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September 15

closed group of operations

Hello. I am interested in the six functions $f(z)=z,~z/(z-1),~1-z,~1/z,~1/(1-z),~1-1/z$ . These seem to be closed under composition and form a group which must be isomorphic to S3 because it is not commutative. It is a subgroup of the Mobius group of transformations. But does it have a special name? Is it the only subgroup of order six? I am looking for a couple of sentences to describe unambiguously its relationship to the Mobius group of transformations. Thanks, Robinh (talk) 08:13, 15 September 2014 (UTC)[reply]

That's only five functions ... (I've added some spaces after your commas to ease counting.) -- SGBailey (talk)

(OP) corrected, now there are six. It looks better now it's got spaces, but I didn't think math typesetting was sensitive to whitespace. Robinh (talk) 08:47, 15 September 2014 (UTC)[reply]

(I think for WP math mode, as in LaTeX, any whitespace >1 renders the same as whitespace=1, but whitespace=0 is different. I took the liberty of testing/confirming this in your original question, as you can see if you look at the source now :) SemanticMantis (talk) 14:18, 15 September 2014 (UTC)[reply]

In the context, Robinh's suspicion seems to have been correct: whitespace only has significance in LaTeX if it changes the parsing. It does not change the display spacing; for that you need an explicit spacing (e.g. ~, which I've taken the liberty of including) —Quondum 15:24, 15 September 2014 (UTC)[reply]

It is not a normal subgroup, so cannot be the only subgroup of order six. However, within

PSL(2,\mathbb {Z} )

this is the group of automorphisms of the quadratic form

x^{2}+y^{2}+xy

whose zeros (projectively) are the cube roots of unity. This group is called the anharmonic group. Sławomir Biały (talk) 12:18, 15 September 2014 (UTC)[reply]

As for uniqueness, you can generate an infinite family of 6-function sets with the same group structure by taking the one that you found and using the mapping from the Mobius group to 2-dimensional matrix groups and changing basis. For example, here's one that (I hope) also gives you S3:

f(z)=z,-z,{\frac {3+z}{1-z}},{\frac {3+z}{z-1}},{\frac {3-z}{1+z}},{\frac {z-3}{1+z}};

For subgroups of order 6 which are unrelated to the S3 that you found, you can quite easily find representations of Z6 (e.g. the group generated by

f(z)=e^{i\pi /3}z

). 129.234.186.11 (talk) 16:50, 15 September 2014 (UTC)[reply]

(OP) thanks everyone. So it's not normal (Sławomir, how did you discover this? Did you compose a general element of the Mobius group with one of the transformations above and observe that it made a function not in my set of six? Or is it obvious to you in a more direct way?). And therefore not unique because I can consider $g^{-1}Hg$ for some $g\in M$ . It says in Cross-ratio that "The anharmonic group is the group of order 6 generated by λ ↦ 1/λ and λ ↦ 1 − λ. It is abstractly isomorphic to S3 . . .". This is close to what I'm seeking. But what does "abstractly isomorphic to S3" mean? Does this statement have connotations not implied by plain old "isomorphic"? Robinh (talk) 20:13, 15 September 2014 (UTC)[reply]

I take "abstractly isomorphic" to mean that there is no particular implied realization of the group as a group of permutations of three objects. However, it is easy to see that they do, in fact, act as permutations on the points {0,1,∞}, so this gives an explicit isomorphism of the group with a symmetric group and it might be helpful if the article cleared that up. As for the non-normality, I just did it by conjugating by some random thing in SL(2,Z) (any element of infinite order will work). But if you believe what I said about the quadratic form, it is also "obvious" from that point of view. Sławomir Biały (talk) 21:57, 15 September 2014 (UTC)[reply]

OK thanks Sławomir. I see what abstractly isomorphic means now. I am having difficulty understanding the relationship between the Mobius group and your quadratic form

x^{2}+y^{2}+xy

. Also, would it help to consider the group of six functions and their group action on the set {0,1,∞}? Robinh (talk) 22:17, 15 September 2014 (UTC)[reply]

The quadratic form can be determined from the fixed points of the order three elements. These are the primitive sixth roots of unity, and the order two elements either fix or exchange them as well. This tells us the eigenvectors of the corresponding matrices in SL(2,Z), from which the quadratic form is built. Sławomir Biały (talk) 13:14, 16 September 2014 (UTC)[reply]

Graph theory:

Is there a special name, or a recognized short expression/description, for characterizing the following directed graph G ?

Every vertex v - from which there is an edge in G - and to which there is an edge in G, satisfies the following two conditions:

For every u there is w, such that if G has an edge from u to v, then G has both an edge from v to w and an edge from u to w.
For every w there is u, such that if G has an edge from v to w, then G has both an edge from u to v and an edge from u to w.

HOOTmag (talk) 14:25, 15 September 2014 (UTC)[reply]

You could call it "a digraph where every edge which leads into or out of a vertex which is neither a sink nor a source is part of a directed 3-cycle", but I suspect that you want something simpler. -- ToE 00:36, 17 September 2014 (UTC)[reply]

Oops. Not a directed 3-cycle, but a 3-edged transitive closure. -- ToE 11:34, 17 September 2014 (UTC)[reply]

So why did you strike out the whole previous comment? You should have only struck out the words "directed 3-cycle", shouldn't you? HOOTmag (talk) 12:54, 17 September 2014 (UTC)[reply]

Because 1) I didn't think that it was as interesting without the directed 3-cycles and 2) I am uncertain of my vocabulary. The best I can come up with in words is: "A digraph where every edge which leads into/out-of a rich vertex (one which is neither a sink nor a source) is part of a three edge transitive closure with another edge which leads out-of/into that vertex." I don't think that "rich vertex" is a common term, I'm not sure "three edge transitive closure" is appropriate as I think of closure involving a superset, and I'd hope for an easier way of indicating that the vertex in question is in the middle of the three edge transitive closure than pairing "into/out-of" with "out-of/into".

That is the direction in which I think of the question when posed for digraphs, but when put in terms of relations, I think of "surrogate transitivity", where when three elements are related such that if they fail transitivity, there are two surrogate elements which can be substituted, one at a time, for the end points, to satisfy transitivity. That is what I think when I write uRv ∧ vRw -> ∃ u', w' s.t. u'Rv ∧ u'Rw ∧ vRw' ∧ uRw', but it doesn't sound as good when I try to put my thoughts into words.

Searching online, I see the term "semi-transitive graph", but I've not been able to pull up a full definition. From this abstract: we first consider graphs allowing symmetry groups which act transitively on edges but not on darts (directed edges). We see that there are two ways in which this can happen and we introduce the terms bi-transitive and semi-transitive to describe them. That doesn't sound like what you are asking. -- ToE 21:25, 17 September 2014 (UTC)[reply]

if U and W are disjoint, then U, {v}, and W are a tripartite graph that could be called topologically ordered, although the ordering isn't complete. There is a kind of transitive closure here if one thinks of the graph as representing a topological ordering, but "transitivity" in graph theory is more about symmetry than ordering. --Mark viking (talk) 15:04, 18 September 2014 (UTC)[reply]

Relation theory:

Is there a special name, or a recognized short expression/description, for characterizing the following binary relation R ?

Every v for which there are both b satisfying bRv and c satisfying vRc, satisfies the following two conditions:

For every u there is w, such that if uRv then both vRw and uRw.
For every w there is u, such that if vRw then both uRv and uRw.

HOOTmag (talk) 14:25, 15 September 2014 (UTC)[reply]

It sounds as if you are asking for something similar to a transitive relation, but less restrictive.

Whereas transitivity says that if uRv and vRw then uRw, I think that you are asking for the property where if uRv and vRw then there exists u' and w' with u'Rv and vRw', where u'Rw and uRw'. Is that correct? -- ToE 17:17, 16 September 2014 (UTC)[reply]

Yes, my original condition is a bit similar to transitivity. However, my original condition is not equivalent to the condition you've presented on my behalf. For example, look at the following relation: ( (1,0), (0,2), (1,2), (3,0), (0,4), (3,4) ). It satisfies my original condition, yet not the condition you've presented on my behalf. HOOTmag (talk) 19:41, 16 September 2014 (UTC)[reply]

I think that our conditions are equivalent. Note that I do not require u to necessarily be distinct from u'. From your example, 1R0 and 0R4, and while it is false that 1R4, we do have 2 such that 0R2 and 1R2, and 3 such that 3R0 and 3R4. But if we start with 1R0 and 0R2, then we have transitivity with those very elements. Perhaps it is not so important, but if you still feel that your example fails with my condition (which I believe to be equivalent to yours), I'd like to see how it fails.

I attempted to rephrase your condition as I found its original form confusing, but perhaps my form was no better. I think that your more recent rephrasing of the question, using b and c instead of the initial u and w is a big improvement, as part of my confusion was over what relation those initial u and w had to those that followed. (Answer: none, unlike in my reformulation.)

I think that it would be even better by rewriting your conditions from:

For every u there is w, such that if uRv then both vRw and uRw

to:

For every u where uRv, there is w, such that vRw and uRw

I find the former wording a bit confusing, as my first impression is that w is chosen given u, but irrespective of the choice of v, but this is not the case given the initial, "Every v". I think that the rewording emphasizes that the choice of w is dependent on both u and v.

And after all that, sorry, I don't know what this is called. I hope that, with the clarified wording, others will be able to give you a better answer. -- ToE 22:37, 16 September 2014 (UTC)[reply]

Yes, I now think that your first proposal is equivalent to mine, and I also think that it's even better than mine, because it's shorter. However, I don't agree with your claim that the wording - "for every u there is w" - makes one think w is chosen irrespective of the choice of v : The choice of w depends on v as well, just because the wording - "for every u there is w" - appears after the words: "every v...satisfies", So no confusion may arise. HOOTmag (talk) 00:22, 17 September 2014 (UTC)[reply]

And I suppose that you could take things one step less restrictive, where if uRv and vRw then there exists u' and w' with u'Rv and vRw' such that u'Rw'. -- ToE 18:15, 16 September 2014 (UTC)[reply]

Your condition is too weak. For example, look at the following relation: ( (1,0), (0,2), (1,2), (3,0), (0,4) ). This relation satisfies your condition, yet not mine. HOOTmag (talk) 20:01, 16 September 2014 (UTC)[reply]

Agreed. This condition is weaker than yours. -- ToE 22:37, 16 September 2014 (UTC)[reply]

I should have clarified that it was intentionally offered as weaker than what I understood your condition to be. Your condition reminded me of transitivity but weaker, and this was the next logical step even weaker. -- ToE 12:42, 21 September 2014 (UTC)[reply]

determine interior angle of triangle give the coordinates of the vertices.

So I was asked to calculate the interior angles of a triangle given the x coordinates P(2,0) Q(0,3) and R(3,4)

Though this is a basic geometry question, this was in the context of a chapter where we are learning the dot product of vectors.

Anyway, I feel like I understand how to solve the problem up to a point, but I must have made mistakes somewhere.

First thing I did was calculate the length and slope of the sides PQ, QR, and RP.
PQ length = sqrt (13) with slope of -2/3

QR length = sqrt (10) with slope of 1/3

RP length = sqrt (17) with slope of 4

Next I used the slopes of each sides to convert the sides to vectors. I called

vector based on PQ vector A <3, -2>

vector based on QR vector B <3, 1>

vector based on RP vector C <1, 4>

Then I used the identity that says that the dot product of vectors A and B = |A||B| cos theta, where theta is the angle formed by the two vectors

The dot product of A and B is 7 (associated with vertex Q of the triangle)

The dot product of B and C is 7 (associated with vertex R of the triangle)

The dot product of C and A is -5 (associated with vertex P of the triangle)

Therefore, I thought, that

the cosine of theta for angle Q is 7/sqrt (130) - > theta = approx 52 degrees

the cosine of theta for angle R is 7/sqrt (170) - > theta = approx 57 degrees

the cosine of theta for angle P is -5/sqrt (221) - > theta = approx 109.7 degrees

These are not the correct answers! they don't even add up to 180

I looked in the back of my book and apparently my value for angle R = approx 57 degrees is correct, but the other two are wrong. So maybe I did something wrong with my vector arithmetic or coordinate geometry?--Jerk of Thrones (talk) 21:36, 15 September 2014 (UTC)[reply]

Your first vector had the i and j components switched. I also don't think you should use slope to determine vectors. After all, slope can be infinite, and vectors representing the sides of a triangle can't. Instead, find vectors as the difference between each pair of coords:

P = (2,0) 
Q = (0,3) 
R = (3,4)

P-Q vector = (2-0,0-3) = (2,-3)
R-Q vector = (3-0,4-3) = (3,1)
R-P vector = (3-2,4-0) = (1,4)

You could also use the Q-P, R-Q, and P-R vectors, which would just point in the opposite directions. Now, which directions the vectors point are rather critical when using the dot products. If one vector is pointing the wrong way at a point, you will get the supplementary angle at that point. If your angles still don't add up to 180, then that's probably the issue. I don't think this particular triangle will have any angles over 90, so if you get any, try subtracting it from 180 to get the correct angle.

(BTW, the "R-P vector" would normally just be called the "RP vector", but I wanted to make it a bit clearer by adding the minus sign.) StuRat (talk) 00:55, 16 September 2014 (UTC)[reply]