Projecting vector field from $\mathbb{R}^n$ to $T^n$ Let $T^n = S^1 \times ...\times S^1$. $T^n$ is Lie group and the map $\pi:\mathbb{R}^n \to T^n$ defined by $\pi(x_1,..x_n) = (e^{ix_1},..,e^{ix_n})$ gives a homomorphism between the additive group to multiplicative group $T^n$.

which vector fields on $\mathbb{R}^n$ project to vector fields on
  $T^n$ under $d\pi$,

I'm having a hard time understanding this question.
I don't have good background in differentiable manifolds; I have the definition that a vector field $X$ on a Lie group $G$ is a linear map $X: C^{\infty}(G) \to C^{\infty}(G)$ s.t $X(fg) = fX(g) + gX(f)$.
My question:
What is formally meant by $d\pi$ and how does it send a vector field to another?
I know that if $X$ is a vector field on $\mathbb{R}^n$, and $x \in \mathbb{R}^n$ then we have a tangent vector $X_x: C^{\infty}(\mathbb{R}^n) \to \mathbb{R}$ where $X_x(f) = (Xf)(x)$. Then $d\pi_x(X_x)(f) = X_x(f \circ \pi)$ is a tangent vector itself.
That is, $d\pi_x : T_x\mathbb{R}^n \to T_{\pi(x)}T^n$. So how does $d\pi$ operate on vector fields, and what do I need to check in this case?
Thanks!
 A: Functions on $\mathbb{T}^n$ give functions on $\mathbb{R}^n$, i.e. you have a natural map (precomposing by $\pi$) $\pi^*: C^\infty(\mathbb{T}^n)\to C^\infty(\mathbb{R}^n)$. 
The question is : for which $X: C^\infty(\mathbb{R}^n)\to C^\infty(\mathbb{R}^n)$ is there $\tilde{X} : C^\infty(\mathbb{T}^n)\to C^\infty(\mathbb{T}^n)$ making the obvious diagram commute; that is such that $X\circ \pi^* = \pi^*\circ \tilde{X}$ ? 
Phrasing the question as "which vector fields on $\mathbb{R}^n$ project to vector fields on $\mathbb{T}^n$ under $\mathrm{d}\pi$ ?" suggests that the author was thinking of vector fields in terms of sections of the tangent bundle instead of as derivations of the function algebra. 
Indeed, for a manifold $M$ you have a tangent bundle $p:TM\to M$, and a smooth map $f:M\to N$ induces $\mathrm{d}f : TM\to TN$ defined by $\mathrm{d}f (v,x) = \mathrm{d}f_x(v)$. We call a vector field on $M$(this is where the terminology comes from by the way) a section of $p$, that is a map $s: M\to TM$ such that $p\circ s= id_M$. This last condition simply means that for all $x$, $s(x)\in T_xM$, that is $s$ is a smooth map that associates to each point a tangent vector. 
Now if you are given a vector field $X$ according to this second definition, you get a derivation of $C^\infty(M)$ the following way : given $f:M\to \mathbb{R}$, define $\mathcal{L}_Xf : M\to \mathbb{R}$ by $\mathcal{L}_Xf (x) = \mathrm{d}f(X(x),x)$. This is clearly a smooth function, and since you have a natural identification $T\mathbb{R}\cong \mathbb{R\times R}$, under this natural identification, you get that $\mathcal{L}_Xf : M\to \mathbb{R}$ is smooth, and it's a standard fact that $\mathcal{L}_X$ is a derivation of $C^\infty(M)$, it's called the Lie derivative along $X$. 
Ok now it's a theorem that the map $\mathcal{L}: \Gamma(M,TM) \to \mathrm{Der}(C^\infty(M))$ is a Lie algebra isomorphism (well actually, a vector space isomorphism, which allows you to give $\Gamma(M,TM)$ the structure of Lie algebra deduced from the one on $\mathrm{Der}(C^\infty(M))$); where $\Gamma(M,TM)$ denotes the sections of the tangent bundle, and $\mathrm{Der}(A)$ is the Lie algebra of derivations of the algebra $A$.
This allows you to see a vector field as either a derivation (as you defined it) or as a section of the tangent bundle. Now with the point of view of sections, the question becomes easier to understand with this phrasing. 
Indeed you have a diagram $\require{AMScd} \begin{CD}
T\mathbb{R}^n @>{\mathrm{d}\pi}>> T\mathbb{T}^n\\ @V{p}VV @VV{p}V\\
\mathbb{R}^n @>>{\pi}> \mathbb{T}^n
\end{CD}$ where $\mathrm{d}\pi$ appears explicitly. Now if you are given a section $X$ on the right side, you may be interested in finding out when there is a section $\tilde{X}$ on the left side which makes this commute :
$\require{AMScd} \begin{CD}
\mathbb{R}^n @>{\pi}>> \mathbb{T}^n\\ @V{X}VV @VV{\tilde{X}}V\\
T\mathbb{R}^n @>>{\mathrm{d}\pi}> T\mathbb{T}^n
\end{CD}$, that is "when is there $\tilde{X}$ that lifts $\mathrm{d}\pi \circ X$; that is the projection of $X$ under $\mathrm{d}\pi$". 
So "project under $\mathrm{d}\pi$" now makes sense ! 
So you're asking for $\tilde{X}$ such that $\mathrm{d}\pi\circ X = \tilde{X}\circ \pi$. By some usual manifold nonsense, this is the same as saying that for any $f:\mathbb{T}^n \to \mathbb{R}$, and any $x\in\mathbb{R}^n$, $\mathrm{d}f_{\pi(x)}\circ \mathrm{d}\pi_x(X(x)) = \mathrm{d}f_{\pi(x)}(\tilde{X}(\pi(x)))$; but the LHS is just $\mathrm{d}(f\circ\pi)_x(X(x)) = \mathcal{L}_X(f\circ\pi)(x)$ and the RHS is just $(\mathcal{L}_{\tilde{X}}f)(\pi(x))$. 
Asking that they're equal for all $f,x$, is just asking that $\mathcal{L}_X\circ\pi^* = \pi^*\circ\mathcal{L}_{\tilde{X}}$; which, if you use the identification between sections and derivations, is exactly the equation I gave at the very beginning. 
Now I don't know if you also wanted a hint to solve the question, but I think it's way easier to understand what's happening with vector fields seen as sections rather than as derivations, it's way more geometric and the condition for a vector field to project to another one under $\mathrm{d}\pi$ becomes clear. 
