existence of unipotent linear operator of order k

Question

If $X$ is a given vector space (over the field of real numbers), we say that a linear operator $T:X\rightarrow X$ is unipotent of order $k\in\mathbf{N}$, $k\geq 2$, if it holds that $T^k=I$ and $T^{k-1}\neq I$, where $I:X\rightarrow X$ is the identity operator. Consider the vector space $V:=\{g:\mathbf{R}\rightarrow\mathbf{R}\}$. The question is: can we provide an example of unipotent operator $T:V\rightarrow V$ for any given $k\geq 2$? If $k=2$, we can define $T(g):=-g$, $g\in V$, but what if $k\geq 3$? I am not even sure that such unipotent operators exist in the case $k\geq 3$. Remark. If $X:=\mathbf{R}^n$ (i.e., if $X$ is finitely dimensional), then we know from the undergraduate course in linear algebra that linear operators $T$ can be represented by a square matrix of type $n\times n$, which solves the problem since we can construct unipotent matrices of order $k$. This argument probably extends to the case of separable infinitely-dimensional Hilbert spaces (since such spaces allow Schauder basis). But the problem with $V$ is, as I see it, that the basis for $V$ is unknown, so we have no way to represent the action of $T$.

Answer 1

If you are fine with finite dimensional, then we can solve for infinite dimensional. Let $\{x_\alpha\}_{\alpha \in I}$ be a basis. Fix $\alpha_1, \alpha_2, \cdots, \alpha_n \in I$. Then we can extend an operator $T$ on $\mathbb{R}^n$ by noting $I = (I\setminus \{\alpha_1,\dots, \alpha_n\}) \cup (\{\alpha_1,\dots, \alpha_n\})$ and we simply identify $x_{\alpha_1},\dots, x_{\alpha_n}$ as a basis of $\mathbb{R}^n$ and apply $T$ there, then extend $T$ by being the identity. This works for a "linear algebra" basis and a "Hilbert Space" basis.

Now, perhaps this is slightly unsatisfying because it is not a natural way to define such a $T$ on $X$. For instance, it works just fine if $X$ is the set of all functions since the most natural basis of $X$ would be the set $\{f_x: \mathbb{R} \to \mathbb{R}\}$ that is defined as $f_x(y) = \delta_{x,y}$. However, if we required the $f$ in $X$ to be continuous, this might be slightly annoying. Well, I do not think that it gets much better.

There are simpler examples though to solve your problem: Consider the interval $[1, k)$, we can break $\mathbb{R}$ up into translates of this interval: $\mathbb{R} = [1, k) \cup [k, 2k) \cup \cdots $. We can then break each interval up as $[1,k) = [1,2) \cup [2,3) \cup \cdots \cup [k-1,k)$ and then cyclically permute these sub half-intervals: $t: [i,i+1) \mapsto [i+1,i+2)$ for $i < k-1$ and $t:[k-1, k) \mapsto [1,2)$. This is an operation with order $k-1$. We then given a function $f \in X$, break $f$ into a sum $f = \sum_{k\in \mathbb{Z}}f \chi_{[k,k+1)}$, where $\chi_A$ is the function that equals $1$ on the set $A$ and zero outside $A$. Now, having that decomposition, we can permute the intervals as above to get an operator $T$ of order k-1$. You can check that this operator is linear. (This is what @daw just wrote.)

For continuous function, you can try something like a partition of unity (usually partitions of unity are smooth, but let's do a continuous version). Let $f_n$ be the function that is piecewise linear, equaling zero at $n-1$ and $n+1$ and equal to $1$ at $n$. Then we have the property that $\sum_{n\in\mathbb{Z}}f_n = 1$ and so we can do the permuting game with the $f_n$ instead of $\chi_[k,k+1)$. Note that since the $f_n$ are continuous, unlike the characteristic function $\chi$, we see that this decomposition works if we were to require that all the functions $f$ in $X$ were continuous. One might try to see if they can extend this to "smooth" partitions of unity.

Note that this approach does not work if the functions considered satisfy some sort of rigidity properties like analyticity. For more general classes of functions, one might just stick with the first method that picks a basis.

Answer 2

My idea is to consider transformation on the parameter space that satisfy the unipotent property.

Define the the nonlinear coordinate transform $$ f(x) = ((\lfloor x\rfloor + 1 ) \mod n ) + (x-\lfloor x\rfloor). $$ This $f$ maps $[0,1)$ to $[1,2)$, $\dots$, $[n-1,n)$ to $[0,1)$. Then $(f^n) (x)=x$, and $$ Tg:=g\circ f $$ fulfills $T^n=I$.

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If you are willing to consider complex functions, i.e., $$ V = \{ g: \mathbb C \to \mathbb C\}, $$ then a nice construction is possible: Let $\omega$ be a $n$-th root of unity, i.e., $\omega^n=1$, $\omega\ne 1$. Then define $T$ by $$ (Tg)(x):= g(\omega x), \quad T g = g \circ (\omega id) $$ then $T^ng=g$.