# A question on algebraic torus

Let $T$ be an algebraic torus defined over $\mathbb Q$, $T_\infty$ be its real points and $\pi_0(T_\infty)$ be the group of connected components of $T_\infty$.

Why is the homomorphism $T(\mathbb Q)\to \pi_0(T_\infty)$ surjective?

Thanks

-
Are you sure this is true? This would be a consequence of weak approximation, but weak approximation does not hold for all tori. (See the Counterexamples section of aimath.org/WWN/qptsurface2/articles/html/9a .) – David Speyer Feb 6 '12 at 16:11
In this question mathoverflow.net/questions/50570/… , it is attributed to Serre that $T(\mathbf{Q})$ is always dense in $T(\mathbf{R})$ for any algebraic torus $T/\mathbf{Q}$, but I don't know the precise reference. – François Brunault Feb 6 '12 at 17:14
It is proved in Sansuc's classical paper digizeitschriften.de/main/dms/img/?IDDOC=505419, Cor. 3.5(iii), that for any connected linear algebraic group $G$ over $\mathbf{Q}$, the group of $\mathbf{Q}$-points $G(\mathbf{Q})$ is dense in $G(\mathbf{R})$. Sansuc writes at the end of the proof of Cor. 3.5 that for tori this result is due to Serre, but gives no reference. Voskresenskii in his book "Algebraic tori" (in Russian) writes on page 178 that Serre worked on weak approximation for tori and that his results were not published. – Mikhail Borovoi Mar 7 '12 at 19:04
The proof of real approximation for tori in English can be found in Voskresenskii's book in English "Algebraic Groups and Their Birational Invariants", Theorem 11.5. – Mikhail Borovoi Mar 7 '12 at 19:27
Voskresenskii proves that for a $k$-torus $T$ defined over a global field $k$ and splitting over a finite Galois extension $L/k$ with Galois group $\Pi$, and for a finite set $S$ of places $v$ of $k$ with cyclic decomposition groups in $\Pi$, the group $T(k)$ is dense in $\prod_{v\in S}T(k_v)$. (This result is also due to Serre). Now take $k=\mathbf{Q}$, $S=\{\infty\}$, then a decomposition group of $\infty$ is of order 2, hence cyclic, whence we obtain that $T(\mathbf{Q})$ is dense in $T(\mathbf{R})$. – Mikhail Borovoi Mar 7 '12 at 19:42