Gauge orbit orthogonality in a gauged linear sigma model on $\mathbb{C}P^{N}$ I am back with another question from the book Mirror Symmetry, this time from Section $15.1.1$.
Consider the gauged linear sigma model for $N$ complex scalar fields and the Lagrangian:
$$
L=-\sum_{i=1}^{N}
|D_{\mu}\phi_{i}|^{2}
-
\frac{e^{2}}{2}\left(\sum_{i=1}^{N}|\phi_{i}|^{2}-r\right)
$$
where $r>0$, and $D_{\mu}\phi_{i}=(\partial_{\mu}+iv_{\mu})\phi_{i}$ for the $U(1)$ gauge field $v_{\mu}$.
This Lagrangian is invariant under $U(1)$ gauge transformations acting as:
$$
\phi_{i}\mapsto e^{i\gamma(x)}\phi_{i}\text{ , and }
v_{\mu}\mapsto v_{\mu}-\partial_{\mu}\gamma(x)
$$
The vacuum manifold consists of field configurations for which $\sum_{i=1}^{N}|\phi_{i}|^{2}=r$, and since these fields are also related by local phase rotations, this is nothing other than $\mathbb{C}P^{N-1}$.
Since there is no kinetic term for the gauge field, it can easily be integrated out using the equation of motion:
$$
\sum_{i=1}^{N}
\left(
\overline{D_{\mu}\phi^{i}}\phi_{i}
-
\overline{\phi_{i}}D_{\mu}\phi_{i}
\right)
=
0
$$
In the low energy ($e\to\infty$), limit, a fixed field configuration $\{\phi_{i}(x)\}$ determines a map from the worldsheet (where $x$ lives), to $\mathbb{C}P^{N-1}$. We can now pick a tangent vector $\xi^{\mu}$ on the worldsheet, and use this map to obtain a tangent vector on $\mathbb{C}P^{N-1}$. The claim is then that the resulting vector is orthogonal to the gauge group orbit. We now arrive at my question.
Firstly, it seems to me that the resulting tangent vector should be the push-forward of $\xi^{\mu}$ along $\phi$, namely $\xi^{\mu}\frac{\partial\phi^{i}}{\partial x^{\mu}}$. Instead, the book seems to state that it is either $\xi^{\mu}$ itself, or even $\xi^{\mu}D_{\mu}\phi^{i}$.
Secondly, the gauge group orbit is clearly $\delta\cdot(\phi_{i},\bar{\phi}_{i})=(i\phi_{i},-i\bar{\phi}_{i})$, so the claim is that $\langle\xi^{\mu}D_{\mu}\phi,i\phi\rangle=0$. This is apparently able to be derived from the $v_{\mu}$ equation of motion, but all this equation of motion says to me is that:
$$
\langle\xi^{\mu}D_{\mu}\phi^{i},i\phi\rangle
=
-
\overline{\langle\xi^{\mu}D_{\mu}\phi^{i},i\phi\rangle}
$$
i.e. that this expression is purely imaginary. (Note that I have also had to assume that $\xi^{\mu}$ is real to get to this point, I am not quite sure this is the case either).
Any pointers would be much appreciated.
 A: I do not know if you still care about this question but I tried something and decided that maybe is worth sharing. To focus better on the geometric picture, I went for a simplification and decided to try to derive a simplified model, where instead of a world-sheet a world-line is used. I found the model fairly instructive, so maybe you would too and maybe it can give you some intuition into the world-sheet case. According to it, you cannot expect the tangent vector to be pushed forward to a vector orthogonal to the $\text{U}(1)-$orbits (in fact that non-orthogonality is where the gauge field $v$ arises). Furthermore, the covariant derivative of the motion however is orthogonal to the  $\text{U}(1)-$orbits.
Analysis of a world-line rather than a world-sheet. To simplify things for my intuition, I represent $\mathbb{C}^N$ together with the standard Hermitian dot product as $\mathbb{R}^{2N}$ together with the standard real dot product paired with the standard block-digonal symplectic matrix $J$. Then for any $\phi \, \in \, \mathbb{C}^N = \mathbb{R}^{2N}$, we have the following identity between complex and real notations $i \, \phi_{\mathbb{C}} \, = \, J \, \phi_{\mathbb{R}}$. Furthermore, there is a dot product plus symplectic product interpretation of the Hermitian dot product $$\bar{\phi}_{\mathbb{C}}^{\,T} \psi_{\mathbb{C}} \, =\, \phi_{\mathbb{R}}^T \psi_{\mathbb{R}} \, + \, \phi^T_{\mathbb{R}} J \, \psi_{\mathbb{R}}$$ From now on, we can think that the norm is simply the real norm (wich is equal to the Hermitian one anyway) $$| \,\phi \,|^2 \,=\, \phi^T \phi$$ We are interested in the dynamics of the lagrangian $$L \, = \, \frac{1}{2} \left|\,\frac{d\phi}{dt} + v J \, \phi \,\right|^2 \, + \, \frac{\lambda}{2}\big(\,|\phi|^2 - 1\,\big)$$ where $v$ is real valued scalar function.  The Euler-Lagrange equations of this Lagrangian are
\begin{align}
&\frac{d}{dt}\left(\frac{d\phi}{dt} + vJ\phi \right) \, =\, -\,vJ\left(\frac{d\phi}{dt} + vJ\phi \right) \, + \, \lambda \phi \,\,\,\,\,\text{ (the equation for $\phi$)}\\
&|\phi|^2 \, = \, \phi^T \phi \,=\,1 \,\,\,\,\,\text{ (the equations for $\lambda$)}\\
&\left(\frac{d\phi}{dt} + vJ\phi \right)^TJ\,\phi \, =\, 0  \,\,\,\,\,\text{ (the equations for $v$)}
\end{align}
Because $J\phi^T J\phi = |J\phi|^2 = |\phi|^2 = 1$, we can rewrite the last equation as $v = -\,\frac{d\phi}{dt}^T J\phi$.
Aslo,
I am going to use the notation $$\frac{\nabla}{dt} \, =\, \frac{d}{dt} \, +\, vJ$$ Then, the Euler-Lagrange equations can be written as
\begin{align}
&\frac{\nabla^2 \phi}{dt^2}  \, = \, \lambda \phi \,\,\,\,\,\text{ (the equation for $\phi$)}\\
&|\phi|^2 \, = \, \phi^T \phi \,=\,1 \,\,\,\,\,\text{ (the equations for $\lambda$)}\\
&v = -\,\frac{d\phi}{dt}^T J\phi  \,\,\,\,\,\text{ (the equations for $v$)}
\end{align}
Differentiate the second equation with respect to $t$
$$0 \, =\, \frac{d}{dt} \big(\,\phi^T\phi\,\big) \, =\, 2\, \phi^T\frac{d\phi}{dt}$$ which represent the fact that the trajectories' velocities are always tangent to the unit sphere. Due to the fact that $J$ is symplectic, $$\phi^TJ\,\phi \, = \, 0$$ and so we can rewrite the latter equation as
$$0 \, =\, \phi^T\left(\frac{d\phi}{dt} + vJ\phi \right) \, =\, \phi^T\frac{\nabla \phi}{dt}$$ Now, differentate covariantly the latter identity
$$0 \, =\, \frac{\nabla}{dt} \left( \phi^T\frac{\nabla \phi}{dt}\right) \, =\, \left|\frac{\nabla \phi}{dt}\right|^2 \, +\,  \phi^T\frac{\nabla^2 \phi}{dt^2}$$
Plugging the first Euler-Lagrange equation is this latter identity, we get an equation for $\lambda$
$$0 \, =\, \left|\frac{\nabla \phi}{dt}\right|^2 \, +\,  \phi^T\frac{\nabla^2 \phi}{dt^2} \, =\, \left|\frac{\nabla \phi}{dt}\right|^2 \, +\,  \phi^T \big(\lambda \phi\big)  \, =\, \left|\frac{\nabla \phi}{dt}\right|^2 \, +\,  \lambda  \big(\phi^T\phi\big)$$
and since $\phi^T\phi=1$
$$\lambda \, =\,  -\,\left|\frac{\nabla \phi}{dt}\right|^2$$
Now, the Euler-Lagrange equations take the form
\begin{align}
&\frac{\nabla^2 \phi}{dt^2}  \, = \, -\,\left|\frac{\nabla \phi}{dt}\right|^2 \phi \\
&\\
&\text{where }\\
&v \,=\, -\,\frac{d\phi}{dt}^T J\phi\\
&\frac{\nabla}{dt} \, =\, \frac{d}{dt} \, +\, vJ
\end{align}
with the unit-sphere constriant $|\phi|^2 \,=\,1$ authomatically satisfied as long as the intital filed (at $t=0$) is on the unit sphere, i.e. $|\phi_0|^2 = 1$. This is because with the choice of $\lambda$ we arrenged for $|\phi|^2$ to be a conserved quantity.
Now, let us look at the geometric consequences. From the action of $\text{U}(1)$ on the unit sphere as $e^{i\theta} \phi$ in the complex picture, we can see, by simply differentating with respect to $\theta$, that the tangent vector to the unit sphere that is also tangent to the circular orbit $e^{i\theta} \phi$ at the point $\phi$ is $i\phi$. In the real picture, the latter tangent vector is $J\phi$.  As you can see, $$v \,=\, -\,\frac{d\phi}{dt}^T J\phi$$  tells us that in general the trajectories of the motion are not perpendicular to the circular orbits of the $\text{U}(1)$ action. In fact, since $|\phi| = 1$, scalar function $v$ is negative of the magnitude of the projection of $\frac{d\phi}{dt}$ on the $\text{U}(1)$ orbits. On the other hand,
$$\frac{\nabla \phi}{dt}^T J \phi \, = \, \left(\frac{d\phi}{dt} + vJ\phi\right)^T J\phi \,=\, \frac{d\phi}{dt}^T J\phi + v \, \phi^TJ^T J\phi \, =\,  \frac{d\phi}{dt}^T J\phi + v \, = \, 0$$ so the covariant derivative $\frac{\nabla \phi}{dt}$ is orthogonal to the $\text{U}(1)$ orbits. In fact I can rewrite the covariant derivative of $\phi$ as follows
\begin{align}
\frac{\nabla \phi}{dt} \, =\, \frac{d\phi}{dt} \, +\, v\,  J\phi \, =& \, \frac{d\phi}{dt} \, -\,\left(\frac{d\phi}{dt}^T J\phi \right)\,  J\phi \, =\,  \frac{d\phi}{dt} \, -\,\left( \big(J\phi\big)^T  \frac{d\phi}{dt} \right)\,  J\phi \\
 =& \, \frac{d\phi}{dt} \, -\, \left( J\phi \right)\,(J\phi\big)^T   \frac{d\phi}{dt} \\
=& \, \Big(\, \text{Id} \, -\,  \big(J\phi\big)\,\big(J\phi\big)^T \,\Big)  \frac{d\phi}{dt}
\end{align}
where the linear operator $\text{Id} \, -\,  \big(J\phi\big)\,\big(J\phi\big)^T $ is the orthogonal projector onto the orthogonal complement of the tangent vector $J\phi$ to the $\text{U}(1)-$orbit $e^{i\theta}\phi$. So basically, the gauge field $v$ is chosen so that we differentiate the trajectory with respect to time and then projecti it orthogonally to the $\text{U}(1)-$orbit.
