Computing central isogenies of reductive groups

Question

Is there any easy way to figure out what the groups $GL_n/ \mu_k$ ($\mu_k$ central) are depending on $n$ modulo $k$, or something? For example $GL_3/ \{ \pm \mathrm{Id}_3 \}$ I suspect it is $\cong GL_3$ because the fundamental group of $SL_3$ is order $3$ and $3$ is relatively prime with $2$. Is that enough to conclude? What about other orders?

Answer 1

EDIT: There is a mistake in the below solution (see the comments below)! I'll leave it up in case anyone can still learn something from it!

$\newcommand{\GL}{\mathrm{GL}}$$\newcommand{\SL}{\mathrm{SL}}$ Note that we have a short exact sequence of algebraic groups

$$0\to N\to Z\times\SL_n\xrightarrow{\varphi}\GL_n\to 0$$

where $Z$ is the center of $\GL_n$, $N$ is $\mu_n$ is embedded as $x\mapsto (xI,x^{-1}I)$, and $\varphi$ is the multiplication map. Note then that for each $k$ we get a short exact sequence

$$0\to N\varphi^{-1}(\mu_k)\to Z\times \SL_n\to\GL_n/\mu_k\to 0$$

Now, note that if $\varphi(z,g)=zg\in \mu_k$ then $g$ is evidently central. So, it's clear that

$$\varphi^{-1}(\mu_k)=\left\{ (abI,b^{-1}I):a\in\mu_k, b\in\mu_n)\right\}$$

If $\mathrm{gcd}(k,n)=1$ then $\mu_k\cap \mu_n=1$ and thus we see that the expression $abI$ with $a\in\mu_k$ and $b\in\mu_n$ is unique. Thus, we deduce that $\varphi^{-1}(\mu_k)=N(\mu_k\times \{I\})$. Thus, we see that

$$\GL_n/\mu_k\cong (Z\times \SL_n)/(N(\mu_k\times \{I\}))=(Z\times\SL_n/(\mu_k\times\{I\}))/N$$

Now, note that $Z/\mu_k\cong Z$ via the map $f:Z\to Z:z\mapsto z^k$. So, we see that

$$(Z\times\SL_n/(\mu_k\times\{I\}))/N\cong (Z\times\SL_n)/(\{(x^kI,x^{-1}I):x\in\mu_n)\}$$

Consider the map $\psi:Z\times\SL_n\to\GL_n$ given by $(z,g)\mapsto z^\ell g$ where $\ell k=1\mod n$. This is surjective since $z\mapsto z^\ell$ is surjective on $Z$. Moreover, the kernel is the set of $(z,g)$ such that $z^\ell g=1$ this means that $g=x^{-1}I$ and $z^\ell=x$, but this has only one solution which is $z=x^k$. Thus, $\ker\psi=\{(x^kI,x^{-1}I)\}$ and thus, in conclusion, we get an isomorphism

$$\GL_n/\mu_k\cong (Z\times\SL_n)/N\varphi^{-1}(\mu_k)\cong (Z/\SL_n)/\{(x^kI,x^{-1}I):x\in\mu_n)\}\cong \GL_n$$

I leave it to you to figure out what happens when $\mathrm{gcd}(k,n)>1$.

EDIT: Just to point out $\SL_3$ is simply connected. I think you meant that $Z(\SL_3)=\pi_1(\mathrm{PGL}_3)$ is of order $3$.

Answer 2

In this post I will show that $GL_n/\mu_k\cong GL_n$ if and only if $k\equiv \pm 1 \pmod n$.

We do this by classifying surjective homomorphisms from $GL_n$ to $GL_n$. Every surjective homomorphism from $GL_n$ to $GL_n$ arises from an irreducible representation of $GL_n$ of dimension $n$. These are classified.

In characteristic zero, every such homomorphism is conjugate to one of the form $g\mapsto g(\det(g))^m$ or $g\mapsto{} ^tg^{-1}(\det(g))^m$ for some $m\in \mathbb{Z}$. In positive characteristic, there are more due to Frobenius twists, but it turns out we can ignore them since their kernels are not of the form $\mu_k$.

It remains to compute the kernels of these homomorphisms. The answer is that the kernels are $\mu_{|mn\pm 1|}$ (Write an arbitrary element as something central times something in $SL_n$ to see this). This computation completes the proof.