3 ms·

Yeah, there really are more, provided you define "more" right. They form a topological space, and the rank-2 tensors are a 9-dimensional space versus a 6-dimens

by gjm11 12y ago

Yeah, there really are more, provided you define "more" right. They form a topological space, and the rank-2 tensors are a 9-dimensional space versus a 6-dimensional space for the outer products. Alternatively, any matrix whose determinant isn't zero is a non-dyad.

sampo 12y ago
But the cardinality of the set of all points in a 9-dimensional space is not larger than the cardinality of the set of points in a 6-dimensional space, or even between 0 and 1 in the 1-dimensional real line. And the set theoretic cardinality is the normal way to define what "more" means, or to compare sets of an infinite amount of objects. How would you define "more", to get e.g. R² have "more" points than R?
- wetmore 12y ago
  You can say that R^2 has "more" points than R by inclusion: there is a natural injection of R into R^2 as a proper subspace. Think of the odd numbers in the integers, for example. "More" is a fairly loose concept in that there is no strict definition. You can use it to refer to cardinality, or some other sense of order.
- gjm11 12y ago
  There are lots of ways to define "more" (or "bigger" or whatever term one chooses to use). Which one is most appropriate depends on what kind of things you're working with. Working with vector spaces over a given field: "bigger" means "higher dimension". Working with manifolds: "bigger" means "higher dimension". Working with sets: "bigger" usually means "larger cardinality" but could sometimes mean "proper superset". In general, you can take "X is bigger than Y" to mean "Y is isomorphic to some substructure of X, but not vice versa". If "isomorphic" means "in bijection with" and "substructure" means "subset" then this is cardinality. If they mean "isomorphic as vector spaces over F" and "vector subspace" then it's dimension. If they mean "homeomorphic" and "topological subspace" then actually it isn't true that lower-dimensional manifolds are always "smaller than" higher-dimensional ones (so maybe we actually want a different definition) but in this particular case the rank-1 tensors are a lower-dimensional proper subspace of all the tensors so the definition gives the "right" answer.

by gjm11 12y ago

sampo 12y ago

But the cardinality of the set of all points in a 9-dimensional space is not larger than the cardinality of the set of points in a 6-dimensional space, or even between 0 and 1 in the 1-dimensional real line. And the set theoretic cardinality is the normal way to define what "more" means, or to compare sets of an infinite amount of objects. How would you define "more", to get e.g. R² have "more" points than R?

wetmore 12y ago
You can say that R^2 has "more" points than R by inclusion: there is a natural injection of R into R^2 as a proper subspace. Think of the odd numbers in the integers, for example. "More" is a fairly loose concept in that there is no strict definition. You can use it to refer to cardinality, or some other sense of order.
gjm11 12y ago
There are lots of ways to define "more" (or "bigger" or whatever term one chooses to use). Which one is most appropriate depends on what kind of things you're working with. Working with vector spaces over a given field: "bigger" means "higher dimension". Working with manifolds: "bigger" means "higher dimension". Working with sets: "bigger" usually means "larger cardinality" but could sometimes mean "proper superset". In general, you can take "X is bigger than Y" to mean "Y is isomorphic to some substructure of X, but not vice versa". If "isomorphic" means "in bijection with" and "substructure" means "subset" then this is cardinality. If they mean "isomorphic as vector spaces over F" and "vector subspace" then it's dimension. If they mean "homeomorphic" and "topological subspace" then actually it isn't true that lower-dimensional manifolds are always "smaller than" higher-dimensional ones (so maybe we actually want a different definition) but in this particular case the rank-1 tensors are a lower-dimensional proper subspace of all the tensors so the definition gives the "right" answer.