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Let $G$ be a simple Lie group of dimension $n$ (connected or even simply connected). Let $T$ be a maximal torus of dimension $d$. Notice that $\frac{n}{d}$ is an integer which I will denote by $m$. Let $g_{1},\dots g_{m}$ be $m$ elements of $G$ such that for $i\neq j$ we have $$ g_{i}T\cap g_{j}T=\emptyset$$

Notation: $g_{i}T=T_{i}$.

We define a smooth map. $$V:T_{1}\times \dots \times T_{m}\rightarrow G$$ $$(x_{1},\dots ,x_{m})\mapsto x_{1}x_{2}\dots x_{m}$$

My question is the following: is the map $V$ a local diffeomorphism around the point $(g_{1},\dots ,g_{m})$ i.e. the linear map (derivation) $T_{(g_{1},\dots ,g_{m})}V$ is of rank $n$ ?

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  • $\begingroup$ It's best to define your terms and notation explicitly since the theory of Lie groups (starting with real manifolds) has been infiltrated by the theory of linear algebraic groups defined over various fields. For example, what do you mean by "maximal torus"? This is a term coming from algebraic groups, which doesn't even occur in the index of Knapp's 1996 Birkhauser text. While "torus" in its older sense occurs in the study of compact Lie groups, a Lie group in general was originally studied by passing to the complexified Lie algebra and its (conjugate) Cartan subalgebras. $\endgroup$ Commented Nov 4, 2015 at 18:00

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No. Take $g_2=1$, $g_1\notin T$, then $g_1 T\cap g_2 T=g_1T\cap T=\emptyset$. The differential of the map $$\phi\colon T\times T\to G,\quad (t_1,t_2)\mapsto g_1 t_1 g_2 t_2=g_1t_1t_2$$ at the point $(1,1)\in T\times T$ (or at any other point of $T\times T$) is clearly of rank $d=\mathrm{dim}\, T$, and not $2d$, which implies the answer NO to the original question.

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