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surjection A map φ : A → B which is sur- operation. So what is left over are the elements
jective. which are present in only one of the two sets, for
each of the two sets. See also set difference.
surjective Let A and B be two sets, with
A being the domain and B the codomain of a symmetric operator An operator T ,
function f . Then the function f is surjective if, densely defined on a Hilbert space, satisfying
for any y ∈ B, there is at least one element < T u, v> = <u, T v> for all u, v in the
x ∈ A with f(x) = y. Thus, the range of domain of T .
a surjective functions is equal to its codomain.
Surjective functions are also said to be onto B, symplectic manifold A pair (P, ω) where
and may be many-to-one. See also bijection and P is a manifold and ω is a closed non-degenerate
injection. 2-form on P . As a consequence P is of even
dimension 2n.
λ
symmetric difference Given two sets, A Canonical coordinates are coordinates (q ,
and B, their symmetric difference is defined as p ) such that the local expression of ω is of the
λ
λ
A ⊗ B = (A − B) ∪ (B − A). form ω = dp ∧ dq . Canonical coordinates
λ
Comment: To visualize this more clearly, always exist on a symplectic manifold (Darboux
think about the elements of the two difference theorem).
sets, A − B and B − A. An element d ∈ A − B
1
if and only if d ∈ A and d ∈ B. This must syntax For a symbol σ, a unique and precise
1
1
be true for all of the elements of A − B, by the definition of the form of the term and all terms
definition for set difference. Similarly, for an composing it.
element d ∈ B − A. So the intersection of the Comment: Syntax defines such things as
2
two difference sets, A − B and B − A must whether a term is a constant or a variable, how
be empty: any elements shared between A and many arguments it has, and the syntax of those
B would already be removed by the set difference arguments.

© 2003 by CRC Press LLC
© 2003 by CRC Press LLC