2
$\begingroup$

I want to prove that $\Bbb R$ isa homomorphic image of $\mathrm{GL}_2(\Bbb R)$.

I have found a subgroup $\left\{\begin{bmatrix}1&a \\ 0&1\end{bmatrix} : a \in \Bbb R\right\}$ of $\mathrm{GL}_2(\Bbb R)$ which is isomorphic to $\Bbb R$. How to prove the requires statement using this subgroup or any other simpler way?

Thank you.

$\endgroup$
4
  • $\begingroup$ The subgroup you have found is nice and simple, but is it a homomorphic image of $GL_2(\Bbb R)$? $\endgroup$
    – lhf
    Apr 24, 2020 at 14:43
  • $\begingroup$ I wonder whether the subgroups of $GL_2(\Bbb R)$ that are isomorphic to $\Bbb R$ are known. $\endgroup$
    – lhf
    Apr 24, 2020 at 14:45
  • $\begingroup$ @lhf is there more than one subgroup that is isomorphic to $\Bbb R$? $\endgroup$ Apr 24, 2020 at 14:46
  • $\begingroup$ I don't know. See math.stackexchange.com/questions/3641778/… $\endgroup$
    – lhf
    Apr 24, 2020 at 14:55

2 Answers 2

7
$\begingroup$

Try $\det$, which is a homomorphism $\mathrm{GL}_2(\Bbb R)\to \Bbb R^\times$, followed by $\operatorname{abs}$, which is a homomorphism between multiplicative groups $\Bbb R^\times \to \Bbb R_{>0}$, followed by $\ln$, which is a homomorphism $\Bbb R_{>0}\to\Bbb R$ (from multiplicative to additive). $$ (\mathrm{GL}_2(\Bbb R),\cdot)\stackrel\det\longrightarrow(\Bbb R^\times,\cdot)\stackrel{|\ \cdot\ |}\longrightarrow(\Bbb R_{>0},\cdot)\stackrel\ln\longrightarrow(\Bbb R,+),$$ where we have surjectivtiy (and also, e.g. continuity) at each step. To make the whole thing "more algebraic", one may prefer to use squaring instead of absolute value in the middle

$\endgroup$
2
  • $\begingroup$ Thank you. is this homomorphism is also continuous? $\endgroup$ Apr 24, 2020 at 14:34
  • $\begingroup$ @Wanttolearn it is the composition of continuous homomorphisms $\endgroup$ Apr 24, 2020 at 14:35
0
$\begingroup$

How about the determinant function?

$\endgroup$
3
  • $\begingroup$ Thank you. I am also thought of determinant. But it goes to $\Bbb R \backslash \{0\}$. From this how to proceed further? Using Log map? The resulting homomorphism is also continuous? $\endgroup$ Apr 24, 2020 at 14:31
  • $\begingroup$ You're right. Let's think on it more. You didn't mean $(\Bbb R^*,×)$ did you? $\endgroup$
    – user403337
    Apr 24, 2020 at 14:39
  • $\begingroup$ Ok. You're right. Log. $\endgroup$
    – user403337
    Apr 24, 2020 at 14:40

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .