8 votes
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Necessary and sufficient mathematical structure for spacetime continuum

First, let me remark that there is no such thing as a “necessary and sufficient condition” for modelling spacetime. It’s a model, so it can’t be an exact description (and Physics doesn’t answer the ...
peek-a-boo's user avatar
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8 votes
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Is smoothness of multiplication redundant in the definition of Lie Group?

As suggested, I’m turning my comment (I was worried I was missing something) into an answer. The statement is wrong. Consider the classical manifold $\mathbb{R}$, with the non-smooth “addition” law ...
Aphelli's user avatar
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7 votes

Prove that flow map $\theta_t$ is orientation preserving

Here is another answer exploiting the "group" property of flows: $$\theta_{s+t} = \theta_{s} \circ \theta_{t}.$$ Now given $\theta_t$, write it as $$\theta_{t} = \theta_{t/2} \circ \theta_{t/...
Mohith Nagaraju's user avatar
7 votes
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The hyperbolic space is a conformally compact Einstein manifold

The fact that the hyperbolic space is Einstein is a standard fact, so I'll focus on the conformal compactification. Let me emphasize the fact that conformal compactifications are built to bring "...
Didier's user avatar
  • 18.9k
6 votes
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On the homology of manifolds with boundary

The inclusion $\mathrm{int}(M)\rightarrow M$ is a homotopy equivalence. This follows from the existence of a collar neighborhood of $\partial M$ in $M$. Thus, $H_i(\mathrm{int}(M))\rightarrow H_i(M)$ ...
Thorgott's user avatar
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6 votes
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Submanifold of ball is entire ball

First note, by compactness, there is a $ \varepsilon > 0 $ such that the $ \varepsilon $-neighborhood of $ \partial \mathbb{B}^n $ (the set of points within $ \varepsilon $ of $ \partial \mathbb{B}^...
Jake Mirra's user avatar
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6 votes

Can a differentiable manifold be defined the same as a manifold but with the requirement of being diffeomorphic to $\mathbb{R}^n$?

Your definition is fine if you only care about differentiable manifolds that are subsets of $\mathbb{R}^n$. The point being made in the Wikipedia article is that there they are starting with an ...
Eric Wofsey's user avatar
6 votes

Is $\mathcal{O}(-1)\cong\mathcal{O}(1)$?

One may prove that any complex vector bundle, viewed as a real, unoriented vector bundle is isomorphic to its complex dual as follows: Let $\xi$ be a complex vector bundle over a base $B$ (assume this ...
J.V.Gaiter's user avatar
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6 votes

Can adding a single element to a Lie group make it infinite-dimensional?

There is no bound on the dimensionality of $G$. In fact, a single element can generate a Lie group of arbitrarily large dimension. Thus the $n$-dimensional torus $\mathbb T^n$ is the closure of the ...
Robert Israel's user avatar
6 votes
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Lee Smooth Manifolds: Fundamental Theorem of Flows (9.12) Details

$\theta^{(p)} : \mathcal D^{(p)} \to M$ is the unique maximal integral curve starting at $p$. We don't need Zorn's lemma here. $\mathcal D^{(p)}$ is defined to be the union of all open intervals ...
Kenny Wong's user avatar
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6 votes
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Why does the adjoint representation appear in the definition of a connection 1-form?

Like I said in the comments, I think the natural starting point is that of a horizontal distribution on $\pi:P\to M$, i.e. $TP=\ker d\pi\oplus\mathcal{H}$. The notion of a principal $G$-bundle $P$ is ...
Quaere Verum's user avatar
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6 votes
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Why is this map from a $G$-torsor to $G$ smooth?

One has to be slightly careful here regarding the notion of smoothness, particularly the smooth structure on the orbit of a point. Suppose a Lie group $G$ acts smoothly (not necessarily freely) on a ...
peek-a-boo's user avatar
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5 votes
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Can a real-valued function on the sphere have exactly 2 critical points which are not antipodal?

Take the function $f(x_1, \dots, x_{d}) = x_{d}$ defined on $\mathbb S^{d-1}$. Let $\phi : \mathbb S^{d-1 }\to \mathbb S^{d-1}$ be a diffeomorphism such that $\phi (0,\cdots, 0,1) = (0,\cdots, 0,1)$ ...
Arctic Char's user avatar
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5 votes

Compactly supported harmonic forms

It's true, as @ArcticChar proved, that if $\alpha$ is harmonic and compactly supported then $d\alpha = \delta\alpha = 0$. But this is not very useful on a noncompact manifold, because if $\alpha$ is ...
Jack Lee's user avatar
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5 votes
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Intuitive understanding of inward and outward vector definitions

Before talking of a boundary point, let's look at the more general picture of an interior point $p$. A tangent vector $v \in T_pM$ can be thought of as a vector tangent to a curve $\gamma \colon (-\...
Didier's user avatar
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5 votes
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Why can we only integrate top forms?

The central idea motivating integration is to provide a way of "summing up" all the values of some function over a given region. This must be done through some limiting process, e.g. a ...
CJ Dowd's user avatar
  • 1,559
5 votes

Critical points theorem for regular surfaces

No; it is possible for a fiber to be regular "by accident" even if it contains critical points. For instance, consider the function $f(x,y,z)=(x^2+y^2+z^2-1)^2$. Then every point of $f^{-1}...
Eric Wofsey's user avatar
5 votes
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Explicit Lie group embedding of $O(n)$ into $SO(n+1)$

I’ll just write up an answer with a few ways of attacking the problem, and let you fill in some of the details (which can easily be found if you search the site or any decent book). First, let me make ...
peek-a-boo's user avatar
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5 votes
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Differential of conjugation map is smooth

This is basically by definition of smoothness: smooth means infinitely differentiable, so the derivative is still smooth. More precisely, suppose $M$ and $N$ are smooth manifolds and $f:M\to N$ is ...
Eric Wofsey's user avatar
5 votes

Trivial principal bundles and curvature.

In the non-abelian case, the curvature forms $F_{\mathcal{M}}^{A}$ and $F_{s}$ are generally not equal. This is because the connection form $A$ and the section $s$ transform differently under gauge ...
FreeMind's user avatar
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5 votes
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If $\alpha\wedge d\alpha$ is a volume form, there exists a vector field $X$ such that $i_X\alpha\equiv1$ and $i_X (d\alpha)\equiv0$.

If you're not already aware, the vector field $X$ is called the Reeb vector field associated to $\alpha$. The key piece of the construction is Jacobi's theorem, which states that the determinant of an ...
Kajelad's user avatar
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5 votes
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Quotient of cohomology groups with different coefficents

Universal coefficients gives, for any abelian group $A$ and any $X$, an isomorphism $$H^1(X, A) \cong \text{Hom}(H_1(X), A)$$ (since $H_0(X)$ is free, so the Ext term vanishes). If in addition $H_1(X)$...
Qiaochu Yuan's user avatar
5 votes
Accepted

Group laws of the flow of time-dependent vector field

You're mixing the notations. With $\psi_t$ you'd be fine to add them. However this is $\psi_{t_0,t_1}$ which is actually the curve $\beta_{t_1-t_0}$ as explained on page 238. Note that $$\psi_{t_0,t_1}...
CyclotomicField's user avatar
5 votes

Relation between trilinear forms, Jacobi identity and closed $2$-forms

The relevant formula is the "conceptual" formulation of the exterior derivative. If $\omega$ is a $2$-form and $X,Y,Z$ are vector fields (here, on $\Bbb R^3$, we have \begin{multline*} d\...
Ted Shifrin's user avatar
5 votes
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Why is the signature of a manifold homotopy invariant?

The definition you've given for the signature is phrased in terms of de Rham cohomology and is indeed not obviously homotopy invariant. However, the pairing you've given has a "lift" to ...
Connor Malin's user avatar
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4 votes

Reference: the number of smooth structures on a topological $n$-manifold, $n\geq 5$ is finite

There are not really as many structures on $T^n$ as that: typically one mods out $H^3(T^n;\mathbb{Z}/2)$ by the action of $GL(n,\mathbb{Z}/2)$. This is discussed by Wall in chapter 15A of Surgery on ...
Dave Davidson's user avatar
4 votes

Can every Lie group be realized as conformal group of smooth manifolds

First of all, one cannot talk about conformal transformations of a smooth manifold: For the notion of conformality to be defined you need an extra structure besides a smooth atlas. I will work with ...
Moishe Kohan's user avatar
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4 votes
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Show a operator is not a tensor field on $\mathbb{R}^3$

Note: I found this one quite hard to explain, and ended up making two major edits. Edit 1 was to highlight that the crux of the matter is that $T|_x(v_1, v_2, v_3)$ depends on the partial derivatives ...
Kenny Wong's user avatar
  • 32.1k
4 votes

The embedded submanifolds of a smooth manifold (without boundary) of codimension 0 are exactly the open submanifolds

There is a proposition that characterizes embedded submanifolds of codimension $k$ as follows. Let $N\subset M$ denote a codimension $k$ embedded submanifold of $M$. If $p\in N$, then there exists a ...
Alekos Robotis's user avatar
4 votes
Accepted

Calculating the differential of the map $A \mapsto A A^T$, $A \in M(n\times n, \mathbb{R})$

Regarding your 2nd question: the tangent space of the space of matrices $M(n\times n, \mathbb R)$ at a point $A$ is, as you said, identified with $M(n\times n, \mathbb R)$ itself. Regarding your 1st ...
T.P.'s user avatar
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