I'm pretty unsure about this, because I don't know much about manifolds and other difficult words. So I hope I learn something here too.
I use your equation with all parameters set to simple values as Chris did
eqn = w^4 - w^2 ((kx^2 + kz^2) + 1/4) + 2 kx^2 == 0;
Your second equation k^2==kx^2+kz^2 defines a tube with different radii k. We could for instance look at both equations and choose k==3
contPlot = ContourPlot3D[{w^4 - w^2 ((kx^2 + kz^2) + 1/4) + 2 kx^2 == 0,
kx^2 + kz^2 == 9}, {kx, -5, 5}, {w, -5, 5}, {kz, -5, 5},
ContourStyle -> {Directive[Opacity[0.3], Red], Automatic}]

We are interested in the intersection of both contours. What we can do here, is to parametrize k^2==kx^2+kz^2 differently. A tube along w with radius k can expressed as the parametric equation
$$f(w,\phi)=\{k \cos(\phi), w,k \sin(\phi)\}$$
therefore we define a transformation rule
rule = Thread[{kx, w, kz} :> {k*Cos[phi], w, k*Sin[phi]}]
(* {kx :> k Cos[phi], w :> w, kz :> k Sin[phi]} *)
We can now apply this transformation and solve your initial equation for w. With this we get solutions for w which only depend on phi and k. With those solutions, we have an explicit parametrization of the curve in 3d
sol = Solve[eqn /. rule, w];
paramPlot =
ParametricPlot3D[{kx, w, kz} /. rule /. sol /. k :> 3, {phi, 0,
2 Pi}, PlotStyle -> Red] /. Line[pts_] :> Tube[pts, 0.1]

And we can of course combine them to see whether it fits with our imagination
Show[{contPlot, paramPlot}]

What you want to have now is a plot where on the first axis is k and on the second w
ParametricPlot[Evaluate[{k, w} /. sol], {k, 0, 9}, {phi, -Pi, Pi}]

\FractioBoxand\SuperscriptBox? – m0nhawk Oct 29 '12 at 10:04Read the FAQs! 3) When you see good Q&A, vote them up byclicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. ALSO, remember to accept the answer, if any, that solves your problem,by clicking the checkmark sign` – chris Oct 29 '12 at 10:13kxin the (omega,k) plane which are solution toF(omega,k,kx)=0above? – chris Oct 29 '12 at 10:16