# A differentiable manifold of class $\mathcal{C}^{r}$ tangent to $E^{\pm}$ and representable as graph

The stable manifold theorem tell us:

• A local stable manifold $W^{s}_{loc}(x^{*})$, which is a differentiable manifold of class $\mathcal{C}^{r}$ and dimension $n_{-},$ tangent to the stable subspace $E^{-}$ at $x^{*}$, and which can be representable as graph.

• A local unstable manifold $W^{u}_{loc}(x^{*})$, which is a differentiable manifold of class $\mathcal{C}^{r}$ and dimension $n_{+},$ tangent to the stable subspace $E^{+}$ at $x^{*}$, and which can be representable as graph.

I don’t get it very well what does it mean about A differentiable manifold of class $\mathcal{C}^{r}$ tangent to $E^{\pm}$ and representable as graph.

Given the below imagen, how can it be interpreted with such definition?

#### Solutions Collecting From Web of "A differentiable manifold of class $\mathcal{C}^{r}$ tangent to $E^{\pm}$ and representable as graph"

Since stable and unstable subspaces complement each other (I mean, $\mathbb{R}^2 = E^{+} \oplus E^{-}$ and any vector $v = \pi_{E^{+}} v + \pi_{E^{-}} v$), the two dimensional case could be interpreted this way:

There exists a (1) $C^r$-function $\psi\, \colon E^{+} \rightarrow E^{-}$ such that (2) $\psi(0) = 0$, (3) $D\psi = 0$ and set (4) $x + \psi ( x)$ is a local unstable manifold.

• (1) tells you that set $x + \psi(x)$ would be a $C^r$ differentiable manifold; also, since this manifold has special analytical form ($x + \psi(x)$, $x \in E^{+}$, $\psi(x) \in E^{-}$), it’s called graph
• (2) tells that this manifold passes through the origin
• (3) means that manifold is tangent to $E^{+}$ at origin
(it’s very easy to see through straightforward differentiation)
• (4) just means that all trajectories that start at $x + \psi(x)$ in some small neighbourhood of origin
tend to origin in backward time

Hope this illustration will help: