Symplectic forms and symplectic groups: Part 1

We follow Chapter 2 of a paper by Garton (and perhaps some other sources, for example these notes written by Elkies), but I will also write bunch more gibberish so that I can learn better. Please correct me if I have made mistakes!

Let $R$ be a commutative ring and fix $n \in \mathbb{Z}_{\geq 1}.$ A bilinear from $$\langle \cdot, \cdot \rangle : R^{n} \times R^{n} \rightarrow R$$ is

• symmetric if $\langle v, w \rangle = \langle w, v \rangle$ for all $v, w \in R^{n};$
• alternating if $\langle v, v \rangle = 0$ for all $v \in R^{n};$ 
• nondegenerate if two maps $R^{n} \rightarrow (R^{n})^{\vee}$ given by $v \mapsto \langle v, - \rangle$ and $w \mapsto \langle -, w \rangle$ are injective;
• symplectic if it is alternating and nondegenerate. 

Remark. If the bilinear form is alternating, then $$0 = \langle v + w, v + w \rangle = \langle v, w \rangle + \langle w, v\rangle,$$ so $$\langle w, v \rangle = -\langle v, w\rangle.$$ This condition is generally weaker than the alternating condition, unless (the image of) $2$ is invertible in $R.$ Then we have $$\langle v, v \rangle = \frac{1}{2}(\langle v, v \rangle + \langle v, v \rangle),$$ so the above condition with $w = v$ gives the alternating condition back.

If the given bilinear form is symmetric or alternating, we only need to check one of the two maps $R^{n} \rightarrow (R^{n})^{\vee}$ is injective when it comes to checking nondegeneracy.

Matrix representation. Abstract definition of bilinear forms may sound sleek, but to show certain forms exist, it is helpful to use matrix representation of the forms. In particular, we want to see whether we can construct a symplectic bilinear form on $R^{n}.$ Denote by $\mathrm{Bi}(R^{n})$ the set of all bilinear forms $R^{n} \times R^{n} \rightarrow R$ and $\mathrm{Mat}_{n}(R)$ the set of all $n \times n$ matrices over $R.$ Then we have a bijection $$\mathrm{Bi}(R^{n}) \leftrightarrow \mathrm{Mat}_{n}(R)$$ given by $$\langle \cdot, \cdot \rangle \mapsto [\langle e_{i}, e_{j} \rangle]_{1 \leq i, j \leq n},$$ and whose inverse is given by $$A \mapsto ((v,w) \mapsto v^{T}Aw).$$ This bijection is not additive but it preserves the multiplication by $R.$

Remark. Here, the correspondence depends the standard basis $e_{1}, \dots, e_{n}$ for $R^{n}.$ If $B : R^{n} \rightarrow R^{n}$ induces an $R$-linearly free elemenets $Be_{1}, \dots, Be_{n},$ then we cannot use the same matrix for bilinear forms on $R^{n}.$ If the orignal matrix is $A,$ the new matrix is $B^{T}AB.$ If it happens that $\det(B) \in R^{\times},$ which is always true when $R$ is a field, then this observation shows that we get a self-bijection $\mathrm{Bi}(R^{n}) \leftrightarrow \mathrm{Bi}(R^{n})$ given by $$\langle \cdot, \cdot \rangle \mapsto \langle B(\cdot), B(\cdot) \rangle.$$ Since $(B^{T}AB)^{T} = B^{T}A^{T}B$ and $\det(B^{T}AB) = \det(B)^{2} \det(A),$ we see that

• $\langle \cdot, \cdot \rangle$ is symmetric if and only if $\langle B(\cdot), B(\cdot) \rangle$ is symmetric;
• $\langle \cdot, \cdot \rangle$ is alternating if and only if $\langle B(\cdot), B(\cdot) \rangle$ is alternating;
• $\langle \cdot, \cdot \rangle$ is nondegenerate if and only if $\langle B(\cdot), B(\cdot) \rangle$ is nondegenerate,

when $\det(B) \in R^{\times},$ or equivalently $B \in \mathrm{GL}_{n}(R).$

Lemma. Let $A$ be the matrix corresponding to a bilnear form $$\langle \cdot, \cdot \rangle : R^{n} \times R^{n} \rightarrow R.$$ The bilinear form is

• symmetric if and only if $A^{T} = A;$
• alternating if and only if $A^{T} = -A.$

Proof. We assume $A^{T} = \epsilon A$ for some $\epsilon \in R.$ We have $$\begin{align*} \langle w, v \rangle &= w^{T}Av \\ &= (v^{T}A^{T}w)^{T} \\ &= v^{T}A^{T}w \\ &= \epsilon v^{T}Aw \\ &= \epsilon \langle v, w \rangle. \end{align*}$$ This proves the two statements. $\Box$


Lemma. Let $A$ be the matrix corresponding to a bilnear form $$\langle \cdot, \cdot \rangle : R^{n} \times R^{n} \rightarrow R.$$ The following are equivalent:

• $v \mapsto \langle v, -\rangle$ is injective;
• $\ker(A^{T}) = 0.$

Moreover, the following are equivalent as well:

• $w \mapsto \langle -, w\rangle$ is injective;
• $\ker(A) = 0.$

In particular, if $\det(A) \in R^{\times}$ (i.e., $A$ is invertible in $\mathrm{Mat}_{n}(R)$ by Cramer's rule), then the form is nondegnerate.

Proof. Saying that $v \mapsto \langle v, -\rangle$ is injective means that if $v^{T}Ae_{1} = \cdots = v^{T}Ae_{n} = 0,$ then $v = 0.$ Equivalently, this means that if $v^{T}A = 0,$ then $v = 0,$ or better yet, if $A^{T}v = 0,$ then $v = 0.$ This is equivalent to saying that $\ker(A^{T}) = 0.$

Saying that $w \mapsto \langle -, w\rangle$ is injective means that if $e_{1}^{T}Aw = \cdots = e_{n}^{T}Aw = 0,$ then $w = 0.$ Equivalently, this means that if $Aw = 0,$ then $w = 0.$ This is equivalent to saying that $\ker(A) = 0,$ and we are done $\Box$

Symplectic forms force parity in $n$ over a field. Let $k$ be a field with characteristic $\neq 2.$ For any $n \in \mathbb{Z}_{\geq 1},$ the following are equivalent:

• there is a symplectic bilinear form on $k^{n};$
• $n$ is even. 

Proof. We prove the converse first. Assume that $n$ is even so that we can write $n = 2g$ for some $g \geq 1.$ Then consider the matrix $$A = \begin{bmatrix} 0 & I_{g} \\ -I_{g} & 0 \end{bmatrix},$$ where $I_{g} \in \mathrm{Mat}_{g}(R)$ is the identity matrix. We have $A^{T} = -A$ and $$\det(A) = (-1)^{g(g+1)/2} \in R^{\times},$$ so the bilinear from on $k^{n}$ corresponding to $A$ is symplectic. Note that this direction works for any commutative ring in place of $k.$

Now, assume that there is a symplectic bilinear from $\langle \cdot, \cdot \rangle : k^{n} \times k^{n} \rightarrow k.$ Denote by $A = (a_{i,j})_{1 \leq i, j \leq n}$ the corresponding matrix. Since $A^{T} = -A,$ we have $a_{i,i} = -a_{i,i},$ which implies that $a_{i,i} = 0$ because $\mathrm{char}(k) \neq 2.$ Since $A$ must be invertible, observation already says $n \geq 2.$ We have so far only used that $2$ is invertible in $k,$ and we did not use the assumption that $k$ is a field.

Fix any $v_{1} \neq 0$ in $k^{n}.$ Since the form is nondegenerte, there must be some $w_{1} \neq 0$ in $k^{n}$ such that $\langle v_{1}, w_{1} \rangle \neq 0.$ We are over a field, so we may renormalize so that $\langle v_{1}, w_{1} \rangle = 1.$ Being over a field also implies that $v_{1}, w_{1}$ are linearly independent. Hence, we can construct an even dimensional subspace $H_{1} := kv_{1} + kw_{1} \subset k^{n}.$ Consider $$H_{1}^{\perp} := \{x \in k^{n} : \langle x, H_{1} \rangle = 0\}.$$ We claim that $k^{n} = H_{1} \oplus H_{1}^{\perp}.$

To see this, note that the bilinear form on $k^{n}$ restricts to $H_{1}$ with the matrix $$\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix},$$ with respect to the basis $v_{1}, w_{1}.$ We know this restirction gives a symplectic form on $H_{1}.$ In particular, it is non-degenerate, and this is enough to show that $H_{1} \cap H_{1}^{\perp} = 0,$ as follows: fix any $v \in H_{1} \cap H_{1}^{\perp}.$ Then $\langle v, v_{1} \rangle = \langle v, w_{1} \rangle = 0.$ Since the form restricted to $H_{1}$ is already nondegenerte, this ensures that $v = 0,$ as desired.

Since the form on $k^{n}$ is nondegenerate, we are given with the injection $k^{n} \hookrightarrow (k^{n})^{\vee}$ given by $v \mapsto \langle v, - \rangle.$ Since we have $\dim_{k}((k^{n})^{\vee}) = n = \dim_{k}(k^{n}),$ this injection must be an isomorphism, which uses the fact that $k$ is a field. Under this isomorphism, we have $$H_{1}^{\perp} \simeq \{f : V \rightarrow k : f(H_{1}) = 0\},$$ and the right-hand side has dimension $n - \dim_{k}(H_{1}) = n - 2.$ Since $H_{1} \cap H_{1}^{\perp} = 0,$ we have $$H_{1} + H_{1}^{\perp} = H_{1} \oplus H_{1}^{\perp} = k^{n},$$ due to the matching dimensions. 

Note that the form from $k^{n}$ restricts to $H_{1}^{\perp}$ as a symplectic form, so we may repeat our argument to $H_{1}^{\perp}$ and then to their subspaces to deduce that $n$ is even. $\Box$

Existence of symplectic forms for even $n$. If we are over a general commutative ring $R,$ the proof above at least ensures that there exists a specific symplectic form given by the matrix $$\begin{bmatrix} 0 & I_{g} \\ -I_{g} & 0 \end{bmatrix},$$ when $n = 2g$ is even.

Question. For general $R,$ does the existence of symplectic form on $R^{n}$ forces $n$ to be even?

Remark. Note that if $2 = 0$ in $R$ (e.g., $R = \mathbb{F}_{2}$), then the $n \times n$ identity matrix represents a symplectic form for every $n \geq 1,$ not just even ones. Thus, we can at least tell that the characteristic assumption on $R$ is crucial.

Equivalence among forms. We say two bilinear forms $\langle \cdot, \cdot, \rangle, \langle \cdot, \cdot, \rangle' \in \mathrm{Bi}(R^{n})$ are equivalent if there is $g \in \mathrm{GL}_{n}(R)$ such that $\langle v, w, \rangle' = \langle gv, gw \rangle$ for all $v, w \in R^{n}.$

It is conceptually easier to think about the action of $\mathrm{GL}_{n}(R)$ on $\mathrm{Bi}(R^{n})$ given by $$(gF)(v, w) := F(gv, gw),$$ where $g \in \mathrm{GL}_{n}(R)$ and $F \in \mathrm{Bi}(R^{n}).$ With this action, two bilinear forms are equivalent if and only if they are in the same orbit. The last proof also shows that when $R = k$ is a field with characteristic $\neq 2,$ all symplectic forms are equivalent, meaning they all lie in the same orbit. We might try to classify symplectic forms over other commutative rings next time.

In any case, we are able to define the notion we would like to think about. Fix the symplectic form $F \in \mathrm{Bi}(R^{2g})$ given by the matrix above. Then the symplectic group of genus $g$ is defined to be $$\mathrm{Sp}_{2g}(R) := \{g \in \mathrm{GL}_{n}(R) : g F = F\}.$$ The general symplectic group of genus $g$ is defined to be $$\mathrm{GSp}_{2g}(R) := \{g \in \mathrm{GL}_{n}(R) : (g F)/F \in R^{\times}\},$$ where the condition means that there is an element $m(g) \in R^{\times}$ such that $(gF)(v, w) = m(g) F(v, w)$ for all $v, w \in R^{2g}.$ We immediately note that $$\mathrm{Sp}_{2g}(R) \subset \mathrm{GSp}_{2g}(R).$$ The map $m : \mathrm{GSp}_{2g}(R) \rightarrow R^{\times}$ defines a group homomorphism.