# Automorphisms of Schemes and their $A$-points

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 $$\textbf{PGL}(n)$$.

• 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. – Nadia Apr 29 '19 at 0:46
• I thought $\mathbb{P})k^1$ was a $k$-scheme by definition.. – Nadia Apr 29 '19 at 1:37
• 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 $\mathbb{P}_k^1$? – Nadia Apr 29 '19 at 1:40
• Ok. It just seems like we switched to the category of schemes over $S$ when we compute the $S$-points of $Aut_k(X)(S)=Aut_S(\mathbb{P}_S^n)$. – Nadia Apr 29 '19 at 2:36
• (I have deleted my previous comments and made them into an answer) – Qfwfq Apr 29 '19 at 15:10

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 $$S$$.

• 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 $\operatorname{Sch}_S$? – Nadia Apr 29 '19 at 19:09
• 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 $\mathbb{P}^n_k\in\mathrm{Sch}_{\mathbb Z}$. – Qfwfq Apr 29 '19 at 19:24

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 $$k$$.

• 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? – Nadia Apr 29 '19 at 18:35
• @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. – Qfwfq Apr 29 '19 at 19:28
• @user192302: the automorphism object tells you what parameterized families of automorphisms look like. – Qiaochu Yuan May 2 '19 at 1:31

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 $$\operatorname{Aut}(X)$$.