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

2 n

.

Table of contents

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

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.

Symplectic parabolic subgroup

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.

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

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}

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} .

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 ( $q^{m^{2}} \prod_{i = 1}^{m} (q^{2 i} - 1)$ )

Counting successive symplectic pairs gives

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

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

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

Tangent space of the symplectic group

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

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

and its generalization the indefinite orthogonal group has:

O^{T} SO = I

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

Y^{T} A Y = I

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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