Given connections $\nabla$ and $\bar{\nabla}$ as connections on $\mathbb{R}^4$ and the 3-sphere of radius $r$: $\mathbb{S}^3(r)$, the vector fields $X,Y$ tangent to $\mathbb{S^3}(r)$, how do I obtain the expression: $$\nabla_{X} Y = \bar\nabla_{X} Y - \dfrac{1}{r^2}<X,Y>p,$$ where $p : \mathbb{S}^3 \to \mathbb{R}^4$ represents the position vector.
I came across this expression in a paper which mentions it as a standard result. However I was unable to derive it from my own limited knowledge.
The Levi-Civita connection of an embedded manifold is obtained by an orthogonal projection of the Levi-Civita connection of the ambient manifold (see comment). In particular, $$\begin{aligned} \nabla_XY&=\left[\nabla_XY\right]^\top+\left[\nabla_XY\right]^\perp\\ & =\bar\nabla_XY+\left\langle\nabla_XY,\frac{p}{r}\right\rangle\frac{p}{r}\\ & =\bar\nabla_XY+\frac{1}{r^2}\left\langle\nabla_XY,p\right\rangle p\\ \end{aligned}$$ Now, since the $\mathbb R^4$ is equipped with the flat metric, this means that $\nabla_XY=X(Y)$. Hence you obtain $$ \langle\nabla_XY,p\rangle = \underbrace{X\langle Y,p\rangle}_{=0} -\langle Y,X(p)\rangle=-\langle X,Y\rangle, $$ since $p$ is the inclusion map.