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Visualizing higher dimensional spheres

  1. Jun 26, 2014 #1
    Visualizing a higher-dimensional sphere seems impossible to me. The best that can be done is to come up with several different representations, each of which is inaccurate in a different way.

    A property of n-spheres is that they have n/2 planes of rotation, each at right angles to the other. Amazingly even in a rigid rotation the planes can have completely different periods of rotation.

    The easiest case is the 3-sphere, which is embedded in Euclidian four-space. To simplify things, let's set the two periods of rotation to be equal. Instead of a plane of rotation we can consider a circle of rotation, which is just the intersection of the plane of rotation with the sphere. The key properties are

    • Two circles of rotation
    • The two circles of rotation are at right angles to one another
    • Each point on each circle is at the same distance from the each point on the other circle.
    • Every point on the 3-sphere revolves in a circle. Each such circle has the same radius

    Imagine an ordinary cylinder in 3D, with the usual axis of rotation drawn as a line. There is a circle at each end where the surface takes its right angle bend. Identify each circle with one of the two 4D circles of rotation, and identify that 3D axis of rotation line with the point at the center of the 3-sphere. Now to get the planes at right angles, curve the cylinder 90 degrees. If we measure the distance between the circles in angular units, then we've satisfied all four criteria.

    The model is geometrically meaningless and doesn't scale up to higher dimensions. Let's try for something a bit more geometric. In a 30-sphere the area near the circles of rotation is small and the maximum area is at a sort of "latitude" which is half way between them. We can make a model that is at least roughly accurate like this.

    Imagine an ordinary cylinder in 3D. There is a circle at each end where the surface takes its right angle bend. Identify each circle with one of the two 4D circles of rotation.
    Then shrink both of these circles to a point, thus transforming the cylinder into a 2-sphere. Note that every plane that passes through the origin of this sphere defines an orbit in 4D, except for the planes that pass through the circles of rotation which we forced to be points. Since these two points are each zero dimensional, no angle between them is defined, so we can say that they are at right angles to one another :-).

    This model has the "advantage" that it looks like the familiar 2-sphere, with the difference that there are only 90 degrees of latitude instead of 180. We have a well-defined many-to-one mapping between the between the 3 and 2 spheres, because we can describe each point in each with three polar coordinates. Each orbit in the 2-sphere has the same tilt and a phase as the corresponding orbit in the 3-sphere. Details later. It isn't an isomorphism because the circles can intersect. We can fiddle with the model to get an isomorphism by giving each orbit a unique altitude above the surface of the 2-sphere. This altitude can be in a tiny range so that it looks identical. But there is another way. Instead we can animate the model so that while points can occupy the same space momentarily it will still be obvious which point is which by the way they move.

    Now let'svisualize the two rotations with different periods. Each orbit can rotate in two different ways. Each orbit is a circle so it can revolve around its own center. Each orbit can also revolve around the axis of the sphere, which is called precession. The periods of those rotations can be anything, and those rotations can even be in opposite directions.

    "Each orbit in the 2-sphere has the same tilt and a phase as the corresponding orbit in the 3-sphere." Instead of R^4 let's use two unit discs in C^2. Each point is x and y. x = ze^it, y = (1-z^2)e^iu, with 0<=z<=1. For each circular orbit t is radians along the orbit, u is phase, and tilt is arcos(z).
  2. jcsd
  3. Jun 26, 2014 #2
    There are some attempts to visualize a 3-sphere, here
  4. Jul 3, 2014 #3
    Here's a better way to visualize rotation of a 3-sphere.

    The 3-sphere can be parameterized with polar coordinates as sin(x)e^it, cos(x)e^iu, with 0<=x<pi/2, -pi<=t,u<pi. Map to the three axis of R^3 using x, sin(x)t, and cos(x)u.

    One may think of this as specifying a rectangle for each value of x, with the rectangles stacked along the x axis. For each such rectangle rotation causes the points to move in a diagonal direction. This direction is the same for every point in that rectangle. Each rectangle is topologically a torus, so movement over an edge means reentry over the opposite edge.

    I can't imagine what this would look like.
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