We now turn to the opposite extreme from inner products: forms for which every vector is orthogonal to itself.
Definition 9.7.1 (Symplectic form). A symplectic form on a vector space over a field is a nondegenerate alternating bilinear form on . When , alternating forms are the same as skew-symmetric ones, so a symplectic form is just a nondegenerate skew-symmetric bilinear form.
Example 9.7.2 (The standard symplectic form). The form on (Example 9.1.4) is symplectic. More generally, on with coordinates , the standard symplectic form is Its matrix is block diagonal, It is alternating (the terms of cancel in pairs) and nondegenerate ().
Note. When , there is a quick reason why symplectic forms exist only in even dimension: the matrix of a skew-symmetric form on a space of dimension satisfies , so if is odd then and . The next theorem proves much more, in every characteristic.
Theorem 9.7.3 (Symplectic bases). Let be a symplectic form on a finite-dimensional vector space over any field . Then is even, and has a basis such that for all . In such a symplectic basis, the matrix of is .
Proof. Induction on . If , take and the empty basis. Otherwise pick any nonzero . Since is nondegenerate, the functional is nonzero, so there is with ; set , so that , and because alternating forms are skew-symmetric. The vectors are linearly independent: if , then .
Let . Then : if satisfies , then and . Since is nondegenerate, (Lemma 9.1.30), so
The restriction of to is alternating, and it is nondegenerate: if satisfies for all , then, since also for all , vanishes on , so . By induction, is even, and has a basis with the required relations among themselves. The remaining relations for , , hold because and is skew-symmetric. So is a symplectic basis, and . ∎
Corollary 9.7.4 (Uniqueness of symplectic forms). Any two symplectic forms on vector spaces of the same finite dimension are equivalent. In matrix terms, if , every invertible skew-symmetric matrix satisfies for some invertible .
Proof. Map a symplectic basis of one space to a symplectic basis of the other. For the matrix statement, apply the theorem to on and use Proposition 9.1.6, with the matrix whose columns are the symplectic basis. ∎
Note. Compare with symmetric forms: over they are classified by the signature, and over the classification involves number theory. Nondegenerate alternating forms, by contrast, have a single normal form over every field, determined by the dimension alone.
Definition 9.7.5 (Symplectic group). Let be a symplectic form on with . The symplectic group of is The symplectic group is .
Proposition 9.7.6 (The symplectic group). is a subgroup of , and any linear with for all is automatically invertible. Choosing a symplectic basis identifies with .
Proof. If , then for all , so ; thus is injective, hence invertible. The identity preserves ; composites of maps preserving preserve ; and . Finally, if is the matrix of in a symplectic basis and are coordinate vectors, then , and this equals for all iff . So , which is an isomorphism , carries onto . ∎
Example 9.7.7 (). For , a direct computation gives So iff . Geometrically, over , a linear map of the plane preserves the signed-area form iff it preserves signed area. In every dimension, taking determinants in gives for .
A complex vector space can be regarded as a real vector space of twice the dimension. Under this change of viewpoint, a Hermitian form splits into a symmetric form and an alternating form, and the unitary group appears as an intersection of real groups.
Let be a finite-dimensional complex vector space, and let denote regarded as a real vector space. If is a basis of over , then is a basis of over (since for real ), so . Let be multiplication by ; it is -linear with . An -linear map is -linear iff , since .
Proposition 9.7.8 (Decomposition of Hermitian forms). Let be a Hermitian form on . Define by so that . Then:
- is a symmetric bilinear form on .
- is an alternating bilinear form on .
- , , , and for all .
- is nondegenerate iff is nondegenerate iff is nondegenerate.
- If is a Hermitian inner product, then is an inner product on and is a symplectic form on .
Proof. (i) is additive in each variable, and for real ; since is -linear, is -bilinear. Symmetry: .
(ii) Similarly is -bilinear, and since is real.
(iii) , which gives the first two identities. Next, , so . Finally , so .
(iv) Replacing by in gives . Suppose is nondegenerate. If , then for all , so and . If , then for all , so again and . Conversely, if is nondegenerate and , then , so ; the same argument works for .
(v) for , so is an inner product, and is alternating by (ii) and nondegenerate by (iv). ∎
Proposition 9.7.9 (Two out of three). Let be a nondegenerate Hermitian form on , with and as above, and let be -linear. Consider the conditions:
- preserves : for all ;
- preserves ;
- is -linear: .
Any two of them imply the third. Moreover, preserves iff (a) and (b) hold.
Proof. (a) and (c) imply (b): .
(b) and (c) imply (a): .
(a) and (b) imply (c): first, is injective (if , then for all , so since is nondegenerate), hence surjective. For all , So for every , and by nondegeneracy of .
The last statement holds because with and real-valued. ∎
Corollary 9.7.10 (The unitary group as an intersection). Identify with by , where . For the standard Hermitian inner product , the form is the standard inner product of and is the standard symplectic form of Example 9.7.2. Regarding as the subgroup of of matrices commuting with ,
Proof. With and , so is the dot product on and is the standard symplectic form. Thus and are the groups of real-linear maps preserving and , respectively. The group consists of the -linear maps preserving , i.e., those satisfying (a), (b), and (c) of Proposition 9.7.9, and by that proposition any two of the three conditions suffice. ∎
Exercise 9.7.11 (Bilinear and symplectic forms).
- Find the signature of the quadratic form on , by constructing an orthogonal basis as in Theorem 9.6.6.
- Let be a symplectic form on a vector space of dimension . A subspace is isotropic if for all , i.e., . Show that every isotropic subspace has dimension at most , and that isotropic subspaces of dimension exist.
Show solution
(1) The symmetric bilinear form is Take , with . Then , so . In this plane take , with , and , with . They are orthogonal: So in the orthogonal basis the matrix of is , and the signature is . (Equivalently, the matrix has eigenvalues .)
(2) By Lemma 9.1.30, . If , then , so . In a symplectic basis , the subspace is isotropic, since for all , and has dimension .