6
$\begingroup$

How to construct an injection of $PGL(n,k)$ in $PGL(n+1,k)$ if $GL(n,k)$ injects in $GL(n+1,k)$. I think it depends on the field k. For example, if we put $\varphi:GL(n,k)\longrightarrow GL(n+1,k)$ defined by :

$$\varphi(g)=\left( \begin{array}{cc} g & 0 \cr 0 & \chi(g) \cr \end{array} \right)$$

where $\chi$ is a caracter of $GL(n,k)$, that is a morphism of groups of $GL(n,k)$ to $k^{\times}$, then $\chi$ has the form $\chi=\phi\circ det$, where $\phi$ is an endomorphism of $k^{\times}$.

Then, $\varphi$ induces a morphisme of groups of $PGL(n,k)$ in $PGL(n+1,k)$ if and only if the morphism $\phi$ satisfies : For all $x\in k^{\times}$, $\phi(x)^{n}=x$.

For $k=\mathbb{R}$ that is imposible.

Improve this question
$\endgroup$
3
  • $\begingroup$ What's wrong about just $1$ instead of $\chi(g)$? $\endgroup$
    – Marc Palm
    Commented Mar 16, 2012 at 8:40
  • $\begingroup$ @pm: It doesn't preserve the $k^*$ action, so does not lift to PGL. $\endgroup$
    – Martin Brandenburg
    Commented Mar 16, 2012 at 8:57
  • $\begingroup$ Exactly, that doesn't induced a morphism. $\endgroup$
    – Rajkarov
    Commented Mar 16, 2012 at 11:39

1 Answer 1

Reset to default
12
$\begingroup$

You can't do it in general. A quick computer calculation (I used Magma) shows that ${\rm PGL}(4,4)$ has no subgroup isomorphic to ${\rm PGL}(3,4)$. (It does have one isomorphic to ${\rm SL}(3,4)$.)

I suspect that there is an embedding ${\rm PGL}(2,K) \to {\rm PGL}(3,K)$, but that is coming from the irreducible orthogonal action of ${\rm GL}(2,K)$ on a 3-dimensional module.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Thank you Mr Holt for precision. I'm interested in this injection for $k$ non archimedean local field. You say that $PGL(2,K)$ embeds in $PGL(3,K)$ but is it true that $PGL(2n,K)$ embeds in $PGL(2n+1,K)$ ? $\endgroup$
    – Rajkarov
    Commented Mar 16, 2012 at 11:49
  • 1
    $\begingroup$ I think the embedding $\mathrm{PGL}(2,K) \to \mathrm{PGL}(3,K)$ can be seen as follows : let $\mathrm{GL}_(2,K)$ act naturally on the space of binary quadratic forms over $K$. If $g \in \mathrm{GL}_(2,K)$ is such that $Q \circ g =\lambda Q$ for every binary quadratic form $Q$, then by taking squares of linear forms we see that $g$ fixes every elements of $\mathbf{P}^1(K)$, so that $g$ is scalar. $\endgroup$
    – François Brunault
    Commented Mar 16, 2012 at 11:56
  • 3
    $\begingroup$ Rajkarov: No. For example ${\rm PGL}(5,3)$ has no subgroup isomorphic to ${\rm PGL}(4,3)$. The 2- to 3-dimensional embedding is exceptional. $\endgroup$
    – Derek Holt
    Commented Mar 16, 2012 at 15:16

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .