Transvections

Assume $n \ge 2$ throughout,

Definition A linear functional on $V$ is a linear map from $V$ to $\mathbb F_q$. The set of these is the dual space $V^*$. Given $f \in V^*$ write $V_f$ for the kernel of $f$, if $f \not = 0$ then $V_f$ is a subspace of dimension $n-1$ (any such space of codimension 1 we call a hyperplane).

Lemma If you have $f,f' \in V$ with the same hyperplane then there exists $\lambda$ such that $f' = \lambda f$.
proof: The result is clear if $V_f=V$ assume not so $f,f'\not = 0$. Take $v \in V \setminus V_f$ (i.e. not in the common hyperplane) so $vf, vf' \not = 0$ and put $\lambda = (vf')(vf)^{-1}$ then $f' - \lambda f \in V^*$ and its kernel is contained in $\langle V_f, v \rangle = V$ so it equals zero.

Definition A linear automorphism $\tau \in GL(V)$ is called a transvection with direction $d \in V^\#$ if $\tau$ fixes $d$ and $$v \tau - v$$ is a scalar multiple of $v$ for all $v$.

Clearly $1$ is a transvection of any direction.

Lemma If $\tau$ is a transvection with direction $d$ then the vectorspace $\operatorname{fix}(\tau)$ is a hyperplane containing $d$.
proof: Define $f : V \to \mathbb F_q$ by $(vf)d = v \tau - v$ i.e. $f$ is that scalar multiple which a transvection defines. Since $\tau$ is linear $f \in V^*$. The result is clear for $\tau = 1$ and if not $f \not = 0$ so $d \in \operatorname{fix}(\tau) = V_f$.

Corollary Any transvection can be written as $\tau_{f,d}$ mapping $v$ to $v + (vf)d$ for some $f \in V^*$, $d \in (V_f)^\#$.

Lemma For $f,f' \in V^*$, $d \in V^\#$, $g \in GL(V)$ we have

• $\tau_{f,g}^g = \tau_{g^{-1}\circ f,dg}$
• $\tau_{f,d} \tau_{f',d} = \tau_{f+f',d}$

proof: first we need to check $(dg)(g^{-1} \circ f) = 0$ and $d(f+f')=0$ for the notation to be meaningful. Now the results follow by computation.

Definition For a direction $d \in V^\#$ set $\mathscr T(d) = \{\tau_{f,d}|f \in V^*, d \in V_f\}$ and $\mathscr T$ the union over all directions (all transvections).

Proposition $\mathscr T^\#$ is a single conjugacy class in $GL(V)$ and lies in $SL(V)$. If $n\ge 3$ then $\mathscr T^\#$ is a single conjugacy class in $SL(V)$.
proof: By the calculation lemma previous, we know that $\mathscr T^\#$ is closed under conjugation. Let $\tau_{f,d},\tau_{f',d'} \in \mathscr T^\#$ and write $e_1=d,e'_1=d'$ then take bases $e_1,\ldots,e_{n-1}$ and $e'_1,\ldots,e'_{n-1}$ of the hyperplanes $V_f,V_{f'}$, choose $e_n,e'_n$ such that $e_n f = e_n' f' = 1$. Now we have bases for $V$ so there is a GL map $g$ from one to the other.
For $i < n$ we have $e'_i(g^{-1} \circ f) = e_i f = 0$ so $V_{g^{-1} \circ f}$ is the space spanned by $\langle e'_1,\ldots,e'_{n-1} \rangle = V_{f'}$ so they are scalar multiples of each other, let $\lambda$ be such that $f' = \lambda (g^{-1} \circ f)$ and $$1 = e'_n f' = \lambda e'_n (g^{-1} \circ f) = \lambda e_n f = \lambda$$ so since $dg = d'$ it follows that $\tau_{f,d}^g = \tau_{f',d}$!
If they are all conjugate they all have the same determinant $\delta$, now $\det(\tau_{f,d}\tau_{f',d}) = \det(\tau_{f+f',d})$ so $\delta^2 = \delta$ proves they lie in $SL$.
For $n \ge 3$ we can use the $\mu$ trick as before to get them in $SL$.

Proposition If $d \in V^\#$ then $\mathscr T(d)$ is an abelian normal subgroup of the stabilizer $SL(V)_d$; $\mathscr T(d)$ are all conjugates in $SL(V)$.
proof: Certainly elements of $\mathscr T(d)$ stabilize $d$, by the computational lemma before we see that it is an abelian group (commutative and closed, therefore has identity and inverses). If $g \in SL(V)$ with $dg=d$ then $\mathscr T(d)^g = \mathscr T(d)$ (by the previous) so $\mathscr T(d) \unlhd SL(V)_d$. Given $d,d' \in V^\#$ by the transitivity lemma (that requires dimension > 1) exists $g \in SL(V)$ which takes $d$ to $d'$ then $\mathscr T(d)^g = \mathscr T(d')$.

Proposition The set $\mathscr T$ generates $SL(V)$.
proof: Elementary matrices.

Definition A group if perfect if $G' = G$. This is equivalent to there being no nontrivial abelian quotients: Clearly if $G/[G,G] \not = 1$ then $G\not = [G,G]$. Conversely if $G/N \simeq A = \{Ng\}$ then we always have $Ngg' = Ng'g$ i.e. there is some $n$ such that $gg' = ng'g$. So every commutator $[g,g']$ is an element of $N$.

Proposition If $n \ge 2$ the group $SL_n(q)$ is perfect provided $(n,q)$ isn't $(2,2)$ or $(2,3)$.
proof: We just need to show that each $\tau \in \mathscr T^\#$ is a commutator since that set generates our group. If $\tau$ has direction $d$ take some $\sigma \in \mathscr T(d)^\#$ not equal to $\tau^{-1}$, then $\sigma \tau \in \mathscr T(d)^\#$ so there is a $g \in SL_n(q)$ that conjugates $\sigma \tau = \sigma^g$, whence $\tau = \sigma^{-1} g^{-1} \sigma g= [\sigma,g]$.
For $n=2$ we will use $2 \times 2$ matrices directly, in some basis $\tau = \begin{pmatrix} 1 & \gamma \\ 0 & 1 \end{pmatrix}$ (nonzero $\gamma$), now for any nonzero $\lambda$ and $\mu \in \mathbb F_q$ we have $$\begin{pmatrix} \lambda & 0 \\ 0 & \lambda^{-1} \end{pmatrix}\begin{pmatrix} 1 & \mu \\ 0 & 1 \end{pmatrix}\begin{pmatrix} \lambda^{-1} & 0 \\ 0 & \lambda \end{pmatrix}\begin{pmatrix} 1 & -\mu \\ 0 & 1 \end{pmatrix}=\begin{pmatrix} 1 & \mu(\lambda^2-1) \\ 0 & 1 \end{pmatrix}$$ so for $q>3$ take $\lambda \not = 0,1,-1$ then $\lambda^2-1\not = 0$ so let $\mu = \gamma(\lambda^2-1)^{-1}$.

Proposition

• $Z(GL(V)) = \{\lambda 1 | \lambda \in \mathbb F_q^\# \}$
• $Z(SL(V)) = \{\lambda 1 | \lambda \in \mathbb F_q^\#, \lambda^n = 1\}$

proof: Certainly these are contained in the center, suppose $g \in Z(GL(V))$ or $Z(SL(V))$ then for each nonzero vector $d \in V^\#$ take a transvection $\tau_{f,d} \in \mathscr T(d)$, this should be unaffected by conjugation but we know $\tau_{f,d}^g = \tau_{g^{-1} \circ f,dg}$ so $dg = \lambda_d d$. Then $$\lambda_d d + \lambda_v d = (d+v)g = \lambda_{d+v}(d+v) = \lambda_{d+v}d + \lambda_{d+v}v$$ so $\lambda_d = \lambda_v = \lambda$ and $g = \lambda \cdot I$. For $SL$ we also require $\det(g) = \lambda^n = 1$. This tells us that $Z(GL_n(q)) \simeq \mathbb F_q^\#$ and $Z(SL_n(q)) \simeq C_{(n,q-1)}$.