Is the "op" lattice $\mathscr{L}^\perp_{\mathscr H}$ of subspaces of a Hilbert space $\mathscr H$ also atomistic with the covering property?

Question

Let $\mathscr{L}_{\mathscr H}$ denote the usual lattice of subspaces of a Hilbert space $\mathscr H$, where for $p,q\in\mathscr H$ we write $p\leq q$ iff $p$ is a subspace of $q$. Then, as discussed in, e.g., Beltrametti&Cassinelli https://books.google.com/books?id=yWoq_MRKAgcC&pg=PA98, this lattice is atomistic with the covering property. (And another discussion of atomistic lattices with the covering property is in Section 4.2.5 (pages 4-10 and 4-11) of https://arxiv.org/abs/1211.5627)

(That is, very briefly, $0$ is the weakest element, i.e., $\forall p\in{\mathscr L_H}:0\leq p$, and pure states $a\in{\mathscr L_H}$ (one-dimensional subspaces) are "atoms", i.e., $\forall p\in{\mathscr L_H}:0\leq p\leq a\implies 0=p\mbox{ .or. }p=a$. And then the covering property means $\forall p\in{\mathscr L_H}: a\not\leq p\implies \not\exists q|p\leq q\leq p\sqcup a$.)

Now consider the "op" lattice $\mathscr{L}^\perp_{\mathscr H}$, where $q\leq p$ in $\mathscr{L}_{\mathscr H}$ means $p\leq q$ in $\mathscr{L}^\perp_{\mathscr H}$. Then, e.g., $0^\perp={\mathscr H}$ is now the weakest element in $\mathscr{L}^\perp_{\mathscr H}$. So can we now say that...
    (a) the orthocomplements of pure states are the atoms of $\mathscr{L}^\perp_{\mathscr H}$, and
    (b) they also possess the corresponding covering property with respect to this lattice?
And can you prove it, or even better (and presumably easier) cite a proof? Note that by "prove", I mean assume it's already been proven for $\mathscr{L}_{\mathscr H}$ (which it has), and then you only need prove it must also be true for $\mathscr{L}^\perp_{\mathscr H}$.

Answer 1

This depends on whether $\mathcal L_{\mathcal H}$ is the lattice of all subspaces of $\mathcal H$, or just closed subspaces.

In the first cas, the result is not true : the orthocomplements of pure states are closed hyperplanes of $\mathcal H$, while atoms of $\mathcal L_{\mathcal H}^\perp$ are maximal proper subspaces, ie hyperplanes, but not necessarily closed.

If we consider only closed subspace, the result is true.

For $p,q\in\mathcal L_\mathcal H$, you have : $$p\leq q^\perp \Longleftrightarrow p^\perp \geq q$$ This means that we have a monotone Galois connection $\mathcal L_{\mathcal H}\leftrightarrows \mathcal L_{\mathcal H}^\perp$, whose closure operator is the toplogical closure. Since this is the identity on closed subspaces, we have an isomorphism of lattices $\mathcal L_{\mathcal H} \simeq \mathcal L_{\mathcal H}^\perp$ which maps pure states (which are the atoms of $\mathcal L_{\mathcal H}$) to closed hyperplanes (which therefore are the atoms of $\mathcal L_{\mathcal H}^\perp$). The covering property is also preserved.