Julia: Exterior algebra: `⋆` is not a unary operator

[rleegates]

I'm currently writing a library for exterior algebra on manifolds and need to define the Hodge star operator. In particular, the Hodge star maps a k-form on an n-dimensional manifold to an n-k-form. Is there any use-case which disagrees with ⋆ being a unary operator? If not and it's a quick fix, could fulfilling this request be part of the 0.5 release?

[simonbyrne]

You can call the function form of *, which accepts a single argument, by enclosing it in parentheses

julia> (*)(1)
1

[oxinabox]

We are talking about \star right?

What does work for \star is, like what @simonbyrne said for *, using it as a function.

julia> ⋆(x)=2*x
⋆ (generic function with 1 method)


julia> ⋆(4)
8

I believe the current unary operators are defined here.

[Keno]

I'm fine with letting \star be both unary and binary, but from a syntax perspective people may prefer to have it behave like *

[TotalVerb]

What about making unary * syntax for unsafe_load? (Joking. That would be terrible.)

On a more relevant note, how about making \bigcup, \bigcap, \prod, \sum, \bigvee, and \bigwedge unary? These symbols (⋃ ⋂ ∏ ∑ ⋁ ⋀) are fairly commonly used unary in mathematics.

[stevengj]

@TotalVerb, I think @JeffBezanson's perspective from way back was that \cup would be a binary operator and \bigcup (etc.) would just be an identifier name, so you can basically use it as a unary operator but you have to add parens.

[simonbyrne]

now we can pass comprehensions in functions, these should be reduction operators, i.e. \bigcup (foo(i) for i in X)

[TotalVerb]

I can get behind aliases like

const ∑ = sum
const ∏ = prod
⋀(x) = reduce(∧, x)  # all?
⋁(x) = reduce(∨, x)  # any?
⋃(x) = reduce(union, x)
⋂(x) = reduce(intersect, x)

though the empty case is troublesome, as usual.

[JeffreySarnoff]

wrt sets, Union( ∅, ∅ ) is Union( ∅ ) is (∅) which I am free to write this way () # math, not as a tuple
similarly Intersection, so how about

⋃(x) = reduce(union, x)
⋂(x) = reduce(intersect, x)

⋃() = Void()
⋂() = Void()

# and if it is useful to cover reducing over an empty collection, here are two
const emptyTuple = ()
const zeroElements = []

⋃(emptyTuple) = emptyTuple
⋂(emptyTuple) = emptyTuple
⋃(zeroElements) = zeroElements
⋂(zeroElements) = zeroElements

[c42f]

Fixed in #31604