# Is Deligne's braiding functorial?

$$\newcommand{\ssc}{{\rm sc}} \newcommand{\ad}{{\rm ad}} \newcommand{\Fbar}{{\overline F}}$$ Let $$F$$ be a field and $$\Fbar$$ be a fixed algebraic closure of $$F$$. Let $$G$$ be a (connected) reductive group over $$F$$. Let $$G^\ssc$$ denote the universal cover of the commutator subgroup $$[G,G]$$ of $$G$$. Following Deligne, Variétés de Shimura: interprétation modulaire et techniques de construction de modèles canoniques, Proc. Sympos. Pure Math 33, Part 2, 1979, Section 2.0.11, we consider the composite homomorphism $$\rho\colon\ G^\ssc\to [G,G]\to G.$$

Deligne (loc. cit., Section 2.0.2) noticed that the commutator map $$[\ ,]\colon\ G\times G\to G,\quad\ g_1,g_2\mapsto [g_1,g_2] := g_1 g_2 g_1^{-1} g_2^{-1}$$ lifts to a certain map (morphism of $$F$$-varieties) $$\lbrace\ , \rbrace \colon\ G\times G\to G^\ssc,\quad\ g_1,g_2\mapsto \lbrace g_1,g_2 \rbrace$$ as follows. The commutator map $$G^\ssc\times G^\ssc\to G^\ssc,\quad\ s_1,s_2\mapsto [s_1,s_2]:= s_1 s_2 s_1^{-1} s_2^{-1}$$ clearly factors via a morphism of $$F$$-varieties $$(G^\ssc)^\ad\times (G^\ssc)^\ad\to G^\ssc$$ where $$(G^\ssc)^\ad=G^\ssc/Z_{G^\ssc}$$ and $$Z_{G^\ssc}$$ denotes the center of $$G^\ssc$$. Identifying $$(G^\ssc)^\ad$$ with $$G^\ad:= G/Z_G$$, we obtain the desired morphism of $$F$$-varieties $$\lbrace\ ,\rbrace\colon\ G\times G\to G^\ad\times G^\ad\to G^\ssc.$$ On $$\Fbar$$-points, if $$g_1,g_2\in G(\Fbar),\ g_1=\rho(s_1) z_1,\ g_2=\rho(s_2) z_2$$ where $$s_1,s_2\in G^\ssc(\Fbar),\ z_1,z_2\in Z_G(\Fbar)$$, then $$\lbrace g_1,g_2\rbrace=[s_1,s_2].$$ The constructed map $$\lbrace\ ,\rbrace$$ has nice properties, in particular, $$\rho\big(\lbrace g_1,g_2\rbrace\big)=[g_1,g_2]\qquad\text{and} \qquad \lbrace g_1,g_2\rbrace=\lbrace g_2,g_1\rbrace^{-1}.$$ Actually, $$\lbrace\ ,\rbrace$$ is a symmetric braiding of the crossed module $$(G^\ssc\to G)$$. We call it Deligne's braiding.

Now let $$\varphi\colon G\to H$$ be a homomorphism of reductive $$F$$-groups. It induces a homomorphism $$\varphi^\ssc\colon G^\ssc\to H^\ssc$$. The maps $$[\ ,]\colon\ G\times G\to G,\ g_1,g_2\mapsto [g_1,g_2]\quad\text{and} \quad [\ ,]\colon\ G^\ssc\times G^\ssc\to G^\ssc,\ s_1,s_2\mapsto [s_1,s_2]$$ are functorial in $$G$$: $$\varphi\big([g_1,g_2]\big)=\big[\varphi(g_1),\varphi(g_2)\big]\quad\text{and} \quad \varphi^\ssc\big([s_1,s_2]\big)=\big[\varphi^\ssc(s_1),\varphi^\ssc(s_2)\big].$$

Question. Is Deligne's braiding functorial? In other words, is it true that for any homomorphism $$\varphi\colon G\to H$$, we have $$\varphi^\ssc\big(\lbrace g_1,g_2\rbrace\big)= \big\lbrace\varphi(g_1),\varphi(g_2)\big\rbrace\quad \text{for all}\ \ g_1,g_2 \in G\ ?$$

The answer is Yes when homomorphism $$\varphi$$ is normal, that is, $$\varphi(G)$$ is normal in $$H$$. Indeed, then $$\varphi$$ induces homomorphisms $$Z_G\to Z_H,\quad Z_{G^\ssc}\to Z_{H^\ssc},\quad G^\ad\to H^\ad.$$ In general I expect the answer No, but cannot construct a counter-example.

$$\newcommand{\ssc}{{\rm sc}} \newcommand{\sss}{{\rm ss}} \newcommand{\ad}{{\rm ad}} \newcommand{\wh}{\widehat} \newcommand{\wt}{\widetilde} \newcommand{\pitil}{\tilde\pi} \newcommand{\rhotil}{\tilde\rho} \newcommand{\gtil}{\tilde g} \newcommand{\stil}{\tilde s}$$The answer is Yes.

Proposition. Let $$\varphi\colon G\to H$$ be a homomorphism of connected reductive groups over an algebraically closed field $$F$$. Let $$\varphi^\ssc\colon G^\ssc\to H^\ssc$$ denote the induced homomorphism. Then for any $$g_1,g_2\in G(F)$$ we have $$\big\lbrace \varphi(g_1),\varphi_(g_2)\big\rbrace _H=\varphi^\ssc\big\lbrace g_1,g_2\big\rbrace _G$$ where for simplicity we write $$\varphi^\ssc\big\lbrace g_1,g_2\big\rbrace _G$$ instead of $$\varphi^\ssc\big(\lbrace g_1,g_2\rbrace _G\big)$$.

Proof. Since the map $$\big\lbrace \ ,\big\rbrace _H$$ factors via $$H^\ad$$, we may and shall assume that $$H=H^\ad$$.

Consider the homomorphisms $$\varphi\colon G\to H$$ and $$\rho_H\colon H^\ssc\to H$$. The fiber product $$\wh G=G\times_H H^\ssc$$ is endowed with two homomorphisms $$\hat \pi_G\colon \wh G\to G\quad\text{and}\quad \hat\pi_H\colon\wh G\to H^\ssc.$$ Since the homomorphism $$\rho_H\colon H^\ssc\to H$$ is surjective with finite kernel, so is the homomorphism $$\hat \pi_G\colon \wh G\to G$$.

Let $$\wt G$$ denote the identity component of $$\wh G$$. Let $$\pitil_G\colon\wt G\to G\quad\text{and}\quad \pitil_H\colon \wt G\to H^\ssc$$ denote the restrictions to $$\wt G$$ of $$\hat \pi_G$$ and $$\hat \pi_H$$, respectively. Then $$\pitil_G$$ is a surjective homomorphism with finite kernel. It follows that $$\wt G$$ is a connected reductive $$F$$-group.

Write $$G=C\cdot G^\ssc$$ where $$C$$ is the radical (largest central torus) of $$G$$, and $$G^\sss=[G,G]$$ is the commutator subgroup of $$G$$. Similarly, write $$\wt G=\wt C\cdot \wt G^\ssc$$ where $$\wt C$$ is the radical of $$\wt G$$, and $$\wt G^\sss=[\wt G,\wt G]$$. Then we have surjective homomorphisms $$\pitil^\sss\colon \wt G^\sss\to G^\sss$$ and $$\pitil_C\colon \wt C\to C$$ with finite kernels. It follows that there exists a unique surjective homomorphism with finite kernel $$\rhotil\colon G^\ssc\to\wt G^\sss$$ such that $$\pitil^\sss\circ\rhotil=\rho_G\colon\ G^\ssc\to G^\sss.$$

From the commutative diagram $$\require{AMScd}$$ $$\begin{CD} G^\ssc @>\rhotil>> \wt G^\sss @>\pitil_H>> H^\ssc\\ @| @VV\pitil^\sss V @VV\rho_H V\\ C^\ssc @>\rho_G>> G^\sss @>\varphi>> H \end{CD}$$ we see that $$\varphi^\ssc=\pitil_H\circ \rhotil\colon\ G^\ssc\to H^\ssc.$$

Now let $$g_i\in G(F)$$, $$i=1,2$$. Write $$g_i=c_i\cdot s_i \qquad\text{where}\ \ c_i\in C(F),\ s_i\in G^\sss(F).$$ We lift $$c_i$$ to some $$\tilde c_i\in\wt C(F)$$ and $$s_i$$ to some $$\stil_i\in \wt G^\sss(F)$$. We set $$\gtil_i=\tilde c_i\cdot \stil_i\in \wt G (F)$$; then $$\pitil_G(\gtil_i)=g_i$$. We lift $$\stil_i$$ to some $$s^\ssc_i\in G^\ssc(F)$$.

Set $$h_i^\ssc=\pitil_H(\gtil_i)\in H^\ssc(F)$$. Then $$\rho_H(h_i^\ssc)=\varphi(g_i)$$, and we have \begin{align*} \big\lbrace \varphi(g_1),\varphi(g_2)\big\rbrace _H :&=[h_1^\ssc,h_2^\ssc]= \pitil_H[\gtil_1,\gtil_2]\\ &=\pitil_H[\stil_1,\stil_2]= \pitil_H\rhotil [s_1^\ssc,s_2^\ssc]=\varphi^\ssc[s_1^\ssc,s_2^\ssc]. \end{align*} Since $$g_i=c_i\cdot s_i=c_i\cdot\pitil_H(\stil_i)=c_i\cdot \rho_G(s_i^\ssc)$$, we see that $$\big\lbrace g_1,g_2\big\rbrace _G=[s_1^\ssc,s_2^\ssc]\in G^\ssc(F)$$ and $$\varphi^\ssc[s_1^\ssc,s_2^\ssc]=\varphi^\ssc\big\lbrace g_1,g_2\big\rbrace _G.$$ Thus $$\big\lbrace \varphi(g_1),\varphi(g_2)\big\rbrace _H=\varphi^\ssc\big\lbrace g_1,g_2\big\rbrace _G,$$ as required.