Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let

  • $GL_{n}(\mathbb{R}) = \{ A \in M_{n}(\mathbb{R}) \ | \ \text{det}(A) \neq 0 \}$

  • $SL_{n}(\mathbb{R}) = \{ A \in M_{n}(\mathbb{R}) \ | \ \text{det}(A) = 1 \}$

We know that $$GL_{n}(\mathbb{R})/SL_{n}(\mathbb{R}) \cong \mathbb{R}^{\ast}$$

I am interested to know the answer for this question.

  • Does there exist a subgroup $H$ such that $GL_{n}(\mathbb{R})/H \cong \mathbb{R}$?
share|improve this question

2 Answers 2

up vote 11 down vote accepted

$\mathbb{R}^{\ast} \simeq \{ \pm 1 \} \times \mathbb{R}$: the isomorphism is gotten by looking at sign and taking the logarithm of the absolute value. So $\mathbb{R}^{\ast}$ has a quotient isomorphic to $\mathbb{R}$, and the kernel of the corresponding composition (the group of matrices with determinant $\pm 1$) provides the desired $H$.

share|improve this answer
5  
You think very fast. –  anonymous Jan 22 '11 at 20:43
1  
do you mean logarithm of the absolute value? –  Dylan Wilson Jan 23 '11 at 0:39
    
@Dylan: yes, sorry, that's what I meant. –  Qiaochu Yuan Jan 23 '11 at 0:45

Well, another way of thinking about this.

  • Consider the mapping from $\text{GL}_{n}(\mathbb{R}) \to (\mathbb{R}_{+},\times)$, $A \mapsto (\text{det}\:(A))^{2}$ and note that $(\mathbb{R},+) \cong (\mathbb{R}_{+},\times)$.

  • Consider the mapping from $\text{GL}_{n}(\mathbb{R}) \to (\mathbb{R}_{+},\times)$, $A \mapsto |\text{det}\:(A)|$ and note that $(\mathbb{R},+) \cong (\mathbb{R}_{+},\times)$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.