3 ms·

From my experience, in literature you will usually found (external) direct sum of spaces defined as Cartesian product. But as comment before `Proposition 6.1` s

by ubavic 4y ago

From my experience, in literature you will usually found (external) direct sum of spaces defined as Cartesian product. But as comment before `Proposition 6.1` says, there is a bijection between definition from book and 'standard' definition, so it doesn't really matter as long we are consistent with choice. (i guess this nonstandard choice was made because it behaves nicely from the point of inner sums) It is unfortunate coincidence, but Haskell sum types don't correspond to sums of spaces. Sum type is disjoint union of types, while sum of (vector) spaces is isomorphic to Cartesian product of those spaces (and therefore corresponds to product types in Haskell).

whatshisface 4y ago
>It is unfortunate coincidence, but Haskell sum types don't correspond to sums of spaces. Since types on computers are discrete, it makes more sense to name them according to the fact that if |x| is the number of values that a variable of type x can take on, |x+y|=|x|+|y| and |x*y|=|x|*|y|.
knappa 4y ago
> It is unfortunate coincidence, but Haskell sum types don't correspond to sums of spaces. It depends on the perspective. Both are a form of coproduct https://en.wikipedia.org/wiki/Coproduct https://en.wikipedia.org/wiki/Coproduct which forms a disjoint set for types (and sets and topological spaces) while forming a direct product for vector spaces. Confusingly, the vector space sum is a product of sets. (but you can see that it is an addition when you look at the resulting dimensions) The product operation for vector spaces is the tensor product.
- nextaccountic 4y ago
  > Confusingly, the vector space sum is a product of sets. But why?
  - knappa 4y ago
    It's the unique thing that makes the commutative diagram from the wikipedia article work. As in: You want V⊕W to be a vector space that contains the vector spaces V and W. You want that for any linear maps V → U and W → U, you get a linear map V⊕W → U which agrees on the inclusions. Since these maps can be arbitrary, you know that dim(V⊕W)≥dim(V)+dim(W) since you can't have any colinearities between the images of V and W in V⊕W. Plus you want that the map V⊕W → U is unique. This means that the images of V and W span V⊕W. Otherwise you have another vector that you can send to arbitrary places. This all means that V⊕W must be a vector space that has dim(V⊕W)=dim(V)+dim(W). Now all you have to do is provide a concrete candidate for V⊕W and the set of ordered pairs (v,w)∈V×W with coordinate-wise addition (+etc.) works.
nextaccountic 4y ago
> From my experience, in literature you will usually found (external) direct sum of spaces defined as Cartesian product. But as comment before Proposition 6.1 says, there is a bijection between definition from book and 'standard' definition I don't get it, does this mean that the direct sum and the cartesian product are in some sense equivalent?
- knappa 4y ago
  There are different sum- and product-like operations for different types of things: For sets (and topological spaces), the sum operation is the disjoint union and the product is the Cartesian product. For vector spaces, the sum operation is the Cartesian product and the product operation is the tensor product. If you think of finite dimensional vector spaces as functions on finite sets, then these all match up. i.e. if V (vector space) is the functions on S (set) and W (vector space) is the functions on T (set), then the functions on S⊔T are V⊕W because you just list the values of the functions on S and then on T. Further, you can see that functions on S×T are V⊗W since the natural basis for V⊗W consists of vectors e_s⊗e_t that pick out elements (s,t) ∈ S×T.
  - nextaccountic 4y ago
    Whoa, thanks! That's confusing. And actually.. since a vector space has an underlying set, it seems to me that to do a direct sum on a vector space, you must first do a direct sum on the set (that is, do a disjoint union), and then do other things to map the rest of the structure of a vector space into the sum. So, somehow, a disjoint union (of the underlying set) becomes a cartesian product (of the whole vector space)? Also: the direct sum of an abelian group is also a cartesian product, right? Of any algebraic structure, not only vectors? Why are sets so special to have a different direct sum than algebraic structures built on top of sets? (Is there somewhere to read about this to gain intuition?)
    - bmacho 4y ago
      > And actually.. since a vector space has an underlying set, it seems to me that to do a direct sum on a vector space, you must first do a direct sum on the set (that is, do a disjoint union), and then do other things to map the rest of the structure of a vector space into the sum. No. > So, somehow, a disjoint union (of the underlying set) becomes a cartesian product (of the whole vector space)? No. If anything it is the other way: the cartesian product of the underlying sets becomes the (equivalent of) "disjoint union" of the vector spaces. > (Is there somewhere to read about this to gain intuition?) I don't think you want that. But there definitely are some category theory textbooks, and introductory pdfs. E.g. Tom Leinster Basic Category Theory is a textbook, and it aims to give you intuition. https://arxiv.org/abs/1612.09375 https://arxiv.org/abs/1612.09375

by ubavic 4y ago

whatshisface 4y ago

>It is unfortunate coincidence, but Haskell sum types don't correspond to sums of spaces. Since types on computers are discrete, it makes more sense to name them according to the fact that if |x| is the number of values that a variable of type x can take on, |x+y|=|x|+|y| and |x*y|=|x|*|y|.

knappa 4y ago

> It is unfortunate coincidence, but Haskell sum types don't correspond to sums of spaces. It depends on the perspective. Both are a form of coproduct https://en.wikipedia.org/wiki/Coproduct https://en.wikipedia.org/wiki/Coproduct which forms a disjoint set for types (and sets and topological spaces) while forming a direct product for vector spaces. Confusingly, the vector space sum is a product of sets. (but you can see that it is an addition when you look at the resulting dimensions) The product operation for vector spaces is the tensor product.

nextaccountic 4y ago
> Confusingly, the vector space sum is a product of sets. But why?
- knappa 4y ago
  It's the unique thing that makes the commutative diagram from the wikipedia article work. As in: You want V⊕W to be a vector space that contains the vector spaces V and W. You want that for any linear maps V → U and W → U, you get a linear map V⊕W → U which agrees on the inclusions. Since these maps can be arbitrary, you know that dim(V⊕W)≥dim(V)+dim(W) since you can't have any colinearities between the images of V and W in V⊕W. Plus you want that the map V⊕W → U is unique. This means that the images of V and W span V⊕W. Otherwise you have another vector that you can send to arbitrary places. This all means that V⊕W must be a vector space that has dim(V⊕W)=dim(V)+dim(W). Now all you have to do is provide a concrete candidate for V⊕W and the set of ordered pairs (v,w)∈V×W with coordinate-wise addition (+etc.) works.

nextaccountic 4y ago

> From my experience, in literature you will usually found (external) direct sum of spaces defined as Cartesian product. But as comment before Proposition 6.1 says, there is a bijection between definition from book and 'standard' definition I don't get it, does this mean that the direct sum and the cartesian product are in some sense equivalent?

knappa 4y ago
There are different sum- and product-like operations for different types of things: For sets (and topological spaces), the sum operation is the disjoint union and the product is the Cartesian product. For vector spaces, the sum operation is the Cartesian product and the product operation is the tensor product. If you think of finite dimensional vector spaces as functions on finite sets, then these all match up. i.e. if V (vector space) is the functions on S (set) and W (vector space) is the functions on T (set), then the functions on S⊔T are V⊕W because you just list the values of the functions on S and then on T. Further, you can see that functions on S×T are V⊗W since the natural basis for V⊗W consists of vectors e_s⊗e_t that pick out elements (s,t) ∈ S×T.
- nextaccountic 4y ago
  Whoa, thanks! That's confusing. And actually.. since a vector space has an underlying set, it seems to me that to do a direct sum on a vector space, you must first do a direct sum on the set (that is, do a disjoint union), and then do other things to map the rest of the structure of a vector space into the sum. So, somehow, a disjoint union (of the underlying set) becomes a cartesian product (of the whole vector space)? Also: the direct sum of an abelian group is also a cartesian product, right? Of any algebraic structure, not only vectors? Why are sets so special to have a different direct sum than algebraic structures built on top of sets? (Is there somewhere to read about this to gain intuition?)
  - bmacho 4y ago
    > And actually.. since a vector space has an underlying set, it seems to me that to do a direct sum on a vector space, you must first do a direct sum on the set (that is, do a disjoint union), and then do other things to map the rest of the structure of a vector space into the sum. No. > So, somehow, a disjoint union (of the underlying set) becomes a cartesian product (of the whole vector space)? No. If anything it is the other way: the cartesian product of the underlying sets becomes the (equivalent of) "disjoint union" of the vector spaces. > (Is there somewhere to read about this to gain intuition?) I don't think you want that. But there definitely are some category theory textbooks, and introductory pdfs. E.g. Tom Leinster Basic Category Theory is a textbook, and it aims to give you intuition. https://arxiv.org/abs/1612.09375 https://arxiv.org/abs/1612.09375