A symplectic form is alternating and nondegenerate
Objective: Does ω(v,v)=0 imply ω is nondegenerate?
A symplectic form measures oriented pairs of directions rather than lengths. Nondegeneracy means every nonzero vector has some partner with nonzero pairing. Before calculating, declare the manifold or vector space, dimension, coefficient field, symplectic-form sign, orientation, regularity, compactness, boundary, time interval, and whether the conclusion is linear, local, global, topological, variational, or only valid after choosing auxiliary data.
Read every object in words: On a real vector space V, a bilinear form ω is alternating when ω(v,v)=0 for every v, equivalently ω(u,v)=−ω(v,u). It is nondegenerate when the map ω♭:V→V*, v↦ω(v,·), has zero kernel. The exact statement used here is: A finite-dimensional real vector space carrying a nondegenerate alternating form has even dimension. The pair (V,ω) is then a symplectic vector space. Separate an algebraic identity from a theorem that needs closedness, nondegeneracy, compactness, transversality, regularity, convexity, monotonicity, or a compactness theorem.
The complete proof or honestly bounded proof route is: 1. Choose nonzero v and use nondegeneracy to find w with ω(v,w)≠0. 2. Normalize the pair and split off their two-dimensional span using the symplectic orthogonal. 3. Induct on the remaining nondegenerate complement, removing two dimensions each time. Record every sign, pullback, primitive, isotopy, boundary term, quotient, regular value, gauge, perturbation, orientation, compactification, and equality case instead of hiding it behind a theorem name.
Reconstruct the model “For ω((q,p),(Q,P))=qP−pQ on R², find ω♭(q,p) and test nondegeneracy.”. The checked result is ω♭(q,p)=(−p,q) in dual coordinates, and its kernel is {0}; hence ω is nondegenerate. The nearest failure boundary is: A skew-symmetric matrix in odd dimension has determinant zero, so an odd-dimensional alternating form cannot be nondegenerate. A phase portrait, finite orbit sample, truncated complex, numerical action, mesh, or small residual is evidence only for the recorded model and tolerance; it is not a proof of global integrability, persistence, compactness, transversality, non-squeezing, or a Floer-theoretic invariant.
A finite-dimensional real vector space carrying a nondegenerate alternating form has even dimension. The pair (V,ω) is then a symplectic vector space.
Choose nonzero v and use nondegeneracy to find w with ω(v,w)≠0.
Normalize the pair and split off their two-dimensional span using the symplectic orthogonal.
Induct on the remaining nondegenerate complement, removing two dimensions each time.
For ω((q,p),(Q,P))=qP−pQ on R², find ω♭(q,p) and test nondegeneracy.
- Fix (q,p) and read the functional of (Q,P) as −pQ+qP.
- Its coordinate row in the dual basis is (−p,q).
- This row vanishes only when q=p=0.
Result: ω♭(q,p)=(−p,q) in dual coordinates, and its kernel is {0}; hence ω is nondegenerate.