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For example, how do I know that with:

$$f(x_1,x_2,x_3,x_4)=\frac{x_1 x_2+x_3 x_4-x_2 x_3-x_1 x_4}{x_1 x_2+x_3 x_4-x_1 x_3-x_2 x_4}$$

$f$ has the property:

$$f(x_1,x_2,x_3,x_4)=f(x_2,x_1,x_4,x_3)=f(x_4,x_3,x_2,x_1)=f(x_3,x_4,x_1,x_2)$$

I mean it's not that easy to discover all of them, right? I myself used to know only the first part of that symmetry until I find the completed one today. Is there any general method to examine the symmetry of a given function?

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12  
Wonderful. Now you disguised the cross ratio by multiplying it out... The symmetries are much better hidden this way, yes. By the way: you found it? In your deleted question somebody gave you a link... –  t.b. Jul 2 '11 at 10:50
    
@Theo Buehler: Okay, okay, I admit I shouldn't ask such unthoughtful questions, but at least I've learned to change my question to make it more acceptable, that's wonderful progress right? –  Voldemort Jul 2 '11 at 11:09
    
@Theo Buehler: btw, thought I've now get the direct answer, I still can't figure out how the answer is deduced. Try to discover that f(x1,x4,x3,x2)=λ/λ-1 seems impossible for me. –  Voldemort Jul 2 '11 at 11:19

3 Answers 3

Okay, I'll answer your question on the cross ratio:

First of all, recall its definition

$$\lambda = \lambda(z) = [z,z_2,z_3,z_4] = \frac{z-z_3}{z - z_4} : \frac{z_2 - z_3}{z_2 - z_4}$$

which is the unique Möbius transformation $z \mapsto \lambda(z)$ sending the ordered triple $(z_2,z_3,z_4)$ to $(1,0,\infty)$—it is a general and easy fact that a Möbius transformation is determined by the images of three distinct points.

In view of the symmetries you display in your question, the only thing we have to figure out are the Möbius transformations that arise from permuting the three entries $(z_2,z_3,z_4)$ in the cross ratio. For this we need only post-compose our initial cross ratio $\lambda$ with the unique Möbius transformation sending $(1,0,\infty)$ to some permutation of $(1,0,\infty)$.

For instance, the cross-ratio $[z,z_4,z_3,z_2]$ is obtained from $\lambda$ by postcomposing it with the Möbius transformation sending the triple $(1,0,\infty)$ to $(\infty,0,1)$. This is easy: Since $1$ is sent to $\infty$, we must have a transformation of the form $\lambda \mapsto \frac{a\lambda + b}{\lambda - 1}$ (with $a \neq -b$ but we won't need that additional information). As $0$ is sent to $0$, we must have $b = 0$ and since $\infty$ is sent to $1$ we must have $a = 1$, and our sought transformation is

$$[z,z_4,z_3,z_2] = \frac{\lambda}{\lambda - 1}.$$

Similar considerations lead to the entire list of symmetries given on Wikipedia.

In order to save you some work, I'll give one more indication: swapping the last two entries corresponds to $(1,0,\infty) \mapsto (1,\infty,0)$, so simply $\lambda \mapsto \frac{1}{\lambda}$. Hence, there really only remains one more thing to do: figure out the transformation sending $(1,0,\infty)$ to $(\infty,1,0)$ and this is just as easy as the case I just did. You'll get the expression for $[z,z_3,z_4,z_2]$ and after that hardly any further computation is required.

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Thanks. Using composed transformations is indeed much easier than guessing and comparing the terms. I think this is what Ahlfors intended. –  Voldemort Jul 9 '11 at 4:35
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Just a word of warning: The exercises in Ahlfors tend to be quite tricky, there quite are a few that took me a huge amount of time to figure out. –  t.b. Jul 9 '11 at 7:59

Not a general answer, but in this particular case you can write $$ f(x_1,x_2,x_3,x_4)= \frac{(x_1-x_3)(x_2-x_4)}{(x_1-x_4)(x_2-x_3)} $$ which makes it easier to see the symmetries.

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Hans: The OP is aware of that. S/he asked a lazy question about an hour ago on the symmetries of the cross ratio (that's the reason for my comment above). –  t.b. Jul 2 '11 at 10:54

You have four symbols $x_1,x_2,x_3$ and $x_4$. If you have two different permutations which fix your function, then they will fix it if applied consecutively. The identity fixes the function, as does the inverse of any permutation which fixes it. The permutations fixing the function therefore form a group.

Since there are four objects being permuted, the group will be a subgroup of the symmetric group $S_4$ which has order 24.

You have one permutation of order 2 - an even permutation consisting of the product of two transpositions. Are there any others?

If you put $x_1=x_3$ and see that the function has the value 0 in this case, any permutation which fixes the expression has to preserve the value zero. Looking at the factorised expression, you have to preserve the two factors in the numerator:

ie $(x_1,x_3)$ has to map to itself, or $(x_3,x_1)$ or $(x_2,x_4)$ or $(x_4,x_2)$ so four possibilities, and it is easy to see that each can be extended uniquely to a permutation fixing the function.

Then observe that the group $V$ fixing the function is a Normal Subgroup of $S_4$ so there will be a natural action of $S_4/V$, a non-abelian group of order 6, on f. This has two elements of order 3 (slightly unexpected), and analysing this phenomenon gives you your "hard to discover" value.

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In addition, I should perhaps say that I encountered this strange group of rational functions of order 6 before I encountered the cross product (I didn't do enough geometry when I was young).It is one of those neat things I keep tucked away for a rainy day. The cross product made sense of the functions, which seemed otherwise inexplicable. So there are some interesting facts here, but more important to understand what is going on - and the projective geometry link in the deleted question helped with that. –  Mark Bennet Jul 2 '11 at 20:14

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