Simple computation with the $n$-form $dz_1\wedge...\wedge dz_n$ in $\mathbb{C}^n$

Question

Let $z_j=x_i+iy_j$ be the coordinates for $\mathbb{C}^n$ and consider the $n$-form $\eta:=dz_1\wedge...\wedge dz_n$.

I've just read the following (the contex is probably not important):

Let $N\subset\mathbb{C}^n$ be a submanifold with $\dim_\mathbb{R}N=n$ and take a local orthonoral frame $\epsilon_1,...\epsilon_n$ for $TN$.

If $e_1,...,e_n$ is the standard basis for $\mathbb{R}^n\subset\mathbb{C}^n$, then $e_1,...,e_n,Je_1,...,Je_n$ is a basis for $\mathbb{C}^n$. Define the $\mathbb{R}$-linear map $T:\mathbb{C}^n\to\mathbb{C}^n$ with $Te_j=\epsilon_j$ and $T(Je_j)=J\epsilon_j$.

By construction, $T$ is in fact $\mathbb{C}$-linear. Then: \begin{align*} \eta(\epsilon_1,...,\epsilon_n)&=dz_1\wedge...\wedge dz_n(Te_1,...,Te_n)\\ &=\det{}_{\mathbb{C}}(T) \end{align*} where $\det{}_\mathbb{C}(T)$ is the determinant of the matrix in $GL(n,\mathbb{C})$ representing $T$.

I don't understand how $\det{}_\mathbb{C}(T)$ showed up. I would be ok with it if the last equation was $$dz_1\wedge...\wedge dz_n(Te_1,...,Te_n)=\det{}_{\mathbb{C}}(T)\cdot dz_1\wedge...\wedge dz_n(e_1,...,e_n)$$

What am I missing?

Answer 1

You're right. But $dz_1\wedge\dots\wedge dz_n (e_1,\dots,e_n) = 1$. Note, for example, that in $\Bbb C$, we have $dz(e_1) = (dx+i\,dy)(e_1) = dx(e_1)+i\,dy(e_1)=1+i\cdot 0 = 1$. Extrapolating, $dz_1\wedge\dots\wedge dz_n(e_1,\dots,e_n) = dz_1(e_1)dz_2(e_2)\cdots dz_n(e_n) = 1$, since $dz_k(e_j)=0$ when $k\ne j$.

Answer 2

If you have a n-dimensional manifold, then any n-form is a multiple of the det function, since the n-forms are a 1-dimensional space and det is an alternating function. If you take the dual standard basis, the n-form $dz_1 \wedge \ldots \wedge dz_n(v_1\ldots,v_n)$ is the det of the matrix whose columns are $v_1,\ldots v_n$.

Answer 3

The determinant pops out of an application of the definitions and properties of the wedge product, and does not have much to do with differential geometry at all. I asume you know the definition of tensor product andalternating operator. Let $V$ be an $\mathbb R\ (\text{or}\ \mathbb C)$- vector space of dimension $n.$ For a $k$-covector $f$ and an $\ell$-covector $g$ on a vector space $V$, the wedge product $f\wedge g$ is by definition the $k+ℓ$-covector $f\wedge g=\frac{1}{k!\ell!}A(f\otimes g)$ where $A$ is the alternating operator.

One shows that $f_1\wedge\cdots \wedge f_m=\frac{1}{k_1!\wedge\cdots\wedge k_m!}A(f_1\otimes\cdots \otimes f_m)$ for $k_i$- covectors $f_i.$ In particular, if each $f_i$ is a $1$-covector (a linear functional on $V)$, then $f_1\wedge\cdots \wedge f_m=A(f_1\otimes\cdots \otimes f_m).$

So, using the definition of $A$, with $\sigma\in S_m,$ the group of permutations on $m$ elements, and $\{v_1,\cdots, v_m\}\subset V,$ we compute

$f_1\wedge\cdots \wedge f_m(v_1,\cdots, v_m)=A(f_1\otimes\cdots \otimes f_m)(v_1,\cdots, v_m)=$

$\sum_{\sigma\in S_m}(\text{sgn} \sigma) f_1(v_{\sigma (1)})\cdots f_m(v_{\sigma (m)})=\det f_i(v_j).$