Recall that a Boolean algebra is a complemented distributive lattice. The set of subspaces of a vector space comes very close to being a boolean algebra. It satisfies all the required properties, except being distributive. Even then it is a [modular lattice](https://ncatlab.org/nlab/show/modular+lattice), which is almost the distributivity property. This raises the question of constructing a Boolean algebra from this complemented modular lattice. So,


* Does there exist a left (or right) adjoint to the forgetful functor $F$ from the category of Boolean algebras (and bounded-lattice homomorphisms) to that of complemented modular lattices?
* If so, what does it look like?



Boolean algebra of the lattice of subspaces of a vector space?
I'm assuming "bounded-lattice homomorphism" means what I would call "0,1-homomorphism", i.e., it takes top to top and bottom to bottom? Then there is no adjoint, left or right.


Let $L$ be the lattice of subspaces of a two-dimensional vector space $V$ (over any field). I claim that there is no 0,1-homomorphism from $L$ into any Boolean algebra. This is because we can find three one-dimensional subspaces of $V$ any two of which are complements in $L$. But then the same is true of their images under any 0,1-homomorphism, and you cannot find three elements of any Boolean algebra with this property.


So, no left adjoint to the forgetful functor because every Boolean algebra has at least one 0,1-homomorphism into the two-element Boolean algebra $\{0,1\}$ and thus there is no Boolean algebra $B$ with the property that ${\rm Hom}(B, \{0,1\}) \cong {\rm Hom}(L, \{0,1\})$.


Now let $M$ be the lattice of subspaces of the vector space $\mathbb{F}\_2^3$, the three-dimensional vector space over the two-element field. Letting $B\_4$ be the four-element Boolean algebra, if I have calculated correctly\* there are exactly fifty-eight 0,1-homomorphisms from $B\_4$ into $M$. But the number of 0,1-homomorphisms from $B\_4$ into any Boolean algebra is either a power of two or infinite. So there is no Boolean algebra $B$ with ${\rm Hom}(B\_4, B) \cong {\rm Hom}(B\_4, M)$ and thus no right adjoint.




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${}^\*$Because $M$ has sixteen elements, two of which (top and bottom) have exactly one complement, and fourteen of which have exactly four complements. So there are $2 + 14\cdot 4 = 58$ 0,1-homomorphisms from $B\_4$ into $M$.


