Canonical isomorphism between $\Lambda^2 V$ and $\mathbb{R}$

Question

Let us take a two-dimensional real vector space $V$, and define $$\det \colon V \times V \rightarrow \mathbb{R} \colon (u,v) \mapsto \det(u\vert v).$$ Since this map is bilinear, the universal property of the tensor product induces a unique linear map $\overline{\det}\colon V \otimes V \rightarrow \mathbb{R}$ such that $\overline{\det}(u\otimes v) = \det(u\vert v)$. Now $\det (u\vert u) = 0$, so the latter function gives me an isomorphism that I still call $\det \colon \Lambda^2 V \rightarrow \mathbb{R}$.

My question is the following: is that a canonical isomorphism? I am not really choosing any basis in this construction, so my answer is yes. My doubt arises when I write down the matrix $(u\vert v)$, but I am not using any basis to do it either.

Answer 1

Let $e_1, e_2$ be a basis. If $u = ae_1 + ce_2$ and $v = be_1 + de_2$ are in $V$, then $(u|v)$ is the matrix

$$(u|v) = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$$

This matrix representation depends on your choice of basis. If you use another basis, say $e_1' = \frac{1}{2}e_1, e_2' = \frac{1}{2}e_2$, and you take the same $u$ and $v$ as before, they are now written as $u = 2ae_1' + 2ce_1'$ and $v = 2be_1' + 2de_2'$, and now you would write

$$(u|v) = \begin{pmatrix} 2a & 2b \\ 2c & 2d \end{pmatrix}$$

which is not the same determinant.

Answer 2

If $V$ is an $n$-dimensional space over the field $F$, then the $k$-th component $$ \operatorname{\Lambda}^kV $$ of the exterior algebra is an $F$-vector space of dimension $\dbinom{n}{k}$. In particular, $\operatorname{\Lambda}^nV$ is a one-dimensional vector space. But a one-dimensional vector space is not $F$. It is only isomorphic to $F$ by the choice of a basis.

In the case of the exterior algebra over $V$ with basis $\{e_1,\dots,e_n\}$, the standard choice for the basis of $\operatorname{\Lambda}^nV$ is $e_1\wedge\dots\wedge e_n$. This way, an endomorphism $f\colon V\to V$ induces an endomorphism $$ \operatorname{\Lambda}^nf\colon\operatorname{\Lambda}^nV \to\operatorname{\Lambda}^nV $$ so $\operatorname{\Lambda}^nf(e_1\wedge\dots\wedge e_n)= c e_1\wedge\dots\wedge e_n$ and $c=\det(f)$. One can show this scalar is independent of the choice of the basis: it's essentially the uniqueness of the determinant for matrices.

This doesn't mean that the isomorphism $\operatorname{\Lambda}^nV\to F$ you defined is independent of the basis, because you need a basis for associating a matrix to an $n$-tuple of vectors in $V$.