In characteristic 0 or good prime characteristic, there are standard ways to relate the unipotent variety $\mathcal{U}$ of a simple algebraic group $G$ and the nilpotent variety $\mathcal{N}$ of its Lie algebra $\mathfrak{g}$.  Recall that $p$ is *good* just when it fails to divide any coefficient of the highest root (the root system being irreducible), *bad* otherwise.  The only possible bad primes are $2,3,5$.  In characteristic 0, algebraic versions of the exponential and logarithm maps  provide Ad $G$-equivariant isomorphisms in both directions, whereas in good characteristic $p>0$, the less direct arguments of Springer <a href="https://mathscinet.ams.org/mathscinet-getitem?mr=0263830">*here*</a> yield similar results.    [The isogeny type of $G$ adds some complications here.]    

There is scattered literature on the varieties $\mathcal{U}$ and $\mathcal{N}$ when $p$ is bad, often treated case-by-case, e.g., four papers by Lusztig posted on arXiv starting <a href="https://arxiv.org/abs/math/0503739">*here*</a>, along with papers by his student T. Xue.  A serious challenge when $p$ is bad is to find a uniform explanation for the failure of the numbers of unipotent classes and nilpotent orbits to agree in some cases: the details were worked out by Holt-Spaltenstein and others.   In spite of this breakdown in $G$-equivariance, a natural question can be raised:

> Are the two varieties $\mathcal{U}$ and $\mathcal{N}$ isomorphic in all characteristics, for example when $G$ is simply connected? 

The answer does not seem to be written down explicitly (?), but for example one can see indirectly that both varieties have the same dimension in all characteristics: the number of roots.  A paper <a href="https://mathscinet.ams.org/mathscinet-getitem?mr=887203">*here*</a> by S. Keny, a former student of Steinberg, showed case-by-case that *regular* nilpotent elements always exist and form a dense orbit in $\mathcal{N}$.
By definition, the isotropy group in $G$ of such an element has dimension equal to the rank of $G$.