Galois cohomology of projective linear group I am currently trying to compute Galois cohomology $H^{1}(\overline{k}/k, PGL_2(\overline{k}))$. As far as I know these cocycles correspond to isomorphism classes of smooth genus-$0$ curves over $k$. Any curve of genus $0$ is either isomorphic to $\mathbb{P}^{1}(k)$ or to a quadric $ax^2+by^2+cz^2=0$ which has no $k$-points. How to understand this from general tools such as exact sequences etc.?
For instance, sequence $0\longrightarrow k^{*} \longrightarrow GL_2{(k)}\longrightarrow PGL_2(k)\longrightarrow 0$ gives rise to a long exact sequence of cohomology. However I don't know how to obtain the result I am interested in.
I will be very grateful for any help or reference.
 A: As Alex Wertheim has mentioned in the comments, the key idea here is twisted forms. The (informal) general idea is the following: suppose we are given a category $\mathcal C_K$ of structures defined over a field $K$ (for example, $\mathcal C_K$ may be affine $K$-varieties, projective $K$-varieties, $K$-algebras, etc.). Let $L/K$ be a Galois extension with Galois group $G$, and let $\mathcal C_L$ be the corresponding category of structures defined over $L$. Then $G$ acts on each object of $L$ by automorphisms.
The main theorem is the following. Given an object $X$ of $\mathcal C_K$, we can obtain its base change $X_L$, an object of $\mathcal C_L$. Then there will be a bijection $$\{K\text{-isomorphism classes of }\mathcal C_K\text{ objects isomorphic to }X_L\text{ after base change}\}\longleftrightarrow H^1(G,\text{Aut}_{\mathcal C_L}(X_L))$$ where $\text{Aut}_{\mathcal C_L}(X_L)$ has the $G$-conjugation action induced by the action of $G$ on $X_L$. The objects on the left hand side of this correspondence are the so-called "twisted forms." This correspondence probably easiest to see in the case of affine varieties, or equivalently, (reduced f.g.) commutative $K$-algebras. Given an affine $K$-variety $X$ with coordinate ring $K[X]$ and its base change $X_L$ with coordinate ring $L[X]$, a $1$-cocycle $H^1(G,\text{Aut}_L(L[X]))$ canonically defines a semilinear action on $L[X]$. The theory of faithfully flat descent says that the category of $L$-algebras with semilinear $G$-actions is equivalent to the category of $K$-algebras, and so this canonically gives a $K$-algebra which becomes isomorphic to $L[X]$ after base change. A good reference for this fact about twisted forms is Springer's book on algebraic groups.
To explain the correspondence between conics and cocycles, we apply this to the category of projective varieties over $K$ and $X = \mathbb P^1_K$. The left-hand side of our correspondence is then simply the $K$-isomorphism classes of smooth curves of genus $0$, since we are base changing to an algebraically closed field. The right hand side of the correspondence is $H^1(G,\text{Aut}_{\bar K}(\mathbb P^1_{\bar K})) = H^1(G, \text{PGL}_2(\bar K))$. Tracing through the bijection explicitly will tell you which conic corresponds to which cocycle.
As an example, let's consider the simple case of $\mathbb C/\mathbb R$. Then $\mathbb P^1_{\mathbb R}$ embeds in $\mathbb P^2_{\mathbb R}$ as the conic $C : y^2-xz=0$ via $[s:t]\mapsto [s^2:st:t^2]$. Now one checks that there are only two cocycles in $H^1(G,\text{PGL}_2(\mathbb C))$, namely the trivial cocycle and the one which sends complex conjugation to $\left(\begin{smallmatrix} 0 & -1 \\ 1 & 0 \end{smallmatrix}\right)$. This is the automorphism of $\mathbb P^1$ given by $[s:t]\mapsto [-t:s]$ which corresponds to the automorphism of $C$ given by $[x:y:z]\mapsto [z:-y:x]$. To find the conic in $\mathbb P^2_{\mathbb R}$ corresponding to this cocycle, we need to find a conic for which complex conjugation interchanges $x$ and $z$ and negates $y$. So we make the change of variables $x = x'+iz'$, $y=iy'$, $z=x'-iz'$. Then complex conjugation interchanges $x$ and $z$ while negating $y$. The equation of the resulting conic is $$0 = xz-y^2 = x'^2+y'^2+z'^2.$$ So the conic $C' : x'^2 + y'^2 + z'^2 = 0$ is the desired conic, which is an instance of the sort of conic you mention in your question.
As Alex also mentioned, this is all tied up with quaternion algebras, but I won't get too far into that here. The reason is basically that $\text{PGL}_2(\bar K)$ is also the automorphism group of the matrix algebra $M_2(\bar K)$. So 4-dimensional central simple algebras defined over $K$ are also classified by $H^1(G,\text{PGL}_2(\bar K))$. A conic $ax^2+by^2+cz^2$ will correspond to a quaternion algebra easily expressible in terms of the $a,b,c$.
