Let be a vector field with . Then the origin is a fixed point of the generated flow on . Let be the splitting into stable, center and unstable directions. Moreover, there are three invariant manifolds (at least locally) passing through and tangent to the corresponding subspaces at .

Theorem (Pliss). For any , there exists a center manifold .

Generally speaking, the size of the center manifold given above depends on the pre-fixed regularity requirement. Theoretically, there may not be a center manifold, since could shrink to as . An explicit example was given by van Strien (here). He started with a family of vector fields . It is easy to see that is a fixed point, with . The center manifold can be represented (locally) as the graph of .

**Lemma.** For , , is at most at .

Proof. Suppose is at , and let be the finite Taylor expansion. The vector field direction always coincides with the tangent direction along the graph , which leads to

.

Note that . Then up to an error term , the right-hand side in terms of : ; while the left-hand side in terms of :

.

So for : , ;

: , ;

: , .

Note that if , we evaluate the last equation at to conclude that . This will force for all , which contradicts the second estimate that . Q.E.D.

Consider the 3D vector field . Note that the singular set are two lines , (in particular it contains the origin ). Note that . Hence a cener manifold through is tangent to plane , and can be represented as . We claim that (at least locally).

Proof of the claim. Suppose on the contrary that for some . Note that , and is flow-invariant. However, there is exactly one flow line passing through : the line . Therefore , which contradicts the fact that is tangent to plane at . This completes the proof of the claim.

The planes are also invariant under the flow. Let’s take the intersection . Then we check that is a (in fact *the*) center manifold of the restricted vector field in the plane . We already checked that is not , so is .