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\begin{document}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{center}
{\Large\bf  Topologia das Variedades \\[15mm]
\small Cohomology, lecture 1: Grassmann algebra}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\\[14mm]

{\small\bf Misha Verbitsky  
\\[2mm]
April 24, 2018\\[4mm]

{\tiny\bf IMPA, Tuesdays and Thursdays, 10:30, Sala 224}
}
\end{center}

\newpage

{\bf \blue Analysis situs}


\centerline{\epsfig{file=Analysis-situs.JPEG,width=0.42\linewidth}}

{\it\small
"Analysis Situs" is a seminal mathematics paper that Henri
Poincar\'e published in 1895. Poincar\'e published five
supplements to the paper between 1899 and 1904.}


\newpage

{\bf \blue Bilinear maps}

\definition
Let  $U$, $V$, $W$ be vector spaces over $k$.
A map $U\times V \overset{\mu}{\to} W$, $u, v \mapsto
\mu(u, v)$ is called {\bf \blue bilinear}
if for all $u\in U$ and $v\in V$ the maps
$\mu(u, \cdot):\; V \to W$ É $\mu(\cdot, v):\; U \to W$
are linear.

\remark Clearly, a linear combination of bilinear maps is bilinear.
Then {\bf \purple the space $\Bil(U\times V, W)$
of bilinear maps $U\times V \to W$ is a vector space}.

\definition
{\bf \blue Bilinear form} on $V$ is a bilinear map
$V\times V \overset{\mu}{\to} k$. {\bf \blue Bilinear symmetric form}
is a form which satisfies $\mu(x,y) = \mu(y,x)$ for all $x, y\in V$.
{\bf \blue Bilinear anti-symmetric form}m or
{\bf\blue bilinear skew-symmetric form} is a form
which satisfies $\mu(x,y) = -\mu(y,x)$.
We denote the first by $\Sym^2 V^*$,
and the second by  $\Lambda^2 V^*$.

\newpage


{\bf \blue Tensor product}

\definition
Let $S$ be a set. Define {\bf \blue vector space, freely generated by $S$}
as the space of functions $\psi:\; S \arrow k$ which are equal zero
outside of a finite subset $\Sup_\psi\subset S$.

\definition
Let $V,V'$ be vector spaces over $k$, and $W$ a vector space
freely generated by $v\otimes v'$, with $v\in V, v'\in V'$,
and $W_1\subset W$ a subspace generated by combinations
$a v \otimes v'-v\otimes av'$, $a (v \otimes v')-(av)\otimes v'$,
$(v_1+ v_2)\otimes v'-
v_1 \otimes v' - v_2 \otimes v'$ and $v\otimes (v'_1+ v'_2)-
v\otimes v'_1 - v\otimes v'_2$, where $a\in k$.
Define {\bf\blue the tensor product} $V \otimes_k V'$
as a quotient vector space $W/W_1$.

\proposition
{\bf\red For any vector spaces $V, V'$, $R$, there is a natural identification
 $\Hom(V\otimes_k V', R)= \Bil(V\times V', R)$.}

\proof
Clearly, any bilinear map $\rho \in \Bil(V\times V', R)$
defines a linear map $\tilde \rho:\; W \arrow R$, and
$\tilde \rho$ vanishes on $W_1$. This 
gives a map $\Bil(V\times V', R)\arrow \Hom(V\otimes_k V', R)$.
Inverse map takes  $\tau \in \Hom(V\otimes_k V', R)$ and interprets 
it as a bilinear map in $\Bil(V\times V', R)$.
\endproof


\corollary {\bf \purple For finite-dimensional $V, V'$, one has
$V\otimes_k V'=\Bil(V\times V', k)^*$.}

\newpage

{\bf \blue Dimension of the tensor product}

\claim {\bf \red Dimension of $\Bil(V\times V', k)$
is equal to $\dim V \dim V'$. }

\pstep Let $\{\lambda_i\}$ be a basis in $V^*$ and
$\{\lambda'_i\}$ a basis in $V'$.  Denote by  $\{v_i\}$
 $\{v_i'\}$ the dual basis in $V, V'$. 
Then $\lambda_i \lambda'_j$
can be interpreted as vectors in $\Bil(V\times V', k)$.
These vectors are clearly linearly independent:
indeed \[\sum_{i,j} a_{ij} \lambda_i \lambda'_j(v_p, v_q')= a_{pq}.\]
This gives $\dim \Bil(V\times V', k)\geq\dim V \dim V'$.

{\bf \green Step 2:} 
On the other hand, $\dim V\otimes V'\leq \dim V \dim V'$,
because it is generated by $v_p \otimes v_q$, hence
$\dim \Bil(V\times V', k)\leq\dim V \dim V'$. 
\endproof

\corollary Let $\{x_i\}$ and $\{y_i\}$ be bases in $V, W$.
{\bf \red Then $\{x_i \otimes y_j\}$ is a basis in $V\otimes_k W$.}
\endproof


\newpage

{\bf \blue Algebra over a field}

Fix a ground field $k$. Recall that 
a map $(V_1\times V_2) \overset{\mu}{\arrow}
V_3$ of vector spaces is called {\bf \blue bilinear} of
for any  $v_1\in V_1$, $v_2 \in V_2$,
the maps
$\mu(v_1, \cdot ):\; V_2 \arrow V_3, \ \mu(\cdot, v_2):\; V_1 \arrow V_3$
(one element is fixed) is $k$-linear.

To express this, we use the tensor product sign, and write
$\mu:\; V_1\otimes V_2 \arrow V_3.$

\definition
Let $A$ be a vector space over $k$, and 
 $\mu:\; A \otimes A \arrow A$ a bilinear map
(called {\bf \blue ``multiplication''}).
The pair $(A, \mu)$ is called {\bf \blue
algebra over a field $k$} if $\mu$ is {\bf \blue associative}:
$\mu(a_1, \mu(a_2, a_3)) = \mu(\mu(a_1, a_2), a_3)).$
The product in algebra is written as $a\cdot b$ or $ab$.
If, in addition, there is an element $1\in A$ such that
$\mu(1,a) = \mu(a,1) =a$ for all $a\in A$, 
this element is called {\bf \blue unity}, a
and $A$ {\bf \blue an algebra with unity}.

\definition
A {\bf \blue homomorphism} of algebras
$r:\; A \arrow A'$ is a linear map which
is compatible with a product. {\bf \blue
Isomorphism} of algebras is an invertible 
homomorphism. {\bf \blue Subalgebra} of
an algebra $A$ is a vector subspace which
is closed under multiplication.

\newpage

{\bf \blue Tensor algebra}

Let $V \times W$ map to $V\otimes W$ by putting
$v, w$ to $v \otimes w$. Clearly, this map is bilinear.
Similarly, one has a bilinear map
$V^{\otimes m}\times V^{\otimes n}\arrow V^{\otimes {m+n}}$
putting $x_1 \otimes ... \otimes x_n, y_1 \otimes ... \otimes y_n$
to $x_1 \otimes ... \otimes x_n\otimes y_1 \otimes ... \otimes y_n$.


\definition
Let $V$ be a vector space over $k$.
{\bf \blue Tensor algebra}, or {\bf \blue free algebra
generated by $V$} is $T(V):=\bigoplus_i V^{\otimes^i}$
equipped with the multiplicative structure defined above,

\exercise 
Prove that {\bf \purple $T(V^*)$ is the algebra of polylinear forms on
$V$ defined above.}

\remark If $x_1, ..., x_r$ is a basis in $V$,
then {\bf\purple the basis in $V^{\otimes^n}$ is formed by all different
monomials} of the form $x_{i_1} \otimes x_{i_2}\otimes ... \otimes x_{i_n}$.

\newpage

{\bf \blue Two-sided ideals }

\definition
Let $A$ be an algebra and $J\subset A$ its subspace.
Then $J$ is caled {\bf \blue left ideal}
if for all $a\in A, j\in J$, one has $ja\in J$, and
{\bf \blue right ideal} if one has $aj\in J$.
$J$ is called {\bf \blue two-sided ideal} if 
is is both right and left ideal.

\remark
Let $J\subset A$ be a two-sided ideal, 
$x, y \in A/J$ some vectors, and $\tilde x, \tilde y$
 Define the product $xy \in A/J$ by 
putting $x \cdot y$ to the class represented by
$\tilde x, \tilde y$. Since $ja\in J$
and $aj\in J$, {\bf \purple this gives a bilinear map 
$A/J \otimes A/J\arrow A/J$}, defining
an associative multiplicative structure
on $A/J$.

\claim
In these assumptions, {\bf \red $A/J$ is an algebra,}
with the product defined as above.

\exercise
{\bf \purple Prove it.}

\newpage

{\bf \blue Algebra defined by generators and relations}


\definition
Let $V$ be a vector space over $k$ {\bf \blue (``the space of generators'')},
and $W\subset T(V)$ another vector space {\bf \blue (``the space of relations'')}.
Consider a quotient $A$ of $T(V)$ by the subspace  $T(V) W T(V)$
generated by the vectors $v\otimes w \otimes v'$, where
$w\in W$ and $v, v' \in T(V)$. 

\claim 
{\bf \red There is a natural product structure on the space
$A:= \frac{T(V)}{T(V) W T(V)}$.}

\proof $T(V) W T(V)$ is a 2-sided ideal. \endproof

\definition
In this situation, we say that $A$ {\bf \blue is an algebra
defined by generators and relations}.

\exercise
{\bf \purple Prove that any algebra can be defined by 
generators and relations.}

\definition
An algebra is called {\bf \blue finitely generated}
if it can be defined by generators and relations,
and the space of generators is finitely-dimensional.
An algebra is called {\bf \blue finitely presented}
if the space $W$ of relations is finitely-dimensional.


\newpage

{\bf \blue Graded algebras}

\definition
An algebra $A$ is called {\bf\blue graded} if $A$ is 
represented as $A= \bigoplus A^i$, where $i\in \Z$,
and the product satisfies $A^i \cdot A^j
\subset A^{i+j}$. Instead of $\bigoplus A^i$
one often writes  $A^*$, 
where $*$ denotes all indices together.
Some of the spaces $A^i$ can be zero, but the 
ground field is always in $A^0$, so that it is non-empty. 

\example
{\bf \purple The tensor algebra 
$T(V)$ and the polynomial algebra $\Sym^*(V)$
are obviously graded.}


\definition
A subspace $W\subset A^*$ of a graded algebra
is called {\bf\blue graded} if $W$ is a direct sum of components
$W^i \subset A^i$. 

\exercise
Let $W\subset  T(V)$ be a graded subspace.
Prove that then {\bf \purple the algebra generated by $V$ with relation
space $W$ is also graded.}

\newpage

{\bf \blue The Grassmann algebra}

\definition
Let $V$ be a vector space, and $W\subset V\otimes V$
a subspace generated by vectors $x\otimes
y + y\otimes x$ and $x\otimes x$, for all $x, y\in V$.
A graded algebra defined by the generator space $V$ and
the relation space $W$ is called {\bf\blue Grassmann algebra},
or {\bf\blue exterior algebra}, and denoted
$\Lambda^*(V)$. The space $\Lambda^i(V)$
is called {\bf\blue $i$-th exterior power} of $V$, and
the multiplication in $\Lambda^*(V)$ -- 
{\bf\blue exterior multiplication}. Exterior multiplication
is denoted $\wedge$.


\exercise
{\bf \purple Prove that $\Lambda^1 V$ is isomorphic to $V$.}


\definition
An element of Grassmann algebra is called {\bf\blue even}
if it lies in $\bigoplus_{i\in \Z} \Lambda^{2i}(V)$
and {\bf \blue odd} if it lies in $\bigoplus_{i\in \Z} \Lambda^{2i+1}(V)$.
For an even or odd $x\in \Lambda^*(V)$, we define a number
$\tilde x$ called {\bf\blue parity} of $x$.
The parity of $x$ is 0 for even $x$ and 1 for odd.

\claim
{\bf \purple In Grassmann algebra, 
$x\wedge y = (-1)^{\tilde x \tilde y}y\wedge x$.}


\newpage

{\bf \blue Antisymmetric tensors}

\definition
Let $V^{\otimes^n}$ be $n$-th product of $V$ with itself,
equipped with the natural symmetric group $\Sigma_n$-action exchanging
the tensor components. A tensor 
$\psi \in V^{\otimes^n}$ is called {\bf \blue antisymmetric}
if for any permutation $\sigma \in \Sigma_n$
we have $\sigma(\psi)=(-1)^{\tilde \sigma}\psi$,
and {\bf \blue symmetric} if $\sigma(\psi)=\psi$.
We denote the space of all antisymmetric tensors by 
$\tilde \Lambda^n V$ and the space of symmetric
tensors by $\widetilde \Sym^n V$.

{\bf \green Theorem 1:}
Let $V$ be a vector space,
$\Lambda^n V$ the $n$-th component of 
its Grassmann algebra, and $\Sym^n V$ the
$n$-th component of its symmetric algebra. 
Then {\bf \red $\Lambda^n V$ is naturally 
identified with $\tilde \Lambda^n V$, and
$\Sym^n V$ with $\widetilde \Sym^n V$:}
the projection from $V^{\otimes n}$ to 
 $\Lambda^n V$ (or $\Sym^n V$) induces an
isomorphism from $\tilde \Lambda^n V$ (or $\widetilde\Sym^n V$)
to $\Lambda^n V$ (or $\Sym^n V$).

\newpage

{\bf \blue Antisymmetrization}

The natural map from $\Lambda^n V$ to $\tilde \Lambda^n V$
is given by {\bf \blue antisymmetrization}
\[\Alt(x_1\otimes ...\otimes x_n):= 
\frac 1{n!}\sum_{\sigma\in \Sigma_n}(-1)^{\tilde \sigma}
x_{\sigma(1)}\otimes ...\otimes x_{\sigma(n)}.
\]
{\bf \green Its properties:}

1. Clearly, $\im \Alt$ lies in the space of antisymmetric 
tensors, and $\Alt(\eta - (-1)^{\tilde \sigma} \eta)=0$,
hence {\bf \purple $\Alt$ defines a map from Grassmann algebra
to the space of antisymmetric tensors.}

2. The natural {\bf \purple projection 
from antisymmetric tensors to the Grassmann algebra is
inverse to $\Alt$.}

\newpage

{\bf \blue Grassmann algebra: dimension of components}

\remark
For linearly independent vectors $x_1, ..., x_k$,
the antisymmetrization 
$x_1 \wedge x_2 \wedge ... \wedge x_k :=\frac {1}{k!} \sum (-1)^{\tilde \sigma}
x_{\sigma(1)}\otimes ...\otimes x_{\sigma(k)}$
is non-trivial. Indeed, the monomials
$x_{\sigma(1)}\otimes ...\otimes x_{\sigma(k)}$
are linearly independent in $V^{\otimes k}$.
{\bf \purple This implies $\dim \Lambda^k V\geq \binom{\dim V}{k}$}.

\claim {\bf \red $\dim \Lambda^k V= 
\binom{\dim V}{k}$,  and
$\dim \Lambda^* V=2^{\dim V}$.}

\proof Let $x_1, ..., x_n$ be a basis in $V$.
Then {\bf \purple the space $\Lambda^k V$ is generated
by antisymmetric tensors $x_{i_1} \wedge x_{i_2}  \wedge ... \wedge x_{i_k}$,
$i_1 < i_2 < ... < i_k$,
which are all linearly independent.}
\endproof

\newpage

{\bf \blue Tensor product of vector bundles}


\definition
Let $V,V'$ be $R$-modules, $W$ a free abelian group
generated by $v\otimes v'$, with $v\in V, v'\in V'$,
and $W_1\subset W$ a subgroup generated by combinations
$rv \otimes v'-v\otimes rv'$, $(v_1+ v_2)\otimes v'-
v_1 \otimes v' - v_2 \otimes v'$ and $v\otimes (v'_1+ v'_2)-
v\otimes v'_1 - v\otimes v'_2$.
Define {\bf\blue the tensor product} $V \otimes_R V'$
as a quotient group $W/W_1$.

\exercise
Show that $r \cdot v\otimes v'\mapsto (rv)\otimes v'$
{\bf \purple defines an $R$-module structure on $V \otimes_R V'$.}


\definition
{\bf \blue Tensor product} of vector bundles
is a tensor product of the corresponding sheaves of modules.

\exercise 
 Let $B$ and $B'$ ve vector bundles on $M$,
$B\restrict x$, $B'\restrict x$ their fibers, and
$B\otimes_{C^\infty M} B'$ their tensor product.
{\bf \purple
Prove that $B\otimes_{C^\infty M} B'\restrict x = 
B\restrict x\otimes_\R B'\restrict x$.}

\newpage

{\bf \blue Cotangent bundle}

\definition
Let $M$ be a smooth manifold, 
$TM$ the tangent bundle, and $\Lambda^1 M = T^*M$ its dual bundle.
It is called {\bf \blue cotangent bundle} of $M$.
Sections of $T^*M$ are called {\bf \blue 1-forms}
or {\bf \blue covectors} on $M$.
For any $f\in C^\infty M$, consider a functional
$TM\arrow C^\infty M$ obtained by mapping $X\in TM$ to
a derivation of $f$: $X \arrow D_X(f)$. Since
this map is linear in $X$, it defines a section
$df\in T^*M$ called {\bf \blue the differential} of $f$.

\claim {\bf \red $\Lambda^1 M$ is
generated as a $C^\infty M$-module by
$d(C^\infty M)$.}

\proof Locally in coordinates $x_1, ..., x_n$ this is clear,
because the covectors $dx_1, ..., dx_n$ give a basis
in $T^* M$ dual to the basis $\frac d{dx_1}, ..., \frac d{dx_n}$ in 
$TM$. To obtain the same statement globally, use the
partition of unity. 
\endproof


\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 $\Lambda^0 M=C^\infty M$.




\end{document}





