Let $X$ be a smooth projective variety over $\mathbb{C}.$ On page 3 in this preprint of Simpson, it is stated that
Notice first of all that the algebraic de Rham theory is not going to work well in the case of higher cohomology with coefficients in the multiplicative group scheme, i.e. when $T=K(\mathbb{G}_m,n)$ for $n\ge 2.$ I won't go into the explanation of that here!
My question is: Why is it not going to work well? Is it explained somewhere?
On page 69 of the same preprint, it is written
The idea behind the definition of “very presentable” is that we want to require the higher homotopy groups to be unipotent. Note that if we don’t require $π_1$ to be affine, or $π_i$ to be unipotent $(i ≥ 2),$ the comparison between algebraic and analytic de Rham cohomology (announced in [Si]) is no longer true.
Maybe this is the reason? However, I don't understand why it is no longer true. Theorem 9.2 of [Si] shows that if $T$ is a very presentable homotopy sheaf, e.g, $T=K(\mathbb{G}_a,2),$ then the analytification of the algebraic de Rham cohomology with coefficient in $T$ is homotopy equivalent to the analytic de Rham cohomology with coefficient in $T^{an}.$ Is there an example where this fails when $T$ is not very presentable, e.g. $T=K(\mathbb{G}_m,2)?$
Any help is greatly appreciated!
Reference: [Si] C. Simpson. Homotopy over the complex numbers and generalized de Rham cohomology. Moduli of Vector Bundles, M. Maruyama (Ed.), Lecture Notes in Pure and Applied Mathematics 179, Marcel Dekker (1996), 229-263.
