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The general projective transformation of the $x$-$y$ plane is given by $$\tilde{x}=\frac{a_1x+a_2 y+a_3}{a_7x+a_8y+a_9},\quad\tilde{y}=\frac{a_4x+a_5y+a_6}{a_7x+a_8y+a_9}$$ for some constants $a_i\in\mathbb{R}$. From this, we find the infinitesimal generators $$\begin{array}{lll} X_1=x\partial_x,& X_2=y\partial_x,& X_3=\partial_x, \\ X_4=x\partial_y,& X_5=y\partial_y,& X_6=\partial_y, \\ X_7=-x^2\partial_x-xy\partial_y,& X_8=-xy\partial_x-y^2\partial_y,& X_9=-x\partial_x-y\partial_y \end{array} $$ corresponding respectively to each parameter $a_i$. With the commutator $$[X_i,X_j]=X_iX_j-X_jX_i,$$ the vector space generated by $\{X_1,\ldots,X_9\}$ forms a Lie algebra. Let us denote this vector space by $\mathfrak{g}$. A basis for that space could be $\{X_1,\ldots,X_8\}$ since $X_9=-X_1-X_5$.

I want to show that it is isomorphic to the Lie algebra $$\mathfrak{sl}(3,\mathbb{R}):=\{A\in\mathbf{M}_{3\times 3}(\mathbb{R}):\text{trace}(A)=0\}$$ with commutator $[A,B]=AB-BA$, by explicitly giving an isomorphism $$f:(\mathfrak{sl}(3,\mathbb{R}),[\cdot,\cdot])\longrightarrow(\mathfrak{g},[\cdot,\cdot])$$

I made a few attempts that all failed. Since $X_1+X_5+X_9=0$, we can see a sort of correspondence with traceless matrices, but I don't see how to use that.

EDIT: To get the generators $X_i$, we differentiate with respect to each $a_i$ both $\tilde{x}$ and $\tilde{y}$ and evaluate at the identity transformation, that is at $a_1=1,a_5=1,a_9=1$ and $a_2=a_3=a_4=a_6=a_7=a_8=0$. For example, to get $X_2$, we have $$\frac{d\tilde{x}}{da_2}=\frac{y}{a_7x+a_8y+a_9},\quad\frac{d\tilde{y}}{da_2}=0$$ which evaluates to $y$ and $0$ respectively so that $X_2=y\partial_x+0\partial_y=y\partial_x$.

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    $\begingroup$ Can you show how you obtain the formulas for the infinitesimal generators, say $X_2$? $\endgroup$ – Sammy Black May 30 '13 at 17:26
  • $\begingroup$ @SammyBlack Done. $\endgroup$ – Spenser May 30 '13 at 17:41
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I think that some slightly better notation solves your problem. Let $$ \begin{array}{lll} Y_{1,1} = x\partial_x, & Y_{1,2} = x\partial_y, & Y_{1,3} = -x^2\partial_x - xy\partial_y \\ Y_{2,1} = y\partial_x, & Y_{2,2} = y\partial_y, & Y_{2,3} = -xy\partial_x - y^2\partial_y\\ Y_{3,1} = \partial_x, & Y_{3,2} = \partial_y, & Y_{3,3} = -x\partial_x - y\partial_y. \end{array} $$ This is essentially the transpose of your setup, but indexed in as a matrix with a pair of coordinates. In other words, $$ Y_{i,j} = \partial_{a_{i,j}} = \frac{\partial x}{\partial a_{i,j}} \partial_x + \frac{\partial y}{\partial a_{i,j}} \partial_y \qquad \textrm{for each } 1 \le i,j \le 3. $$ Now, the generators $Y_{i,j}$ of $\mathfrak{g}$ behave like matrix units under the commutator bracket. In other words, $$ [Y_{i,j}, Y_{k,l}] = \delta_{j,k}Y_{i,l} - \delta_{l,i}Y_{k,j}. $$ Your isomorphism is just the identity map.

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Over any field $k$, the group $G=GL_{n+1}(k)$ acts on $Pk^n$ via linear maps on $k^{n+1}$, and the quotient map $q:k^{n+1}-\{0\} \rightarrow Pk^n$, whose fibers are lines (with $0$ removed). In matrices, on the affine subset identifiable with homogeneous coords $\{x_i\}$ with $x_{n+1}=1$, this is $$ \{g_{ij}\}\cdot \pmatrix{x_1 \cr x_2 \cr \vdots \cr x_n \cr 1} \;=\; \pmatrix{ g_{11}x_1+g_{12}x_2+\ldots+g_{1,n}x_n+g_{1,n+1}\cdot 1 \cr g_{21}x_1+g_{22}x_2+\ldots+g_{2,n}x_n+g_{2,n+1}\cdot 1 \cr \vdots \cr g_{n+1,1}x_1+g_{n+1,2}x_2+\ldots+g_{n+1,n}x_n+g_{n+1,n+1}\cdot 1 } $$ Renormalizing to get the representative with last coordinate $1$, this is $$ g\cdot x \;=\; \pmatrix{ {g_{11}x_1+g_{12}x_2+\ldots+g_{1,n}x_n+g_{1,n+1}\cdot 1 \over g_{n+1,1}x_1+g_{n+1,2}x_2+\ldots+g_{n+1,n}x_n+g_{n+1,n+1}\cdot 1 } \cr \vdots \cr {g_{n,1}x_1+g_{n,2}x_2+\ldots+g_{n,n}x_n+g_{n,n+1}\cdot 1 \over g_{n+1,1}x_1+g_{n+1,2}x_2+\ldots+g_{n+1,n}x_n+g_{n+1,n+1}\cdot 1 } \cr 1 } $$ Visibly the center acts trivially here, and $SL_{n+1}(k)$ maps finite-to-one to $PGL_{n+1}(k)$, so the Lie algebras are the same. That is, the projective-affine transformation group $PGL_{n+1}(k)$ has the same Lie algebra as $SL_{n+1}(k)$, described however one likes in further detail.

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Rough idea: Consider the projective plane $PR^2$. A protective transformation on $R^2$ corresponds to a projective tranformation on $PR^2$. Since $Aut(PR^2)= \{T\in M_{3\times 3}| \det T=1\}$, you get $\mathfrak{aut}(PR^2)=\{t\in M_{3\times 3}| \operatorname{tr}(t)=0\}$.

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    $\begingroup$ That looks about right, but can the isomorphism be made explicit? $\endgroup$ – Sammy Black May 30 '13 at 18:17
  • $\begingroup$ Thanks @SammyBlack. My question is indeed to give an explicit isomorphism. $\endgroup$ – Spenser May 30 '13 at 18:18

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