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Notation: $\mathbb{P}^2$ denotes complex projective plane.

We have an action $$SL_3 \times \mathbb{P}^2 \to \mathbb{P}^2, \ (A,[v])\mapsto [Av]$$ with kernel $\mathbb{C}^*.Id\cap SL_3 \cong C_3$ where $C_3$ is cyclic group of order $3$ generated by a third root of unity. $SL_3 / C_3 = PSL_3$ acts effectively on $\mathbb{P}^2$. Hence, we have an injective homomorphism $$ \phi: PSL_3 \hookrightarrow Aut(\mathbb{P}^2).$$

Is it true that $PSL_3 \cong Aut(\mathbb{P}^2)$ ? How to prove this? I know that $Aut(\mathbb{P}^2) \cong PGL_3$? But I am not sure how does this help?

  • Please write $\mathbb{P}^2$, not $\mathbb{P}_2$. – Martin Brandenburg Dec 03 '14 at 15:49
  • sorry! I will edit. –  Dec 03 '14 at 17:06
  • It's not too hard to see that $PGL_3=PSL_3$. Projectivization makes all scalar matrices equivalent to the identity, so that there's a special linear matrix in every equivalence class-multiply by 1 over the determinant. – Kevin Carlson Dec 03 '14 at 17:19
  • But scalars in $PGL_3$ are arbitrary while in $PSL_3$ are only third roots of unity. Maybe I did not understand what you mean. –  Dec 03 '14 at 18:00
  • This fact (or actually, a more general statement) is known as the "Fundamental Theorem of Projective Geometry". You can find it as Theorem 1.5 in this book, which is available online. What you are looking for is the Corollary 1.7. – Jesko Hüttenhain Dec 03 '14 at 19:43
  • Great! Thanks @JeskoHüttenhain. –  Dec 03 '14 at 20:57
  • @JeskoHüttenhain, btw could you look into this question of mine http://math.stackexchange.com/questions/1049937/how-does-molien-series-describe-polynomial-invariants. I saw in your profile that you work on invariant theory. Please suggest an answer to the question, if you have time. Thanks in advance. –  Dec 03 '14 at 20:59
  • Sure thing. I left an answer for you, maybe it can help. – Jesko Hüttenhain Dec 03 '14 at 21:33

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