determinant as wedge product of 1-forms

Question

I am wondering if the determinant $\det \colon \underbrace{V \times \dots \times V}_{n-times} \to \mathbb R$ can be written as wedge product of 1-forms. (Given that $V$ is an $n$-dimensional vector space with the standard basis $e_1,...,e_n$ and the dual basis of the 1-forms $\sigma_1,...,\sigma_n.$)

Answer 1

Yes, in fact this is sometimes taken as the definition of the determinant. You may have seen the determinant defined as the unique alternating, $n$-linear map such that $\det(e_1,\dots,e_n) = 1$. Rephrased, we can say this succinctly as "the unique $n$-form such that $\omega (e_1,\dots,e_n) = 1$.

One way to approach this is to define the wedge product of $k\in\Lambda^k(V^*)$ and $l \in \Lambda^l(V^*)$ by $$k\wedge l = \frac{1}{k!l!} \mathrm{Asym}(k \otimes l)$$ Where $\mathrm{Asym}$ is the antisymmetrization operator: $$\mathrm{Asym}(T)(x_1,\dots,x_k) = \sum_{\sigma \in S_n} \mathrm{sgn}(\sigma) T(x_{\sigma(1)},\dots,x_{\sigma(k)}).$$

Since $\dim(\Lambda^k(V^*)) = \binom{n}{k}$ (for $\dim(V) = n$), we clearly see that the space of $n$-forms has dimension $\binom{n}{n} = 1$. All you have to do now if find an $n$-form such that $\omega(e_1,\dots,e_n) = 1$ (you seem to have the right idea on how to find that), and voila, we have produced the (unique) determinant operator.

For more details I highly recommend chapter 3 of Loring Tu's Introduction to Manifolds available in pdf form online.

Answer 2

If $n = 2$, you can see that $\det(X,Y) = \sigma_1 \wedge \sigma_2 (X,Y)$. Generalize to any $n$.