TLDR: How does one compute $\partial_X Y$, where $\partial$ is the Levi-Civita connection on $\mathbb{R}^n$ and $X,Y$ are vector fields on $\mathbb{R}^n$ (or possibly only defined on a submanifold of $\mathbb{R}^n$)?

For some context, I'm working on the following exercise (which appears as Exercise 7.4 in these lecture notes).

For $\theta \in (0,\pi/2)$, define $$ T_\theta^2 = \{ (e^{i\alpha}\cos\theta, e^{i\beta}\sin\theta) \,\mid\, \alpha,\beta\in \mathbb{R} \} \subset \mathbb{C}^2 \cong \mathbb{R}^4. $$ Determine the values of $\theta$ for which $T_\theta^2$ is a minimal submanifold of the 3-sphere $S^3$.

That a submanifold $M^m \subset N^n$ is minimal here means that $$ \mathrm{trace}\;B = \sum_{i=1}^m B(X_i, X_i) \equiv 0, $$ where $X_j$ form an orthonormal frame of $TM$ and $B$ is the second fundamental form of $M$ in $N$, defined as $$ B(X,Y) = (\nabla_X Y)^\perp, $$ where $\nabla$ is the Levi-Civita connection of the ambient manifold $N$, and $\perp$ denotes the normal component of the vector field.

After some work, I managed to show that $$ \text{trace}\, B = (\langle \partial_{E_1}E_1, E_3 \rangle + \langle \partial_{E_2}E_2, E_3 \rangle) E_3, $$ where $\partial_{X}Y$ now denotes the Levi-Civita connection of $\mathbb{R}^4$, and where $$ E_1 = (ie^{i\alpha},0), \quad E_2 = (0,ie^{i\beta}) \quad\text{and}\quad E_3 = (-e^{i\alpha}\sin\theta, e^{i\beta}\cos\theta). $$ The vectors $E_1$ and $E_2$ form an orthonormal frame for the tangent bundle of $T_\theta^2$, and $E_3$ spans its normal space in $S^3$. So the problem boils down to determining the values of $\theta$ for which $$ \langle \partial_{E_1}E_1, E_3 \rangle + \langle \partial_{E_2}E_2, E_3 \rangle \equiv 0. $$

My question: How does one actually compute the vector fields $\partial_X Y$, i.e. in my case $\partial_{E_1}E_1$ and $\partial_{E_2}E_2$? As far as I know these are supposed to be directional derivatives, but I've never worked this in practice. I would guess that there should be a simple formula for doing this in Euclidean spaces.

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    $\begingroup$ Do you know how to do directional derivatives of a scalar function? Just apply this component by component to the vector field. $\endgroup$ – Ted Shifrin Apr 29 at 17:20
  • $\begingroup$ I do know how to do that, given that the scalar function is defined on $\mathbb{R}^n$ (in which case I would compute the gradient and dot it with the direction). I guess what confuses me in this case is that the vector fields $E_i$ are defined only on a submanifold, i.e. on the torus, so I don't really know what how to perform the computation. $\endgroup$ – MisterRiemann Apr 29 at 17:23
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    $\begingroup$ Remember that if the vector field $X$ is tangent to $M$, you only need values of $Y$ on $M$ to compute $(\partial_X Y)(p) = (Y\circ\alpha)'(0)$ where $\alpha$ is a curve in $M$ with $\alpha(0) = p$ and $\alpha'(0) = X(p)$. $\endgroup$ – Ted Shifrin Apr 29 at 18:02
  • $\begingroup$ Ah, that final comment helped me figure it out. I made it more difficult for myself than I should have. Thank you for helping, Ted! If you wish, you can post your comments as an answer and I will accept it. $\endgroup$ – MisterRiemann Apr 29 at 18:21

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