felixwellen / synthetic-zariski

Latex documentation of our understanding of the synthetic /internal theory of the Zariski-Topos
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Automorphisms of affine and projective space #4

Closed felixwellen closed 2 months ago

felixwellen commented 1 year ago

We should write down what we know about those automorphism groups.

dwarn commented 1 year ago

I don't have time to write this up properly at the moment, so I just sketch the argument for automorphisms of A^1 as I remember it from discussions with Ingo. Much of it should generalise to other spaces. An endomorphism of A^1 is given by a polynomial. The claim is that it is an isomorphism iff this polynomial is not not linear with invertible leading coefficient. If it is not not of this form, we can write down a candidate for its inverse, by inverting the linear part. One can use "Hensel lifting" (5.4 in diffgeo) to improve this to an actual inverse. Importantly, the derivative is invertible at each point, since to be invertible is not-not stable.

If our polynomial is an isomorphism we show it is not not linear with invertible leading coefficient. Our polynomial is not not a unit times a product of linear factors, so suppose it is. We claim the number of linear factors is 1. Of course it cannot be 0. It cannot be more than 1 because then there is (up to not not) multiple roots or a double root. So it is 1, as needed.

felixwellen commented 10 months ago

Here is an update: By now we expect that $\mathbb P^n$ has the classical automorphism group $\mathrm{PGL}_{n+1}$. David proved this for $n=1$, here is a sketch of my parallel, less direct proof attempt (which also works): We can take a Möbius transformation to turn an automorphism $\mathbb P^1$ into a map fixing 0,1, $\infty$ . Then it is an automorphism of $\mathbb A^1$ on each chart, given by polynomials of the form $P=\alpha X+\epsilon_2 X^2+\dots \epsilon_n X^n$ and $Q=\beta Y+\delta_2 Y^2+\dots \delta_l Y^l$ (where all the $\epsilon$ s and $\delta$ s are nilpotent). By pushout-recursion for $\mathbb P^1$ and writing everything down carefully, we get the relation $P(X)\cdot Q(\frac{1}{X})=1$. This implies all the $\epsilon$ s and $\delta$ s are 0 and $\alpha$ and $\beta$ are 1. This does not directly generalize to all $n$, because we don't know that automorphisms preserve hyperplanes (we used that for n=1, points are preserved).

felixwellen commented 2 months ago

The general case for automorphisms of $\mathbb P^n$ is written down in here