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            lecture 10 \hfil
  \tiny M. Verbitsky }}
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\begin{document}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\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}


