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Consider the set of all invertible $2\times 2$ matrices over $\mathbb R$ (I think we can do it over $\Bbb C$ but I didn't look at it) $${\bf GL}(2,\Bbb R)=\left\{\left(\begin{matrix} a&b \\c &d\end{matrix}\right):a,b,c,d\in\Bbb R,ac-bd\neq 0 \right\}$$

Consider now the set $\bf M$ of all functions of the form

$$f(x)=\frac{ax+b}{cx+d}$$

again with $a,b,c,d\in\Bbb R,ad-bc\neq 0 $. If we identify each of these with a matrix $$\left(\begin{matrix} a&b \\c &d\end{matrix}\right)$$ then we can define an isomoprhism between $\rm GL$ and $(\bf M,\circ)$ as groups with operations of matrix multiplications and functional composition, respectively, and identities$$\left(\begin{matrix} 1&0 \\0 &1\end{matrix}\right)=e$$ $$x=\frac{1x+0}{0x+1}=e'$$ since if $$\eqalign{ & f = \frac{{ax + b}}{{cx + d}} \cr & g = \frac{{ex + f}}{{gx + h}} \cr} $$ then $$f \circ g = \frac{{\left( {ae + bg} \right)x + af + bh}}{{\left( {ce + dg} \right)x + dh + cf}}$$ which corresponds to

$$\left(\begin{matrix} a&b \\b &c\end{matrix}\right)\left(\begin{matrix} e&f \\g &h\end{matrix}\right)=\left(\begin{matrix} ae + bg&af + bh \\ce + dg &dh + cf\end{matrix}\right)$$ and similarily for inversion, $${f^{ - 1}} = \frac{1}{{ac - bd}}\frac{{dx - b}}{{ - cx + a}}$$

and $$\left(\begin{matrix} \frac d\Delta& \frac {-b}\Delta \\\frac{-c}\Delta &\frac a\Delta \end{matrix}\right)=e$$

I include the determinant $\Delta=ac-bd$ inside the matrix to avoid any multiplication by scalars considerations.

My question is: How can this be generalized to ${\bf GL}(n,K)$, and what is the theoretical relevance of this? I know Möbius Transformations are important in Complex Analysis, for instance, but I haven't seen any "higher dimesional" equivalent around.

ADD As users noted, for every $a\in \Bbb R$,$$\left(\begin{matrix}a&0\\0&a\end{matrix}\right)\mapsto x$$

so the isomoprhism is actually obtained by quoting ${\bf GL}(2,\Bbb R)$ by $I=\{aI_2:a\in\Bbb R\}$

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You don't have an isomorphism but just a surjective map $f : \textrm{GL}_2(\Bbb{R}) \longrightarrow \textbf{M}$. The kernel of this map is all scalar multiples of the identity (as you can check) and so by the first isomorphism theorem,

$$\textbf{M} \cong \textrm{GL}_2(\Bbb{R})/ \{ aI\}$$

where $a$ ranges over all non-zero real numbers. The group on the right is called $\textrm{PGL}_2(\Bbb{R})$ (if you are working over $\Bbb{R}$). The generalisation of this is of course when you consider $\textrm{GL}_n(\Bbb{C})/\{aI\}$ that (as you may have guessed) is called $\textrm{PGL}_n(\Bbb{C})$.

I don't really know much about this group except that if you have $\textrm{GL}_n$ acting on say $\Bbb{R}^n$, then you can ask what the associated action will be on the quotient $\Bbb{R}\textrm{P}^{n-1}$ space. Now recall that real projective space $\Bbb{R}P^{n-1}$ is the space of all lines through the origin in $\Bbb{R}^n$. Suppose we let a scalar multiple of the identity act on one of these lines. Then the line just get sent to itself yes? Rather, every element on the line is sent to a scalar multiple of itself. However in terms of projective space, that line is considered as just one point and so any scalar multiple of the identity matrix becomes the "identity matrix" on $\Bbb{R}P^{n-1}$. That's why we quotient it out to get that the "actual group" acting on $\Bbb{R}P^{n-1}$ is $\textrm{PGL}_{n}$.

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    Yes, I missed the fact that $$\left(\begin{matrix}a&0\\0&a\end{matrix}\right)\mapsto x$$ for each $a$. Could you add a little about ${\bf PGL}_n(K)$?2012-10-28
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    @PeterTamaroff But you do have an isomorphism between the quotient and the group of mobius transformations.2012-10-28
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    What do you mean when you say action? Do you mean a function of the form $\star:K\times V\to V$?2012-10-28
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    @PeterTamaroff The action of $\textrm{GL}_n$ on the vector space is the usual one given by left multiplication of matrices.2012-10-28
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    @PeterTamaroff See the edit.2012-10-28
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This map isn't an isomorphism; all scalars are sent to the identity. The generalization goes by the name of the projective general linear group; this naturally acts on projective space, hence the name, and is the natural target for projective representations, which are important for example in quantum mechanics.

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    I have just started linear algebra this quatrimester, so I'd appreciate if you explain some of the above. I know the basics about groups, rings, integral domains and fields to get to define vector spaces, but not much more of abstract algebra than that. For example, I don't understand what "naturally acts on", "the natural target for" mean.2012-10-28
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    @Peter: can you ask a more specific question? What I mean is that the action of $\text{GL}(V)$ on $V$ induces various other actions functorially, e.g. the action of $\text{GL}(V)$ on the projective space $\mathbb{P}(V)$ of lines in $V$. This action factors through the projective general linear group $\text{PGL}(V)$ because scalars all act trivially on lines. The second use of "natural" may be regarded as a definition of projective representation.2012-10-28
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    Oh, well. My point is I know too little algebra to understand what you're telling me. What does "factors through" mean, for example? I don't about "functors" either, so maybe I'll have to wait a little till I come back to look at this. Also, I now realize that one thing is not clear in my question. I wanted to know what functions correspond to the $n$th dimensional generalization. Maybe a multivariate version of Möbius' Transformations?2012-10-28
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    @Peter: it means that the map $\text{GL}(V) \to \text{Aut}(\mathbb{P}(V))$ is the composition of a pair of maps $\text{GL}(V) \to \text{PGL}(V), \text{PGL}(V) \to \text{Aut}(\mathbb{P}(V))$. Again, all I am really saying is that the scalars act trivially on projective space. The functions you want are, as I said, precisely the elements of the projective linear group. These are the appropriate generalization of Mobius transformations to higher dimensions.2012-10-28
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    @Peter: in three dimensions elements of the projective linear group may be written down explicitly as pairs of rational functions $\left( \frac{ax + by + c}{dx + ey + f}, \frac{gx + hy + i}{dx + ey + f} \right)$, and this generalizes to higher dimensions. But this gets less useful the higher up you go; it really is better to think about projective space.2012-10-28