4
$\begingroup$

Let $F$ be a finite field with many elements, say more than 7 for example, and $X$ be the corresponding projective plane $\mathbb{P}^2(F)$.

For a set of points in $X$, if any three of them are noncollinear (so they are automatically different points), we say this set of points are noncollinear.

Define the following configuration complex: $C_n$ is the free abelian group generated by all such noncollinear $n$-tuples $(x_0,x_1,\dots,x_n)\in X^{n+1}$ (i.e. points $x_0,x_1,\dots,x_n$ are noncolinear); the differentials are defined as $$ d(x_0,\dots,x_n)=\sum_{i=0}^n(-1)^i(x_0,\dots,\hat{x}_i,\dots,x_n). $$ So we get a chain complex $$ C_*:\dots\mathop{\to}\limits^{d} C_1\mathop{\to}\limits^{d} C_0\mathop{\to}\limits^{\epsilon}\mathbb{Z}\to0, $$ where $\epsilon$ is the augmentation map.

My question is whether this chain complex is almost acyclic, i.e. its homology is always zero except for one term (the last nonzero term).

Or if $|F|$ is sufficiently large can we deduce that $H_n(C_*)=0$ for $n\ll|F|$?

Here is something I know about the last nonzero term: if $|F|=q$ is odd, then the longest arc of $X$ is a conic going through $q+1$ points. So now $C_q\neq0, C_{q+1}=0$. If $|F|=q$ is even, then the longest arc of $X$ is a $(q+2)$-arc. So now $C_{q+1}\neq0, C_{q+2}=0$.

Improve this question
$\endgroup$
5
  • 2
    $\begingroup$ It would be helpful to say something about where this problem comes from. I assume you're aware of e.g. Goncharov's "Geometry of configurations, polylogarithms and motivic cohomology". $\endgroup$
    – Dan Petersen
    Commented May 11, 2022 at 5:38
  • 2
    $\begingroup$ Interesting question! The reverse problem, where $(x_0, x_1, \dots, x_n)$ form a simplex if all the $x_i$ ARE collinear has been studied -- see mathscinet.ams.org/mathscinet-getitem?mr=772475 and mathscinet.ams.org/mathscinet-getitem?mr=3482438 . But I can't find anyone who is looked at this version. $\endgroup$
    – David E Speyer
    Commented May 11, 2022 at 13:53
  • 2
    $\begingroup$ @DanPetersen: The one-dimention-less analogue of this question is the complex of injective words, so I think this version would also be right (at least when $n=1,2,3,4$). On the other hand if we replace finite fields by infinite fields this statement would be trivially right. BTW I know Goncharov's article but I never read it. $\endgroup$
    – XYC
    Commented May 17, 2022 at 17:43
  • $\begingroup$ Why is the statement trivially right for infinite fields? $\endgroup$
    – Richard Stanley
    Commented Aug 31, 2022 at 0:02
  • 1
    $\begingroup$ @RichardStanley : Because if $F$ has infinite elements then for any cycle $z=\sum_in_i(x_0^i, \dots, x_p^i)\in C_p$ we can always find an element $x\in X$ such that $x$ is non-collinear to any two of $x_n^j$'s, and insert $x$ to all the base elements in $z$ at the first slot we have $z=d[\sum_in_i(x,x_0^i, \dots, x_p^i)]$. Because in $d[\sum_in_i(x,x_0^i, \dots, x_p^i)]$ all base elements with $x$ at the first slot cancels out since $z$ is a cycle. See Lemma 1 in Section 1 also Lemma 3.1 in Suslin's paper- $K_3$ of a field and the Bloch group. $\endgroup$
    – XYC
    Commented Sep 8, 2022 at 8:30

0

Reset to default

You must log in to answer this question.