\magnification=2000 \overfullrule=0pt \parindent=0pt \nopagenumbers \input amstex %\voffset=-.6in %\hoffset=-.5in %\hsize = 7.5 true in %\vsize=10.4 true in \voffset=1.8 true in \hoffset=-.6 true in \hsize = 10.2 true in \vsize=8 true in \input colordvi \def\cltr{\Red} % Red VERY-Approx PANTONE RED \def\cltb{\Blue} % Blue Approximate PANTONE BLUE-072 \def\cltg{\PineGreen} % ForestGreen Approximate PANTONE 349 \def\cltp{\DarkOrchid} % DarkOrchid No PANTONE match \def\clto{\Orange} % Orange Approximate PANTONE ORANGE-021 \def\cltpk{\CarnationPink} % CarnationPink Approximate PANTONE 218 \def\clts{\Salmon} % Salmon Approximate PANTONE 183 \def\cltbb{\TealBlue} % TealBlue Approximate PANTONE 3145 \def\cltrp{\RoyalPurple} % RoyalPurple Approximate PANTONE 267 \def\cltp{\Purple} % Purple Approximate PANTONE PURPLE \def\cgy{\GreenYellow} % GreenYellow Approximate PANTONE 388 \def\cyy{\Yellow} % Yellow Approximate PANTONE YELLOW \def\cgo{\Goldenrod} % Goldenrod Approximate PANTONE 109 \def\cda{\Dandelion} % Dandelion Approximate PANTONE 123 \def\capr{\Apricot} % Apricot Approximate PANTONE 1565 \def\cpe{\Peach} % Peach Approximate PANTONE 164 \def\cme{\Melon} % Melon Approximate PANTONE 177 \def\cyo{\YellowOrange} % YellowOrange Approximate PANTONE 130 \def\coo{\Orange} % Orange Approximate PANTONE ORANGE-021 \def\cbo{\BurntOrange} % BurntOrange Approximate PANTONE 388 \def\cbs{\Bittersweet} % Bittersweet Approximate PANTONE 167 %\def\creo{\RedOrange} % RedOrange Approximate PANTONE 179 \def\cma{\Mahogany} % Mahogany Approximate PANTONE 484 \def\cmr{\Maroon} % Maroon Approximate PANTONE 201 \def\cbr{\BrickRed} % BrickRed Approximate PANTONE 1805 \def\crr{\Red} % Red VERY-Approx PANTONE RED \def\cor{\OrangeRed} % OrangeRed No PANTONE match \def\paru{\RubineRed} % RubineRed Approximate PANTONE RUBINE-RED \def\cwi{\WildStrawberry} % WildStrawberry Approximate PANTONE 206 \def\csa{\Salmon} % Salmon Approximate PANTONE 183 \def\ccp{\CarnationPink} % CarnationPink Approximate PANTONE 218 \def\cmag{\Magenta} % Magenta Approximate PANTONE PROCESS-MAGENTA \def\cvr{\VioletRed} % VioletRed Approximate PANTONE 219 \def\parh{\Rhodamine} % Rhodamine Approximate PANTONE RHODAMINE-RED \def\cmu{\Mulberry} % Mulberry Approximate PANTONE 241 \def\parv{\RedViolet} % RedViolet Approximate PANTONE 234 \def\cfu{\Fuchsia} % Fuchsia Approximate PANTONE 248 \def\cla{\Lavender} % Lavender Approximate PANTONE 223 \def\cth{\Thistle} % Thistle Approximate PANTONE 245 \def\corc{\Orchid} % Orchid Approximate PANTONE 252 \def\cdo{\DarkOrchid} % DarkOrchid No PANTONE match \def\cpu{\Purple} % Purple Approximate PANTONE PURPLE \def\cpl{\Plum} % Plum VERY-Approx PANTONE 518 \def\cvi{\Violet} % Violet Approximate PANTONE VIOLET \def\clrp{\RoyalPurple} % RoyalPurple Approximate PANTONE 267 \def\cbv{\BlueViolet} % BlueViolet Approximate PANTONE 2755 \def\cpe{\Periwinkle} % Periwinkle Approximate PANTONE 2715 \def\ccb{\CadetBlue} % CadetBlue Approximate PANTONE (534+535)/2 \def\cco{\CornflowerBlue} % CornflowerBlue Approximate PANTONE 292 \def\cmb{\MidnightBlue} % MidnightBlue Approximate PANTONE 302 \def\cnb{\NavyBlue} % NavyBlue Approximate PANTONE 293 \def\crb{\RoyalBlue} % RoyalBlue No PANTONE match %\def\cbb{\Blue} % Blue Approximate PANTONE BLUE-072 \def\cce{\Cerulean} % Cerulean Approximate PANTONE 3005 \def\ccy{\Cyan} % Cyan Approximate PANTONE PROCESS-CYAN \def\cpb{\ProcessBlue} % ProcessBlue Approximate PANTONE PROCESS-BLUE \def\csb{\SkyBlue} % SkyBlue Approximate PANTONE 2985 \def\ctu{\Turquoise} % Turquoise Approximate PANTONE (312+313)/2 \def\ctb{\TealBlue} % TealBlue Approximate PANTONE 3145 \def\caq{\Aquamarine} % Aquamarine Approximate PANTONE 3135 \def\cbg{\BlueGreen} % BlueGreen Approximate PANTONE 320 \def\cem{\Emerald} % Emerald No PANTONE match %\def\cjg{\JungleGreen} % JungleGreen Approximate PANTONE 328 \def\csg{\SeaGreen} % SeaGreen Approximate PANTONE 3268 \def\cgg{\Green} % Green VERY-Approx PANTONE GREEN \def\cfg{\ForestGreen} % ForestGreen Approximate PANTONE 349 \def\cpg{\PineGreen} % PineGreen Approximate PANTONE 323 \def\clg{\LimeGreen} % LimeGreen No PANTONE match \def\cyg{\YellowGreen} % YellowGreen Approximate PANTONE 375 \def\cspg{\SpringGreen} % SpringGreen Approximate PANTONE 381 \def\cog{\OliveGreen} % OliveGreen Approximate PANTONE 582 \def\pars{\RawSienna} % RawSienna Approximate PANTONE 154 \def\cse{\Sepia} % Sepia Approximate PANTONE 161 \def\cbr{\Brown} % Brown Approximate PANTONE 1615 \def\cta{\Tan} % Tan No PANTONE match \def\cgr{\Gray} % Gray Approximate PANTONE COOL-GRAY-8 \def\cbl{\Black} % Black Approximate PANTONE PROCESS-BLACK \def\cwh{\White} % White No PANTONE match \loadmsbm \input epsf \def\ctln{\centerline} \def\u{\underbar} \def\ssk{\smallskip} \def\msk{\medskip} \def\bsk{\bigskip} \def\hsk{\hskip.1in} \def\hhsk{\hskip.2in} \def\dsl{\displaystyle} \def\hskp{\hskip1.5in} \def\lra{$\Leftrightarrow$ } \def\ra{\rightarrow} \def\mpto{\logmapsto} \def\pu{\pi_1} \def\mpu{$\pi_1$} \def\sig{\Sigma} \def\msig{$\Sigma$} \def\ep{\epsilon} \def\sset{\subseteq} \def\del{\partial} \def\inv{^{-1}} \def\wtl{\widetilde} %\def\lra{\Leftrightarrow} \def\del{\partial} \def\delp{\partial^\prime} \def\delpp{\partial^{\prime\prime}} \def\sgn{{\roman{sgn}}} \def\wtih{\widetilde{H}} \def\bbz{{\Bbb Z}} \def\bbr{{\Bbb R}} {\bf Computing homology groups:} Computing simplicial homology groups for a finite $\Delta$-complex is, in the end, a matter of straightforward linear algebra. First the punchline! Start with the chain complex \ssk \ctln{ $\cdots\ra C_{n+1}(X){\buildrel{\del_{n+1}}\over\ra}C_n(X) {\buildrel{\del_n}\over\ra}C_{n-1}(X)\ra\cdots$ } \ssk for the simplicial homology of a finite $\Delta$-complex $X$, with matrices $\Delta_n$ representing the boundary maps $\del_n$ (in the standard bases given by the $(n+1)$- and $n$-simplices). We can, by row and column operations over $\bbz$ (that is, we add integer multiples of a row or column to another, permute rows or columns, and multiply a row or column by $-1$) put the matrix $\Delta_n=(d_{ij})$ into {\it Smith normal form} $A_n=(a_{ij})$, which means that its entries are $0$ except along the ``diagonal'' $a_{ii}=b_i$; and further, $b_i|b_{i+1}$ for all $i$ (where everything is assumed to divide $0$ and $0$ divides only $0$). We sketch the proof below. \msk Then if we let $M_n$ = the number of columns of zeros of $A_n$ and $m_n$ = the number of {\it non-zero} rows of $A_{n+1}$ [note the change of subscript!], and $b_1,\ldots b_k$ the non-zero diagonal entries of $A_{n+1}$ [note that $m_n=k$], then \ssk \ctln{ $H_n(X)\cong\bbz^{M_n-m_n}\oplus Z_{b_1}\oplus\cdots\oplus Z_{b_k}$ } \msk The point is that the matrices $A_n$ are representatives of the boundary maps $\del_n$, with respect to some very carefully chosen (and compatible) bases for $C_n(X)$. Essentially, $Z_n(X)\cong Z^{M_n}$ with basis the last $M_n$ elements of our basis for $C_n(X)$, and $B_n(X)\cong b_1\bbz\oplus\cdots\oplus b_k\bbz$, mapping into the last $k$ coordinates. \vfill \eject Specifically, we show that we can decompose each $C_n(X)=U_n\oplus V_n\oplus W_n$, where $Z_n(X)=\{0\}\oplus V_n\oplus W_n$ (so $\del_n$ is injective on $U_n$), and $\del_{n}(U_{n})\subseteq W_{n-1}$ (with finite index). Furthermore, we may choose bases so that the matrix of $\del_{n+1}:U_{n+1}\ra W_{n}$ is the diagonal matrix diag$(b_1,\ldots,b_k)$. Note that, choosing bases for the $V_n$ to fill out a basis for $C_n(X)$ (using the bases for $U_n$ and $W_n$ implied by the above statement), the matrix for $\del_{n+1}$ has the block form consisting of all $0$'s, except for the ``$U_{n+1}$ to $W_n$'' part, which is the diagonal matrix. (This is a slightly permuted version of Smith normal form.) Then \ssk \ctln{$B_n(X)=$im$(\del_{n+1})= b_1\bbz\oplus\cdots\oplus b_k\bbz \subseteq \{0\}\oplus W_n\subseteq V_n\oplus W_n$} \ssk so $H_n(X)=Z_n(X)/B_n(X)\cong (V_n\oplus W_n)/(\{0\}\oplus b_1\bbz\oplus\cdots\oplus b_k\bbz)\cong V_n\oplus Z_{b_1}\oplus\cdots\oplus Z_{b_k}$. The dimension of $V_n$ can be determined from the normal forms of $\del_{n+1}$ and $\del_n$ as $\dim(V_n)=\dim(V_n\oplus W_n)-\dim(W_n)=\dim(Z_n(X))-\dim(U_{n+1})$ = number of non-pivot columns of $\del_n$ $-$ number of pivot rows of $\del_{n+1}$ = number of zero columns of $A_n$ $-$ number of non-zero rows of $A_{n+1}$, as desired. \msk It remains only to show that the matrices $\Delta_n$ can be put into Smith normal form, and that this {\it implies} the existence of the decompositions of $C_n(X)$ as described. The basic idea is that row and column operations on a matrix can really be interpreted as choosing different (ordered) bases for the domain or codomain, to describe a linear transformation $L:V\ra W$. Working with integer matrices and operations over $\bbz$ means that we can work with bases of $\bbz^n$. Since $Lu_i=\sum_j a_{ij}w_j$ for bases $\{u_i\}$ and $\{w_j\}$, interchanging rows or columns corresponds to interchanging basis elements in $W$ or $V$, repsectively (to make the actual vectors described by the summation remain the same). Multiplication of a row or column by $-1$ multiplies the corresponding basis element by $-1$. And adding a multiple $m$ of one row, $k$, to another, $\ell$ amounts to replacing the basis vector $u_\ell$ with $u_\ell+mu_k$ (since the new row describes, in terms of the $w_j$, the image of that vector in $V$); adding a multiple $m$ of one column $k$ to another, $\ell$, is reflected in the bases as \ssk \ctln{$a_{i1}w_1+\cdots +a_{im}w_m= a_{i1}+\cdots a_{ik}(w_k-mw_\ell)+ \cdots +(a_{i\ell}+ma_{ik})w_\ell+\cdots +a_{im}w_m$} \ssk so it can be interpreted as a change of basis in the codomain. \msk Using these operations to arrive in Smith normal form is a fairly straighforward matter of induction. Given an integer matrix $A=(a_{ij})$, Let $N(A)$ be the minimum, among all non-zero entries of $A$, of their absolute value. If $|a_{ij}|=N(A)$ does not divide some entry $a_{kl}=b$ of $A$, we can by row and column operations decrease $N(A)$. If $i=k$ or $j=\ell$, then adding the right multiple of the column or row containing $a_{ij}$ to the one containing $b$ will yield the remainder of $b$ upon division by $a_{ij}$ in $b$'s place, lowering $N(A)$. Otherwise, we may assume that both $a_{i\ell}=\alpha a_{ij}$ and $a_{kj}=\beta a_{ij}$, and then adding $1-\alpha$ times the $i$-th row to the $k$-th yields a row with $a_{ij}$ in the $j$-th spot and $a_{k\ell}+(1-\alpha)\beta a_{ij}$ in the $\ell$-th, which $a_{ij}$ still doesn't divide. Then we apply the previous operation to reduce $N(A)$. \ssk Since this cannot continue indefinitely [$N(A)\in \bbz^+$], eventually $N(A)=|a_{ij}|$ {\it does} divide every entry of $A$. By swapping rows and columns, we may assume that $N(A)=|a_{11}|$, and then by row and column operations we can zero out every other entry in the first row and column. Striking out this row and column, we have a minor matrix $B$ satisfying $N(A)|N(B)$. Note that then this will remain true under any subsequent row or column operation on $B$. By induction, there are row and column operations for $B$ (which we interpret as operations on $A$ not using the first row or column) putting $B$ into Smith normal form; essentially, we just continually use the process above on smaller and smaller matrices. Since $N(A)|N(B)$ at the start, this remains true throughout the row and column operations, so each diagonal entry divides the ones that come later. All other entries are eventually zeroed out. \msk After we have put our matrices representing the boundary maps $\del_n$ into Smith normal form, this provides us with bases $\{u_i^n\},\{w_j^{n-1}\}$ for $C_n(X)$ and $C_{n-1}(X)$ so that $\del_nu_i^n = b_{i,n}w_i^{n-1}$ for $i\leq k_n$ and $\del_nu_i^n=0$ for $i>k_n$. In order to make these bases ``compatible'', as we desire, we need to use the fact that $\del_n\circ \del_{n+1}=0$. We may assume that $b_{i,n}\neq 0$ for all $i\leq k_n$ (otherwise we just shift $k_n$). \ssk The point is that $\del_{n+1}u_i^{n+1}=b_{i,n+1}w_i^{n}$ implies that $0=\del_n\del_{n+1}u_i^{n+1}=b_{i,n+1}\del_nw_i^{n}$, so $\del_nw_i^{n}=0$ for every $i\leq k_n$ (since $b_{i,n+1}\neq 0$). Also, $\del_n$ is injective on the span of $u_1^n,\ldots ,u_{k_n}^n$, since their images, $b_{i,n}w_i^n$, are linearly independent (as non-zero multiples of basis elements). This also means that they span a subspace complementary to $Z_n(X)$. We therefore wish to choose $\{u_i^n : 1\leq i\leq k_n\}$ as our basis for $U_n$ and $\{w_j^n : 1\leq j \leq k_{n+1}\}$ as our basis for $W_n$, and {\it show} that this can be extended to a basis for $C_n(X)$ by choosing vectors lying in $Z_n(X)$; the added vectors will be a basis for $V_n$. Once we do this, we know that in these bases our boundary maps will have the form we desire, since the added vectors $v_k$ map to $0$ under $\del_n$, as do the $w_j^n$ we've kept, and $\del_nu_i=b_{i,n}w_i^{n-1}$. For notational convenience in what follows, we will denote the vectors $u_i,w_j$ that we have elected {\it not} to keep for our basis by $u_i^\prime,w_j^\prime$. \vfill \eject To finish building our basis for $C_n(X)$ we essentially need to show that the span $W_n$ of the $w_j$'s form a direct summand of $Z_n(X)=\ker(\del_n)$, that is, the $w_j$ form the subset of some basis for $Z_n(X)$. To show that $W_n$ is a summand, we show that $Q_n=Z_n(X)/W_n$ is a free abelian group; choosing a basis $\{v_k+W_n\}$ for $Q_n$, it then follows that $\{v_k\}\cup\{w_j\}$ is a basis for $Z_n(X)$; given $z\in Z_n(X)$, $z+W_n$ can be expressed as a combination of the $v_k+W_n$; $z$ minus that combination therefore lies in $W_n$, so is a combination of the $w_j$. That shows that they span. Writing $0$ as a linear combination and then projecting to $Q_n$ shows that the coefficients of the $v_k$ are $0$ (since the $v_k+W_n$ are linearly independent in $Q_n$), and so the coeficients of the $w_j$ are zero (since the $w_j$ are linearly independent in $W_n$), proving linear independence. \msk And to show that $Q_n=Z_n(X)/W_n$ is a free abelian group, since it is finitely generated, it suffices to show that it contains no torsion elements. That is, if $z\in Z_n(X)$ and $cz\in W_n$ for some $c\neq 0$, then $z\in W_n$. But this is immediate, really; expressing $z=\sum c_iw_i+\sum c_i^\prime w_i^\prime$ (which is unique, since the $w_i,w_i^\prime$ form a basis for $C_n(X)$), then $cz=\sum cc_iw_i+\sum cc_i^\prime w_i^\prime\in W_n$ implies that the $cc_i^\prime=0$ (since this expression is still unique and the $w_i$ form a basis for $W_n$), so the $c_i^\prime=0$ and $z=\sum c_iw_i\in W_n$, as desired. \msk The fact that the $u_i$ and $Z_n(X)$ (hence the $u_i$ and the basis $\{v_k,w_j\}$ for $Z_n(X)$) span $C_n(X)$ follows from the fact that the $u_i^\prime$ all lie in $Z_n(X)$. This implies that $C_n(X)\cong$span$\{u_i\}\oplus$span$\{v_k\}\oplus$span$\{w_j\}$, finishing our proof. \vfill \end Therefore the $w_j$ are (uniquely) $\bbz$-linear combinations of the $u_i^\prime$; since the $w_j,w_j^\prime$ form a basis for $C_n(X)$, the $u_i^\prime$ can also be (uniquely) expressed as linear combinations of these. Expressing this in matrix form $w_j=\sum e_{ji}u_i^\prime$, $u_i^\prime=\sum f_{ij}W_j$, the matrices $E=(e_{ji}),F=(f_{ij})$ satisfy $EF=I$. This implies that the matrix $E$ can be row-reduced (over $\bbz$) This also means that the $u_i^\prime$ form a basis for $Z_n(X)$. [If a linear combination of the $u_i,u_i^\prime$ is sent to $0$, then the coefficients of the $u_i$ are zero, so the vector is really a linear combination of the $u_i^\prime$, so these vectors span $Z_n(X)$.]