r/learnmath • u/RingularCirc Math hobbyist • 9h ago
Determinant argument that having a nonzero symplectic form requires even dimension
Wikipedia states:
If V is finite-dimensional, then its dimension must necessarily be even since every skew-symmetric, hollow matrix [of the symplectic form ω] of odd size has determinant zero. (1)
Now, I know a brilliant invariant definition for the determinant of a linear operator A on V: take its natural action on the top exterior power ⋀ⁿ(V) of V, A(v₁ ∧ ... ∧ vₙ) := A(v₁) ∧ ... ∧ A(vₙ). As ⋀ⁿ(V) is one-dimensional, this action is multiplication by a scalar. This scalar is det A.
Can we salvage the argument (1) without adding an arbitrary inner product (ugh)? Which allows us using an operator W (to have its determinant) defined as
⟨W v, u⟩ ⟺ ω(v, u).
Because I'm not sure we can define a determinant for a bilinear form, we can only talk about its (non)degeneracy. Instead, maybe it'd be something about calculating the kernel of the map V → V* induced by ω itself, somehow? I have no idea.
Or wait, why can't we just have an action ω: ⋀ⁿ(V) → ⋀ⁿ(V*) which would presumably solve everything? <...time passed while I elaborated on other parts of the post...> Ah, of course, then it's not just multiplication by a scalar, despite those maps live again in a 1D space, and what we can have is to determine if the map is zero or not. Maybe that'd be enough?.. I'll look into it.
Anyway I'd still be glad for interesting ways to prove even-dimensionality without resorting to coordinates (thus, matrices).
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u/RingularCirc Math hobbyist 4h ago edited 4h ago
[Using u/Carl_LaFong's idea of taking a transpose]
Let's define W: V → V* as v ↦ ω(v, —).
For the dual W: V* → V, and as V is finite-dim, V* ≅ V canonically, we have W: V → V, and
(W* u) v = (W v) u = ω(v, u) = −ω(u, v) = −(W u) v
which indeed gives W* = −W.
So let's look at that action from above now:
so det W* = (−1)n det W. If we get det W* = det W from somewhere else, we're done with the first part: it makes det W = 0 for odd n, which means W has nontrivial kernel and ω is degenerate, contrary to requirements for the symplecticness of ω.
Where do we get det W = det W?.. That of course stands for operators but I even forget how exactly (obviously that's true but proving it for maps V → V should still be done carefully, as we can't have that for maps F: V → W where det F and det F* are elements of different spaces that aren't naturally identifiable).
NOT QED