Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
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 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 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 at 7:06

2 Answers 2

up vote 12 down vote accepted

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.

share|improve this answer
the determinant of my matrix is 2. –  Peter McNamara May 5 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 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 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 $\mathbb Z\to \mathbb Z/2$.

share|improve this answer
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 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 at 22:25

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.