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Chapter 6, at page 163, says that direct sums can't be built directly from the cartesian product because, instead of defining ordered pairs (a, b) != (b, a), th

Chapter 6, at page 163, says that direct sums can't be built directly from the cartesian product because, instead of defining ordered pairs (a, b) != (b, a), they are unordered, so E ∐ F = F ∐ E However, direct sums in general are defined as being ordered, see https://math.stackexchange.com/questions/2694051/what-is-the-direct-sum https://math.stackexchange.com/questions/2694051/what-is-the... and https://en.wikipedia.org/wiki/Direct_sum https://en.wikipedia.org/wiki/Direct_sum How to make sense of that? Actually I'm confused because Wikipedia says "The direct sum of two abelian groups A A and B B is another abelian group A ⊕ B consisting of the ordered pairs (a, b) where a ∈ A and b ∈ B . To add ordered pairs, we define the sum (a, b) + (c, d) = (a + c, b + d)", and, while vector spaces have a little more structure than abelian groups, the definition should match up; and this definition seems just wrong, since it implies that direct sums are just cartesian products! PS: in my understanding, direct sums in math are analogous to sum types in programming (like in Haskell, data X = One Float | Another Float Float, X is a direct sum between ℝ and ℝ²), or better yet, anonymous sum types like the polymorphic variants of OCaml.

bmacho 4y ago
I believe that this upenn text doesn't use the 2 variable direct sum operation at all (which is not commutative), but abuses its notation to denote the direct sum of an indexed family. (But then he can't talk about E ∐ F which was his original goal :/)
- nextaccountic 4y ago
  > denote the direct sum of an indexed family. Indeed he says so, but yes he keeps using E ∐ F Why is the direct sum of an indexed family different in terms of commutativity?
  - bmacho 4y ago
    The direct sum of an indexed family is not a function of two variables, so it is meaningless to talk about its commutativity.
knappa 4y ago
Yes, this looks like an error. You don't want unordered pairs, which do not form a vector space, but an isomorphism of under order permutation of the terms. Roughly speaking, V⊕W is not the same as W⊕V, but if you swap the order everywhere then math won't notice.
- deleted 4y ago
  [deleted]
- bmacho 4y ago
  I think it's correct, and also seems pretty deliberately. I also think that his way is inferior to the common way (I can't think of 1 single advantage of it, really).
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.

bmacho 4y ago

I believe that this upenn text doesn't use the 2 variable direct sum operation at all (which is not commutative), but abuses its notation to denote the direct sum of an indexed family. (But then he can't talk about E ∐ F which was his original goal :/)

nextaccountic 4y ago
> denote the direct sum of an indexed family. Indeed he says so, but yes he keeps using E ∐ F Why is the direct sum of an indexed family different in terms of commutativity?
- bmacho 4y ago
  The direct sum of an indexed family is not a function of two variables, so it is meaningless to talk about its commutativity.

knappa 4y ago

Yes, this looks like an error. You don't want unordered pairs, which do not form a vector space, but an isomorphism of under order permutation of the terms. Roughly speaking, V⊕W is not the same as W⊕V, but if you swap the order everywhere then math won't notice.

deleted 4y ago
[deleted]
bmacho 4y ago
I think it's correct, and also seems pretty deliberately. I also think that his way is inferior to the common way (I can't think of 1 single advantage of it, really).

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.