Symplectic group ( $Sp (2 n, R)$ )

The real symplectic group consists of matrices

A

satisfying

A^{T} J A = J

. It is an embedded matrix Lie group of dimension

2 n^{2} + n

in size

2 n

Symplectic group over a field ( $Sp (V, B)$ )

 0  0

For a finite-dimensional vector space over any field with a nondegenerate alternating bilinear form

B

, the symplectic group over a field consists of invertible linear maps preserving

B

. In a symplectic basis its matrices satisfy

g^{T} J g = J

. Matching two symplectic bases gives a unique such linear map, and a symplectic basis exists by successively splitting off paired vectors. This includes arbitrary-characteristic and infinite-field versions, beyond the real Lie group and the symplectic group over a finite field.

Symplectic matrix ( $A^{T} J A = J$ )

 2  0

An invertible real matrix preserving the standard symplectic form. With

J = (0 - I I 0)

, it satisfies

A^{T} J A = J

, equivalently

A J A^{T} = J

. The first identity describes preservation of the form; the second describes preservation of coordinate Poisson brackets.

Reciprocal spectrum of a symplectic matrix ( $χ_{M} (x) = x^{2 m} χ_{M} (x^{- 1})$ )

 0  0

For a symplectic matrix

M

of size

2 m

with determinant one, the defining identity

M^{T} J M = J

gives

M^{- 1} = J^{- 1} M^{T} J

. Since similarity and transposition preserve the characteristic polynomial,

χ_{M^{- 1}} = χ_{M}

. Factoring

x I - M = M (x M^{- 1} - I)

then proves the displayed reciprocal identity for

x \neq = 0

. If

χ_{M} (z) = (z - λ)^{d} R (z)

with

R (λ) \neq = 0

, this identity gives

χ_{M} (x) = x^{2 m - d} (1 - λ x)^{d} R (1/ x)

near

x = λ^{- 1}

. Thus reciprocal eigenvalues have the same algebraic multiplicity. The fixed points

\pm 1

require the stronger parity argument in even multiplicities at fixed points of reciprocal root pairing.

Symplectic parabolic subgroup

 0  0

The stabilizer of a totally isotropic

k

-space in a finite symplectic space also stabilizes its orthogonal complement. Its block decomposition has a unipotent normal subgroup

Q

and a block-diagonal complement

L ≅ GL_{k} (q) \times Sp_{2 m - 2 k} (q)

. The paired

k

-space carries the contragredient action. A row-vector convention gives upper-triangular blocks; a column-vector convention transposes the arrangement. The unipotent radical count for a symplectic parabolic subgroup yields its order.

Symplectic Lagrangian stabilizer ( $P_{m} = Sym_{m} (q) ⋊ G L_{m} (q)$ )

 0  0

With a Lagrangian subspace first in a paired symplectic basis, its stabilizer consists of

(A 0 A S A^{- T})

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 A S A^{T}

. Its index in

S p_{2 m} (q)

\prod_{i = 1}^{m} (q^{i} + 1)

, the number of Lagrangian subspaces.

Symplectic point stabilizer ( $P_{1} = U ⋊ (F_{q}^{\times} \times S p_{2 m - 2} (q))$ )

 0  0

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

e

w

w - B (w, b) e

and

f

f + b + t e

. Its multiplication is

(b, t) (b^{'}, t^{'}) = (b + b^{'}, t + t^{'} + B (b, b^{'}))

, giving order

q^{2 m - 1}

. Its commutator is central with parameter

2 B (b, b^{'})

. 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

Unipotent radical count for a symplectic parabolic subgroup ( $∣ Q ∣ = q^{2 k (m - k) + k (k + 1) /2}$ )

 0  0

In row-block order

(W^{'}, U, W)

with dual bases, a unit-diagonal symplectic flag stabilizer satisfies

E = J_{U} D^{T}

and

F - F^{T} = D J_{U} D^{T}

. The matrix

D

is arbitrary. The latter right side is alternating, so each off-diagonal pair of entries of

F

gives one free choice and each diagonal entry is free. Consequently

∣ Q ∣ = q^{2 k (m - k) + k (k + 1) /2} .

(1)

Reversed dual-basis order inserts a reversal matrix but does not change the count. The argument holds in characteristic two without division by two.

Finite symplectic group order ( $q^{m^{2}} \prod_{i = 1}^{m} (q^{2 i} - 1)$ )

 0  0

Counting successive symplectic pairs gives

∣ Sp_{2 m} (q) ∣ = q^{m^{2}} \prod_{i = 1}^{m} (q^{2 i} - 1) .

(1)

For each nonzero first vector there are

q^{2 m - 1}

partners with pairing one, after which its nondegenerate plane splits off. The exact defining-characteristic part is

q^{m^{2}}

, realized by an upper symplectic unitriangular subgroup.

Symplectic group as a regular level set

 0  0

The map

F : M_{2 n} (R) \to Skew_{2 n} (R)

given by

F (A) = A^{T} J A

has

Sp (2 n, R) = F^{- 1} (J)

. At a symplectic

A

D F_{A} (H) = H^{T} J A + A^{T} J H

(1)

is surjective, so the regular level set theorem makes the symplectic group an embedded submanifold.

Tangent space of the symplectic group

 0  0

At the identity, the symplectic Lie algebra consists of

X

satisfying

X^{T} J + J X = 0

; tangent spaces elsewhere are its left translates.

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Symplectic group by

Ciro Santilli  40 Updated 2025-07-16

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Intuition, please? Example? mathoverflow.net/questions/278641/intuition-for-symplectic-groups The key motivation seems to be related to Hamiltonian mechanics. The two arguments of the bilinear form correspond to each set of variables in Hamiltonian mechanics: the generalized positions and generalized momentums, which appear in the same number each.

Seems to be set of matrices that preserve a skew-symmetric bilinear form, which is comparable to the orthogonal group, which preserves a symmetric bilinear form. More precisely, the orthogonal group has:

O^{T} I O = I

(1)

and its generalization the indefinite orthogonal group has:

O^{T} SO = I

(2)

where S is symmetric. So for the symplectic group we have matrices Y such as:

Y^{T} A Y = I

(3)

where A is antisymmetric. This is explained at: www.ucl.ac.uk/~ucahad0/7302_handout_13.pdf They also explain there that unlike as in the analogous orthogonal group, that definition ends up excluding determinant -1 automatically.

Therefore, just like the special orthogonal group, the symplectic group is also a subgroup of the special linear group.

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