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.

I read somewhere that $PGL_2(\mathbb{C})=SL_2(\mathbb{C})/N$ where $N$ is the normal subgroup consisting of $\pm \left( \begin{matrix} 1 & 0\\ 0 & 1 \end{matrix}\right)$. It is unclear to me how this follows from the fact that $PGL_2(\mathbb{C})=GL_2(\mathbb{C})/aI$ where $aI$ are the scalar matrices ($I$ the identity and $a\in\mathbb{C}^*$). However, I did see somewhere the statement that $SL_2(\mathbb{C})/N$ is called $PSL_2(\mathbb{C})$ and that $PGL_2(\mathbb{C})\cong PSL_2(\mathbb{C})$.

Two questions:

1) How does one show that $PGL_2(\mathbb{C})\cong PSL_2(\mathbb{C})$?

2) Is there another (perhaps easier) way of seeing that $PGL_2(\mathbb{C})=SL_2(\mathbb{C})/N$?

Thanks.

share|improve this question
add comment

2 Answers

up vote 3 down vote accepted

Hint: If $M \in GL_2(\mathbb{C})$, let $\alpha$ be a square root of $\det(M)$. Then $\frac{1}{\alpha}M \in SL_2(\mathbb{C})$.

Notice that the equality is false over $\mathbb{R}$.

share|improve this answer
    
Ah, of course -- thanks! –  Nilay Kumar Jun 12 '13 at 3:27
add comment

Hints: Given an element $ M \in {\rm GL}(2, \mathbb{C})$, you can find a complex number $\lambda$ so that $\lambda M$ has determinant $1$. Note that the image of $M$ is the same as the image of $\lambda M$ in ${\rm PGL}(2,\mathbb{C}).$

share|improve this answer
    
Got it, thanks! –  Nilay Kumar Jun 12 '13 at 3:27
add comment

Your Answer

 
discard

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

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