# automorphism of the projective space $\mathbb{P}_A^n$

In exercise 16.4.B of Vakil's notes, he establishes that the group of automorphisms of $\mathbb{P}_k^n$ is $PGL_{n+1}(k)$. This I can manage to show, but in the remarks following the exercise he asks why this does not work over an arbitrary base ring $A$, and I have a hard time to see why. I have three questions and would be grateful if someone can comment on them:

1) First, is it true that $\pi^*O_{\mathbb{P}_A}(1)\simeq O_{\mathbb{P}_A}(1)$ if $\pi:\mathbb{P}_A^{n}\to\mathbb{P}_A^{n}$ is an automorphism? Over a field $k$, this holds because an automorphism induces an isomorphism on Picard groups. So a generator must be sent to a generator, but I do not understand the Picard group of $\mathbb{P}_A^{n+1}$, so I can't see if this generalizes. In any case, if it is true, please give an argument. If it's not, please give a counter-example.

2) Does it make sense to define $PGL_{n+1}(-)$ as the functor that sends $A$ to the set of $(n+1)\times (n+1)$ invertible matrices over $A$ modulo multiples of identity? If so, is it representable (possibly in some sub-category of schemes over $A$)?

3) If $PGL_{n+1}(A)$ does not parametrize automorphisms of $\mathbb{P}_A^{n+1}$, what does it parametrize?

4) Is there a good description of automorphisms of $\mathbb{P}_A^n$?

• The Picard group of $\mathbb{P}^n_A$ is $\mathbb{Z}$ when $A$ is a factorial ring. If $A$ is not factorial, there is no hope to get such a simple group. Just let $n=0$. – Martin Brandenburg Oct 18 '14 at 20:58
• See proposition 6.12 in Hida's "p-Adic Automorphic Forms on Shimura Varieties" : If $S = Spec(R)$ is connected, then $PGL_n(R) = Aut(P^{n-1}_R)$. – Watson Nov 7 '18 at 19:01

Ch. 0, §5 of Mumford's Geometric Invariant Theory (3rd ed.) answers 1) and 4) well, I think. Let $PGL(n+1) = \{\operatorname{det} (a_{ij}) \ne 0\} \subset \operatorname{Proj}\mathbb{Z}[a_{00},\ldots,a_{nn}]$ denotes the projective general linear group which acts on $\mathbb{P}^n_\mathbb{Z}$ in the usual way. Let $\mathscr{PGL}(n+1)$ denote the functor $S \mapsto \operatorname{Hom}(S,PGL(n+1))$ in the category of noetherian schemes. Let $\operatorname{Aut}(\mathbb{P}^n)$ denote the functor sending $S$ to the group of automorphisms $\operatorname{Aut}_S(\mathbb{P}^n_S)$ of $\mathbb{P}^n_S$ over $S$.

Theorem. The functors $\mathscr{PGL}(n+1)$ and $\operatorname{Aut}(\mathbb{P}^n)$ are isomorphic.

I don't know of a better description in general (i.e. for non-noetherian schemes), but this seems to answer 4).

Moreover, his proof answers 1): the invertible sheaves on $\mathbb{P}^n_S = \mathbb{P}^n_\mathbb{Z} \times S$ are all of the form $p_1^*(\mathcal{O}_{\mathbb{P}^n_\mathbb{Z}}(k)) \otimes p_2^*(L)$ where $L$ is an invertible sheaf on $S$, and $p_1,p_2$ are the projection morphisms. So, if $\alpha \in \operatorname{Aut}_S(\mathbb{P}^n_S)$, then $\alpha^*(\mathcal{O}_{\mathbb{P}^n_S}(1)) \cong p_1^*(\mathcal{O}_{\mathbb{P}^n_\mathbb{Z}}(k)) \otimes p_2^*(L)$ for some $k$ and $L$. Then, since $\alpha$ is an automorphism over $S$, we must have

$$\bigoplus_{i=0}^n X_i \cdot \mathcal{O}_S \cong p_{2*}p_1^*\mathcal{O}_{\mathbb{P}^n_\mathbb{Z}}(1) \cong p_{2*}(p_1^*(\mathcal{O}_{\mathbb{P}^n_\mathbb{Z}}(k)) \otimes p_2^*(L)) \cong \bigoplus_{r_0+\cdots+r_n=k}^n (X_0^{r_0}\cdots X_n^{r_n}) \cdot L$$

so $k=1$ and in fact by [EGAII, 4.2] every isomorphism $\alpha$ can be obtained by choosing an invertible sheaf $L$ and an isomorphism $\bigoplus_{i=0}^n X_i \cdot \mathcal{O}_S \overset{\sim}{\to} \bigoplus_{i=0}^n X_i \cdot L$. What's nice is that before, $PGL(n+1)$ consisted of $(n+1) \times (n+1)$ matrices of sections of $\mathcal{O}_{\mathbb{P}^n_\mathbb{Z}}(1)$ with non-vanishing determinant; now the isomorphisms $\bigoplus_{i=0}^n X_i \cdot \mathcal{O}_S \overset{\sim}{\to} \bigoplus_{i=0}^n X_i \cdot L$ are given by $(n+1) \times (n+1)$ matrices of sections of $L$ with non-vanishing determinant.

In particular, to answer your comment on Qiaochu's answer, it looks like $S$ having trivial Picard group is sufficient went to answer 3) affirmatively, but I'm not convinced it's necessary: by the proof it looks like what we need is that for all invertible sheaves $L$ on S such that $\bigoplus_{i=0}^n X_i \cdot \mathcal{O}_S$ and $\bigoplus_{i=0}^n X_i \cdot L$ are isomorphic, we also have that $\mathcal{O}_S$ and $L$ are isomorphic.

• What do the $X_i$ denote? – user7090 Dec 20 '17 at 19:55
• @Jadwiga The $X_i$ are the variables on $\mathbb{P}^n$. My apologies for forgetting to define it! – Takumi Murayama Dec 20 '17 at 20:07
• Ok no problem. Your answer has been very helpful regardless! – user7090 Dec 20 '17 at 20:08

2) No. Here is a topological analogy, which I learned from MO: instead of commutative rings, let's think about C*-algebras of continuous functions on compact Hausdorff spaces $X$. Then $\text{GL}_n(C(X))$ is equivalently the space of continuous functions $X \to \text{GL}_n(\mathbb{C})$, for example. The "correct" definition of $\text{PGL}_n(C(X))$, e.g. the one that makes it representable by a space, is therefore the space of continuous functions $X \to \text{PGL}_n(\mathbb{C})$.

This is not the same as $\text{GL}_n(C(X))$ mod its center; the discrepancy is described up to homotopy by an exact sequence

$$H^{0}(X, \mathbb{C}^{\times}) \to H^{0}(X, \text{GL}_n(\mathbb{C})) \to H^{0}(X, \text{PGL}_n(\mathbb{C})) \to H^{1}(X, \mathbb{C}^{\times}) \to \cdots$$

associated to the short exact sequence

$$1 \to \mathbb{C}^{\times} \to \text{GL}_n(\mathbb{C}) \to \text{PGL}_n(\mathbb{C}) \to 1$$

of topological groups, where by $H^0(X, G)$ I mean homotopy classes of continuous maps $X \to G$ and by $H^1(X, G)$ I mean homotopy classes of continuous maps $X \to BG$, where $BG$ is the delooping of $G$. Here $H^1(X, \mathbb{C}^{\times}) \cong \text{Pic}(X) \cong H^2(X, \mathbb{Z})$ is the Picard group of continuous line bundles on $X$, so the exact sequence tells us that the obstruction to lifting an element in $\text{PGL}_n(C(X))$ to an element of $\text{GL}_n(C(X))$, up to homotopy, is given by a line bundle on $X$; roughly speaking this is because there is a $\mathbb{C}^{\times}$s worth of lifts locally but no guarantee that it is possible to patch them up globally into a consistent global lift.

• this was cool! thanks. – adrido Oct 19 '14 at 6:46