Finite homogeneous geometries

TL;DR

This paper provides a classification proof for finite homogeneous geometries of high dimension that avoids reliance on the classification of finite simple groups, contributing to geometric and algebraic understanding.

Contribution

It offers a new proof of the classification of finite homogeneous geometries that does not depend on the classification of finite simple groups.

Findings

Classification of non-trivial, finite homogeneous geometries of high dimension.

Proof does not rely on finite simple groups classification.

Reproduces part of the author's DPhil thesis.

Abstract

This paper reproduces the text of a part of the Author's DPhil thesis. It gives a proof of the classification of non-trivial, finite homogeneous geometries of sufficiently high dimension which does not depend on the classification of the finite simple groups.

Equations

$$\frac{s_r-s_{r-2}}{s_{r-1}-s_{r-2}}\ge \frac{s_{r-1}-s_{r-3}}{s_{r-2}-s_{r-3}}$$

$$s_r\ge s_{r-2}+(s_{r-1}-s_{r-2})\frac{s_{r-1}-s_{r-3}}{s_{r-2}-s_{r-3}}.$$

$$s_r-s_{r-1}\ge \frac{(s_{r-1}-s_{r-2}{)}^2}{s_{r-2}-s_{r-3}}.$$

$$s_r-s_{r-1}\ge (s_2-s_1{)}^{r-1}/(s_1-1{)}^{r-2}$$

$$s_r\le \frac{{\phi }^2(\theta \phi -4\alpha s_1\psi {)}^2-\phi }{4\alpha s_1}$$

$$a_{443}=(s_1-1){\alpha }^2/s_1$$

$$\phi ={\alpha }^2+s_1^2(s_1-1{)}^2-2\alpha s_1(s_1+1)=(\alpha -s_1(s_1-1){)}^2-4\alpha s_1.$$

$$\begin{gathered} (\theta \phi -4\alpha s_1\psi {)}^2 \\ =(\alpha -s_1(s_1-1){)}^2(\theta (\alpha -s_1(s_1-1))+4\alpha s_1^2(s_1-1){)}^2 \\ =(\alpha -s_1(s_1-1){)}^2((s_1-s_1^2+2\alpha s_1-\alpha )(\alpha -s_1(s_1-1))+4\alpha s_1^2(s_1-1){)}^2 \\ =(\alpha -s_1(s_1-1){)}^2(\alpha (2s_1-1)+s_1(s_1-1){)}^2(\alpha +s_1(s_1-1){)}^2 \\ <(\alpha +s_1(s_1-1){)}^4(\alpha (2s_1-1)+s_1(s_1-1){)}^2. \end{gathered}$$

$$\frac{{\phi }^2(\theta \phi -4\alpha s_1\psi {)}^2}{4\alpha s_1}<\frac{(\alpha +s_1(s_1-1){)}^8(\alpha (2s_1-1)+s_1(s_1-1){)}^2}{4\alpha s_1}.$$

$$s_2-s_1=(s_1-1)\alpha +(s_1-1{)}^2\ge \alpha +s_1(s_1-1)$$

$$\begin{gathered} \frac{{\phi }^2(\theta \phi -4\alpha s_1\psi {)}^2}{4\alpha s_1}<\frac{(s_2-s_1{)}^8(\alpha (2s_1-1)+s_1(s_1-1){)}^2}{4\alpha s_1} \\ =\frac{(s_2-s_1{)}^8(2\alpha +\alpha /(s_1-1)+s_1{)}^2(s_1-1{)}^2}{4\alpha s_1} \\ <\frac{25}{16}\frac{(s_2-s_1{)}^8(\alpha +s_1{)}^2(s_1-1{)}^2}{\alpha s_1} \\ =\frac{25}{16}\frac{(s_2-s_1{)}^8(s_2-1{)}^2}{\alpha s_1}<(s_2-s_1{)}^{10}. \end{gathered}$$

$$(s_2-s_1{)}^{10}\le \frac{(s_2-s_1{)}^{19}}{(s_1-1{)}^{18}}.$$

$$s_r\le \frac{(4\beta s_1+(s_1^2-\beta {)}^2{)}^2(s_1^2-\beta {)}^2(s_1^2+\beta {)}^2(s_1^2-\beta +2s_1\beta {)}^2}{4\beta s_1},$$

$$a_{452}=(s_1-1)\beta /s_1$$

$$(4\beta s_1+(s_1^2-\beta {)}^2)=s_1^4+{\beta }^2-\beta s_1(2s_1-4).$$

$$(4\beta s_1+(s_1^2-\beta {)}^2{)}^2<2(s_1^4+{\beta }^2{)}^2.$$

$$0<s_1^2-\beta +2s_1\beta <(s_1+\beta {)}^2,$$

$$s_r<\frac{(s_1^4+{\beta }^2{)}^2(s_1^2-\beta {)}^2(s_1^2+\beta {)}^2(s_1+\beta {)}^2}{2\beta s_1}=\frac{(s_1^8-{\beta }^4{)}^2(s_1+\beta {)}^2}{2\beta s_1}.$$

$$s_r<\frac{(s_2-s_1{)}^{10}}{2\beta s_1(s_1-1{)}^2}$$

$$s_r<\frac{(s_2-s_1{)}^{16}}{2(s_1-1{)}^{14}\beta s_1}<\frac{(s_2-s_1{)}^{16}}{(s_1-1{)}^{15}}.$$

$${\hat{s}}_2=1+(\hat{\alpha }+{\hat{s}}_1)({\hat{s}}_1-1)=1+{\hat{s}}_1^2({\hat{s}}_1-1)=s_1^6-s_1^4+1.$$

$$a^2=t^6-t^4+1=(t^3-\frac{1}{2}t{)}^2-\frac{t^2}{4}+1$$

$$(2t^3-t+2a)(2t^3-t-2a)=t^2-4.$$

$$\frac{s_3-1}{s_1-1}={\hat{s}}_2=1+(\hat{\alpha }+{\hat{s}}_1)({\hat{s}}_1-1)=1+(s_1^2+s_1+1{)}^2(s_1^2+s_1)$$

$$s_3/s_1=1+(s_1^2+s_1+1{)}^2(s_1-1{)}^2=s_1^6+2s_1^5+2s_1^4-2s_1^2-2s_1.$$

$$g(x)=x^3+x^2+\frac{x}{2}-\frac{1}{2}\text{mbox}andh(x)=\frac{5}{4}{x}^2+\frac{3}{2}x+\frac{1}{4}.$$

$$(2t^3+2t^2+t-1-2a)(2t^3+2t^2+t-1+2a)=5t^2+6t+1.$$

$$\frac{s_3-1}{s_1-1}={\hat{s}}_2=({\hat{s}}_1+\hat{\alpha })(\hat{s}-1)+1=s_1(s_1+1)({\hat{s}}_1^2+{\hat{s}}_1+1)+1$$

$$s_3/s_1=1+(s_1^2-1)({\hat{s}}_1^2+{\hat{s}}_1+1)=s_1^6+2s_1^5+3s_1^4+s_1^3-s_1^2-3s_1-2.$$

$$g(x)=x^3+x^2+x-\frac{1}{2}\text{mbox}andh(x)=x^2+2x+\frac{9}{4}.$$

Full text

2010 Mathematics Subject Classification: Primary 05B25, 20B20, 20B25; Secondary 51A05, 51A15.

Preamble

This note is essentially a reproduction of Appendix I of the Author’s doctoral thesis [1], though there are a few new, minor corrections. In the thesis, the Appendix was a sequel to Chapters 3 - 6 of [1] which were published as [2]. Thus, the text below is a complement to [2] and we use the notation and terminology of that paper without further comment. The main changes we have made to Appendix I of [1] are to adapt the references, so that we refer to results in [2] rather than the corresponding results in [1].

The main result of [2] was a proof of the classification of infinite, locally finite, homogeneous geometries which did not depend on the classification of finite simple groups. A different proof, also not relying on the classification of finite simple groups, had previously been given by Zilber (see the references in [2]). In particular, the geometries considered were of infinite dimension. By contrast, Theorem 1 below assumes only that the geometry is of sufficiently large finite dimension (and so can be finite). As the complements of Jordan sets for a finite primitive Jordan group form a homogeneous geometry (with the Jordan group acting as a group of automorphisms), Theorem 1 also gives a classification (not relying on the classification of the finite simple groups) of the finite primitive Jordan groups which are not $3$-transitive and which have ‘enough’ (at least 24) different sizes of Jordan sets.

The final version of the thesis [1] contains a number of corrections written in by hand and is not available in electronic form. This explains why the Author has re-typed the material rather than making a scanned copy available. The Author thanks Michael Zieve for suggesting that it would be useful to have these results available in a more accessible form.

It should be noted that Zilber’s paper [3] contains a different proof (also not using the classification of the finite simple groups) of Theorem 1 under the weaker hypothesis that the dimension of $\mathcal{X}$ is at least 7. Of course, using the classification of finite simple groups, we know that dimension at least 2 will suffice here.

The Finite Case

We give a proof of:

Theorem 1**.**

Let $\mathcal{X}$ be a locally finite homogeneous geometry of dimension at least 23 with at least 3 points on a line. Then $\mathcal{X}$ is a (possibly truncated) projective or affine geometry over a finite field.

The proof of this result is in a series of lemmas, which are tightenings of results which have appeared in [2]. We use the notation of Section 2.4 of that paper, and assume throughout that $\mathcal{X}=(X,{\rm cl})$ is a locally finite homogeneous geometry of dimension at least 3. In particular, note that the dimension of a closed set is one less than the number of elements in a basis of the set and an $i$-flat is a closed set of dimension $i$. Points, lines and planes are closed sets of dimension 0, 1, 2 respectively. We denote by $s_{i}$ the number of points in an $i$-flat. The parameters $\alpha$ and $\alpha^{\prime}$ are defined in 2.4 of [2].

Lemma 1**.**

If $r\geq 3$, then $s_{r}\geq(s_{2}-s_{1})^{r-1}/(s_{1}-1)^{r-2}$.

Proof.

By ([2], Lemma 3.2.3) we have

[TABLE]

and so

[TABLE]

Thus

[TABLE]

Using this relation repeatedly, we obtain

[TABLE]

which proves the lemma. ∎

Lemma 2**.**

Suppose that $\alpha^{\prime}=0$ and $\alpha\geq 1$ and $s_{1}\geq 3$. If $\theta,\varphi,\psi$ are as in ([2], Theorem 4.3; see also p.319), then

[TABLE]

implies that $r\leq 19$.

Proof.

111Slight changes from final version of thesis

First note that (in the notation of Section 4 of [2])

[TABLE]

and so $\alpha\geq 1$ implies that $\alpha\geq\sqrt{s_{1}}$.

Now,

[TABLE]

If this is positive, then $\varphi<(\alpha-s_{1}(s_{1}-1))^{2}.$ If it is negative, then $\varphi>-4\alpha s_{1}$. So in either case we have $\varphi^{2}<(\alpha+s_{1}(s_{1}-1))^{4}$. Moreover, the $-\varphi$ term in the numerator of the above expression makes a contribution of at most $+1$; the upper bound estimates below are sufficiently crude that we may ignore this.

Also, from the calculations preceding Theorem 4.3 of [2],

[TABLE]

Thus

[TABLE]

Now,

[TABLE]

(as $s_{1}\geq 3$ and $\alpha\geq 2$). So

[TABLE]

As $(s_{2}-s_{1})\geq(s_{1}-1)^{2}$, we have

[TABLE]

Thus, by Lemma 1, if $r$ is as in the statement of the lemma, then $r\leq 19$.

∎

Lemma 3**.**

Suppose that $\alpha^{\prime}=1$ and $\alpha>1$ and $s_{1}\geq 3$. If $\beta=\alpha-1$ and

[TABLE]

then $r\leq 16$.

Proof.

First, note that in the notation of Section 5 of [2],

[TABLE]

and so $s_{1}$ divides $\beta$. Now,

[TABLE]

If this is positive, it is less than $s_{1}^{4}+\beta^{2}$. If it is negative it is greater than $-2s_{1}^{2}\beta$. Thus, in either case,

[TABLE]

Also,

[TABLE]

and so

[TABLE]

Since $s_{2}-s_{1}=\beta(s_{1}-1)+s_{1}(s_{1}-1)$ and $\beta\geq s_{1}$ and $s_{1}\geq 3$, it follows that $s_{2}-s_{1}\geq s_{1}^{2}+\beta$ and so $s_{1}^{8}+\beta^{4}<(s_{2}-s_{1})^{4}$. Thus

[TABLE]

as $(s_{1}+\beta)=(s_{2}-s_{1})/(s_{1}-1)$. Then

[TABLE]

By Lemma 1, it then follows that $r\leq 16$. ∎

Corollary 4**.**

Let $\mathcal{X}$ be a locally finite homogeneous geometry of dimension at least 20 with at least 3 points on a line. Then either $\mathcal{X}$ is a (possibly truncated) projective or affine geometry over a finite field, or one of the following conditions holds in $\mathcal{X}$:

1. (1)

$\alpha^{\prime}=0$, $s_{1}$ is a square, $\alpha=s_{1}(\sqrt{s_{1}}\pm 1)^{2}$ and $s_{i}$ is a square for $i\geq 3$; 2. (2)

$\alpha^{\prime}=0$, $\alpha=s_{1}(s_{1}-1)$ and $(s_{1}-1)(s_{i}-1)$ is a square for $i\geq 3$; 3. (3)

$\alpha^{\prime}=1$, $\alpha=s_{1}^{2}+1$ and $s_{i}/s_{1}$ is a square for $i\geq 3$.

Proof.

This is deduced from the above lemmas and Theorem 3.2.9, Theorem 4.3, and Theorem 5.2 in [2], as in Section 6 of [2]. ∎

Proposition 5**.**

Let $\mathcal{X}=(X,{\rm cl})$ be a locally finite homogeneous geometry and let $Y$ be a closed subset of $X$ such that the localisation $\mathcal{X}_{Y}$ has dimension at least 3. Let $z\in X\setminus Y$ and $Z={\rm cl}(Y\cup\{z\})$. Consider the following situations:

1. (a)

condition (1) holds in $\mathcal{X}_{Y}$ and (1) holds in $\mathcal{X}_{Z}$; 2. (b)

condition (1) holds in $\mathcal{X}_{Y}$ and (2) holds in $\mathcal{X}_{Z}$; 3. (c)

condition (2) holds in $\mathcal{X}_{Y}$ and (2) holds in $\mathcal{X}_{Z}$; 4. (d)

condition (3) holds in $\mathcal{X}_{Y}$ and (1) holds in $\mathcal{X}_{Z}$; 5. (e)

condition (3) holds in $\mathcal{X}_{Y}$ and (2) holds in $\mathcal{X}_{Z}$; 6. (f)

condition (3) holds in $\mathcal{X}_{Y}$ and (3) holds in $\mathcal{X}_{Z}$,

where conditions (1), (2) and (3) are as in Corollary 4. Then (a), (b), (d), (e) and (f) are impossible and (c) is impossible if $s_{1}\geq 3$.

Proof.

We use the notation of the corresponding Proposition in Section 6 of [2]. Conditions (a) and (d) were proved to be impossible in that Proposition.

(c) is impossible if $s_{1}\geq 3$: Here $\hat{s}_{1}=s_{1}^{2}$ and $\hat{\alpha}=s_{1}^{2}-\hat{s}_{1}$ and we require that $(s_{1}-1)(s_{3}-1)$ be a square. As $\hat{s}_{2}=(s_{3}-1)/(s_{1}-1)$ we require that $\hat{s}_{2}$ be a square. Now,

[TABLE]

Suppose the polynomial $f(x)=x^{6}-x^{4}+1$ takes the square integer value $a^{2}$ at some $t\in\mathbb{N}$. Then

[TABLE]

so

[TABLE]

If $t>2$ then $t^{2}-4\neq 0$ and $2t^{3}-t>t^{2}-4$. So the above equation has no solution in integers with $t\geq 3$. This proves that (c) is impossible in $s_{1}\geq 3$.

*(e) is impossible: * Here, $\hat{s}_{1}=s_{1}^{2}+s_{1}+1$ and $\hat{\alpha}=\hat{s}_{1}^{2}-\hat{s}_{1}$ and we require that $s_{3}/s_{1}$ be a square. Now,

[TABLE]

so

[TABLE]

Suppose that the polynomial $f(x)=x^{6}+2x^{5}+2x^{4}-2x^{2}-2x$ takes the square integer value $a^{2}$ at some $t\in\mathbb{N}$. One calculates that $f=g^{2}-h$, where

[TABLE]

So

[TABLE]

This equation in integers is impossible if $t\geq 2$, and so (e) is impossible.

(f) is impossible: Here, $\hat{s}_{1}=s_{1}^{2}+s_{1}+1$ and $\hat{\alpha}=\hat{s}_{1}^{2}+1$ and we require that $s_{3}/s_{1}$ be a square. As in the above:

[TABLE]

so

[TABLE]

Suppose the polynomial $f(x)=x^{6}+2x^{5}+3x^{4}+x^{3}-x^{2}-3x-2$ takes the square integer value $a^{2}$ at some $t\in\mathbb{N}$. One calculates that $f=g^{2}-h$ where

[TABLE]

So

[TABLE]

This equation is not soluble in integers if $t\geq 2$, and this proves that (f) is not possible.

(b) is impossible: Consider first the case where $\alpha=s_{1}(\sqrt{s_{1}}+1)^{2}$. Then $\hat{s}_{1}=s_{1}(1+(\sqrt{s_{1}}+1)^{2})$ and $\hat{\alpha}=\hat{s}_{1}^{2}-\hat{s}_{1}$ and we require that $s_{3}$ must be a square (and so $s_{3}/s_{1}$ must be a square). Then:

[TABLE]

so

[TABLE]

Put $t=\sqrt{s_{1}}$. Then $s_{3}/s_{1}$ is equal to

[TABLE]

This is equal to $g^{2}(t)-h(t)$ where

[TABLE]

So, as above, we are required to solve the equation

[TABLE]

with $a,t\in\mathbb{N}$. It is routine (if unpleasant) to check that there are no such solution to this with $t\geq 2$ (for example, $h(x)=0$ has no solutions with $x\geq 1$, and for $x\geq 2$ we have $g(x)>h(x)$). Thus case (b) cannot occur with $\alpha=s_{1}(\sqrt{s_{1}}+1)^{2}$.

Suppose now that $\alpha=s_{1}(\sqrt{s_{1}}-1)^{2}$. Then the calculations are exactly as for the above case, except that we substitute $-t$ for $t$. Thus we end up having to solve the equation

[TABLE]

with $a,t\in\mathbb{N}$. Again it is routine to check that $h(-t)\neq 0$ if $t\geq 2$ and that $g(-t)>h(-t)$ if $t\geq 2$. Thus case (b) cannot occur.

This finishes the proof of the Proposition. ∎

We can now prove the Theorem stated at the beginning.

Let $\mathcal{X}=(X,{\rm cl})$ be a locally finite homogeneous geometry of dimension at least 23 with at least 3 points on a line. Let $p,\ell,\Pi$ be (respectively) a point, line and plane in $\mathcal{X}$ with $p\subseteq\ell\subseteq\Pi$. Suppose $\mathcal{X}$ is not a truncation of a projective or affine geometry over a finite field. By Corollary 4 and Lemma 2.1.1 of [2], one of conditions (1), (2) or (3) of Corollary 4 holds in $\mathcal{X}$, $\mathcal{X}_{p}$, $\mathcal{X}_{\ell}$ and $\mathcal{X}_{\Pi}$ (as each has dimension at least 20). We deduce a contradiction using Proposition 5.

Proposition 5 (parts (d), (e), (f)) implies that condition (3) cannot hold in any of $\mathcal{X},\mathcal{X}_{p}$ or $\mathcal{X}_{\ell}$. If (1) holds in $\mathcal{X}$, then (a) and (b) imply that (3) holds in $\mathcal{X}_{p}$, but this is impossible, by the above. So (2) holds in $\mathcal{X}$ and therefore (1) holds in $\mathcal{X}_{p}$ (by (c) and the above). But now (a) and (b) imply that (3) must hold in $\mathcal{X}_{\ell}$ and we already know that this is not possible, so we have reached a contradiction.

This proves the Theorem.