coordinates $(x^i)$ are harmonic for the metric $g$, and you write the coordinate expression for $\text{Ric}(g)$ in these coordinates, some terms cancel because ...


2

HINTS: Consider the function $f(A) = A^\top A$ on the space of $3\times 3$ matrices. Compute $df_I(B)$ and see what it means for this to be $0$. Next: If you define an orthonormal basis for $T_ISO(3)$, then left-translating it everywhere will give you an orthonormal basis at $T_ASO(3)$ for all $A\in SO(3)$. Does this determine a unique Riemannian metric on ...


2

Qiaochu's comment misses the point: Some metrics have true singularities that cannot be removed by a change of coordinates. I don't know about general metric spaces, but in good-old 4-D space-time, a sufficient condition for a true singularity to exist at point $P$ is that the limit as one approaches the singularity of the trace of the Ricci tensor goes ...


2

I'll do (A). Recall, that in coordinates, we have that $$\Delta_g u=\frac{1}{\sqrt{|g|}}\partial_a\left[g^{ab}\sqrt{|g|}\partial_b[u]\right],$$ where $|g|=\det(g_{ab})$. Note for this case $g_{ab}=e^{2w(x_2)}\delta_{ab}$, where $\delta$ is our usual Euclidean metric (i.e., it's conformally equivalent). Then $g^{ab}=e^{-2w(x_2)}\delta^{ab}$ and ...


1

The $g=dx^2 + dy^2$ notation really means $g = dx \otimes dx + dy \otimes dy$, using the tensor product. Tensor products build up linear maps. That way, you get a map that takes in two vectors to return a scalar--exactly what you would want the metric to do. In other coordinates (like polar), you must remember to keep a distinction between positions and ...


1

can we directly compare $g_X$ and $g_Y$, since they both are defined on $T\mathbb R^2$ Generally, one should think of parameter spaces as different copies of $\mathbb R^2$ floating somewhere in Platonic universe and not interacting with each other at all. But yes, you could check if $g_X$ is a scalar multiple of $g_Y$; if they are, the surfaces are ...


1

No, it's not nearly that simple. The constant $C$ quantifies the connectivity of the manifold. It can be imagined as the severity of traffic jams that occur when all inhabitants of the manifold decide to drive to a random place at the same time. For example, let $M$ be two unit spheres $S^2$ joined by a thin cylinder of radius $r\ll 1$ and length $1$. There ...


1

tentative answer Your question can be phrased as follows: given a metric of the form $e^{2w(x_2)} \left( dx_1^2+dx_2^2 \right)$ on the plane, how to find an isometric immersion of this metric into $\mathbb R^3$? (An isometric immersion is a map $y=f(x)$ such that the relation in your post holds.) One can also go further and ask for isometric embedding. ...


1

No. The torus can be written with an atlas of 9 charts (a 3 x 3 grid) with the transition functions all being translation.


1

I assume that ($g, \nabla$) are induced metric and connection from the embedding $\varphi$. Note that $\nabla \| \varphi\|^2 = 2 \varphi ^\top$, the tangential part of $\varphi$. Thus $$\langle \varphi , \nabla \|\varphi\|^2 \rangle = 2\|\varphi^\top\|^2 \geq 0.$$ (It's true for any immersion $\varphi$) Remark: To show that $\nabla \| \varphi\|^2 = 2 ...


1

The context given makes it clear they are talking about the operator norm. In this context, this can be defined as $$ \|f'(x)\|=\sup\sqrt{\frac{h(f'(x)u,f'(x)u)}{g(u,u)}}, $$ the supremum taken over all nonzero tangent vectors $u$ at $x$.


1

I would guess it is a (unit ?) vector orthogonal to $\frac{\partial}{\partial x_j}$, $j<n$ such that its scalar product with $\frac{\partial}{\partial x_n}$ is negative. The other definition you proposed ($-\frac{\partial}{\partial x_n}$) would depend on the coordinates.



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