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5

Let me try a low tech approach. We start with some definitions from Gang Tian's paper (which means we are in the Euclidean setting). We have $$ A = A_i \mathrm{d}x_i, \qquad A_i:U\to g, $$ where $U\subset\mathbb{R}^n$ is an open set and $g$ is a Lie algebra. Then the Yang-Mills equations for $A$ are $$ \partial_iF_{ij} + [A_i,F_{ij}] = 0, \qquad \qquad (*) ...


5

Not true. Try lines of longitude on a sphere ($\theta=\text{constant}$). Then $\omega = d\phi$ and if you compute $(d\delta+\delta d)\omega$, you don't get $0$.


3

For the Riemannian case this is surely true, you may find it in "Foundations of Differential Geometry" by Kobayashi and Nomizu, volume 1, page 166, proposition 3.5. In the pseudo-Riemannian case this is not true as indicated in a comment above by Zhen Lin (the topology induced by the pseudometric will no longer be Hausdorff, for instance).


2

Roe's book is one of my favorite books, and the place where I first learned some index theory. I know that the lack of a few details can be frustrating sometimes, though. Let me see if I can address some of your questions. $(,)$ is the Hermitian metric on $S$ that Roe refers to in Definition 3.4. You're right that he's not too explicit about this. (I ...


2

The disjoint union of two copies of $S^1$, for example, is a Riemannian manifold which is complete but not connected.


2

Note that Riemann's essay originally was a "habilitation lecture", directed at some larger academic audience. Therefore Riemann tended to avoid too much technical machinery. In the formula at hand he just writes $\sqrt{\sum dx^2}$, but at other places in this lecture he talks about the $dx_i$ and about the fact that the fundamental form has ${n(n+1)\over2}$ ...


2

To Riemann dx was a nilsquare infinitesimal, that is, the most conventional (if controversial) type of infinitesimal. 'The principle of gaining knowledge of the external world from the behaviour of its infinitesimal parts is the mainspring of the theory of knowledge in infinitesimal physics as in Riemann's geometry, and, indeed, the mainspring of all the ...


2

A Riemmanian manifold is called flat if its curvature vanishes everywhere. However, this does not mean that this is is an affine space. It merely means (roughly) that locally it "is like an Euclidean space." Examples of flat manifolds include circles (1-dim), cyclinders (2-dim), the Möbius strip (2-dim) and various other things. Yet, let me add into the ...


2

I think one piece of the puzzle you are missing is that $$ r \partial_r = \sum_i x^i \partial_i.$$ This means that $$ - \frac{1}{\sqrt{g}} \sum_i x^i \partial_i \sqrt{g} = -\frac{1}{\sqrt{g}} r \partial_r (\sqrt{g} ) = -\frac{r}{2g} \partial_r g,$$ as claimed. Edit: I think there is some confusing notation at work here. It seems Roe uses $\nabla$ to mean ...


2

An important fact about tensors is that they depend only on the vector fields at a certain point, i.e. if $T$ is a tensor, $X_1,\ldots,X_n,Y_1,\ldots,Y_n$ are vector fields, and $p$ is a point such that $X_i(p)=Y_i(P)$ for all $i$, then $T(X_1,\ldots,X_n)|_p=T(Y_1,\ldots,Y_n)|_p$. In your example, we wish to evaluate the curvature of given vector fields at ...


1

I figured it out myself: the answer is 'always'. Consider the mapping $\hat{\nabla}$ defined by, $\hat{\nabla}:\lambda^{bc}_d\mapsto \varphi^*(\tilde{\nabla}_a \varphi_* \lambda^{bc}_d).$ The first step is to show that this mapping is a derivative operator. It obviously commutes with addition, index substitution and contraction because all three maps do ...


1

If $G < Isom (\mathbb{R}^n) \cong O(n) \rtimes \mathbb{R}^n$ is a subgroup of isometries of some Euclidean space $\mathbb{R}^n$ acting freely and properly discontinuously on it, then the quotient will be a flat Riemannian manifold. Conversely, any flat Riemannian manifold arises in this way. In the realm of surfaces, this gives you planes, disks, ...


1

I believe I've found the answer in this text and more details in Kosinski's differential manifolds, p. 53. In a nutshell: for a neat submanifold $N$ of $M$, one can define the normal bundle abstractly as the quotient $(T_N M)/TN$. Any choice of Riemannian metric induces an "orthogonal" representation of the normal bundle. Choosing a vector field $X$ on ...


1

There is a rather significant difference. I'm not sure how comfortable you are with various vector bundle constructions, but both 2-forms and metrics are special elements of $E = T^*M \otimes T^*M$ the second dual tensor power bundle, which is identified with the space of smooth varying bilinear maps (that is to say a section $q$ of the bundle $E$ is a ...


1

For infinitesimal lengths, hyperbolic and Euclidean metrics are the same. (This follows from the fact that the ratio between squared lengths and Gaussian curvature becomes zero.) So perhaps the author meant the following argument: for infinitesimal lengths, the hyperbolic metric and the Euclidean one agree, and the latter is proportional to first order, so ...


1

The pseudosphere is such a surface. However this can only be done locally. The global metric in the upperhalfplane cannot be embedded in 3-space by a difficult old result of Bernstein.


1

Assume there exists precisely one geodesic segment from $p$ to $q$, say $\gamma : [0,l] \to M$, parametrized by arc length. By compactness we can extend $\gamma$ to the intervall $[0,\infty[$. Now consider a sequence $\gamma_n : [0,l_n] \to M$ of minimal arc length geodesics from $p$ to $\gamma (l + \frac 1 n)$. Since limits of minimal geodesics are minimal ...


1

Thanks wspin. However, let me elaborate a bit, some detail is still not clear to me. For any $s \in (-\epsilon,\epsilon)$, let $\gamma_s(t)$ be a minimal geodesic connecting $\gamma(0)$ with $\gamma(t+s)$. Let $V(t) :=\partial_s \gamma_s(t)|_{s=0}$ be the associated vector field along the original geodesic $\gamma$. By construction $V(0) = 0$ and $V(1) = ...


1

First of all, I assume that you are asking about forms which are only locally Lipschitz, otherwise the answer will be different (the function $x^2$ on the real line is only locally Lipschitz). Then the simplest thing to do is to write your form in local coordinates and require all the component functions to be locally Lipschitz. There are alternative ...



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