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Let $X$ be a smooth irreducible algebraic variety over the field of complex numbers ${\mathbb{C}}$. Let $x\in X({\mathbb{C}})$. Let $\tau$ be an automorphism of ${\mathbb{C}}$ (not necessarily continuous), and let $\tau X$ denote the $\tau$-conjugated ${\mathbb{C}}$-variety obtained from $X$ by transport of structure (i.e. by action of $\tau$ on the coefficients of equations defining $X$). We consider the topological fundamental groups $\pi_1(X({\mathbb{C}}),x)$ and $\pi_1((\tau X)({\mathbb{C}}),\tau x)$.

In the papers of Serre, Exemples de variétés projectives conjuguées non homéomorphes, C. R. Acad. Sci. Paris 258 (1964), 4194–4196, and of Milne and Suh, Nonhomeomorphic conjugates of connected Shimura varieties, one can find examples of $X$ and $\tau$ such that $\pi_1((\tau X)({\mathbb{C}}),\tau x)$ and $\pi_1(X({\mathbb{C}}),x)$ are not isomorphic. The authors conclude that in these cases the topological spaces $(\tau X)({\mathbb{C}})$ and $X({\mathbb{C}})$ are not homeomorphic.

In my very recent preprint with Cyril Demarche (excuse me for advertising my own work!) we consider the following situation. Let $X=G/H$, where $G$ is a connected linear algebraic group over ${\mathbb{C}}$, and $H\subset G$ any algebraic subgroup, not necessarily connected. Set $x:=eH\in X({\mathbb{C}})$. We prove that in this case $\pi_1((\tau X)({\mathbb{C}}),\tau x)$ and $\pi_1(X({\mathbb{C}}),x)$ are canonically isomorphic. I am trying to understand, what this really means.

Question. For a homogeneous space $X=G/H$ over ${\mathbb{C}}$ as above, and for $\tau\in {\rm Aut}({\mathbb{C}})$, is it always true that
(1) $(\tau X)({\mathbb{C}})$ and $X(\mathbb{C})$ are homotopically equivalent, or even
(2) $(\tau X)({\mathbb{C}})$ and $X(\mathbb{C})$ are homeomorphic, or even
(3) $\tau X$ and $X$ are isomorphic ${\mathbb{C}}$-varieties?

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Have you got in mind examples of $X$ not isomorphic to $\tau X$ as homogeneous spaces? I think that we can find $G$ unipotent and not isomorphic to $\tau G$ but this is far to any negative answer to your questions. – YCor Jan 8 '13 at 13:37
@Yves: I wanted to know, whether our result that the fundamental groups are isomorphic is trivial because $\tau X$ and $X$ are isomorphic $\mathbb{C}$-varieties. – Mikhail Borovoi Jan 8 '13 at 18:11
What is known when $H=1$? – Tom Goodwillie Jan 9 '13 at 13:40
What is known when $X$ is connected? What is known when $X$ is finite? – Tom Goodwillie Jan 9 '13 at 22:05

I answer the question in the comment of Tom Goodwillie: What is known when $H=1$?

Theorem. Let $G$ be a connected linear algebraic group over ${\mathbb{C}}$. Let $\tau$ be an automorphism of ${\mathbb{C}}$. Then the complex varieties $G$ and $\tau G$ are isomorphic.

Note that I do not claim that the algebraic groups $G$ and $\tau G$ are always isomorphic, see the comment of Yves Cornulier.

Proof. It suffices to show that as a variety $G$ can be defined over $\mathbb Q$.

Write $G^{\rm u}$ for the unipotent radical of $G$, and set $G^{\rm red}:=G/G^{\rm u}$, then $G^{\rm red}$ is a connected reductive ${\mathbb{C}}$-group. By Mostow's theorem $G\simeq G^{\rm u}\rtimes G^{\rm red}$, hence $G\simeq G^{\rm u}\times G^{\rm red}$ as a ${\mathbb{C}}$-variety. Using the exponential map, one sees easily that $G^{\rm u}$ is isomorphic to an affine space (the Lie algebra of $G^{\rm u}$) as a variety, hence as a variety it can be defined over ${\mathbb{Q}}$. Now it suffices to show that the reductive ${\mathbb{C}}$-group $G^{\rm red}$ admits a ${\mathbb{Q}}$-form (as an algebraic group).

Set $G^{\rm ss}=[G,G]$, it is a connected semisimple group. Let $G^{\rm sc}$ denote the universal covering of $G^{\rm ss}$, it is a simply connected semisimple $\mathbb{C}$-group. Let $Z^0$ denote the identity component of the center $Z$ of $G^{\rm red}$, it is a $\mathbb{C}$-torus. We have a canonical epimorphism $\phi\colon G^{\rm sc}\times_{\mathbb{C}} Z^0\to G$ with finite central kernel $\mu$.

Let $G_{1,{\mathbb{Q}}}$ be the direct product over ${\mathbb{Q}}$ of a split ${\mathbb{Q}}$-form (Chevalley's form) of $G^{\rm sc}$ and a split ${\mathbb{Q}}$-form of the torus $Z^0$. We have an epimorphism $\phi\colon G_{1,{\mathbb{C}}}\to G^{\rm red}$. Since $\mu\subset T_{1,{\mathbb{C}}}$ for some split maximal torus $T_{1,{\mathbb{Q}}}\subset G_{1,{\mathbb{Q}}}$, we see that $\mu$ is defined over ${\mathbb{Q}}$ in $T_{1,{\mathbb{C}}}$, i.e. $\mu=\mu_{\mathbb{Q}}\times_{\mathbb{Q}} {\mathbb{C}}$ for some central ${\mathbb{Q}}$-subgroup $\mu_{\mathbb{Q}}\subset G_{1,{\mathbb{Q}}}$.

Now we set $G_{\mathbb Q}^{\rm red}$ to be $G_{1,{\mathbb{Q}}}/\mu_{\mathbb{Q}}$, it is a ${\mathbb{Q}}$-form of $G^{\rm red}$.

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The result in my preprint mentioned in the question was erroneous (the mistake was noticed by a referee). It is possible to construct a quotient $X=G/H$ and and automorphism $\tau$ of $\mathbb{C}$ such that $\pi_1(\tau X(\mathbb{C}))$ is not isomorphic to $\pi_1(X(\mathbb{C}))$, see my preprint with Yves Cornulier based on his answer to this question. In our example $G={\rm SL}(n,\mathbb{C})\times \mathbb{C}^*$ with $n\ge 5$, and $H$ is a finite nonabelian subgroup of order 55. This time the referee noticed no mistakes... Thus the answer to the questions (1), (2), and (3) is NO.

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