Dolbeault cohomology

In mathematics, in particular in algebraic geometry and differential geometry, Dolbeault cohomology (named after Pierre Dolbeault) is an analog of de Rham cohomology for complex manifolds. Let M be a complex manifold. Then the Dolbeault cohomology groups Hp,q(M,C) depend on a pair of integers p and q and are realized as a subquotient of the space of complex differential forms of degree (p,q).

Construction of the cohomology groups

Let Ωp,q be the vector bundle of complex differential forms of degree (p,q). In the article on complex forms, the Dolbeault operator is defined as a differential operator on smooth sections

${\displaystyle {\bar {\partial }}:\Gamma (\Omega ^{p,q})\rightarrow \Gamma (\Omega ^{p,q+1})}$

Since

${\displaystyle {\bar {\partial }}^{2}=0}$

this operator has some associated cohomology. Specifically, define the cohomology to be the quotient space

${\displaystyle H^{p,q}(M,\mathbb {C} )={\frac {{\hbox{ker}}\left({\bar {\partial }}:\Gamma (\Omega ^{p,q},M)\rightarrow \Gamma (\Omega ^{p,q+1},M)\right)}{{\bar {\partial }}\Gamma (\Omega ^{p,q-1})}}.}$

Dolbeault cohomology of vector bundles

If E is a holomorphic vector bundle on a complex manifold X, then one can define likewise a fine resolution of the sheaf ${\displaystyle {\mathcal {O}}(E)}$ of holomorphic sections of E. This is therefore a recollection of the sheaf cohomology of ${\displaystyle {\mathcal {O}}(E)}$.

Dolbeault-Grothendieck lemma

In order to establish the Dolbeault isomorphism we need to prove the Dolbeault-Grothendieck lemma (or ${\displaystyle {\bar {\partial }}}$-Poincaré lemma). First we prove a one-dimensional version of the ${\displaystyle {\bar {\partial }}}$-Poincaré lemma; we shall use the following generalised form of the Cauchy integral representation for smooth functions:

Proposition: Let ${\displaystyle B_{\epsilon }(0):=\lbrace z\in \mathbb {C} \mid |z|<\epsilon \rbrace }$ the open ball centered in ${\displaystyle 0}$ of radius ${\displaystyle \epsilon \in \mathbb {R} _{>0}}$, ${\displaystyle {\overline {B_{\epsilon }(0)}}\subseteq U}$ open and ${\displaystyle f\in {\mathcal {C}}^{\infty }(U)}$, then ${\displaystyle \forall z\in B_{\epsilon }(0)}$

${\displaystyle f(z)={\frac {1}{2\pi i}}\int _{\partial B_{\epsilon }(0)}{\frac {f(\xi )}{\xi -z}}d\xi +{\frac {1}{2\pi i}}\iint _{B_{\epsilon }(0)}{\frac {\partial f}{\partial {\bar {\xi }}}}{\frac {d\xi \wedge d{\bar {\xi }}}{\xi -z}}.}$

Lemma (${\displaystyle {\bar {\partial }}}$-Poincaré lemma on the complex plane): Let ${\displaystyle B_{\epsilon }(0),U}$ be as before and ${\displaystyle \alpha \in {\mathcal {A}}_{\mathbb {C} }^{0,1}(U)}$ a smooth form, then

${\displaystyle {\mathcal {C}}^{\infty }(U)\ni g(z):={\frac {1}{2\pi i}}\int _{B_{\epsilon }(0)}{\frac {f(\xi )}{\xi -z}}d\xi \wedge d{\bar {\xi }}}$

satisfies

${\displaystyle \alpha ={\bar {\partial }}g}$ on ${\displaystyle B_{\epsilon }(0)}$.

Proof. Our claim is that ${\displaystyle g}$ defined above is a well-defined smooth function such that ${\displaystyle f}$ is locally ${\displaystyle {\bar {\partial }}}$-exact. To show this we choose a point ${\displaystyle w\in B_{\epsilon }(0)}$ and an open neighbourhood ${\displaystyle w\in V\subseteq B_{\epsilon }(0)}$, then we can find a smooth function ${\displaystyle \rho \colon B_{\epsilon }(0)\to \mathbb {R} }$ whose support is compact and lies in ${\displaystyle B_{\epsilon }(0)}$ and ${\displaystyle \rho |_{V}\equiv 1}$. Then we can write ${\displaystyle f=f_{1}+f_{2}:=\rho f+(1-\rho )f}$ and define

${\displaystyle g_{i}:={\frac {1}{2\pi i}}\int _{B_{\epsilon }(0)}{\frac {f_{i}(\xi )}{\xi -z}}d\xi \wedge d{\bar {\xi }}}$.

Since ${\displaystyle f_{2}\equiv 0}$ in ${\displaystyle V}$ then ${\displaystyle g_{2}}$ is clearly well-defined and smooth; we note that

{\displaystyle {\begin{aligned}g_{1}&=\int _{B_{\epsilon }(0)}{\frac {f_{1}(\xi )}{\xi -z}}d\xi \wedge d{\bar {\xi }}\\&={\frac {1}{2\pi i}}\int _{\mathbb {C} }{\frac {f_{1}(\xi )}{\xi -z}}d\xi \wedge d{\bar {\xi }}\\&=\pi ^{-1}\int _{0}^{\infty }\int _{0}^{2\pi }f_{1}(z+re^{i\theta })e^{-i\theta }d\theta dr,\end{aligned}}}

which is indeed well-defined and smooth, therefore the same is true for ${\displaystyle g}$. Now we show that ${\displaystyle {\bar {\partial }}g=\alpha }$ on ${\displaystyle B_{\epsilon }(0)}$.

${\displaystyle {\frac {\partial g_{2}}{\partial {\bar {z}}}}={\frac {1}{2\pi i}}\int _{B_{\epsilon }(0)}f_{2}(\xi ){\frac {\partial }{\partial {\bar {z}}}}{\Big (}{\frac {1}{\xi -z}}{\Big )}d\xi \wedge d{\bar {\xi }}=0}$

since ${\displaystyle (\xi -z)^{-1}}$ is holomorphic in ${\displaystyle B_{\epsilon }(0)\setminus V}$ .

{\displaystyle {\begin{aligned}{\frac {\partial g_{2}}{\partial {\bar {z}}}}=&\pi ^{-1}\int _{\mathbb {C} }{\frac {\partial f_{1}(z+re^{i\theta })}{\partial {\bar {z}}}}e^{-i\theta }d\theta \wedge dr\\=&\pi ^{-1}\int _{\mathbb {C} }{\Big (}{\frac {\partial f_{1}}{\partial {\bar {z}}}}{\Big )}(z+re^{i\theta })e^{-i\theta }d\theta \wedge dr\\=&{\frac {1}{2\pi i}}\iint _{B_{\epsilon }(0)}{\frac {\partial f_{1}}{\partial {\bar {\xi }}}}{\frac {d\xi \wedge d{\bar {\xi }}}{\xi -z}}\end{aligned}}}

applying the generalised Cauchy formula to ${\displaystyle f_{1}}$ we find

${\displaystyle f_{1}(z)={\frac {1}{2\pi i}}\int _{\partial B_{\epsilon }(0)}{\frac {f_{1}(\xi )}{\xi -z}}d\xi +{\frac {1}{2\pi i}}\iint _{B_{\epsilon }(0)}{\frac {\partial f}{\partial {\bar {\xi }}}}{\frac {d\xi \wedge d{\bar {\xi }}}{\xi -z}}={\frac {1}{2\pi i}}\iint _{B_{\epsilon }(0)}{\frac {\partial f}{\partial {\bar {\xi }}}}{\frac {d\xi \wedge d{\bar {\xi }}}{\xi -z}}}$

since ${\displaystyle f_{1}|_{\partial B_{\epsilon }(0)}=0}$, but then ${\displaystyle f=f_{1}={\frac {\partial g_{1}}{\partial {\bar {z}}}}={\frac {\partial g}{\partial {\bar {z}}}}}$ on ${\displaystyle B_{\epsilon }(0)}$. QED

Now are ready to prove the Dolbeault-Grothendieck lemma; the proof presented here is due to Grothendieck[1]. We denote with ${\displaystyle \Delta _{\epsilon }^{n}(0)}$ the open polydisc centered in ${\displaystyle 0\in \mathbb {C} ^{n}}$ with radius ${\displaystyle \epsilon \in \mathbb {R} _{>0}}$.

Lemma (Dolbeault-Grothendieck): Let ${\displaystyle \alpha \in {\mathcal {A}}_{\mathbb {C} ^{n}}^{p,q}(U)}$ where ${\displaystyle {\overline {\Delta _{\epsilon }^{n}(0)}}\subseteq U}$ open and ${\displaystyle q>0}$ such that ${\displaystyle {\bar {\partial }}\alpha =0}$, then there exists ${\displaystyle \beta \in {\mathcal {A}}_{\mathbb {C} ^{n}}^{p,q-1}(U)}$ which satisfies

${\displaystyle \alpha ={\bar {\partial }}\beta }$ on ${\displaystyle \Delta _{\epsilon }^{n}(0)}$.

Before starting the proof we note that any ${\displaystyle (p,q)}$-form can be written as

${\displaystyle \alpha =\sum _{IJ}\alpha _{IJ}dz_{I}\wedge d{\bar {z}}_{J}=\sum _{J}{\Big (}\sum _{I}\alpha _{IJ}dz_{I}{\Big )}_{J}\wedge d{\bar {z}}_{J}}$

for multi-indices ${\displaystyle I,J,|I|=p,|J|=q}$, therefore we can reduce the proof to the case ${\displaystyle \alpha \in {\mathcal {A}}_{\mathbb {C} ^{n}}^{0,q}(U)}$.

Proof. Let ${\displaystyle k>0}$ be the smallest index such that ${\displaystyle \alpha \in (d{\bar {z}}_{1},\dots ,d{\bar {z}}_{k})}$ in the sheaf of ${\displaystyle {\mathcal {C}}^{\infty }}$-modules, we proceed by induction on ${\displaystyle k}$. For ${\displaystyle k=0}$ we have ${\displaystyle \alpha \equiv 0}$ since ${\displaystyle q>0}$; next we suppose that if ${\displaystyle \alpha \in (d{\bar {z}}_{1},\dots ,d{\bar {z}}_{k})}$ then there exists ${\displaystyle \beta \in {\mathcal {A}}_{\mathbb {C} ^{n}}^{0,q-1}(U)}$ such that ${\displaystyle \alpha ={\bar {\partial }}\beta }$ on ${\displaystyle \Delta _{\epsilon }^{n}(0)}$. Then suppose ${\displaystyle \omega \in (d{\bar {z}}_{1},\dots ,d{\bar {z}}_{k+1})}$ and observe that we can write

${\displaystyle \omega =d{\bar {z}}_{k+1}\wedge \psi +\mu }$

where ${\displaystyle \psi ,\mu \in (d{\bar {z}}_{1},\dots ,d{\bar {z}}_{k})}$. Since ${\displaystyle \omega }$ is ${\displaystyle {\bar {\partial }}}$-closed it follows that ${\displaystyle \psi ,\mu }$ are holomorphic in the variables ${\displaystyle z_{k+2},\dots ,z_{n}}$ and smooth in the remaining ones on the polydisc ${\displaystyle \Delta _{\epsilon }^{n}(0)}$. Moreover we can apply the ${\displaystyle {\bar {\partial }}}$-Poincaré lemma to the smooth functions ${\displaystyle z_{k+1}\mapsto \psi _{J}(z_{1},\dots ,z_{k+1},\dots ,z_{n})}$ on the open ball ${\displaystyle B_{\epsilon _{k+1}}(0)}$, hence there exist a family of smooth functions ${\displaystyle g_{J}}$ which satisfy ${\displaystyle \psi _{J}={\frac {\partial g_{J}}{\partial {\bar {z}}_{k+1}}}}$ on ${\displaystyle B_{\epsilon _{k+1}}(0)}$; ${\displaystyle g_{J}}$ are also holomorphic in ${\displaystyle z_{k+2},\dots ,z_{n}}$. Define ${\displaystyle {\tilde {\psi }}:=\sum _{J}g_{J}d{\bar {z}}_{J}}$ then

{\displaystyle {\begin{aligned}\omega -{\bar {\partial }}{\tilde {\psi }}&=d{\bar {z}}_{k+1}\wedge \psi +\mu -\sum _{J}{\frac {\partial g_{J}}{\partial {\bar {z}}_{k+1}}}d{\bar {z}}_{k+1}\wedge d{\bar {z}}_{J}+\sum _{j=1}^{k}\sum _{J}{\frac {\partial g_{J}}{\partial {\bar {z}}_{j}}}d{\bar {z}}_{j}\wedge d{\bar {z}}_{J\setminus \lbrace j\rbrace }\\&=d{\bar {z}}_{k+1}\wedge \psi +\mu -d{\bar {z}}_{k+1}\wedge \psi +\sum _{j=1}^{k}\sum _{J}{\frac {\partial g_{J}}{\partial {\bar {z}}_{j}}}d{\bar {z}}_{j}\wedge d{\bar {z}}_{J\setminus \lbrace j\rbrace }\\&=\mu +\sum _{j=1}^{k}\sum _{J}{\frac {\partial g_{J}}{\partial {\bar {z}}_{j}}}d{\bar {z}}_{j}\wedge d{\bar {z}}_{J\setminus \lbrace j\rbrace }\in (d{\bar {z}}_{1},\dots ,d{\bar {z}}_{k}),\end{aligned}}}

therefore we can apply the induction hypothesis to it, there exists ${\displaystyle \eta \in {\mathcal {A}}_{\mathbb {C} ^{n}}^{0,q-1}(U)}$ such that

${\displaystyle \omega -{\bar {\partial }}{\tilde {\psi }}={\bar {\partial }}\eta }$

on ${\displaystyle \Delta _{\epsilon }^{n}(0)}$ and ${\displaystyle \zeta :=\eta +{\tilde {\psi }}}$ ends the induction step. QED

The previous lemma can be generalised by admitting polydiscs with ${\displaystyle \epsilon _{k}=+\infty }$ for some of the components of the polyradius.

Lemma (extended Dolbeault-Grothendieck) If ${\displaystyle \Delta _{\epsilon }^{n}(0)}$ is an open polydisc with ${\displaystyle \epsilon _{k}\in \mathbb {R} \cup \lbrace +\infty \rbrace }$ and ${\displaystyle q>0}$, then

${\displaystyle H_{\bar {\partial }}^{p,q}(\Delta _{\epsilon }^{n}(0))=0}$.

Proof. We consider two cases: ${\displaystyle \alpha \in {\mathcal {A}}_{\mathbb {C} ^{n}}^{p,q+1}(U),q>0}$ and ${\displaystyle \alpha \in {\mathcal {A}}_{\mathbb {C} ^{n}}^{p,1}(U)}$.

Let ${\displaystyle \alpha \in {\mathcal {A}}_{\mathbb {C} ^{n}}^{p,q+1}(U),q>0}$, and we cover ${\displaystyle \Delta _{\epsilon }^{n}(0)}$ with polydiscs ${\displaystyle {\overline {\Delta _{i}}}\subset \Delta _{i+1}}$, then by the Dolbeault-Grothendieck lemma we can find forms ${\displaystyle \beta _{i}}$ of bidegree ${\displaystyle (p,q-1)}$ on ${\displaystyle {\overline {\Delta _{i}}}\subseteq U_{i}}$ open such that ${\displaystyle \alpha |_{\Delta _{i}}={\bar {\partial }}\beta _{i}}$; we want to show that

${\displaystyle \beta _{i+1}|_{\Delta _{i}}=\beta _{i}}$.

We proceed by induction on ${\displaystyle i}$: for ${\displaystyle i=1}$ it is true by the previous lemma. Let the claim be true for ${\displaystyle k>1}$ and take ${\displaystyle \Delta _{k+1}}$ with ${\displaystyle \Delta _{\epsilon }^{n}(0)=\bigcup _{i=1}^{k+1}\Delta _{i}}$ and ${\displaystyle {\overline {\Delta _{k}}}\subset \Delta _{k+1}}$, then we find a ${\displaystyle (p,q-1)}$-form ${\displaystyle \beta '_{k+1}}$ defined in an open neighbourhood of ${\displaystyle {\overline {\Delta _{k+1}}}}$ such that ${\displaystyle \alpha |_{\Delta _{k+1}}={\bar {\partial }}\beta _{k+1}}$. Let ${\displaystyle U_{k}}$ be an open neighbourhood of ${\displaystyle {\overline {\Delta _{k}}}}$ then ${\displaystyle {\bar {\partial }}(\beta _{k}-\beta '_{k+1})=0}$ on it and we can apply again the Dolbeault-Grothendieck lemma to find a ${\displaystyle (p,q-2)}$-form ${\displaystyle \gamma _{k}}$ such that ${\displaystyle \beta _{k}-\beta '_{k+1}={\bar {\partial }}\gamma _{k}}$ on ${\displaystyle \Delta _{k}}$. Now, let ${\displaystyle V_{k}}$ open with ${\displaystyle {\overline {\Delta _{k}}}\subset V_{k}\subsetneq U_{k}}$ and ${\displaystyle \rho _{k}\colon \Delta _{\epsilon }^{n}(0)\to \mathbb {R} }$ a smooth function such that ${\displaystyle \operatorname {supp} (\rho _{k})\subset U_{k}}$, ${\displaystyle \rho |_{V_{k}}=1}$ and ${\displaystyle \rho _{k}|_{\Delta _{\epsilon }^{n}(0)\setminus U_{k}}=0}$: ${\displaystyle \rho _{k}\gamma _{k}}$ is then a well-defined smooth form on ${\displaystyle \Delta _{\epsilon }^{n}(0)}$ which satisfies ${\displaystyle \beta _{k}=\beta '_{k+1}+{\bar {\partial }}(\gamma _{k}\rho _{k})}$ on ${\displaystyle \Delta _{k}}$, hence the form ${\displaystyle \beta _{k+1}:=\beta '_{k+1}+{\bar {\partial }}(\gamma _{k}\rho _{k})}$ satisfies

{\displaystyle {\begin{aligned}\beta _{k+1}|_{\Delta _{k}}&=\beta '_{k+1}+{\bar {\partial }}\gamma _{k}=\beta _{k}\\{\bar {\partial }}\beta _{k+1}&={\bar {\partial }}\beta '_{k+1}=\alpha |_{\Delta _{k+1}}.\end{aligned}}}

If instead ${\displaystyle \alpha \in {\mathcal {A}}_{\mathbb {C} ^{n}}^{p,1}(U)}$, we cannot apply the Dolbeault-Grothendieck lemma twice; we take ${\displaystyle \beta _{i}}$ and ${\displaystyle \Delta _{i}}$ as before, we want to show that

${\displaystyle \lVert ({\beta _{i}}_{I}-{\beta _{i+1}}_{I})|_{\Delta _{k-1}}\rVert _{\infty }<2^{-i}}$.

Again, we proceed by induction on ${\displaystyle i}$: for ${\displaystyle i=1}$ the answer is given by the Dolbeault-Grothendieck lemma. Next we suppose that the claim is true for ${\displaystyle k>1}$. We take ${\displaystyle \Delta _{k+1}\supset {\overline {\Delta _{k}}}}$ such that ${\displaystyle \Delta _{k+1}\cup \lbrace \Delta _{i}\rbrace _{i=1}^{k}}$ covers ${\displaystyle \Delta _{\epsilon }^{n}(0)}$, then we can find a ${\displaystyle (p,0)}$-form ${\displaystyle \beta '_{k+1}}$ such that

${\displaystyle \alpha |_{\Delta _{k+1}}={\bar {\partial }}\beta '_{k+1}}$,

which also satisfies ${\displaystyle {\bar {\partial }}(\beta _{k}-\beta '_{k+1})=0}$ on ${\displaystyle \Delta _{k}}$, i.e. ${\displaystyle \beta _{k}-\beta '_{k+1}}$ is a holomorphic ${\displaystyle (p,0)}$-form wherever defined, hence by the Stone-Weierstrass theorem we can write it as

${\displaystyle \beta _{k}-\beta '_{k+1}=\sum _{|I|=p}(P_{I}+r_{I})dz_{I}}$

where ${\displaystyle P_{I}}$ are polynomials and ${\displaystyle \lVert r_{I}|_{\Delta _{k-1}}\rVert _{\infty }<2^{-k}}$, but then the form ${\displaystyle \beta _{k+1}:=\beta '_{k+1}+\sum _{|I|=p}P_{I}dz_{I}}$ satisfies

{\displaystyle {\begin{aligned}{\bar {\partial }}\beta _{k+1}&={\bar {\partial }}\beta '_{k+1}=\alpha |_{\Delta _{k+1}}\\\lVert ({\beta _{k}}_{I}-{\beta _{k+1}}_{I})|_{\Delta _{k-1}}\rVert _{\infty }&=\lVert r_{I}\rVert _{\infty }<2^{-k}\end{aligned}}}

which completes the induction step; therefore we have built a sequence ${\displaystyle \lbrace \beta _{i}\rbrace _{i\in \mathbb {N} }}$ which uniformly converges to some ${\displaystyle (p,0)}$-form ${\displaystyle \beta }$ such that ${\displaystyle \alpha |_{\Delta _{\epsilon }^{n}(0)}={\bar {\partial }}\beta }$. QED

Dolbeault's theorem

Dolbeault's theorem is a complex analog[2] of de Rham's theorem. It asserts that the Dolbeault cohomology is isomorphic to the sheaf cohomology of the sheaf of holomorphic differential forms. Specifically,

${\displaystyle H^{p,q}(M)\cong H^{q}(M,\Omega ^{p})}$

where Ωp is the sheaf of holomorphic p forms on M.

A version for logarithmic forms has also been established.[3]

Proof

Let ${\displaystyle {\mathcal {F}}^{p,q}}$ be the fine sheaf of ${\displaystyle C^{\infty }}$ forms of type ${\displaystyle (p,q)}$. Then the ${\displaystyle {\overline {\partial }}}$-Poincaré lemma says that the sequence

${\displaystyle \Omega ^{p,q}{\xrightarrow {\overline {\partial }}}{\mathcal {F}}^{p,q+1}{\xrightarrow {\overline {\partial }}}{\mathcal {F}}^{p,q+2}{\xrightarrow {\overline {\partial }}}\cdots \,}$

is exact. Like any long exact sequence, this sequence breaks up into short exact sequences. The long exact sequences of cohomology corresponding to these give the result, once one uses that the higher cohomologies of a fine sheaf vanish.

Explicit example of calculation

The Dolbeault cohomology of the ${\displaystyle n}$-dimensional complex projective space is

${\displaystyle H_{\bar {\partial }}^{p,q}(P_{\mathbb {C} }^{n})={\begin{cases}\mathbb {C} &p=q\\0&{\text{otherwise}}\end{cases}}}$

We apply the following well-known fact from Hodge theory:

${\displaystyle H_{\scriptscriptstyle {\rm {dR}}}^{k}(P_{\mathbb {C} }^{n},\mathbb {C} )=\bigoplus _{p+q=k}H_{\bar {\partial }}^{p,q}(P_{\mathbb {C} }^{n})}$

because ${\displaystyle P_{\mathbb {C} }^{n}}$ is a compact Kähler complex manifold. Then ${\displaystyle b_{2k+1}=0}$ and

${\displaystyle b_{2k}=h^{k,k}+\sum _{p+q=2k,p\neq q}h^{p,q}=1.}$

Furthermore we know that ${\displaystyle P_{\mathbb {C} }^{n}}$ is Kähler, and ${\displaystyle 0\neq [\omega ^{k}]\in H_{\scriptscriptstyle {\bar {\partial }}}^{k,k}(P_{\mathbb {C} }^{n})}$, where ${\displaystyle \omega }$ is the fundamental form associated to the Fubini-Study metric (which is indeed Kähler), therefore ${\displaystyle h^{k,k}=1}$ and ${\displaystyle h^{p,q}=0}$ whenever ${\displaystyle p\neq q}$, which yields the result.

Footnotes

1. ^ http://www.numdam.org/item?id=SHC_1953-1954__6__A18_0
2. ^ In contrast to de Rham cohomology, Dolbeault cohomology is no longer a topological invariant because it depends closely on complex structure.
3. ^ Navarro Aznar, V. (1987), "Sur la théorie de Hodge-Deligne", Inventiones Mathematicae, 90 (1): 11–76, doi:10.1007/bf01389031, Section 8

References

• Dolbeault, P. (1953). "Sur la cohomologie des variétés analytiques complexes". C. R. Acad. Sci. Paris. 236: 175–277.
• Wells, R.O. (1980). Differential Analysis on Complex Manifolds. Springer-Verlag. ISBN 0-387-90419-0.
• Gunning, R.C. (1990). Introduction to Holomorphic Functions of Several Variables, Volume 1. Chapman and Hall/CRC. p. 198. ISBN 9780534133085.
• Griffiths, Phillip; Harris, Joseph (2014). Principles of Algebraic Geometry. John Wiley & Sons. p. 832. ISBN 9781118626320.