Modules & Representation Theory
Modules & Representation Theory
A module over a ring $R$ generalizes vector spaces (where $R$ is a field) to rings. Structure theorems for modules over PIDs classify finitely generated abelian groups (as $\mathbb{Z}$-modules) and canonical forms of linear maps (as $k[x]$-modules). Representation theory studies group actions on vector spaces via matrix representations; characters (traces of matrices) are powerful invariants that classify irreducible representations.
Modules & Structure Theorem
An $R$-module $M$ is an abelian group with scalar multiplication $R\times M\to M$ satisfying $r(m+n)=rm+rn$, $(r+s)m=rm+sm$, $(rs)m=r(sm)$, $1m=m$. The structure theorem for finitely generated modules over a PID: $M\cong R^r\oplus R/d_1\oplus\cdots\oplus R/d_k$ where $d_1|d_2|\cdots|d_k$ (invariant factor form), or $M\cong R^r\oplus\bigoplus R/p_i^{n_i}$ (elementary divisor form). Application to $R=\mathbb{Z}$: every finitely generated abelian group $\cong\mathbb{Z}^r\oplus\mathbb{Z}_{d_1}\oplus\cdots\oplus\mathbb{Z}_{d_k}$.
Characters of Group Representations
A representation of $G$ is a homomorphism $\rho:G\to GL_n(k)$. The character $\chi_\rho:G\to k$ is $\chi_\rho(g)=\text{tr}(\rho(g))$. Characters are class functions ($\chi(hgh^{-1})=\chi(g)$) and irreducible representations have orthogonal characters: $\frac{1}{|G|}\sum_{g\in G}\chi_i(g)\overline{\chi_j(g)}=\delta_{ij}$ (over $\mathbb{C}$). The number of irreducible representations equals the number of conjugacy classes. The regular representation decomposes as $\bigoplus (\dim V_i)V_i$.
Example 1
Classify finitely generated abelian groups of order 12.
Solution: $12=2^2\cdot 3$. Invariant factor form: $d_1|d_2$ and $d_1 d_2=12$: (a) $\mathbb{Z}_{12}$ ($d_1=1$ is trivial); (b) $\mathbb{Z}_2\oplus\mathbb{Z}_6$ ($d_1=2,d_2=6$). Elementary divisors: $\mathbb{Z}_4\oplus\mathbb{Z}_3\cong\mathbb{Z}_{12}$; or $\mathbb{Z}_2\oplus\mathbb{Z}_2\oplus\mathbb{Z}_3\cong\mathbb{Z}_2\oplus\mathbb{Z}_6$. So exactly 2 abelian groups of order 12: $\mathbb{Z}_{12}$ and $\mathbb{Z}_2\times\mathbb{Z}_6$.
Example 2
Compute the character table of $S_3$.
Solution: $S_3$ has 3 conjugacy classes: $\{e\}$, $\{(12),(13),(23)\}$, $\{(123),(132)\}$. 3 irreducible representations: $\chi_1$ (trivial): $(1,1,1)$; $\chi_2$ (sign): $(1,-1,1)$; $\chi_3$ (standard, dim 2): by orthogonality and $\sum(\dim\chi_i)^2=|G|=6$: $1+1+4=6$ ✓. $\chi_3(e)=2$; $\chi_3((12))$ by orthogonality to $\chi_1$: $1\cdot 2+3\chi_3((12))+2\chi_3((123))=0$ and orthogonality to $\chi_2$: $1\cdot 2-3\chi_3((12))+2\chi_3((123))=0$. Adding: $4+4\chi_3((123))=0$, so $\chi_3((123))=-1$; then $2+3\chi_3((12))=2$, so $\chi_3((12))=0$.
Practice
- Prove that any finitely generated $\mathbb{Z}$-module (abelian group) has the structure given by the structure theorem.
- Show $\mathbb{Z}[x]$ is not a PID (hence the structure theorem does not apply).
- Find all irreducible representations of $\mathbb{Z}/4\mathbb{Z}$ over $\mathbb{C}$.
- Prove Maschke's theorem: every representation of a finite group over a field of characteristic not dividing $|G|$ is completely reducible.
Show Answer Key
1. Every finitely generated abelian group is isomorphic to $\mathbb{Z}^r\oplus\mathbb{Z}/d_1\oplus\cdots\oplus\mathbb{Z}/d_k$ where $d_1|d_2|\cdots|d_k$ (invariant factor decomposition). Proof uses the Smith normal form of the presentation matrix.
2. $\mathbb{Z}[x]$ is not a PID because the ideal $(2,x)=\{2f(x)+xg(x):f,g\in\mathbb{Z}[x]\}$ is not principal: any generator would need to divide both 2 and $x$, forcing it to be $\pm1$, but $1\notin(2,x)$ since elements of $(2,x)$ have even constant term.
3. The irreducible representations of $\mathbb{Z}/4\mathbb{Z}=\{0,1,2,3\}$ over $\mathbb{C}$ are 1-dimensional (abelian group). They are $\rho_k(n)=i^{kn}$ for $k=0,1,2,3$, i.e., the characters $1,i^n,(-1)^n,(-i)^n$.
4. If $V$ is a representation and $W\le V$ is a subrepresentation, define $W^\perp$ using an averaged inner product $\langle v,w\rangle_G=\frac{1}{|G|}\sum_{g}\langle gv,gw\rangle$, which is $G$-invariant (char $k\nmid|G|$ ensures averaging works). Then $V=W\oplus W^\perp$ and $W^\perp$ is also a subrepresentation. Iterate to decompose completely.