Two dimensions and beyond (or at least between)

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I fear that this will be a shaggy dog post. One of my students posed an interesting question the other day, and I found myself surprised by the answer. So here’s the quick version of today’s post: What’s the dimension of a tetrix? A tetrix is also known as a Sierpinski tetrahedron, and indeed it is like a three-dimensional version of the Sierpinski triangle, except that, being a fractal, its dimension is less than three. There’s a picture below (which I think of as upside-down), and also a cool java applet version here which you can spin around.

As background, the dimension of a fractal is usually pretty weird — they have so many holes that they’re not quite two or three dimensional. So to calculate its dimension, we have to think about exact what dimensions are. If you take a square or a triangle, for example, and double the length of the outside edge then you can fit 4 (which is 22) of the original pieces in the new object. And if you triple the length of the outside edge, you can fit 9 (that is, 32) of the original object in there.

In other words, assuming the original side length is 1, if the new outside edge has length S and you can fit N of the original objects in there, it turns out that N=S^2. The 2 in the exponent is related to the dimension being two.

Now look at three-dimensional objects. If you double the length of the outside edge, you can fit 8=2^3 little cubes in there. And if you triple the length of the outside edge, you can fit 27=3^3 cubes in there.

Once again, treating the original cube as being 1×1×1, if the new outside edge has length S and you can fit N of the original objects in there, it turns out that N=S^3. And this time the 3 in the exponent is because it is three-dimensional. In general, if D is the dimension, then N=S^D. This description of dimension is known as the Hausdorff dimension.

So what about fractals? Let’s look at the famous Sierpinski Triangle, below on the left. It’s made from smaller versions of itself, but because it has a big old hole in the middle, you only put three Sierpinski triangles together to form a larger version. That’s easier to see in the colored version on the right.

In other words, if S=2 then N=3 and we’re faced with 3=2^D. Taking the natural log (or any log) of both sides gives ln(3)=ln(2^D), but that’s just D \cdot ln(2), giving a dimension of D=\frac{ln(3)}{ln(2)} \approx 1.58. So even though the Sierpinski Triangle lives in the two-dimensional world, it has so many holes in it that it’s almost half a dimension shy of that.

Now look at the Sierpinski Pyramid. It looks like the Pyramids of Giza, except with lots of holes.

You make a larger version by putting together five smaller versions: if S=2 then N=5. And that leads to 5=2^D. As before, taking the natural log (or any log) of both sides gives ln(5)=ln(2^D)= D \cdot ln(2) so D=\frac{ln(5)}{ln(2)} \approx 2.32.

Which brings us back to the tetrix/Sierpienski tetrahedron. Larger versions are formed by putting together four smaller versions (one on each corner). So what’s the dimension?

Like I said, a bit of a shaggy dog post. But I still found it harder to wrap my brain around than the dimension of 2.32 in the pyramid. And it made me wonder: what’s the smallest possible dimension for a fractal that lives in the third dimension?

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5 Responses to “Two dimensions and beyond (or at least between)”

  1. Batman Says:

    Well, the Cantor dust has Hausdorff dimension log(8)/log(3), or about 1.89.

  2. Ξ Says:

    Cool! I was also thinking that it should be possible to take a line and turn it into a fractal (like Cantor Dust), but make it bendy so the line lives in 3D. Thinking about the Cantor Dust, by varying that (breaking it into more pieces per side, like 33 or 151 and only picking the corners) it should be possible to get a dimension of epsilon.

    Hmmm….what if the fractal were connected? (Is the Sierpinski Triangle/Pyramid/Tetrahedron connected? I think the portion that’s removed would need to be an open set for that to happen.)

  3. TwoPi Says:

    I am horrid at geometry, and have very little spacial intuition, so forgive me if this is daft.

    I’m trying to wrap my brain around Hausdorff dimension epsilon, for a nonplanar figure. I can see 1+epsilon, perhaps, but struggle with the idea of dimension below 1 for a nonplanar figure.

    Can a Banach-Tarsky style construction yield a nonplanar set with Hausdorff dimension 0?

  4. Ξ Says:

    I’m not sure about the last question, but here’s how to make 3d figure that’s less than 1d (so to speak).

    Cantor dust is formed by taking a cube, breaking it into 27 (3x3x3) pieces, and removing all but the 8 corners. Then do the same thing to each of those corners, etc. The dimension part comes in because when the side length is multiplied by 3, there are 8 copies of the original (the 8 corners), so the dimension is ln(8)/ln(3).

    But now think of starting with a cube and breaking it into 1000 (10x10x10) pieces and removing all but the 8 corners. Then do the same thing to each of those corners. This means that if you make a copy of the original that is 10 times as large on each side, there will only be 8 copies of the original (in the corners) so the dimension is ln(8)/ln(10), which is less than 1. There’s nothing special about the 10, either — it could be larger.

    Now that I think about it, by breaking it into 512 (8x8x8 ) pieces and proceeding as above, you’d end up with something of Hausdorff dimension ln(8)/ln(8)=1. And that’s weird to me too — I feel like it should mean that if I take a thin (1-d) pieces of wire, I should be able to make that entire figure out of it.

  5. TwoPi Says:

    Thank you, that makes perfect sense to me.

    Now that I’ve finished my coffee and paced a bit, let me recast my second question. The Banach-Tarski paradox shows (e.g.) that one can assemble two unit spheres into a single unit sphere, using only rigid motions. Why doesn’t this imply that the unit sphere has Hausdorff dimension infinity? Because it seems to say that one can use two spheres to build a sphere which is 1 times the size of the original, leading to 2 = 1^D.

    Further, I seem to recall that the B-T construction can also be used to reassemble a sphere of radius r into another sphere of radius R, for any r and R. That would seem to imply that the Hausdorff dimension of the sphere is 0.

    Didn’t Hausdorff himself prove that one can take the unit interval [0, 1], partition it into countably many sets, and reassemble them via rigid motions to get the interval [0, 2]? If so, I wonder what Hausdorff’s take on that example was? Does this construction raise concern about the dimensionality of [0, 1]?

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