Image of the pushforward $dF_{p}$ with $F(p)=(p,p,p)$ in $S^{1}$

Question

Let $F\colon S^{1}\to S^{1}\times S^{1}\times S^{1}$ be given by $F(p)=(p,p,p)$. One can show that this map is differentiable, but I'm trying to find the image of $dF_{p}\colon T_{p}S^{1}\to T_{F(p)}(S^{1}\times S^{1}\times S^{1})$ (the differential or pushforward). The pushforward is defined by $dF_{p}(X)(f) = X(f\circ F)$, recall that $X\colon C^{\infty}(S^{1})\to \mathbb{R}$ are the derivations. Clearly $(f\circ F)(p) = f(p,p,p)$, but I don't quite understand what they want me to find in the image, i.e. if there will be a special form. Any suggestions? I was thinking maybe look at the behavior of the basis of the tangent space.

Answer 1

You can always introduce the standard coordinates on $S^1$. Define $$\phi: (0,2\pi) \to S^1$$ $$\phi(\theta)=e^{i\theta}$$

Meanwhile you introduce a chart on $S^1 \times S^1 \times S^1$ as follows:

$$\psi: (0,2\pi) \times (0,2\pi) \times (0,2\pi) \to S^1 \times S^1 \times S^1$$

$$\psi(\theta_1,\theta_2,\theta_3)=(e^{i\theta_1},e^{i\theta_2},e^{i\theta_3})$$

Then the coordinate representation of $F$ will be :

$$\hat{F}=\psi^{-1} \circ F \circ \phi $$

$$\hat{F}(\theta)=(\theta,\theta,\theta)$$

Hence $$F_\star\bigg(\dfrac{\partial}{\partial \theta}\bigg)=\dfrac{\partial}{\partial \theta_1}+\dfrac{\partial}{\partial \theta_2}+\dfrac{\partial}{\partial \theta_3}$$