# Is GL2( R ) - > PGL2( R ) surjective?

Consider $GL_2$ as the affine group scheme with coordinate ring ${\mathbb Z}[x_1,x_2,x_3,x_4,y]/(\det\left(\begin{array}{cc}x_1& x_2\\ x_3& x_4\end{array}\right)y-1)$. The group scheme $PGL_2$ is then given by the subring $S$ of $GL_1$-invariants, which is the subring generated by all monomials of the form $x_ix_iy^2$. In this way, we define $PGL_2( R )=Hom(S,R)$ for any ring $R$. By Hilbert 90 we know that the sequence $$0\to GL_1( R )\to GL_2( R )\to PGL_2( R )\to 1$$ is exact if $R$ is a field. From that one can derive that it stays exact if $R$ is factorial. But what for a general commutative ring with unit? Is it always exact? If not, is there a handy description of all rings for which it is?

• This boils down to whether it holds when $R$ is the coordinate ring of $PGL_2$, and a positive answer is equivalent to whether there is a setwise section $PGL_2\to GL_2$ defined over $\mathbf{Z}$, if I don't miss anything. – YCor May 5 '14 at 12:42
• The diagram $1\rightarrow{\rm{GL}}_1\rightarrow {\rm{GL}}_n\rightarrow{\rm{PGL}}_n\rightarrow 1$ of smooth affine $R$-group schemes is exact for the etale topology, so the obstruction to surjectivity on $R$-points is the triviality of the induced map between the first two etale-topology H$^1$'s, which by descent theory is the map ${\rm{Pic}}(R)\rightarrow {\rm{Vec}}_n(R)$ carrying a line bundle $L$ to $L^{\oplus n}$. So short-exactness on $R$-points is equivalent to $L^{\oplus n}\simeq R^n$ iff $L\simeq R$. Considering det, this holds if ${\rm{Pic}}(R)[n]=0$ and conversely for Dedekind $R$. – user76758 May 6 '14 at 7:03
• By the way, since PGL$_n$ represents the automorphism functor of projective $(n-1)$-space (by deformation theory to bootstrap from the classical case on field-valued points), failure of surjectivity is exactly the condition of $\mathbf{P}^{n-1}_R$ admitting an $R$-automorphism which does not arise from an invertible $n \times n$ matrix over $R$. – user76758 May 6 '14 at 7:06

The answer is no. Here is an explicit example. Let $R=\mathbb{Z}[\sqrt{-5}]$.

Consider the matrix $$\left(\begin{array}{cc}1+\sqrt{-5}& 2\\ 2& 1-\sqrt{-5}\end{array}\right).$$

It represents an element of $PGL_2(R)$ that is not in the image of $GL_2(R)$.

The motivation for this example is that the ideal $(2,1+\sqrt{-5})$ is not principal in $R$ and this should be relevant because the next term in the long exact sequence is $H^1(R,\mathbb{G}_m)$ (and so the sequence written in the question will be exact whenever this $H^1$ vanishes).

I would love to see a more conceptual proof of the failure of $GL_2(R)\to PGL_2(R)$ to be surjective.

• the determinant of my matrix is 2. – Peter McNamara May 5 '14 at 13:29
• In general the map will be surjective when Pic(R) is trivial. Brian Conrad's homework on group schemes walks you through this to a certain extent. If R as a Dedekind domain you can think of the elements of $PGL_n(R)$ as being pseudomatrices of fractional ideals all of whose $n$-fold products which go into the determinant are principal. If the class group is trivial, then all the entries have to be principal anyway and you're left with a standard matrix in $GL_n(R)$ up to unit multiple in $R$. – stankewicz May 5 '14 at 13:53
• Slightly better: the pointed set $PGL_2(R)/GL_2(R)$ is identified with the fibre over "0" of $\eta:H^1(X,GL_1) \to H^1(X,GL_2)$, where $X = \mathrm{Spec}(R)$. Interpreting $H^1(X,GL_n)$ as isomorphism classes of rank $n$ bundles, the map $\eta$ sends a line bundle $L$ to the rank $2$ bundle $L \oplus L$. Thus, the question is: can we have a non-trivial line bundle $L$ such that $L \oplus L \simeq \mathcal{O}_X^{\oplus 2}$? The answer is yes: any order $2$ element in $\mathrm{Pic}(X)$ for $X$ a Dedekind domain does the trick (as in the example above). – anonymous May 5 '14 at 15:08

For a topological perspective, take $R$ to be the tensor product of $\mathbb R$ with the coordinate ring of $PGL_2$. surjectivity would imply that the map of topological spaces $GL_2(\mathbb R)\to PGL_2(\mathbb R)$ has a right inverse, which it does not because the induced map of fundamental groups $\mathbb Z\to \mathbb Z$ is $x\mapsto 2x$. Or you can argue similarly over $\mathbb C$, where the map of fundamental groups is $\mathbb Z\to \mathbb Z/2$.

• alternatively (essentially equivalently, but maybe even more topologically), $R$ can be taken to be taken to be the ring of continuous functions from $PGL_2(\mathbf{R})$ to $\mathbf{R}$. Then $PGL_2(R)$ is the group of continuous self-maps of $PGL_2(\mathbf{R})$ (with the law on the target). Then the identity self-map of $PGL_2(\mathbf{R})$ is not in the image, by the same argument. – YCor May 5 '14 at 18:53
• I'd also like to point out that $\pi_1(GL_2(\mathbb R))$ and $\pi_1(GL_2(\mathbb C))$ are two different infinite cyclic groups. The one comes from $SL_2$ and the other does not. – Tom Goodwillie May 5 '14 at 22:25