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A linear map $A:(V,{\omega }_V)\to (W,{\omega }_W)$ is symplectic when it pulls back the target symplectic form to the source form. Nondegenerate bilinear form makes such a map injective. In equal finite dimensions it is an isomorphism. The automorphisms form the symplectic group and preserve the symplectic orientation on the ambient space and on every symplectic subspace.

A rank-$n$ vector subbundle of a rank-$2n$ symplectic vector bundle is Lagrangian when each fibre is a Lagrangian subspace. If it exists, a compatible almost complex structure supplies a complementary Lagrangian subbundle by applying $J$. Existence is a global condition: the area-oriented tangent bundle of the two-sphere has no real line subbundle, by triviality of real line bundles over the sphere and the Hairy ball theorem.

For $H={\sum }_j(q_j^2+p_j^2)/2$ and $E>0$, the oscillator flow is a free circle action on the sphere of radius $\sqrt{2E}$. Its quotient is Complex projective space. For two modes, the reduced form in $w=z_2/z_1$ is $iEdw\wedge d\bar{w}/(1+|w{|}^2{)}^2$, of total area $2\pi E$. Reduction at zero energy is singular instead of a smooth quotient of the asserted dimension.

For a planar cotangent bundle, rotation has moment map $\ell =x{p}_y-y{p}_x$. At $\ell \ne 0$ its level is regular and the circle action is free. Polar canonical coordinates give $\omega =dr\wedge d{p}_r+d\theta \wedge d\ell$. Fixing $\ell$ and quotienting $\theta$ leaves $T^{\ast }{\mathbb{R}}_{>0}$ with the displayed form. The unit-mass free Hamiltonian becomes $p_r^2/2+{\ell }^2/(2r^2)$.

At a regular free circle group level of a moment map, the inverse image of a Lagrangian submanifold of the reduced space is a Lagrangian submanifold of the original space. Its restricted symplectic form is the pullback of the reduced form and hence vanishes. Its dimension is one more than that of the reduced Lagrangian, exactly half the ambient dimension. It is a circle bundle over the original Lagrangian.

For a line $⟨e⟩$, complete $(e,f)$ to a symplectic basis with complementary symplectic space $W$. The unipotent radical has elements $u(b,t)$ sending $e$ to $e$, $w$ to $w-B(w,b)e$ and $f$ to $f+b+te$. Its multiplication is $(b,t)(b^{\prime },t^{\prime })=(b+b^{\prime },t+t^{\prime }+B(b,b^{\prime }))$, giving order $q^{2m-1}$. Its commutator is central with parameter $2B(b,b^{\prime })$. It is Heisenberg-type in odd characteristic and abelian in characteristic two. The full stabilizer also scales $e,f$ inversely and acts symplectically on $W$.

With a Lagrangian subspace first in a paired symplectic basis, its stabilizer consists of $\left(\begin{array}{cc}A& AS\\ 0& A^{-\mathsf{T}}\end{array}\right)$ with $S$ symmetric. The additive symmetric-matrix unipotent radical has order $q^{m(m+1)/2}$; the Levi subgroup is $G{L}_m(q)$, acting by $S\mapsto AS{A}^{\mathsf{T}}$. Its index in $S{p}_{2m}(q)$ is ${\prod }_{i=1}^m(q^i+1)$, the number of Lagrangian subspaces.

The scalar center has order $\gcd (2,q-1)$. The quotient is simple except at $(m,q)=(1,2),(1,3),(2,2)$. The last exception is $S{p}_4(2)\cong S_6$.

This map preserves the alternating form and has inverse $T_{v,-c}$. Transvections along a fixed line form an abelian normal subgroup of its stabilizer. They generate the symplectic group by aligning successive symplectic basis pairs.

For $q>3$, embedded perfect $S{L}_2(q)$ groups contain every symplectic transvection. For $q=3,m\ge 2$, four coefficient-one transvections on the lines of an isotropic plane multiply to the identity; their abelianized class is annihilated by 3 and 4. For $q=2,m\ge 3$, seven transvections on an isotropic three-space multiply to the identity; their class is annihilated by 2 and 7. Thus all nonexceptional cases are perfect.

For even n, the even-weight subspace of $F_2^n$ has radical $⟨1⟩$ for the dot product. Its quotient is a nondegenerate alternating bilinear form space of dimension n−2. The symmetric group acts by coordinate permutations. At n=6 this action is faithful and gives $S_6\cong S{p}_4(2)$.

The standard symplectic cylinder restricts one conjugate coordinate pair to a disk and leaves the remaining symplectic coordinates unrestricted. Its normalized symplectic capacity is $\pi R^2$, despite its infinite symplectic volume when $n>1$.

For $U^{T}JU=J$ with real orthogonal skew-symmetric $J$, the dual $\tilde{\rho }=J{\rho }^T{J}^{†}$ transforms as $\widetilde{U\rho U^{†}}=U\tilde{\rho }{U}^{†}$. Hence $I_J(\rho )$ is constant on the reachable unitary orbit of a density operator. Different values can obstruct density operator controllability even when the ordinary spectra agree. For a four-dimensional $J$ pairing indices $1,4$ and $2,3$, the two diagonal arrangements $(a,a,b,b)$ and $(a,b,b,a)$ have invariants $4ab$ and $2(a^2+b^2)$.

For a closed $2n$-dimensional symplectic manifold, a symplectic blowup of size $\epsilon =\pi R^2$ removes the symplectic volume of the radius-$R$ ball. Its exceptional divisor has lower dimension and contributes no ambient volume. The displayed volume loss shows that different size choices can give forms that are not related by a symplectomorphism.

For a symplectic blowup of a $2n$-dimensional symplectic manifold, the displayed formula gives its First Chern class. The exceptional divisor has normal bundle $\mathcal{O}(-1)$; the discrepancy coefficient is $n-1$. In dimension four it is one.

For arrays of values in $[0,1]$, let $U_N$ average a product of $d$ entries over all ordered index tuples from $N$ coordinates, and let $V_N$ average it over the tuples with distinct indices. Decompose the first average as $U_N=\alpha V_N+(1-\alpha )W_N$, where $\alpha =(N{)}_d/N^d$ and the repeated-index average $W_N$ lies in $[0,1]$. This gives $|U_N-V_N|\le 1-\alpha$. For uniformly sampled indices, a union bound over the $\left(\genfrac{}{}{0ex}{}{d}{2}\right)$ pairs bounds the collision probability by $\left(\genfrac{}{}{0ex}{}{d}{2}\right)/N$. This deterministic estimate is the bridge between empirical-product limits and the De Finetti theorem.

On the Dahlquist test equation, the symmetric two-step two-derivative formula has amplification polynomial $F_z(w)=(z^2-7z+15)w^2-16zw-(z^2+7z+15)$. For every real $z<0$, its leading coefficient is positive and $F_z(-1)=2z<0$, whereas $F_z(w)\to +\infty$ as $w\to -\infty$. The intermediate value theorem therefore gives a real root below minus one. Thus the method is not A-stable, without any appeal to a multistep order barrier.

On the Dahlquist test equation, write the amplification polynomial as $D{\xi }^2-16z\xi -A$, where $D=z^2-7z+15$ and $A=z^2+7z+15$. The identity $|A{|}^2-|D{|}^2=28\mathrm{Re}z(|z{|}^2+15)$ shows that both roots of a polynomial can lie inside the open unit disk only if $\mathrm{Re}z<0$. But the complex quadratic Schur criterion would then require $32|\mathrm{Re}z|(|z{|}^2+15)<28|\mathrm{Re}z|(|z{|}^2+15)$, which is impossible. Thus the strict linear stability domain is empty. This concerns robust damping of all modes, rather than convergence on fixed time intervals or the bounded-root stability set of the symmetric two-derivative formula.

For the symmetric two-step two-derivative formula, the closed-disk root condition for a multistep method fails off the imaginary axis. Positive real part makes the product of root moduli exceed one; negative real part fails the complex quadratic Schur criterion, as in the empty strict decay domain of the symmetric two-derivative formula proof. On $z=iy$, write $D=15-y^2-7iy=|D|e^{i\delta }$ and set $\xi =i{e}^{-i\delta }\eta$. The amplification polynomial becomes ${\eta }^2-16y\eta /|D|+1=0$. Its two roots are distinct and on the unit circle exactly when $|16y/|D||<2$, or $y^4-45y^2+225>0$. Thus the bounded-root set consists of the imaginary intervals $|y|<\sqrt{(45-15\sqrt{5})/2}$ and $|y|>\sqrt{(45+15\sqrt{5})/2}$, including $y=0$ and excluding the double-root endpoints. In particular, the set is unbounded. Unit-modulus oscillations are bounded but do not tend to zero.

A symmetric tensor invariant under rotations about a fixed nonzero vector $y$ has the form $\alpha I+\beta y{y}^T$. Choose the third coordinate axis parallel to $y$: rotation by a half-turn makes the mixed axial-transverse entries vanish, and all transverse rotations make the transverse block a scalar multiple of the identity. Its two transverse eigenvalues are $\alpha$ and its axial eigenvalue is $\alpha +\beta |y{|}^2$. The trace and the quadratic contraction $y^{T}Ty$ determine these two coefficients. Full isotropy holds exactly when $\beta =0$.