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18.7: Fractional linear transformations


Exercise (PageIndex{1})

Watch video “Möbius transformations revealed” by Douglas Arnold and Jonathan Rogness. (It is available on YouTube.)

The complex plane (mathbb{C}) extended by one ideal number (infty) is called the extended complex plane. It is denoted by (hat{mathbb{C}}), so (hat{mathbb{C}}=mathbb{C} cup{infty})

A fractional linear transformation or Möbius transformation of (hat{mathbb{C}}) is a function of one complex variable (z) that can be written as

(f(z) = dfrac{acdot z + b}{ccdot z + d},)

where the coefficients (a), (b), (c), (d) are complex numbers satisfying (acdot d - bcdot c ot= 0). (If (acdot d - bcdot c = 0) the function defined above is a constant and is not considered to be a fractional linear transformation.)

In case (c ot=0), we assume that

(f(-d/c) = infty quad ext{and} quad f(infty) = a/c;)

and if (c=0) we assume

(f(infty) = infty.)


Non-linear transformations of age at diagnosis, tumor size, and number of positive lymph nodes in prediction of clinical outcome in breast cancer

Background: Prognostic factors in breast cancer are often measured on a continuous scale, but categorized for clinical decision-making. The primary aim of this study was to evaluate if accounting for continuous non-linear effects of the three factors age at diagnosis, tumor size, and number of positive lymph nodes improves prognostication. These factors will most likely be included in the management of breast cancer patients also in the future, after an expected implementation of gene expression profiling for adjuvant treatment decision-making.

Methods: Four thousand four hundred forty seven and 1132 women with primary breast cancer constituted the derivation and validation set, respectively. Potential non-linear effects on the log hazard of distant recurrences of the three factors were evaluated during 10 years of follow-up. Cox-models of successively increasing complexity: dichotomized predictors, predictors categorized into three or four groups, and predictors transformed using fractional polynomials (FPs) or restricted cubic splines (RCS), were used. Predictive performance was evaluated by Harrell's C-index.

Results: Using FP-transformations, non-linear effects were detected for tumor size and number of positive lymph nodes in univariable analyses. For age, non-linear transformations did, however, not improve the model fit significantly compared to the linear identity transformation. As expected, the C-index increased with increasing model complexity for multivariable models including the three factors. By allowing more than one cut-point per factor, the C-index increased from 0.628 to 0.674. The additional gain, as measured by the C-index, when using FP- or RCS-transformations was modest (0.695 and 0.696, respectively). The corresponding C-indices for these four models in the validation set, based on the same transformations and parameter estimates from the derivation set, were 0.675, 0.700, 0.706, and 0.701.

Conclusions: Categorization of each factor into three to four groups was found to improve prognostication compared to dichotomization. The additional gain by allowing continuous non-linear effects modeled by FPs or RCS was modest. However, the continuous nature of these transformations has the advantage of making it possible to form risk groups of any size.

Keywords: Breast cancer Categorical Continuous Fractional polynomials Prognostic Splines.

Conflict of interest statement

Ethics approval and consent to participate

The study was approved by the Regional Ethical Review Board at Lund University, Lund Sweden (LU 240–01), and carried out in accordance with the code of ethics of the World Medical Association.

Since the study handles archival paraffin material, often several decades old, informed consent was not possible to retrieve from all patients. However, all data was analyzed and presented anonymously. In addition a note was published in the local paper, informing all previous breast cancer patients about the possibility to contact the research group if they did not want their tumor tissue to be used in scientific studies. This procedure was accepted by the Regional Ethical Review Board.

Consent for publication

See the Ethics approval and consent to participate section above.

Competing interests

The authors declare that they have no competing interests.

Publisher’s Note

Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.


Proof that there is a unique linear fractional transformation that maps three distinct points to three distinct points in the extended complex plane.

The following is a theorem and a proof from Complex Variables by Herb Silverman. The bolded are parts that I don't understand in the proof.

Theorem: Given three distinct points, $z_1,z_2, z_3$ in the extended $z$ plane and three distinct points $w_1,w_2,w_3$ in the extended $w$ plane, there exists a unique bilinear transformation $w=t(z)$ such that $T(z_k)=w_k$ for $k=1,2,3$.

Proof. We first assume that none of the six points is $infty$. Let $w=T(z)=frac$. We wish to solve for $a,b,c,$ and $d$ in terms of $z_1,z_2,z_3,w_1,w_2,$ and $w_3$. This sounds more complicated than it is. For $k=1,2,3,$ we have $w-w_k=frac-frac=frac<(ad-bc)(z-z_k)><(cz+d)(cz_k+d)>(3.9)$

Multiplying (3.10) by (3.11) we have $frac<(w-w_1)(w_2-w_3)><(w-w_3)(w_2-w_1)>=frac<(z-z_1)(z_2-z_3)> <(z-z_3)(z_2-z_1)>(3.12)$ Solving for $w$ in terms of $z$ and the six points gives the desired transformation. If one of the points were the point at $infty$, say $z_3=infty$, (3.12) would be modified by taking the limit as $z_3$ approached $infty$. In this case, we would have $frac<(w-w_1)(w_2-w_3)><(w-w_3)(w_2-w_1)>=frac (3.13)$

Corollary. Given three distinct points, $z_1,z_2,z_3$ in the extended $z$ plane there exists a unique bilinear transformation $w=T(z)$ such that $T(z_1)=0, T(z_2)=1, T(z_3)=infty$ and it is given by $w=frac<(z-z_1)(z_2-z_3)><(z-z_3)(z_2-z_1)>$.

Firstly, I don't see how to obtain the desired transformation. I did find an expression for $w$ in terms of all other variables (which looks complicated) and found $T(z_1)=w_1$ and $T(z_2)=w_2$. However, I don't get $T(z_3)=w_3$. In fact, if we look at (3.12) and plug in $z_3$ in place of $z$, the fraction on the right hand side isn't even defined since the denominator is $. So I don't see how we can get the desired result from this expression. Actually, I don't understand why the author sets out to find an expression for $w$ by multiplying fractions. What may be the reasoning behind this?

Moreover, I don't understand the part when one of the points is $infty$. It makes sense to take the limit as $z_3 o infty$, but how does this bring the expression (3.13)?

Finally, for the corollary, again this is similar to the first question, I think we're supposed to use (3.12) and simply plug in the appropriate $w_k$ values but in case of $T(z_3)=infty$, the expression just doesn't make sense to me. How does $frac<(w-w_1)(w_2-infty)><(w-infty)(w_2-w_1)>$ make sense?

I would greatly appreciate it if anyone clarifies the above questions to me, I'm having trouble reading this page because of these.


3 Answers 3

Explaining someone's motivation is more of a psycholical question than a mathematical one but here are some remarks. If we write a linear transformation in the projective line over a field, any field not necessarily the complex numbers, as a matrix eqation and then pull back the projective points (x:y) and (x':y') to the affine line as x/y and x'/y', we have x'/y'=(a(x/y) + b)/(c(x/y) + d).The complex projective line is the same as the Argand plane,which is the real Euclidean plane with one extra point corresponding to the point(1:0) on the projective line. It is also the same, by stereographic projection,as the surface of the sphere in real Euclidean 3-space. Getting back to the projective line, this transformation preserves cross-ratio, from which flow many other results. Hope this helps with the motivation!

For a simply connected domain $Omega$ , due to the Riemann mapping theorem it is biholomorphic (or conformally equivalent) to either

Fractional linear transformations are the elements in the automorphism groups $mathrm(hat>) = Bigg< frac : ad-bc eq 0 Bigg> $ $ mathrm(mathbb) = Bigg < : a eq 0 Bigg> $ $mathrm(mathbb) = Bigg< frac<arz+ ar> : |a|^2 - |b|^2 = 1 Bigg> $

Our mathematical ancestors discovered that the map $z o 1/z$ has an interesting property: It preserves the family of lines and circles. Since any map of the form $z o az + b$ also has the same property, compositions of such maps preserve this family. Such compositions are precisely the so-called fractional linear transformations.


1 Answer 1

Under the stereographic projection, we can identify points on the riemman sphere $hat>=mathbbcup$ with points on the unit sphere, $S^2$ . If two points $x,yinhat>$ are mapped to $p,qin S^2$ respectivley, the chordal metric $d(x,y)=|p-q|$ with the usual euclidean metric on $mathbb^3$ for a proof you can look at these great notes (which also explain and prove the formulas for the stereographic projection, which I'll use implicitly later on).

Denote the stereographic projection by $pi:hat> o S^2$ . As the distance is taken with respect to points on $S^2$ , the statement is equivalent to saying that, for a given mobius transformation $M:hat> ohat>$ , the function $picirc Mcircpi^<-1>$ is continuous with respect to the euclidean distance. We can write every mobius transformation as a composition of linear maps $z o az$ (Denoted $H_a$ ), an inversion $z o 1/z$ (denoted $I$ ) and a translation $z o z+b$ (Denoted $M_b$ ). So, if we know that $picirc H_acirc pi^<-1>$ , $picirc M_bcirc pi^<-1>$ , and $picirc Icirc pi^<-1>$ are continuous with respect to the euclidean distance- we may finish, cause we have:

$picirc (Mcirc N)circ pi^<-1>=picirc Mcirc pi^<-1>circpicirc Ncirc pi^<-1>$

And the composition of continuous maps is continuous. So we only need to check any of these cases. Right now I'm short in time so I'll give only the example of $I$ , I hope to expand tomorrow on how to do the other cases, but they can be done similarly. Maybe, also, I'll have a way with less calculations - I'm sure there's a much more elegant to do it.

So, fot the $I$ case- write a point on $S^2$ as $(t,u,v)$ . Then $pi^<-1>(t,u,v)=frac<1-v>$ . (Ignore now from the point $(0,0,1)$ - the north pole - which is mapped to $infty$ ). Now, $I(pi^<-1>(t,u,v))=frac<1-v>(t-iu)$ . Now applying $pi$ gives the point $(2cdotfrac<(1-v)t>,-2cdotfrac<(1-v)u>,frac<(1-v)^2 t^2+(1-v)^2 u^2-(t^2+u^2)^2><(1-v)^2 t^2+(1-v)^2 u^2+(t^2+u^2)^2>)=(t, - u,-v) $ . Tracking where the point $infty$ goes, we see that $(0,0,1)$ goes to $(0,0,-1)$ . This, therefore, is a continuous mapping (it is continuous in any of its coordinates- the case where we approach to the north pole also works). Geometrically, this is a rotation around the x-axis, so this is another way to see that it is continuous.

EDIT: I've been trying to think about ways to complete the other cases without messy calculations (which can be done). I came up with a nice interpretation only for special kind of the linear maps- you can decompose these further, to multiplication by an element of the form $z o e^z$ for real $ heta$ , and $z o az$ for $a>0$ . If you think about it, the first kind of translations correspond to rotations around the z-axis: if you rotate the plane, the projection will just rotate.


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Mapping Circles by a Linear Fractional Transformation

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A linear fractional transformation (or Mཫius transformation) in the complex plane is a conformal mapping that has the form , where , , , and are complex, with . The transformation transforms circles in the plane into circles in the plane, where straight lines can be considered to be circles of infinite radius.

In this Demonstration, the red circle is transformed into the blue circle of the form .

Contributed by: Izidor Hafner  (February 2016)
Open content licensed under CC BY-NC-SA


∑idiot's Blog

The Only LFTs You’ll Ever Need

Let us consider linear fractional transformations where the coefficients are all integers and . Since multiplying all of the coefficients by a constant doesn’t change the transformation, we may multiply everything by -1 whenever we want (for instance, to make a chosen coefficient positive). Recall, also, that writing the transformation as a matrix, is not an entirely unreasonable thing to do, as composites of transformations are correctly determined by matrix multiplication. I’ll be talking about sequence of these matrices, so let my original matrix be

If , then the requirement means that , and we may choose . Then our transformation is simply . Since is an integer, we could write this as .

If , then there is a so that . I’ll leave you to check the following:

Writing this right-most matrix as , we are looking at another matrix with integer coefficients whose determinant is 1. So we may repeat the process, finding integers and LFTs with . The process must stop eventually, because are all non-negative integers. Eventually, one of them will be 0, and we already dealt with transformations like that in the previous paragraph.

Let , and note that , which we’ve used before. Also, let (you might note that is the identity matrix). Then our relation above is . That is,

The collection of matrices that we have been working with, 2ࡨ matrices with integer coefficients whose determinant is 1, tend to get called the modular group and we have just shown that and can be multiplied together to give you any element of the group. More officially, they “generate” the group. You might fuss and say that technically these matrices are really the special linear group , and note that the modular group is really the quotient of this group where you say two matrices are the same if one is -1 times the other.

One interesting thing about this process is that it is really just the process you go through to find the greatest common divisor of and . The standard algorithm (Euclid’s) is to note that, if , there is an integer such that where . The next step is to repeat the process, writing , with . Iterating, you find so that , and eventually a is 0, and the process stops.

You might be a little worried (justifiably) that this process only seems to look at and , leaving and out of the picture. If you look at what we did with the matrices, our process took and wrote it as , and the and depended on and . In the final step of the process, where the -coefficient is 0, we ended up with a -coefficient of 1, but the -coefficient will be a function of the values and , so those values do show up and get used, eventually. Of course, since , knowing one of or is enough to determine the other (assuming you know and ). You should be thinking that the and are the extra variables that run around in Euclid’s alrgorithm when you do the “extended” version.

We’ve been talking about all of this in terms of matrices, but remember that the matrices represent linear fractional transformations. The matrix is the transformation , and the power is then . These are just horizontal translations. The matrix is the transformation . With our relations among the above, we see that we could write

Hurray for continued fractions! Of course, I feel like I set something up backwards. Plugging in will give me a continued fraction for , but I was saying that the come from thinking about the greatest common divisor of and . Ah well.

[Update 20091203: The error I’m concerned about above stems from some loose dealings with the end of the iterative procedure. Letting , we will get to something like where , i.e. . So we end up saying . And that does depend on the initial , presumably in a manner such that plugging in makes the fraction make sense.]

What I’ve said is that every 2ࡨ matrix with coefficients in and determinant 1 can be written in terms of and . However, you might want to use instead of . Going through essentially the same process as before, with some minus signs sprinkled in appropriately, one can determine that and can be used, instead of and , to generate any of the matrices we are considering. What is the advantage of over ? One way to state the advantage, for our story, is that applying to points in the upper half plane leaves them in the upper half plane (and similaly for the lower half), whereas flips points in upper half plane to the lower half plane. We should think of this an advantage, because all of our Ford circles lie in the upper half plane. If you go back to yesterday’s discussion, you’ll note that I used in the factorization.

So, anyway, and are the only LFTs you need. You can quickly check that and also (not quite as quickly) that . If you write , then I guess you get to say that the modular group is generated by and , and can establish an isomorphism between the modular group and the free product of the cyclic group of order 2 with the cyclic group of order 3. That’s something.

LFTs and Ford Circles

Given 4 complex numbers, , we may consider the linear fractional transformation (LFT)

Well, 4 numbers are enough to make a matrix, . Is there any better reason to relate the linear fractional transformation with this matrix?

Suppose you have two matrices, and . Then the product is as follows:

If you take the composite of the two linear fractional transformations, i.e.,

and then play around simplifying that expression for a few minutes, you obtain the LFT

which is precisely the LFT corresponding to the product matrix above. So, if nothing else, writing LFTs as matrices this way won’t lead us astray when thinking about composites.

This idea is not without its confusion, for me anyway. Generally when you think about a 2ࡨ matrix of complex values, you are thinking about that matrix as a linear map , which is not what we are doing above. Instead, I guess we are saying that the “monoid” (group without inverses) of 2ࡨ matrices, , acts (in the technical sense) on as linear fractional transformations. My guess is that there are even better ways to say what is going on.

I think it is also important to keep in mind that two different matrices may correspond to the same LFT. For example, represents the same LFT as . More generally, if is any complex value (nonzero), then represents the same LFT as . I guess one can think of as a -vector space (isomorphic to ), and then think of its projective space (the quotient where two “vectors” (matrices here) are the same when they differ by a scalar (complex) multiple), which I’ll denote . Then I think I’m saying that the action of on actually is an action of the quotient, . I’m not sure if this is a useful viewpoint (or, indeed, correct).

Yesterday, when I was talking about how to picture what an LFT does to , I wrote down a factorization of the LFT as a composite. Our new notation gives us another way to write that factorization (recall , and that we had assumed ):

As is frequently useful, we will assume that (indeed, this factorization seems to require it – I think I’m missing something somewhere, anybody see it?). Notice that is the determinant of the matrix representing our LFT. We may then multiply all entries in our matrix (without changing the LFT, as discussed above) by , and obtain a matrix with determinant 1. Let’s do that, making .

Yesterday, when I was working on the factorization above, I only had something like idea where I was going. I think today I’ve got about twice that, so I want to re-write the factorization. Let me write it as

So what’s the connection with Ford circles? Recall that for a reduced fraction , the associated Ford circle is the circle centered at with radius . Following Rademacher (and, presumably, others), let us say that the “fraction” also gets a Ford “circle”, the line in the plane. This isn’t such a nasty thing to do, as it has the tangency properties I talked about when talking about Ford circles. Anyway, let us think about applying our transformation , as the composite given above, and see what happens to this line . We’ll assume that , and are all integers.

The first step, is the linear translation . Since is real (since are integers), this translation is a horizontal shift, which leaves unchanged.

Next up, , which is . Thinking of a point on the line , you can quickly determine that its polar coordinates are . The transformation is the composite of: (1) inversion with respect to the unit circle (the point becomes ), (2) reflection across the horizontal axis (giving ), and finally (3) multiplication by -1 (giving , since these are polar coordinates). This final point is . As varies through , also varies through this interval, and so we get the graph of the polar curve . If you know your polar curves, you know what this looks like…

So, the first two transformations take the line to the circle with center (0,1/2) and radius 1/2. The next in our composite is multiplication by , which is just a scaling (since ). This scaling takes our circle to the circle with center and radius . Finally, the last transformation is another horizontal translation, leaving our circle centered at . We recognize this as the Ford circle for the fraction (as long as that fraction is reduced).

Wasn’t that fun? If you want to think about it some more, you might convince yourself that any point above the line will get moved to a point inside the Farey circle resulting from this process.

Anyway, enough out of me. Hopefully tomorrow I’ll have slightly more of an idea what I’m talking about. Don’t count on it though.


Algorithm for S-box

This section mainly deals with the structure of our S-box. Before we discuss the constituent algorithm, we need to go through some fundamental facts.

A function (f: >_<2>^ ightarrow >_<2>) is called a Boolean function. We define a vectorial Boolean function (F: >_<2>^ ightarrow >_<2>^) as

where (x=(x_<1>,, x_<2>,ldots ,x_)in >_<2>^) and each of (f_) ’s for (1le ile m) is a Boolean function referred to as coordinate Boolean function. An (n imes n) S-box is precisely defined as a vectorial Boolean function (S: >_<2>^ ightarrow >_<2>^) .

At this stage, it seems quite practical to understand the structural properties of the Galois field used to construct an S-box. Generally for any prime p, Galois field (GF(p^)) is given by the factor ring (>_

[X]/ <eta (x)>) where (eta (x)in >_

[X]) is an irreducible polynomial of degree n.

For an (8 imes 8) S-box, we use (GF(2^<8>)) . In advanced encryption standards (AES), the construction of (GF(2^<8>)) is based on the degree 8 irreducible polynomial (eta (x)=x^<8>+x^<4>+x^<3>+x+1) . In Hussain et al. (2013b), (eta (x)=x^<8>+x^<4>+x^<3>+x^<2>+x+1) is used as the generating polynomial. Here we choose (eta (x)=x^<8>+x^<6>+x^<5>+x^<4>+1) as the irreducible polynomial that generates the maximal ideal (<eta (x)>) of the principal ideal domain (>_<2>[X]) . It is important to note that we may choose any degree 8 irreducible polynomial for constructing (GF(2^<8>)) however the choice of generating polynomial may affect our calculations as the binary operations are carried modulo the used polynomial (see Benvenuto 2012 for details).

Generally the construction of an S-box requires a nonlinear bijective map. In literature many algorithms based on such maps or their compositions are presented to synthesize cryptographically strong S-boxes. We present the construction of S-box based on an invertible nonlinear map known as the fractional linear transformation. It is a function of the form (frac) generally defined on the complex plain (>) such that a, b, c and (d in >) satisfy the non-degeneracy condition (ad-bc e 0) . The set of these transformations forms a group under the composition. The identity element in this group is the identity map and the the inverse (frac<-cz+a>) of (frac) is assured by the condition (ad-bc e 0) . One can easily observe that the algebraic expression of this map has a combined effect of inversion, dilation, rotation and translation. The nonlinearity and algebraic complexity of the fractional linear transformation motivates the idea to employ this map for byte substitution.

For the proposed S-box we apply fractional linear transformation g on the Galois field discussed above, i.e. (g:GF(2^<8>) ightarrow GF(2^<8>)) given by (g(t)=frac) , where (a,, b,, c) and (din GF(2^<8>)) such that (ad-bc e 0) and t varies from 0 to (255 in GF(2^<8>)) . We may choose any values for parameters a, b, c and d that satisfy the condition (ad-bc e 0) . Here, for calculations, we take (a=29=00011101,, b=15=00001111,,c=8=00001000) and (d=9=00001001) . One may observe that as we are working on a finite field, g(t) needs to be explicitly defined at (t=47) (at which denominator vanishes), so in order to keep g bijective we may define the transformation as given below

Following the binary operations defined on the Galois field (Benvenuto 2012), we calculate the images of g as shown in Table 1.

Thus the images of the above defined transformation yield the elements of the proposed S-box (see Table 2).

It is important to mention that an (8 imes 8) S-box has 8 constituent Boolean functions. A Boolean function f is balanced if () and () have same cardinality or the Hamming weight HW ((f)=2^) . The significance of the balance property is that the higher the magnitude of a function’s imbalance, the more likelihood of a high probability linear approximation being obtained. Thus, the imbalance makes a Boolean function weak in terms of linear cryptanalysis. Furthermore, a function with a large imbalance can easily be approximated by a constant function. All the Boolean functions (f_,,i le i le 8) , involved in the S-box as shown in Table 2 satisfy the balance property. Hence, the proposed S-box is balanced. It might be of interest that in order to choose feasible parameters leading to balanced S-boxes satisfying all other desirable properties (as discussed in the next section), one can use constraint programming, a problem solving strategy which characterises the problem as a set of constraints over a set of variables (Kellen 2014 Ramamoorthy et al. 2011).

An S-box is used to convert the plain data into the encrypted data, it is therefore essential to investigate the comparative performance of the S-box. We, in the next section, analyse the newly designed S-box through various indices to establish the forte of our proposed S-box.


Output feedback control of linear fractional transformation systems subject to actuator saturation

In this paper, the control problem for a class of linear parameter varying (LPV) plant subject to actuator saturation is investigated. For the saturated LPV plant depending on the scheduling parameters in linear fractional transformation (LFT) fashion, a gain-scheduled output feedback controller in the LFT form is designed to guarantee the stability of the closed-loop LPV system and provide optimised disturbance/error attenuation performance. By using the congruent transformation, the synthesis condition is formulated as a convex optimisation problem in terms of a finite number of LMIs for which efficient optimisation techniques are available. The nonlinear inverted pendulum problem is employed to demonstrate the effectiveness of the proposed approach. Moreover, the comparison between our LPV saturated approach with an existing linear saturated method reveals the advantage of the LPV controller when handling nonlinear plants.


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