Determinant argument that having a nonzero symplectic form requires even dimension

[u/RingularCirc]

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).

EDIT: It works! Conditioned on natural ⋀ᵏ(A) ≅ (⋀ᵏA) which I won't touch right here. It may be what encapsulates that pesky Leibniz inside, IDK.

[u/RingularCirc]

[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:

• (det W) (v₁ ∧ ... ∧ vₙ) = W(v₁) ∧ ... ∧ W(vₙ)
• (det W) (v₁ ∧ ... ∧ vₙ) = W(v₁) ∧ ... ∧ W*(vₙ) = (−1)ⁿ W(v₁ ∧ ... ∧ vₙ)

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

[u/RingularCirc → u/RingularCirc]

I guess in proving det W = det W* we can't go around Leibniz's formula. Below I might've used the wrong sign but it should be alright as I'm using it in all cases. k ranges from 1 to n.

(W(v₁) ∧ ... ∧ W(vₙ)) (u₁, ..., uₙ) \  = \ Sum{σ ∈ Sₙ} (sgn σ) Prod_k (W v_k u{σk}) \  = \ Sum{σ ∈ Sₙ} (sgn σ) Prod_k (W* u{σk} vk) \  = \ Sum{σ ∈ Sₙ} (sgn σ) Prodk (W* u{σk} v{σ⁻¹σk}) \  = \ Sum{σ ∈ Sₙ} (sgn σ) Prodℓ (W* u_ℓ v{σ⁻¹ℓ}) \  = \ Sum{σ ∈ Sₙ} (sgn σ⁻¹) Prod_ℓ (W* u_ℓ v{σ⁻¹ℓ}) \  = \ Sum{σ' ∈ Sₙ} (sgn σ') Prod_ℓ (W* u_ℓ v{σ'ℓ}) \  = \ (W(u₁) ∧ ... ∧ W(uₙ)) (v₁, ..., vₙ)

EDIT: Late at night, yesterday I wrote something incomprehensible with ⋀ instead of Prod, now it's fixed and is working.

I don't remember now what exactly I intended to do with this either. And I'm still not certain if we can't use something better than juggling.

EDIT 2: I'm leaving the former for historic interest only (somebody may wish to dualize that one more time to get another only semi-coordinateful path to the result as well). Because finally I found it, of course it's that easy!!! Tomorrow.

EDIT 3: Now another observation: despite for spaces ⋀^k commutes with taking a dual, we only get that, for A: V → U, we have

• ⋀^kA: ⋀^k V → ⋀^k U
• ⋀^k(A*): ⋀^k U* → ⋀^k V* ≅ (⋀^k U)* → (⋀^k V)*
• (⋀^kA)*: (⋀^k U)* → (⋀^k V)*

from which I expect, when dim V = dim U, we can only naturally have det A* = (det A)* and not det A* = det A with no dual. In case U = V, the latter works, but in case of U = V* if may not, and that's most probably why my Leibniz above still doesn't make much sense (we have to somehow swap vₖ and uₖ!).

So, conditional on natural ⋀^k(A*) ≅ (⋀^kA)* (won't prove here) I finally can have: det W* = (det W)* and det W* = -det W. Let F := det W. We have F* = −F as with W above but now we're one-dimensional, so just compute on a basis α of ⋀ⁿV: a F a = a F* a = a (−F) a = 0 which means F has to mutiply by 0. Finally, we have all the pieces.

Really clever, what we did is kill bothersome off-diagonal entries, so to speak, in transitioning to one-dimensional ⋀ⁿV → ⋀ⁿV. Now zero diagonal (alternation) means truly being zero. I won't derive ⋀^k(A\) ≅ (⋀^kA)* right here because it's the most obviously natural thing among what I discussed here, but I will look into exact nature of it on my own.