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A group $G$ acts $k$-transitive on some set $X$ if for every two $k$-tupels $(x_1, \ldots, x_k)$ and $(y_1, \ldots, y_k)$ there exists some $g \in G$ such that $$ g\cdot x_1 = y_1, \ldots, g\cdot x_k = y_k. $$ (if $k = 1$ we simple say $G$ acts transitively).

Let $K$ some field, and $V$ be a finite dimensinoal vector space over $K$. Then denote by $SL(V)$ the space of all linear transformations with determinant $1$, the so called special linear group, and further denote by $PSL(V)$ the projective special linear group

Now I am asked to prove that $SL(V)$ and $PSL(V)$ act $2$-transitively on every linear subspace of dimension $1$, but I think this could be sharpened, but I nowhere find this result so I am asking if I overlooked something. My claim is:

Proposition: Let $n = \dim(V)$, then if $n > 1$ the group $SL(V)$ as well as $PSL(V)$ act $n$-transitively on $\{ U \le V : \dim U = 1 \}$, and if $n = 1$ these groups acts $k$-transitively for every $k$.

Proof: i) Let $n > 1$ and let $U_1, \ldots, U_n$ be distinct one-dimensional subspaces of $V$, then they are generated by vectors $u_1, \ldots, u_n$ which are linear independent because these subspaces are distinct, further let $W_1, \ldots, W_n$ be another set of distinct lineary independent one-dimensional subspaces with linear independent generating vectors $w_1, \ldots, w_n$. Then define the linear map $A : V \to V$ by $$ A(u_i) = w_i $$ and set $\alpha := \det(A)$. By the linear independence it is $\alpha \ne 0$. Now define $$ A'(u_1) = \frac{1}{\alpha} w_1 \quad \mbox{and} \quad A'(u_i) = w_i, ~ i=2,\ldots, n. $$ Then if fix $u_1, \ldots, u_n$ as a basis for $V$, and consider the matrix $[A] = (a_{ij})$ with respect to that basis, by definition we have for $$ [A'] = \left\{ \begin{array}{ll} \frac{1}{\alpha} a_{ij} & i = 1 \\ a_{ij} & i \ne 1 \end{array}\right. $$ so that by the linearity of the determinant function in its rows we have $$ \det(A') = \frac{1}{\alpha} \det(A) = \frac{\det A}{\det A} = 1 $$ therefore $A' \in SL(V)$.

Further in the projective special linear space, two $n$ distinct vectors correspond two $n$ distinct linear $1$-dimensional subspaces, and by the above $SL(V)$ act $n$-transitively on them, so $PSL(V)$ too.

For the case $n = 1$, all $1$-dimensional subspaces equal $V$, and so the identity map maps all $k$-tupel of linear subspace onto each other. $\square$

So is my proof correct, and why everywhere I find just the result that they act $2$-transitively?

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  • $\begingroup$ I'm not sure if I understand the Proposition, but how can a regular matrix take, let's say, $n$ "independent" lines to $n$ lines contained in a $2$-plane, for example? $\endgroup$ – Peter Franek Jun 23 '14 at 22:49
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"Let $n > 1$ and let $U_1, \ldots, U_n$ be distinct one-dimensional subspaces of $V$, then they are generated by vectors $u_1, \ldots, u_n$ which are linear independent because these subspaces are distinct". The mistake is highlighted. When you have two non-zero vectors from distinct lines they are automatically linearly independent, but when you have three or more this is false. For example $i,j$ and $i+j$ span distinct lines, but they are not linearly independent.

And this is why any matrix group can be no better than 2-transitive on lines, tuples of lines that span subspaces of different dimensions can not be related by a non-singular matrix.

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The action will not be $n$ transitive, for instance, if $n\geq 3$ and if you take a $k$-uple, $k\geq 3$ of lines that are all contained in a $2$ dimensional subspace of $V$, then any translates by $SL(V)$ will have that same property.

However, if the ground field $\Bbbk$ is algebraically closed, $PSL(V)$ will act simply transitively on all circuits of lines, that is $(n+1)$-uples of lines such that any $n$-uple is linearly independent. This is because if $(\ell_1,\dots,\ell_{n},\ell_{n+1})$ is such a circuit, then taking a nonzero vector $x$ in $\ell_{n+1}$, it decomposes as $$x=x_1+\cdots+x_n$$ for some nonzero $x_i\in \ell_i$. If $(\ell_1^0,\dots,\ell_{n}^0,\ell_{n+1}^0)$ is a reference circuit (and $x^0=x_1^0+\cdots+x_n^0$ are defined similarly) then the linear automorphism $g$ defined by sending the $x_i^0$ to the $x_i$ (for $i=1,\dots,n$) will also send $x^0$ to $x$, and thus will have $g\cdot(\ell_1^0,\dots,\ell_{n}^0,\ell_{n+1}^0)=(\ell_1,\dots,\ell_{n},\ell_{n+1})$. To convert this $g$ to an element $g'$ of $SL(V)$ you consider $g'=\lambda g$ where $\lambda^n\det(g)=1$. Such a $\lambda$ always exists because $\Bbbk$ is assumed to be algebraically closed.

Simple transitivity is easy enough to prove, I'll just leave this hanging for now unless you're interested.

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You mention that if $U_1,\ldots,U_n$ are distinct 1-dimensional subspaces of $V$, then the vectors $u_1,\ldots,u_n$ that span them are linearly independent, but this need not be the case. For example, in $\mathbb{R}^2$, the three 1-dimensional subspaces $\langle (1,0) \rangle, \langle (0,1) \rangle$, and $\langle (1,1) \rangle$ are distinct, but the three vectors spanning them are not linearly independent.

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