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Misha Verbitsky
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When our hyperk"ahler 4-fold is no longer compact, is its instantons moduli space(maybe with some constriant about decaying condition) still hyperk"ahler?

The "decay constraint" condition is called "gravitational instanton". Gravitational instanton is a complete hyperkahler metric on a 4-manifold with L^2-integrable curvature. There is a lot of literature on gravitational instantons and on instanton bundles on gravitational instantons. The gravitational instantons are classified according to the asymptotic growth of a volume of a ball: when it grows as a 4-th degree (like on Euclidean space), it's ALE instanton, degree 3 is ALF, and the classification becomes very detailed and intricate for smaller degrees. ALF and ALE spaces can be constucted as hyperkahler quotients of flat hyperkahler spaces, for lower degrees it's probably false.

The space of instantons on ALE space is hyperkahler: P.B. Kronheimer, H. Nakajima, Yang-Mills instantons and ALE gravitational instantons. Math. Ann. 288 (1990), 263-307.

The same is true for instantons on Taub-NUT space, which is ALF:

Cherkis, Sergey A. Moduli spaces of instantons on the Taub-NUT space. Comm. Math. Phys. 290 (2009), no. 2, 719–736.

and on arbitrary ALF space:

Cherkis, Sergey A. Instantons on gravitons. Comm. Math. Phys. 306 (2011), no. 2, 449–483.

There is no framing in these results, you need to consider instantons with L^2-integrable curvature.

I am pretty much sure one can prove the same result for all gravitational instantons in place of ALE and ALF, and in fact I have a proof (which is pretty easy), but I am sure it's already published somewhere. The literature is a mess.

For some noncompact k"ahler surface that are not hyperk"ahler, for example blowup of C2 at origin, does its framed instantons moduli space still admit a hyperk"ahler structure?

Mostly not, in fact, it can easily be odd-dimensional. One could imagine results like that (for instance, $CP^2$ is not hyperkahler, but instantons on $CP^2$ are hyperkahler, if the framing is chosen right), but this would probably mean that the complement to the curve where you have chosen framing is hyperkahler.

Misha Verbitsky
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