# Exterior powers and choice

Under the assumption that any vector space has a basis (so under the assumption of the axiom of choice), we can prove the following algebraic statements :

1) If $$\varphi:V\to W$$ is an injective linear map between vector spaces, then the exterior powers $$\Lambda^k \varphi : \Lambda^k V\to \Lambda^k W$$ are injective. (See Corollary 5.9 here)

2) If $$v_1,\cdots,v_k$$ are linearly independent vectors in a vector space $$V$$ then $$v_1\wedge \cdots\wedge v_k\neq 0$$. (See Theorem 7.1 here)

My question is (similar to this question about tensor products) :

Are those statements (1 and 2) still true without the help of the axiom of choice? Or do they imply a form of choice in some sense ?

• Jeremy Rickard gave an efficient way to deal with such questions in your previous question you link at (mathoverflow.net/questions/325037). What have you tried? does his approach fail here and why? – YCor May 20 '19 at 19:23
• I tried to apply the finite-dimensional reduction but since we are not dealing with the same type of map (in this case we are dealing with exterior powers, and in the tensor case, it was not about tensor powers), I didn't see how to transpose the arguments (at least it is not trivial to me how to do this, but I might be wrong), and it didn't lead to anything. – Phil-W May 20 '19 at 19:53
• I am pretty sure the statements (1) and (2) can be checked to be $\Pi_1$ sentences (in the Lévy hierarchy). The point is that they hold if and only if they hold in all transitive models of a sufficiently large fragment of ZF. By Schoenfield absoluteness, since they are provable in ZFC, they are provable in ZF. (The relevant form of Schoenfield absoluteness is (2) of this question: mathoverflow.net/questions/269682/…) – Gabe Goldberg May 21 '19 at 1:32
• Note that (2) is a particular case of (1) with $(V,W)$ replaced with $(F^k,V)$ and $F$ the underlying field. – YCor May 21 '19 at 4:50
• An element $x$ in the kernel of $\Lambda^k\varphi$ can be written in terms of finitely many $v\in V$. As well as the $\varphi(v)$, a proof that $\varphi(x)=0$ using the relations in the exterior power only uses finitely many other elements $w\in W$. Doesn't this reduce (1) to a question about finite dimensional spaces? – Jeremy Rickard May 21 '19 at 10:05

As YCor notes in comments, (2) is a special case of (1), so I'll only address (1).

Suppose $$\Lambda^k\varphi:\Lambda^kV\to\Lambda^kW$$ is not injective, and let $$x\neq0$$ be in the kernel.

Then $$x$$ can be written in the form $$x=\sum_{i=1}^mv_{i1}\wedge\dots\wedge v_{ik}.$$ Let $$V'$$ be the finite dimensional subspace of $$V$$ spanned by $$\{v_{ij}\mid1\leq i\leq m,1\leq j\leq k\}$$, and $$\varphi':V'\to W$$ the restriction of $$\varphi$$ to $$V'$$.

Let $$x'=\sum_{i=1}^mv_{i1}\wedge\dots\wedge v_{ik},$$ considered as an element of $$\Lambda^kV'$$. Then $$x'\neq0$$, since it is sent to $$x$$ by the map induced by the inclusion of $$V’$$ into $$V$$. Also $$\Lambda^k\varphi'(x')=\Lambda^k\varphi(x)=0,$$ so $$x'$$ is a nonzero element of the kernel of $$\Lambda^k\varphi'$$.

Hence we may as well assume that $$V$$ is finite dimensional.

The fact that $$\Lambda^k\varphi(x)=0$$ follows from a finite number of the relations defining the exterior power $$\Lambda^kW$$, involving only finitely many elements of $$W$$. If we replace $$W$$ by the finite dimensional subspace $$W''$$ spanned by the image of $$\varphi$$ and these finitely many elements, then $$\varphi$$ induces a map $$\varphi'':V\to W''$$, and $$\Lambda^k\varphi''(x)=0$$, since the same relations that implied $$\Lambda^k\varphi(x)=0$$ in $$\Lambda^kW$$ also imply that $$\Lambda^k\varphi''(x)=0$$ in $$\Lambda^kW''$$.

Hence we can also assume that $$W$$ is finite dimensional, and what remains is a problem about finite dimensional vector spaces that can easily be answered without choice by choosing bases.

• Thank you for your detailed answer. What are the "relations defining the exterior power $\Lambda^k W$", and how they can imply the fact that $\Lambda^k \varphi (x)=0$ ? Isn't the latter the hypothesis to prove injectivity ? I would have thought we would have to remplace $W$ by the finite dimensional $\varphi(V)$ where $V$ is finite-dimensional. I guess I don't see where those "finitely many elements" come from if not from $\varphi(V)$. Or do they come from the fact that a representant of $\Lambda^k \varphi(x)$ in the tensor power of $W$ is in the ideal defining the exterior power of $W$ ? – Phil-W May 21 '19 at 20:04
• @Phil-W $\Lambda^kW$ is generated as a vector space by elements $w_1\wedge\dots\wedge w_k$ subject to relations saying that $w_1\wedge\dots\wedge w_k$ is a multilinear and skew-symmetric function of $(w_1,\dots,w_k)$. It’s those relations that I mean. A linear combination of the elements $w_1\wedge\dots\wedge w_k$ is zero in $\Lambda^kW$ if and only if you can prove that it is zero using those relations. – Jeremy Rickard May 22 '19 at 2:11