Entries in a one-parameter unipotent subgroup in $\text{SL}(n,\mathbb R)$ are polynomials of bounded degree Let $G=\text{SL}(n,\mathbb R)$ and $(u_x)_{x\in \mathbb R}$ be a one-parameter unipotent (eigenvalues of $u_x$ are all $1$) subgroup of $G$.
From the papers I have read, it seems that the entries of $u_x$ are polynomials in $x$ and the degrees should bounded by $n$. But why?
I personally can only come up with trivial example where entries are linear polynomials in $x$, and can't exclude the posibility of non-polynomial entries.
 A: Standard Lie theory implies that every one-parameter subgroup has the form $\exp(Mx)$ where $M \in \mathfrak{sl}_n(\mathbb{R})$. The eigenvalues of $\exp(Mx)$ are $\exp(\lambda x)$ where $\lambda$ are the eigenvalues of $M$, so for all of them to be equal to $1$ for all values of $x$ the eigenvalues of $M$ must all be zero, so $M$ is nilpotent. (This is an if-and-only-if, so we've completely classified one-parameter unipotent subgroups.) This gives $M^n = 0$, so
$$\exp(Mx) = \sum_{i=0}^{n-1} M^i \frac{x^i}{i!}$$
has entries which are polynomials in $x$ of degree at most $n-1$.
For examples you can take $M$ to be a nilpotent Jordan block; in general $M$ is similar to a direct sum of nilpotent Jordan blocks. The smallest example with polynomials of degree greater than $1$ occurs when $n = 3$; we can take the Jordan block $M = \left[ \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array} \right]$ which gives
$$\exp(Mx) = I + M x + M^2 \frac{x^2}{2} = \left[ \begin{array}{ccc} 1 & x & \frac{x^2}{2} \\ 0 & 1 & x \\ 0 & 0 & 1 \end{array} \right].$$
