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At first, I figured that an automorphism of a scheme $X$ would be a homeomorphism $f:|X| \to |X|$ of topological and an isomorphism of sheaves $f^{\#}: \mathcal{O}_X \to f_*(\mathcal{O}_X)$.

However, if $X= \mathbb{P}_k^n$, then the automorphisms of $X$ are actually defined as follows:

Let $\textbf{Aut}(\mathbb{P}^n)$ denote the functor taking a scheme $S=\operatorname{Spec}A$. where $A$ is a commutative algebra over an algebraically closed field $k$, to the group of automorphisms $\operatorname{Aut}_A(\mathbb{P}_A^n)$ of $\mathbb{P}_A^n$ over $A$.

I think the definition means that we need to look at the "automorphisms of all $A$-points of $\mathbb{P}_k^n$". I'm not sure if this last phrase is correct, I've just heard it being said.

However, how does this naturally follow from the "naive" interpretation of what an automorphism of a scheme $X$ would be? I can't figure out why or when the $A$-points of a scheme are of significance. For example, when Hartshorne discusses the automorphism group of $\mathbb{P}_k^n$ he determines it to be $PGL(n, k)$, i.e he really only considers the $k$-points of the group scheme $\textbf{PGL}(n)$.

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  • $\begingroup$ I actually get a bit confused at exactly the meaning of $\mathbb{P}_k^n$. I know that $\mathbb{P}^n$ is a scheme representing a particular functor. I thought that then $\mathbb{P}_k^n$ denotes the scheme representing that functor restricted to $k$-algebras. $\endgroup$ – user7090 Apr 29 '19 at 0:46
  • $\begingroup$ I thought $\mathbb{P})k^1$ was a $k$-scheme by definition.. $\endgroup$ – user7090 Apr 29 '19 at 1:37
  • $\begingroup$ Are you saying that the definition I gave for an automorphism is for a plain scheme $\mathbb{P}^n$ but the group Hartshorne computes is for the $k$-scheme $\mathbb{P}_k^1$? $\endgroup$ – user7090 Apr 29 '19 at 1:40
  • $\begingroup$ Ok. It just seems like we switched to the category of schemes over $S$ when we compute the $S$-points of $Aut_k(X)(S)=Aut_S(\mathbb{P}_S^n)$. $\endgroup$ – user7090 Apr 29 '19 at 2:36
  • $\begingroup$ (I have deleted my previous comments and made them into an answer) $\endgroup$ – Qfwfq Apr 29 '19 at 15:10
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I just organized my comments into an answer/review (though the existing answers are already perfect).

Schemes over $k$.

The category $Sch_S$ of schemes over a given scheme $S$ is by definition the comma category over $S$, i.e. objects are morphisms of schemes $\pi:X\to S$ and arrows are morphisms of schemes making the resulting diagram commute. For $S=\mathrm{Spec}k$ we get the category $Sch_k$ of $k$-schemes. We will denote by $Sch$ the category of plain schemes over the integers (though in real life I'd not dare touching anything that is not, say, over $\mathbb C$...).

There is a forgetful functor $u:Sch_k \to Sch$, $(X\to \mathrm{Spec}k)\mapsto X$ sending a $k$-scheme to its underlying ($\mathbb Z$-) scheme. Notice that while e.g. $Spec\mathbb{C}\in Sch_\mathbb{C}$ is a one point scheme, $u(Spec\mathbb{C})\in Sch$ is a pretty messy object.

There is also a base change functor $b_k:Sch\to Sch_k$, $X\mapsto (X\times_{Spec \mathbb Z}Spec k\to Spec k)$. Notice that $b_k$ is definitely not the inverse of $u$ (e.g. $\mathbb{C}\otimes_{\mathbb Z}\mathbb{C}\neq \mathbb C$, and $\mathbb{Z}\otimes_{\mathbb Z}\mathbb C\neq \mathbb Z$).

Given $X$ over $k$, there are the two corresponding functors of points, $h_{X/k}:Sch_k^{op}\to Set$ and $h_{X/\mathbb Z}:Sch^{op}\to Set$. The former is not "the restriction" of the latter: $Sch_k$ is not a subcategory of $Sch$, rather a comma category.

Notice that $h_{X/k}\neq h_{X/\mathbb Z}\circ u$ (where I've still called $u:Sch_k^{op}\to Sch^{op}$): take e.g. $X=Speck$ and evaluate both functors at $T=Speck$. What you can say in general is $h_{X/k}\subseteq h_{X/\mathbb Z}\circ u$ is a subfunctor. Likewise, $h_{X/\mathbb Z}\neq h_{X/k}\circ b_k$: take e.g. $X=Speck$ and evaluate at $T=Spec\mathbb Z$.

What's the relationship between $\mathbb{P}^n_{\mathbb Z}$ and $\mathbb{P}^n_k$? The latter is just the image of the former under the functor $b_k$. But the functor of points of $\mathbb{P}^n_k$ is not a restriction of the functor of points of $\mathbb{P}^n_{\mathbb Z}$, it's really another functor: it gives the set of only morphisms over $k$ as opposed to all the morphisms.

Notice that the automorphism group of a $k$-scheme $X$ as a $k$-scheme i.e. in $Sch_k$ may be completely different from the automorphism group of $X$ as a plain scheme. Consider for example $X=Spec\mathbb{C}$: $Aut_{\mathbb C}(X)$ is trivial, while $Aut_{\mathbb Z}(X)$ is all the ring automorphisms of the $\mathbb Z$-algebra $\mathbb C$.

The same happens for projective space. Consider the $k$-scheme $\mathbb{P}^n_k$. If we consider it in $Sch_k$, then its $Aut$ is $PGL_n(k)$. I don't know what happens over the integers, but if we consider $\mathbb{P}^n_k$ as a scheme over the prime field $k_0$ of $k$, then $Aut$ is a semidirect product of $PGL_n(k)$ and the Galois group of $k$ over $k_0$. Apparently, the $Aut_{k_0}$ is the automorphism group of the abstract projective geometry (in the sense of incidence structures) induced by $\mathbb{P}^n_k$.


Automorphisms functor.

This is an aspect orthogonal to the above one. Let's consider everything in $Sch_k$ for simplicity.

The automorphisms functor of $X\in Sch_k$ is defined by $\mathbf{Aut}_k(X):S\mapsto Aut_S(X\times_k S)$, where $Aut_S$ denotes automorphisms in $Sch_S$ for every $S\in Sch_k$.

Now you see that, whether this is representable or not, we can talk about the $k$-points of $\mathbf{Aut}_k(X)$ and it is obvious from the definition that $(\mathbf{Aut}_k(X))(k)=Aut_k(X)$ as abstract groups. So, the automorphisms functor is designed to have as its $k$-points precisely the automorphisms of $X$ as an object of $Sch_k$.

What about more general points of $\mathbf{Aut}_k(X)$? You can think of $X\times_k S$ as a trivial $X$-bundle over $S$. An $S$-point $\sigma$ of $\mathbf{Aut}_k(X)$ is an isomorphism $\sigma:X\times S\to X\times S$ that commutes with projections to the base $S$. So you can see $\sigma$ as a "bundle automorphism" of $X\times_k S$, or as a family of automorphisms of $X$ parametrized by $S$.

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  • $\begingroup$ Sorry I am still very confused by the statement "consider the $k$-scheme $\mathbb{P}_n^k$, If we consider it in $\operatorname{Sch}_k.$." Isn't $\mathbb{P}_n^k$, in $\operatorname{Sch}_k$ be definition? Does it make sense to consider $\mathbb{P}_n^k$ in $\operatorname{Sch}_S$? $\endgroup$ – user7090 Apr 29 '19 at 19:09
  • $\begingroup$ I have been kind of pedantic throughout the answer; that statement was an innocuous abuse of language: I meant, consider $(\pi:\mathbb{P}^n_k\to\mathrm{Spec}(k))\in\mathrm{Sch}_k$ as opposed to just $\mathbb{P}^n_k\in\mathrm{Sch}_{\mathbb Z}$. $\endgroup$ – Qfwfq Apr 29 '19 at 19:24
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The object you've defined is not the group of automorphisms of $\mathbb{P}^n$; among other things, it is a group-valued functor, not a group. Here is a simpler example of this sort of thing:

In any category $C$, if $X, Y$ are two objects you can consider the set $\text{Hom}(X, Y)$ of morphisms $X \to Y$. If $C$ has finite products, then one can furthermore consider the presheaf sending an object $Z$ to the set $\text{Hom}(X \times Z, Y)$, which one can equivalently think of as $\text{Hom}_Z(X \times Z, Y \times Z)$, where $\text{Hom}_Z$ means the hom is taken in the slice category over $Z$. If this presheaf is representable, its representing object is called an exponential object $Y^X$. The exponential object is strictly more information than just the homset, which one recovers by taking global points $\text{Hom}(1, Y^X) \cong \text{Hom}(X, Y)$.

Similarly, in any category $C$, if $X$ is an object one can consider the group $\text{Aut}(X)$ of automorphisms of $X$, and if $C$ has finite products, then one can consider the group-valued presheaf sending an object $Z$ to the group $\text{Aut}_Z(X \times Z)$. If this presheaf is representable, we might call its representing object an "automorphism object" of $X$. Again it contains strictly more information than just the automorphism group of $X$, which again one recovers by taking global points.

$\mathbb{P}^n$ as a $k$-scheme has $PGL_n(k)$ as its group of automorphisms, and if I'm not mistaken its automorphism object furthermore exists and is the group scheme $PGL_n$ over $k$. Similarly one can construct the group scheme $GL_n$ over $k$ by considering automorphisms of base changes of $k^n$ as a $k$-vector space object in affine schemes over $k$.

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  • $\begingroup$ Thank you for the nice answer. However, I am still confused as to what exactly the automorphism object of a scheme encodes. That is, why would one want to look at this object? What information about the scheme $X$ does it provide? $\endgroup$ – user7090 Apr 29 '19 at 18:35
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    $\begingroup$ @user192302: the automorphisms object (provided it exists) tells you how the automorphisms of $X$ fit together geometrically, as opposed to just being a set with an abstract group operation. The algebraic group $PGL_{n,k}$ is an algebraic variety and a group (in a compatible way), not just a group. $\endgroup$ – Qfwfq Apr 29 '19 at 19:28
  • $\begingroup$ @user192302: the automorphism object tells you what parameterized families of automorphisms look like. $\endgroup$ – Qiaochu Yuan May 2 '19 at 1:31
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I am not sure I understand your problem, but maybe what you are missing is that for a projective variety $X$, the functor $A\mapsto \operatorname{Aut}(X_A) $ is representable - that is, there is a group scheme $\operatorname{\underline{Aut}}_X$ such that $\operatorname{\underline{Aut}}_X(A)=\operatorname{Aut}(X_A) $ (this follows directly from the theory of the Hilbert scheme). In particular, the group of $k$-points $\operatorname{\underline{Aut}}_X(k)$ is the usual automorphism group $\operatorname{Aut}(X) $.

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