If you have taken a (proof-based) course in linear algebra, chances are that you've encountered the following abstract description of the determinant:
DEF Let $R$ be a ring. A map $δ\colon M_n(R)\to R$ is called a normed determinant map if it is
If furthermore $δ$ is normed, i.e. $δ(\mathbb 1_n) = 1$, we call it a normed determinant map.
In any standard treatise, the notion of a determinant map requires being normed, however we will deviate from that for reasons that will become appearent later.
This description is quite elegant, since one can then prove the usual properties of the determinant (such as invariance under conjugation and transposition), as well as uniqueness. Furthermore, we can see that this definition matches with the explicit construction using permutations.
However, I personally like starting with the following definition, which – in my opinion – outlines the intuitive meaning more clearly, and is easier to remember:
DEF If $R=\mathbb R$ and we identify $T\in M_n(R)$, with the linear map w.r.t. the canonical basis of $ℝ^n$, the determinant is the signed volume of the image of the unit hypercube under $T$.
This compelling definition is missing an important part: What does “volume” mean? To make things rigorous, we must be able to measure the volume of a parallelepiped spanned by $n$ vectors.
In what follows, $(e_i)_i$ is the canonical basis of $ℝ^n$.
DEF A map $ω\colon {(ℝ^n)}^n\to ℝ$ is called a volume form of $ℝ^n$ if it is $ℝ$-linear in each argument and alternating. The vector space of all volume forms (with pointwise addition and scalar multiplication) is denoted by $Λ^n(ℝ^n)$.
Personally, I knew both what a normed determinant map is and what a volume form is – however, it never occurred to me that they were basically describing the same thing. The reason becomes clear with the following correspondence:
LEM Let $\mathcal B = (b_1,\ldots,b_n)$ be an ordered basis of $\mathbb R^n$. The map $$ φ_{\mathcal B}\colon δ\mapsto \left((v_1,\ldots,v_n)\mapsto δ(T(\mathcal B, v_1,\ldots,v_n)\right), $$ where $T(\mathcal B, v_1,\ldots,v_n)$ is the unique linear map mapping sending $b_i$ to $v_i$, is a bijection from determinant maps to volume forms.
Proof TODO
Note that this definition does not require $ω(e_1,\ldots,e_n) = 1$ – in other words, we can choose what volume the unit cube shall have. This is of course not an accident – it allows us to have the vector space structure on $Ω^n(ℝ^n)$, and we can see the following:
LEM Every volume form is the scalar multiple of $ω_1:=\left((e_1,\ldots,e_n)\mapsto 1\right)$. In particular, $Ω^n(ℝ^n)\simeq ℝ$.
Proof. First, note that due to alternation we have $ω(\ldots,e_i,e_i,\ldots) = 0$. Furthermore, we can observe that for any permutation $π\in S_n$, we have $ω(e_{π(1)},\ldots,e_{π(n)})) = \mathrm{sgn}(π)ω_1$. It follows straightforwardly that for any $v_1,\ldots,v_n\in ℝ^n$ with components $v_i=\sum_j v_ij e_j$, the volume form computes to $$ ω(v_1,\ldots,v_n) = \sum_{π\in S_n} (\prod_i v_{iπ(i)}) ω_1. $$