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An anti-involution of $\mathbf{P}^n (\mathbf{C})$ is an anti-homography $\phi$ such that $\phi^2 = \mathrm{Id}$. Is it true that all anti-involutions are induced by anti-involutions on $\mathbf{C}^{n+1}$?

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1 Answer 1

Not quite. Look at the map $\tau: (w:z) \mapsto (-\overline{z}: \overline{w})$ from $\mathbb{P}^1$ to itself. So $\tau^2 : (w:z) \mapsto (-w:-z)$, which is the same point as $(w:z)$, so $\tau$ is involutary. Any lift of this to $GL_2(\mathbb{C})$ would look like $\sigma: (w,z) \mapsto (-\lambda \overline{z}, \lambda \overline{w})$ for some nonzero scalar $\lambda$. (I am using colons for homogenous coordinates and commas for ordinary coordinates.) Then $\sigma^2: (w,z) = (- \lambda \overline{\lambda} w, - \lambda \overline{\lambda} z)$. Since $\lambda \overline{\lambda} >0$, it is impossible that $\sigma^2$ be the identity, in other words, it is impossible for $\sigma$ to be an involution.


This is the only issue.

Let $\rho$ be the standard anti-homography $(z_1: \ldots: z_n) \mapsto (\overline{z_1}: \ldots:\overline{z_n})$. Let $\tau$ be any other anti-homography. Then $\rho \circ \tau$ is a homography, and is hence induced by a linear map $A \in GL_{n+1}(\mathbb{C})$. So every anti-homography is of the form $z \mapsto \overline{Az}$.

The hypothesis that $\tau$ is an involution means that $\overline{A} A$ induces the identity on $\mathbb{P}^n$. So $\overline{A} A = \lambda \mathrm{Id}$ for some scalar $\lambda$. If $\lambda$ is a positive real, then we can multiply $A$ by $\sqrt{\lambda}^{-1}$ to make $\overline{A} A = \mathrm{Id}$. If $\lambda$ is not a positive real, we lose.

I think there are probably some more things to say here, but this should give you a good start.

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Nice answer, but I'm really supposed to downvote you. I'm across the ocean right now, so it is after 5 pm. :) –  Ted Shifrin Aug 1 '13 at 15:36

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