Nakajima & Yoshioka [1] showed that 
\begin{equation}
F^{inst}(\epsilon\_1,\epsilon\_2,\mathbf{a},\mathbf{q}) = \sum\_{n = 1}^\infty \mathbf{q}^nF^{inst}\_n(\epsilon\_1,\epsilon\_2,\mathbf{a}) := \epsilon\_1\epsilon\_2\log Z^{inst}(\epsilon\_1,\epsilon\_2,\mathbf{a},\mathbf{q})
\end{equation}
is regular at $\epsilon\_1,\epsilon\_2 = 0$. If I'm not mistaken, `regular' in this case means that $F^{inst}(\epsilon\_1,\epsilon\_2,\mathbf{a},\mathbf{q})$ is analytic as a function of $\epsilon\_1,\epsilon\_2$ near $\epsilon\_1 = \epsilon\_2 = 0$ ([2] said the same thing in Theorem 5.14). However I struggle to understand why this result is true. 


The proof of this result is given in Proposition 7.3 [1]. My understanding of the argument is that they comparing the coefficients of $\mathbf{q}^n$ in the blow-up formula for each $n$ to recursively relates $F^{inst}\_n(\epsilon\_1,\epsilon\_2,\mathbf{a})$ as a regular function of $F^{inst}\_1(\epsilon\_1,\epsilon\_2,\mathbf{a}),...,F^{inst}\_{n-1}(\epsilon\_1,\epsilon\_2,\mathbf{a})$. By induction we can determine $F^{inst}\_n(\epsilon\_1,\epsilon\_2,\mathbf{a})$ and argue that it must be regular at $\epsilon\_1 =\epsilon\_2 = 0$ for all $n$. What I fail to see is why this is sufficient to conclude that $F^{inst}(\epsilon\_1,\epsilon\_2,\mathbf{a},\mathbf{q})$ must be regular at $\epsilon\_1 = \epsilon\_2 = 0$ as well since I have no idea about the nature of convergence of the sum $\sum\_{n = 1}^\infty \mathbf{q}^nF^{inst}\_n(\epsilon\_1,\epsilon\_2,\mathbf{a})$. 


I also cannot convince myself intuitively why this should be true. The partition function $Z^{inst}(\epsilon\_1,\epsilon\_2,\mathbf{a},\mathbf{q})$ is a sum over a vector of Young diagrams $\mathbf{k} = (k\_1,...,k\_N)$:
\begin{align}
Z^{inst}(\epsilon\_1,\epsilon\_2,\mathbf{a},\mathbf{q}) = \sum\_{\mathbf{k}}\mathbf{q}^{|\mathbf{k}|}\prod\_{l,m = 1}^N\Big[&\prod\_{(i,j)\in k\_l}((i - k'\_{mj})\epsilon\_1 + (k\_{li} - j + 1)\epsilon\_2 + a\_l - a\_m)^{-1}\\
\times&\prod\_{(i,j)\in k\_m}((k'\_{lj} - i + 1)\epsilon\_1 + (j - k\_{mi})\epsilon\_2 + a\_l - a\_m)^{-1}\Big].
\end{align}
For this to converge we must require that $\mathbf{a}$ satisfies $a\_l - a\_m \notin \mathbb{Z}\epsilon\_1 + \mathbb{Z}\epsilon\_2$ for all $l \neq m$. When $\epsilon\_1,\epsilon\_2$ are small, the terms involving $\mathbf{k}$ with small $|\mathbf{k}|$ becomes analytic since $|(...)\epsilon\_1 + (...)\epsilon\_2| << a\_l - a\_m$. When $l = m$ we get $1/(\epsilon\_1\epsilon\_2)^n$ factors but this should be suppressed after taking $\epsilon\_1\epsilon\_2\log(...)$ so it makes sense to me why $F^{inst}\_n(\epsilon\_1,\epsilon\_2,\mathbf{q})$ are analytic at $\epsilon\_1,\epsilon\_2 = 0$ for all $n$. But we can always choose a sequence $\epsilon\_1 = -\epsilon\_2 = h\_n \rightarrow 0, k = 1,2,3,...$ such that there exists a Young diagram vector $\mathbf{k}\_n$ (ofcourse $|\mathbf{k}\_n|\rightarrow \infty$) such that the denominator of the corresponding term goes to zero as fast as we wanted $|(...)\epsilon\_1 + (...)\epsilon\_2 + a\_l - a\_m| = |(...)h\_n - (a\_l - a\_m)| \rightarrow 0$. I cannot see why $\lim\_{n\rightarrow \infty}-h^2\_n\log Z^{inst}(h\_n,-h\_n,\mathbf{a},\mathbf{q})$ should converge which means $\lim\_{\epsilon\_1,\epsilon\_2\rightarrow \infty}\epsilon\_1\epsilon\_2\log Z^{inst}(\epsilon\_1,\epsilon\_2,\mathbf{a},\Lambda)$ shouldn't even exists (I see people write this all the time, so I need to understand this).


What did I misunderstand? Could someone clarify this for me please, thank you!


References:


[1] <https://arxiv.org/abs/math/0306198>


[2] <https://arxiv.org/pdf/0808.0884.pdf>



Nekrasov Partition Function: $F^{inst}(\epsilon_1,\epsilon_2,\mathbf{a},\mathbf{q})$ analytic at $\epsilon_1 = \epsilon_2 = 0$?
We meant that each coefficient of $\mathfrak q^n$ in $\varepsilon\_1 \varepsilon\_2 \log Z^{\mathrm{inst}}(\varepsilon\_1,\varepsilon\_2, \vec{a}, \mathfrak q)$ is regular at $\varepsilon\_1 = \varepsilon\_2 = 0$.


