Timeline for Riemannian metric on a level set of a smooth function on a manifold

Current License: CC BY-SA 3.0

11 events

when toggle format	what		by	license	comment
Jan 8, 2018 at 20:56	comment	added	Deane Yang		If you assume that the metric on $M$ is known, then the metric on $L$ is, for each $x \in L$, the restriction of the metric on $T_xM$ to the subspace $T_xL$. So the only issue is identifying the subspace $T_xL$. This, however, is the space of vectors $v \in T_xM$ such that $\langle dG(x),v\rangle = 0$. Another way to say this is that it is the subspace of $T_xM$ that is orthogonal to $\nabla G(x)$. This, I believe, is as explicit as you can get with the metric on $L$.
Jan 8, 2018 at 4:49	comment	added	Deane Yang		Maybe you’re ultimately interested in more than just the metric? What’s more interesting is how to compute the second fundamental form and Riemann curvature of the submanifold using $G$.
S Jan 7, 2018 at 18:36	history	suggested	Ali Taghavi		I add a tag
Jan 7, 2018 at 17:38	review	Suggested edits
S Jan 7, 2018 at 18:36
Apr 13, 2017 at 12:19	history	edited	CommunityBot		replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
Apr 3, 2016 at 9:44	comment	added	Learning math		@BenMcKay Yes assume I know that $DG$ has full rank at all points, so we cna apply the implicit function theorem, so we can express the level set locally as a graph.
Apr 3, 2016 at 7:52	comment	added	Ben McKay		Or use the implicit function theorem to write out coordinates $(x,y)$ in which your map is $G(x,y)=x$. Do you know if $G'$ has full rank?
Apr 3, 2016 at 7:51	answer	added	Raziel		timeline score: 3
Apr 3, 2016 at 7:50	comment	added	Ben McKay		You might try using a moving frame of orthonormal vector fields, with the first so many of them tangent to the level sets of $G$, and the rest perpendicular.
Apr 3, 2016 at 6:42	history	edited	Learning math	CC BY-SA 3.0	added 11 characters in body
Apr 3, 2016 at 6:37	history	asked	Learning math	CC BY-SA 3.0