If $\operatorname{dim}(V)=n$, then $\operatorname{dim}(\wedge^n (V))=1$

Question

Context

Suppose $V$ is a $n$-dimensional vector space over $\mathbb{R}$. In our differential geometry class, we define the tensor product \begin{align} \overbrace{V\otimes \cdots \otimes V}^{k\text{ copies}}\text{ to be the dual space of the space of multilinear forms on $\overbrace{V\times \cdots \times V}^{k\text{ copies }}$} \end{align} and define the tensor algebra $T(V)$ to be \begin{align} T(V)\triangleq \bigoplus_{k=1}^{\infty} \overbrace{V\otimes \cdots \otimes V}^{k\text{ copies}}\text{ where vector multiplication is the linear extension of }(v)(w)\triangleq v\otimes w \end{align} It is easy to check that $T(V)$ forms a ring. We then can let $I(V)\subseteq T(V)$ to be the ideal generated by $\{ v\otimes v \in T(V):v \in V\}$, and define the exterior algebra $\bigwedge^*(V)$ to be the quotient algebra of $T(V)$ by $I(V)$. Denote \begin{align} \wedge^k(V)\triangleq \pi (\overbrace{V\otimes \cdots \otimes V}^{k\text{ copies}})\text{ where $\pi:T(V)\rightarrow \wedge^*(V)$ is the quotient map } \end{align}

Question

I understand what is happening so far, but I am having a hard time proving that $\wedge^n (V)$ is really 1-dimensional. More specifically, I can't prove that $w_1 \wedge \cdots \wedge w_n \in \wedge^n(V)$ is non-zero when $\{w_1,\dots ,w_n\}$ is basis for $V$. As a side note for any helper, I can prove that $\wedge^n (V)$ is spanned by $\{w_1\wedge \dots \wedge w_n \}$, so all I need is a proof that $w_1\wedge \dots \wedge w_n$ is non-zero.

Conceptually, I understand that $\wedge^n(V)$ is somehow the dual space of the space of alternating $n$-linear forms, but I don't know how to implement this mathematically. I believe proving that $w_1\wedge \cdots \wedge w_n$ requires us to consider the determinant function, but again I don't know how. Maybe the trouble lies in that I can't give a concrete characterization of $I(V)$, so I can't really say $w_1\otimes \cdots \otimes w_n $ is not an element of $I(V)$.

The following are posts that I think relate to my problem but aren't really duplicates. Thanks in advance for helps in all form.

This post asked almost the same question, but it is still not a duplicate since the first answer practically assumed what I wanted to prove in the first place.

This post that proves $\operatorname{dim}\wedge^k(V)=\binom{n}{k}$ is also not a duplicate since the definition used in its answer is different from mine: His or her post's answer defines $w_1\wedge \cdots \wedge w_n$ to be a multilinear form on $V^n$ while mine is a subset of the dual space of multilinear form on $V^k$, not a multilinear form on $V^k$.

Answer 1

Just posting this to tie together and elaborate on some of the points already mentioned in the comments:

1. A key observation is that the ideal $I(V)$ is not just any ideal in $T(V)$, it's a graded ideal which means $$I(V)=\bigoplus_k I^k(V)$$ where $I^k(V)=I(V)\cap T^k(V)$ (with $T^k(V)=\underbrace{V\otimes\cdots\otimes V}_k$). Hence the exterior algebra $\bigwedge V=T(V)/I(V)$ is (isomorphic to) a graded algebra $$\bigwedge V=\bigoplus_k\bigwedge^k V$$ where $\bigwedge^k V=T^k(V)/I^k(V)$. So in particular your $\bigwedge^n(V)$ is (isomorphic to) $T^n(V)/I^n(V)$.

2. By composing the universal properties for $T^k(V)$ and the quotient construction, it's possible to show that $\bigwedge^k V=T^k(V)/I^k(V)$ satisfies a universal property for alternating $k$-linear maps on $V$: any alternating $k$-linear map on $V$ factors uniquely through this space.

3. Therefore to show that $\bigwedge^n V\ne 0$, more specifically that if $v_1,\ldots,v_n\in V$ are linearly independent then $0\ne v_1\wedge\cdots\wedge v_n\in\bigwedge^n V$, it suffices to find an alternating $n$-linear map on $V$ which maps $v_1,\ldots,v_n$ to a nonzero thing. But clearly the determinant does this.

4. It's also possible to view this through the lens of duality. If $V^*$ is dual to $V$ under the pairing $\langle-,-\rangle:V^*\times V\to\mathbb{R}$, then $\bigwedge^k V^*$ is dual to $\bigwedge^k V$ under the pairing satisfying $$\langle v^*_1\wedge\cdots\wedge v^*_k,v_1\wedge\cdots\wedge v_k\rangle=\det(\langle v^*_j,v_i\rangle)$$ where $v^*_j\in V^*$ and $v_i\in V$ are arbitrary. This also extends to a duality between $\bigwedge V^*$ and $\bigwedge V$. Now if $v_1,\ldots,v_n$ is a basis of $V$ and we take $v^*_1,\ldots,v^*_n$ to be the dual basis in $V^*$, then it is immediate that $$\langle v^*_1\wedge\cdots\wedge v^*_n,v_1\wedge\cdots\wedge v_n\rangle=1$$ so $v_1\wedge\cdots\wedge v_n\ne 0$.

5. Since $V$ is finite-dimensional, the duality relationship shows that $\bigwedge^k V^*$ is isomorphic to the space of linear forms on $\bigwedge^k V$, which by the universal property is isomorphic to the space of alternating $k$-linear forms on $V$. Of course by symmetry $\bigwedge^k V$ is isomorphic to the space of alternating $k$-linear forms on $V^*$. So you can also view the above argument as talking about alternating multilinear forms.

As noted by @Kaira, this is all nicely covered in Chapter 4, Section 19 of Loring Tu's book Differential Geometry. If you want to go deeper on the multilinear and exterior algebra itself, you might also take a look at the classic books by Werner Greub, as well as those by Marvin Marcus.

Answer 2

See Theorem 4.2 here and the paragraphs after it, which are expressed in the setting of a (finite) free module $M$ over a commutative ring and proves a basis property of all the exterior powers $\Lambda^k(M)$, not just the top exterior power. Let me quote the paragraph immediately after the proof:

Exterior powers are closely connected to determinants, and most proofs of Theorem 4.2 for finite free M use the determinant. What we used in lieu of theorems about determinants is our knowledge of bases of tensor powers of a free module. For aesthetic reasons, we want to come back later [in Section 6] in and prove properties of the determinant using exterior powers, so we did not use the determinant directly in the proof of Theorem 4.2. However, the linear map $\alpha_{k,M}$ looks a lot like a determinant.

Section 7, in particular Theorem 7.1, applies Theorem 4.2 to finite-dimensional real vector spaces.