$\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.


1 Answer 1


$\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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .