Finitely-Generated Free Abelian Groups

An Interactive Primer on Homomomorphisms between Free Abelian Groups

Algebraic Topology
Interactive Tutorial
Author

Sushovan Majhi

Published

July 26, 2021

Let \((G,+)\) and \((G',+)\) be finitely-generated free (additive) abelian groups of dimension \(m\) and \(n\), respectively. We choose arbitrary ordered bases \(\mathcal{E}=(e_1,e_2,\ldots,e_m)\) and \(\mathcal{E}'=(e_1',e_2',\ldots,e_n')\) for \(G\) and \(G'\), respectively.

Matrix of A Homomorphism

A homomorphism \(F:G\to G'\) is a linear function, i.e., \(F(x+y)=F(x)+F(y)\) for all \(x,y\in G\).

As a consequence, \(F\) is uniquely determined by its values on the elements of \(\mathcal{E}\). For each \(1\leq j\leq m\), the image \(F(e_j)\) can be written uniquely as a (formal) linear combination of the elements of \(\mathcal{E}'\). Let \[ F(e_j) = \sum\limits_{i=1}^{n} a_{ij}e_i'. \]

The matrix \(A:=\{a_{ij}\}\in\mathcal{M}_{n,m}(\mathbb Z)\) is called the matrix of \(F\) w.r.t. the (ordered) bases \(\mathcal{E},\mathcal{E}'\), and is denoted by \([F]_{\mathcal{E},\mathcal{E'}}\). In matrix notation, we have \[ \bigg[F(e_1),\ldots,F(e_j),\ldots,F(e_m)\bigg]= \bigg[e_1',\ldots,e_i',\ldots,e_n'\bigg] \begin{bmatrix} a_{11} & \ldots & a_{1j} & \ldots & a_{1m} \\ \ldots & \ldots & \ldots & \ldots & \ldots \\ a_{i1} & \ldots & a_{ij} & \ldots & a_{im} \\ \ldots & \ldots & \ldots & \ldots & \ldots \\ a_{n1} & \ldots & a_{nj} & \ldots & a_{nm} \\ \end{bmatrix}. \] Tolerating a mild abuse of notation, we can also express the above relation as \(F(\mathcal{E})=\mathcal{E'}[F]_{\mathcal{E},\mathcal{E'}}\).

Example:

Let \(F:\mathbb{Z}^3\to \mathbb{Z}^2\) be a homomorphism defined by \[F(a,b,c)=(3a+4b-2c,2b+5c).\]

A natural choice of ordered bases is \(\mathcal{E}=\bigg((1,0,0),(0,1,0),(0,0,1)\bigg)\) and \(\mathcal{E}'=\bigg((1,0),(0,1)\bigg)\). In order to compute \([F]_{\mathcal{E},\mathcal{E}''}\), the matrix of \(F\) w.r.t. the chosen bases, we note that \[ \begin{aligned} F(1,0,0) &= (3,0) = 3(1,0) + 0(0,1) \\ F(0,1,0) &= (4,2) = 4(1,0) + 2(0,1) \\ F(0,0,1) &= (-2,5) = -2(1,0) + 5(0,1) \end{aligned} \] So, the matrix of \(F\) is \[ \begin{bmatrix} 3 & 4 & -2\\ 0 & 2 & 5 \end{bmatrix}. \]

Question What would the matrix of \(F\) be if we chose the bases \(\bigg((1,1,0),(1,0,1),(0,1,1)\bigg)\) and \(\bigg((1,0),(1,1)\bigg)\)?

Following the technique (basically the definition) of the above example we can compute the matrix of \(F\) under the new set of bases.

Change of Basis

Let \(\mathcal{B}=(e_1,e_2,\ldots,e_m)\) be an ordered basis of a finitely-generated free abelian group \(G\). Now we consider an new basis \(\mathcal{C}=(f_1,f_2,\ldots,f_m)\) of \(G\). The basechange matrix encodes the change the basis from (old) \(\mathcal{B}\) to (new) \(\mathcal{C}\). The basechange matrix facilitates the computation of the matrix of a homomomorphism w.r.t. a set of new bases.

The starting point is to write the elements of the new basis as a linear combination of those of the old basis. For \(j=1,2,\ldots,m'\), \[ f_j = \sum\limits_{i=1}^{m} p_{ij}e_i. \]

The matrix \(P\) is called the basechange matrix the order bases \(\mathcal{B}\text{ and }\mathcal{C}\). In matrix notation, \[\mathcal{C}=\mathcal{B}P.\]

As we can immediately see, the basechange matrix is unique, and it is also invertible [artin], i.e., \(P\in GL_{m}(\mathbb Z)\). Given an invertible, integer matrix, one can also use the above matrix relation to compute the new basis.

Example:

For this example, I let you interact with the tutorial. We choose the dimension of \(G\) using the following slider. Upon choosing a value, you readily see the initial basis:

and randomly generated basechange matrix \(P\).

Now, we take a random invertible matrix of the chosen dimension:

The new basis of \(G\) is:

Application to Homomorphisms

Let \(F:G\to G'\) be a homomorphism. Let \(A\) denote the matrix of \(F\) w.r.t. a pair of old ordered bases \((\mathcal{B},\mathcal{B}')\) for \(G\) and \(G'\), respectively. If a new pair of order bases \((\mathcal{C},\mathcal{C}')\) chosen, one would be interested to compute the matrix of \(F\) w.r.t. the new bases.

Let \(P\in\mathcal{M}_m(\mathbb Z)\) and \(Q\in\mathcal{M}_n(\mathbb Z)\) denote the basechange matrices for \(G\) and \(G'\), respectively. So, \[\mathcal{C}=\mathcal{B}P\text{, and }\mathcal{C}'=\mathcal{B}'Q.\]

By the definition of \(A\), note that \(F(\mathcal{B})=\mathcal{B}'A\). Therefore, \[ \begin{aligned} F(\mathcal{B})P &=(\mathcal{B}'A)P \\ &= (\mathcal{C}'Q^{-1})AP \\ &= \mathcal{C}'(Q^{-1}AP) \end{aligned} \]

Using the fact that \(F\) is a homomomorphism and letting \(\mathcal{B}=(e_1,e_2,\ldots,e_m)\) and \(\mathcal{C}=(f_1,f_2,\ldots,f_m)\), we simplify the LHS. \[ \begin{aligned} F(\mathcal{B})P &= \big[F(e_1),\ldots,F(e_j),\ldots,F(e_m)\big]P \\ &= \left[\sum_{j=1}^mF(e_j)p_{1j},\ldots,\sum_{j=1}^mF(e_j)p_{mj}\right] \\ &= \left[F\left(\sum_{j=1}^m p_{1j}e_j\right),\ldots,F\left(\sum_{j=1}^m p_{1j}e_j\right)\right] \\ &= \left[F\left(f_1\right),\ldots,F\left(f_m\right)\right] \\ &= F(\mathcal{C}). \end{aligned} \]

Therefore, \[F(\mathcal{C})=\mathcal{C}'\left(Q^{-1}AP\right).\]

We conclude that \(\left(Q^{-1}AP\right)\) is the matrix of \(F\) w.r.t. the new pair of basis \((\mathcal{C},\mathcal{C}')\).