New answers tagged

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I would recommend An Introduction to Mathematical Relativity (2021) by Natário. It also contains many suggestions for further reading. A free version of it is available on arXiv.


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Using the connection, the tangent bundle $T S(\wedge^2M)$ splits as a direct sum into the vertical part $$\mathcal VS(\wedge^2M):=\ker d\pi,$$ where $\pi\colon S(\wedge^2M)\to M$ is the projection, and into the horizontal part $$\mathcal H S(\wedge^2M).$$ We have $$\pi^* TM\cong\mathcal H S(\wedge^2M)$$ via $d\pi.$ Then, $\tilde\gamma'(0)\in \mathcal H S(\...


3

You can get a map which is continuous outside a set of lower dimension. Let $K\subset L$ be compact Lie groups and let $s:K\backslash L\to L$ be a section to the projection $L\to K\backslash L$. Now this section can be chosen continuous outside a set of lower dimension, since the projection is a fibre bundle. For $x\in L$ we write $s(x)$ for $s(Kx)$. We ...


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For any $m<n$ the $n\times n$ unitary matrix $\Omega$ has the block decomposition $$\Omega=\begin{pmatrix} A&B\\ C&D\end{pmatrix},$$ where $A$ has dimensions $m\times m$, $D$ has dimensions $(n-m)\times(n-m)$, $B$ has dimensions $m\times(n-m)$ and $C$ has dimensions $(n-m)\times m$. Up to a set of measure zero, the matrix $D$ will not have a unit ...


6

You need $X_0$ to be closed inside $X$, otherwise the theorem is false (there are simple counterexamples). So let us suppose that $X_0$ is closed. Inclusions give homomorphisms $$\pi_1(X_0) \longrightarrow \pi_1(X\setminus K) \longrightarrow \pi_1(X)$$ To prove that the first homomorphism is injective it suffices to prove that the composition $\pi_1(X_0) \to ...


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This is not a real answer but just a suggestion for where to browse. What I am aware of is that the sign of the first eigenvalue is related to the (non)existence of variations that preserve the volume but increase scalar curvature, see the recent preprint https://arxiv.org/pdf/2109.09556.pdf by M. Dahl and K. Kroencke. In particular, they cite the following ...


1

This is an old thread, but I figured I would add a short comment that may explain some of the reasoning behind the 'wishful thinking' that Neves refers to. Let us, for all integer $p \geq 1$, write $\omega_p$ for the $p$-width of $\mathbf{S}^3$, that is the min-max value associated to a $p$-parameter Almgren-Pitts construction. Then \begin{equation} \omega_1 ...


5

In Kahler geometry, there is a long-standing open problem of determining which manifolds admit metrics of constant scalar curvature (cscK). In particular, there is a conjecture due to Yau, Tian and Donaldson which states that for particular class of Kahler manifolds, the existence of a cscK metric is equivalent to an algebro-geometric condition known as K-...


0

As a simple example, consider the curve $y=f(x)$, $x,y\in\mathbb{R}$, with measure given by the arclength $ds=(1+f'^2)^{1/2}dx$. The curvature is $\kappa=f''(1+f'^2)^{-3/2}$, for a uniform distribution we want $d\kappa=cds$, with $c>0$. This gives for the curve the differential equation $$(1+f'(x))f'''(x)-3f'(x)f''(x)^2=c(1+f'(x))^3.$$ Here is a numerical ...


1

Any surface, compact or otherwise, admits a metric with constant scalar curvature, so we can always find metrics with uniform density. Furthermore, any compact manifold admits a metric of constant scalar curvature in its conformal class, by the solution to the Yamabe problem. On the other hand, the scalar curvature measures the deviation of the size of ...


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As already pointed out by Will Jagy there is an example in this paper by Karcher-Pinkall-Sterling. It is build from a tetrahedral tesselation of $S^3$ with dihedral angles $\tfrac{\pi}{2},$ $\tfrac{\pi}{2},$ $\tfrac{\pi}{2},$ $\tfrac{\pi}{3},$ $\tfrac{\pi}{2},$ $\tfrac{\pi}{5},$ see the attached figure. The property of dividing the volumes into two equal ...


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I haven't read all of DCCT so take this with a grain of salt, but after having spent a lot of time with it, this is how I would recommend getting started on the abstract stuff. First one must learn classical Grothendieck Topos Theory. Chapter 1.2 of DCCT gives a pretty good motivation and some nice examples of sheaves on $\mathsf{Cart}$, but I would ...


3

I would say that understanding traditional differential cohomology is a reasonable prerequisite. There are multiple good sources: Ulrich Bunke: Differential cohomology Diferential Cohomology. Categories, Characteristic Classes, and Connections. Further suggestions can be found in this answer: References for differential cohomology and differential ...


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Contrary to my earlier comments, I now believe (but I haven't completely checked) that this is false in general. (I prefer not to comment on the paper you linked.) Suppose that $X$ is a Kähler manifold. The form $\alpha$ defines a class in $H^1(S,\Omega_S^1)$. The existence of $\beta$ should imply that $\alpha$ lifts to a class $\alpha'\in H^1(S, \Omega_{S'}^...


1

I do not there is a strong relation between the two notions in the general case. Curvature is obviously a local object. On the other hand, the behaviour of the first Dirichlet eigenfunction near a boundary point $p$ is highly non-local. For example, if $\Omega$ is the union of two balls of different radii connected by a narrow channel, then the normal ...


1

Positive scalar curvature implies that if $\textrm{index}(\Sigma_j)\leq I$ then $\Sigma_j$ have bounded area and genus. This is proven here https://arxiv.org/pdf/1509.06724.pdf (Theorem 1.3). That paper also contains some other examples related to the Colding--Minicozzi looping example. A natural generalization is whether or not one can generalize this to a ...


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I think that they are not Quaternion Kahler unless they are quaternion projective spaces. Quaternionic Grassmanians are examples of Riemannian symmetric spaces. The Riemannian symmetric spaces admitting Quaternion Kahler structures (called Wolf spaces) are classified completely. See the table of https://en.wikipedia.org/wiki/Quaternion-K%C3%...


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If your foliation is "regular", which roughly means the leaves have constant dimension, then every chart of the manifold has the form $L \times T$, where $L$ is the longitudinal (leafwise) direction and $T$ the transeversal direction. Moreover, the change of coordinates is of the form $(x,y) \mapsto (f(x,y),h(y))$. (This is thanks to the Frobenius ...


4

In addition to Nik Weaver's references, let me just sketch the proof which is in fact not very difficult: A construction of Kaplansky (Rings of operators, Thm 26) shows that if $\mathcal{A}$ is a $*$-algebra with the property that for every $A \in M_n(\mathcal{A})$ the matrix $1 + A^*A$ is invertible, then every idempotent in $M_n(\mathcal{A})$ is ...


3

Yes, every finitely generated Hilbert module comes from a hermitian complex vector bundle in this way. In fact more is true: arbitrary Hilbert modules over $C(X)$ correspond to continuous (in an appropriate sense) bundles of Hilbert spaces over $X$. Unfortunately, the only place I've seen this proven is in a dissertation (Takehashi, Fields of Hilbert Modules,...


1

Question. Is there a holomorphic tubular neighbourhood of S in M? I have a counterexample. Let $\phi:\; Z^2 \to Aut(C^n)$ be a group homomorphism, with $Aut(B)$ denote the group of holomorphic automorphisms of $C^n$. We assume that the image of $\phi$ preserves 0 and acts trivially on $T_0 C^n$. To construct $\phi$ which has infinite image take, for example,...


4

Note first that every pseudo-Riemmanian manifold admits a null vector field which is not identically $0$ (just construct one locally and multiply it by a bump function). So by "non-zero vector field" I assume you mean "nowhere vanishing". Let $(M,g)$ be a pseudo-Riemannian manifold of signature $(p,q)$. The tangent bundle $TM$ always ...


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Claim. The thief $T$ can escape if $C$ is a circle, with a simple strategy of dribbling left and right each policeman at a time in such a way that he is left out of reach of the thief no matter what the future dribbles will be. Proof. The key insight is due to Pietro Majer: the thief can approach $C$ in such a way that its shadow $S$ (closest point) on $C$ ...


2

Such manifolds are precisely the Riemannian ones and their “opposites”. Suppose $M$ is a pseudo-Riemannian manifold whose signature is not trivial in one direction or the other. We work in a neighbourhood of a point and will define a vector field with arbitrarily small support, so we may as well consider $M=\mathbb R^d$. Suppose also, after a linear change ...


1

On the generalization of the Wigner semicircle law to real symmetric tensors studies the complex eigenvalues and eigenvectors of a real symmetric tensor $T\in\otimes^p\mathbb{R}^N$, subject to the eigenvector normalization $x^2=1$, not $x\bar{x}=1$. As discussed in The number of eigenvalues of a tensor, the normalization without complex conjugation ensures ...


0

The cops should always win. Here is a sketch of a proof. One cop (for short) trying to catch the thief crossing a segment: the cop can run on the segment at top speed towards the thief (provided the thief starts far enough and the cop's reaction time is 0). Clearly they will meet. Likewise a dense set of cops on a convex curve $C$: they can run towards the ...


0

In fact the short exact sequence above $\textit{doesn't}$ splits even in the case of retraction and $k=1$. What I mean is the following. Indeed, it does split as a sequence of $\mathcal{O}_S$-modules. Now take $S$ to be the diagonal in $M=S\times S$. Put $k=1$ and consider the sequence of corresponding $\mathcal{O}_M$-modules. This ext is so called Atiyah's ...


1

If the manifold $Y$ admits a global (meromorphic) vector field $W$, then around a regular point $p$ (that is, $W(p)\notin \{0,\infty\}$) one has a local holomorphic rectifying chart as you point out. Hence both $W$ and the starting vector field are locally conjugate near $p$. Therefore your question reduces to knowing whether $Y$ admits at least a global ...


3

The following book has recently appeared: Araminta Amabel, Arun Debray, Peter J. Haine (eds.), Differential Cohomology: Categories, Characteristic Classes, and Connections. Based on Fall 2019 talks at MIT's Juvitop seminar by: A. Amabel, D. Chua, A. Debray, S. Devalapurkar, D. Freed, P. Haine, M. Hopkins, G. Parker, C. Reid, and A. Zhang. (arXiv:2109.12250)


1

Maybe this one: A Mathematical Introduction to General Relativity


4

The isometric immersion that you describe above is the higher dimensional pseudosphere. Now, concerning your final question, I presume that you need to search about isometric immersions of the hyperbolic space $\mathbb H^n$ by means of a warped product representation (of $\mathbb H^n$) into the Euclidean space. Now, some additional things that you might be ...


2

Consider a 2-torus. The second homotopy is trivial, so the spheres can't feel the homology, or cohomology. The same for any manifold obtained by quotienting a simply connected manifold by a cocompact discrete group action, so all surfaces of genus 2 or more.


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