I'm trying to understand the proof of the holomorphic version of the Frobenius integrability theorem given in p. 51-52 of Voisin's text "Hodge Theory and Complex Algebraic Geometry I".

Statement: Let $X$ be a complex manifold of complex dimension $n$ and $E \subset T_X$ an involutive holomorphic subbundle of complex rank $r$. Then $E$ is integrable in the holomorphic sense. Meaning locally there exist holomorphic functions $\varphi : U \to \mathbb{C}^{n-r}$ with $ker(d\varphi_*)=E|_U$.

The idea of the proof there (If I understand it correctly) goes as follows:

  1. Use the real Frobenius theorem for the real part of the distribution $E_\mathbb{R}$ to get locally functions $\varphi : U \to V \subset \mathbb{R}^{2(n-r)}$ with $ker(d\varphi_*)= E_{\mathbb{R}}|_U$
  2. The almost complex structure $I : T_{U,\mathbb{R}} \to T_{U,\mathbb{R}}$ descends to an endomorphism $I : T_{U,\mathbb{R}}/E_{\mathbb{R}}|_U \to T_{U,\mathbb{R}}/E_{\mathbb{R}}|_U$ which itself descends to an almost complex structure on $V$.
  3. There exists a complex submanifold transverse to the fibers of $\varphi$ (possibly restricting $U$) which is locally isomorphic to $V$ and whose complex structure agrees with the almost complex structure from (2). In other words there's locally a section of $\varphi$ whose image is a complex manifold whose almost complex structure agrees with the one from (2).
  4. Due to (3) we can put a complex structure on $V$ making $\varphi : U \to V$ a holomorphic map


There are several things I'm having trouble understanding.

Firstly step (2) isn't elaborated on in the text and it isn't so clear to me why the almost complex structure should descend down to $V$.

Secondly given step (3) it's unclear to me why we need step (2). If there's a section whose image is a holomorphic manifold then there's a unique complex structure on $V$ s.t. $\varphi$ is a holomorphic (isn't there?).

Thirdly and perhaps the most crucial is that step (3) seems dangerously circular. Part of the conclusion of the theorem is that $E$ has an involutive complement in $T_X$ meaning $E|_U \oplus F|_U = T_X|_U$ for some involutive holomorphic subbundle $F|_U$. The existence of the section in (3) is stronger than that assertion. It may be that (3) relies on a weaker statement but it's unclear to me how to deduce it from what was done up until now in the book.

Fourthly Nowhere in the proof is it claimed that the almost complex structure on $V$ makes it isomorphic to an open subset of $\mathbb{C}^{n-r}$. It is not claimed that the almost complex structure is the standard one, nor is it obvious to me that it should be from this perspective.

Lastly There's the issue of what the proof uses which I raised in a comment. It doesn't seem very likely to me that one can prove the holomorphic Frobenius from the real Frobenius theorem alone. The real Frobenius is a consequence of the smooth Poincare lemma (closed form is locally exact) therefore a proof of this sort would give the Holomorphic Poincare lemma as a corollary of the smooth Poincare lemma which doesn't sound reasonable to me. The question is therefore what is used in the proof above which goes beyond the smooth Poincare lemma. It most certainly seems to me that this happens somewhere around (3).

  • $\begingroup$ One idea that might help you understand tht theorem is simply reproving Frobenius theorem while checking that if you start with holomorphic vector bundles the local submersions you get are holomorphic. $\endgroup$
    – Omar
    Apr 23, 2017 at 10:00
  • $\begingroup$ @Omar I don't think this can be done without using some form of holomorphic Poincare lemma (ensuring that solutions to holomorphic first order systems are holomorphic). In particular I'm not sure that one can prove holomorphic frobenius from real frobenius alone as the latter doesn't know about holomorphic Poincare lemma. $\endgroup$ Apr 23, 2017 at 10:03
  • 1
    $\begingroup$ Much as the holomorphic Poincare Lemma follows from the smooth one because the Cauchy-Riemann equations imply holomorphicity of output when input is holomorphic, why so skeptical for Frobenius' theorem? Multivariable differential calculus works in the holomorphic setting as in the smooth setting by the same proofs (e.g., inverse function theorem), so a $C^{\infty}$-submanifold of a complex manifold is a complex submanifold iff its tangent space is a $\mathbf{C}$-subspace at all points (see para. 2 of p. 10 of math.stanford.edu/~conrad/papers/rhtalk.pdf); that is what you're overlooking. $\endgroup$
    – nfdc23
    Apr 23, 2017 at 14:02

2 Answers 2


The usual proofs work in the holomorphic category directly. No need for almost complex structures. Take a holomorphic subbundle $V \subset TM$ of the tangent bundle of a complex manifold $M$. In local holomorphic coordinates $x,y$ we can arrange by linear algebra at some point, say at the origin, that $V_{(0,0)}$ is $dy=A \, dx$ for some matrix $A$. Then we will see that nearby, $V=(dy-f(x,y)dx)$, some $f(x,y)$ holomorphic. Consider the map $\phi(x,y,v,w)=(x,y,v)$ for $x,v \in \mathbb{C}^p$ and $y,w \in \mathbb{C}^q$. This map is a submersion $\phi \colon TM \to M \times \mathbb{C}^p$. Restricted to $V$, it is (by dimension count) a local biholomorphism, so has an inverse, $w=f(x,y)v$, linear in $v$ since the fibers $V_{(x,y)}$ are linear. Take each holomorphic vector field $X(x)$ in the $x$-plane, and lift it to a unique holomorphic vector field $\hat{X}(x,y)$ which projects via $(x,y) \mapsto x$ to $X$. This holomorphic vector field is $\hat{X}(x,y)=(X(x),f(x,y)X(x))$, clearly by linear algebra. Because the holomorphic lifts project to the origin vector fields, their flows project to their flows, so their brackets project to their brackets. So if you lift commuting vector fields, like constant ones, you get ones with vertical brackets. If $V$ is bracket closed, you get zero brackets of those lifts. So commuting vector fields lift to commuting vector fields. The flows of the lifts then give holomorphic integral manifolds, by the holomorphic implicit function theorem. There is a unique one through each point $(x,y)=(0,y)$, and then the flow times along the fields together with the initial $y$ value make holomorphic coordinates in which the fields are coordinates translations.

  • $\begingroup$ It's not clear to me how one can get that $V$ is $dy=0$. This seems to me like the bulk of the proof. That is showing that an involutive subbundle is locally the kernel of exact holomorphic 1-forms. I think linear algebra gives you that it is locally the kernel of closed holomorphic 1-forms see here: math.stackexchange.com/questions/1419718/… $\endgroup$ Apr 23, 2017 at 10:52
  • $\begingroup$ I only ask that $V_0$ is $dy=0$, at the origin. Since $V_0 \subset T_0 M$ is some linear subspace, a linear change of local coordinates will arrange that the various $dy$ 1-forms are linearly independent on $V_0$, and that is actually all we use here: $dy$ is a linear function of $dx$ on $V_0$. $\endgroup$
    – Ben McKay
    Apr 23, 2017 at 10:55
  • $\begingroup$ Aha, you were talking about the fiber I see. Sorry again but, it's not clear to me what version of the holomorphic implicit function theorem you're using. $\endgroup$ Apr 23, 2017 at 10:59
  • $\begingroup$ Also It's not clear to me why the lifts of holomorphic vector fields via the projection are unique. $\endgroup$ Apr 23, 2017 at 11:05
  • $\begingroup$ @SaalHardali: there is only one version of the holomorphic implicit function theorem; its proof is identical to the one in the $C^{\infty}$-setting (via reduction to the inverse function theorem, whose proof is the same too). Perhaps your skepticism on various things could be cleared up if you state what you know about equivalences among various possible definitions of holomorphicity for functions of several complex variables (say assuming continuity, to avoid digressing into irrelevant measure-theoretic issues). Have you tried speaking with an expert in complex analysis at your university? $\endgroup$
    – nfdc23
    Apr 23, 2017 at 14:30

Looking at the proof in the book, I'd say that only the independence of complex structure on $T_v V$ from the point in the fiber is lacking a proof. This is indeed a bit technical, but here is a sketch:

Take $w \in T_v V$ and consider its lifts $w' \in T_{u'}(U_{\mathbb{R}}), \ w'' \in T_{u''}(U_{\mathbb{R}})$ for two points $u', u''$ in the fiber $\phi^{-1}(v)$. We need to show that $T\phi (Iw')=T\phi (Iw'')$ so that we can define $Iw$ by choosing a lift $w'$ and declaring $Iw=T\phi (I w')$ independently of the choice of the lift.

The fiber $\phi^{-1}(v)$ is a smooth submanifold of $U$ which is tangent to the holomorphic distribution $E$, and a smooth submanifold of a complex manifold $U$ is complex submanifold if and only if its tangent space at each point is a complex subspace of $T_u(U)$. This is satisfied, so $\phi^{-1}(v)$ is a complex submanifold of $U$. Shrinking $U$ if necessary, we can assume that $U$ is biholomorphic to a polydisk where $\phi^{-1}(v)$ is given by $z_{k+1}=...=z_{n}=0$ and points have coordinates $u'=(0,...0)$ and $u''=(\overline{a_1}, ..., \overline{a_k},0,..0)$. Consider a constant holomorphic vector field on $\phi^{-1}(v)$ given by $$\overline{a_1} \frac{\partial}{\partial z_1}|_a+...+\overline{a_k}\frac{\partial}{\partial z_k}|_a.$$ By construction it is a section of our holomorphic distribution $E$ and therefore can be written as $$f_1(a)X_1|_{(a,0)}+...+f_k(a)X_k|_{(a,0)}$$for some local frame $X_1, ..., X_k$ for $E$ on $U$ and functions $f_i(a)$, holomorphic on the fiber $\phi^{-1}(v)$. Trivially extend it to holomorphic section of $E$ on the whole $U$ by $$X|_{(a, b)}=f_1(a)X_1|_{(a,b)}+...+f_k(a)X_k|_{(a,b)}.$$

Now consider the flow of $X$ as a real vector field (using the identification of $TU$ with $T(U_{\mathbb{R}}$)). It is a fact (for example, see here or here) that at each time $t=t_0$ the flow $\Phi_{t_0}$ is a biholomorphism (on its domain of definition). Note that the curve $$\theta: [0,1] \to U$$ given by $$t \mapsto (\overline{a_1}t, ..., \overline{a_k} t, 0 ,...,0)$$is integral for $X$ and connects point $u'$ with point $u''$, so the flow $\Phi_{t=1}$ is defined in the neighborhood of $u'$ and maps to a neighborhood of $u''$ biholomorphically.

Now, recall the points in the holomorphic tangent space $T_u'(U)$ are represented by germs of complex curves $[z \mapsto \gamma(z)]$, while points of the smooth tangent space $T_u'(U_{\mathbb{R}})$ are represented by germs of smooth curves $[t \mapsto \theta(t)]$ and the identification (as real vector spaces) $T_{u'}(U) \to T_{u'}(U_{\mathbb{R}})$ simply goes like this $$[z \mapsto \gamma(z)] \mapsto [t \mapsto \gamma(t)].$$ Multiplication by $i$ acts on $T_{u'}(U)$ like $$i \cdot [z \mapsto \gamma(z)]=[z \mapsto \gamma(iz)],$$while the almost complex structure on $T_{u'}(U_{\mathbb{R}})$ induced by $T_{u'}(U) \to T_{u'}(U_{\mathbb{R}})$ acts by $$I[t \mapsto \gamma (t)]=[t \mapsto \gamma(it)].$$

So choose a complex germ $[z \mapsto \gamma(z)]$, such that the corresponding real curve $[t \mapsto \gamma (t)]$ represents the lift $w'$. Applying $\phi_{t=1}$, we get a holomorphic germ $[z \mapsto \Phi_{t=1} (\gamma (z))]\in T_{u''}(U)$. We claim that the inderlying smooth curve $[t \mapsto \Phi_{t=1} (\gamma (t))]$ is again a lift of $w$ so can be taken to be $w''$. For that consider the map $$(s, t) \mapsto \Phi_s (\gamma (t)).$$ For fixed $t$ the curve is integral for the vector field $X$, tangent to the distribution $E$, so $$\phi \circ \Phi_s \circ \gamma (t)$$ is a constant function of $s$ for fixed $t$. So $$w=T \phi (w')=[\phi (\gamma (t)]=[\phi \circ \Phi_{t=0} \circ \gamma (t)]= [\phi \circ \Phi_{t=1} \circ \gamma (t)]=T \phi [\Phi_{t=1} (\gamma (t))],$$as claimed.

Finally, since the construction was essentially holomorphic, it respects the almost complex structure! The vectors $$I w' = [t \mapsto \gamma (it)], \ \ Iw''=[t \mapsto \Phi_{t=1} (\gamma (it))$$are mapped to the same vector by $T\phi$, because we can repeat the argument above to show that $$(s,t) \mapsto \phi \circ \Phi_s \circ \gamma(it)$$is independent of $s$.

If points $u'$ and $u''$ do not lie in a single submanifold chart, one can repeat this construction several times along some path connecting $u'$ and $u'$ inside the fiber.

PS. I agree that probably proving the holomorphic Frobenius theorem from the first principles of analysis of complex variables would be easier than taking this route through the real Frobenius theorem. But this proof is instructive of the beautiful interplay between holomorphic and smooth geometries, underlying a complex manifold.


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