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Let $K$ be a field and consider the surjective determinant homomorphism $\mathrm{GL}_n(K)\to K^\times$. Since the kernel is the special linear group $\mathrm{SL}_n(K)$ we obtain a short exact sequence

$$\require{AMScd}\begin{CD} 1 @>>> \mathrm{SL}_n(K) @>i>> \mathrm{GL}_n(K) @>\det>> K^\times @>>> 1\\ \end{CD}$$

Since the determinant map has a section $s:K^\times\to\mathrm{GL}_n(K)$ defined by $$s(\alpha):=\begin{pmatrix} \alpha&&&\\ &1&& \\ &&\ddots& \\ &&&1\end{pmatrix},$$ we conclude from the splitting lemma that $\mathrm{GL}_n(K)\approx\mathrm{SL}_n(K)\rtimes K^\times$. Here's a question:

Under what conditions on $K$ and $n$ does the inclusion homomorphism $i:\mathrm{SL}_n(K)\to\mathrm{GL}_n(K)$ have a retraction?

Example: If $K=\mathbb{R}$ and $n$ is odd, then every $\alpha\in\mathbb{R}^\times$ has a unique real $n$-th root, so we obtain a group homomorphism $\sqrt[n]{\cdot}:\mathbb{R}^\times\to\mathbb{R}^\times$, and we can use this to define a retraction $r:\mathrm{GL}_n(\mathbb{R})\to\mathrm{SL}_n(\mathbb{R})$ by $$r(A):=\frac{1}{\sqrt[n]{\det(A)}}\cdot A.$$ Now it follows from the splitting lemma that $\mathrm{GL}_n(\mathbb{R})\approx \mathrm{SL}_n(\mathbb{R})\times\mathbb{R}^\times$. Here's another question:

Is there a topological/geometric explanation for this example? Is there a topological/geometric obstruction when $n$ is even?

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    $\begingroup$ It's if and only if the set of $n$-th roots of unity in $K^*$ has a direct summand in $K^*$. $\endgroup$
    – YCor
    Commented Oct 30, 2015 at 17:53
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    $\begingroup$ Just an observation that if $K$ has characteristic $p> 0$ and $n$ is a power of $p$ then $GL_n(K) = SL_n(K) \times Z(GL_n(K))$. $\endgroup$
    – Jay Taylor
    Commented Oct 30, 2015 at 17:58
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    $\begingroup$ @YCor The square roots of unity are a direct summand of $\mathbb{R}^\times$: $\mathbb{R}^\times = \{\pm 1\}\oplus \mathbb{R}^\times_{>0}$, but there can be no retraction $\varphi:GL_2(\mathbb{R})\to SL_2(\mathbb{R})$ because $A:=[-1,0;0,1]$ and $B:=[0,1;1,0]$ both square to $1$, every element of $SL_2(\mathbb{R})$ which squares to $1$ is central, so we would need $1=[\varphi(A),\varphi(B)]=\varphi([A,B])=\varphi([-1,0;0,-1])=[-1,0;0,-1]$. This same argument seems to work for any $K$ of characteristic not $2$. $\endgroup$
    – Julian Rosen
    Commented Oct 30, 2015 at 22:54
  • $\begingroup$ @JulianRosen thanks, you're right. My claim in the comment is not correct (only one implication holds); this is fixed in the answer below. $\endgroup$
    – YCor
    Commented Oct 30, 2015 at 23:55

1 Answer 1

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$\mathrm{GL}_n(K)\to\mathrm{SL}_n(K)$ has a retraction iff the following two conditions hold

  1. The subgroup of $n$-root of unity in $K^*$ has a direct summand in $K^*$.
  2. $x\mapsto x^n$ is surjective on $K$

Let us first check that (1.) is equivalent to the existence of a retraction in restriction to $K^*\mathrm{SL}_n(K)$ (the subgroup generated by homotheties and unimodular matrices). Clearly it implies it (consider the set of scalar matrices whose diagonal entry belongs to this direct summand). Conversely, if there is a retraction, its kernel has trivial intersection with $\mathrm{SL}_n(K)$, hence contained in its centralizer, which is reduced to scalar matrices, so it should form the set of scalar matrices with diagonal entry in some direct summand of the set $n$-roots of unity in $K^*$.

Now it is clear that $K^*\mathrm{SL}_n(K)$ equals $\mathrm{GL}_n(K)$ if and only if (2.) holds; so if both (1.) and (2.) hold it follows that we have a retraction; conversely if we have a retraction, its kernel is a normal subgroup with trivial intersection with $\mathrm{SL}_n(K)$, hence contained in its centralizer, which is reduced to scalar matrices, so $\mathrm{GL}_n(K)$ should be generated by unimodular and scalar matrices, i.e. (2.) holds, and then the first verification shows that (1.) holds.

Edit: as noticed by Julian Rosen, (2.) together with (1.) implies something much stronger than (1.), namely that the subgroup of $n$-roots of unity is actually trivial, which means that $x\mapsto x^n$ is injective on $K$. To conclude:

$\mathrm{GL}_n(K)\to\mathrm{SL}_n(K)$ has a retraction (in the category of groups) iff $x\mapsto x^n$ is a permutation of $K$

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  • $\begingroup$ By the way Condition (1.) should be characterized in a more convenient way. Call $n^*$-root of unity, any $m$-root of unity where all prime divisors of $m$ divide $n$. Then (1.) implies that (X) any $n^*$-root of 1 in $K$ is also a $n$-root of 1. I'm wondering whether conversely (X) implies (1.). (X) is equivalent to the following even simpler one from a model-theoretic point of view: every $n^2$-root of unity is a $n$-root of unity. $\endgroup$
    – YCor
    Commented Oct 31, 2015 at 0:02
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    $\begingroup$ The map $x\mapsto x^n$ kills the $n$-th roots of unity, so if this map is surjective and the $n$-th roots of unity are a summand, then the $n$-th roots of unity should be trivial. Actually is looks like conditions (1) and (2) together are equivalent to "the map $x\mapsto x^n$ is bijective on $K$". $\endgroup$
    – Julian Rosen
    Commented Oct 31, 2015 at 0:39

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