\documentclass{slides} \usepackage{amssymb, amsmath, amscd, color, epsfig} %\usepackage[matrix,arrow]{xy} \newcommand{\green}{\color[rgb]{0,0.4,0}} \newcommand{\purple}{\color[rgb]{0.4,0,0.4}} \newcommand{\red}{\color[rgb]{0.7,0,0}} \newcommand{\blue}{\color{blue}} \def\eqref#1{(\ref{#1})} \newcommand{\goth}{\mathfrak} \newcommand{\g}{{\frak g}} \newcommand{\arrow}{{\:\longrightarrow\:}} \newcommand{\Z}{{\Bbb Z}} \newcommand{\C}{{\Bbb C}} \newcommand{\R}{{\Bbb R}} \newcommand{\Q}{{\Bbb Q}} \renewcommand{\H}{{\Bbb H}} \newcommand{\6}{\partial} \def\1{\sqrt{-1}\:} \newcommand{\restrict}[1]{{\left|_{{#1}}\right.}} \newcommand{\cntrct} % contraction with a vector field {\hspace{2pt}\raisebox{1pt}{\text{$\lrcorner$}}\hspace{2pt}} \def\Bbb#1{\mathbb #1} \newcommand{\calo}{{\cal O}} \newcommand{\cac}{{\cal C}} % Correcting TeX... %\let\oldtilde=\tilde %\renewcommand{\tilde}{\widetilde} \renewcommand{\bar}{\overline} \renewcommand{\phi}{\varphi} \renewcommand{\epsilon}{\varepsilon} \renewcommand{\geq}{\geqslant} \renewcommand{\leq}{\leqslant} % Operatornames \newcommand{\Ob}{\operatorname{{\cal O}b}} \newcommand{\Mor}{\operatorname{{\cal M}or}} \newcommand{\Der}{\operatorname{Der}} \newcommand{\even}{{\rm even}} \newcommand{\ev}{{\rm even}} \newcommand{\odd}{{\rm odd}} \newcommand{\const}{{\it const}} \newcommand{\diam}{{\sf diam}} \newcommand{\fl}{{\rm fl}} \newcommand{\im}{\operatorname{im}} \newcommand{\End}{\operatorname{End}} \newcommand{\Sym}{\operatorname{Sym}} \newcommand{\Hol}{\operatorname{{\cal H}ol}} \newcommand{\Tot}{\operatorname{Tot}} \newcommand{\Id}{\operatorname{Id}} \newcommand{\id}{\operatorname{\text{\sf id}}} \newcommand{\Vol}{\operatorname{Vol}} \newcommand{\Hom}{\operatorname{Hom}} \newcommand{\Aut}{\operatorname{Aut}} \newcommand{\Alt}{\operatorname{Alt}} \newcommand{\Iso}{\operatorname{Iso}} \newcommand{\Sec}{\operatorname{Sec}} \newcommand{\Can}{\operatorname{Can}} \newcommand{\Sing}{\operatorname{Sing}} \newcommand{\Spin}{\operatorname{Spin}} \newcommand{\codim}{\operatorname{codim}} \newcommand{\coim}{\operatorname{coim}} \newcommand{\coker}{\operatorname{coker}} \newcommand{\slope}{\operatorname{slope}} \newcommand{\rk}{\operatorname{rk}} \newcommand{\Def}{\operatorname{Def}} \newcommand{\Lie}{\operatorname{Lie}} \newcommand{\Tw}{\operatorname{Tw}} \newcommand{\Tr}{\operatorname{Tr}} \newcommand{\Spec}{\operatorname{Spec}} \newcommand{\Sup}{\operatorname{Sup}} \newcommand{\Diff}{\operatorname{Diff}} \renewcommand{\Re}{\operatorname{Re}} \renewcommand{\Im}{\operatorname{Im}} \newcommand{\inbfpare}[1]{{% \mbox{\tt (}\hspace{-5pt}\mbox{\tt (} #1 % \mbox{\tt )}\hspace{-5pt}\mbox{\tt )}% }} \newcommand{\comment}[1]{{}} \def\blacksquare{\hbox{\vrule width 10pt height 10pt depth 0pt}} \def\endproof{\blacksquare} \def\shortdash{\mbox{\vrule width 4.5pt height 0.55ex depth -0.5ex}} \makeatletter %\@ifundefined{Bbb} % {\newcommand{\Bbb}[1]{{\mathbb #1}}}% %{}% {\edef\Bbb#1{{\Bbb #1}}} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Pagestyle % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \newcommand{\ps@verbit}{% \renewcommand{\@oddhead}{% \scriptsize {\it \small Geometry of manifolds, lecture 10 \hfil \tiny M. Verbitsky }} \renewcommand{\@evenhead}{\@oddhead} \renewcommand{\@oddfoot}{\hfil\thepage\hfil} \renewcommand{\@evenfoot}{\@oddfoot}} \pagestyle{verbit} \setlength\paperheight {10in}% \setlength\paperwidth {13.5in} \setlength{\textwidth}{0.8\paperwidth} \setlength{\textheight}{0.8\paperheight} \setlength{\pdfpageheight}{\paperheight} \setlength{\pdfpagewidth}{\paperwidth} \addtolength{\topmargin}{-20mm} \addtolength{\leftmargin}{-25mm} \addtolength{\rightmargin}{-25mm} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Lemma, sublemma, corollary, proposition, theorem, % % definition,example defined there: % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \newcounter{section} \newcounter{Mycounter}[section] \newcounter{lemma}[section] \setcounter{lemma}{0} \renewcommand{\thelemma}{\noindent{Lemma \thesection.\arabic{lemma}}} \newcommand{\lemma}{% \setcounter{lemma}{\value{Mycounter}} \refstepcounter{lemma} \stepcounter{Mycounter} {\bf \green LEMMA:\ }} \newcounter{claim}[section] \setcounter{claim}{0} \renewcommand{\theclaim}{\noindent{Claim \thesection.\arabic{claim}}} \newcommand{\claim}{% \setcounter{claim}{\value{Mycounter}} \refstepcounter{claim} \stepcounter{Mycounter} {\bf \green CLAIM:\ }} \newcounter{corollary}[section] \setcounter{corollary}{0} \renewcommand{\thecorollary}{\noindent{Corollary \thesection.\arabic{corollary}}} \newcommand{\corollary}{% \setcounter{corollary}{\value{Mycounter}} \refstepcounter{corollary} \stepcounter{Mycounter} {\bf \green COROLLARY:\ }} \newcounter{theorem}[section] \setcounter{theorem}{0} \renewcommand{\thetheorem}{\noindent{Theorem \thesection.\arabic{theorem}}} \newcommand{\theorem}{% \setcounter{theorem}{\value{Mycounter}} \refstepcounter{theorem} \stepcounter{Mycounter} {\bf \green THEOREM:\ }} \newcounter{conjecture}[section] \setcounter{conjecture}{0} \renewcommand{\theconjecture}{\noindent{Conjecture \thesection.\arabic{conjecture}}} \newcommand{\conjecture}{% \setcounter{conjecture}{\value{Mycounter}} \refstepcounter{conjecture} \stepcounter{Mycounter} {\bf \green CONJECTURE:\ }} \newcounter{proposition}[section] \setcounter{proposition}{0} \renewcommand{\theproposition} {\noindent{Proposition \thesection.\arabic{proposition}}} \newcommand{\proposition}{% \setcounter{proposition}{\value{Mycounter}} \refstepcounter{proposition} \stepcounter{Mycounter} {\bf \green PROPOSITION:\ }} \newcounter{definition}[section] \setcounter{definition}{0} \renewcommand{\thedefinition} {\noindent{Definition~\thesection.\arabic{definition}}} \newcommand{\definition}{% \setcounter{definition}{\value{Mycounter}} \refstepcounter{definition} \stepcounter{Mycounter} {\bf \green DEFINITION:\ }} \newcounter{example}[section] \setcounter{example}{0} \renewcommand{\theexample}{\noindent{Example \thesection.\arabic{example}}} \newcommand{\example}{% \setcounter{example}{\value{Mycounter}} \refstepcounter{example} \stepcounter{Mycounter} {\bf \green EXAMPLE:\ }} \newcounter{remark}[section] \setcounter{remark}{0} \renewcommand{\theremark}{\noindent{Remark \thesection.\arabic{remark}}} \newcommand{\remark}{% \setcounter{remark}{\value{Mycounter}} \refstepcounter{remark} \stepcounter{Mycounter} {\bf \green REMARK:\ }} \newcounter{observation}[section] \setcounter{observation}{0} \renewcommand{\theobservation}{\noindent{Question \thesection.\arabic{observation}}} \newcommand{\observation}{% \setcounter{observation}{\value{Mycounter}} \refstepcounter{observation} \stepcounter{Mycounter} {\bf \green OBSERVATION:\ }} \newcounter{question}[section] \setcounter{question}{0} \renewcommand{\thequestion}{\noindent{Question \thesection.\arabic{question}}} \newcommand{\question}{% \setcounter{question}{\value{Mycounter}} \refstepcounter{question} \stepcounter{Mycounter} {\bf \green QUESTION:\ }} \newcommand{\exercise}{{\bf \green EXERCISE:\ }} \newcommand{\proof}{{\bf \green Proof:\ }} \begin{document} \setcounter{page}{1} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{center} {\Large\bf Geometry of manifolds \\[15mm] \small Lecture 10: de Rham algebra} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \\[14mm] {\small Misha Verbitsky } \\[20mm] {\tiny\bf Math in Moscow and HSE \\[2mm] April 22, 2013 } \end{center} \newpage {\bf \blue K\"ahler differentials} \definition Let $R$ be a ring over a field $k$, and $V$ an $R$-module. A $k$-linear map $D:\; R \arrow V$ is called {\bf\blue a derivation} if it satisfies {\bf\blue the Leibnitz identity} $D(ab) = a D(b) + b D(a)$. The space of derivations from $R$ to $V$ is denoted $\Der_k(R, V)$. \remark {\bf \purple $\Der_k(R, V)$ is an $R$-module}, with a natural $R$-action. \definition Let $R$ be a ring over a field $k$. Define an $R$-module $\Omega^1_k R$ {\bf \blue (the module of K\"ahler differentials)} with the following generators and relations.\\ \phantom{xx} * {\bf \red Generators} of $\Omega^1_k R$ are indexed by elements of $R$; for each $a\in R$, the corresponding generator of $\Omega^1_k R$ is denoted $da$. \\ \phantom{xx} * {\bf \red Relations} in $\Omega^1_k R$ are generated by expressions $d(ab) = a db + b da$, for all $a, b \in R$, and $d\lambda =0$ for each $\lambda\in k$. \exercise Prove that the map $d:\; R \arrow \Omega^1_k R$ mapping $a$ to $da$ {\bf \purple is a derivation}. \newpage {\bf \blue Universal property of K\"ahler differentials} \claim Let $V$ be an $R$-module, and $D\in \Der_k(R, V)$ a derivation. Then {\bf \purple there exists a unique $R$-module homomorphism $\phi_D:\; \Omega^1_k R \arrow V$ mapping $bda$ to $bD(a)$.} \remark Consider a category $\cac$ of $R$-modules equipped with a derivation $(V, D:\; R \arrow V)$, and define morphisms in $\cac$ as morphisms of $R$-modules which commute with the derivation map. Then {\bf \purple $\Omega^1 R$ is an initial object in this category.} This is called {\bf \blue the universal property of the module of K\"ahler differentials}. \claim {\bf \red $\Der_k(R,V)=\Hom_R(\Omega^1 R, V)$.} \proof A composition of a derivation $R \stackrel d \arrow \Omega^1 R$ and an $R$-module homomorphism $\Omega^1 R \arrow V$ lies in $\Der_k(R,V)$. On the other hand, {\bf \purple any derivation $\xi\in \Der_k(R,V)$ is obtained this way,} by the universal property. \endproof \corollary {\bf \purple $\Der_k(R)= (\Omega^1 R)^*$,} where $V^*:= \Hom(V, R)$. \newpage {\bf \blue K\"ahler differentials over polynomials} \remark Unlike the derivations, {\bf \blue the K\"ahler differentials are functorial on $R$.} \claim Let $R\stackrel \phi \arrow R'$ be a ring homomorphism. Consider $\Omega^1 R'$ as an $R$-module, using the action $r, a \arrow \phi(r)a$. Then {\bf \purple there exists an $R$-module homomorphism $\Omega^1 R \arrow \Omega^1 R'$, mapping $dr$ to $d\phi(r)$. } \claim Let $R= k[t_1, ..., t_n]$ be a polynomial ring over a field of characteristic 0. {\bf \red Then $\Omega^1_k R$ is a free $R$-module generated by $d t_1, dt_2, ..., dt_n.$} {\bf \green Proof. Step 1:} For each polynomial $P\in R$, the element $dP$ can be expressed as a sum $\sum \frac {d P}{d t_i} dt_i$. {\bf \purple Therefore, each $\alpha\in \Omega^1 R$ cam be written as $\sum_i Q_i dt_i$.} {\bf \green Step 2:} This expression $\alpha=\sum_i Q_i dt_i$ is unique, because the pairing $\Der_k(R)\times \Omega^1 R\arrow R$ maps $\frac d{dt_k}\times \sum_i Q_i dt_i$ to $Q_k$, hence {\bf \purple the coefficients $Q_k\in R$ are determined unambiguously.} \endproof \newpage {\bf \blue Cotangent bundle} \definition Let $A$, $B$ be finitely generated $R$-modules, and $\nu:\; A\times B \arrow R$ a bilinear pairing. Define {\bf\blue the annihilator of $\nu$ in $B$} as a submodule consisting of all elements $b \in B$ for which the homomorphism $\nu(\cdot, b):\; A \arrow R$ vanishes. \definition Let $M$ be a smooth manifold, $R:= C^\infty M$ the ring of smooth finctions, and $\nu:\; \Der(R)\times \Omega^1 R \arrow R$ the pairing obtained from an isomorphism $\Der(R)=(\Omega^1 R)^*$. Consider its annihilator $K\subset \Omega^1 R$. Define {\bf \blue the cotangent bundle} as $\Lambda^1 M:= \Omega^1 R /K$. \claim {\bf \red $\Lambda^1 M$ is generated as a $C^\infty M$-module by $d(C^\infty M)$.} \claim In these assumptions, {\bf \red $\Lambda^1 M$ is an image of the tautological map $\Omega^1 M \stackrel \tau\arrow (\Omega^1 M)^{**}$. Moreover, $\Lambda^1(M)=\Der(R)^*$.} {\bf\green Proof. Step 1:} By construction, {\bf \purple $\Lambda^1 M$ is a quotient of $\Omega^1 M$ by a kernel of a map $\Omega^1 M \stackrel \tau \arrow\Der(R)^*=(\Omega^1 M)^{**}$.} {\bf \purple Therefore, $\Lambda^1 M= \im \tau$.} {\bf \green Step 2:} Let $V\subset \Der(R)^*$ be a $C^\infty M$-submodule generated by the symbols $df\in \Der(R)^*$ which are paired with vector fields $v$ as $\langle v, df\rangle$. {\bf \red Clearly, $V=\im \tau$.} {\bf \purple If $M$ admits coordinates $t_1,..., t_n$, one has $V=\Der(R)^*$,} because $\Der(R)^*$ is freely generated by $\frac d{dt_i}$. \newpage {\bf \blue Cotangent bundle (cont.)} \claim In these assumptions, {\bf \red $\Lambda^1 M$ is an image of the tautological map $\Omega^1 M \stackrel \tau\arrow (\Omega^1 M)^{**}$. Moreover, $\Lambda^1(M)=\Der(R)^*$.} {\bf\green Proof. Step 1:} By construction, {\bf \purple $\Lambda^1 M$ is a quotient of $\Omega^1 M$ by a kernel of a map $\Omega^1 M \stackrel \tau \arrow\Der(R)^*=(\Omega^1 M)^{**}$.} {\bf \purple Therefore, $\Lambda^1 M= \im \tau$.} {\bf \green Step 2:} {\bf \purple If $M$ admits coordinates $t_1,..., t_n$, one has $V=\Der(R)^*$,} because $\Der(R)^*$ is freely generated by $\frac d{dt_i}$. {\bf \green Step 3:} {\bf \purple Using partition of unity, we obtain that $V$ is a subsheaf of $\Der(R)^*=(TM)^*$.} Indeed, suppose that for some covering $\{U_i\}$ one has $\alpha\restrict {U_i} = \sum f_j(i) dg_j(i)$, $\phi_i$ a partition of unity subordinate to $\{U_i\}$, and $\psi_i$ functions with support in $U_i$ satisfying $\psi_i\restrict{\sup \phi_i}=1$. Then $\alpha=\sum \phi_i f_j(i)d(\psi_ig_j(i))$. {\bf \green Step 4:} {\bf \purple Locally in $M$, $V=\Der(R)^*$,} because locally $M$ admits coordinates. Therefore, the sheaves $V$ and $\Der(R)^*$ coincide. \endproof \corollary {\bf \red $\Lambda^1 M$ is a locally free sheaf of $C^\infty M$-modules. Moreover, $\Lambda^1 M=TM^*$. } \corollary {\bf \red $\Lambda^1 M$ is generated as a $C^\infty M$-module by $d(C^\infty M)$.} \newpage {\bf \blue De Rham algebra} \definition Let $M$ be a smooth manifold. {\bf \blue A bundle of differential $i$-forms on $M$} is the bundle $\Lambda^i T^* M$ of antisymmetric $i$-forms on $TM$. It is denoted $\Lambda^i M$. \remark $\Lambda^0 M=C^\infty M$. \definition Let $\alpha\in (V^*)^{\otimes i}$ and $\alpha\in (V^*)^{\otimes j}$ be polylinear forms on $V$. Define the {\bf\blue tensor multiplication} $\alpha \otimes \beta$ as\\ \phantom{xxxxxxx} $\alpha \otimes \beta(x_1, ..., x_{i+j}):= \alpha(x_1, ..., x_j) \beta(x_{i+1}, ..., x_{i+j}).$ \definition Let $\bigotimes_k T^*M \stackrel \Pi \arrow \Lambda^k M$ be the antisymmetrization map, \[ \Pi(\alpha)(x_1, ..., x_n):= \frac 1{n!} \sum_{\sigma\in \Sym_n} (-1)^\sigma \alpha(x_{\sigma_1}, x_{\sigma_2}, ..., x_{\sigma_n}). \] Define {\bf \blue the exterior multiplication $\wedge:\; \Lambda^i M\times \Lambda^j M \arrow \Lambda^{i+j} M$} as $\alpha \wedge \beta := \Pi (\alpha \otimes \beta)$, where $\alpha \otimes \beta$ is a section $\Lambda^i M\otimes \Lambda^j M \subset \bigotimes_{i+j} T^*M$ obtained as their tensor multiplication. \remark The fiber of the bundle $\Lambda^* M$ at $x\in M$ {\bf \purple is identified with the Grassmann algebra $\Lambda^* T^*_x M$.} This identification is compatible with the Grassmann product. \newpage {\bf \blue Coordinate monomials} \definition Let $t_1, ..., t_n$ be coordinate functions on $\R^n$, and $\alpha\in \Lambda^* \R^n$ a monomial obtained as a product of several $dt_i$: $\alpha = dt_{i_1}\wedge dt_{i_2} \wedge ... \wedge dt_{i_k}$ $i_1 < i_2 < ...< i_k$. Then $\alpha$ is called {\bf \blue a coordinate monomial}. \claim {\bf \purple $\Lambda^*\R^n$ is a trivial bundle, and coordinate monomials are free generators of $\Lambda^*\R^n$.} \definition An associative algebra $A^* = \oplus_{i\in \Z}A^i$ is called {\bf \blue a graded algebra} if for all $a\in A^i$, $b\in A^j$, the product $ab$ lies in $A^{i+j}$. \example De Rham algebra is a graded algebra. \newpage {\bf \blue De Rham differential} \definition {\bf\blue De Rham differential} $d:\; \Lambda^*M \arrow \Lambda^{*+1}M$ is an $\R$-linear map satisfying the following conditions.\\ \phantom{xxx} * For each $f \in \Lambda^0M=C^\infty M$, $d(f)\in \Lambda^1 M$ is equal to the image of the K\"ahler differential $df\in \Omega^1 M$ in $\Lambda^1 M = \Omega^1 M/K$.\\ \phantom{xxx} * {\bf \blue (Leibnitz rule)} $d(a\wedge b) = da \wedge b + (-1)^j a\wedge db$ for any $a\in \Lambda^i M, b \in \Lambda^j M$.\\ \phantom{xxx} * $d^2=0$. \remark A map on a graded algebra which satisfies the Leibnitz rule above is called {\bf \blue an odd derivation}. \remark The following two lemmas are needed to prove uniqueness of de Rham differential. \lemma Let $A=\bigoplus A^i$ be a graded algebra, $B\subset A$ a set of multiplicative generators, and $D_1, D_2:\; A \arrow A$ two odd derivations which are equal on $B$. {\bf \purple Then $D_1=D_2$.} \endproof \lemma {\bf \red $\Lambda^* M$ is generated by $C^\infty M$ and $d(C^\infty M)$.} \proof By definition, $\Lambda^* M$ is generated by $\Lambda^0 M = C^\infty M$ and $\Lambda^1 M$. However, $d(C^\infty M)$ generate $\Lambda^1 M$, as shown above. \endproof \newpage {\bf \blue De Rham differential: uniqueness and existence} \theorem\\ {\bf \red De Rham differential is uniquely determined by these axioms.} \proof De Rham differential is an odd derivation. Its value on $C^\infty M$ is defined by the first axiom. On $d(C^\infty M)$ de Rham differential valishes, because $d^2=0$. \endproof \definition Let $t_1, ..., t_n$ be coordinate functions on $\R^n$, $\alpha_i$ coordinate monomials, and $\alpha:= \sum f_i \alpha_i$. Define $d(\alpha):= \sum_i \sum_j\frac{df_i}{dt_j}dt_j \wedge \alpha_i$. \exercise\\ {\bf \purple Check that $d$ satisfies the properties of de Rham differential.} \corollary {\bf \red De Rham differential exists on any smooth manifold.} \proof Locally, de Rham differential $d$ exists, as follows from the construction above. Since $d$ is unique, it is compatible with restrictions. {\bf \purple This means that $d$ defines a sheaf morphism.} Restricting this sheaf morphism to global sections, we obtain de Rham differential on $\Lambda^* M$. \endproof \newpage {\bf \blue Superalgebras} \definition Let $A^* = \oplus_{i\in \Z}A^i$ be a graded algebra over a field. It is called {\bf\blue graded commutative}, or {\bf\blue supercommutative}, if $ab = (-1)^{ij} ba$ for all $a\in A^i, b \in A^j$. \example {\bf \purple Grassmann algebra $\Lambda^* V$ is clearly supercommutative.} \definition Let $A^*$ be a graded commutative algebra, and $D:\; A^* \arrow A^{*+i}$ be a map which shifts grading by $i$. It is called a {\bf\blue graded derivation} if $D(ab) = D(a) b + (-1)^{ij} a D(b)$, for each $a \in A^j$. \remark If $i$ is even, graded derivation is a usual derivation. If it is even, it an odd derivation. \definition Let $M$ be a smooth manifold, and $X\in TM$ a vector field. Consider an operation of {\bf \blue convolution with a vector field} $i_X:\; \Lambda^i M \arrow \Lambda^{i-1}M$, mapping an $i$-form $\alpha$ to an $(i-1)$-form $v_1, ..., v_{i-1} \arrow \alpha(X, v_1, ..., v_{i-1})$ \exercise {\bf \purple Prove that $i_X$ is an odd derivation.} \newpage {\bf \blue Supercommutator} \definition Let $A^*$ be a graded vector space, and $E:\; A^*\arrow A^{*+i}$, $F:\; A^*\arrow A^{*+j}$ operators shifting the grading by $i, j$. Define {\bf\blue the supercommutator} $\{E, F\}:= EF - (-1)^{ij} FE$. \definition An endomorphism which shifts a grading by $i$ is called {\bf \blue even} if $i$ is even, and {\bf \blue odd} otherwise. \exercise Prove that a supercommutator satisfies {\bf graded Jacobi identity,} \[ \{ E, \{F, G\}\} = \{\{ E, F\}, G\} + (-1)^{\tilde E \tilde F} \{ F, \{E, G\}\} \] where $\tilde E$ and $\tilde F$ are 0 if $E, F$ are even, and 1 otherwise. \remark There is a simple mnemonic rule which allows one to remember a superidentity, if you know the commutative analogue. Each time when in commutative case two letters $A$, $F$ are exchanged, in supercommutative case one needs to multiply by $(-1)^{\tilde E\tilde F}$. \exercise {\bf \purple Prove that a supercommutator of superderivations is again a superderivation.} \newpage {\bf \blue Lie derivative} \definition Let $B$ be a smooth manifold, and $v\in TM$ a vector field. An endomorphism $\Lie_v:\; \Lambda^* M \arrow \Lambda^* M$, preserving the grading is called {\bf\blue a Lie derivative along $v$} if it satisfies the following conditions.\\ \phantom{XX} (1) On functions $\Lie_v$ is equal to a derivative along $v$. (2) $[\Lie_v, d]=0$.\\ \phantom{XX} (3) $\Lie_v$ is a derivation of the de Rham algebra. \remark The algebra $\Lambda^*(M)$ is generated by $C^\infty M=\Lambda^0(M)$ and $d(C^\infty M)$. The restriction $\Lie_v\restrict{C^\infty M}$ is determined by the first axiom. On $d(C^\infty M)$ is also determined because $\Lie_v(df) =d(\Lie_v f)$. {\bf \red Therefore, $\Lie_v$ is uniquely defined by these axioms.} \lemma {\bf \purple $\{d, \{d, E\}\}=0$ for each $E\in \End(\Lambda^* M)$.} \proof By the super Jacobi identity, $\{d, \{d, E\}\}=-\{d, \{d, E\}\}+\{\{d, d,\} E\}\}$, however, $\{d, d\} =2d^2 =0$. \endproof \theorem {\bf \blue (Cartan's formula)} Let $i_v$ be a convolution with a vector field. {\bf \red Then $\{d, i_v\}$ is a Lie derivative along $v$.} \proof $\{d, \{d, i_v\}\}=0$ by the lemma above. A supercommutator of two derivations is a derivation. Finally, $\{d, i_v\}$ acts on functions as $i_v(df)=\langle v, df\rangle.$ \endproof \end{document}