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!


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

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


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 $\varepsilon_1 = \varepsilon_2 = 0$.

  • $\begingroup$ Hi, thank you for your reply. So $F^{inst}(\epsilon_1,\epsilon_2,\mathbf{a},\mathbf{q})$ may not be regular at $\epsilon_1 = \epsilon_2 = 0$? Then how do we know that $\lim_{\epsilon_1,\epsilon_2\rightarrow 0}\epsilon_1\epsilon_2\log Z^{inst}(\epsilon_1,\epsilon_2,\mathbf{a},\mathbf{q})$ exists? $\endgroup$ – user113988 Jun 16 '18 at 16:08
  • $\begingroup$ It exists as a formal power series in $\mathfrak q$ at this stage. In the next step we compare it with Seiberg-Witten prepotential as a formal power series. $\endgroup$ – Hiraku Nakajima Jun 17 '18 at 22:29
  • $\begingroup$ Hello Prof. Nakajima, thank you for the clarification. Is this also how I should think of the work of Nekrasov & Okounkov (arxiv.org/pdf/hep-th/0306238.pdf) on the same result? They said that as $\epsilon_1 = -\epsilon_2 = \hbar \rightarrow 0$, $Z(\epsilon_1,\epsilon_2,\mathbf{a},\mathbf{q})$ is dominated by the limit shape term (maximumizing of the functional (4.39)) because the num of Young diagrams of size $n$ grows as $O(e^{\sqrt{n}})$. I struggle to understand the result since I don't see why the limit should exist. Would you mind comment on that as well? Thank you. $\endgroup$ – user113988 Jun 19 '18 at 4:38
  • $\begingroup$ ^^^ as in, should I also think of $Z(\epsilon_1,\epsilon_2,\mathbf{a},\mathbf{q})$ as a formal series in $\mathbf{q}$ (or $\Lambda$) too? I don't think they said anything about the series being formal. I could understand their result if number of Young diagrams $r = 1$ but otherwise we get a lot of factors like $1/(a_l - a_m + \hbar(...))$. I find it unconvincing that nothing will go wrong no matter how we take $\hbar \rightarrow 0$. $\endgroup$ – user113988 Jun 19 '18 at 4:59
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    $\begingroup$ It is one of main results in our paper. So I think that reading through our paper is the only way. $\endgroup$ – Hiraku Nakajima Jun 21 '18 at 7:19

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