\documentclass[11pt ]{article} \setlength{\textwidth}{14 true cm} \setlength{\textheight}{20 true cm} \usepackage{graphicx} \usepackage{hangul} \usepackage{amscd,amsmath} \usepackage{amsfonts} \usepackage{amssymb,theorem} \usepackage{longtable} \newcommand{\we}{\wedge} \newcommand{\vts}{v_1\otimes\cdot\cdot\cdot\ot v_k} \newcommand{\vs}{V_{1}\times\cdot\cdot\cdot\times V_k} \newcommand{\ot}{\otimes} \newcommand{\pauo}{\frac{\partial}{\partial u_1}} \newcommand{\paui}{\frac{\partial}{\partial u_i}} \newcommand{\pauj}{\frac{\partial}{\partial u_j}} \newcommand{\wid}{\widetilde} \newcommand{\ml}{\mathcal{L}} \newcommand{\hs}{\hspace} \newcommand{\inv}{^{-1}} \newcommand{\vphi}{\varphi} \newcommand{\pax}{\frac{\partial}{\partial x}} \newcommand{\pay}{\frac{\partial}{\partial y}} \newcommand{\paz}{\frac{\partial}{\partial z}} \newcommand{\paxi}{\frac{\partial}{\partial x_i}} \newcommand{\payj}{\frac{\partial}{\partial y_j}} \newcommand{\paxj}{\frac{\partial}{\partial x_j}} \newcommand{\payi}{\frac{\partial}{\partial y_i}} \newcommand{\li}{[X,Y]} \newcommand{\dis}{\displaystyle} \newcommand{\disi}{\displaystyle{\sum_{i=1}^n}} \newcommand{\disj}{\displaystyle{\sum_{j=1}^n}} \newcommand{\pa}{\partial} \newcommand{\Aff}{\mbox{\it Aff}} \newcommand{\aff}{\mbox{\it aff}} \newcommand{\cc}{\mathcal{C}} \newcommand{\dc}{\mathcal{D}} \newcommand{\fc}{\mathcal{F}} \newcommand{\ccn}{\mathcal{C}^{\infty}} \newcommand{\sbb}{\mathbb{S}} \newcommand{\rb}{\mathbb{R}} \newcommand{\cb}{\mathbb{C}} \newcommand{\rc}{\mathcal{R}} \newcommand{\alp}{\alpha} \newcommand{\bet}{\beta} \newcommand{\del}{\delta} \newcommand{\gam}{\gamma} \newcommand{\vep}{\varepsilon} \newcommand{\eps}{\epsilon} \newcommand{\lam}{\lambda} \newcommand{\kap}{\kappa} \newcommand{\sig}{\sigma} \newcommand{\ome}{\omega} \newcommand{\Gam}{\Gamma} \newcommand{\Ome}{\Omega} \newcommand{\Sig}{\Sigma} \newcommand{\Del}{\Delta} \newcommand{\Lam}{\Lambda} \newcommand{\pr}{(\,\,,\,\,)} \newtheorem{thm}{정리} \newtheorem{cor}[thm]{따름정리} \newtheorem{lem}[thm]{보조정리} \newtheorem{prop}[thm]{명제} \newtheorem{cl}{Claim} {\theorembodyfont{\rm} \newtheorem{ex}{예} \newtheorem{que}{질문} \newtheorem{notation}{Notation}[section] \newtheorem{defn}{정의} \newtheorem{rem}{주} \newtheorem{note}{Note} } \renewcommand{\thenote}{} \renewcommand{\therem}{} \newenvironment{proof}{{\bf 증명}}{\hfill\framebox[2mm]{}} \newenvironment{proof1}{{\bf 정리증명}}{\hfill\framebox[2mm]{}} \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 $\{e_{i_1}\we\cdot\cdot\we e_{i_r},i_1<\cdot\cdot$ 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