Let $G$ be an algebraic group defined over a char 0
local field $k$. Following Borel and Tits (73) we define
the group $G^+(k)$ or $G^+$ by the subgroup of $G(k)$
generated by the unipotent elements of $G(k)$. 


Suppose $G$ is generated by a finite set of unipotent 
$k$-subgroups, say $U\_1,\cdots, U\_n$. Is it true that 
the group generated by $U\_1(k), \cdots, U\_n(k)$ is 
$G^+$? 


I feel the answer is positive but do not know how to prove it.
It seems that the ideas of the original paper of Borel and Tits can 
help, but I still do not read French (which I always plan to learn) yet. 



The group G^+ of algebraic groups over local fields
I haven't really checked the details, but I guess an argument works as follows. First, there exist $(k\_i)$ such that the product map $\prod U\_{k\_i}\to G$ is onto. I think this implies that its differential is surjective on a Zariski open set. Therefore it is onto at some point in $\prod U\_{k\_i}(K)$; it follows (by the constant rank theorem, applies between $p$-adic manifolds) that the image of $\prod U\_{k\_i}(K)$ in $G(K)$ has non-empty interior for the norm topology. It follows that the subgroup $V$ generated by the $U\_i(K)$ is open for the norm topology.


Now first assume $G$ is simple. Then a classical result (Prasad, Bull Soc Math France 1982, who attributes it to Tits) is that any open subgroup of $G(K)$ is either compact or contains $G^+$. Since $V$ is not compact (because any of the $U\_i(K)$, if not trivial, is noncompact), it follows that $V=G^+$. 


In the semisimple case, the same holds: since $V$ is an open subgroup and all its projection to $K$-simple factors are noncompact, $V=G^+$. Now since $G$ is generated by unipotents, it is semisimple modulo the unipotent radical. At this point the argument is complete for $G$ reductive but still needs a little more in general. If $H$ is the quotient by the unipotent radical $W$, it is probably true, that an open subgroup of $G(K)$ whose projection to the reductive quotient $H(K)$ contains $H(K)^+$, has to contain $N(K)^+$, where $N$ is the largest perfect normal subgroup in $G$. This should allow to conclude modulo $N$, i.e. reduce to the case when $G$ itself is unipotent, which is easy.


[Edit: I was more careful with the argument that the $U\_i(K)$ generate an open subgroup] 


