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March 27[edit]

Lattice of quotient groups[edit]

Instead of using normal subgroups, we'll think of all partitions of a group corresponding to congruence relations. If I partially order them by Q1 is smaller than Q2 iff Q1 is finer than Q2, does the set form a complete lattice? Subgroups do... Money is tight (talk) 07:06, 27 March 2010 (UTC)[reply]

A congruence relation on a group is the same thing as a normal subgroup, isn't it? Algebraist 10:48, 27 March 2010 (UTC)[reply]

A partition is uniquely determined by the normal subgroup of elements congruent to the identify, so I think we can just look at those. A partition is finer than another one if the corresponding normal subgroups are contained in each other (one-line proof omitted). The converse is obviously true (the normal subgroup is one of the parts). Therefore, the set of partitions of a group will be a complete lattice iff the set of normal subgroups of a group is one, which it is. So the answer to your question is: yes. --Tango (talk) 20:29, 27 March 2010 (UTC)[reply]

maruschka doll and half-life - fractals?[edit]

Are these things fractals? What if you find two dolls within each maruschka doll, would it be a fractal? And the half-life of radioactive materials, if I plot it, is it a fractal?--Quest09 (talk) 18:32, 27 March 2010 (UTC)[reply]

I'd say the nesting dolls are fractal, especially if each is identical to the others, but they aren't always like that: [1]. I'm not quite following the half-life argument. Are you talking about the decay of radioactive elements ? There has to be some pattern which repeats at different scales, and I'm not sure if a radioactive decay curve is complex enough to be called a "pattern". After all, a straight line could be considered fractal if you wanted to include it. StuRat (talk) 18:43, 27 March 2010 (UTC)[reply]

I think a fractal is a set whose fractal dimension is not an integer. 66.127.52.47 (talk) 01:53, 28 March 2010 (UTC)[reply]

The first four iterations of the Koch snowflake

I don't agree with that. If we look at the illustration to the right, that shows the first 4 iterations of a fractal, with ~~dimensions~~ iterations 1-4 (or is it 0-3) ? I wouldn't normally call just an equilateral triangle a fractal, because there isn't yet any repetition of the pattern on a smaller scale. But the shapes after that are fractals, IMHO, and each has an integer fractal dimension. StuRat (talk) 14:16, 28 March 2010 (UTC)[reply]

All iterations have fractal dimension 1, and the limiting curve has dimension 1.26... . I think usually only the limiting curve is considered a fractal, not any finite iteration. Taking e.g. the 3rd iteration, it has no self similarity - it contains no smaller copies of itself, but rather of the 2nd iteration. -- Meni Rosenfeld (talk) 14:37, 28 March 2010 (UTC)[reply]

How is that 1.26 calculated ? StuRat (talk) 19:28, 28 March 2010 (UTC)[reply]

In each iteration, the scale is decreased x3 and the relative length is increased x4, so it's

\log _{3}4\;\!

. Some more information is available in fractal dimension and Koch snowflake. -- Meni Rosenfeld (talk) 19:44, 28 March 2010 (UTC)[reply]

More specifically, the Koch snowflake has a Hausdorff dimension of log(4)/log(3), which is approximately 1.26186. As pointed out below, having non-integer Hausdorff dimension is not sufficiently inclusive to be a defining characteristic of fractals, as there are fractals with an integer Hausdorff dimension - see list of fractals by Hausdorff dimension for examples. But self-similarity is too inclusive, as a straight line and an exponential curve are self-similar but would not usually be considered to be fractals. Gandalf61 (talk) 09:37, 29 March 2010 (UTC)[reply]

I thought that the Peano curve was considered by most to be a fractal, though it has an integer dimension in the limit. -- 174.31.194.126 (talk) 21:08, 28 March 2010 (UTC)[reply]

Rather than fractional Hausdorff dimension, a better working definition of a fractal is that its Hausdorff dimension is strictly larger than its topological dimension. This includes all sets with fractional Hausdorff dimension because the topological dimension is always an integer less than or equal to the Hausdorff dimension. It still excludes some of the examples (plane-filling curves, because the property does not distinguish between a curve and its image, or the Mandelbrot set, because it has nonempty interior), but it does include several fractals with integer dimension (such as the Smith–Volterra–Cantor set or the boundary of the Mandelbrot set).—Emil J. 15:59, 29 March 2010 (UTC)[reply]