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\begin{document}
 \parindent=0cm
  \section*{1. Tensor product.}
Let $V_1,V_2,\cdot\cdot\cdot,V_k$ be vector spaces over $\rb$ or
$\cb$.

$V_1\otimes\cdot\cdot\cdot\ot V_k:=M/N$, where $M=$ vector space
generated by the set $V_1\times\cdot\cdot\cdot\times V_k$,

$N=$ subspace generated by elements of type
$(x_1,\cdot\cdot,x_i+x_i',\cdot\cdot,x_k)-(x_1,\cdot\cdot,x_i,\cdot\cdot,x_k)-(x_1,\cdot\cdot,x_i',\cdot\cdot,x_k)$
 and
 $(x_1,\cdot\cdot,ax_i,\cdot\cdot,x_k)-a(x_1,\cdot\cdot,x_i,\cdot\cdot,x_k)$.

 이 때 $(x_1,\cdot\cdot\cdot,x_n)$의 equivalence class를 $[(x_1,\cdot\cdot,x_k)]=x_1\ot\cdot\cdot\cdot,\ot x_k$ 라고
 쓴다.\\

 $<$Universal property$>$

 아래 그림에서와 같이 multilinear $f$가 주어지면 diagram을 commute하는 linear transformation $\overline{f}$가
 유일하게 존재한다.\\

$\hs{2em}\vs\hs{1em}\overset{\pi}{\longrightarrow}\hs{1em}\bigotimes
V_i$

$\hs{0.5em}multilinear\,\,f\searrow\hs{3em}\swarrow\exists
!\overline{f}\hs{3em}\overline{f}(\vts)=f(v_1,\cdot\cdot\cdot,v_k).$

$\hs{9em}W$\\

Let $V$ be a vector space with basis
$\{e_1,\cdot\cdot\cdot,e_n\}$.

$T^r(V):=\bigotimes_{i=1}^r V$, $T^0(V):=\rb\,\,$ and
$\{e_{i_1}\ot\cdot\cdot\ot e_{i_r}\}$ becomes a basis of
$T^r.$\\

Let $V,W$ be vector spaces with basis
$\{e_1,\cdot\cdot,e_n\},\{b_1,\cdot\cdot,b_m\}$ . Then

$V\ot W$=vector space with basis $\{e_i\ot
b_j\,|\,i=1,\cdot\cdot,n,j=1,\cdot\cdot, m\}$

(증명은 $V\bigotimes(\bigoplus_iW_i)\cong\bigoplus_i(V\bigotimes
W_i )$, $\rb\ot
\rb\cong \rb $ 을 이용해서 보일 수 있다.)\\

이제 graded algebra구조를 갖는 tensor algebra $T(V)$를 정의하자.

$T(V)=\bigoplus_{r=0}^{\infty}T^r(V)$: tensor algebra with obvious
associative product.

 $T(V)\cong P$ : noncommutative polynomial algebra generated by $\{e_1,\cdot\cdot,e_n\}$

 $T$ is a functor :
 $\{vector\,\,space\}\rightarrow\{graded\,\,algebra\}$\\

 {\bf 숙제 14-1.} Show $v\ot\cdot\cdot\ot v_k=0\Leftrightarrow
 v_1=0$ , $\cdot\cdot\cdot $, or $v_k=0$ .\\

\section*{2. Alternating algebra.}

$\Lambda^r(V)=T^r(V)/\textbf{a}_r$, $\textbf{a}_r=$ subspace
generated by elements of type $x_1\ot\cdot\cdot\cdot\ot x_r$ where
$x_i=x_j$ for some $i\neq j$. Denote the equivalence class of
$x_1\ot\cdot\cdot\cdot\ot x_r$ by  $[x_1\ot\cdot\cdot\cdot\ot
x_r]=x_1\wedge\cdot\cdot\cdot\wedge x_r$. Then
$x_1\we\cdot\cdot\we x_i\we\cdot\cdot\we x_j\we\cdot\cdot\we
x_r=-x_1\we\cdot\cdot\we x_j\we\cdot\cdot
x_i\we\cdot\cdot x_r $.\\

$<$ Universal property $>$

아래 그림에서와 같이 multilinear alternating $f$가 주어지면
diagram을 commute하는 $\overline{f}$가
 유일하게 존재한다.\\


$\hs{2em}V\times\cdot\cdot\times V
\hs{1em}\overset{\pi}{\longrightarrow}\hs{1em}\Lambda^r(V)$

$\hs{4.5em}\,\,f\searrow\hs{3em}\swarrow\exists
!\overline{f}\hs{3em}\overline{f}(v_1\we\cdot\cdot\we v_k
)=f(v_1,\cdot\cdot,v_k).$

$\hs{8em}W$\\

$\Lambda (V)=\bigoplus_{r=0}^{\infty}\Lambda^r(V)$:
Exterior(Alternating) algebra with obvious associative product
$\we$.

$\Lambda$ is a functor :
$\{vector\,\,spaces\}\rightarrow\{graded\,\,algebra\}$

$\Lam^r(V)$ has a basis $\{e_{i_1}\we\cdot\cdot\we
e_{i_r},i_1<\cdot\cdot<i_r\}$ , and  $dim\Lam^r(V)=_nC_r$.\\

{\bf 숙제 14-2.} Show $v_1\we\cdot\cdot\we v_k\neq
0\Leftrightarrow
 $ $\{v_1,\cdot\cdot\cdot ,v_k\}$ is linearly independent .\\


\section*{3. Symmetric algebra.}

$S^r(V)=T^r(V)/\textbf{b}_r$, $\textbf{b}_r=$ subspace generated
by elements of type


$x_1\ot\cdot\cdot\cdot\ot x_r-x_{\sigma(1)}\ot\cdot\cdot\cdot\ot
x_{\sigma(r)}$, $\sigma\in Perm(r)= S_r$.\\

$<$ Universal property $>$

아래 그림에서와 같이 multilinear symmetric $f$가 주어지면
diagram을 commute하는 linear $\overline{f}$가
 유일하게 존재한다.\\


$\hs{2em}V\times\cdot\cdot\times V
\hs{1em}\overset{\pi}{\longrightarrow}\hs{1em}S^r(V)$

$\hs{4.5em}\,\,f\searrow\hs{3em}\swarrow\exists
!\overline{f}\hs{3em}\overline{f}(v_1\cdot\cdot v_k
)=f(v_1,\cdot\cdot,v_k).$

$\hs{8em}W$\\

$S^r(V)$ has a basis $\{e_{i_1}\cdot\cdot
e_{i_r},i_1\leq\cdot\cdot\leq i_r\}$ , $dim S^r(V)=_nH_r$.



$S(V)=\bigoplus_{r=0}^{\infty}S^r(V)$: Symmetric algebra with
obvious product $\cdot$.

$\hs{2.7em}\cong$ commutative polynomial algebra in $n-$variables,
$\rb[x_1,\cdot\cdot,x_n]$.

$S$ is a functor :
$\{vector\,\,spaces\}\rightarrow\{graded\,\,algebra\}$\\

{\bf 숙제 14-3.} Show $v\cdot\cdot\cdot v_k=0\Leftrightarrow
 v_1=0$, $\cdot\cdot\cdot$, or $v_k=0$ .\\


\section*{4. $T,\Lam,S$ as functors.}

For $f:V\rightarrow V',\,\,g:W\rightarrow W'$, linear, the tensor
product of $f$ and $g$ ,

$f\ot g:V\ot W\rightarrow V'\ot W'$ is induced and given by $(f\ot
g)(v\ot w)=f(v)\ot g(w)$.

Let $f=(f_{ij})$ with respect to basis $e=\{e_1,\cdot\cdot,e_m\}$
and $e'=\{e_1',\cdot\cdot,e_n'\}$,

$g=(g_{ij})$ with respect to basis $b=\{b_1,\cdot\cdot,b_p\}$ and
$b'=\{b_1',\cdot\cdot,b_q'\}$ . Then

the matrix of $f\ot g$ with respect to $\{e_i\ot b_j\}$ and
$\{e_i'\ot b_j'\}$ is given as the following :

$e_i\ot b_j\mapsto f(e_i)\ot g(b_j)=(\sum_k f_{ki}e_k')\ot(\sum_l
g_{lj} b_l' )=\sum_{k,l} f_{ki}g_{lj}e_k'\ot b_l'$\\

**그림 24**\\


 A linear $f:V\rightarrow W$ induces $T^r(f),\Lam^r(f),S^r(f)$
such that

$T^r(f):T^r(V)\rightarrow T^r(V),\,\,v_1\ot\cdot\cdot\cdot\ot v_r
\mapsto f(v_1)\ot\cdot\cdot\cdot\ot f(v_r)$

$\Lam^r(f):\Lam^r(V)\rightarrow
\Lam^r(V),\,\,v_1\we\cdot\cdot\cdot\we v_r\mapsto
f(v_1)\we\cdot\cdot\cdot\we f(v_r)$

$S^r(f):S^r(V)\rightarrow S^r(V),\,\,v_1\cdot\cdot\cdot v_r\mapsto
f(v_1)\cdot\cdot\cdot f(v_r)$

Let $e=\{e_1,\cdot\cdot,e_n\}$,$b=\{b_1,\cdot\cdot,b_m\}$  be
basis for $V$, $W$ and $f=(f_{ij})$ with respect to

$e$ and $b$. i.e., $f(e_i)=\sum_if_{ji}b_j$. Then

$\dis{T^r(f):e_I=e_{i_1}\ot\cdot\cdot\ot e_{i_r}\mapsto
f(e_{i_1})\ot\cdot\cdot\ot
f(e_{i_r})=\sum_{j_1}f_{j_1i_1}b_{j_1}\ot\cdot\cdot\ot
\sum_{j_r}f_{j_ri_r}b_{j_r}}$

$\dis{\hs{20.8em}=\sum_{j_1,\cdot\cdot,j_r}(f_{j_1 i_1 }\cdot\cdot
f_{j_ri_r})\underbrace{b_{j_1}\ot\cdot\cdot\ot b_{j_r}}}$

$\hs{32em}b_J$\\

$\Lam^r(f):e_I=e_{i_1}\we\cdot\cdot\we e_{i_r}\mapsto
f(e_{i_1})\we\cdot\cdot\we
f(e_{i_r})=\dis{\sum_{j_1,\cdot\cdot,j_r}(f_{j_1 i_1 }\cdot\cdot
f_{j_ri_r})(b_{j_1}\we\cdot\cdot\we b_{j_r})}$

$\dis{\hs{2em}I=(i_1<\cdot\cdot<i_r)\,,\,J=(j_1<\cdot\cdot< j_r)
)\hs{2em}=\sum_J(\underbrace{\sum_{\sigma\in
S_r}sgn(\sigma)f_{\sigma(j_1) i_1 }\cdot\cdot
f_{\sigma(j_r)i_r}})b_J}$

$\hs{30em}det\,f_{IJ}$\\

**그림 25**









  \end{document}
