Let A be the set of matrix of type \begin{bmatrix} a&0\\ 0 & a^{-1} \end{bmatrix} where a is a positive number. How to show that if $\phi$ is a Group homomorphism from A to $\mathbb{R}_{+}$ with multiplication, then there exists some x $\in $ $\mathbb{R}$ such that $\phi(a_t)=e^{tx}$, where $a_{t}$=\begin{bmatrix} e^{t}&0 \\ 0 & e^{-t}\\ \end{bmatrix}

I understand that every element of A can be represented in the form of $a_t$, but I am not getting why a continuous group homomorphism is of type this.


1 Answer 1


Assuming that you are using the subspace topology on $A$, you have mutually inverse group homeomorphisms $$f_A: (\mathbb{R}, +) \to A, \, t \mapsto a_t \text{ and } g_A: A \to (\mathbb{R},+), \, M \mapsto \ln(\pi_{1,1}(M))$$ where $\pi_{1,1}: A \to \mathbb{R}_+$ denotes the projection onto the top-left entry. Similarly there are mutually inverse group homeomorphisms $$f_+: (\mathbb{R}, +) \to (\mathbb{R}_+, \cdot), \, t \mapsto e^t \text{ and } g_+: (\mathbb{R}_+, \cdot) \to (\mathbb{R},+), \, x \mapsto \ln(x).$$

With this we can transform your question to a well-known problem: Let $\phi : A \to (\mathbb{R}_+, \cdot)$ be a continuous group homomorphism. Then $\psi := g_+ \circ \phi \circ f_A$ is a continuous group homomorphism $(\mathbb{R}, +) \to (\mathbb{R}, +)$, so it is given by $$\psi(t) = t \cdot \psi(1) \text{ for all } t \in \mathbb{R}.$$ This is immediate for $t \in \mathbb{Z}$ by the homomorphism property, then follows for $t \in \mathbb{Q}$ by clearing denominators and lastly follows for $t \in \mathbb{R}$ since $\mathbb{Q}$ is dense in $\mathbb{R}$ and limits in $\mathbb{R}$ are unique.

Now we have $\phi = f_+ \circ \psi \circ g_A$ and hence $$\phi(a_t) = \phi \circ f_A(t) = f_+ \circ \psi(t) = f_+(t \cdot \psi(1)) = e^{t \cdot \psi(1)} \text{ for all } t \in \mathbb{R}.$$


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