@A.J.Pan-Collantes: What I had in mind is this: A system of $r$ first-order equations on a single unknown function $f:U\to\mathbb{R}$ can be thought of as defining a submanifold $\Sigma\subset J^1(U,\mathbb{R})$ of codimension $r$, and a local solution is a function $f:V\to\mathbb{R}$ for which the section $j^1(f):V\to J^1(U,\mathbb{R})$ has image in $\Sigma$. The submanifold $\Sigma$ is involutive if the Pfaff rank of the form $\Sigma^*\omega$ is $n{-}r$, where $\omega$ is the contact form on $J^1(U,\mathbb{R})$. The system $(1)$ is that your $\Sigma$ is involutive, so solutions exist....

@HamidrezaFasihi_Ramandi: By 'generic', I mean outside a set defined by the satisfaction of some nontrivial identities on the curvature and its covariant derivatives up to some order. For example, a 'generic' Riemannian metric in dimension $2$ has $\mathrm{d}K$ vanishing only at isolated points, where $K$ is the Gauss curvature of the metric. Usually, when we say that a condition on a geometric structure is 'generic', we mean that the opposite of that condition is characterized by some set of nontrivial equations on such geometric structures, which may not be explicitly described.

Is the convex body $M$ the set of vectors that satisfy $\phi(v)\le 1$? Otherwise, you haven't said how $M$ is related to anything else, so I'm guess that this is what you mean. Let me know if you mean something different.

There's something missing in your description of your problem. For example, you don't assume any connection between $S$ and $M$ or put any condition on $Q$. If you let $S$ and $M$ be arbitrary and just take $Q(v,w) = 0$ for all $v$ and $w$, then your hypotheses are satisfied but $M$ need not be an ellipsoid. (This is a counterexample to your claim even in dimension $3$.) Moreover, since there is no connection between $S$ and $M$ assumed, for the basis $\{e_1,e_2\}$ that you first construct in your 'proof', there is no reason to suppose that $e_1$ and $e_2$ to belong to $M$.

@BenWebster: Yah, I didn't think it would really change the answer, but I didn't want the OP to be confused. Thanks for fixing it. Also, for some reason the OP asked for 4-dimensional complex representations, so I guess that you'd want to include representations such as $\rho(A) = e^{\lambda\,\log|\det(A)|} A$ where $\lambda$ is a complex number with nonzero imaginary part.

I know that this is being picky, but it's not true that $\mathrm{GL}(4,\mathbb{R})\simeq \mathrm{SL}(4,\mathbb{R})\times \mathbb{R}^{\times}$. Where would $A = \mathrm{diag}(-1,1,1,1)$ go under such an isomorphism?

Well, an Hermitian connection on the tangent bundle that was without torsion would already imply that the almost complex structure was integrable. Anyway, I have a better example that I will add to my above answer.

Unless you tell us more about how your curve $\gamma:[0,1]\to\mathbb{R}^n$ is specified and what sorts of information about it can be easily computed, it's hard to give advice about an 'easy-to-check condition'. For example, it's a simple matter to construct examples of smooth $\gamma$ that don't lie in a proper subspace of $\mathbb{R}^n$ but for which uniform sampling from $[0,1]^m$ to test $m$ points $\gamma(t_1),\ldots,\gamma(t_m)$ for linear independence will have a 99.99999% probability of 'yes', but still $\gamma([0,1])$ does not lie in a proper subspace.

@PaulCusson: Yes, since $[X_i,X_j]=-[X_j,X_i]$ and since the $X_k$ are linearly independent, it follows from the definition of the $\alpha^k_{ij}$ in your question that $\alpha^k_{ij}=-\alpha^k_{ji}$.

By the way, you need to modify your definition of conical Lagrangian a little bit because the sentence implies that $\phi(CT) = C$, and that is clearly not what you want. Maybe there is some confusion between your use of $C$ and $CT$.