# Question about definition of a hypersurface

• A
tm33333
In my notes on general relativity, hypersurfaces are defined as in the image. What confuses me is that if f=constant, surely the partial differential is going to be zero? I'm not sure if I'm missing something, but surely the function can't be equal to a constant and its partial differential be non-zero?

thanks.

#### Attachments

• Screenshot 2021-05-11 at 21.05.25.png
11.2 KB · Views: 49

## Answers and Replies

2022 Award
It's requiring that ##\partial_af## be non-zero everywhere, then saying the subset of points with the same value of ##f## define a hypersurface. Analogously, you can define a function ##f(x,y)## on a two dimensional Euclidean plane and the lines of constant ##f## are the contour lines (1d analogues to 3d hypersurfaces). The gradient on a contour isn't zero, it is perpendicular to the contour line.

(Note that geographical contour lines can close, but a closed contour line encloses at least one point where the gradient is zero, so the definition of a hypersurface excludes this possibility).

tm33333
Staff Emeritus
Homework Helper
Gold Member
Well, the requirement that ##\partial_a f=0## everywhere is a bit strict. It is sufficient that it is non-zero at the hypersurface being described by the particular constant. (Although you will need the full requirement if you intend to make a foliation of the manifold.)

As an example, consider the sphere in standard Euclidean space with ##f = x^2 + y^2 + z^2##. For ##R>0##, ##f = R^2## defines a sphere of radius ##R##, which is a level surface of ##f## in ##\mathbb R^3##.

tm33333 and Ibix
tm33333
Thank you both. That definitely clarifies things!

Orodruin
Mentor
Moderator's note: Thread title edited to be more descriptive of the specific question.