Now, we see a connection with the sign of a permutation: it’s the only nontrivial way we know (and in fact it’s the only way to do it at all!) to assign a scalar value to a permutation, which in this special case we know the determinant must do.
Huh? Off the top of my head, here’s another way to assign a scalar value to a permutation: multiply together the lengths of all the cycles it contains. (No idea whether this is useful for anything. Taking the least common multiple of the lengths of all the cycles tells you the order of the permutation, i.e. how many times you have to apply it before you get the identity, though.)
Thanks. Yeah, I knew there was some qualifier missing that would make it true, I just couldn’t intuit exactly what it was.
Edited to add:
Actually I would say that the determinant distributes through multiplication. Commutativity: a⨁b=b⨁a. Distributivity: a⨀(b⨁c)=(a⨀b)⨁(a⨀c). Neither is a perfect analog, because the determinant is a unary operation, but distributivity at least captures that there are two operations involved. But unlike my other comment, this one doesn’t actually impair comprehension, as there’s not really a different thing you could be trying to say here using the word “commutes”.
Huh? Off the top of my head, here’s another way to assign a scalar value to a permutation: multiply together the lengths of all the cycles it contains. (No idea whether this is useful for anything. Taking the least common multiple of the lengths of all the cycles tells you the order of the permutation, i.e. how many times you have to apply it before you get the identity, though.)
The assignment has to commute with multiplication, and your proposed assignment would not. Just consider, say, (12)(23)=(123).
I’ve edited the post to make this clearer, thanks for the comment.
Thanks. Yeah, I knew there was some qualifier missing that would make it true, I just couldn’t intuit exactly what it was.
Edited to add: Actually I would say that the determinant distributes through multiplication. Commutativity: a⨁b=b⨁a. Distributivity: a⨀(b⨁c)=(a⨀b)⨁(a⨀c). Neither is a perfect analog, because the determinant is a unary operation, but distributivity at least captures that there are two operations involved. But unlike my other comment, this one doesn’t actually impair comprehension, as there’s not really a different thing you could be trying to say here using the word “commutes”.