# Intrinsic curvature.

1. Jan 25, 2009

### matheinste

Hello all.

I understand that a two dimensional surface can have curvature without it being referred to a higher dimension. So that a surface such as that of a sphere does not need to refer to a third dimension to determine its own intrinsic curvature and so on for higher dimensions.

Can a one dimensional space/surface such as a line have intrinsic curvature.

Matheinste.

2. Jan 25, 2009

### Hurkyl

Staff Emeritus
All one-dimensional manifolds are homeomorphic to a line or to the circle*.

Let's start with the line with a Riemann metric. You can choose a base-point and a direction, parametrize the line by directed arclength. Making the change-of-coordinates to use arclength as a coordinate, the metric is the constant tensor field ds ds. Therefore, all Riemann lines are isometric to a (possibly infinite) open interval in R.

For the circle, I believe the exact same idea shows that all Riemann circles are isometric to a circle in R².

In other words, the only diffeomorphism-invariant properties of a one-dimensional Riemann manifold are
1. Whether it's a line or a circle
2. Its length
and if it's an infinite line,
3. Whether it's infinite on both ends, or just one end

* There's also the long ray and the long line, depending on your exact technical assumptions, but let's ignore those.

Last edited: Jan 25, 2009
3. Jan 26, 2009

### matheinste

I have digested what you said and understand it.

My non rigorous train of thought was that i believe given a surface/2 dimensional manifold we can decide if it is curved by drawing a circle and measuring its diameter and its circumference, all measurements being taken upon the surface, and comparing the ratio with pi.

Is there a test of this nature for a line.

Matheinste.

4. Feb 4, 2009

### wofsy

curvature is a two dimensional concept - curvature for curves is an extrinsic idea and may not be derived from the metric. There is no idea of intrinsic curvature for a curve

5. Feb 4, 2009

### matheinste

Hello wofsy.

That's what i thought but was not sure of.

Thanks.

Matheinste.