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Let $G$ be an algebraic group, and $G_{Id}$ the connected component of the identity. Then $G_{Id}$ is a normal subgroup of $G$ and $G/G_{Id}$ is the component group of $G$.

Let $G_{c}\subset G$ be another connected component of $G$. Is is possible to define a group structure on $G_c$?

Assume we know that for any connected component $G_c$ of $G$ we have an injective morphism $f_c:G_c\rightarrow H\times F$ where $H$ is a group and $F$ a finite group. Now, for any $g\in g$ there exists a unique connected component $G_c$ such that $g\in G_c$ and we may define a map $$f:G\rightarrow H\times F,\: g\mapsto f_c(g).$$ If $f$ is surjective can we conclude that $f$ is an isomorphism?

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  • $\begingroup$ The answer to your first question is certainly yes, since there is a bijection between the set $G_c$ and the connected component of the identity (which is usually denoted $G^\circ$ or the like). The answer to the second question is presumably no, but the formulation is not quite clear to me. $\endgroup$
    – Jim Humphreys
    Commented Mar 24, 2015 at 22:10
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    $\begingroup$ If the base field isn't separably closed, it's possible that a non-identity connected component might not have a rational point, in which case it can't possibly be given the structure of a group scheme over the base field. For example, $\mu_3=\mathrm{Spec}(\mathbf{Q}[X]/(X^3-1))$ has two components, both single points, and the non-identity one is not a $\mathbf{Q}$-rational point (and incidentally splits into two components over $\mathbf{Q}(\zeta_3)$). But maybe you want your base field to be algebraically closed? $\endgroup$
    – Keenan Kidwell
    Commented Mar 24, 2015 at 22:23
  • $\begingroup$ Yes, I am assuming the base field to be algebraically closed. $\endgroup$
    – user58018
    Commented Mar 25, 2015 at 17:40

2 Answers 2

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Each connected component is (in a natural way) a torsor under the identity component. The choice of a rational point (if there is one) defines an isomorphism with the identity component, and makes the component into an algebraic group.

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  • $\begingroup$ Thanks a lot to both of you for the answer. I will rephrase my second question in a more precise way. $\endgroup$
    – user58018
    Commented Mar 25, 2015 at 17:12
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To expand my short comments, it's possible (though not usually interesting) to place a group structure on any component $G_c$ using an obvious bijection between this set and the connected component of the identity.

Concerning the second question, look at a maximal torus $T$ in a connected simple algebraic group such as $\mathrm{SL}_n(\mathbb{C})$. Usually the normalizer $N:=N_G(T)$ isn't a direct product of a finite group and $T$ even though you could write each connected component as $\{n\} \times T$ for an element $n$ of the finite (hence algebraic) Weyl group $N/T$. This sort of thing happens for example in special linear groups, where the Weyl group is a symmetric group.

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