\magnification 1400
\parskip 10pt
\parindent 0pt
\hoffset -0.3truein
\hsize 7truein

\voffset -0.4truein
\vsize 10truein

\def\emph{}

%%\def\S{\Sigma_{i=1}^n}
\def\Sm{\Sigma_{i=1}^m}
\def\S{\sum_{i=1}^n}

\def\s{{\sigma}}
\def\ep{\epsilon}
\def\f{\vskip 10pt}
\def\h{\hskip 10pt}
\def\u{\vskip -8pt}
\def\bh{\hfil \break}

\def\N{{\bf N}}
\def\S{\Sigma_{i=1}^n}

\def\0{{\bf 0}}
\def\Z{{\bf Z}}
\def\R{{\bf R}}
\def\C{{\bf C}}
\def\x{{\bf x}}
\def\a{{\bf a}}
\def\y{{\bf y}}
\def\e{{\bf e}}
\def\i{{\bf i}}
\def\j{{\bf j}}
\def\k{{\bf k}}
\def\p{{\bf p}}
\def\f{\phi}
\def\s{\sigma}
\def\ep{\epsilon}
\def\f{\vskip 10pt}
\def\u{ \vskip -5pt }
\def\h{\vskip -5pt \hskip 20pt}
\def\grad{ \nabla }

\def\tens{ \otimes }
\def\l{\vskip 5pt
\hrule
\vskip -5pt}

Let $v_1, ..., v_m$ be a basis for $V$.

The exterior power of $V =$ $$\Lambda^pV = \{\Sigma a_{i_1}...a_{i_p} v_{i_1} \wedge ... \wedge 
v_{i_p} ~|~ a_{i_1}...a_{i_p} \in \R , i_1 < ... < i_p\}.$$

I.e., $\Lambda^pV$ is the vector space over $\R$ with basis consisting of the
$ \left(\matrix{n \cr p}\right)$ elements $v_{i_1} \wedge ... \wedge 
v_{i_p}$.

Recall  $v \wedge v = 0$ implies $v_i \wedge v_j = v_j \wedge v_i$



$\Lambda^*V = \cup_{p = 0}^\infty \Lambda^pV$ is a graded associative algebra with addition + and multiplication $\wedge$.

If $v \in \Lambda^pV$ then $v$ has degree $p$ (i.e., it's in the $p$th grade).

$\Lambda^0V = \R$ which is generated by 1, the multiplicative identity for this algebra.  

The wedge product is bilinear:
{$(c_1 v_1 + c_2v_2) \wedge w = c_1 (v_1 \wedge w) + c_2 (v_2 \wedge w)$} 
and \bh
 $v \wedge (c_1w_1 + c_2w_2) = c_1(v \wedge w_1) +  c_2(v \wedge w_2)$

Prop 5.2:  $\alpha \in \Lambda^pV$, $\beta \in  \Lambda^q V$, then $\alpha \wedge \beta = (-1)^{pq} \beta \wedge \alpha$.

Prop 5.6: If $A:  V \rightarrow W$ is a linear transformation, then 
$\Lambda^pA: \Lambda^pV \rightarrow
\Lambda^pV$ defined by   $\Lambda^pA(v_1 \wedge ... \wedge v_p) =   
Av_1 \wedge ... \wedge Av_p$ is a linear transformation which is independent of the choice of basis vectors.    


Prop 5.7:  If $dim V = n$,  $\Lambda^nA: \Lambda^nV \rightarrow
\Lambda^nV$ is given by   $\Lambda^nA(c v_1 \wedge ...  \wedge v_n) =   
(det~A)(c v_1 \wedge ... \wedge v_n)$.    

\l

Defn:  The dual of $T_pM= T_p^*M$ is the {\it cotangent space} to $M$ at $p$.

If $ {\partial \over \partial x_1}, ..., {\partial \over \partial x_m}$ is a basis for $T_pM$, then the  dual basis will be denoted $dx_1, ..., dx_m$. 

$\Lambda^pT_x^*M$ is the p-fold exterior power of the vector space $T_x^*M$  

\eject

The {\it bundle of p-forms} = $\Lambda^pT^*M = \cup_{x \in M} \Lambda^pT_x^*M$ 
  a manifold of dimension $m + {\left(\matrix{m \cr p}\right)}$ . 

Let $\pi: \Lambda^pT^*M \rightarrow M$, $\pi(x, v) = x$.

If $(U, \phi)$ is a chart for $M$, then the following is a chart for $\Lambda^pT^*M$

$\Phi: \pi^{-1}(U) \rightarrow \phi(U) \times \Lambda^p \R^m \subset \R^m \times
 \R^{\left(\matrix{m \cr p}\right)}$ 

$\Phi(x, v) = (\phi(x), y_1, ..., y_{\left(\matrix{m \cr p}\right)})$  where $y_i$ are the coefficients of $v$ in terms of a given basis for  $\Lambda^pT_x^*M$ 


A {\it differential p-form} is a section of $\Lambda^pT^*M$

I.e, a vector field = smooth function $s: M \rightarrow \Lambda^pT^*M$, $s(x) = (x, v_x)$
 

Ex:  0-form = smooth function.

1-form: $\Sigma a_i(x) dx_i$, locally defined.

\l

6.2:  Partitions of unity

Suppose $f:  M \rightarrow \R$.

Defn:  $supp ~f = \overline{\{\x \in M ~|~ f(x) \not= 0\}}$

Defn:  $\{D_\alpha\}_{\alpha \in A}$ is locally finite if $\forall x \in M$, there exists $U$ open such that $U \cap D_\alpha = \emptyset$  for all but finitely many $\alpha$. 

Defn:  A partition of unity on $M$ is a countable collection 
$\{f_i\}_{i \in I}$ of smooth functions such that 

1.)  $f_i \geq 0$
\hfill
2.)  $\{supp ~f_i ~|~ i \in I\}$ is locally finite
\hfill
3.) $\Sigma_i f_i = 1$

Thm 6.1:  Given any open covering $\{V_\alpha \}$ of a manifold $M$, there 
exists a partition of unity $\{f_i\}$ such that $supp ~f_i \subset 
V_{\alpha_i}$ for some ${\alpha_i}$.

Such a partition of unity is said to be {\it subordinate} to the covering $\{V_\alpha \}$.

\l

Prop: Change of basis:  $dy_{i_k} = \Sigma {\partial y_{i_k} \over 
\partial 
x_j} dx_j$

\eject

6.4  The exterior derivative.


Let $\Omega^p(M)$ = vector space of all $p$-forms on $M$

$d: \Omega^p(M) \rightarrow \Omega^{p+1}(M)$

If $f \in \Omega^0(M)$ (i.e., $f$ is smooth):

 $df = \S {\partial f \over\partial x_i} dx_i$  = $({\partial f \over\partial x_1} , ..., {\partial f \over\partial x_m} )$

where basis = $\{dx_1, ..., dx_n \}$

If $\alpha = \S a_{i_1...i_p}(x) dx_{i_1} \wedge ... \wedge dx_{i_p} \in  \Omega^p(M) $,

let $d\alpha =  \S da_{i_1...i_p}(x)\wedge dx_{i_1} \wedge ... \wedge dx_{i_p} \in  \Omega^p(M)$
 
Note $d^2 \alpha = 0$ and $d(\alpha \wedge \beta) =   d\alpha \wedge 
\beta 
+
(-1)^p \alpha \wedge d\beta$ for $\alpha \in  \Omega^p(M)$

\l
 
Ex:  If $\alpha =  a_1(x) dx_1 + a_2(x) dx_2 +  a_3(x) dx_3$

$d\alpha =  da_1(x) dx_1 + da_2(x) dx_2 + da_3(x) dx_3
=$

$ [ {\partial a_1 \over\partial x_1}dx_1 + {\partial a_1 \over\partial x_2}dx_2 +  
{\partial a_1 \over\partial x_3}dx_3 ]
\wedge dx_1 + 
[ {\partial a_2 \over\partial x_1}dx_1 + {\partial a_2 \over\partial x_2}dx_2 +  
{\partial a_2 \over\partial x_3}dx_3 ]
\wedge dx_2 +
[ {\partial a_3 \over\partial x_1}dx_1 + {\partial a_3 \over\partial x_2}dx_2 +  
{\partial a_3 \over\partial x_3}dx_3 ]
\wedge dx_3$


$ [ 0 + {\partial a_1 \over\partial x_2}dx_2 \wedge dx_1  +  
{\partial a_1 \over\partial x_3}dx_3 
\wedge dx_1] + 
[ {\partial a_2 \over\partial x_1}dx_1 \wedge dx_2+ 0 + 
{\partial a_2 \over\partial x_3}dx_3 
\wedge dx_2] +
[ {\partial a_3 \over\partial x_1}dx_1 \wedge dx_3 
+ {\partial a_3 \over\partial x_2}dx_2 
\wedge dx_3 + 0 ]$


$ [ 
{\partial a_3 \over\partial x_2}
 -
{\partial a_2 \over\partial x_3}]dx_2 
\wedge dx_3 +
[{\partial a_1 \over\partial x_3} - {\partial a_3 \over\partial x_1}] dx_3 
\wedge dx_1 +
[ {\partial a_2 \over\partial x_1} -{\partial a_1 \over\partial x_2}] dx_1 \wedge dx_2  $


$ ( 
{\partial a_3 \over\partial x_2}
 -
{\partial a_2 \over\partial x_3}, {\partial a_1 \over\partial x_3} - {\partial a_3 \over\partial x_1}, {\partial a_2 \over\partial x_1} -{\partial a_1 \over\partial x_2})
= \nabla \times (a_1, a_2, a_3)$

Note $d^2 \alpha = 0$

Ex:

$d([ 
{\partial a_3 \over\partial x_2}
 -
{\partial a_2 \over\partial x_3}]dx_2 
\wedge dx_3 +
[{\partial a_1 \over\partial x_3} - {\partial a_3 \over\partial x_1}] dx_3 
\wedge dx_1 +
[ {\partial a_2 \over\partial x_1} -{\partial a_1 \over\partial x_2}] dx_1 \wedge dx_2)  $

$= 
d[{\partial a_3 \over\partial x_2}
 -
{\partial a_2 \over\partial x_3}]dx_2 
\wedge dx_3 +
d[{\partial a_1 \over\partial x_3} - {\partial a_3 \over\partial x_1}] dx_3 
\wedge dx_1 +
d[ {\partial a_2 \over\partial x_1} -{\partial a_1 \over\partial x_2}] dx_1 \wedge dx_2)  $


$= 
[\S {\partial^2 a_3 \over\partial x_2 \partial x_i}
 -
{\partial^2  a_2 \over\partial x_3 \partial x_i} dx_i] \wedge dx_2 
\wedge dx_3 +
[\S {\partial^2  a_1 \over\partial x_3 \partial x_i} - {\partial^2  a_3 \over\partial x_1 \partial x_i} dx_i] \wedge dx_3 
\wedge dx_1 +
[\S  {\partial^2  a_2 \over\partial x_1 \partial x_i} -{\partial^2  a_1 \over\partial x_2 \partial x_i} dx_i] \wedge dx_1 \wedge dx_2  $
= 0




\end


























Let $V$ be a finite-dimensional vector space over $\R$.

The dual of $V$ = $V^* = \{f:  V \rightarrow \R ~|~ f$ linear $\}$

Note $V^*$ is a vector space.  The elements of $V^*$ are called {\it covectors}.


If $e_1, ..., e_n$ basis for $V$, then $w_1, ..., w_n$ basis for $V^*$ where 
$w_i: V \rightarrow \R$ where $w_i(e_j) = \delta_{ij} = \cases{1 & $i = j$ \cr 0 & $i \not= j$}$


Let $F_*:  V \rightarrow W$ be a linear map between vector spaces \bh
The dual map map is $F^*:  W^* \rightarrow V^*$, $F(g) = g \circ F$.

$F_*$ is injective implies $F^*$ injective 

$F_*$ is surjective implies $F^*$ surjective 


$(G_* \circ F_*)^* = F^* \circ G^*$.


$d:  V \rightarrow (V^*)^*$, $d(v) = h$ where $h: V^* \rightarrow R$, $h(f) = f(v)$.

Thus $ (V^*)^*$ is naturally isomorphic to $V$.

\l

Defn:  The dual of $T_pM= T_p^*M$ is the {\it cotangent space} to $M$ at $p$.

If $ {\partial \over \partial x_1}, ..., {\partial \over \partial x_m}$ is a basis for $T_pM$, then the  dual basis will be denoted $dx_1, ..., dx_m$. 

%%Defn:  The cotangent bundle of M 

%%$B(v_1 + v_2, w) = B(v_1, w) + B(v_2, w)$
%%$B(v, w_1 + w_2) = B(v, w_1) + B(v, w_2)$
\eject

$B$ is bilinear if \bh
$B(cv_1 + dv_2, w) = cB(v_1, w) + dB(v_2, w) $ \bh
$B(v, cw_1 + dw_2) = cB(v, w_1) + dB(v, w_2)$

Thus



$B( (v_1, w_1)  + (v_2, w_2)) = 
B(v_1 + v_2, w_1 + w_2)\bh = B(v_1, w_1 + w_2)  + B(v_2, w_1 + w_2) \bh
= B(v_1, w_1) + B(v_1, w_2)  + B(v_2, w_1)  + B(v_2, w_2) $

\l
$B$ is linear if \bh
 $B( (v_1, w_1)  + (v_2, w_2)) =  B( (v_1, w_1))  + B((v_2, w_2))$ \bh
$B( c(v_1, w_1)) = cB( (v_1, w_1) $   
\l





\l

Let $( V \times W )^b = \{B: V \times W \rightarrow \R ~|~ B$ bilinear $\}$

$V \tens W = [( V \times W )^b ]^* = \{h: ( V \times W )^b \rightarrow \R ~|~ h$ linear $\}$

If $v \in V, w \in W$, let $v \tens w:  ( V \times W )^b \rightarrow \R, ~  
(v \tens w)(B) = B(v,w)$



Prop:  $v \tens w \in V \tens W$.

Proof:  $(v \tens w)(c_1B_1 + c_2B_2) =  (c_1B_1 + c_2B_2) (v,w) = c_1B_1(v,w) + c_2B_2(v,w)= 
c_1(v \tens w)(B_1) + c_2(v \tens w)(B_2) $.

Prop:  $\S r_i(v_i \tens w_i)  \in V \tens W$.


Proof:  $ (\S r_i(v_i \tens w_i)) (c_1B_1 + c_2B_2) = \S r_i(v_i \tens w_i) (c_1B_1 + c_2B_2) =$ \bh
$\S c_1r_i(v_i \tens w_i)(B_1) + c_2r_i(v_i \tens w_i)(B_2) =$
$c_1(\S r_i( v_i \tens w_i))(B_1) + c_2 ( \S r_i(v_i \tens w_i))(B_2)$.

Note $ (\S r_i(v_i \tens w_i)) (B) = \S r_i(v_i \tens w_i) (B) =
\S r_i B(v_i, w_i)$

Prop:  The product $\tens$ is bilinear:

Proof:  $( (c_1 v_1 + c_2v_2) \tens w)(B) = B(c_1 v_1 + c_2v_2, w) =  c_1B(v_1, w) + c_2B(v_2, w) $
\bh and ...

Claim:  $V \tens W = < v \tens w ~|~ v \in V, w \in W, $

\rightline{$(c_1 v_1 + c_2v_2) \tens w = c_1 (v_1 \tens w) + c_2 (v_2 \tens w),$} 

\rightline{$v \tens (c_1w_1 + c_2w_2) = c_1(v \tens w_1) +  c_2(v \tens w_2)>$}

$= < v \tens w ~|~ v \in V, w \in W, $
{$( v_1 + v_2) \tens w = v_1 \tens w + v_2 \tens w,$} 

\rightline{$v \tens (w_1 + w_2) = v \tens w_1 + v \tens w_2, (cv) \tens w = c(v \tens w) = v \tens (cw)>$}

\eject

Let $v_1,..., v_m$ be a basis for $V$, $w_1,..., w_n$ basis for $W$.  If $B$ is a bilinear form, then $B$ is uniquely determined by the $nm$ values $B(v_i,w_j)$. 

 Thus  
$dim( V \times W )^b = nm = V \tens W $.

Hence a basis for $ V \tens W $ is $\{v_i \tens w_j ~|~ i =1,..., m, j = 1, ..., n\}$


Thus $V \tens W = \{ \Sigma c_{ij} v_i \tens w_j ~|~$ {$( v_1 + v_2) \tens w = v_1 \tens w + v_2 \tens w,$} 

\rightline{$v \tens (w_1 + w_2) = v \tens w_1 + v \tens w_2, (cv) \tens w = c(v \tens w) = v \tens (cw)\}$}
\l
The universal mapping property:  There exists $\phi$, $V \tens W$ such that $\phi$ is bilinear and given any bilinear map:  $B:  V \times W \rightarrow U$, there exists a map $A:  V \tens W\rightarrow U$ such that $A \circ \phi = B$.  Moreover  $\phi$ and $V \tens W$ are unique in the sense that if $X$ and $\psi$ satisfy the universal mapping property, then there exists an isomorphism $f:  V \tens W \rightarrow X$ such that $f \circ \phi = \psi$ 

Fix $B:  V \times W \rightarrow U$ and take $f: U \rightarrow \R \in U^*$ 

Then $f \circ B:   V \times W \rightarrow \R$ is bilinear:

$(f \circ B)(c_1v_1 + c_2v_2, w) = f(c_1B(v_1, w) + c_2B(v_2, w) )
= c_1(f \circ B)(v_1, w) + c_2(f \circ B)(v_2, w)$


Similarly $(f \circ B)(v, c_1w_1 + c_2w_2) = c_1(f \circ B)(v, w_1) + c_2(f \circ B_(v, w_2)$

Thus $f \circ B \in ( V \times W )^b$

$\alpha: U^*  \rightarrow ( V \times W )^b$, $\alpha(f) = f \circ B$


Thus $\beta:  V \tens W \rightarrow (U^*)^*$,  $\beta$
 

\l

The tensor algebra = $T(V) = \oplus_{k=0}^\infty (\tens^k V)$

$c_0 + c_1v_i +   \Sigma c_{ij} v_i \tens v_j + ... + \Sigma c_{i_1...i_p} v_{i_1} \tens ... \tens v_{i_p}$


$(  v_{1} \tens ... \tens v_{p})$$(  u_{1} \tens ... \tens u_{q})$
$= v_{1} \tens ... \tens v_{p}\tens u_{1} \tens ... \tens u_{q}$

A subring $I$ of a ring $R$ is an {\it ideal} of $R$ if $ar \in I$ and $ra \in I$ for all $a \in I$, $r \in R$.



Let $I(V) =$ ideal generated by $\{v \tens v ~|~ v \in V\}$

The exterior algebra of $V$ is the quotient $\Lambda^*V = T(V) /I(V)$


Let $\pi: T(V) \rightarrow  \Lambda^*V$ be the quotient map.

The {\it $p$-fold exterior power of $V$} is $\Lambda^pV = \pi (\tens^pV) = (M(v_1, ..., v_n))^*$, the dual of the space of multilinear forms.

The {\it exterior product} of $\alpha = \pi(a) \in  \Lambda^pV$ and $\beta = \pi(b) \in \Lambda^qV is $$\alpha \wedge \beta = \pi(a \tens b)$$ 


\l