2
$\begingroup$

Thrall states ("On the Projective Structure of a Modular Lattice", 1951, https://scholar.google.com/scholar?cluster=4998496641867146321):

One measure of the complexity of a modular lattice is how much it lacks being distributive. Each triple of elements in $L$ for which the distributive law does not hold generates a lattice which has a five element sublattice $P = \{r, s, t, u, w\}$ in which $r$ is the meet and $w$ is the join of each pair selected from $s, t, u$. We shall call such a five element lattice a projective root (p.r.) and shall use the notation $P = [r; s, t, u; w]$ to indicate that $P$ is a projective root. If any one (and therefore all) of the six quotients [intervals] $s/r, t/r, u/r, w/s, w/t, w/u$ is prime [has height 1] in $L$, we say that $P$ is a prime projective root (p.p.r.) n $L$.

It is easily seen (we omit the proof) that if $L$ contains any projective root, then it contains a prime projective root. Indeed, if $P = [r;s,t,u;w]$ is a p.r. with dimension $s/r = k$, then $L$ contains $k$ p.p.r.'s $P_i = [r_i; s_i, t_i, u_i; w_i], i= 1, \ldots, k$, such that $r=r_1, w_1 = r_2, \ldots, w_{k-i} = r_k, w_k = w$.

My assumption is that the proof takes $P$ and dissects out of the interval $[r, w]$ such a chain of $P_i$. So the value of the result is showing that the signatures of non-distributivity, the p.r.s, are constructed from p.p.r.s.

I'm just learning lattice theory and I can't prove this result, or even see what the intended method of attack is. I've checked Google Scholar, and none of the references seem to pay attention to this result. Similarly, I've looked at a few of the standard references, and none of them discuss this result, or even this approach toward determining the "structure" of a non-distributive modular lattice.

$\endgroup$

1 Answer 1

1
$\begingroup$

Theorem. If $P=[0;s,t,u;1]$ is a projective root in the modular lattice $L$ with $\dim(s/0)=k>1$, then $L$ contains projective roots $P_0=[0;s_0,t_0,u_0;\Omega]$ with $\dim(s_0/0) = 1$ and $P^+=[\Omega;s^+,t^+,u^+;1]$ with $\dim(s^+/\Omega) = k-1$.

This produces a first prime factor $P_0$ of $P$, and you may apply the theorem to $P^+$ to obtain further prime factors.

Proof. (I will write $xy$ for '$x$ meet $y$' and $x+y$ for '$x$ join $y$'.)

Since $\dim(s/0)>1$, there exists an element $s_0$ satisfying $0<s_0 <s$ such that $\dim(s_0/0)=1$. Define $t_0 = t(s_0+u)$, $u_0 = u(s_0+t)$, and $\Omega = (s_0+t)(s_0+u)$.

Note that since $0<s_0\leq s$ and $s$ is a complement of $u$ in $P$, $\dim((s_0+u)/u)=\dim(s_0/0)=1$. Since $t$ is also a complement of $u$, $\dim(t_0/0)=\dim(t(s_0+u)/tu)=\dim((s_0+u)/u)=1$. Similarly $\dim(u_0/0)=1$.

Claim 1. $P_0=[0;s_0,t_0,u_0;\Omega]$ is a prime projective root.

Proof. Since $s, t, u$ pairwise meet to $0$ and $0<s_0\leq s, 0<t_0\leq t, 0<u_0\leq u$, the elements $s_0, t_0, u_0$ pairwise meet to $0$. We calculate that $s_0+t_0 = s_0+t(s_0+u) =_{\textrm{mod}} (s_0+t)(s_0+u)=\Omega$. (Here ``$=_{\textrm{mod}}$'' indicates that I am using the modular law.) The same argument shows that $s_0+u_0=\Omega$. Finally, $t_0+u_0 = \Omega$ by a dimension count. Both $\dim(t_0/0)$ and $\dim(u_0/0)$ equal $1$. Since $t_0+u_0\leq (s_0+t)(s_0+u) = \Omega = s_0+u_0$, and the first and last of these has dimension $2$ over $0$ we have equality throughout, giving $t_0+u_0 = \Omega$. (This dimension count can be replaced by a modularity argument.) Altogether, we have that $[0;s_0,t_0,u_0;\Omega]$ is a prime projective root. \\\

Claim 2. If $s^+=s+\Omega, t^+=t+\Omega$, and $u^+=u+\Omega$, then $P_0=[\Omega;s^+,t^+,u^+;1]$ is a projective root (necessarily with $\dim(s^+/\Omega)=k-1$).

Proof. Since $s, t, u$ pairwise join to $1$ and $s\leq s^+\leq 1, t\leq t^+\leq 1, u\leq u^+\leq 1$, the elements $s^+, t^+, u^+$ pairwise join to $1$. Since each of $s^+, t^+, u^+$ are above $\Omega$ we have the first inequality in $$\Omega\leq s^+t^+ = (s+\Omega)(t+\Omega) =_{\textrm{mod}} t(s+\Omega)+\Omega.$$ Since $\Omega=s_0+t_0$ and $s_0\leq s, t_0\leq t$, this simplifies to $$t(s+\Omega)+\Omega = t(s+s_0+t_0)+\Omega = t(s+t_0)+\Omega =_{\textrm{mod}} ts + t_0 +\Omega = 0+t_0+\Omega=\Omega.$$ The same calculation shows that $s^+u^+=\Omega$ and $t^+u^+=\Omega$, hence $P^+$ is a projective root. To see why $\dim(s^+/\Omega)=k-1$ is necessarily $k-1$, note that $$ \dim(1/0) = 2k = \dim(1/\Omega)+\dim(\Omega/0) = 2\dim(s^+/\Omega)+2\dim(s_0/0), $$ hence $\dim(s^+/\Omega)=k-1$. \\\

$\endgroup$
2
  • $\begingroup$ Thanks! I am going to write up a somewhat more general version of this and I will link to it here. $\endgroup$
    – Dale
    Oct 15, 2023 at 15:41
  • 1
    $\begingroup$ It turns out that my generalization (adding more elements to the set $\{s, t, u\}$ isn't true. $\endgroup$
    – Dale
    Oct 20, 2023 at 3:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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