properly discontinuously造句
造句与例句手机版
- A hyperbolic 3-manifold is a 3-manifold equipped with a properly discontinuously.
- This reduces the problem of studying space form to studying isometries \ Gamma of M _ K which act properly discontinuously.
- A subgroup \ Gamma \ subset G acts freely, properly discontinuously on X if and only if it is discrete and torsion-free.
- Classically, a Kleinian group was required to act properly discontinuously on a non-empty open subset of the Riemann sphere, but modern usage allows any discrete subgroup.
- Additionally, if \ gamma is loxodromic, then \ langle \ gamma \ rangle acts properly discontinuously and cocompactly on M \ setminus \ operatorname { Fix } _ M ( \ gamma ).
- More generally we see that any group G which acts properly discontinuously on a locally finite tree ( in this context this means exactly that the stabilizers in G of the vertices are finite ) is hyperbolic.
- In geometry, a Clifford Klein form is a double coset space, where is a reductive Lie group, is a closed subgroup, and is a discrete subgroup ( of ) that acts properly discontinuously on the homogeneous space.
- Let ? be a Fuchsian group of the first kind : a discrete subgroup of the M鯾ius group preserving the unit circle . acting properly discontinuously on the unit disk " D " and with limit set the unit circle.
- Note an interesting feature of this example : it acts properly discontinuously on a hyperbolic space ( the hyperbolic plane ) but the action is not cocompact ( and indeed G is " not " quasi-isometric to the hyperbolic plane ).
- In particular, if \ Gamma is any discrete subgroup in G ( so that it acts properly discontinuously by left-translations on G ) the quotient \ Gamma \ backslash G is a Riemannian manifold locally isometric to G with the metric g.
- It's difficult to see properly discontinuously in a sentence. 用properly discontinuously造句挺难的
- The Svarc-- properly discontinuously and with compact quotient ( such an action is often called " geometric " ) on a proper length space Y then it is finitely generated, and any Cayley graph for G is quasi-isometric to Y.
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