I couldn't find this on the literature, but in some simple cases it is possible to compute differential complex forms using the determinant formula, just as the real case. One simple example is given by $$dz_k \wedge d\overline{z_l} \left(\frac{\partial}{\partial z_k}, \frac{\partial}{\partial \overline{z_l}}\right) = \det \left[ \begin{array}{cc} dz_k\left( \frac{\partial}{\partial z_k} \right) & d\overline{z_l}\left( \frac{\partial}{\partial z_k} \right)\\ dz_k\left( \frac{\partial}{\partial \overline{z_l}} \right) & d\overline{z_l}\left( \frac{\partial}{\partial \overline{z_l}} \right) \end{array} \right] = 1.$$

I wonder if this is valid in general. More precisely, is valid the (more general) formula below?

$$dz_{i_1} \wedge \ldots \wedge dz_{i_p} \wedge d\overline{z_{j_1}} \wedge \ldots \wedge d\overline{z_{j_q}}(v_1, \ldots, v_{p+q}) = $$ $$ = \det \left[ \begin{array}{cccccc} dz_{i_1}(v_1) & \ldots & dz_{i_p}(v_1) & d\overline{z_{j_1}}(v_1) & \ldots & d\overline{z_{j_q}}(v_1)\\ \vdots & & \vdots & \vdots & & \vdots\\ dz_{i_1}(v_{p+q}) & \ldots & dz_{i_p}(v_{p+q}) & d\overline{z_{j_1}}(v_{p+q}) & \ldots & d\overline{z_{j_q}}(v_{p+q})\\ \end{array} \right]$$

Thank you for you help.


I believe this determinant formula is quite often taken as the definition of the wedge product of differential forms. It doesn't matter whether they are complex or real here. However one should be careful, because sometimes wedge product with different normalization is used, and your formula should then be modified by some combinatorial factor.


Let $\omega = \omega^1 \wedge ... \wedge \omega^n$ be an $n-$form constructed as wedge product of $1-$forms $\omega^i$. Then by definition of the wedge product we have $$ \omega(X_1...X_n) = \mathrm{det} \left( \omega^i ( X_j ) \right)_{i,j=1...n}. $$ Note that I didn't use any coordinates or particular basis to write this formula down. Every form can be written down as a sum of product forms as above, so action of every $n$-form can be evaluated. Sometimes it is introduce to (locally) introduce a basis on the tangent and cotangent planes. We can pick arbitrary vector fields $e_i$ and $1-$ forms $f^j$ such that $$ f^j (e_i) = \delta^j_i .$$ Often one uses some specific coordinates $x^i$ and picks $e_i = \frac{\partial}{\partial x^i}$, $f^j = d x^j$. Note however, that there is nothing special about this choice - any other will do. For example on a complex manifold we use $dz^i$ and $d \bar z^i$, even though they surely can't be regarded as independent real coordinates.

  • $\begingroup$ I'm not taking this as definition. I consider it more natural to write $$\displaystyle\frac{\partial}{\partial z_j} = \frac{1}{2}\left( \frac{\partial}{\partial x_j} - i \frac{\partial}{\partial y_j} \right)$$ $$\displaystyle\frac{\partial}{\partial \overline{z_j}} = \frac{1}{2}\left( \frac{\partial}{\partial x_j} + i \frac{\partial}{\partial y_j} \right)$$ $$dz_j = dx_j + idy_j$$ $$d\overline{z_j} = dx_j - i dy_j$$ then expand everything and make the calculations. I want to know if the two approaches coincide. $\endgroup$ – Integral Dec 6 '17 at 16:25
  • $\begingroup$ For the real differential forms I do consider the determinant definition. But for the complex case, I think it's relevant to know if the determinant definition coincide with the definition I mentioned above. Thanks. $\endgroup$ – Integral Dec 6 '17 at 16:27
  • $\begingroup$ Hint: consider for a moment real differential forms on a real manifold. Suppose that you use determinantal definition, but you work strictly in terms of coordinates $x^i$, so you have basis vectors $\frac{\partial}{\partial x^i}$ and basic covectors $d x^i$. Then you change coordinates to $y^j$ and ask whether all your previous formulas look the same now. In fact they do - how would you prove this? Can this method be extended to complex case? $\endgroup$ – Blazej Dec 6 '17 at 17:03
  • $\begingroup$ There is the Jacobian to consider there right? I'm not sure where this is going to. $\endgroup$ – Integral Dec 6 '17 at 18:01
  • $\begingroup$ Note that two matrices will appear: $\frac{\partial x^i}{\partial y^j}$ and also its inverse, $\frac{\partial y^j}{\partial x^i}$. One is from transformation law of the basis vectors, the other is from the transformation law of the covectors. In the full determinant they will cancel, because $\mathrm{det}M^{-1}=\frac{1}{\mathrm{det}M}$ for any matrix $M$. Thus the form of determinant formula is left unchanged in the second coordinate system. Let me know if this is still too vague, I will try to write a more complete answer. $\endgroup$ – Blazej Dec 6 '17 at 19:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.