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The automorphism group of the algebra of $n$-dimensional matrices over a field $K$ is $PGL_n(K)$. The automorphism group of $n-1$-dimensional projective space over $K$ is also $PGL_n(K)$. Therefore, twists of them over $\bar{K}$ are both classified by the same Galois cohomology group, $H^1(PGL_n(K))$.

Twists of $M_n(K)$ are central simple algebras of rank $n^2$. Twists of $K\mathbb P^{n-1}$ are projective varieties geometrically isomorphic to $\mathbb P^{n-1}$.

There should be a bijection between these two sets.

How explicit and geometrically/algebraically nice can we make it?

What properties of one object correspond to properties of the other?

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References:

Artin, M. Brauer-Severi varieties. Brauer groups in ring theory and algebraic geometry (Wilrijk, 1981), pp. 194–210, Lecture Notes in Math., 917, Springer, Berlin-New York, 1982.

Knus, Max-Albert; Merkurjev, Alexander; Rost, Markus; Tignol, Jean-Pierre The Book of Involutions. (English summary) With a preface in French by J. Tits. American Mathematical Society Colloquium Publications, 44. American Mathematical Society, Providence, RI, 1998

Let $A$ be a central simple $K$-algebra of degree $n$ (i.e., of rank $n^2$). Let $Gr(n, A)$ be the subspace of the grassmannian of rank $n$ dimensional subspaces. Let $X \subset Gr(n, A)$ be the subvariety of subspaces which are also right ideals of $A$. Then $X$ is the corresponding space, called the Severi-Brauer, Brauer-Severi, or Chatelet variety of $A$, and is denoted here by $SB(A)$.

When $A = End_K(V)$ is trivial, every right ideal of $A$ of dimension n (as a $K$-vector space) is of the form $\operatorname{Hom}_K(V, U)$, where $U$ is a 1-dimensional vector space. So the space is $\mathbb{P}_K(V)$.

I am not sure of an explanation in the other direction as nice. It is the case that for a Severi-Brauer variety $SB(A)$, you can consider the following subbundle $\mathcal{I}$ of the constant bundle $SB(A) \times A$. The bundle $\mathcal{I}$ consists of pairs $(I, a)$, where $a \in I$. Then $\mathcal{I}$ is a locally free of rank $n$ and the global endomorphism ring $\operatorname{End}_{SB(A)}(\mathcal{I})$ is isomorphic to $A$ (or the opposite algebra of $A$, I can't remember which).

However, this doesn't answer your question of how to start with a twist of $\mathbb{P}^{n-1}$ and nicely produce a central simple algebra of rank $n^2$. I think for that you may have to argue by galois descent.

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    $\begingroup$ One possible construction of the algebra: given a twisted form P of P^{n-1}, let J be the unique (up to non-unique isomorphism) non-split extension of O_P by Omega^1_P (it's O(-1)^n geometrically). The global sections of End(J) is a central simple algebra A. Moreover, using J, one can build an isomorphism between P and the space P_A defined by P_A(S) = all triples (V,phi,f) where V is a rank n vector bundle on S, phi is an isomorphism End(V) =~ A_S, and f is a quotient V --->> O_S. One can also check that P_A is a Brauer-Severi variety, giving an alternative to the SB(A) construction above. $\endgroup$
    – Bhargav
    Commented Apr 12, 2012 at 3:47
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    $\begingroup$ @Bhargav -- The first part of your comment is correct: the standard way to reconstruct the central simple algebra from the Severi-Brauer variety is to consider the central extension of the algebra of vector fields on the Severi-Brauer variety. The second part of your comment is not quite correct, depending on how you are defining the moduli space of triples: your construction will typically give a $\mu_n$-gerbe over the Severi-Brauer variety (consider the case where $A$ is already $\text{End}(W)$). The coarse moduli space is the Sever-Brauer variety, of course. $\endgroup$
    – Jason Starr
    Commented Apr 12, 2012 at 12:09
  • $\begingroup$ @Jason: Oops, thanks for catching that! $\endgroup$
    – Bhargav
    Commented Apr 12, 2012 at 16:29
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    $\begingroup$ @Bhargav: Now that I understand better what you were proposing, I realize that my objection is wrong. I believe your original proposal is correct as stated. $\endgroup$
    – Jason Starr
    Commented Apr 14, 2012 at 0:49
  • $\begingroup$ @Bhargav: thanks for your addition. It's what I didn't want to write, because I wasn't as familiar with the descent argument. My bundle I and your bundle J are the same, right? $\endgroup$
    – mark
    Commented Apr 15, 2012 at 14:00

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