# Tag Info

3

Of course, rays in $\mathbb R^n$ typically do not stay within bounded distance from each other, so that interpretation isn't available. The inequality with three products and fixed $\delta$ no longer speaks of the shape of triangles as much as of their size, since the Gromov product is multiplied by $\lambda$ under the scaling $x\mapsto \lambda x$. (Such ...

3

$U(p,q)$ is the symmetry group of the Hermitian form of type $(p,q),$ defined by $$|| (z_1,\ldots,z_n)||^2 = |z_1|^2 + \cdots + |z_p|^2 - |z_{p+1}|^2 - \cdots - |z_{p+q}|^2.$$ Let's consider the case $p = n, q = 1$. Any vector in $\mathbb C^{n+1}$ can have positive, zero, or negative norm, and pretty clearly this depends only on the complex line spanned by ...

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I have a version without any integrals: four acute angles related by four equations, but not an easy system to solve. Then your area is a simple function of those angles. If it will help, i can type it in. By the way, the calculation is intrinsic. The upper half plane or disc just makes everything worse. Meanwhile, looking at your hyperbolic network ...

2

You will not make it that way, what your professor showed you is how to make an equiangular hexagon (all angles are equal in measure) in a Beltrami Klein disk model , you can transform that into an equiangular hexagon in a Poincare disk model, but that is a later worry. The first step is to get a regular hexagon in a Beltrami Klein disk model devide the ...

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Yes, the geometric classification of surfaces tells us that a simply connected Riemannian surface $S$ must be (up to diffeomorphism) the sphere $S^2$, the complex plane $\mathbf{C}$, or the hyperbolic plane $\mathbf{H}$. Given that $\mathbf{H}$ is the only one of these with negative curvature, $S$ must be the hyperbolic plane.

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I'll choose the half plane model for this. Wikipedia already tells us A circle (curve equidistant from a central point) with center $\langle x,y\rangle$ and radius $R$ is modeled by a circle with center $\langle x,y\cosh R\rangle$ and radius $y\sinh R$. I'll choose hyperbolic centers $(0, 1)$ and $(0, \cosh d + \sinh d)$, so I have these euclidean ...

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The unitary group of signature $(n,m)$ is the group of matrices preserving a Hermitian form of signature $(n,m)$, and is isomorphic to this group preserving a particularly simple such form (a diagonal one): \operatorname{U}(n,m) = \left\{ \begin{bmatrix} A & B \\ C & D \end{bmatrix} \in \operatorname{GL}_{n+m}(\mathbb{C}) : AA^* - BB^* = I_n,~ ...

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$\frac{1}{y}(dx^2+dy^2)$ is shorthand for $\frac{1}{y} (dx \otimes dx + dy\otimes dy)$. This is a bilinear form on the tangent space (Really it lives in a tensor bundle but this seems a bit grand for such a simple object). It eats two tangent vectors and spits out a number, and this assignment is linear with respect to each of the two tangent vectors. ...

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A hyperbolic metric on n-dimensional smooth manifold M is a Riemannian metric g in M which is locally isometric to the standard Riemannian metric on the hyperbolic n-space $H^n$. Locally isometric here means that every point in M admits a neighborhood U so that the restriction of g to U is isometric to an open subset of the hyperbolic n-space. Such metric g ...

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