Is $PGL(n,\Bbb{R}) \cong SL(n+1, \Bbb{R})$ for even n?

Question

Is the following claim correct?

Claim: If n is even, $PGL(n,\Bbb{R}) \cong SL(n+1, \Bbb{R})$.

Proof:

Recall $PGL(n,\Bbb{R}) \cong GL(n+1, \Bbb{R})/Z$, where $Z = \{M | M=\alpha I, \alpha \in \Bbb{R}\}$.

Define a homomorphism $ \phi : GL(n+1,\Bbb{R}) \to SL(n+1, \Bbb{R})$ by $\phi (M) = \frac{1}{{\det(M)}^{\frac{1}{n+1}}}M$ (the root is well defined since n+1 is odd). Clearly the image of this map lies in $SL(n+1, \Bbb{R})$, since $\det(\frac{1}{{det(M)}^{\frac{1}{n+1}}}M) = {(\frac{1}{{\det(M)}^{\frac{1}{n+1}}})}^{n+1}\det(M)=1$; in fact it is all of $SL(n+1, \Bbb{R})$ since it acts as the identity on $SL(n+1, \Bbb{R})$.

The kernel of $\phi$ is exactly those matrices where $\frac{1}{{\det(M)}^{\frac{1}{n+1}}}M = I$, which clearly are matrices in $Z$; In fact it is all of $Z$, because it is easily see that any scalar matrix satisfies this condition. The claim now follows from the first isomorphism theorem:

$PGL(n,\Bbb{R}) \cong GL(n+1, \Bbb{R})/Z \cong GL(n+1, \Bbb{R})/\ker(\phi) \cong Im(\phi) \cong SL(n+1, \Bbb{R})$

Answer 1

The proof looks good to me.

Another way of framing this picture is to observe that for a finite-dimensional vector space $V$ over the field $\Bbb F$, the sequences $$0 \to Z(SL(V)) \to SL(V) \to PSL(V) \to 0$$ and $$0 \to PSL(V) \to PGL(V) \to \Bbb F^* / (\Bbb F^*)^{\dim V} \to 0$$ are exact. (A sequence $0 \to A \stackrel{f}{\to} B \stackrel{g}{\to} C \to 0$ is exact iff

• $f$ is injective,
• $\operatorname{im} f = \operatorname{ker} g$, and
• $g$ is surjective.)

In our situation $\Bbb F = \Bbb R$ and $V = \Bbb R^n$. If $n$ odd, then $Z(SL(V)) = \{I\}$ and $\Bbb R^* = (\Bbb R^*)^n$, so in fact $$SL(V) \cong PSL(V) \cong PGL(V) .$$ If $n$ even, then $Z(SL(V)) = \{\pm I\}$, and so $SL(V) \to PSL(V)$ is a double cover, and $(\Bbb R^*)^n = \Bbb R^+$, so $\Bbb R^* / (\Bbb R^*)^n \cong \{\pm 1\}$ and thus $PSL(V) \subset PGL(V)$ is a subgroup of index $2$.