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A lattice $(L, \le)$ is a modular lattice @ wiki if the following modular law holds: $$\forall x \in L: a \le b \implies a \lor (x \land b) = (a \lor x) \land b.$$

The diamond isomorphism theorem @ wiki on modular lattice says that

For any two elements $a,b$ of a modular lattice, one can consider the intervals $[a \land b, b]$ and $[a, a \lor b]$. They are connected by order-preserving maps $$\varphi: [a \land b, b] \to [a, a \lor b], \psi: [a, a \lor b] \to [a \land b, b]$$ that are defined by $\varphi(x) = x \lor a$ and $\psi(y) = y \land b$. Then, $\varphi$ and $\psi$ are both isomorphisms between $[a \land b, b]$ and $[a, a \lor b].$

I tried to prove this theorem by first showing that $\varphi$ preserves the operations $\land$ and $\lor$ as follows:

$$ \forall x_1, x_2 \in [a \land b, b]: \\ \varphi(x_1 \land x_2) = (x_1 \land x_2) \lor a\\ \varphi(x_1) \land \varphi(x_2) = (x_1 \lor a) \land (x_2 \lor a) = (a \lor x_1) \land (x_2 \lor a) =_{\text{modular law}} a \lor (x_1 \land (x_2 \lor a)) $$

What is the next to show that $\varphi(x_1 \land x_2) = \varphi(x_1) \land \varphi(x_2)$?


Edit 1: Note that $a \lor (x_1 \land (x_2 \lor a)) \ge a \lor (x_1 \land x_2)$. How to show the other direction? (Maybe this relies on the fact that $\varphi$ and $\psi$ are bijections.)

Edit 2: The bijection $\varphi$ is a lattice isomorphism can also be verified by showing that $\varphi$ is order-preserving, which is quite easy. A theorem connecting the order-theoretic definition and the algebraic definition of lattice isomorphism may be helpful to show that $\varphi(x_1 \land x_2) = \varphi(x_1) \land \varphi(x_2)$. Any ideas?

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  • $\begingroup$ The first part is not finished. Yes, the second part should follow by duality. Finally show the function is a bijection. $\endgroup$ – William Elliot Mar 4 '18 at 8:15
  • $\begingroup$ @WilliamElliot I failed to complete the first part by showing that $\varphi(x_1 \land x_2) = \varphi(x_1) \land \varphi(x_2)$. Any ideas? I am able to show that the function is a bijection. $\endgroup$ – hengxin Mar 4 '18 at 8:28
  • $\begingroup$ Apply the modular law again to the part in brackets. $\endgroup$ – Paul K Mar 4 '18 at 8:40
  • $\begingroup$ @PaulK Thanks. However, I don't understand how to proceed. We don't know the ordering among $a, x_1,$ and $x_2$. $\endgroup$ – hengxin Mar 4 '18 at 8:48
  • $\begingroup$ @PaulK Maybe the equality to show relies on the fact that $\varphi$ and $\psi$ are bijections and $\varphi^{-1} = \psi$. (I have no idea how to make use of this fact yet.) $\endgroup$ – hengxin Mar 4 '18 at 9:21
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Let (M) denote the modular law in included in your question.
Notice that the path you're following doesn't really use the fact that $x \in [a \wedge b, b]$.
Perhaps the best approach is to prove first that $\varphi$ and $\psi$ are inverse of each other.
For example, for $x \in [a \wedge b, b]$, \begin{align} \psi\varphi(x) &= \psi(x \vee a)\\ &= (x \vee a) \wedge b\\ &= x \vee(a \wedge b)\tag{$\because x \leq b$ and (M)}\\ &= x.\tag{$\because a \wedge b \leq x$} \end{align} Dually, $\varphi\psi(y)=y$, whenever $y \in [a,a\vee b]$. Hence $\varphi$ and $\psi$ are bijective maps.

Now, it is clear that both $\varphi$ and $\psi$ are order-preserving maps.
Hence the intervals $[a \wedge b, b]$ and $[a, a \vee b]$ are order-isomorphic.
But an order-isomorphism between lattices is necessarily an isomorphism, and therefore $\varphi$ and $\psi$ are isomorphisms.

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  • $\begingroup$ Thanks. This works (upvote +1). However, my question is how to show that $\varphi(x_1 \land x_2) = \varphi(x_1) \land \varphi(x_2)$. I have come up with an argument (just posted as an answer). Would you mind reviewing it? Thanks again. $\endgroup$ – hengxin Mar 4 '18 at 12:21
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I come up with the following argument (please review it):

First, we can verify that $\varphi$ and $\psi$ are bijections and that $\varphi^{-1} = \psi$ and $\psi^{-1} = \varphi$.

Next, it is easy to show that $\psi$ preserves the operation $\land$: $$\psi(y_1 \land y_2) = y_1 \land y_2 \land b = (y_1 \land b) \land (y_2 \land b) = \psi(y_1) \land \psi(y_2).$$

Then, $$ \psi\big(\varphi(x_1) \land \varphi(x_2) \big) = \psi\varphi(x_1) \land \psi\varphi(x_2) = x_1 \land x_2. $$

Therefore, we have $$ \varphi(x_1 \land x_2) = \varphi\Big(\psi\big(\varphi(x_1) \land \varphi(x_2) \big)\Big) = \varphi(x_1) \land \varphi(x_2). $$

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  • $\begingroup$ Looks good to me! I think there is no reason not to mark this answer as the correct one. $\endgroup$ – amrsa Mar 4 '18 at 12:22

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