The Symplectic Sum

The Symplectic Sum: An Infuriating Construction

Never Let Anybody Chase Your Diagrams for You

The symplectic sum construction upsets me. It is canonical, but not blessed by the categorical dialectic.

Definition 1.  A symplectic vector space is a pair (V,\omega) where  V is a vector space and  \omega is a non-degenerate anti-symmetric bilinear form.

Example 2. Consider \mathbb{R}^n \oplus (\mathbb{R}^n)^\ast. Then,

 \omega((v,\ell),(v',\ell')) := \ell'(v)-\ell(v')

is symplectic.

Exercise 3. Show that a symplectic vector space must have even dimension.

Construction 4. Let (V_1,\omega_1),(V_2,\omega_2) be symplectic vector spaces. Then,

 \omega((v_1,v_2),(w_1,w_2)) := \omega_1(v_1,w_1) + \omega(w_2,v_2)

makes (V_1 \oplus V_2, \omega) into a symplectic vector space.

Proof. We show non-degeneracy. Fix (v_1,v_2) \in V_1 \oplus V_2. Suppose that

 \omega((v_1,v_2),(w_1,w_2)) := \omega_1(v_1,w_1) + \omega_2(v_2,w_2) =0

for all 0 \neq (v_2,w_2) \in V_1\otimes V_2. Take v_2 \neq 0 and  w_2 = 0. Then,  (v_2,w_2) \neq 0. But, we get

\omega_1(v_1,w_1) = -\omega_2(v_2,0) = 0

for all v_2 \in V_2. So, because \omega_1 is symplectic, we have v_1=0. Thus, \omega_2(v_2,w_2) = 0 for all \omega_2 \neq 0. By non-degeneracy of \omega_2 we have v_2=0 as well. This completes the proof.

Construction 4. is, in an intuitive sense, the canonical symplectic product form. Yet, (V_1 \oplus V_2, \omega_1 \oplus \omega_2)  is not the categorical product. 

Definition 5. (The Symplectic Category) Let (V_1,\omega_1), (V_2,\omega_2) be symplectic vector spaces. A linear map T:V_1 \to V_2 is symplectic if  \omega_2(Tv,Tw) = \omega_1(v,w). A symplectomorphism is an invertible symplectic map.

Exercise 6. Check that the composition of symplectic maps is symplectic and that the identity is symplectic. So, there is a category Symp of symplectic spaces and symplectic linear maps.

Exercise 7. Show that every symplectic vector space is symplectomorphic to the one given as Example 2.

Exercise 8. Show that symplectomorphisms preserve volume. (Hint: Use non-degeneracy to show that \omega^n is a volume form).

“Counter”-example 9. The projection  \pi : (V_1 \oplus V_2,\omega=\omega_1 \oplus \omega_2) \to (V_1,\omega_1) is not symplectic.

Proof. Compute  \omega_1(\pi v,\pi w) = \omega_1(v_1,w_1) \neq \omega(v,w) whenever \omega_2(v_2,w_2) \neq 0. This completes the proof.


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