Find all group homomorphisms from $\mathbb{Q}$ to $GL_n(\mathbb{C})$

Question

I have been looking at finding group homomorphisms from various groups into $GL_n(\mathbb{C})$. For example, I think I understand all homomorphisms from $\mathbb{Z} \to GL_n(\mathbb{C})$. I believe that one just picks an invertible matrix $A$ and map $1$ to this. Since $\mathbb{Z}$ is a cyclic group this gives all homomorphisms.

Now I am wondering how to do this for $\mathbb{Q}$ and $\mathbb{R}$ (both groups under addition). I thought I had worked this out for $\mathbb{Q}$. I thought that if one knew $\phi(1)$ then one would know $\phi(m)$ for any $m\in\mathbb{Z}$ and then this would extend to $\mathbb{Q}$, but I see now that this probably wouldn't fix say $\phi(1/m)$.

Answer 1

In order to appreciate the complexity of the set of homomorphisms ${\mathbb Q}\to GL(n, {\mathbb C})$, consider the simpler problem of describing homomorphisms of additive groups ${\mathbb Q}\to {\mathbb R}$. By regarding ${\mathbb R}$ as a rational vector space and choosing (using the Axiom of Choice) a basis (of the cardinality of continuum), we obtain: $$ {\mathbb R} \cong \bigoplus_{\alpha \in {\mathbb R}} {\mathbb Q}. $$
Every nonzero group homomorphism $h: {\mathbb Q}\to {\mathbb R}$ then corresponds to finitely many nonzero group homomorphisms $$ h_i: {\mathbb Q}\to {\mathbb Q}, i=1,...,n, $$ compositions of $h$ with coordinate projections
$$ \bigoplus_{\alpha \in {\mathbb R}} {\mathbb Q}\to {\mathbb Q}.$$ Each nonzero homomorphism $h_i: {\mathbb Q}\to {\mathbb Q}$ has the form $$ h_i(q)= a_i q, q\in {\mathbb Q}, a_i\in {\mathbb Q}^\times, $$ see here. Thus, describing homomorphisms ${\mathbb Q}\to {\mathbb R}$ amounts to describing tuples of nonzero rational numbers and finite subsets of ${\mathbb R}$.

Answer 2

• The homomorphism $\phi : \Bbb{Q}\to GL_n(\Bbb{C})$ will be of the form $$\phi(1/k!)= P \pmatrix{T_1(k) \\ & T_2(k)\\ && \ldots \\ &&& T_L(k)}P^{-1}\tag{1}$$ where each $T_l(k)$ is upper triangular $\in GL_{n_l}(\Bbb{C})$, and their elements on the same diagonal are equal, ie. $T_l(k)_{i,j} = T_l(k)_{i+1,j+1}$, and if $n_l\ne 1$ then $T_l(1)_{1,2}\ne 0$.

• Once we know $T_l(k)$ and $T_l(k+1)_{1,1}\in \Bbb{C}^*$ there is a unique matrix satisfying $T_l(k+1)^{k+1} = T_l(k)$ and the above condition. To prove $(1)$ exists use that all the $\phi(1/k!)$ commute and the simultaneous diagonalization theorems.

• Since $|T_l(k)_{1,1}| = |T_l(1)_{1,1}|^{1/k!}$ is already determined by $T_l(1)$, then $\phi$ is fully determined by $P$, the $T_l(1)$ and the sequences $a_l(k)=\frac1{2\pi} arg(T_l(k)_{1,1})\in \Bbb{R/Z}$ .

• $m/k! \to m a_l(k)$ is an homomorphism $\Bbb{Q\to R/Z}$. Let $b_l\in [0,1)$ such that $a_l(1)-b_l\in \Bbb{Z}$, then $m/k!\to m (a_l(k)-b_l/k!)$ is an homomorphism $\Bbb{Q/Z\to Q/Z}$. Since $\Bbb{Q/Z}\cong \prod_p' \Bbb{Z[p^{-1}]/Z}$ the homomorphisms $f:\Bbb{Q/Z\to Q/Z}$ are of the form $f(x) = \sum_p \{a_p x\}_p$ with $a_p\in \Bbb{Z}_p$ (the $p$-adic integers) and $\{.\}_p$ is the fractional part isomorphism $\Bbb{Q_p/Z_p\to Z[p^{-1}]/Z}$.

• The last question is given the $T_l$, can we change $P$ without changing $\phi$. The obvious modification is to change the ordering of the $T_l$, changing $P$ accordingly.

  Then there are 2 kind of modifications : if two sequences $T_l,T_{l'}$ are equal then we can mix their corresponding part of $P$ with a $GL_2(\Bbb{C})$ transformation without changing $\phi$.

  Otherwise if $Q_l$ commutes with all the $T_l(k))_{k\ge 1}$ then we can multiply the corresponding part of $P$ with $Q_l$ without changing the result. I think this happens when $Q_l$ is upper triangular and its elements on the same diagonal are equal.