Let $R$ be a real von Neumann algebra of bounded operators on the real Hilbert space $H$ and ${\cal L}(R)$ the lattice of orthogonal projectors in $R$.
Is it true that (similarly to what happens in the complex case) ${\cal L}(R)''= R$?
The problem is that the spectral decomposition is not allowable for antiselfadjoint operators in real Hilbert spaces and some of them are necessarily present in $R$. If, for instance, all of them are obtained in terms of weak limits of commutators of selfadjoint operators the answer is YES, but I am not very familiar with real von Neumann algebras.
Addendum
The answer is negative as it easily arises from the case of $R$ given by all of real $2\times 2$ matrices of the form $aI+ bJ$ with $a,b \in \mathbb R$ and $J$ a fixed matrix with $JJ=-I$ and $J^*=-J$. In this case, the only orthogonal projectors of $R $ are $0$ and $I$ and the von Neumann algebra generated by them is just made of the real matrices of the form $aI$. All nonvanishing antiselfadjoint elements $bJ$ cannot be included.
Nevertheless, in more complicated cases the von Neumann algebra generated by the lattice of projectors of a given (real) non commutative von Neuman algebra includes some antiselfadjoint elements: the commutators of selfadjoint elements. Are there conditions assuring that these operators are all of antiselfadjoint operators of the initial von Neumann algebra?