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Can it only be done in Euclidean space? Doesn't Euclidean space only model extrinsic curvature?

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No, Euclidian space is not necessary. You can "model" intrinsic curvature using the beautiful language of Riemannian geometry, whose great triumph was formulating a vocabulary that lets you talk about the curvature of a space without making reference to an extrinsic space in which the curved space is embedded: hence the term intrinsic.

This is crucial to general relativity, since embedding a curved 4-space in flat Euclidean space requires that the Euclidean space be ten-dimensional, and dealing with that embedding would suck.

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I'm not quite sure what you mean by the term "model" in this context, but:

If a space is a Euclidean space, in the sense that it has a Euclidean metric, then its Levi Civita connection (the connection compatible with its metric) has no intrinsic curvature (for example a flat plane is like this). However, it may be given some extrinsic curvature by means of an embedding into a higher dimensional space (the flat plane may be rolled up into a cylinder in $\mathbb{R}^3$).

But if you were a two dimensional organism living on the cylinder, you couldn't detect this extrinsic curvature by locally measuring angles and distances.

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  • $\begingroup$ Very clear. How can one relate both metrics? $\endgroup$ – stringparser Apr 11 '12 at 13:27

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