Habiro lecture 5

V5A2-Habiro-Rings/Habiro_Rings_Notes.tex

@@ -56,6 +56,7 @@ \section*{Preliminaries}
 \include{Lecture 2}
 \include{Lecture 3}
 \include{Lecture 4}
+\include{Lecture 5}
 \newpage
 \printbibliography
 \end{document}

V5A2-Habiro-Rings/Lecture 2.tex

@@ -57,7 +57,7 @@ \section{Lecture 2 -- 18th October 2024}\label{sec: lecture 2}
     \end{equation} Applying the same manipulation to $f(qt)=\frac{f(qt)-f(q^{2}t)}{qt}$ we have $f(qt)=(1-qt)f(q^{2}t)$ which by induction and substituting into (\ref{eqn: q Pochhammer expansion of exponential}) we get $f(t)=(t;q)_{\infty}^{-1}$, yielding the claim. 
 \end{proof}
 This gives an expression for $(t;q)_{\infty}^{-1}$. 
-\begin{corollary}
+\begin{corollary}\label{corr: expansion of t q Pochhammer}
     There is an equality
     $$(t;q)_{\infty}^{-1}=\sum_{n\geq0}\frac{t^{n}}{(q;q)_{n}}$$
     in $\ZZ[q^{\pm},\frac{1}{1-q},\frac{1}{1-q^{2}}, \dots][[t]]$.

V5A2-Habiro-Rings/Lecture 4.tex

@@ -89,11 +89,13 @@ \section{Lecture 4 -- 15th November 2024}\label{sec: lecture 4}
     \end{align*}
     as desired. 
 \end{proof}
-We can modify the Nahm sum to rid ourselves of the $q^{a/2}$ factor in (\ref{eqn: q-difference equation of 1x1 Nahm sum}). 
+We can modify the Nahm sum to rid ourselves of the $q^{a/2}$ factor in (\ref{eqn: q-difference equation of 1x1 Nahm sum}).\marginpar{The exposition that follows is drawn from the erratum discussed at the begining of lecture 5. The discussion below is likely extremely error-prone and the reader is encouraged to consult \cite{NahmAsymptotics} for full details.}
 \begin{corollary}\label{corr: q-difference equation of modified 1x1 Nahm sum}
-    Let $f_{a}^{\mathrm{mod}}(t,q)=f_{a}(q^{-a/2}t,q)$. Then the modified Nahm sum satisfies the $t$-deformed $q$-difference equation 
+    Let 
+    $$f_{a}^{\mathrm{mod}}(t,q)=f_{a}(q^{-a/2}t,q)=\sum_{n\geq0}(-1)^{an}\frac{q^{\frac{1}{2}an^{2}-\frac{1}{2}an}}{(q;q)_{n}}t^{n}$$
+    Then the modified Nahm sum satisfies the $t$-deformed $q$-difference equation 
     \begin{equation}\label{eqn: q-difference equation}
-        f_{a}^{\mathrm{mod}}(t,q) - f_{a}^{\mathrm{mod}}(qt,q) = t\cdot f_{a}^{\mathrm{mod}}(q^{a}t,q).
+        f_{a}^{\mathrm{mod}}(t,q) - f_{a}^{\mathrm{mod}}(qt,q) = (-1)^{a}t\cdot f_{a}^{\mathrm{mod}}(q^{a}t,q).
     \end{equation}
 \end{corollary}
 \begin{proof}
@@ -123,6 +125,13 @@ \section{Lecture 4 -- 15th November 2024}\label{sec: lecture 4}
     (\ref{eqn: V functional equation for 1x1 Nahm ansatz}) of \Cref{prop: V functional equation for 1x1 Nahm ansatz} should be thought of as a multiplicative analogue of the Taylor expansion. 
 \end{remark}
 Using this, we can compute the ratio of the leading factors of \Cref{eqn: 1x1 Nahm sum ansatz,eqn: qt 1x1 Nahm sum ansatz}. 
+\begin{proposition}\label{prop: asymptotic expansion of modifed 1x1 Nahm sum at 1}
+    The Nahm sum $f_{a}(t,q)$ satisfies the asymptotic formula 
+    \begin{equation}\label{eqn: asymptotic expansion of modified 1x1 Nahm sum at 1}
+        f_{a}(t,q)\sim\exp\left(\frac{V(t)}{h}\right)\cdot O(h)
+    \end{equation}
+    as $q\to 1$ and $O(h)\in\QQ[[t,h]]$. 
+\end{proposition}
 \begin{corollary}\label{corr: exponent ratio for 1x1 Nahm sum antsatz}
     Let $V(t)$ be as in the Nahm equation ansatz. The ratios of the exponential factors satisfy 
     $$\exp\left(\frac{V(qt)}{h}\right)/\exp\left(\frac{V(t)}{h}\right)\sim\exp((\partial^{\log}V)(t))(1+O(h)).$$ 
@@ -138,7 +147,7 @@ \section{Lecture 4 -- 15th November 2024}\label{sec: lecture 4}
 Of special interest to us will be the final terms of the product above, where we denote $Z(t)=\exp((\partial^{\log}V)(t))$ and $\widetilde{Z}(t,q)=\exp\left((\partial^{\log}V)(t)\right)\exp\left(\frac{h}{2}((\partial^{\log})^{2}V)(t)\right)$.
 This allows us to produce a functional equation of the $g_{a}$ appearing in \Cref{eqn: 1x1 Nahm sum ansatz,eqn: qt 1x1 Nahm sum ansatz} which we will soon revisit. Note specializing at $q-1$ produces 
 \begin{equation}\label{eqn: specialized Z functional equation}
-    1-Z(t)=t\cdot Z(t)^{a}
+    1-Z(t)=(-1)^{a}t\cdot Z(t)^{a}
 \end{equation}
 
 We now explicitly describe $V(t)$. As suggested by the preceding discussion, it suffices to solve the differential equation $V'(t)=\frac{1}{t}\log(Z(t))$. 
@@ -152,12 +161,11 @@ \section{Lecture 4 -- 15th November 2024}\label{sec: lecture 4}
 \begin{proof}
     We compute 
     \begin{align*}
-        V'(t) &= -\frac{Z'(t)}{1-Z(t)}\log(Z(t)) - a\cdot\log(Z(t))\frac{Z'(t)}{Z(t)} \\
-        &= \frac{(Z(t)+a(1-Z(t)))Z'(t)}{(1-Z(t))Z(t)}\log(Z(t))
+        V'(t) &= -\frac{Z'(t)}{1-Z(t)}\log(Z(t)) - a\cdot\log(Z(t))\cdot\frac{Z'(t)}{Z(t)} \\
+        &= \log(Z(t))\cdot\frac{\left(Z(t)+a(1-Z(t))\right)Z'(t)}{(1-Z(t))Z(t)}
     \end{align*}
-    so taking the derivative of (\ref{eqn: specialized Z functional equation}), we have 
-    $$-Z'(t)=Z(t)^{a}+at\cdot Z(t)^{a-1}Z'(t)=Z'(t)\left(1+at\cdot Z(t)^{a-1}\right)$$
-    but $t\cdot Z(t)^{a-1}=\frac{1-Z(t)}{Z'(t)}$ so $-Z'(t)=\frac{Z'(t)}{Z(t)}(Z(t)+a(1-Z(t)))=-\frac{1-Z(t)}{t}$ which on substitution into the final line of the first series of displayed equations above yields the claim. 
+    and differentiating (\ref{eqn: specialized Z functional equation}) we get 
+    $-Z'(t)=(-1)^{a}Z(t)^{a}+(-1)^{a}at\cdot Z(t)^{a-1}Z'(t)$. Now note $(-1)^{a}at\cdot Z(t)^{a-1}=\frac{1-Z(t)}{Z'(t)}$ so $-Z'(t)=\frac{Z'(t)}{Z(t)}\left(Z(t)+a(1-Z(t))\right)=-\frac{1-Z(t)}{t}$. Substituting this into the equation for $V'(t)$ we get the desired result. 
 \end{proof}
 We can also discuss asymptotics of Nahm sums at roots of unity $\zeta_{m}$. In parallel to \Cref{prop: asymptotics of q t Pochhammer at root of unity}, we have the following result for Nahm sums. 
 \begin{theorem}\label{thm: Nahm sum asymptotics at roots of unity}
@@ -169,11 +177,31 @@ \section{Lecture 4 -- 15th November 2024}\label{sec: lecture 4}
 \end{theorem}
 \begin{proof}[Proof Outline]
     We use the ansatz 
-\begin{equation}\label{eqn: Nahm sum at roots of unity ansatz}
-    f_{a}(t,q) = \exp\left(\frac{V(t^{m})}{m^{2}h}\right)g_{a,m}(t,h)
-\end{equation}
-where we want to show that $g_{a,m}(t,h)\in\QQ(\zeta_{m})[[t,h]]$. As before we have 
-$$\exp\left(\frac{V((qt)^{m})}{m^{2}h}\right)=\exp\left(\frac{V(e^{mh}t^{m})}{m^{2}h}\right)$$
-and in the Taylor expansion we have 
-$$\exp\left(\frac{V(t^{m})}{m^{2}h}+\frac{1}{m}\log(Z(t^{m}))+O(h)\right).$$
+    \begin{equation}\label{eqn: Nahm sum at roots of unity ansatz}
+        f_{a}(t,q) = \exp\left(\frac{V(t^{m})}{m^{2}h}\right)g_{a}(t,h)
+    \end{equation}
+    where we want to show that $g_{a}(t,h)\in\QQ(\zeta_{m})[[t,h]]$ where \emph{a priori}, $g_{a}(t,q)\in\QQ(\zeta_{m})((h))[[t]]$. 
+
+    By \Cref{prop: V t dilogarithm equation}, we have that $t\cdot V'(t)=\log(Z(t))$ so we can write
+    $$\widetilde{Z}(t^{m},h)=\frac{\exp\left(\frac{V(q^{m}t^{m})}{m^{2}h}\right)}{\exp\left(\frac{V(t^{m})}{m^{2}h}\right)}=Z(t^{m})^{1/m}\cdot(1+O(h)).$$
+    and thus 
+    \begin{equation}\label{eqn: expansion of g function}
+        g_{a}(t,h)-\widetilde{Z}(t^{m},h)g_{a}(\zeta_{m}e^{h}t,h)=(-1)^{a}t\cdot\prod_{j=0}^{a-1}\widetilde{Z}(e^{\pi i h}t^{m},h)\cdot g_{a}(\zeta_{m}^{a}e^{ah}t,h).
+    \end{equation}
+    shwoing specialization at $h=0$ is well-defined. As such, we can write $g_{a}(t,0)$ as 
+    $$\sum_{j=0}^{a-1}t^{j}h_{j}(t^{m})$$
+    and $h_{j}$ a function in $t$. As such, by (\ref{eqn: expansion of g function}) we have
+    \begin{align*}
+        \sum_{j=0}^{a-1}t^{j}h_{j}(t^{m}) - Z(t^{m})^{1/m}\sum_{j=0}^{a-1}\zeta_{m}^{j}t^{j}h_{j}(t^{m}) &= (-1)^{a}t\cdot Z(t^{m})^{a/m}\cdot\sum_{j=0}^{a-1}\zeta_{m}^{aj}t^{j}h_{j}(t^{m})
+    \end{align*}
+    and we separate the sum for each $j$ by taking residue classes of exponents of $t$ where we have 
+    \begin{align*}
+        h_{j}(t^{m})(1-\zeta_{m}^{j}Z(t^{m})^{1/m})&=(-1)^{a}\zeta_{m}^{a(j-1)}Z(t^{m})^{a/m}\cdot h_{j-1}(t^{m}) && j>0 \\
+        h_{0}(t^{m})(1-Z(t^{m})^{1/m})&=(-1)^{a}\zeta_{m}^{a(m-1)}t^{m}Z(t^{m})^{a/m}h_{m-1}(t^{m}) && j=0
+    \end{align*}
+    where 
+    $$\prod_{j=0}^{m-1}(1-\zeta_{m}^{j}Z(t^{m})^{1/m})=(-1)^{am}\prod_{j=0}^{m-1}\zeta_{m}^{aj}t^{m}Z(t^{m})^{a}$$
+    holds as it simplifies to 
+    $$1-Z(t^{m})=(-1)^{a}t^{m}\cdot Z(t^{m})^{a}$$
+    by factorization results for cyclotomic polynomials. 
 \end{proof}

V5A2-Habiro-Rings/Lecture 5.tex

@@ -0,0 +1,48 @@
+\section{Lecture 5 -- 22nd November 2024}\label{sec: lecture 5}
+We continue our discussion of $q$-series and in particular a property of the modified Nahm sum considered in \Cref{corr: q-difference equation of modified 1x1 Nahm sum}.\marginpar{As it stands, the proofs and structure for this lecture are more rough than usual, and will be updated in due course.}
+
+The following definition is due to Konsevich-Soibelman \cite{DTInvariants}.
+\begin{definition}[Admissable Series]\label{def: admissable series}
+    A series $f\in\ZZ((q))[[t]]$ is admissable if it can be written as 
+    $$f=\prod_{n\geq1}\prod_{i\in\ZZ}(q^{i}t^{n};q)_{\infty}^{a_{n,i}}$$
+    such that for each $n$ only finitely many $a_{n,i}$ are nonzero. 
+\end{definition}
+\begin{remark}
+    These $a_{n,i}$'s are precisely Donaldson-Thomas invariants. 
+\end{remark}
+Admissable series force an algebraicity condition on the $q$, allowing $f$ to be written as an element of $\ZZ[q][[t]]$. Up to a condition on the residue of the series $f$ mod $(t)$, series in $\ZZ((q))[[t]]$ admit such an expansion. 
+\begin{proposition}
+    Let $f\in\ZZ((q))[[t]]$. If $f\equiv1\pmod{(t)}$ then $f$ admits a unique expansion as an admissable series. 
+\end{proposition}
+This result is in fact much more general and it can be shown that the modified Nahm sum 
+$$f_{a}(t,q)=\sum_{n\geq0}(-1)^{an}\frac{q^{\frac{1}{2}an^{2}-\frac{1}{2}an}}{(q;q)_{n}}t^{n}$$
+as previously defined is admissable. The original proof is highly involved, and we will instead offer a simpler exposition of the same result. Recall from \Cref{prop: logarithm at worst simple poles at roots of unity}, we have 
+\begin{equation}\label{eqn: modified Nahm sum as exponent of sum}
+    (q^{i}t;q)_{\infty}=\exp\left(-\sum_{\ell\geq 1}\frac{1}{\ell}\cdot\frac{q^{i\ell}t^{n\ell}}{1-q^{\ell}}\right)
+\end{equation}
+and further recall that $\ZZ((q))[[t]]$ is a $\lambda$-ring -- admits an action with $\NN$ as a multiplicative monoid -- and admits Adams operations $\psi_{n}:\ZZ((q))[[t]]\to\ZZ((q))[[t]]$ by $t\mapsto t^{n},q\mapsto q^{n}$. The Adams operations allow us to rewrite (\ref{eqn: modified Nahm sum as exponent of sum}) as 
+\begin{equation}\label{eqn: modified Nahm sum as exponent of Adams sum}
+    \exp\left(-\sum_{\ell\geq1}\psi_{\ell}\left(\frac{q^{i}t^{n}}{1-q}\right)\right).
+\end{equation}
+We introduce the notion of the plethystic exponential.
+\begin{definition}[Plethystic Exponential]\label{def: Plethystic exponential}
+    Let $A$ be a $\lambda$-ring and $\sum_{\ell\geq 1}\frac{a_{\ell}}{\ell}$ a convergent series in $A$. The plethystic exponential is of the series is given by 
+    $$\exp\left(\sum_{\ell\geq 1}\frac{1}{\ell}\psi_{\ell}(a_{\ell})\right).$$
+\end{definition}
+Now taking 
+$$\phi(t,q)=-\sum_{n\geq1}\sum_{i\in\ZZ}a_{n,i}q^{i}t^{n}\in\ZZ((q))[[t]]$$
+writing $\frac{\phi(t,q)}{1-q}$ as the plethystic logarithm of $f_{a}(t,q)$, inverse to the plethystic exponential, the coefficients of the expansion as an admissable series will be those coefficients of the plethystic logarithm since the plethystic exponential gives an isomorphism $t\QQ[q^{\pm}][[t]]\to 1+t\QQ[q^{\pm}][[t]]$. It thus suffices to show that the plethystic logarithm of $f_{a}$ is a function that is a sum $\frac{\phi_{0}(t)}{1-q}$ with an element of $\QQ[q^{\pm}][[t]]$. 
+
+Now using the ansatz
+\begin{equation}\label{eqn: plethystic exponential ansatz}
+    f_{a}(t,q)=\exp\left(\sum_{\ell\geq 1}\frac{1}{\ell}\frac{\phi_{0}(t^{\ell})}{1-q^{\ell}}\right)g_{a}(t,q)
+\end{equation}
+we show that there exists a chioce of function $\psi_{0}(t)$ so that the remainder $g_{a}(t,q)\in 1+t\QQ[q^{\pm}][[t]]$ and from which the result would follow by application of the Plethystic exponential. But a choice of $\phi_{0}\in t\cdot\QQ[[t]]$ can be made  such that $\sum_{\ell\geq 1}\frac{\phi_{0}(t^{\ell})}{\ell^{2}}=-V(t)$. 
+
+This is gives the desired result as stated below. 
+\begin{theorem}[Kontsevich-Soibelman, Efimov]\label{thm: modified 1x1 Nahm sum is admissable}
+    The $q$-series 
+    $$f_{a}(t,q)=\sum_{n\geq0}(-1)^{an}\frac{q^{\frac{1}{2}an^{2}-\frac{1}{2}an}}{(q;q)_{n}}t^{n}$$
+    is admissable. 
+\end{theorem}
+

V5A2-Habiro-Rings/refs.bib

@@ -149,6 +149,7 @@ @Article{NahmAsymptotics
 
 @Book{ZagierDilogarithm,
  Editor = {Cartier, Pierre and Julia, Bernard and Moussa, Pierre and Vanhove, Pierre},
+ Chapter = {The Dilogarithm Function},
  Title = {Frontiers in number theory, physics, and geometry {II}. {On} conformal field theories, discrete groups and renormalization. {Papers} from the meeting, {Les} {Houches}, {France}, {March} 9--21, 2003},
  ISBN = {978-3-540-30307-7},
  Year = {2007},
@@ -161,4 +162,22 @@ @Book{ZagierDilogarithm
  chapter = {The Dilogarithm Function},
  author = {Zagier, Don},
  shorthand = {Zag07}
+}
+
+@Article{DTInvariants,
+ Author = {Kontsevich, Maxim and Soibelman, Yan},
+ Title = {Cohomological {Hall} algebra, exponential {Hodge} structures and motivic {Donaldson}-{Thomas} invariants},
+ FJournal = {Communications in Number Theory and Physics},
+ Journal = {Commun. Number Theory Phys.},
+ ISSN = {1931-4523},
+ Volume = {5},
+ Number = {2},
+ Pages = {231--352},
+ Year = {2011},
+ Language = {English},
+ DOI = {10.4310/CNTP.2011.v5.n2.a1},
+ Keywords = {14N35,14D21,14D07,81T30},
+ zbMATH = {6055373},
+ Zbl = {1248.14060},
+ shorthand = {KS11}
 }