\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 $V$ 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 $\Lambda^*V = T(V) /I(V)$

\l



\end