A bilinear form $\Omega$ on a real vector space $V$ is symplectic iff it is alternating ($\Omega(v,v) = 0$) and nondegenerate (its left radical is trivial).
- alt : Ω.IsAlt
- nondeg : LinearMap.SeparatingLeft Ω
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The symplectic orthogonal $E^\Omega = \{v \in V : \Omega(v, e) = 0 \text{ for all } e \in E\}$.
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A subspace $E$ is symplectic with respect to $\Omega$ iff $E \cap E^\Omega = \{0\}$ (Definition 4 in the textbook).
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A subspace $E$ is Lagrangian iff $E^\Omega = E$, equivalently isotropic and half-dimensional (Definition 5 in the textbook).
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An alternating bilinear form is skew-symmetric: $\Omega(x, y) = -\Omega(y, x)$.
For a symplectic form on a finite-dimensional space, the orthogonal of the whole space is trivial: $V^\Omega = \{0\}$ (equivalent to nondegeneracy).
Dimension formula for the symplectic orthogonal: $\dim E + \dim E^\Omega = \dim V$.
$E$ is a symplectic subspace iff $E$ and $E^\Omega$ are complementary: $V = E \oplus E^\Omega$.
A Lagrangian subspace has half the dimension of the ambient symplectic space: $2 \dim E = \dim V$.
The linear map $V \to E^*$ sending $v \mapsto \Omega(\cdot, v)|_E$, used to identify $V/E$ with the dual of a Lagrangian.
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The kernel of symplRestrict Ω E is exactly the symplectic orthogonal $E^\Omega$.
For a Lagrangian subspace $E$, the symplectic form induces a canonical isomorphism $V/E \simeq E^*$.
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Recursive construction of the $n$-fold wedge power $\Omega^{\wedge n}$ as an alternating $2n$-form, built by inductively wedging with the $2$-form $\Omega$.
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Wedge power $\Omega^{\wedge n}$ packaged as an alternating $m$-form: equals
wedgePowBilinFormAux when $m = 2n$ is even, and zero otherwise.
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The symplectic volume form $\frac{1}{n!}\Omega^{\wedge n}$ where $\dim V = 2n$ (Definition 6 in the textbook).