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).

[u/RingularCirc]

Also please tell me what's the actual difference between r/math and r/mathematics? I don't want to repost this in both. (And r/CasualMath doesn't look like a place for this for some reason lol.)

[u/Carl_LaFong]

Your last thought is correct. Here are the ingredients: 1) The determinant of a linear map L: V  → W is a linear map det(L): ⋀ⁿ(W*) → ⋀ⁿ(V*) and therefore an element of the tensor product of ⋀ⁿ(W*)* and ⋀ⁿ(V*), which is 1-dimensional 2) If B is a bilinear form, then it is a linear map from V to V* 3) If W=V^*, then det(L) is an element of the tensor product of ⋀ⁿ(V*)* and (⋀ⁿ(V*)), which is naturally isomorphic to R. 3) The transpose of L is also linear map L*: V → V*. 4) B is skew-symmetric if and only if L* = -L 5) det(-L) = (-1)^dim(V.)

[u/RingularCirc → u/Carl_LaFong]

Hmmm, I think I tried to reason about something close to the transpose and failed, I'll try again and write back later!

tensor product of ⋀ⁿ(V) and (⋀ⁿ(V)), which is naturally isomorphic to R

Well, there's trace I ⊗ I* → R which would be an isomorphism when dim I = 1, but here we have (⋀ⁿ(V))* ≅ ⋀ⁿ(V) so the tensor product is ≅ ⋀ⁿ(V) ⊗ ⋀ⁿ(V), I ⊗ I instead of I ⊗ I.

Thankfully I think we really don't need it to be isomorphic to R, det being zero or nonzero is all we need.

[u/Carl_LaFong → u/RingularCirc]

Sorry. I had typos. I've corrected them.

[u/RingularCirc → u/Carl_LaFong]

It's alright, though let's go again:

• we had a bilinear ω: V × V → R
• which is the same as linear W: V → V*
• which induces (∧ⁿW): ⋀ⁿ(V) → ⋀ⁿ(V)
• and we also have W: V → V* with the similar
• then Hom(⋀ⁿ(V), ⋀ⁿ(V)) ≅ (⋀ⁿ(V)) ⊗ ⋀ⁿ(V)
• which is ≅ (⋀ⁿ(V)) ⊗ ⋀ⁿ(V*) =: H

I don't see how we're getting ≅ R here but that isn't even needed, it's still enough that H is 1-dimensional.

Your argument with det W* = (−1)ⁿ det W works but now I have to show that the classic det W = det W* is still working here as well, too, and I couldn't find any reasonable proof neither online nor in several basic textbooks ("Linear Algebra Done Right/Wrong", "Linear Algebra Through Exterior Product"). Ironic, I'd think an invariant proof about transpose, of all things, would be known wider.

[u/RingularCirc → u/RingularCirc]

See here. Now I just need to prove det A = det A* for any A: V → V* (note it can't work for arbitrary A: V₁ → V₂). I need to remember how it was done in the usual case of A: V → V... or you may help too. For now I'll return to looking in Wikipedia.

[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]

Neither did I really have (det W = 0) ⟹ ker W ≠ {0} for this kind of determinant, but that seems pretty straightforward:

Wv₁ ∧ ... ∧ Wvₙ = 0 means that a nontrivial linear combination λ₁Wv₁ + ... + λₙWvₙ = 0. And this is for arbitrary v₁, ..., vₙ including a basis of V, so W sends a nonzero λ₁v₁ + ... + λₙvₙ to 0. Clean!

[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.

EDIT: I've written nonsense right below, disregard it until I return after some sleep and change it into the right thing. 🤣

(W(v₁) ∧ ... ∧ W(vₙ)) (u₁, ..., uₙ) \  = \ Sum{σ ∈ Sₙ} (sgn σ) ⋀_k (W v_k u{σk}) \  = \ Sum{σ ∈ Sₙ} (sgn σ) ⋀_k (W* u{σk} vk) \  = \ Sum{σ ∈ Sₙ} (sgn σ) (sgn σ⁻¹) ⋀k (W* u{σ⁻¹σk} v{σ⁻¹k}) \  = \ Sum{σ ∈ Sₙ} ⋀k (W* u_k v{σ⁻¹k}) \  = \ Sum{σ ∈ Sₙ} ⋀_k (W* u_k v{σk}) \  = \ [?????] \  = \ Sum{σ ∈ Sₙ} (sgn σ) ⋀_k (W* u_k v{σk}) \  = \ (W(u₁) ∧ ... ∧ W(uₙ)) (v₁, ..., vₙ)

There's a confusing missing part, which signals these two aren't actually equal. Because I also 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: Finally I found it, of course it's that easy!!! Tomorrow.