Skip to main content
NinjaDarth's user avatar
NinjaDarth's user avatar
NinjaDarth's user avatar
NinjaDarth
  • Member for 3 years, 11 months
  • Last seen more than a month ago
comment
If d/dx is an operator, on what does it operate?
It doesn't "stop being a mathematical operation". The syntax, itself, is a mathematical structure that can be, and is normally, formalized as a magma or as its generalization to different signatures, which are called term algebras for that signature. The elements of a free object in a term algebra are, in fact, one and the same as an abstract syntax tree over that signature, and it is here that you formalize such notions as abstract syntax tree. This is where the "syntax" or "meta level" - as you refer to it - actually lives.
comment
If d/dx is an operator, on what does it operate?
This gets more directly to the question: what type of functional or operator is this: $λx(\_)$; i.e. $y ↦ λx(y)$, when we want bound $x$'s in $y$ to be regarded as such? I think you need to fall back to term algebras, and to magmas (or whatever the generalization of magma's is called) to formalize that as a function that respects the handling of bound variables. It's just easier to skip the formalization and describe it as just a syntactic functional, instead; somewhat like Landin's "let (_) = (_) in (_)".
comment
If d/dx is an operator, on what does it operate?
You need to be more specific. This one: $\frac{d}{dx}(y) = Dλx(y)$? The $(\_)$ place-holder is not an argument in the usual sense, because it can take bound variables. So it has to be read as something at the "alpha level", so to say, rather than "beta level"; and the place holder is more akin to being of what's referred to as a "reference type" in programming languages. It is a syntactic functional, maybe formalized as such in the Magma (or generalization thereof) underlying the Term algebra.
Loading…
awarded
revised
Meaning of the coadjoint representation and its orbits
Upgraded the layout (\hspace 1em becomes \quad), removed the editorial remarks. You can find them in the earlier edits.
Loading…
comment
Meaning of the coadjoint representation and its orbits
Ok. I'll remove the editorial, if you wish.
revised
If d/dx is an operator, on what does it operate?
Redoing the adjustment - to compensate for an apparent bug in the math rendering software being used on this site.
Loading…
answered
Loading…
awarded
awarded
revised
Meaning of the coadjoint representation and its orbits
Additional remarks on the Cartan-Maurer forms and co-tangent bundle.
Loading…
awarded
revised
Loading…
Loading…
awarded
answered
Loading…
answered
Loading…