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By reading an introduction to symplectic geometry the author of the text said that the standard symplectic form $w_0 = \sum_{j=1}^{n} dx_j \wedge dy_j$ on $\mathbb{R}^{2n}$ equals the form $w = \frac{i}{2} \sum_{j=1}^{n} dz_j \wedge d\bar{z}_j$ on $\mathbb{C}^n$ under the identification $\phi: \mathbb{C}^n \to\mathbb{R}^{2n}$ defined by $(z_1,...,z_n) = (x_1 + iy_1,...,x_n + iy_n) \mapsto (x_1,...,x_n,y_1,...,y_n)$. When I understood this correctly this means that $\phi$ is a symplectomorphism, i.e. $\phi^{*}w_0 = w$. Now I would really like to understand how one can compute this pullback:

$\phi^{*}w_0 = \phi^{*}(\sum_{j=1}^{n} dx_j \wedge dy_j) = \sum_{j=1}^{n}\phi^{*}(dx_j) \wedge \phi^{*}(dy_j)$, so show that

$\phi^{*}(dx_j) \wedge \phi^{*}(dy_j) = \frac{i}{2}(dz_j \wedge d\bar{z}_j)$

I absolutely do not see why this equality holds. Thanks a lot for any help!

Donut
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    Your $\phi$ is not quite correct. I think you mean $(x_1, y_1, \dots, x_n, y_n)$ or $(x_1, \dots, x_n, y_1, \dots, y_n)$ rather than $(x_1 + iy_1, \dots, x_n + iy_n)$. – Michael Albanese Nov 20 '16 at 16:45
  • Oh yes, you are absolutely right. I corrected it, so it makes sense! Thank you! – Donut Nov 26 '16 at 08:54

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If we just solve the last expression: \begin{equation} \begin{split} \frac{i}{2}(dz_j \wedge d\bar{z}_j)&=\frac{i}{2}(dx_j+idy_j)\wedge(dx_j-idy_j)\\ &=\frac{i}{2}(dx_j\wedge dx_j-idx_j\wedge dy_j+idy_j\wedge dx_j+dy_j\wedge dy_j)\\ &=\frac{i}{2}(-idx_j\wedge dy_j+idy_j\wedge dx_j)\\ &=\frac{i}{2}(-idx_j\wedge dy_j-idx_j\wedge dy_j)\\ &=\frac{1}{2}(dx_j\wedge dy_j+dx_j\wedge dy_j)\\ &=dx_j\wedge dy_j. \end{split}\end{equation}

MattG88
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  • Thanks a lot for this computation. I actually had problems in understanding the pullback correctly. So the question was why is $\phi^{}(dx_j) \wedge \phi^{}(dy_j) = dx_j \wedge dy_j$, but i got it now. – Donut Nov 26 '16 at 08:59