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January 24

E8 (mathematics)

Can someone explain in simple terms what this is? I'm particularly interested in the maths behind the left image (not digital!) and whether the image on the right (which I made) has any underlying similarity. Thanks SmartSE (talk) 00:12, 24 January 2012 (UTC)[reply]

The picture on the left shows a "mystic rose" with 30 points (see complete graph, though the rose has the diameters omitted, and seems to have smaller roses superimposed — — hereis the simple complete graph). Your construction is also a complete graph, but I can't see any significant similarities -- perhaps someone else can? Dbfirs 07:53, 24 January 2012 (UTC)[reply]

The picture on the left actually has 240 points corresponding to root vectors of the E8 root system. Its a projection from eight dimensional space to two dimensions. You could also think of it as a projection of the 4_21_polytope. Actually the Petrie polygon projection is used here which shows the eight rings of 30 points. --Salix (talk): 09:13, 24 January 2012 (UTC)[reply]

Ah ok, well that's something at least - now I know that it is a complete group. Why are all our examples circles and not squares? Do complete square groups have their own name? Can you explain how it is "a projection from eight dimensional space to two dimensions" - just the thought of that makes my head hurt! SmartSE (talk) 17:45, 24 January 2012 (UTC)[reply]

Are you sure you meant complete group? The phrase used about was complete graph. Clink the link t see the article. — Fly by Night (talk) 18:43, 24 January 2012 (UTC)[reply]

(edit conflict) It is difficult to be sure what counts as "simple terms" without knowing how much group theory you are familiar with. E₈ is a type of group called a simple Lie group. It can be associated with a finite group with 696729600 members. This finite group is also the symmetry group of a semi-regular polytope in 8 dimensions called the 4₂₁ polytope, which has 240 vertices and 6720 edges. And the left-hand image of the pair is a projection of vertices and edges of the 4₂₁ polytope onto 2 dimensions, in which the 240 vertices are arranged in 8 concentric rings of 30 vertices each. Gandalf61 (talk) 09:35, 24 January 2012 (UTC)[reply]

Ah, yes, thanks for the correction. I missed that! I'm not used to thinking in eight dimensions! Dbfirs 10:28, 24 January 2012 (UTC)[reply]

Sorry, I should have made that clearer - I know absolutely nothing about group theory and don't find our articles particularly helpful given that. I don't really get what's so special about E₈ and especially why it "can be associated with a finite group with 696729600 members". Does that means there are 696729599 similar patterns that can be made?! How's that calculated? SmartSE (talk) 17:45, 24 January 2012 (UTC)[reply]

You're at a real disadvantage then. The reason it (and some of its cousins) are special is because of certain group theoretical properties. Most people spend three or four years at university and still don't feel group theory. The honest, and brutal, answer is that you won't be able to understand or appreciate the details until you know quite a lot about group theory. Given that you don't know anything about group theory then you're in a bit of a bind. It's like asking someone in Norway to explain Norwegian literature to you in Norwegian. You have to be able to speak Norwegian first, and even those that speak Norwegian fluently might not appreciate certain types of Norwegian literature. (Mathematical theories are like literature: not everyone finds group theory beautiful. Some very able professional mathematicians don't particularly like group theory.) — Fly by Night (talk) 18:58, 24 January 2012 (UTC)[reply]

Hmm. Good analogy! I'll just have to carry on thinking them as pretty patterns and not much else then! SmartSE (talk) 16:27, 25 January 2012 (UTC)[reply]

I guess the best place to start is Group (mathematics) for basics of group theory. Rather that dive straight into looking at E8 it might help to look at a simpler ones like A₂. There are many different things called A₂ which are all related (see ADE classification). The Lie group is a set of 3 by 3 matrices with special properties (actually an infinite number of matrices), the Lie algebra is a different set of 3 by 3 matrices (which all have trace zero). From these you get a root system, for A₂ this just two vectors at 120° to each other. Now these two vectors at 120° define two planes at 120°, The reflections in these planes define a group with six elements: the identity, three reflections and rotations by ±120° this is the Weyl group and is the group of symmetries of a triangle. For E₈ the root system has 240 vectors in 8 dimensions, these define 240 planes and the Weyl group of the reflections in these planes has 696729600.--Salix (talk): 23:44, 24 January 2012 (UTC)[reply]

That makes some sense, but is still a little over my head... thanks for trying though! SmartSE (talk) 16:27, 25 January 2012 (UTC)[reply]

See David Madore's writeup on this topic which has appeared just two days after your question. – b_jonas 19:27, 28 January 2012 (UTC)[reply]

Note: he links to the interactive demonstration he's created which also has some explanation under it. – b_jonas 19:29, 28 January 2012 (UTC)[reply]

Looking for a graph theory term...

In an undirected graph with no self-loops, this is a path which:

is simple, except that it may (but is not required to be) a cycle,
has length at least 1, and
each vertex in the path has degree 2 if and only if it is not an end vertex.
- For the purposes of this definition, a cycle is considered to have either no end vertices or one end vertex—two end vertices would require two distinct paths of this type.

Is there a standard term for such a thing? « Aaron Rotenberg « Talk « 17:30, 24 January 2012 (UTC)[reply]

To help visualize the question a bit, this is basically a "straight line" in a graph, with no splits. « Aaron Rotenberg « Talk « 18:03, 24 January 2012 (UTC)[reply]

This is sort of like an ear in ear decomposition. The last ear added to the graph has these properties, although previous ones get more ears added on to them so they no longer have all vertices of degree 2. Rckrone (talk) 02:29, 26 January 2012 (UTC)[reply]

I don't know, but if there is, it could be mentioned as a link from Homeomorphism (graph theory). – b_jonas 19:22, 28 January 2012 (UTC)[reply]

Odds question

This is probably very elementary for someone who's knowledgeable in probability. Over the years (I'm not a regular player), I've bought 10-11 scratch tickets of one particular game. I looked on the back of the ticket I just played today and the odds of winning (including break even) are 1 in 4.442. I have never won. What are the odds of buying 10 tickets of a game whose odds are 1 in 4.442 and not winning any? Low, I would think, but I'm wondering if my outcome is very uncommon. 20.137.18.53 (talk) 20:16, 24 January 2012 (UTC)[reply]

The chances of not winning any are about 1 in 12.813, assuming you have given the numbers correctly, which is not clear, because at the very least you are misusing the word "odds". (Odds are given in the form "1 to X", not "1 in X".) Looie496 (talk) 20:26, 24 January 2012 (UTC)[reply]

Please forgive my terminology usage faux pas. Just looked at the ticket again and the word printed on it is "chances" but I misremembered it when I typed "odds" in the question above. 20.137.18.53 (talk) 20:36, 24 January 2012 (UTC)20.137.18.53 (talk) 20:30, 24 January 2012 (UTC)[reply]

For the mathematics behind Looie's figures, if your chance of winning an individual draw is 1 in 4.442, then the probability winning any one draw is p_w = 1 / 4.442 = 0.2251, and the probability of losing any one draw is 1 - p_w = 0.7749. So the probability losing any ten in a row is (1 - p_w)¹⁰ = 0.07804 = 1 / 12.81. Thus your luck so far has been strong, but not excessively so. -- ToE 01:14, 25 January 2012 (UTC) "If it weren't for bad luck I'd have no luck at all!"[reply]

I do not understand Looie's figures. If 1 out of 4442 scratch tickets is a winner, then the mean number of times you win is 1/4442 per ticket, and that is 10/4442=1/444 for 10 tickets. So every 444 times you buy 10 scratch tickets you will lose about 443 times and win about once. Bo Jacoby (talk) 23:20, 24 January 2012 (UTC).[reply]

The OP's IP geolocates to the United States, where "." is a radix point (or decimal mark) and not a thousands separator. -- ToE 00:09, 25 January 2012 (UTC)[reply]

There's also the fact that Looie and you are making different assumptions about the overlap of events. Looie is assuming that the tickets are independent, while you're assuming that there's no overlap --- it's impossible to win with more than one ticket. Since the OP mentioned that the tickets are spread out over years, the implication is that they're from separate ticket runs, and so Looie's assumption is likely correct.--195.37.234.132 (talk) 07:57, 25 January 2012 (UTC)[reply]