# eigenfunctions -> characteristic factors

In Universal Characteristic Factors and Furstenberg Averages, Tamar Ziegler talks about "unconventional ergodic averges" which were used in Furstenberg's proof of Szemeredi's theorem. I had trouble understanding the notion of "characteristic factor" in dynamical system, so I focused on some of the earlier examples in her paper.

Let $X = (X, \mathcal{B}, \mu, T)$ be a measure-preserving dynamical system. I will not worry about the σ-algebra $\mathcal{B}$ or the measure $\mu$ (which is Lebesgue). So we can just write the dynamical system $T: X \to X$ preserving $\mu$ on $\mathcal{B}$.

This induces a map on the function spaces taking $T:\psi(x) \mapsto \psi(Tx)$. Ziegler looks at averages $$\frac{1}{N} \sum_{n=1}^N f(T^n x)g(T^{2n} x)h(T^{3n}x) \to \int f \, d\mu \cdot \int g \, d\mu \cdot \int h \, d\mu.$$ If $X$ is weak mixing the left and right sides are equal.

In general, $f,g,h$ at the three points $x,T^n x, T^{2n}x$ are not ''independent" and this average is another dynamical system based on $T$. One obstruction to these averages becoming constant is an eigenfunction $\psi(Tx) = \lambda \psi(x)$. Then $$\frac{1}{N} \sum_{n=1}^N \psi(T^n x)^2\psi^{-1}(T^{2n} x) = \psi(x).$$ In this case, $T$ restricted to eigenfunctions behaves like rotations $z \mapsto z + \alpha$ in an abelian group.

Zieglier says that if you have a second-order eigenfunction $T \phi = \psi \phi, T\psi = \lambda \psi$ the average becomes a nilsystem. Then you get a relation $$\phi(T^n x)^3 \phi(T^{2n} x)^{-3} \phi(T^{3n}x) = \phi(x).$$ (This is a bit like the identity $n^2 - 3(n+1)^2 + 3(n+2)^2 - (n+3)^2 = 0$. Finally $$\frac{1}{N} \sum_{n=1}^N \phi(T^n x)^3 \phi(T^{2n} x)^{-3} \phi(T^{3n}x) = \phi(x).$$

I guess I am wondering to what extent these 'characteristic factors' generalize the notion of 'eigenfunction' for the operators $T$ and how they are related to nilsystems. Is it possible to find invariant nilsystems in dynamical systems arising from ordinary differential equations in several variables?

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