Restricting a symplectic form to a subspace #
A symplectic form restricts to a subspace on which it stays nondegenerate. A subspace L
complementary to its symplectic complement L^ω is automatically of that kind, as is L^ω, and
then ω is the product of the two restrictions under the linear equivalence supplied by the
splitting. A pair of vectors with nonzero symplectic pairing spans such a subspace.
Main declarations #
TauCeti.SymplecticForm.restrict: the restriction ofωto a subspace on which it stays nondegenerate, again a symplectic form.TauCeti.SymplecticForm.nondegenerate_restrict_orthogonal_of_isCompl: the splittingV = L ⊕ L^ωalready forcesωto restrict nondegenerately toL^ω(forLitself this is Mathlib'sLinearMap.BilinForm.nondegenerate_restrict_of_disjoint_orthogonal).TauCeti.SymplecticForm.disjoint_span_pair_orthogonal: a pair with nonzero symplectic pairing spans a subspace disjoint from its symplectic complement.TauCeti.SymplecticForm.isSymplectomorphism_prodEquivOfIsCompl: the equivalence associated toV = L ⊕ L^ωis a symplectomorphism from the product of the restricted forms toω.
The restriction of a symplectic form to a subspace on which it remains nondegenerate.
Nondegeneracy is genuinely a hypothesis: ω restricts to 0 on any isotropic subspace.
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The symplectic complement of a subspace complementary to it also carries a nondegenerate
restriction of ω: a vector of L^ω orthogonal to L^ω is orthogonal to L as well, hence to
L ⊔ L^ω = V, and nondegeneracy of ω makes it zero. Unlike
LinearMap.BilinForm.restrict_nondegenerate_iff_isCompl_orthogonal this needs no
finite-dimensionality.
Along the splitting V = L ⊕ L^ω, the symplectic form is the product of its restrictions to
the two summands: the cross terms vanish by the very definition of the symplectic complement.
A pair with nonzero symplectic pairing spans a symplectic plane: it is disjoint from its own
symplectic complement, so ω restricts to it nondegenerately.