We'll use symplectic space for a finite-dimensional vector space V over a field F, equipped with a bilinear pairing (.,.) that is alternating and nondegenerate. Recall that ``alternating'' means that (v,v)=0 for all v in V, from which it follows that (v,w) = -(w,v) for all v,w in V; and ``nondegenerate'' means that the only vector v such that (v,w=0) for all v in V is the zero vector. [The same condition is usually required from the right, but we have seen that for a pairing on a single finite-dimensional space the two are equivalent, and here the equivalence follows directly from the identity (v,w) = -(w,v).]

We shall show: every symplectic space is even-dimensional, and all symplectic spaces of the same dimension are isomorphic.

It is clear at any rate that the dimension, if positive, must be at least 2, since the only alternating pairing on a 1-dimensional space is the zero pairing, which is degenerate.

We proceed to give some examples of symplectic spaces. The symplest nontrivial example of a symplectic space is the so-called ``hyperbolic plane'': H=F2, with pairing defined by ((a,b), (c,d)) = ad-bc. (This formula should look familiar!) We know already that the alternating pairings on a n-dimensional space constitute a vector space of dimension (n2-n)/2; thus any alternating pairing on F2 is a multiple of the hyperbolic one, and if it is a nonzero multiple r(ad-bc) then scaling one of the unit vectors by r identifies F2 with H as a symplectic space.

If V,W are symplectic spaces, so is their direct sum, with the pairing defined by ((v,w), (v',w')) = (v,v') + (w,w'). [Check that this is not only alternating but also nondegenerate.] Likewise an arbitrary finite direct sum of symplectic spaces has a natural symplectic structure. In particular, the direct sum of m copies of the hyperbolic plane H is a symplectic space Hm of dimension 2m. The pairing can be described as follows: the space has basis vectors e1,...em, f1,...fm; the pairing of any two of them is zero, except that (ei,fi)=1, (fi,ei)=-1 for each i.

We shall show that every symplectic space V of positive dimension is the direct sum of H with another symplectic space W, necessarily of dimension dim(V)-2. By induction on dim(V), it will follow that V has even dimension and is isomorphic with Hm for some positive integer m.

Fix a nonzero vector e1 in V. Since the pairing is nondegenerate, there exists f1 in V such that (e1,f1)=1. Moreover, e1 and f1 are linearly independent: if one was a multiple of the other, (e1,f1) would vanish because the pairing is alternating. Thus e1 and f1 span a hyperbolic plane in V. Call this plane H.

Now let W be the subspace of V consisting of all v such that (v,e1)=(v,f1)=0. [That is, W is what we would call the ``orthogonal complement'' of H if our pairing were an inner product.] We claim:

• W has dimension dim(V)-2;
• The intersection of W with H is {0};
• The pairing on V, restricted to W, is nondegenerate (so makes W a symplectic space too);
• V is the direct sum of W and H, not only as a vector space (which already follows from the first two claims) but also as a symplectic space.
The first claim is clear: any nondegenerate pairing identifies V with its dual; under this identification, W is just the annihilator of H; and we know that the dimensions of a space and its annihilator add up to dim(V). Since H has dimension 2, the claim follows.

The second claim is a consequence of the fact that the pairing on H is nondegenerate. Indeed a vector in both W and H is a vector v in H such that (v,w)=0 for all w in H, and by nondegeneracy the only such v is the zero vector. As already noted, it follows that V is the direct sum of W and H, at least as a vector space.

For the third claim, we must show that if v is a vector of W such that (v,w)=0 for all w in W then v=0. But we already know that (v,w)=0 for all w in H; thus by linearity (v,w)=0 for all w in W+H, which is V. But our pairing (.,.) is nondegenerate on V. Thus v=0 as claimed.

For the final claim, we need only check that for any vectors v,v' in H and w,w' in W we have (v+w,v'+w')= (v,v')+(w,w'). But this is clear, because the cross-terms (v,w') and (w,v') vanish by the definition of V'.

This completes the induction step and the proof.

A linear transformation T of V is said to be symplectic if it preserves the pairing, i.e. if (Tv,Tw)=(v,w) for all v,w in V. Necessarily T is invertible; such T constitute a group, called the symplectic group of V, and written Sp(V).

When F is finite, say of size q, our proof that all symplectic spaces of the same dimension are equivalent also lets us count the symplectic transformations, as we did for linear transformations of a vector space of given dimension over F. We find that there are qm2 times the product of (q2i-1) over i=1,...,m. How many nondegenerate bilinear forms are there on a (2m)-dimensional space over F?

Our results can be interpreted in terms of skew-symmetric matrices. Let Jm be the matrix for the pairing on Hm with respect to the basis e1,...em, f1,...fm. This is a block-diagonal matrix with m 2-by-2 blocks, each block being the matrix J1 with entries

 0 1 -1 0

which gives the matrix for the pairing on H. Of necessity, J1 and (thus) Jm are skew-symmetric.

Now let A be any skew-symmetric matrix. The skew-symmetry condition is equivalent to the requirement that A is the matrix of an alternating form. Our result then says: if A is of odd order, it must be degenerate; if A is of even order 2m, it is either degenerate or of the form SJmS* for some invertible matrix S (giving the change of basis to e1,...em, f1,...fm).

The odd case can be obtained also by considering the determinant: by assumption, det(A)=det(-A*)=det(-A); for a matrix of odd order, this last is -det(A). So, at least in characteristic zero, det(A)=0. We deduce the same result in arbitrary characteristic using a familiar argument: ``det(A)=0'' is a polynomial identity with integer coefficients, so having verified it over the complex numbers we know it is true in general.

Taking determinants in the even case, and noting that det(Jm)= det(J1)m= 1m=1, we find that the determinant of any skew-symmetric matrix of even order is of the form det(S)2. In particular, it is a perfect square. For instance, for order 2, it is the square of a12; for order 4, the square of a12 a34 - a13 a24 + a14 a23. In general, it turns out to be the square of a homogeneous polynomial of degree m in the coefficients of A, linear in each variable separately, called the Pfaffian of A. By now you know enough about permutations, signs and determinants to guess the formula for the Pfaffian of an skew-symmetric matrix of any even order, and then prove that its square equals the determinant of the matrix.

This also yields us to the matrix version of the symplectic group: it consists of those square matrices S of order 2m such that Jm = SJmS*.

An alternative definition of the Pfaffian uses the ``exterior algebra'' of a finite-dimensional vector space, which we have more or less introduced when discussing k-linear alternating forms: these are functionals on the k-th exterior power of V. So, for instance, the determinant of vectors v1, ..., vn in an n dimensional space V can be regarded as v1^v2^...^vn, an elements of the 1-dimensional n-th (a.k.a. top) exterior power of V. Well, a skew-symmetric matrix can be regarded as an element q of the alternating square of V; if V has even dimension 2m, the Pfaffian can be interpreted as (1/m!) q^q^...^q (m times) in the same top exterior power.