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I'm new to this area, so it may very well be possible that I may be missing something easy here.

Let $E$ be a stable complex vector bundle over $X$ of degree $d$ and rank $n$. Then the moduli space $\mathcal{M}^s(n,d)$ of degree $d$ rank $n$ stable vector bundles can be identified with the space of holomorphic structures on $E$.

Further the tangent space to $\mathcal{M}^s(n,d)$ can be seen to be $H^{0,1}(X, End(E))$, where $End(E)$ denotes the endomorphism bundle. Similarly the tangent space to the space of deformations of a complex structure of $E$ can be seen to be $H^{0,1}(E,T^{1,0}E)$.

Given the identification in the previous paragraph, how does one describe the isomorphism between $H^{0,1}(X, End(E))$ and $H^{0,1}(E,T^{1,0}E)$ ?

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  • $\begingroup$ What is the meaning of $T^{1,0}E $? $\endgroup$
    – Will Sawin
    Jun 22, 2021 at 2:47
  • $\begingroup$ It denotes the holomorphic tangent bundle of E. $\endgroup$
    – cr1t1cal
    Jun 22, 2021 at 3:00
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    $\begingroup$ What you are missing is that deformations of $E$ as a manifold will not usually keep the vector bundle structure — in fact, what you denote by $H^{1,0}(E, T^{1,0}E)$ is usually infinite-dimensional. $\endgroup$
    – abx
    Jun 22, 2021 at 3:47
  • $\begingroup$ @cr1t1cal The moduli space $\mathcal{M}^s(n,d)$ is not identified with the holomorphic structure on $E$. In fact $Hol(E)/\sim_{\text{iso}}$ is not even a coarse moduli space. Isomorphism exists between $\mathcal{M}^s(n,d)$ and the space of irreducible Yang-Mills connections with unitary holonomy up to the action of the Gauge group $\mathcal{G}_E:=GL(E)$ $\endgroup$
    – John117
    Jun 22, 2021 at 8:03
  • $\begingroup$ @abx ah yes, that makes sense. Thanks for clarifying my confusion! $\endgroup$
    – cr1t1cal
    Jun 22, 2021 at 13:50

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