It is well known that $\frac{1}{2\pi i}\oint_\gamma \frac{1}{z}dz=1$, where $\gamma$ is a circle around the origin $z=0$. My question is somehow similar to this.

Consider the plane $z_1=0$ in $\mathbb C^2$, and let $\gamma_1$ be a contour around this plane.

Question Show that $\frac{1}{2\pi i}\oint_{\gamma_1} \frac{1}{z_2}dz_1 \wedge dz_2= dz_2$

I attempt to calculate this contour integration as usual, but it is very strange with a two-form being integrated. The key point is what is the precise definition for this kind of contour integration.

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    $\begingroup$ Yeah, this doesn't make sense, this is not a volume form on the contour. There are no $2$-forms on $1$-manifolds. $\endgroup$ Commented Jan 13, 2017 at 19:38
  • $\begingroup$ But I still need an explanation $\endgroup$
    – Hang
    Commented Jan 13, 2017 at 19:43
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    $\begingroup$ I think that is the explanation: automatically $dz_1$ and $dz_2$ are linearly dependent by dimensional considerations, so their wedge is $dz_1\wedge dz_1\bigg|_{\gamma}=0$ so the integral is $0$. $\endgroup$ Commented Jan 13, 2017 at 19:46
  • $\begingroup$ But the answer is $dz_2$ $\endgroup$
    – Hang
    Commented Jan 13, 2017 at 19:47
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    $\begingroup$ That's what they say it is, but you can easily see this is not true: $$dz_1\wedge dz_2 = z_1'(\gamma_1(t))\gamma_1'(t)\,dt\wedge z_2'(\gamma_1(t))\gamma_1'(t)\,dt= z_1'(\gamma_1(t))z_2'(\gamma_1(t))\gamma_1'(t)^2dt\wedge dt=0$$ $\endgroup$ Commented Jan 13, 2017 at 19:50

1 Answer 1


You can interpret the differential 2-form $dz_1 \wedge dz_2$ on $\mathbb{C}^2$ also as a 1-form on $\mathbb{C}^2$ with value in $T^*\mathbb{C}^2$. As such, you would be able to integrate such a form on 1-submanifolds in order to get a 0-form on $\mathbb{C}^2$ with value in $T^*\mathbb{C}^2$, i.e. a usual 1-form.

(To get a sense of what this means, recall from calculus classes the integration of the velocity vector in order to get the displacement of a point particle. How did you defined such an integration, if not as the 'usual' Riemann integration of a vector-valued function instead of a scalar one? Here, over the point $q \in \mathbb{C}^2$, the 1-form takes value in the vector space $T^*_q \mathbb{C}^2 \simeq \mathbb{C}^2 \simeq \mathbb{R}^4$, so the integration of such a 1-form is not so very different from integration in calculus.)

Somewhat put differently, imagine you were to integrate $dz_1 \wedge dz_2$ over the 2-manifold $\gamma_1 \times \gamma_2$ where $\gamma_j \subset [z_j=0]$. In practice, you would compute this integral through two iterated integrals, first performing the integration over each slice $\gamma_1 \times \{p\}$ with $p \in \gamma_2$, then the integral over $\gamma_2$. In some respect, your question concerns the meaning of the first of the two iterated integrals. Since the end result has to be a scalar, the integrand of the second of the two iterated integrals has to be a 1-form on $\gamma_2$. This suggests that the result of integrating the 2-form over each slice is a 1-form. In the present case, since the form $dz_2$ does not depends on $p \in \gamma_2$, you can give a well-defined meaning to your integral (as the value of the integral over any slice), and the result is indeed $dz_2$.

This explanation was more about integration the form $dz_1 \wedge dz_2$ over a curve $\gamma_1 \subset [z_1=0]$, but this is not what your question is about.

Nevertheless, one can still interpret $(1/z_1) dz_1 \wedge dz_2$ (mind the difference with what is written in the question; is it a typo?) as a 1-form over $\mathbb{C}^2 \setminus [z_1 = 0]$ taking value in $T^*\mathbb{C}^2$ (the specific value being $dz_2$). Let $\gamma$ be a closed curve around the plane $[z_1 = 0]$. By (a somewhat generalized) Stokes theorem, we can assume that $\gamma \subset [z_2 = 0]$. In this case, the 'intuitive' explanation I gave tells us more easily that the integral is $((2\pi i)^{-1} \oint_{\gamma} z_1^{-1}dz_1) \wedge dz_2 = dz_2$.

The book Principles of Algebraic Geometry by Griffiths and Harris discusses differential forms with value in more general vector spaces (and vector bundles). This is also applied to complex calculus in higher dimension. The book Differential Forms in Algebraic Topology by Bott and Tu discusses 'fiber integration', where one integrates a $k$-differential form over a $l$-submanifold with $l \le k$ to get a $(k-l)$-differential form with value in the appropriate space.

These references are possibly too technical for your needs, so I preferred to give an informal answer here.

  • $\begingroup$ @AdamHughes The vector space $\mathbb{C}^2$ has a natural metric (and a natural connection), so one can easily identify the many $T^*_q\mathbb{C}^2$'s with $\mathbb{C}^2$ itself. As such, the 2-form $dz_1 \wedge dz_2$ can be interpreted as the 1-form $\alpha$ whose value at $q \in \mathbb{C}^2$ is $\alpha_q = f(q)(dz_1)_q$ with $f(q) \in \mathbb{C}^2$ being identified with the 1-form $(dz_2)_q \in T^*_q\mathbb{C}^2$. The (meaning of the) integration of such a 1-form should not cause any real problem. $\endgroup$ Commented Jan 13, 2017 at 20:06
  • $\begingroup$ I figured it out after I asked, that's why I deleted my original comment, but I wanted to acknowledge your response with gratitude. It seems we came up with different interpretations of the op's question, this was just a more general query since I didn't see a canonical choice, but as you say we can make choices to make the normalization constant $1$ as needed. Cheers! $\endgroup$ Commented Jan 13, 2017 at 20:45
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    $\begingroup$ @AdamHughes Your comment was more than welcome. It made me notice that I had not really answered the question, and I also noticed what I expect to be a typo in the question (so possibly I still don't have given an answer, but then I would argue with you that the integral is zero). It is a difficult thing to understand differential forms and the peculiar way in which they are integrated. I was very pleased when I learnt about their integration over arbitrary submanifolds, as I think this gives a more geometric and closer to usual view of integration. Our comments present both interpretations. $\endgroup$ Commented Jan 13, 2017 at 21:09
  • $\begingroup$ Griffiths and Harris :) $\endgroup$ Commented Jan 15, 2017 at 0:34
  • $\begingroup$ @TedShifrin Thank you! I don't know where my thoughts were! $\endgroup$ Commented Jan 15, 2017 at 17:38

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