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Let $G$ be a semisimple and simply connected linear algebraic group over $\mathbb{C}$. Let $H$ be a connected, Zariski closed and semisimple linear algebraic $\mathbb{Q}$-subgroup of $G$.

Is $H$ a simply connected linear algebraic group?

Here "simply connected" means every central isogeny to $G$ is an isomorphism.

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  • $\begingroup$ No. $H$ need not be simply connected $\endgroup$
    – Venkataramana
    Commented Mar 2, 2015 at 16:54
  • $\begingroup$ @Amir: Your definition of "simply connected" is too narrow for this kind of question, which makes sense and has a similar easy answer over any algebraically closed field. Aside from that, there are already questions on MO which cover similar ground in a more refined style. Search for "simply connected" [algebraic-groups]. $\endgroup$
    – Jim Humphreys
    Commented Mar 2, 2015 at 23:23

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As Venkataramana says, the answer is no. In fact, every linear algebraic $\mathbb{Q}$-group $H$, whether simply connected or not, is (isomorphic to) a Zariski closed $\mathbb{Q}$-subgroup of some $SL_n$ (which is semisimple and simply connected). (By definition of being an algebraic group, $H$ is a Zariski closed $\mathbb{Q}$-subgroup of $GL_k$, for some $k$. Furthermore, it is well known that $GL_k$ can be embedded as a Zariski-closed $\mathbb{Q}$-subgroup of $SL_{k+1}$.)

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