# Is a semi-Euclidean space possible?

Does exists a geometry (3d for example) which is Euclidean in 2 dimensions (x and y coordinates) and non-Euclidean when the third dimension (z) is taken into account? In other words a space where it is possible to construct a square but not a cube?

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An Answer to the first subquestion(v1) is Yes, take e.g. a Cartesian product of a 2D Euclidean plane and a 1D non-Euclidean space. –  Qmechanic Aug 16 '12 at 20:18
What exactly do you mean by non-Euclidean? Do you mean general Riemannian metrics, or a nondefinite Lorentzian one? –  Emilio Pisanty Aug 17 '12 at 0:42
Qmechanic is right, it's a purely mathematical construct. Or did you mean whether such spaces are encountered in physics? –  Yrogirg Aug 17 '12 at 4:07
@Qmechanic "Yes, take e.g. a Cartesian product of a 2D Euclidean plane and a 1D non-Euclidean space" What is a 1D non-Euclidean space? Naively I would thought that there is no geometry in 1D world. –  Leos Ondra Aug 17 '12 at 12:00
@Yrogirg Both - mathematics as well as physics. –  Leos Ondra Aug 17 '12 at 12:02

The question can be interpreted in a few different ways: both what you mean by having the $xy$ slices of the 3-space Euclidean, and what you mean by it being non-Euclidean in the $z$ direction.

The suggestion of taking a Euclidean plan and cross it with a non-Euclidean curve only works if you look at the global structure. Any (smooth) curve is locally Euclidean, i.e. you can straigten it out, so locally the 3-space will be Euclidean. However, if you take $\mathbb{R}^2\times S_1$, i.e. Euclidean plane times a circle, it will not be the same as Euclidean $\mathbb{R}^3$.

However, I suppose what you had in mind was local geometry, not global geometry. Normally, by saying locally Euclidean, we just mean that in an infinitessimal neighbourhood of any point, the space is Euclidean. I'm not completely sure if this necessarily will always allow you extend this to a bigger region (of positive size, although perhaps small).

Another question is if the $xy$ slice itself is Euclidean, i.e. has an Euclidean metric, or if the embedding into the 3-space should also be non-curved.

Let me give you a few examples and discuss them. The way I describe the metric, i.e. distance function, on 3-space parametrised by $(x,y,z)$ is on the form $$ds^2=dx^2+dy^2+dz^2 \iff s'=\sqrt{(x')^2+(y')^2+(z')^2}$$ where $ds$ is the infinitesimal distance between $(x,y,z)$ and $(x+dx,y+dy,z+dz)$, and the length of a path $(x(t),y(t),z(t))$ is computed by integrating $s'$. This is the Euclidean metric on 3-space.

An alternative metric I can make is $$ds^2=f(z)^2\cdot(dx^2+dy^2)+dz^2$$ which means the $xy$ plane is scaled by a factor $f(z)$. Note that if we enter new coordinates $u=f(z)\cdot x$, $v=f(z)\cdot y$ and restrict attention to the surface with constant $z$ (i.e. all $dz$ terms disappear), we find the metric on the surface $S_z$ to be $ds^2=f(z)^2\cdot(dx^2+dy^2)=du^2+dv^2$, so this is still a Euclidean plane. However, in the function $f(z)$ is non-constant, the 3-space is curved. What's more (and perhaps less intuitive) is that the $S_z$ planes lie curved inside the 3-space. One way to see this is that the shortest path between two points in $S_z$ will tend to leave $S_z$ and take a short-cut through the side of $S_z$ in 3-space for which $f(z)$ is lower: on this side, distances are shorter, just as the shortest path between to points on a circle in the plane will pass through the interior of the circle rather move along the circle.

Another metric we can make is $$ds^2=dx^2+dy^2+h(x,y,z)^2\cdot dz^2$$ for some arbitrary positive function $h(x,y,z)$. Again the constant $z$ slices, $S_z$, are Euclidean planes: this time with no scaling involved. Instead, distances along the $z$ direction are scaled. This time, the embedding of the $S_z$ planes itself is non-curved: not only is the metric on $S_z$ equal to $ds^2=dx^2+dy^2$, but there are no short-cuts by passing out of $S_z$.

Mathematically, the curvature is described locally by the Riemann curvature tensor. This is based on the idea of parallell transportation of vectors: i.e. if I start at a point $P$, pick a direction $u$ (described by a tangent vector) and another vector $w$, I can parallell transport $w$ in the $u$ direction: say to a point $P+\epsilon u$. Parallell transport implies that there should be no rotation, and the way this is done depends on the metric in such a way that it is independent of the coordinates chosen. If we pick a second direction, $v$, we can parallell transport $w$ along a closed curve in the $uv$ plane: first in the $u$ direction, then in the $v$ direction, them back to the starting point by going first in the $-u$ direction and then in the $-v$ direction. What happens then is that $w$ may end up rotated. The direction in which the $w$ vector is rotated is denoted $R(u,v)(w)$ where $R(u,v)$ is a rotation acting on the tangent vectors at any point.

E.g. imagine starting at the North Pole with a vector pointing in a particular direction, parallell transport it down to the equator (vector pointing south), then along the equator for a distence (still pointing south), and then back to the North Pole. The vector will have rotated, and the angle it was rotated was the curvature of the Earths surface times the area of the region enclosed by the path. This is what the Riemann curvature measures, but locally.

In both my examples, the metric restricted to the $S_z$ surfaces became Euclidean: i.e. when ignoring the outside space. However, in the first of my examples, the curvature $R(x,y)$ (i.e. for parallell transportation in the $xy$-plane) of the 3-space along $S_z$ is non-zero; and it is not only that vectors in the $xy$-plane get rotated out of it, i.e. in the $z$ direction, but tangent vectors in the $xy$-plane also get rotated within the $xy$-plane. This caught me by surprise since I hadn't expected it. However, it is due to the curvature of the enclosed space. So if these $S_z$ sections are Euclidian is a question of what you meen be Euclidean in the context of a surface embedded in a curved 3-space.

In the last examples, $R(x,y)$ is zero for all tangent vectors. Curvature only becomes apparent when parallell transporting vectors out of the $xy$-plane: i.e. $R(x,z)$ and $R(y,z)$ are non-zero. So these $xy$-planes are Euclidean both in themselves and as subspaces of the 3-space.

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