Choosing a group action to do GIT of hypersurfaces

When studying GIT stability of hypersurfaces $d$ of $\mathbb P^n$ we look at the Hilbert Scheme $H=\mathbb P^N$ parametrizing homogeneous polynomials $f_d(x_0,\ldots,x_n)$ of degree $d$. There is normally a group $G$ acting on $\mathbb P^n$ which induces an action on $H$.

The GIT compactification is then $H^{ss}//G$ where we have restricted to semistable surfaces.

A natural option for $G$ would be to take $G=\mathrm{Aut}(\mathbb P^n)=PGL(n+1)$ so that we identify two hypersurfaces if they are related by an automorphism of $\mathbb P^n$. However in a few sources I have consulted this is never the case. Indeed, they always use either

• $G=GL(n+1)$ (e.g. Mukai's "An Introduction to Invariants and Moduli" in the study of cubic surfaces)

• $G=SL(n+1)$ (e.g. Alcock's "The Moduli Space of Cubic Threefolds")

Q1: Why is $PGL(n+1)$ never used?

I understand that any group used must include $SL(n+1)$ in order to apply Hilbert-Mumford's numerical criterion. So my second question is:

Q2: How to decide if choosing GL or SL? Doesn't using SL imply that we are considering two isomorphic hypersurfaces to be different when taking quotient?

Q3 I understand (see comments) that $SL$ gives a unique linearization of the line bundle, but shouldn't $PGL$ be used in all cases when constructing compactifications of the moduli of hypersurfaces given that it is what determines if two hypersurfaces are isomorphic by isomorphisms of $\mathbb P^n$?

• Although there exists an action of $\textbf{PGL}_{n+1}$ on $\mathbb{P}k[x_0,\dots,x_n]_d$ for every $d$, this only lifts to a linear action on the vector space $k[x_0,\dots,x_n]$ if $n+1$ divides $d$. – Jason Starr Sep 23 '15 at 16:52
• Thank you Jason, that explains Mukai's choice. Do you have a reference for that statement? (Or an idea of where to look for one) – John Sep 23 '15 at 16:58
• In Chapter 7 of Dolgachev's "Lectures on Invariant Theory", he describes the obstruction to linearizing a given action. – Jason Starr Sep 23 '15 at 17:04
• The group $PGL_{N+1}$ is used in www-irma.u-strasbg.fr/~benoist/articles/Thesefinale.pdf and arxiv.org/abs/1505.02249 – Ariyan Javanpeykar Sep 24 '15 at 13:28
• Ariyan is correct. Even in those cases where $\textbf{PGL}_{n+1}$ does not linearize to $\mathcal{O}(1)$ on $\mathbb{P}k[x_0,\dots,x_n]_d$, it does linearize some positive tensor power, e.g., $\mathcal{O}(n)$. So that is not, by itself, a reason one must use $\textbf{SL}_{n+1}$ rather than $\textbf{PGL}_{n+1}$. – Jason Starr Sep 25 '15 at 13:18