I stumbled over a statement on Wikipedia http://en.wikipedia.org/wiki/Duality_%28projective_geometry%29 and would like to ask how this could possibly be true. It states the following

The projective planes $PG(2,K)$ for any division ring $K$ are self-dual.

Is this really true? Or rather: What is wrong with the following argument that (if correct) would contradict the statement:

For a finite dimensional left vector space $V$ over a division ring $K$, we have the dual left vector space $V^{\star}$ over $K^{op}$ (the opposite division ring of $K$). Now, it is certainly true that $P(V^{\star})$ is isomorphic to the dual of $P(V)$. Thus, $P(V)$ is isomorphic to its dual if and only if $P(V) \cong P(V^{\star})$. Since $V$ and $V^{\star}$ have the same dimension, does that not necessarily imply that $K$ and $K^{op}$ are isomorphic by the Second Fundamental Theorem of projective Geometry?

In fact, isn't $PG(2,K)$ (or $PG(n,K)$ for any $n \geq 2$) being self-dual simply eqvuialent to $K \cong K^{op}$ (which is of course is not always the case). I also recall that I have read this somewhere.

So, to put it in one line: What is wrong? The statement on Wikipedia or my reasoning?

EDIT: I found some course notes in which it is also claimed that $PG(n,K)$ is self-dual iff $K$ is self-dual (see 6.1 in http://www.maths.qmul.ac.uk/~pjc/pps/pps6.pdf). This does contradict what Wikipedia says, doesn't it?

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    Your reasoning is very convincing. I think it's a good rule of thumb that when something in Wikipedia looks wrong, and not just at first glance, then it probably is. – Tom Goodwillie Nov 24 '12 at 15:03
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    Looking at the history of the article, that sentence appears to have been introduced by User:Wcherowi on September 25, 2011. en.wikipedia.org/w/… You can also write directly to the interested user on his/her talk-page (link from the above revision) – Pietro Majer Nov 24 '12 at 20:46
  • Thanks for the comment Tom and for your suggestion, Pietro. However, what I still would like to have is a definite answer to my original question. There are surely people on this board who know projective geometry well enough to decide whether PG(2,K) being self-dual really is equivalent to $K \cong K^{op}$ (contradicting the wikipedia article) or not. – E. Vargas Nov 25 '12 at 21:49
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    I can't say I've ever gone through the proof, and I didn't know that it went by one of those opaque numerical names ("Second Theorem of ..."), but I'm sure that there is a theorem to the effect that a division ring can be constructed (up to isomorphism) from the projective plane that it determines. So I am sure you are right. – Tom Goodwillie Nov 25 '12 at 23:02
  • Doesn't Desargues imply Pappus? In which case assuming K is not commutative gives a contradiction. Gerhard "Or Does Pappus Imply Desargues?" Paseman, 2012.11.25 – Gerhard Paseman Nov 26 '12 at 1:10
up vote 5 down vote accepted

I think you are right. There is a theorem that asserts that if two projective spaces $P(V,F)$ and $P(V',F')$ are isomorphic then the skew fields $F$ and $F'$ are isomorphic. Thus if $P(V,F)$ and $P(V^*,F^{op})$ are isomorphic then $F\cong F^{op}$ as skew fields.

I recommand these course notes "Essential Concepts of Projective Geometry" available here (the author gives a lot of references): http://math.ucr.edu/~res/progeom/ in particular look at appendix C.

  • David -- re "There is a theorem that asserts that...": do you have a reference? – algori Nov 26 '12 at 20:52
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    This is theorem IV.17 of "Essential concepts of projective geometry", the author gives a lot of references among them: chapter 6 of R. Hartshorne. "Foundations of Projective Geometry". Benjamin, New York, 1967. – David C Nov 27 '12 at 7:10

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