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

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 the $$\det{}_\mathbb{C}(T)$$ 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?

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 $$k\ne j$$.

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 $$v_1,\ldots v_n$$.

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

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

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