7

On your more general question about differential geometry, i.e. why do people study it? There are many answers, some having little to do with each other. In my opinion differential geometry is perhaps best approached once you have seen the need for formal and computational structure underlying various fairly mundane geometric problems in more than one area....


7

If you are interested in local-to-global results, i.e., collecting local info about the manifold and then patch it together to get a global info then you need tools for the patching part of the process. These often come under the guise of some form of integration. Here is a simple example. If a compact surface admits a metric with negative curvature (local ...


6

One class of examples in a compact, connected Lie group $G$ of rank $r$ are the conjugacy classes of codimension $r$. Taking an element $a\in G$ whose centralizer is a maximal torus (a generic condition), the conjugacy class of $a$ is a submanifold $C_a\subset G$ of codimension $r$ whose normal plane at $a$ is the tangent at $a$ to $Z_a$, the centralizer of ...


4

If you don't read Russian: Rozendorn's construction is presented by Aminov in Extrinsic geometric properties of the Rozendorn surface, which is an isometric immersion of the Lobachevskiĭ plane into $E^5$. Now Aminov's paper was also in Russian, but the article does have an English translation available, which you can find at https://iopscience.iop.org/...


4

This might be my own bias, but I think differential geometry is a really natural area to study. When we look out at the world around us, we see lots of objects that seem smooth, but are not flat. As such, it is only natural to try to study geometry using techniques from calculus, which is the starting point of differential geometry. In this vein, many of the ...


3

I don’t understand modern abstract differential geometry, but the elementary theory of curves and surfaces in $\mathbb{R}^3$, as expounded by Gauss, Euler, Darboux and others, is very useful in engineering and manufacturing. A few examples: The curvature properties of a surface determine how light reflects from it. These reflections are what determine the ...


2

Unless I am terribly mistaken, the answer is yes and the proof strategy is rather simple. The crucial observation is that the definition of bounded geometry depends on quantities that are continuous. Consider first the injectivity radius function of the boundary, $r_{b}\colon\delta X\to \mathbb{R}$, $$r_b(x)=\sup\{\ t>0\mid \exp\colon B_{\delta X}(0_x,t)\...


2

$\newcommand{\R}{\mathbb{R}}$As you state, there is a $SO(6)$-equivariant map $\delta:\operatorname{Sym}^2(\Lambda^3\R^6)\to \Lambda^4 \R^6$ such that $\gamma(\delta(\beta^{\otimes 2})) = \gamma(\beta)^2 - |\beta|^2\operatorname{id}$ on the positive spinors (the central $U(1)\subset\operatorname{Spin}^c(6)$ cancels, so that it suffices to work with the ...


2

Here's how one can construct a specific example to illustrate what can happen: First, recall from my answer to this question that, if you have a smooth map $f:D\to\mathbb{R}^2$ with constant positive singular values, then, letting $(e_1,e_2)$ be the orthonormal frame field on $D$ such that $\bigl(f'(e_1),f'(e_2)\bigr)$ are orthogonal with $|f'(e_i)|=\...


2

The real motivation for the Lie derivative is doing differential calculus with vector fields. If we want fo differentiate the vector field $W$ in the direction of the vector field $V$, we take the flow of $V$ through time, use it to pull back $W$, and take the derivative at $t=0$. To explain better, let $V$ and $W$ be vector fields on a smooth manifold $M$, ...


2

If you don't mind looking at the original Russian version, then the scanned article itself is available at https://arar.sci.am//Content/23775/file_0.pdf I found it by searching https://www.google.com/search?q=розендорн+реализация+метрики which shows also some Russian citations of that work. Update. Perhaps even better, you could read more about this in his ...


2

Let $M$ be a submanifold of a symmetric space $Q$, $G = I^0(Q)$, $\mathfrak{g} = \mathrm{Lie}(G)$. Take $p \in M$ and write $K$ for the isotropy subgroup of $G$ at $p$ and $s_p \in I(Q)$ for the geodesic symmetry at $p$. We have $\Theta = C_{s_p} \colon g \mapsto s_p g s_p$ an involutive automorphism of $G$ and $\theta = \Theta_*$ the corresponding ...


2

EDITED: Added clarification, as pointed out by @TerryTao. Let $g_1$, $g_2$ be Riemannian metrics and $\Delta_1$, $\Delta_2$ their respective Laplacians. Let $\lambda_1$ be an eigenvalue of $\Delta_1$ and $\lambda_2$ an eigenvalue of $\Delta_2$. I think what can be proved is the following: Given an eigenfunction $u_2$ of $\Delta_2$ with eigenvalue $\lambda_2$,...


1

$\newcommand{\spinors}{\mathbb{S}}$The tangent bundle $TZ$ fits into an exact sequence $$ 0\to T_{/M}Z\to TZ\xrightarrow{\pi_*}\pi^*TM\to 0 $$ where the fiberwise tangent bundle $T_{/M}Z$ is two-dimensional and equipped with a canonical complex structure. Explicitly, the fiber of $Z\to M$ over a point $m$ are the (oriented) complex structures on $T_mM$, ...


1

This answer basically just writes out Willie Wong's. In particular, the definition of $\hat{C}$ works for $\omega$ as well; if $\omega$ is a form with values in $E$, then $\hat{\omega}(\eta) = \sum\limits_{i_1, \dots, i_p = 1}^n \eta_{i_1,\dots,i_p}\omega_{i_1,\dots,i_p} \in E$. But then \begin{align} (\hat{\omega} \circ \hat{R_p})(X_1,\dots,X_p) &= \...


1

I think the idea is to think of $\hat{R}_p$ as a mapping from $\Lambda^pM$ to itself, and $\hat{\omega}$ as a mapping from $\Lambda^pM$ to $E$ (the vector bundle in which $\omega$ takes values), and then just compose these two operators (to get something $E$ valued in the end). In the case where $E$ is the trivial (scalar) bundle, it works out to be $\hat{R}...


1

Let me at some additional remarks to RBega2s excellent answer. As pointed out by Otis Chodosh in his comment, the Lawson surfaces $\xi_{1,g}$ are isolated by a result of Kapouleas and Wiygul. Therefore, it is not possible to desingularies two intersecting minimal 2-spheres meeting at an angle $\theta\neq\tfrac{\pi}{2}$, by replacing the circle of ...


1

There is another paper, with a few more details: JONATHAN HOLLAND: A TWO-DIMENSIONAL C 2,1 METRIC WITH NO LOCAL C2 EMBEDDINGIN R3, FOLLOWING POGORELOV ( https://arxiv.org/pdf/1211.4166.pdf ) p.s. I am going through the details myself, so at the moment I can't answers your question directly.


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