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a 2-sphere is a normal sphere. A 3-sphere is

$$ x^2 + y^2 + z^2 + w^2 = 1 $$

My first question is, why isn't the w coordinate just time? I can plot a 4-d sphere in a symbolic math program and animate the w parameter, as w goes from .1 to .9:

4d surface

Isn't that what it means to have a 4th dimension? Just add time?

Apparently not.

This image is from wikipedia,

3-sphere

The caption says that this is a "Stereographic projection of the hypersphere's parallels (red), meridians (blue) and hypermeridians (green)". I don't get that at all. What is a parallel, meridian, hypermeridian? Why can't we just

There is an article here which talks about the 3-sphere in terms of Poincare's Conjecture.

Here there is an image of the "Hopf fibration of the 3-sphere".

hopf

This looks very cool and there are formulas that break down the Hopf fibration into understandable algebra, but what does this mean, at a high level?

Edit: I am looking at the dimensions videos and they are actually very good.

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Time is one way to add another dimension, but not always the most interesting way to visualize it (since you can't see everything at once). That's why certain projections are sometimes used, as in this case. –  Alex Becker May 21 '12 at 1:07
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You could view the unit circle $\Bbb S$ as two dots on the $x$-axis splitting from the origin to the endpoints $\pm 1$ and then coming back, using time as a dimension. But that is hardly "the" way to visualize a circle. Also, you could view a cube as a square standing still for an interval in time. Hardly enlightening, and if you rotate the cube the corresponding "animation" (constructed with cross-sections as frames) will be drastically harder to piece together into a solid figure. "Time" is merely a crutch we stand on to help us visualize. –  anon May 21 '12 at 1:09
    
The square of the space-time "distance" is $x^2+y^2+z^2-c^2t^2$, so it is not directly related to the distance function on $\mathbb{R}^4$. –  André Nicolas May 21 '12 at 1:12
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I think you're missing the fundamental point that a four-dimensional space doesn't actually come equipped with four distinct "dimensions". Being four-dimensional just means that you can always find four linearly independent directions, but there isn't a fixed set of directions that are canonically labeled as "the first dimension", "the second dimension", and so on. So thinking that time is "the fourth dimension" is quite misleading. –  Rahul May 21 '12 at 1:58
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Time isn't the fourth dimension, it's a fourth dimension. –  Blue May 21 '12 at 8:40

2 Answers 2

$4$-manifold topologists routinely think of the fourth dimension as time to help aid visualization. A favorite example of mine is to illustrate that the intersection of two planes in $\mathbb R^4$ can intersect in a single point (which is obvious from the algebra.) Think of a movie with $3$ dimensional frames, and a single line that stays put the whole time. This represents a plane in $\mathbb R^4$, since the line sweeps out a plane as it moves through time. Now put a plane that hits the line in a point in one of the 3D time slices. This gives two planes in $\mathbb R^4$ meeting in a point. On the other hand, while this can be quite useful, it also is highly asymmetric, as you have chosen one direction to be privileged, and from the point of view of 4-dimensional space, no direction is privileged over the others (unlike in Lorentzian geometry). So you can often miss crucial features of what's going on by looking at things as a movie.

Now, as far as $S^3$ is concerned, your animation is just fine. You can think of the $3$ sphere as a movie starting with a point which grows to a unit $2$-sphere which then shrinks back to a point. This can be useful. Usually I think of $S^3$ as the one-point compactification of $\mathbb R^3$. (This is where stereographic projection comes in.) If you delete a point from a circle, you get a space homeomorphic to $\mathbb R$, and similarly deleting a point from $S^2$ yields a space homeomorphic to $\mathbb R^2$. This actually generalizes, and the formulas are not that hard to write down, to show that $S^n\setminus\{pt\}\cong \mathbb R^n$. So from a topological perspective you're not missing much by just looking at $\mathbb R^3$. Geometrically, this projection introduces a lot of distortion, so has to be considered carefully. For example, the Hopf fibration is a way to write $S^3$ as a union of geometric circles all of the same size, but when you look at the Hopf fibration translated into $\mathbb R^3$, the circles appear to be of different sizes, and one even turns into a line going through $\infty$.

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No specific direction is privileged in Lorentzian geometry either. What happens there is you have several classes of directions, e.g. in 3+1 you have two classes of time-like vectors, two classes of light-like vectors, and one class of space-like vectors. –  Hurkyl May 21 '12 at 8:41
    
Yes, I was being imprecise. –  Grumpy Parsnip May 21 '12 at 14:55

I have often played around with the idea of trying to visualize higher-dimensional objects, and I think I've gotten a pretty decent hold on how to think about spheres.

I first suggest starting with the lower dimensional spheres that you're already familiar with. Consider $S^{2}$ for example. It is the set of all points in $\mathbb{R}^3$ that satisfy the relation $$x^2+y^2+z^2=1.$$ Now, if we think of the $z$-dimension as time (I'll now refer to it as $t$), then we would have the relation $$x^2+y^2=1-t^2,$$ which would give us a family of circles (copies of $S^{1}$) with radii of length $\sqrt{1-t^2}$. So, if we consider $t=0$ to be our "initial position" in time, then we have a circle in $\mathbb{R}^2$ of radius $1$. Now, if we move "forward" or "backward" in time, then the quantity $\sqrt{1-t^2}$ will begin to shrink, until at times $t=1$ and $t=-1$ (one unit of "time" forward or backward) when we have the relationship $$x^2+y^2=0,$$ which is only satisfied by a single point in $\mathbb{R}^2$, namely $(0,0)$. So in this way, we can visualize $S^{2}$ as a number of copies of $S^{1}$ all glued together in a higher, third-dimension.

top view of $S^{2}$side view of $S^{2}$

We can also do a similar process to construct $S^{1}$, which is the set of points in $\mathbb{R}^2$ that satisfies the relation $$x^2+y^2=1.$$ Thinking of the $y$-axis now as "time", we are left with the relation, $x^2=1-t^2$ or $$x=\pm\sqrt{1-t^2},$$ a subset of $\mathbb{R}^1$. So again we see that at time $t=0$, we have that $x=\pm 1$, which is exactly what we refer to as $S^{0}$. And again, we see that as we move forward or backward in time we have that $|x|$ becomes smaller, until we reach times $t=1$ and $t=-1$ where we have the only solution being the single value $x=0$. So in this way, $S^1$ can be viewed as a number of copies of $S^0$ being glued together in a higher, second-dimension.

s1 and copies of s0

Finally we come to $S^{3}$, which is the set of points in $\mathbb{R}^4$ satisfying the relation $$x^2+y^2+z^2+w^2=1.$$ Now when we think of the 4th spacial-dimension, $w$, as time, we get $$x^2+y^2+z^2=\sqrt{1-t^2},$$ which is a copy of $S^{2}$ in $\mathbb{R}^3$ of radius $\sqrt{1-t^2}$. So as times moves backward and forward, we find the radius of the sphere shrinking until we are left with the point $(0,0,0)$ at times $t=1$ and $t=-1$. Hence we may think of $S^3$ as nothing more than a number of copies of $S^2$ glued together over some 4th spacial dimension.

We can even get some kind of intuition as to the "look" of $S^4$ and $S^5$. Pretend, if you will, that you have a box which represents $[-1,1]^3$ (the unit cube), and inside this box is a copy of $S^2$. Now, as we move the box left and right, we are moving our portion of $\mathbb{R}^3$ (and our copy of $S^2$) along the $w$-axis. So as the box shifts to the left or right, the radius of the sphere inside the box will begin to shrink, until it becomes only the point $(0,0,0)$ at the origin, when we've moved the box $1$ full unit to either side. This is exactly the same analogy we talked about above. There are an uncountable number of $2$-spheres, each existing at a unique point along the $w$-axis, which "glue" together to give us a $3$-manifold (the $3$-sphere) embedded in $\mathbb{R}^4$.

However we are free to move our box not only left/right, but also forward/backward and up/down. These would correspond to the $v$- and $u$-axes of $\mathbb{R}^6$, with $S^5$ given by the equation $$x^2+y^2+z^2+w^2+v^2+u^2=1.$$ This means that our copy of $S^2$ will shrink, the farther away we move our box from it's initial position, until only the point $(0,0,0)$ is left; which will occur anywhere where $w^2+v^2+u^2=1$ (that is, $1$ unit of distance radially outward from the box's initial position.)

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