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6 votes
2 answers
196 views

Why are they called "screen" distributions?

If $V$ is a vector space and $g$ is a symmetric degenerate bilinear form on $V$, every complementary subspace to the radical ${\rm rad}(V)$ is called a "screen subspace" of $V$: we have an orthogonal ...
Ivo Terek's user avatar
  • 1,163
4 votes
1 answer
132 views

Problem arising in metrizability of connections: Simultaneously skewsymmetrizing matrices

Fact: Let $U$ and $V$ be two $ n \times n$ matrices with determinant $ 1.$ Assume that $S_1,S_2,....S_m$ are linearly independent $n \times n$ matrices such that $U^{-1}S_iU$ and $V^{-1}S_iV$ are ...
Mike Cocos's user avatar
1 vote
1 answer
142 views

Connection of the existence of Killing-Yano tensor and Killing tensor

Stephani states that in 4 dimensions a spacetime admits a non-reducible Killing-Yano tensor only if the Weyl tensor either is of Petrov type D or vanishes. Does this imply that the spacetime also ...
eriugena's user avatar
  • 679
1 vote
0 answers
84 views

A problem of defining addition in a Quotient space

Let $\mathcal{C}$ be the space of all parametric curves $x:[0,1]\rightarrow \mathbb{R}^2$. Let the set of all re-parameterizations of curves is $\Gamma = \{\gamma : [0, 1] \rightarrow [0, 1]| \gamma (...
Janak's user avatar
  • 213
0 votes
0 answers
143 views

A Riemannian manifold with a non-degenerate metric and an inner product $u_{\beta} u^{\beta}=1$

The question is: given a Riemannian manifold with a non-degenerate metric g and an inner product $u_{\beta}u^{\beta}=1$, is $\nabla_{\mu} (u_{\alpha}u_{\beta})=0$ without demanding the trivial ...
Kolten's user avatar
  • 9