Are you asking for examples of Hamiltonian actions that are not Poisson (i.e., Hamiltonian actions for which there is no choice of an equivariant momentum mapping), or are you asking for examples of Poisson actions for which there is an 'interesting' momentum mapping that happens not to be equivariant?

Very nice argument!

Oh, sorry. I momentarily forgot that $J^2$ is the identity, not minus the identity, so the correct formula is indeed $\sigma(A) = JAJ$. (I shouldn't have used the letter $J$, since I usually reserve that for an anti-complex map whose square is minus the identity.) But, yes, any outer involution (non-inner automorphism that squares to the identity) of $SU_h$ is of this form for some $J$. I think this is discussed in Helgason's Differential Geometry, Lie Groups, and Symmetric Spaces.

Well, you do have to choose the real structure $R$ to be real with respect to $h$, i.e., $h(v,w)$ is real for $v,w\in R$. (This can be shown to be possible via induction on the dimension of $V$.) Then $J$ is the identity on $R$ and minus the identity on $iR$. Then an outer involution of $U_h$ is just $\sigma(A) = -JAJ$. The fact is, though, that there's no canonical choice of $\sigma$, the only thing that's 'canonical' is the class of $\sigma$ in the group $\mathrm{Out}(U_h)\simeq\mathbb{Z}_2$.

You don't have to choose a basis of $V$, but you do have to choose a real structure, i.e., an $\mathbb{R}$-subspace $R\subset V$ such that $V = R \oplus iR$. Equivalently, you have to choose an $\mathbb{R}$-linear 'conjugation' $J:V\to V$ such that $J(iv) = -iJ(v)$ and $J^2=\mathrm{id}$. You have to make some choice because 'the' non-inner automorphism is not unique. Only its equivalence class modulo the inner autormorphisms is unique.

@MohammadGhomi: I think it may depend on what you mean by 'small'. In fact, the estimate that this will provide for the first fundamental form will be small if $|F|$ is small, but bounding the difference of the second fundamental form of the approximate surface from the given $l_{ij}$ may require knowing something about the derivatives of $E$, particularly, $E_u$, so you might need to know that $F$ is small in some $C^1$ norm to get $C^0$ estimates on this difference. I'm not sure because I haven't really thought about it that much, but that is what I would expect.

About your question about the textbook, I don't have it in front of me, so I'll have to wait until tomorrow to answer that.

(cont....) For example, consider the exponential map from the tangent space at the north pole $n=(0,0,1)$ of the unit $2$-sphere to the $2$-sphere, so that $\exp_n(r\cos\theta,r\sin\theta,0) = (\sin r\cos\theta,\sin r\sin\theta,\cos r) = (x,y,z)$, then $\exp_n^*(dx^2+dy^2+dz^2) = dr^2+\sin^2r\,d\theta^2$, whereas the flat metric on the tangent plane is $dr^2 + r^2\,d\theta^2$. So the difference is $(r^2-\sin^2r)d\theta^2\ge0$.

Well. $\exp:T_xM\to M$ is the $g$-geodesic exponential map, and $\exp^*(g)$ is the pullback of $g$ under this smooth map, so its quadratic form (with variable coefficients on the vector space $T_xM$ thought of as a smooth manifold. We can think of $g_x$ which is a quadratic form on $T_xM$ as a Riemannian metric with constant coefficients (so it's determined by its value at the origin $0_x\in T_xM$. The statement $exp^*(g)\le g_0$ is the statement that, at every point $v\in T_xM$, the quadratic differential $g_0 - \bigl(exp^*(g)\bigr)_v$ is nonnegative. (cont...)

I guess the generalization to complete Riemannian manifolds $(M,g)$ with nonnegative sectional curvature is that, for any $x\in M$, we have $\exp_x^*(g)\le g_x$ as quadratic forms on $T_xM$, which would then imply that $d\bigl(\exp_x(a),\exp_x(b)\bigr) \le |a-b|$ for $a,b\in T_x$ for which $\alpha(t) = \exp_x(ta)$ and $\beta(t) = \exp_x(tb)$ for $0\le t\le 1$ are $g$-length minimizing geodesics between their endpoints.