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Step 1:\ }} \begin{document} \setcounter{page}{1} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{center} {\Large\bf Complex geometry\\[15mm] \small lecture 2: Cauchy formula } %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \\[14mm] {\small Misha Verbitsky } \\[20mm] {\tiny\bf HSE, room 306, 16:20, \\[2mm] September 26, 2020 } \end{center} \newpage {\bf \blue The Grassmann algebra} \definition Let $V$ be a vector space. Denote by $\Lambda^i V$ the space of antisymmetric polylinear $i$-forms on $V^*$, and let $\Lambda^*V:=\bigoplus \Lambda^i V$. Denote by $T^{\otimes i}V$ the algebra of {\bf \green all} polylinear $i$-forms on $V^*$ (``tensor algebra''), and let $\Alt:\; T^{\otimes i}V\arrow \Lambda^i V$ be {\bf \blue the antisymmetrization}, \[\Alt(\eta)(x_1, ..., x_i):= \frac 1{i!}\sum_{\sigma\in \Sigma_i}(-1)^{\tilde \sigma}\eta(x_{\sigma_1}, ..., x_{\sigma_i}) \] where $\Sigma_i$ is the group of permutations, and $\tilde \sigma=1$ for odd permutations, and 0 for even. Consider the multiplicative operation (``wedge-product'') on $\Lambda^*V$, denoted by $\eta\wedge \nu:= \Alt (\eta \otimes \nu)$. The space $\Lambda^*V$ with this operation is called {\bf \blue the Grassmann algebra}. \remark {\bf \red It is an algebra of anti-commutative polynomials.} {\bf \green Properties of Grassmann algebra:} 1. $\dim \Lambda^i V:= \binom{\dim V}{i}$, $\dim \Lambda^* V=2^{\dim V}$. 2. $\Lambda^*(V \oplus W)= \Lambda^*(V) \otimes \Lambda^*(W)$. \newpage {\bf \blue The Hodge decomposition in linear algebra} \definition Let $(V,I)$ be a space equipped with a complex structure. {\bf \blue The Hodge decomposition} $V\otimes_\R \C:= V^{1,0}\oplus V^{0,1}$ is defined in such a way that $V^{1,0}$ is a $\1$-eigenspace of $I$, and $V^{0,1}$ a $-\1$-eigenspace. \remark Let $V_\C:= V \otimes_\R \C$. The Grassmann algebra of skew-symmetric forms $\Lambda^n V_\C:=\Lambda^n_\R V \otimes _\R C$ admits a decomposition \[ \Lambda^n V_\C= \bigoplus_{p+q=n} \Lambda^p V^{1,0} \otimes \Lambda^q V^{0,1} \] We denote $\Lambda^p V^{1,0} \otimes \Lambda^q V^{0,1}$ by $\Lambda^{p,q}V$. The resulting decomposition $\Lambda^n V_\C= \bigoplus_{p+q=n}\Lambda^{p,q}V$ is called {\bf \blue the Hodge decomposition of the Grassmann algebra}. \remark The operator $I$ induces $U(1)$-action on $V$ by the formula $\rho(t)(v) = \cos t\cdot v + \sin t \cdot I(v)$. We extend this action on the tensor spaces by muptiplicativity. \newpage {\bf \blue $U(1)$-representations and the weight decomposition } \remark {\bf \red Any complex representation $W$ of $U(1)$ is written as a sum of 1-dimensional representations $W_i(p)$}, with $U(1)$ acting on each $W_i(p)$ as $\rho(t)(v) = e^{\1 pt}(v)$. The 1-dimensional representations are called {\bf \blue weight $p$ representations of $U(1)$.} \definition A {\bf \blue weight decomposition} of a $U(1)$-representation $W$ is a decomposition $W= \oplus W^p$, where each $W^p=\oplus_i W_i(p)$ is a sum of 1-dimensional representations of weight $p$. \remark {\bf \red The Hodge decomposition $\Lambda^n V_\C= \bigoplus_{p+q=n}\Lambda^{p,q}V$ is a weight decomposition}, with $\Lambda^{p,q}V$ being a weight $p-q$-component of $\Lambda^n V_\C$. \remark $V^{p,p}$ is the space of $U(1)$-invariant vectors in $\Lambda^{2p}V$. Further on, {\bf \blue $TM$ is the tangent bundle on a manifold, and $\Lambda^iM$ the space of differential $i$-forms.} It is a Grassmann algebra on $TM$. \newpage {\bf \blue Vector fields} \definition Let $X$ be the vector field on a manifold $M$, and $f$ a function. Denote by $\Lie_X f$ {\bf \blue the derivatiive} of $f$ along $X$. \definition {\bf \blue A derivation} on a commutative ring is a map $R \stackrel d \arrow R$ satisfying {\bf \blue the Leibniz identity} $d(xy) = d(x) y + x d(y)$. \theorem Each derivation of the ring $C^\infty M$ of smooth functions on $M$ is given by a vector field $X$; {\bf \red this correspondence is bijective.} \remark This can be used as a definition of a vector field. \exercise Prove that {\bf \red a commutator of two derivations is again a derivation.} \remark Vector fields are the same as derivations of $C^\infty M$. This allows us to define {\bf \blue the commutator of two vector fields} as the commutator of the corresponding derivations. \definition Denote by $TM$ the bundle of vector fields, and by $\Lambda^1 M$ or $T^*$ the dual bundle, called {\bf \blue the bundle of 1-forms}. For any $f\in C^\infty M$, the operation $X \arrow \Lie_X f$ is linear as a function of $X$, hence it defines a section of $T^*M$. We denote this section $df$, and call it {\bf \blue the differential} of $f$. \newpage {\bf \blue De Rham algebra} \definition Let $\Lambda^* M$ denote the vector bundle with the fiber $\Lambda^*T^*_xM$ at $x\in M$ ($\Lambda^*T^*M$ is the Grassman algebra of the cotangent space $T^*_x M$). The sections of $\Lambda^i M$ are called {\bf \blue differential $i$-forms}. The algebraic operation ``wedge product'' defined on differential forms is $C^\infty M$-linear; the space $\Lambda^* M$ of all differential forms is called {\bf \blue the de Rham algebra}. \remark $\Lambda^0 M = C^\infty M$. \theorem {\bf \red There exists a unique operator $C^\infty M\stackrel d \arrow \Lambda^1 M\stackrel d \arrow \Lambda^2 M \stackrel d \arrow \Lambda^3 M \stackrel d \arrow ...$ satisfying the following properties} 1. On functions, $d$ is equal to the differential.\\ 2. $d^2=0$ \\ 3. $d(\eta \wedge \xi) = d(\eta) \wedge \xi + (-1)^{\tilde \eta}\eta \wedge d(\xi)$, where $\tilde \eta=0$ where $\eta\in \lambda^{2i}M$ is {\bf \blue an even form,} and $\eta\in \lambda^{2i+1}M$ is {\bf \blue odd.} \definition The operator $d$ is called {\bf \blue de Rham differential}. \exercise {\bf \purple Prove it}. \definition A form $\eta$ is called {\bf \blue closed} if $d\eta=0$, {\bf \blue exact} if $\eta \in \im d$. The group $\frac{\ker d}{\im d}$ is called {\bf \blue de Rham cohomology} of $M$. \newpage {\bf \blue Cauchy formula in dimension 1 (statement)} \definition Let $U\subset \C^n$ be an open subset, and $f:\; U \arrow \C$ a function of class $C^1$ (differentiable at least once). We say that $f$ is {\bf \blue holomorphic} if the differential $df:\; T_x U \arrow \C$ is complex linear at each $x\in U$. \remark Clearly, $f$ is holomorphic if and only if $df\in \Lambda^{1,0}(U)$, where $\Lambda^{1,0}(U)$ is the Hodge (1,0)-component of the de Rham algebra. {\bf \purple Taylor series decomposition for holomorphic functions in 1 variable is implied by the Cauchy formula:} for any folomorphic function $f$ in disk $\Delta\subset \C$, \[ \int_{\6 \Delta} \frac{f(z)dz}{z-a}= 2\pi\1 f(a), \] where $a\in \Delta$ any point, and $z$ coordinate on $\C$. Indeed, in this case, \[ 2\pi\1 f(a) = \sum_{i\geq 0} a^i\int_{\6 \Delta} f(z) (z^{-1})^{i+1}, \] because $\frac{1}{z-a}=z^{-1}\sum_{i\geq 0} (az^{-1})^i$. \newpage {\bf \blue Cauchy formula in dimension 1 (proof)} Let's prove Cauchy formula, using Stokes' theorem. Since the space $\Lambda^{1,0}\C$ is 1-dimensional, $df\wedge dz=0$ for any holomorphic function on $\C$. This gives \claim A function on a disk $\Delta\subset \C$ {\bf \red is holomorphic if and only if the form $\eta:=f dz$ is closed} (that is, satisfies $d\eta=0$). \endproof Now, let $S_\epsilon$ be a radius $\epsilon$ circle around a point $a\in \Delta$, $\Delta_\epsilon$ its interior, and $\Delta_0:=\Delta\backslash \Delta_\epsilon$. Stokes' theorem gives \[ 0= \int_{\Delta_0} d\left(\frac{f(z)dz}{z-a}\right)= -\int_{S_\epsilon}\frac{f(z)dz}{z-a} + \int_{\6 \Delta} \frac{f(z)dz}{z-a}, \] hence Cauchy formula would follow if we show that $\lim\limits_{\epsilon\rightarrow 0}\int_{S_\epsilon}\frac{f(z)dz}{z-a}=2\pi\1 f(a).$ Assuming for simplicity $a=0$ and parametrizing the circle $S_\epsilon$ by $\epsilon e^{\1 t}$, we obtain \begin{multline*} \int_{S_\epsilon}\frac{f(z)dz}{z}= \int_0^{2\pi} \frac{f(\epsilon e^{\1t})}{\epsilon e^{\1t}} d(\epsilon e^{\1t})=\\ =\int_0^{2\pi} \frac{f(\epsilon e^{\1t})}{\epsilon e^{\1t}} \1 \epsilon e^{\1t}dt =\int_0^{2\pi}f(\epsilon e^{\1t})\1 dt \end{multline*} as $\epsilon$ tends to $0$, $f(\epsilon e^{\1t})$ tends to $f(0)$, and this integral goes to $2\pi\1f(0)$. \newpage {\bf \blue Holomorphic functions on $\C^n$ (reminder)} \theorem Let $f:\; U \arrow \C$ be a differentiable function on an open subset $U\subset \C^n$. {\bf \red Then the following are equivalent.}\\ \ \ (1) {\bf \purple $f$ is holomorphic}.\\ \ \ (2) For any complex affine line $L\in \C^n$, the restriction $f\restrict L=\C$ is {\bf \purple holomorphic as a function of one complex variable.}\\ \ \ (3) $f$ is expressed as a sum of Taylor series around any point $(z_1, ..., z_n)\in U$: for all sufficiently small $t_1, ..., t_n$, one has $f(z_1+t_1, z_2+ t_2, ..., z_n +t_n)= \sum_{i_1, ..., i_n}a_{i_1, ..., i_n} t_1^{i_1}t_2^{i_2}...t_n^{i_n}.$ \proof Equivalence of (1) and (2) is clear, because a restriction of $\theta \in \Lambda^{1,0}(M)$ to a line is a $(1,0)$-form on a line, and, conversely, if $df$ is of type (1,0) on each complex line, it is of type (1,0) on $TM$, which is implied by the following linear-algebraic observation. \lemma Let $\eta\in V^*\otimes \C$ be a complex-valued linear form on a real vector space $(V,I)$ equipped with a complex structure $I$. {\bf \purple Then $\eta\in \Lambda^{1,0}(V)$ if and only if its restriction to any $I$-invariant 2-dimensional subspace $L$ belongs to $\Lambda^{1,0}(L)$.} \\ \exercise {\bf \red Prove it.} (3) clearly implies (1). (1) implies (3) by Cauchy formula (many variables), proven below. \newpage {\bf \blue Cauchy formula (many variables)} \remark Let $U\subset \C^n$ be an open subset, and $z_1, ..., z_n$ complex coordinates. Holomorphicity of $f:\; U \arrow \C$ is equivalent to $df\in \Lambda ^{1,0}(M)$, which is equivalent to $df\wedge dz_1 \wedge dz_1 \wedge ... \wedge dz_n=0$. Denote the form $dz_1 \wedge dz_1 \wedge ... \wedge dz_n$ by $\Phi$. We obtain that {\bf \purple $f$ is holomorphic if and only if the form $f\Phi$ is closed} \theorem {\bf \blue (Cauchy formula in dimension $n$)}\\ Let $\Delta\subset \C^n$ be a polydisk (product of disks) of radius 1, and $\alpha_1, ..., \alpha_n\in \Delta$ complex numbers. Denote by $S\subset \C^n$ the product of circles of radius 1 in variables $z_1, ..., z_n$:, $S= S_1(z_1) \times S_1(z_2) \times ... \times S_1(z_n)$. Let $f$ be a holomorphic function in a polydisk. {\bf \red Then $\int_S V= (2\pi\1)^n f(\alpha_1, ... \alpha_n)$, where \[ V=\frac{f\Phi}{(z_1-\alpha_1) (z_2-\alpha_2) \times ... \times (z_n-\alpha_n)}. \]} \phantom{.\hspace{-20pt}} \pstep Denote by $Z$ the set $\bigcup_{i=1}^n \{(z_1, ..., z_n)\ \ |\ \ z_i=\alpha_i\}$. The complement of $Z$ is the set of definition of the closed differential form $V$. Let $S_\epsilon$ be the product of circles of radius $\epsilon$ with center in $\alpha_1, ..., \alpha_n$. Then $S, S_\epsilon\subset \C^n \backslash Z$, and {\bf \purple the tori $S$, $S_\epsilon$ are homotopy equivalent in the domain $\C^n \backslash Z$, where $V$ is closed.} {\bf \red It remains to show that $\lim_{\epsilon \rightarrow 0}\int_{S_\epsilon} V= (2\pi\1)^n f(\alpha_1, ... \alpha_n)$.} \newpage {\bf \blue Cauchy formula (many variables), part 2} \theorem {\bf \blue (Cauchy formula in dimension $n$)}\\ Let $\Delta\subset \C^n$ be a polydisk (product of disks) of radius 1, and $\alpha_1, ..., \alpha_n\in \Delta$ complex numbers. Denote by $S\subset \C^n$ the product of circles of radius 1 in variables $z_1, ..., z_n$:, $S= S_1(z_1) \times S_1(z_2) \times ... \times S_1(z_n)$. Let $f$ be a holomorphic function in a polydisk. {\bf \red Then $\int_S V= (2\pi\1)^n f(\alpha_1, ... \alpha_n)$, where \[ V=\frac{f\Phi}{(z_1-\alpha_1) (z_2-\alpha_2) ... (z_n-\alpha_n)}. \]} \phantom{.\hspace{-30pt}} \pstep Let $S_\epsilon$ be a product of circles of radius $\epsilon$ with center in $\alpha_1, ..., \alpha_n$. {\bf \purple It remains to show that $\lim_{\epsilon \rightarrow 0}\int_{S_\epsilon} V= (2\pi\1)^n f(\alpha_1, ... \alpha_n)$.} {\bf \green Step 2:} To simplify notation we set $\alpha_i=0$. Parametrize $S_\epsilon$ by the cube $[0, 2\pi]^n$ using the map $t_1, ..., t_n\arrow \epsilon e^{\1 t_1}, ..., \epsilon e^{\1 t_n}$. This gives \begin{multline*} \int_{S_\epsilon} V= \int_{S_\epsilon}f(z) \frac{dz_1}{z_1}\wedge ... \wedge\frac{dz_n}{z_n} =\\ =\int_0^{2\pi} ... \int_0^{2\pi} \frac{f(\epsilon e^{\1t_1 }, \epsilon e^{\1t_2 }, ...,\epsilon e^{\1t_n })}{\epsilon e^{\1t_1}\epsilon e^{\1t_2}... \epsilon e^{\1t_n}}\epsilon^n d\left(e^{\1t_1}\right)d\left(e^{\1t_2}\right) ...d\left(e^{\1t_n}\right) =\\ =(\1)^n\int_0^{2\pi}...\int_0^{2\pi}f(\epsilon e^{\1t_1}, ..., \epsilon e^{\1t_n })dt_1dt_2...dt_n, \end{multline*} which converges to $(2\pi\1)^n f(0, ..., 0)$. \endproof \newpage {\bf \blue Cauchy formula and Taylor expansion} \remark Cauchy formula implies that {\bf \red holomorphic functions defined in a polydisk have Taylor expansion in this polydisk}. Indeed, \[ f(\alpha_1, ... \alpha_n)= \frac1{(2\pi\1)^n} \int_S \frac{fdz_1 \wedge ... \wedge dz_n}{( z_1-\alpha_1) (z_2-\alpha_2) \times ... \times (z_n-\alpha_n)} \] Take the Taylor expansion of $(z_i-\alpha_i)^{-1}$ using \[ \frac 1 {(z_i-\alpha_i)}= \frac {z_i^{-1}} {(1-\alpha_i z_i^{-1})} = \sum_{l=0}^\infty \alpha_i^l z_i^{-l-1}. \] Then \[ f(\alpha_1, ... \alpha_n)= \sum_{i_1=0}^\infty ... \sum_{i_n=0}^\infty \alpha_1^{i_1} .... \alpha_{i_n}^{i_n} \int_{S} f(z_1,..., z_n)z_1^{-i_1-1} ... z_n^{-i_n-1} dz_1 \wedge ... \wedge dz_n. \] \newpage {\bf \blue Complex manifolds (reminder)} \definition {\bf \blue A holomorphic function} on $\C^n$ is a function $f:\; \C^n \arrow \C$ such that $df$ is complex linear, that is $df\in \Lambda^{1,0}(M)$. \remark Holomorphic functions form a sheaf. \definition {\bf \blue A complex manifold} $M$ is a ringed space which is locally isomorphic to an open ball in $\C^n$ with a sheaf of holomorphic functions. \remark In other words, {\bf \purple $M$ is covered with open balls embedded to $\C^n$} and transition functions (being coordinate functions for one ball considered in other coordinate system) {\bf \purple are holomorphic.} \newpage {\bf \blue Complex manifolds (reminder)} \definition {\bf \blue A holomorphic function} on $\C^n$ is a function $f:\; \C^n \arrow \C$ such that $df$ is complex linear, that is $df\in \Lambda^{1,0}(M)$. \remark Holomorphic functions form a sheaf. \definition {\bf \blue A complex manifold} $M$ is a ringed space which is locally isomorphic to an open ball in $\C^n$ with a sheaf of holomorphic functions. \remark In other words, {\bf \purple $M$ is covered with open balls embedded to $\C^n$} and transition functions (being coordinate functions for one ball considered in other coordinate system) {\bf \purple are holomorphic.} \newpage {\bf \blue Integrability of almost complex structures} \definition An almost complex structure $I$ on a manifold is called {\bf\blue integrable} if any point of $M$ has a neighbourhood $U$ diffeomorphic to an open subset of $\C^n$, in such a way that the almost complex structure $I$ is induced by the standard one on $U\subset \C^n$. \claim {\bf \red Complex structure on a manifold $M$ uniquely determines an integrable almost complex structure, and is determined by it.} \proof Complex structure on a manifold $M$ is determined by the sheaf of holomorphic functions $\calo_M$, and $\calo_M$ is determined by $I$ as explained above. Therefore, an integrable almost complex structure defines a complex structure. Conversely, every complex structure gives a sub-bundle in $\Lambda^{1,0}(M)=d\calo_M\subset \Lambda^1(M,\C)$, and {\bf \purple such a sub-bundle defines an almost complex structure by Remark 1.} \endproof \newpage {\bf \blue Frobenius form} \claim Let $B\subset TM$ be a sub-bundle of a tangent bundle of a smooth manifold. Given vector fiels $X,Y\in B$, consider their commutator $[X,Y]$, and lets $\Psi(X,Y)\in TM/B$ be the projection of $[X,Y]$ to $TM/B$. {\bf \red Then $\Psi(X,Y)$ is $C^\infty(M)$-linear in $X$, $Y$: } \[ \Psi(fX,Y)= \Psi(X,fY)=f\Psi(X,Y). \] \proof Leibnitz identity gives $[X,fY]=f[X,Y]+ X(f) Y$, and the second term belongs to $B$, hence does not influence the projection to $TM/B$. \endproof \definition This form is called {\bf \blue the Frobenius form} of the sub-bundle $B\subset TM$. This bundle is called {\bf\blue involutive}, or {\bf \blue integrable}, or {\bf \blue holonomic} if $\Psi=0$. \exercise {\bf \purple Give an example of a non-integrable sub-bundle.} \newpage {\bf \blue Formal integrability} \definition An almost complex structure $I$ on $(M,I)$ is called {\bf\blue formally integrable} if $[T^{1,0}M, T^{1,0}]\subset T^{1,0}$, that is, if $T^{1,0}M$ is involutive. \definition The Frobenius form $\Psi\in \Lambda^2(\Lambda^{1,0}M)\otimes T^{0,1}M$ is called {\bf \blue the Nijenhuis tensor}. \claim {\bf \purple If a complex structure $I$ on $M$ is integrable, it is formally integrable.} \proof Locally, the bundle $T^{1,0}(M)$ is generated by $d/dz_i$, where $z_i$ are complex coordinates. These vector fields commute, hence satisfy $[d/dz_i, d/dz_j]\in T^{1,0}(M)$. This means that the Frobenius form vanishes. \endproof \theorem {\bf \blue (Newlander-Nirenberg)}\\ {\bf\red A complex structure $I$ on $M$ is integrable if and only if it is formally integrable.} \proof (real analytic case) next lecture, probably. \remark {\bf \purple In dimension 1, formal integrability is automatic.} Indeed, $T^{1,0}M$ is 1-dimensional, hence all skew-symmetric 2-forms on $T^{1,0}M$ vanish. \newpage {\bf \blue Possible topics for the next lectures } 1. Proof of Frobenius theorem. 2. Newlander-Nirenberg theorem for real analytic almost complex manifolds. \end{document}