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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.