When I read the paper by Gelfand-Gelfand-Bernstein(1968) and Gelfand-Kirillov on differential operator on base affine spaces and studies of $g$-modules. I am surprised! Because it seems that they are very close to the discovery of the relationship of modern $D$-module theory and representation theory of $g$ (Lie algebra)

It seems that the only thing they did not consider is the graded version of differential operator on base affine space(graded version of cone is projective space) which is exactly the $D$-module theory on flag variety of Lie algebra. In another word, if they further consider this case, maybe this important picture would be named Gelfand-Kirillov theorem not Beilinson-Bernstein theorem. What I am very surprised is that they "miss" this picture and it is discovered and figured out by Beilinson-Bernstein or by Brylinski-Kashiwara 10 years after that.

Actually, using the categorical language, Gelfand-Kirillov constructed the category of $D$-modules on base affine space $G/U$. If we denoted this category by $D-mod_{G/U}$. Then we have a faithful but not exact functor:

$D-mod_{G/U}\rightarrow R*\hat{U(g)}-mod$,where $R$ is algebra of functions on $G/U$, $\hat{U(g)}=U(h)\otimes_{k} U(g)$ for $h$ being cartan subalgebra of $g$ and $R*\hat{U(g)}$ is cross product.

Then, if we further consider graded version of $G/U$, i.e. $G/B$, flag variety. We can get the following exact and fully-faithful functor:

$D-mod_{G/B}\rightarrow U(g)-mod$ and if we single out the $U(g)-mod_{0}$ as $U(g)-mod$ with trivial central character, we get Beilinson Bernstein localization.

However, I heard from my advisor(he attended all the seminars of Kirillov)that Kirillov pushed Bernstein to study the relationship between $D$-module and representation theory of Lie algebra.

Therefore, my question is

Can you tell the similar phenomenon in history of mathematics? Say, some mathematicians were very close to a very promising theory or correct prove for some big conjecture, but they finally missed the last step ?

I hope that this question will not be too boring