Why is addition defined, and not implied, on quotient spaces?

Question

Small question. In chapter 3, section E, page 96 of "Linear Algebra Done Right", addition in quotient vector spaces is defined this way:

I understand why scalar multiplication has to be defined, because multiplying a subset of a vector space with a scalar was not defined. But why can addition of affine subsets be said to work this way, if the sum of subsets of a vector space was already defined?

$v+U, w+U$ are both subsets of a vector space $V$, so $(v+U)+(w+U)$ is the set containing all possible sums of elements of $v+U$ with elements of $w+U$. Shouldn't the statement $(v+U)+(w+U)=(v+w)+U$ be a theorem?

Answer 1

In fact, regardless of what others are saying, the two definitions of addition are equivalent for cosets of a subspace.

We need different notation for the two additions. Let's say $$(a+U)+(b+U)=(a+b)+U$$is the "coset-wise sum", while $$A\oplus B=\{a+b:a\in A, b\in B\}$$ is the "subset sum".

Triviality. If $V$ is a vector space, $U$ is a subspace of $V$ and $a,b\in V$ then $(a+U)+(b+U)=(a+U)\oplus (b+U)$.

Proof: Suppose first that $x\in (a+U)\oplus (b+U)$. The definitions show that there exist $u_1,u_2\in U$ with $x=(a+u_1)+(b+u_2)$; hence $x=(a+b)+(u_1+u_2)\in (a+b)+U=(a+U)+(b+U)$.

Otoh if $x\in (a+U)+(b+U)=(a+b)+U$ then $x=(a+b)+u=(a+u)+(b+0)\in (a+U)\oplus (b+U)$.

Answer 2

The quotient space is given by $V/U = \{v+U\mid v\in V\}$, where $v+U = \{v+u\mid u\in U\}$ is a coset. Note that $v+U = (v+u)+U$ for each $u\in U$ and so the coset representative $v$ is not uniquely determined.

Addition is defined as

$(v+U) + (w+U) := (v+w) + U.$

What needs to be shown is well-definedness, i.e. if you take other representatives of the cosets, $v'+U = v+U$ and $w'+U=w+U$, then the result is the same: $(v+w)+U = (v'+w')+U$.

Answer 3

You are correct that the definition of addition of cosets is a special case of the definition of arbitrary subsets. I have on many occasions encountered what might be referred to as 'non-minimal' definitions, where certain parts are redundant and can be deduced from earlier results.

Sometimes, like in this case, multiple definitions are given when one could easily just refer to the most general definition and point out that the special case follows easily from it. I guess it is a matter of style, but personally I think if overlapping definitions are given in the same text, the author should at the very least explicitly outline their interdependence so the reader knows there if there is any risk of using inconsistent definitions.

The author might have presented it this way in this particular case to keep an individual chapter relatively self-contained, since most readers will not read cover-to-cover. Having said all this, I should point out that Linear Algebra Done Right does seem to be a very good book to learn from :)

Answer 4

The answer of @DavidCUllrich addresses your question of whether the statement should be a theorem; indeed, yes it should.

But you have also asked a subtly different question: Why can addition of affine subsets be said to work this way, if the sum of subsets was already defined?

The short answer is: Commutativity of addition is used strongly to prove that the two definitions are equivalent; whereas Definition 3.86 works even in noncommutative situations where Definition 1.36 does not apply.

To put this another way, Definition 3.86 generalizes to produce operations on quotient sets in other branches of algebra, in situations where Definition 1.36 definitely does not generalize.

In particular, in group theory, consider a group $G$ with group operation written multiplicatively as $g \cdot h$. Given a normal subgroup $N \triangleleft G$, the group operation on the quotient group $G / N$ is defined by $$(*) \qquad\qquad (g \cdot N) \cdot (h \cdot N) = (g \cdot h) \cdot N $$ But in this group theoretical context where the group operation need not be commutative, the operation defined in $(*)$ cannot be made to agree with any generalization of the subset operation in Definition 3.36.

Answer 5

Other answers give several good reasons already; here is one more, not yet mentioned.

Conceptually, we often think of the cosets $v+U$ (or more generally, equivalence classes in a quotient) as a different kind of thing from arbitrary subsets of a space. We’re thinking of the coset/quotient space $V/U$ as a separate object in its own right, not as living inside the power set $\mathcal{P}(V)$, even though formally $V/U \subseteq \mathcal{P}(V)$. So we think of the operations on $V/U$ as defined in themself — we don’t stop to consider whether they’re a special case of the operations on $\mathcal{P}(V)$. In this case the two definitions do agree; but there are other places where the operations you use on a quotient don’t agree with the setwise operations on the power-set (e.g. cosets in a general group/ring/module, such as mod-$n$ arithmetic $\mathbb{Z}/n\mathbb{Z}$).

This way of thinking — viewing different sets as separate objects, even though they may have some elements in common — is widespread and important in maths, but often not explicitly discussed or taught. It’s more explicitly discussed in programming, as a kind of abstraction, or implementation-independence. Formally, unicode-strings (or whatever) are just certain sequences of bytes. But (in many modern programming languages) you declare them a separate type of object, and you deliberately don’t allow byte-sequence functions to act directly on unicode-strings.