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Now, suppose the transformation(in 2d) I am working with has two
separate functions for $x$ and $y$.

That is, the transformation for $x$ is of the form $$ x'=\frac{x}{x+y} $$ and the transformation of $y$ is $$ y'=\frac{y}{x+y} $$ Each is an LFT (The schwarzian derivatives are $0$) but is
the transform as a whole still considered an LFT?

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If you treat $(x,y)$ as a complex variable $z=x+iy$, then $\frac{z}{\Re z+\Im z}$ is no longer a linear fractional transformation. – J. M. Nov 17 '10 at 23:59
But suppose, in my original post, that x and y are strictly positive reals in the typical 2d x/y plane? – omission9 Nov 18 '10 at 0:04
Only component-wise, but remember that two-dimensional geometric transformations can be shown to be equivalent to certain operations on complex numbers. So for the transformation as a whole, I don't think so, due to what I mentioned in my first comment. (I would be happy to be proven wrong.) – J. M. Nov 18 '10 at 0:07
I think I understand, mostly. 2d is just an example though. I was hoping to examine some properties of this under higher dimensions. For example in 3d: x'=x/(x+y+z),y'=y/(x+y+z),z'=z/(x+y+z). Where x,y,z are all positive reals(I am, for other reasons, only concerned with the positive orthant). – omission9 Nov 18 '10 at 0:29

I was under the impression that the term "fractional linear transformation" referred to transformations in one (generally complex) variable. However, it is certainly still valid to extend this idea to more variables; you get elements of the projective special linear groups.

I don't think this is really the concept you're looking for, though, since your transformations have an extra property (they are homogeneous).

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By all conventions I know, this would not quite be even a fairly generalized "linear fractional transformation", for at least one technical-but-critical reason: the image is just the line $X+Y=1$. Possibly you are not asking the question you'd wish to ask.

But this map is "almost" a generalized linear fractional transformation: $g\in GL_3(\mathbb R)$ acts on $\mathbb R^3$ linearly, and then normalizing the third coordinate to $1$ gives the "projective" action of a sort mentioned by @QiaochuYuan: $$ \pmatrix{a & b & c \cr d & e & f \cr g & h & i}\pmatrix{x \cr y & 1} \;=\; \pmatrix{ax+by+c \cr dx+ey+f \cr gx+hy+i}\;\sim\; \pmatrix{{ax+by+c\over gx+hy+i} \cr {dx+ey+f\over gx+hy+i} \cr 1} $$ The problem in your example is that the associated matrix would be $\pmatrix{1 & 0 & 0 \cr 0 & 1 & 0 \cr 1 & 1 & 0}$, which is not invertible.

This leads me to wonder about your larger context, and whether you'd really want to be asking a variant question, to which the answer could be a "generalized" "yes".

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