Maybe I don't understand the definition of 'linear algebraic group'. It seems to me that the subgroup $G_k\subset\mathrm{SL}(2,\mathbb{C})$ defined by the equations $g\begin{pmatrix}1&0\\0&-1\end{pmatrix}=\begin{pmatrix}1&0\\0&-1\end{pmatrix}g$ and $g^k=\begin{pmatrix}1&0\\0&1\end{pmatrix}$ has at least $k$ components. Is this $G_k$ not 'linear algebraic'?

@Partha: However, the usual convention (which is mine also) is that the volume form on $\mathbb{R}\times M$ is $\mathrm{d}t\wedge\mathrm{vol}(M)$. If you look at Harvey and Lawson's definition of the $\mathrm{Spin}(7)$ form in terms of the $\mathrm{G}_2$ form, you'll see that their convention agrees with this; their $4$-form is $\mathrm{d}t \wedge\sigma + {\ast}_\sigma\sigma$ where $\sigma$ is the $3$-form on $M$.

@ArvinRasoulzadeh: Yes, it follows from this: If $g ={ \omega_1}^2 + {\omega_2}^2$ where $\omega_1 = Z^\flat$, and $\kappa(p)$ is the geodesic curvature of the flow line of $Z$ through a point $p$ of the surface, then $\mathrm{d}\omega_1 = \kappa\,\omega_1\wedge\omega_2$ (a consequence of the structure equations). Thus, $\omega_1$ is closed if and only if $\kappa$ vanishes identically, i.e., the flow lines of $Z$ are geodesics.

@ArvinRasoulzadeh: They are not the coordinate vector fields, but they are multiples of the coordinate vector fields. You can always take the coordinate vector fields (which are nonvanishing) and divide by their lengths in the induced metric. The resulting unit vector fields $X$ and $Y$ aren't coordinate vector fields any more because $[X,Y]$ is (usually) nonzero, but that doesn't affect whether they are conjugate with respect to $I\!I$.

@IgorBelegradek: Maybe I misunderstood your first comment, but when you wrote that "The proof is a reduction to the case of a noncompact simple subgroup for which (according to Dani) the answer is well-known", I thought you were insinuating that Dani didn't actually give a proof (or reference) for the case of a noncompact simple subgroup, he just claimed it was 'well known'. Is that the case?

@IgorBelegradek: By the way, if you are satisfied with my answer, you should probably go ahead and accept it, so that it won't keep being brought up periodically (because it doesn't have an accepted answer).