Thoughts on linearity

Where orthodoxy addresses a collection of values which may be added, I'm identifying each of those values, v, with the associated translation through v, (: v+u ←u :). Composing two such translations gives the translation through the sum of the values added by the two translations, (:a+u←u:)&on;(b+u←u) = (:a+b+u←u:). Once a and b are replaced by x = (:a+u←u:) and y = (:b+u←u:), we can write x+y for x&on;y and, indeed, recover the ability to write all orthodox statements about vectors as statements about translations. Because translations are, however, relations we can repeat them, to obtain scaling by the positive natural numbers; in so far as these scalings are monic (on some given collection of translations) we can introduce their reverses to obtain rational scalings – scale up by one positive integer and down by another.

We can define respects addition as a property on relations (from a set of translations, or from a set embeddable in such) by: r respects addition iff r relates u to v and r relates x to y implies r relates u+x to v+y. Any such relation then trivially commutes with every rational scaling. Within the collection of relations that respect addition, one naturally has the positive rational numbers as an abelian multiplicative group with an associated (cancellable abelian associative) addition over which the multiplication (compsition) distributes; in so far as we can find some field – or, at least, some abelian multiplicative group (the positive members of the field) with associated addition over which the multiplication distributes – which subsumes the positive rationals, we can allow its members as further scalings, extending the rationals. I thus chose to leave it to context to select and specify which other relations to deem scalings, provided {scalings} supports abelian multiplication distributing over abelian addition, with the addition cancellable and, aside from the additive identity (if any), the multiplication also cancellable.

So, consider a collection V of relations (e.g. our translations) which is closed under composition and among which composition is abelian - i.e. a&on;b = b&on;a is in V for all a, b in V - and write + for composition's restriction to V, i.e. a+b = a&on;b for a, b in V. For any equivalence I, extend + to act on {relations (V::I)} according to:

Describe a relation (V:f:V) as real linear on V iff: f relates u to x and v to y implies f relates u+v to x+y, i.e. iff f respects addition. Describe a real linear (V:f:V) as a real scaling of V iff f commutes, i.e. f&on;g = g&on;f, with every real linear (V:g:V). Because every real linear respects addition, it also respects repeated addition; so, for each positive natural n, (V:repeat(n):V) is a real scaling of V; hereinafter written as n.v←v using . as a binary operator (to be construed as multiplication of a vector by a scalar).

For example, in a vector space (real or complex), the real scalings are multiplication by real scalars; in the complex case, antilinear and linear maps respect addition but don't commute with one another, aside from the real scalings (which are linear; though the zero map, if V has an additive identity, is also antilinear). I need some way to identify the remaining scalings, when there are any; they commute with all (complex) linear (V:|V) but conjugate-commute with antilinear (V:|V), which seems to depend on having the notion of linearity sewn up already. And they're only present in complex contexts.

The identity on V is manifestly (:repeat(1):) hence real linear and, indeed, a real scaling. If V subsumes U, we can read U as the identity on U, which is a monic mapping (V::V), and ask whether it is real linear: since the identity is (: u←u :U), we can reduce U relates u to x and v to y implies U relates x+y to u+v to u in U and v in U implies u+v in U, which may be read as U is closed under addition. So a collection is real linear iff it is closed under addition.

For any real linear (V:r:V) we have r relates x to u and y to v implies r relates x+y to u+v whence, in particular, any sum of r's left values is a left value of r, likewise for right values; i.e. (:v←v:r) and (r:v←v:) are real linear whenever (V:r:V) is.

So now, given real linear (V:r:V), consider (|r:); it relates u to v iff ({u}&on;r: x←x :) = ({v}&on;r: x←x :) is non-empty; is (|r:) also real linear ? Its collection of values is just (:v←v:r), which we know to be real linear. If (|r:) relates u to v, we have non-empty ({v}&on;r: x←x :) = ({u,v}&on;r: x←x :) = ({u}&on;r: x←x :). If r relates w to z and x is in ({v,u}&on;r: x←x :), then r relates w+v to z+x and w+u to z+x. If r relates w+v to g, not necessarily given to be a sum of members of V (e.g. the collection of positive integers is an additive domain in which 1 is not a sum), can I show that it must also relate w+u to g ? RTP: ({v+w}&on;r: x←x :) = ({u+w}&on;r: x←x :) technical hitch: are there a in ({u+w}&on;r: x←x :) which are not s+t for any s, t ? will it matter if there are ?

In so far as we have an addition on V, we induce one on {(V::A)} for any A via: (V:f:A)+(V:g:A) relates z to a iff: f relates u to a, g relates v to a and u+v = z or one of f, g relates z to a, but a is not a right value of the other. This works regardless of whether we have an addition on A; it is symmetric, associative and cancellable because the addition on V has these properties, so {(V::A)} is a linear context.

Given a linear context, V, describe U as closed under composition iff u, v in U implies u&on;v in U (which we can construe as U respects composition).

Linearity

I'll entertain the possibility that context may wish to deal with some further (V:f|V), which respect addition, as scalings, though I'll shortly impose some restrictions on how far context can push this. The first restriction is simply that context must provide a self-inverse conjugation on scalings, ({scalings}: z* ← z |{scalings}), whose restriction to real scalings is the identity and for which: for any scaling z, (z* + z) and z*&on;z are real scalings. In the absence of non-real scalings, one may use the identity as conjugation.

Describe a relation (V:g|U) which respects addition as: linear precisely if it commutes with all scalings; and as antilinear precisely if, for every scaling z, z*&on;g = g&on;z. If all scalings are real, all relations which respect addition are both linear and antilinear.

The scalings of a linear context form a linear context [I still need to verify that its addition is cancellable] isomorphic to its own {scalings}; I need to introduce a representation of the isomorphism classes of the relevant kind of linear context, via which to abstract scalars with {scalars for V} being a canonical representation of the isomorphism class of which {scalings of V} is a member, thereby establishing a natural (multiplicative) action of scalars on linear contexts, which I need to show will ensure {scalars for V} subsumes {scalars for U} whenever there is some linear (V:|U) of which at least one output is not an additive identity.

I'll describe {linear ({scalars for V}: |V)} as the dual of any given linear context, V, and write it dual(V).

Since several flavours of multiplication are going to come along, I'll distinguish three standard kinds by the binary operator I use to represent them:

.: e.g. n.v
where (at least) one of the values combined (n or v) is a scalar: this is the most mundane (and fundamental) variety. Formally, given the way I've defined scalars, when n is a scalar for V with v in V, n.v is synonymous with n(v).
·: e.g. f·v
where the binary operator is contraction (which I'd better explain later): this generalises the notion of an inner product. Each operand is in a linear context and there is a linear action of one on the other; e.g. when f is a linear map (U:|V) and v is in V, f·v is synonymous with f(v), but were v a linear (V:|W), f·v would be synonymous with f&on;v = (U: f(v(w)) ←w |W).
×: e.g. u×v
denoting the tensor product (which I'd better explain later): this generalises the notion of an outer product. Again, each operand is in a linear context, but there is no contraction involved.

Note that the first – and only the first – of these is always symmetric: n.v = v.n. When one operand is a scalar, all three are synonyms: and I'll usually use the first. The last is seldom symmetric unless synonymous with the first (the exception is when the operands are members of the same one-dimensional space). The second only gets to be symmetric by virtue of special circumstances: when an operand is a scalar; when one operand's linear context is the collection of symmetric linear maps from the other's linear context to this last's dual; there may be further cases, but I trust you get the idea - don't presume symmetry of multiplication except when I write it in the first form.

Note that, despite my notation generally borrowing from programming languages, I use * to denote generally any binary operator, not necessarily presumed to be in any sense multiplicative.

Ramifications

For each v in any linear context V, define 1.v = v and, for every positive integer, n, (1+n).v = v+(n.v). Infer, in particular, that: for every positive integers n, m; (n+m).v = (n.v)+(m.v).

Notice that the identity (V:v←v|V) on V is exactly the relation I use to encode the collection V itself: this saves me having to have a mapping, Identity, which maps each linear context to its identity linear map. The identity is trivially linear in every linear context: indeed, it is trivially a scaling.

Given addition on U, for any A whatsoever, U's addition delivers an addition on {(V:|A)} defined by (V:f|A)+(V:g|A) = (V: f(a)+g(a) ←a |A); for any scaling n of V, there is an associated scaling (: n&on;f ←f :) of {(V:|A)}.

In particular, using g=V (which we already know to be linear) and f=n.V for any positive integer n, we find that (1+n).V is linear whenever n.V is; hence, inductively, for all positive integer n. Further, 1.V = V = (V: v=1.v ←v |V) and, whenever n.V = (V: n.v ←v |V),

so, inductively, n.V means what we might have hoped it would mean, namely (V: n.v ←v |V), for every positive natural n.

Given scalings (V:n|V) and (V:m|V), consider any linear (V:f|V); we're given that it commutes with n and m, so f&on;(n+m) = (V: f(n(v)+m(v)) ←v :V) = f&on;n + f&on;m = n&on;f + m&on;f = (V: n(f(v)) + m(f(v)) ←v :V) = (n+m)&on;f, so n+m commutes with f; likewise n&on;m&on;f = n&on;f&on;m = f&on;n&on;m so n&on;m commutes with f; this being for arbitrary f, we can infer that n+m and n&on;m are scalings. In particular, V being a scaling, n.V is a scaling for every positive integer n.

Given linear (V|f|U), its reverse is (U| u ← f(u) |V) which is trivially linear (though it may be a linear relation rather than a linear mapping - you'll notice the definitions above don't mention mappings). If f is a mapping, f&on;(:u←f(u):) is the identity on V; if its reverse is a mapping (a.k.a. f is monic) then (:u←f(u):)&on;f is the identity on V.

If (V:f|W) is linear and we're given (U:g|V) with g&on;f linear, can we infer that g is linear ? Only on (|f:), but for x, y in (|f:) we have u, v with x=f(u), y=f(v) and can apply g(x+y) = g(f(u)+f(v)) = g(f(u+v)) = g(f(u)) + g(f(v)) = g(x) + g(y); so g is linear if f is epic (i.e. (:f|) = V). Likewise, if g and g&on;f were linear, could we infer that f is linear ? No, but g must be unable to distinguish f(u+v) and f(u)+f(v), so f is linear if g is monic.