text:
\subsection{\texorpdfstring{$\mb{P}$-objects}{P-objects}}
\label{subsection:P-objects}

In this section we consider the case of $\mb{P}$-objects.

Let $\ca{C}$ be a small dg-category over a field $k$ such that $\derived(\ca{C})^c$ has a Serre functor $\mb{S}$ which is a tensor functor: $\mb{S}(-) = - \oL{C} S_{\ca{C}}$.
$\mb{P}^n$-objects were first introduced in \cite{Huyb-Thomas06} where it was proved that given a $\mb{P}^n$-object $P$ in the derived category of a smooth projective variety $X$ one can construct an autoequivalence of $\derived^{\bounded}(X)$ called the $\mb{P}$-twist around $P$.

\begin{rmk}
	The definition of $\mb{P}^n$-objects has been later generalized to that of split $\mb{P}^n$-functors in \cite{Add-symmetries.hyperkahler}, \cite{Cautis-Flops-and-about}, and further to general $\mb{P}^n$-functors in \cite{AL-P-n-func}.
\end{rmk}

Recall the notation $\Hom_{\derived(\ca{C})}^{\bullet}(E,E) = \bigoplus_{n \in \mathbb{Z}} \text{Hom}_{\derived(\ca{C})}(E,E[n])[-n]$ for any $E \in \derived(\ca{C})$.

\begin{dfn}
	\label{def-p-obj}
	An object $P \in \derived(\ca{C})$ is said to be a $\mb{P}^n$-object if the following conditions are satisfied:
	\begin{enumerate}
		\item $P$ is both $\star_{k}$- and $\ca{C}$-perfect.
		\item $\Hom^{\bullet}_{\derived(\ca{C})}(P,P) \simeq k[t] / t^{n+1}$, $\deg(t) = 2$, as graded algebras.
		\item We have an isomorphism $P \oL{C} S_{\ca{C}} \simeq P[2n]$ in $\derived(\ca{C})$.
	\end{enumerate}
\end{dfn}

\begin{rmk}
	If $X$ is a smooth projective variety such that $\derived(\ca{C}) \simeq D_{qc}(X)$, then an object $P \in \derived(\ca{C})$ is a $\mb{P}^n$-object if and only if the corresponding object in $\derived_{\qc}(X)$ is a $\mb{P}^{n}$-object.
\end{rmk}

In \cite{Seg-autoeq-spherical-twist} the author describes two ways to realise the $\mb{P}$-twist around a $\mb{P}$-object as a spherical twist.
The first one considers $P$ as a dg-module over the dg-algebra $k[t]$, $\deg (t) = 2$, and defines the functor\footnote{To define the action of $k[t]$ on $P$ we assume that we replaced $P$ with an h-projective resolution, that we fixed a representative $\tilde{t} \in \text{Hom}_{\ca{C}}(P,P)$ for the generator of degree $2$, and we make $t$ act as $\tilde{t}$.}
\begin{equation}
	\label{P-twist-as-spherical}
	\derived( k[t] ) \xrightarrow{f_{P}(-) = - \stackrel{L}{\otimes}_{k[t]} P} \derived(\ca{C}).
\end{equation}

\begin{rmk}
	Notice that $\star_k$-perfectness implies $\star_{k[t]}$-perfectness because $k[t]$ is smooth, see e.g. \cite[pag. 7]{shklyarov2007serre}.
\end{rmk}

Then, under a technical assumption (which was subsequently shown to be always satisfied \cite{Formality-P-objects}), \cite[Proposition 4.2]{Seg-autoeq-spherical-twist} proves that \eqref{P-twist-as-spherical} is spherical, that its twist is the $\mb{P}$-twist around $P$, and that its cotwist is $[-2n-2]$.

The second construction of \cite{Seg-autoeq-spherical-twist} uses Koszul duality to rewrite \eqref{P-twist-as-spherical} in a different way.
More precisely, the object $k \in \derived(k[t])$ is compact and we have $\mathrm{RHom}_{k[t]}(k,k) \simeq k[\varepsilon] / \varepsilon^2$, with $\deg( \varepsilon) = -1$.
The $k[t]$-module $k[t,e]$, $\deg(e) = 1$, $d(e) = t$, is h-projective and gives a resolution of $k$ as a right $k[t]$-module.
Moreover, $k[t,e]$ carries a left action of $k[\varepsilon] / \varepsilon^2$ via the degree $-1$ map of $k[t]$-modules $k[t,e] \rightarrow k[t,e]$ that sends $p(t) + e q(t)$ to $q(t)$.
Hence, we get a functor
\begin{equation}
	\label{P-twist-as-spherical-2}
	\derived(k[\varepsilon] / \varepsilon^2) \xrightarrow{ - \stackrel{L}{\otimes}_{k[\varepsilon] / \varepsilon^2} P'} \derived(\ca{C})
\end{equation}
where
\[
	P' := k[t,e] \otimes_{k[t]} P = P[e] = \left\{ P[-2] \xrightarrow{t} \underset{\deg. 0}{P} \right\}.
\]
The twist around \eqref{P-twist-as-spherical-2} is the $\mb{P}$-twist around $P$, and its cotwists is given by $[-2n-2]$.

\begin{rmk}
	Construction \eqref{P-twist-as-spherical-2} was generalised to the case of split $\mb{P}^n$-functors in \cite[Theorem 5.1]{AL-P-n-func}.
\end{rmk}

\begin{comment}
Hence, we get a fully faithful functor\footnote{Our discussion only proves that $\derived(k[\varepsilon] / \varepsilon^2)^c \hookrightarrow \derived(k[\varepsilon] / \varepsilon^2) \xrightarrow{\eqref{Koszul-duality}} \derived(k[t])$ is fully faithful; for fully faithfulness on the entire category we use that $k[t,e]$ is $k[t]$-perfect.}
\begin{equation}
	\label{Koszul-duality}
	\derived(k[\varepsilon] / \varepsilon^2) \xrightarrow{ -\stackrel{L}{\otimes}_{k[\varepsilon] / \varepsilon^2} k[t,e]} \derived(k[t]).
\end{equation}
This functor is not an equivalence ($\mathrm{RHom}_{k[t]}(k[t,e], k[t,t^{-1}]) \simeq \mathrm{RHom}_{k[t]}(k,k[t,t^{-1}]) = 0$), however we claim that it restricts to an equivalence between $\mathrm{Tors}(k[t]) = \{ M \in \derived(k[t]) : \dim_k H^{\bullet}(M) < +\infty \}$ and $\derived(k[\varepsilon] / \varepsilon^2)^c$.

Indeed, notice that the right adjoint to \eqref{Koszul-duality} has no kernel when restricted to torsion modules:
\[
	\mathrm{RHom}_{k[t]}(k[t,e], M) = 0 \implies \cdot t : M \xrightarrow{\mathrm{qis}} M[2],
\]
and if $M \in \mathrm{Tors}(k[t])$ then $M$ (as a complex of $k$ vector spaces) is cohomologically bounded, and $M \overset{\mathrm{qis}}{\simeq} M[2]$ can happen if and only if $M$ is zero.
To get the claim is now enough to notice that \eqref{Koszul-duality} and $\mathrm{RHom}_{k[t]}(k[t,e], -)$ restrict to adjoint functors between $\derived(k[\varepsilon] / \varepsilon^2)^c$ and $\mathrm{Tors}(k[t])$.

Now we claim that the functor
\begin{equation}
	\label{P-twist-as-spherical-2}
	g_P : \derived(k[\varepsilon] / \varepsilon^2) \xrightarrow{ -\stackrel{L}{\otimes}_{k[\varepsilon] / \varepsilon^2} k[t,e]} \derived(k[t]) \xrightarrow{\eqref{P-twist-as-spherical}} \derived(\ca{C})
\end{equation}
is spherical and that it gives us the second model to realise a $\mb{P}$ twist as a spherical twist.

Indeed, the above functor is enhanced by tensor product with
\[
	P' := k[t,e] \otimes_{k[t]} P = P[e] = \left\{ P[-2] \xrightarrow{t} \underset{\deg. 0}{P} \right\}
\]
and this bimodule is perfect on both sides.
We see that $P'$ is $\ca{C}$-perfect because it is the convolution of two $\ca{C}$-perfect modules, while it is $k[\varepsilon] / \varepsilon^2$-perfect by \autoref{thm:adjoints} because \eqref{P-twist-as-spherical-2} has a left adjoint.
To see this notice that $- \stackrel{L}{\otimes}_{k[t]} k[t]^{\ast}$ preserves the subcategory of torsion modules and induces on it a Serre functor given by $[-1]$.
Therefore, $\derived(k[\varepsilon] / \varepsilon^2)^c$ has a Serre functor given by $[-1]$, and as the right adjoint to \eqref{P-twist-as-spherical-2} sends $M \in \derived(\ca{C})^c$ to\footnote{Here we use that $\mathrm{RHom}_{\ca{C}}(P,M) \in \mathrm{Tors}(k[t])$ for $M \in \derived(\ca{C})^c$.}
\[
	\mathrm{RHom}_{\ca{C}} ( k[t,e] \otimes_{k[t]} P, M) \simeq \mathrm{RHom}_{k[t]}(k[t,e], \mathrm{RHom}_{\ca{C}}(P,M)) \in \derived(k[\varepsilon] / \varepsilon^2)^c
\]
we can use Serre Duality to deduce that $\derived(k[\varepsilon] / \varepsilon^2)^c \xrightarrow{\eqref{P-twist-as-spherical-2}} \derived(\ca{C})^c$ has a left adjoint, and thus its colimit preserving extension (which is \eqref{P-twist-as-spherical-2}) has a left adjoint.

To conclude that \eqref{P-twist-as-spherical-2} is spherical we will prove that $t_{g_P} \simeq t_{\eqref{P-twist-as-spherical}}$ and $c_{g_P} \simeq c_{\eqref{P-twist-as-spherical}} \simeq [-2n-2]$.

The claim for the cotwist follows from \autoref{lem:twist-cotwist-swapped} because in that lemma we only need $\Phi$ to be fully faithful for the description of the cotwist.

The claim for the twist follows because the action of the twist on $\derived(\ca{C})$ can be recovered from its action on $\derived(\ca{C})^c$, and the latter can be computed by considering the restriction of \eqref{P-twist-as-spherical-2} to
\[
	\derived(k[\varepsilon] / \varepsilon^2)^c \xrightarrow{ - \stackrel{L}{\otimes}_{k[\varepsilon] / \varepsilon^2} k[t,e]} \mathrm{Tors}(k[t]) \xrightarrow{- \stackrel{L}{\otimes}_{k[t]} P} \derived(\ca{C})^c.
\]
However, in this situation we can apply \autoref{lem:twist-cotwist-swapped} because we changed the source category by an equivalent one, and the claim follows.
\end{comment}

\subsubsection{The \texorpdfstring{$k[t]$}{k[t]}-model}

Given $\mb{P}^n$-objects $P_1, \dots, P_m$ we can apply \autoref{twist-twist=twist} to \eqref{P-twist-as-spherical} to obtain a spherical functor
\begin{equation}
	\label{gluing-k[t]}
	\derived(R) \xrightarrow{ - \stackrel{L}{\otimes}_{R} (P_m \oplus \dots \oplus P_1)} \derived(\ca{C})
\end{equation}
whose twist is given by the composition of the $\mb{P}$-twists around $P_m, \dots, P_1$.
Here
\[
	R = \bigoplus_{i=1}^{m} k[t] \, \mathrm{id}_{P_i} \oplus \bigoplus_{i > j} \mathrm{H} \overline{\mathrm{om}}_{\ca{C}}(P_i, P_j)
\]
considered as a subalgebra of $\mathrm{H} \overline{\mathrm{om}}_{\ca{C}}(\oplus_{i=1}^m P_i, \oplus_{i=1}^m P_i)$.

The cotwist cannot be described as in \autoref{thm:cotwist-serre-duality} because $k[t]^{\ast}$ is not a shift of $k[t]$.

\begin{ex}
	A geometric example of this construction can be found in \cite[Theorem 3.3.4]{Barb-Flop-flop}.
\end{ex}

\subsubsection{The \texorpdfstring{$k[\varepsilon] / \varepsilon^2$}{k[e]}-model}

Given $\mb{P}^n$-objects $P_1, \dots, P_m$, let us denote $P'_i := k[t,e] \otimes_{k[t]} P_i = \left\{ P_i[-2] \xrightarrow{t} \underset{\mathrm{deg.} \, 0}{P_i} \right\}$.
Applying \autoref{twist-twist=twist} to \eqref{P-twist-as-spherical-2} we obtain a spherical functor
\begin{equation}
	\label{gluing-k[e]}
	\derived(R) \xrightarrow{ - \stackrel{L}{\otimes}_{R} (P'_m \oplus \dots \oplus P'_1)} \derived(\ca{C})
\end{equation}
whose twist is given by the composition of the $\mb{P}$-twists around $P_m, \dots, P_1$.
Here
\[
	R = \bigoplus_{i=1}^{m} k[\varepsilon] / \varepsilon^2 \, \mathrm{id}_{P'_i} \oplus \bigoplus_{i > j} \mathrm{H} \overline{\mathrm{om}}_{\ca{C}}(P'_i, P'_j)
\]
considered as a subalgebra of $\mathrm{H} \overline{\mathrm{om}}_{\ca{C}}(\oplus_{i=1}^m P'_i, \oplus_{i=1}^m P'_i)$.

\begin{ex}
	A geometric example of this construction can be found in \cite[Theorem 3.3.2]{Barb-Flop-flop}.
\end{ex}

\begin{rmk}
	Our hope was to prove an isomorphism $c_{\eqref{gluing-k[e]}}(-) \simeq - \stackrel{L}{\otimes}_R R^{\ast}[-2n-1]$ in analogy with \autoref{thm:cotwist-serre-duality}.
	Unfortunately, we stumble upon technical issues we do not know how to fix.
	More precisely, the components of the cotwist outside the diagonal are of the form $\mathrm{H} \overline{\mathrm{om}}_{\ca{C}}(P'_j, P'_i)[-1]$, $j < i$, and we would like to use Serre Duality to relate them to $\mathrm{H} \overline{\mathrm{om}}_{\ca{C}}(P'_i, P'_j)^{\ast}[-2n-1]$.
	The problem is that Serre Duality provides us with an isomorphism $\mathrm{Hom}^{\bullet}_{\derived(\ca{C})}(P'_j, P'_i)[-1] \simeq \mathrm{Hom}^{\bullet}_{\derived(\ca{C})}(P'_i, P'_j)[-2n-1]$ of $k[\varepsilon] / \varepsilon^2$-bimodules, but it is not clear whether one can lift this isomorphism to a quasi-isomorphism $\mathrm{H} \overline{\mathrm{om}}_{\ca{C}}(P'_j, P'_i)[-1] \rightarrow \mathrm{H} \overline{\mathrm{om}}_{\ca{C}}(P'_i, P'_j)^{\ast}[-2n-1]$ of $k[\varepsilon] / \varepsilon^2$-bimodules.
\end{rmk}


TITLE: Linear Algebra - linear functionals
QUESTION [0 upvotes]: Let $S$ be a set of vectors in $V$. Define $S^0$ to be the set of linear functional that vanishes on $S$.
Show that $$S^0  = \langle S\rangle^0$$
I am really confused what is meant by $S^0$

REPLY [1 votes]: A linear functional on $V$ is a linear map
$$
f\colon V\to K
$$
where $K$ is the base field (in your case it may be the reals or the complex numbers, please add details), where $K$ is considered as a vector space.
The set of linear functionals on $V$ is usually denoted by $V^*$; so
$$
S^0=\{f\in V^* : f(x)=0,\text{ for all }x\in S\}
$$
Now, set $T=\langle S\rangle$, the subspace spanned by $S$. It is rather easy showing that
$$
T^0\subseteq S^0
$$
because $S\subseteq T$. For the converse inclusion, take $f\in S^0$ and try showing that, for all $x\in T$, it must be $f(x)=0$ (that is, $f\in T^0$).

REPLY [1 votes]: By definition
$$f\in S^0\iff\;\forall\,s\in S\;,\;\;f(s)=0$$
What you have to prove is that a linear functional (i.e., an element of $\;V^*$) vanishes on a set $\;S\;$ iff it vanishes on the vector subspace that set generates, for example
$$f(s)=0\;\;\forall\,s\in S\implies \;\forall\;\text{linear combination}\;\sum_{i=1}^m a_
is_i\;,\;\;a_i\in\Bbb F\;,\;$$
$$s_i\in S\;,\;\;f\left(\sum_{i=1}^m a_is_i\right)=\sum_{i=1}^ma_if(s_i)=0 \;(\text{fill in details})$$

REPLY [0 votes]: $S^0=\{f:V\to\Bbb R \mid \forall s\in S: f(s)=0\}$.
Show that, if $f\in S^0$ then $f\in \langle S\rangle^0$ and the other way around, if
$g\in \langle S\rangle^0$ then $g\in S^0$. (One of these is trivial, the other follows by linearity.)


TITLE: Qualitative study of differential equation
QUESTION [0 upvotes]: Good morning everyone ,
I'd like to ask the following doubt within the exercise :
Study  $y'= \frac{x \cdot e^{y(x)}}{(y(x)-1)} $ .
After separable variables I obtain : $-ye^{-y}=\frac {x^{2}}{2} + C$
The doubt was (in the picture) : I'd like to know after the first two graph on the left (one of $-ye^{-y}$, one of $\frac {x^{2}}{2} + C$) how to properly put them in a sort of bijection, or at least gather them properly so that I could find the graph on the right, that is my final goal.
Help would be appreaciated,
Thanks.

REPLY [1 votes]: With the help of this MATHEMATICA script

gr = Table[
   If[c > 0,
    ContourPlot[y E^-y + x^2/2 + c == 0, {x, -5, 5}, {y, -5, 5},
     ContourStyle -> Red],
    ContourPlot[y E^-y + x^2/2 + c == 0, {x, -5, 5}, {y, -5, 5},
     ContourStyle -> Blue]], {c, -1, 1, 0.1}];
Show[gr]

we can verify the results obtained analytically confirming that
$$
ye^{-y}+\frac{x^2}{2}+c = 0
$$
is an implicit function $y=y(x,c)$ only for $c \ge 0$ (blue curves). In red the results for $c < 0$
I hope this helps.


TITLE: The way into logic, Gödel and Turing
QUESTION [5 upvotes]: I have always read about the geniuses of Alan Turing and Kurt Gödel . Many websites mention their works in logic as revolutionary. I want to understand their works, but I don't exactly know the way through which I should go in order to understand their work. To be precise, I want to know the prerequisites required to learn their theories.  I hope that members here can suggest the correct way.

REPLY [0 votes]: For a very gentle history and introduction to Leibniz, Gödel, and Turing, I recommend Martin Davis' The Universal Computer: The Road from Leibniz to Turing. As I recall, it was accessible to a motivated high school student (U.S.) but managed to convey the basic logic, computability and mathematical ideas. It's a short, easy, fun read even if its target audience is below your level. By the way, Martin Davis is more than qualified to present this material.


TITLE: real matrix fails to have an eigenvalue
QUESTION [0 upvotes]: Wikipedia says:

Having an eigenvalue is an accidental property of a real matrix (since it may fail to have an eigenvalue), but every complex matrix has an eigenvalue.

Yet, IMO, real matrices are subclass of complex ones. So, even without having any mathematical degree I see that this cannot be true.

REPLY [2 votes]: Answer in the comment: The Wikipedia article you linked says it: "Namely, let $V$ be any vector space with some scalar field $K$, and let $T$ be a linear transformation mapping $V$ into $V$. We say that a vector $x$ of $V$ is an eigenvector of $T$ if (and only if) there is a scalar $\lambda\in K$ such that $T(x) = \lambda x$."
($\lambda\in K$) must be in bold


TITLE: Showing that $ZFC$ can't prove the existence of Strongly Inaccessible Cardinals. Why the extra steps?
QUESTION [1 upvotes]: It is possible I am missing something exceedingly obvious. We know that $V_{\kappa} \vDash ZFC$ when $\kappa$ is strongly inaccessible.
Pretty much every proof of $ZFC \not \vdash "\exists$ inaccessible cardinals" goes like:
Suppose $ZFC \vdash "\exists$ strong inaccessibles" (I'll shorten this from now one as $\exists I$). Then pick the least such one, $\kappa$. $V_{\kappa} \vDash "\exists I"$. By standard absoluteness, what $V_{\kappa}$ thinks are strongly inaccessible will indeed be strongly inaccessibles, contradicting the choice of $\kappa$ as the smallest such cardinal.
And this argument is fine to me. But I don't see why we have to go through the trouble of estabilishing absoluteness, picking the smallest one and everything. Why can't we simply do:
Suppose $ZFC \vdash "\exists I"$. Since we may formalize model theory, the truth predicate for set models, consistency statements, etc, we have that $ZFC \vdash "\exists I" \rightarrow Con(ZFC)$ (by exhibiting a set model of $ZFC$). Then we have by basic modus ponens that $ZFC \vdash Con(ZFC)$. This contradicts Godel's incompleteness theorem and then we're done.
This argument seems fine to me, and more natural too. However, most arguments I can see seem to follow the first argument instead. And the first one isn't hard per se, but it makes me wonder why we're doing all these extra steps.
Am I missing something? Does the second argument not work?

REPLY [4 votes]: They both work. However, I don't actually think that the second one is more natural: it involves first proving the second incompleteness theorem, which is nontrivial and not obviously relevant to the specific problem at hand.
By contrast, in the first approach we give a method for building models of $\mathsf{ZFC}$ + "There are no inaccessibles" from models of $\mathsf{ZFC}$. Moreover, this method "preserves niceness" in many senses, e.g. if the starting $\mathsf{ZFC}$-model is well-founded then the $\mathsf{ZFC\neg I}$-model is well-founded as well. Note that this is not something incompleteness can do for us alone - e.g. there is no nice way to take a well-founded model of $\mathsf{ZFC}$ and produce a well-founded  model of $\mathsf{ZFC+\neg Con(ZFC)}$ since there are no well-founded models of $\mathsf{ZFC+\neg Con(ZFC)}$ in the first place.
So I think the first approach actually is the better one.


TITLE: Some conditions which fail when $T$ is not linear.
QUESTION [2 upvotes]: I came across this question in my self study of a little fuctional analysis and I'd help in solving it. I have also attempted it in the hope that I'll be corrected where I made mistakes. Thanks.

Let $X$ and $Y$ be normed linear spaces and $T:X\to Y$ be linear. I want to show the following:
(a) $T$ is continuous if and only if it is continuous at a single point $x_0$ in $X$.
  (b) $T$ is Lipschitz if and only if it is continuous.
   (c) Neither (i) nor (ii) hold in the absence of the linearity assumption on $T$.

Attempts.
(a)
   $(\Rightarrow)$ Obvious, since if $T$ is continuous everywhere in $X$, then it is certainly continuous at a point $x_0\in X$.
$(\Leftarrow)$ Suppose that $T$ is continuous at $x_0 \in X$. Let $\varepsilon \gt 0$ be given. Then there is a $\delta\gt 0$ such that $\|T(x)-T(x_0)\|\lt \varepsilon$ whenever $\|x-x_0\|\lt \delta.$ Let $x_1$ be any arbitrary point in $X$.  Then for $\|x-x_1\| = \|(x + x_0 -x_1)-x_0\|\lt \delta$,
  \begin{align*}
  \|T(x)-T(x_1)\| & = \|T(x)+T(x_0)-T(x_1)-T(x_0)\| \\
  & = \|T(x + x_0 -x_1)-T(x_0)\|\qquad(\because T ~\text{is linear})\\
  & \lt \varepsilon.
  \end{align*}
(b)
$(\Rightarrow)$ Suppose $T$ is continuous. Then $T$ is bounded. So there exists an $M\geqslant 0$ such that for $x,y\in X$ $$\|T(x)-T(y)\| \leq M \|x-y\|,$$ which is precisely the condition for $T$ to be Lipschitz.
$(\Leftarrow)$ Suppose $T$ is Lipschitz. Then $T$ is bounded. Hence $T$ is continuous.
(c) at the moment, I can only think of $T(x) =x^2$ whichs is not linear but continuous everywhere. However, I can't quite deduce which of (a) or (b) fails to hold.

REPLY [2 votes]: $(a)$ and $(b)$ look right. As for $(c)$:
The function
$$
f(x) =
\begin{cases}
0 & x < 0 \\
1 & x \geq 0 \\
\end{cases}
$$
Is non-linear and continuous at $5$ but not continuous everywhere on $\mathbb R$. So this is a counterexample for $(a)$. For $(b)$, Wikipedia gives you $f: \mathbb R \to \mathbb R$, $f(x) = x^2$ as an example of a function that's continuous but not Lipschitz continuous.
To see why $f(x) = x^2$ is not Lipschitz continuous note that "$f : X \to Y$ is Lipschitz" means that there exists a constant $K$ such that for all $x \in X$ we have $\frac{|f(x) - f(x^\prime)|}{|x - x^\prime|} < K$. So in particular, $\frac{d}{dx} f(x) = \lim_{\Delta \to 0} \frac{|f(x) - f(x + \Delta)|}{\Delta} < K$. For $f(x) = x^2$ you have $f^\prime (x) = 2x$. We are on $\mathbb R$ so this means that for every constant $K$ you can find an $x_0$ such that $f(x_0) > K$ so $f$ cannot be Lipschitz.


TITLE: Intersection of 2 tangent lines on a circle
QUESTION [0 upvotes]: Say there are two lines that are tangent to a circle at points A and B, so that they intersect at an external point C, as shown below.

Two congruent right triangles are formed, since the tangent line is perpendicular to the radius. So line AC has the same length as line BC.
If the angle ACB is known, how can the length of line AC be calculated?
For example, if the radius $r$ is 1, and the angle ACB is $60\unicode{xb0}$, what is the length of line AC?

REPLY [3 votes]: If $\angle ACB=60$ then $\angle ACO=30$ and therefore $OC=2$. Now you can use pythagoras and find $AC$.
Using pythagoras : $OC^2=OA^2+AC^2$ and you will have:
$4=1+AC^2$ Therefore
$AC=\sqrt 3$


TITLE: Cauchy sequence and a particular subsequence
QUESTION [0 upvotes]: Let $\{{f_n}\}$ a Cauchy sequence. Then by definition $$\forall k\in\mathbb{N},\quad\exists n(k)\in\mathbb{N}\quad|\quad\forall m,n>n(k)\quad \|f_m-f_n\|<\frac{1}{2^k}$$
I  must prove that we can choose a increasing sequence of natural numbers $\{n_k\}$ for which
$$\|f_{n_{k+1}}-f_{n_k}\|< \frac{1}{2^k}\;\text{for all}\;k\in\mathbb{N}$$
If I choose $n_k> n(k)$ for all $k\in\mathbb{N}$ and then I can suppose without loss of generality that the sequence $\{n_k\}$ is increasing.
Is this formally correct or should we add more?
Thanks!
Let's see if I understand: Let $n_1=n(1)+1$ and $n_2=\max\{n(2)+1,n_1+1\}$. Now $n_2> n_1>n(1)$, then $$\|f_{n_2}-f_{n_1}\|<\frac{1}{2}.$$
Now, let $n_3=\max\{n(3)+1,n_2+1\}$, then $n_3>n_2$ and $$\|f_{n_3}-f_{n_2}\|<\frac{1}{4},$$ since $n_3>n_2\ge n(2)+1>n(2)$
We can proceed in this way for every $k$, right?

REPLY [2 votes]: You should probably explain why you can assume "without loss of generality" that the $n_k$ are decreasing.
If you want to be formal, you can do a "proper" recursive construction:
E.g. let $n_1 = n(1)+1$. Then having defined $n_k$ for some $k$ such that  $$\forall 1 \le l < k: \|f_{n_{l+1}} - f_{n_l}\| < \frac{1}{2^l}\tag{k}$$
(which is voidly obeyed for $k=1$ now) we define $n_{k+1}=\max(n(k+1) + 1, n_k+1))$ and so both ensure the "recursion condition" $(k+1)$ from $(k)$ and the definition of $n(k+1)$, and the strict increasingness of the $n_k$.


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\begin {document}

\baselineskip 16pt
\title{Asymptotic Laplacian-Energy-Like Invariant of Lattices}

\author{\small $  Jia\textrm{-}Bao \ Liu^{1,2},\ \  Xiang\textrm{-}Feng \ Pan^1
 \thanks{Corresonding author. Tel:+0551-6386-1835. \E-mail:~xfpan@ahu.edu.cn(X.-F. Pan),liujiabaoad@163.com(J.-B. Liu),
  hufu@mail.ustc.edu.cn(F.-T. Hu),
 huayanyanhu@163.com(F.-F.Hu).}\ \ ,Fu\textrm{-}Tao \ Hu^{1},\ \  Feng\textrm{-}Feng \ Hu^1
 $\\ \\
\small  $^{1}$ School of Mathematical Sciences, Anhui  University, Hefei 230601, China\\
\small  $^{2}$ Department of Public Courses, Anhui Xinhua
University, Hefei 230088, China\\}

\date{}
\maketitle
\begin{abstract}
Let $\mu_1\ge \mu_2\ge\cdots\ge\mu_n$ denote the Laplacian
eigenvalues of $G$ with $n$ vertices. The Laplacian-energy-like
invariant, denoted by $LEL(G)= \sum_{i=1}^{n-1}\sqrt{\mu_i}$, is a
novel topological index. In this paper, we show that the
Laplacian-energy-like per vertex of various lattices is
independent of the toroidal, cylindrical, and free boundary
conditions. Simultaneously, the explicit asymptotic values of the
Laplacian-energy-like in these lattices are obtained. Moreover,
our approach implies that in general the Laplacian-energy-like per
vertex of other lattices is independent of the boundary
conditions.

\medskip
\noindent {\bf Keywords}: Lattice;  \  \  Laplacian-energy-like
invariant; \  \ Laplacian spectrum
\end{abstract}

\section{ Introduction}
 Throughout this paper we concerned with finite undirected
connected simple graphs. Let $G$ be a graph with vertices labelled
$1,2,\dots,n$. The adjacency matrix $A(G)$ of $G$ is an $n\times
n$ matrix with the $(i,j)$-entry equal to 1 if vertices $i$ and
$j$ are adjacent and 0 otherwise. Supposed
$D(G)=diag(d_1(G),d_2(G),\ldots ,d_n(G))$ be the degree diagonal
matrix of $G$, where $d_i(G)$ is the degree of the vertex $i$,
$i=1,2,\ldots,n$. Let $L(G)=D(G)-A(G)$ is called the Laplacian
matrix of $G$. Then, the eigenvalues of $A(G)$ and $L(G)$ are
called eigenvalues and Laplacian eigenvalues of $G$, denoted by
$\lambda_1, \lambda_2,\dots,\lambda_n$  and $\mu_1,
\mu_2,\dots,\mu_n$, respectively. The underlying graph theoretical
definitions and notations  follow {\rm \cite{CZ}.

Gutman has defined the energy of a graph $G$ with $n$ vertices {\rm
\cite{IG}, denoted by $E(G)$, as $$E(G)=
\sum_{i=1}^{n}|\lambda_i|.$$

Let $G$ be a graph of order $n$ with Laplacian spectrum $\mu_1
\geq \mu_2 \geq \cdots \geq \mu_n.$ The Laplacian-energy-like
invariant of G, LEL for short, is defined as
$$  LEL(G)=\sum_{i=1}^{n-1}\sqrt{\mu_i}.$$
The concept of $LEL(G)$ was introduced by J. Liu and B. Liu ({\rm
\cite{LL}, 2008), where it was shown that it has similar features
as graph energy.

Gutman et al. pointed out in {\rm \cite{GZF} that $LEL(G)$ is more
similar to $E(G)$ than to $LE(G)$.
 The Laplacian-energy-like invariant describes well the properties which are accounted by the
majority of molecular descriptors: motor octane number, entropy,
molar volume, molar refraction, particularly the acentric factor
AF parameter, but also more difficult properties like boiling
point, melting point and partition coefficient Log P. {\rm
\cite{LHY,DA}. D. Stevanovi$\acute{c}$ et al. {\rm \cite{DA} have
proved that $LEL(G)$ has the properties as good as the
Randi$\acute{c}$ $\chi$ index and better than the Wiener index
which is a distance based index. Besides, it is well defined
mathematically and shows interesting relationships in particular
classes of graphs, these recommending $LEL(G)$ as a noval and
powerful topological index.

 The index has
attracted extensive attention due to its wide applications in
physics, chemistry, graph theory, etc. {\rm \cite{GM,LY,MA}.
Details on its theory can be found in the survey on the
Laplacian-energy-like invariant {\rm \cite{LHY}, the recent papers
{\rm \cite{DGCZ,DXG}, and the references cited therein. Due to its
structure, lattices are of special interest, especially the
chemical and physical indices of some lattices were studied
extensively, see for instance {\rm \cite{YLZ,DGMS,ZZ}.

One of reasons we investigated $LEL(G)$ of lattices is that some
physical and chemical indices of lattices have been presented in
 {\rm \cite{YLZ,ZZ,LPC}, however, there is seldom results on $LEL(G)$ of
 lattices.
 Another reason is, there are a great deal of analogies between
the properties of $E(G)$ and $LEL(G)$, asymptotic energy of
lattices has investigated in chemical physics{\rm \cite{YZ,YYZ},
it is natural for us to consider asymptotic $LEL(G)$ of various
lattices.

The energy $E(G)$  of toroidal lattices have been studied in {\rm
\cite{YZ}, the Kirchhoff index of some toroidal lattices has also
investigated in {\rm \cite{YLZ}, a very elementary and natural
question is that if $G$ are lattices, how about $LEL(G)$? It is an
interesting problem to study the Laplacian-energy-like of some
lattices with various boundary condition.  Motivated by results
above, we consider the problem of computation of the $LEL(G)$ of
some lattices in this article.

The rest of the paper is organized as follows. In Section 2, we
propose the asymptotic Laplacian-energy-like of square lattices
and give the related explanations. We provide a detailed
derivation of the asymptotic Laplacian-energy-like change due to
edge deletion in Section 3. The asymptotic Laplacian-energy-like
of hexagonal, triangular, and $3^34^2$ lattices with three
boundary conditions are investigated in Section 4. We present
concluding remarks and conclude the paper in Section 5.

\section{Asymptotic Laplacian-energy-like of square lattices}

 Given graphs $G$ and $H$ with vertex sets $U$ and $V$, the cartesian product
 $G\Box H$ of graphs $G$ and $H$ is a graph such that
the vertex set of $G\Box H$ is the cartesian product $U\Box V$;
and any two vertices $(u,u')$ and $(v,v')$ are adjacent in $G\Box
H$ if and only if either $u = v$ and $u'$ is adjacent with $v'$ in
$H$, or $u' = v'$ and $u$ is adjacent with $v$ in $G$ {\rm
\cite{CZ}.

Let $P_m\Box P_n$, $P_m\Box C_n$, and $C_m\Box C_n$, denote the
square lattices with free, cylindrical and toroidal boundary
conditions, respectively, where $P_n$ and $C_n$ denote the path and
the cycle with $n$ vertices. Obviously, $P_m\Box P_n$ is a sequence
of spanning subgraphs of the sequence $P_m\Box C_n$ of finite
graphs, and $P_m\Box C_n$ is a sequence of spanning subgraphs of the
sequence $C_m\Box C_n$ of finite graphs. Particularly,
\begin{eqnarray} \nonumber  &&\lim_{m,n\to \infty}\frac{\left|\{v\in
V(P_m\Box P_n):~d_{P_m\Box P_n}(v) =
d_{C_m\Box C_n}(v)\}\right|}{\mid V(C_m\Box C_n)\mid} \\
&&=\lim_{m,n\to \infty}\frac{\left|\{v\in V(P_m\Box C_n):~d_{P_m\Box
C_n}(v)=d_{C_m\Box C_n}(v)\}\right|}{\mid V(C_m\Box C_n)\mid}
 =1,\nonumber
 \end{eqnarray}
  that is, almost
all vertices of $C_m\Box C_n$ and $P_m\Box C_n$
 (resp. $C_m\Box C_n$ and $P_m\Box P_n$) have the same degrees.
Let $G_1,G_2$ be graphs with adjacency matrices $A_1,$ $A_2$,
degree matrices $D_1,$ $D_2$ and Laplacan matrices $L_1,$¡¡$L_2,$
respectively. Then $L_1=D_1-A_1,L_2=D_2-A_2.$
 If $\mu_i(G_1),\mu_j(G_2)$ are Laplacian eigenvalues of $G_1,G_2$, then
the Laplacian eigenvalues of $G_1\Box G_2$ are all possible sums
$\mu_i(G_1)+\mu_j(G_2)$, as noted in \cite{DC}. On the other hand,
it is well known that the Laplacian eigenvalues of a path $P_m$
and a cycle $C_m$ are $2-2cos\frac{i\pi}{n} (i=0,1,\dots,n-1)$
 and $2-2cos\frac{2\pi j}{n} (j=0,1,\dots,m-1)$
\cite{DC}, respectively. Consequently, the Laplacian eigenvalues
of $P_m\Box P_n$ (resp. $P_m\Box C_n$ and $C_m\Box C_n$) are
$4-2cos\frac{i\pi}{m}-2cos\frac{j\pi}{n},i=0,1,\dots,n-1;j=0,1,\dots,m-1$
(resp.
$4-2cos\frac{i\pi}{m}-2cos\frac{2j\pi}{n},i=0,1,\dots,n-1;j=0,1,\dots,m-1$
and
$4-2cos\frac{2i\pi}{m}-2cos\frac{2j\pi}{n},i=0,1,\dots,n-1;j=0,1,\dots,m-1$).

Therefore, the Laplacian-energy-like per vertex of $P_m\Box P_n$,
$P_m\Box C_n$, and $C_m\Box C_n$ are defined as
$$1. \ \lim_{m\to\infty} \lim_{n\to \infty}\frac{LEL\Big(P_m\Box P_n\Big)}{\mid V(P_m\Box P_n)\mid}
= \lim_{m\to\infty} \lim_{n\to \infty}\frac{1}{mn}
\sum_{i=0}^{m-1}\sum_{j=0}^{n-1}
\sqrt{4-2cos\frac{i\pi}{m}-2cos\frac{j\pi}{n}} $$
$$=\int_0^{1}\int_0^{1} \sqrt{4-2cos\pi x-2cos\pi y}\ ~dxdy
= \frac{1}{\pi^2}\int_0^{\pi}\int_0^{\pi} \sqrt{4-2cos x-2cos y}\
~dxdy\approx 1.9162.$$

$$2. \ \lim_{m\to\infty} \lim_{n\to \infty}\frac{LEL\Big(P_m\Box C_n\Big)}{\mid V(P_m\Box C_n)\mid}
= \lim_{m\to\infty} \lim_{n\to \infty}\frac{1}{mn}
\sum_{i=0}^{m-1}\sum_{j=0}^{n-1}
\sqrt{4-2cos\frac{i\pi}{m}-2cos\frac{2j\pi}{n}} $$
$$=\int_0^{1}\int_0^{1} \sqrt{4-2cos\pi x-2cos2\pi y}\ ~dxdy
= \frac{1}{2\pi^2}\int_0^{\pi}\int_0^{2\pi} \sqrt{4-2cos x-2cos
y}\ ~dxdy \approx1.9162.$$

$$3. \ \lim_{m\to\infty} \lim_{n\to \infty}\frac{LEL\Big(C_m\Box C_n\Big)}{\mid V(C_m\Box C_n)\mid}
= \lim_{m\to\infty} \lim_{n\to \infty}\frac{1}{mn}
\sum_{i=0}^{m-1}\sum_{j=0}^{n-1} \sqrt{4-2cos\frac{2
i\pi}{m}-2cos\frac{2j\pi}{n}} $$
$$=\int_0^{1}\int_0^{1} \sqrt{4-2cos2\pi x-2cos2\pi y}\ ~dxdy
= \frac{1}{4\pi^2}\int_0^{2\pi}\int_0^{2\pi} \sqrt{4-2cosx-2cos
y}\ ~dxdy \approx 1.9162.$$

The numerical integration value in last line is calculated with
software MATLAB calculation.

\vspace{5pt} \noindent {\bf Remark 2.1}\ By using computer software
MATLAB, we can easily get that the above numerical integration
values, which imply that $P_m\Box P_n$, $P_m\Box C_n$ and $C_m\Box
C_n$ have the same asymptotic Laplacian-energy-like $LEL(P_m\Box
P_n)=LEL(P_m\Box C_n)=LEL(C_m\Box C_n)\approx 1.9162mn$ as $m,n$
approach infinity.

\vspace{5pt} \noindent {\bf Remark 2.2}\ The asymptotic
Laplacian-energy-like $LEL(G)$ of square lattices is independent on
the three boundary conditions, i.e., the free, cylindrical and
toroidal boundary conditions.

The phenomenon above is not accidental, the related explanations
are proposed in next section. Our method implies
 that in general the Laplacian-energy-like per vertex
of lattices is independent of the boundary conditions.

\section{ Graph asymptotic Laplacian-energy-like change due to edge deletion}

We propose a detailed derivation of the asymptotic
Laplacian-energy-like change due to edge deletion, in our proof,
some techniques in \cite{YZ} are referred to. Recall some results
which will be used in our discussion. The authors \rm \cite{LL}
verified the following theorem.
\begin{theorem}\label{3-1}(\cite{LL})
 Let $G$ be a graph on $n$ vertices
with $m$ edges, then
$$\sqrt{2m}\leq LEL(G)\leq \sqrt{2}m.$$
The first equality is attained if and only if $G\cong\overline{K_n}$
or $K_2\cup(n-2)K_1,$ and the second equality is attained if and
only if $G\cong rK_2\cup(n-2r)K_1$, where $0\leq r\leq
\lfloor\frac{n}{2}\rfloor.$
\end{theorem}

 According to Theorem 3.1, one can easily get
 that
 \begin{lemma}\label{3-2}
 Let $G$ be a graph on $n$ vertices
with $m$ edges, then $$LEL(G)\leq \sqrt{2}m < 2m.$$
\end{lemma}

The authors \rm \cite{DS,DWS} investigated how the energy of a
graph changes when edges are removed, and obtained the following
consequence.
\begin{lemma}\label{3-3} (\cite{DS,DWS})
 Let $H$ be an induced
subgraph of a graph $G.$ Then
$$E(G)-E(H)\leq E(G-E(H))  \leq E(G)+E(H).$$
\end{lemma}

In addition, the authors \rm \cite{YZ} obtained that
$$\mid E(G)-E(H)\mid \leq E(G-E(H))  \leq E(G)+E(H).$$

 With a similar method, one
can prove the following result.
\begin{lemma}\label{3-4}
 Let $H$ be a subgraph of a graph $G.$
  Then
  $$\mid LEL(G)-LEL(H)\mid \leq LEL(G-E(H))  \leq LEL(G)+LEL(H).$$
\end{lemma}

Consider two graphs $G$ and $H$ $(V(G)\cap V(H)$ may be disjoint),
denoted by $$\Delta(G,H)=\mid E(G)\mid+ \mid E(H)\mid-2\mid
E(G)\cap E(H)\mid,$$ i.e., $\Delta(G,H)$ equals the number of
edges of symmetric difference of $E(G)$ and $E(H)$.

\begin{theorem}\label{3-5}
 Let $\{G_n\}$ and $\{H_n\}$ be two sequences of graphs such that
$$\lim_{n\to \infty}\frac{\Delta(G_n,H_n)}{LEL(G_n)}=0.$$
Then
$$\lim_{n\to \infty}\frac{LEL(H_n)}{LEL(G_n)}=1.$$
\end{theorem}
{\bf Proof.} Let $F_n$ be the subgraph of $G_n$ or $H_n$ induced by
$E(G)\cap E(H).$ Note that
\begin{eqnarray}
\nonumber \left| \frac{LEL(H_n)}{LEL(G_n)}-1 \right|
&&=\left|\frac{LEL(H_n)-LEL(G_n)}{LEL(G_n)}\right|
=\left|\frac{LEL(H_n)-LEL(F_n)+LEL(F_n)-LEL(G_n)}{LEL(G_n)}  \right|\\
&&\nonumber \leq \left|\frac{LEL(G_n)-LEL(F_n)}{LEL(G_n)}\right|
+\left|\frac{LEL(H_n)-LEL(F_n)}{LEL(G_n)}\right|.\\ \nonumber
\end{eqnarray}
By Lemma 3.2 and Lemma 3.4,
$$\mid LEL(G_n)-LEL(F_n)\mid \leq LEL(G_n-E(F_n))  \leq 2\left|E(G_n)\right|-2\left|E(F_n)\right|,$$
$$\mid LEL(H_n)-LEL(F_n)\mid \leq LEL(H_n-E(F_n))  \leq 2\left|E(H_n)\right|-2\left|E(F_n)\right|.$$
Consequently,
$$\left| \frac{LEL(H_n)}{LEL(G_n)}-1 \right|\leq \frac{2\Big(\mid E(G_n)\mid+ \mid E(H_n)\mid-2\mid
E(F_n)\mid\Big)  }{LEL(G_n)} =\frac{2\Delta(G_n,H_n)
}{LEL(G_n)},$$ implying the theorem holds. \hfill $\blacksquare$

Based on Theorem 3.5, one can straightforwardly arrive to that

\begin{theorem}\label{3-6}
 Let $\{G_n\}$ be a sequence of finite simple graphs with bounded average degree such that
$$\lim_{n\to \infty}\left|V(G_n)\right|=\infty, \lim_{n\to
\infty}\frac{LEL(G_n)}{\left|V(G_n)\right|}=h\neq0.$$ Let $\{H_n\}$
be a sequence of spanning subgraphs of $\{G_n\}$ such that
$$\lim_{n\to\infty}\frac{\left|v \in V(H_n):d_{H_n(v)}=d_{G_n(v)}\right|}{\left|V(G_n)\right|}=1,$$
 then
$$\lim_{n\to \infty}\frac{LEL(H_n)}{\left|V(G_n)\right|}=h.$$
That is, $G_n$ and $H_n$ have the same asymptotic
Laplacian-energy-like.
\end{theorem}

A direct sequence of Theorem 3.6 is that $P_n\square
P_n$,$P_n\square C_n$, and $C_n\square C_n$ have the same asymptotic
Laplacian-energy-like which is shown in the introduction. More
generally, by Theorem 3.6, we have

\vspace{5pt} \noindent {\bf Remark 3.7}\ Let $G_i=P_n$ or
$G_i=C_n$, $i=1,2,\dots,k,$ and $k$ be a constant. If $n$ is
sufficiently large, then the asymptotic Laplacian-energy-like of
the $n$-dimensional lattices
$$LEL(G_1\square G_2 \square  \dots\square G_k)
\approx \frac{n^k}{\pi^k}\int_0^{\pi}\int_0^{\pi}\dots
\int_0^{\pi}\sqrt{\sum_{i=1}^k\Big(2-2cosx_i\Big)} ~dx_1~dx_2\dots
dx_k.$$

\vspace{5pt} \noindent {\bf Remark 3.8}\   Theorem 3.6 provides a
very effective approach to handle the asymptotic
Laplacian-energy-like of a graph with bounded average degree.

 We will use the approach above to
deal with the asymptotic Laplacian-energy-like of some lattices in
the next subsections.

\section{Asymptotic Laplacian-energy-like of some lattices}

\subsection{The hexagonal lattice}

Our notation for the hexagonal lattices follows {\rm
\cite{YZ,SW}}. The hexagonal lattices with toroidal, cylindrical
and free boundary conditions, denoted by $H^t(m,n)$,  $H^c(m,n)$,
and $H^f(m,n)$ are illustrated in Figure 1, respectively, where
$(a_1, b_1), (a_2, b_2), \dots, (a_{m+1}, b_{m+1}), (a_1, c_1^*),
(c_1, c_2^*),(c_2, c_3^*), \dots, (c_{n-1}, c_n^*),\\ (c_n,
b_{m+1})$ are edges in $H^t(m,n)$, and $(a_1, b_1), (a_2, b_2),
\dots, (a_{m+1}, b_{m+1})$ are edges in $H^c(m,n)$(see Figure
1(b)). If we delete edges $(a_1, b_1), (a_2, b_2), \dots,
(a_{m+1}, b_{m+1})$ from $H^c(m,n)$, then the hexagonal lattice,
denoted by $H^f(m,n)$, with free boundary condition is obtained
(see Figure 1(c)).

\begin{figure}[ht]
\centering
  \includegraphics[width=\textwidth]{f1.eps}
\caption{(a) The hexagonal lattice $H^t(m,n)$ with toroidal
boundary condition; (b) the hexagonal lattice $H^c(m,n)$ with
cylindrical boundary condition; (c) the hexagonal lattice
$H^f(m,n)$ with free boundary condition.}
\end{figure}

A recent approach to compute the Laplacian eigenvalues of
$H^{t}(m,n)$ can be found in {\rm \cite{GM,DGMS,JS}. Using
Equation (6.2.2) in {\rm \cite{SW}, the Laplacian matrix
$L(H^{t}(m,n))$ of $H^{t}(m,n)$ is similar to the block diagonal
matrix whose diagonal blocks are
$$B_{ij}=\left(
\begin{array}{cc}
  3 & -1-\omega_{n+1}^{-i}- \omega_{m+1}^{j}\\
 -1-\omega_{n+1}^{i}- \omega_{m+1}^{-j} & 3  \\
 \end{array}
\right)$$
 where $\omega_s=cos\frac{2\pi}{s}+i
sin\frac{2\pi}{s},i=0,1,\dots,m;j=0,1,\dots,n.$ Hence the
Laplacian eigenvalues of $H^{t}(m,n)$ are
$$3\pm \sqrt{3+2cos\frac{2\pi i}{m+1}
+2cos\frac{2\pi j}{n+1}+2cos\Big(\frac{2\pi i}{m+1}+\frac{2\pi
j}{n+1}\Big)}, \  \ i=0,1,\dots,m;j=0,1,\dots,n.$$

By the definition of the Laplacian-energy-like, it is not difficult
to prove the following result:

\begin{theorem}\label{4-1}
For the hexagonal lattices $H^t(m,n), H^c(m,n)$ and $H^f(m,n)$
with toroidal, cylindrical, and free boundary conditions. Then
 $$1.~~\lim_{m\to\infty}
\lim_{n\to\infty}\frac{
 LEL\Big(H^t(m,n)\Big)}{2(m+1)(n+1)} =\lim_{m\to\infty}
\lim_{n\to\infty}\frac{ LEL\Big(H^c(m,n)\Big)}{2(m+1)(n+1)}
=\lim_{m\to\infty} \lim_{n\to\infty}\frac{
 LEL\Big(H^f(m,n)\Big)}{2(m+1)(n+1)}\approx  1.6437.$$\\
  $2.~LEL\Big(H^{t}(m,n)\Big)=LEL\Big(H^{c}(m,n)\Big)=LEL\Big(H^{f}(m,n)\Big)  \approx
3.2714(m+1)(n+1).$
\end{theorem}

{\bf Proof.} By definitions of $H^t(m,n), H^c(m,n)$ and
$H^f(m,n)$, it is obvious that $H^c(m,n)$ and $H^f(m,n)$ are
spanning subgraphs of $H^t(m,n)$. Furthermore, the degree of
almost all vertices of $H^t(m,n), H^c(m,n)$ and $H^f(m,n)$ are 3.
Hence, by Theorem 3.6, one can obtian that
$$\lim_{m\to \infty}\lim_{n\to \infty}\frac{
 LEL\Big(H^t(m,n)\Big)}{2(m+1)(n+1)} =\lim_{m\to \infty}\lim_{n\to
\infty}\frac{ LEL\Big(H^c(m,n)\Big)}{2(m+1)(n+1)} =\lim_{m\to
\infty}\lim_{n\to \infty}\frac{
 LEL\Big(H^f(m,n)\Big)}{2(m+1)(n+1)}.$$

It suffices to prove that
\begin{eqnarray}
&&\nonumber\lim_{m\to\infty} \lim_{n\to
\infty}\frac{ LEL\Big(H^{t}(m,n)\Big)}{2(m+1)(n+1)}\\
  \nonumber &=& \lim_{m\to\infty} \lim_{n\to \infty}
  \frac{1}{2(m+1)(n+1)}\sum_{i=0}^{m}\sum_{j=0}^{n}
\sqrt{ 3+ \sqrt{3+2cos\alpha_i +2cos\beta_j+2cos(\alpha_i+\beta_j)} }\\
  \nonumber & &+\lim_{m\to\infty} \lim_{n\to \infty}\frac{1}{2(m+1)(n+1)}
  \sum_{i=0}^{m}\sum_{j=0}^{n} \sqrt{ 3- \sqrt{3+2cos\alpha_i +2cos\beta_j+2cos(\alpha_i+\beta_j)} }\\
  \nonumber &=&\frac{1}{2}\int_{0}^{1}\int_{0}^{1}
\sqrt{ 3+ \sqrt{3+2cos 2\pi x+2cos 2\pi y+2cos 2\pi(x+y)} } \ ~dxdy \\
  \nonumber & &+\frac{1}{2}\int_{0}^{1}\int_{0}^{1}
\sqrt{ 3- \sqrt{3+2cos 2\pi x+2cos 2\pi y+2cos 2\pi(x+y)} } \ ~dxdy\\
 \nonumber &=&\frac{1}{8\pi^2}\int_{0}^{2\pi}\int_{0}^{2\pi}
\sqrt{ 3+ \sqrt{3+2cos x+2cos y+2cos(x+y)} } \ ~dxdy \\
  \nonumber & &+\frac{1}{8\pi^2}\int_{0}^{2\pi}\int_{0}^{2\pi}
\sqrt{ 3- \sqrt{3+2cos x+2cos y+2cos(x+y)} } \ ~dxdy\\
\nonumber &\approx & 1.6437 .
\end{eqnarray}

The above numerical integration value implies that the hexagonal
lattices $H^t(m,n), H^c(m,n)$ and $H^f(m,n)$ with toroidal,
cylindrical, and free boundary conditions have the same asymptotic
Laplacian-energy-like, i.e.,
$LEL\Big(H^{t}(m,n)\Big)=LEL\Big(H^{c}(m,n)\Big)=LEL\Big(H^{f}(m,n)\Big)
\approx 3.2714(m+1)(n+1)$ as $m,n$ tends to infinity. \hfill
$\blacksquare$

\subsection{The 3.12.12 lattice}
Our notation for the 3.12.12 lattices follows {\rm \cite{LY,ZZ}}.
The 3.12.12 lattice with toroidal boundary condition, denoted by
$J^{t}(m,n)$, is the graph illustrated in Figure 2(a).

\begin{figure}[h]

\subfigure[]{
\begin{minipage}[t]{0.32\linewidth}
\centering
 \includegraphics[width=\textwidth]{f2a.eps}
\end{minipage}}
\subfigure[]{
\begin{minipage}[t]{0.32\linewidth}
\centering
 \includegraphics[width=\textwidth]{f2b.eps}
\end{minipage}}
\subfigure[]{
\begin{minipage}[t]{0.32\linewidth}
\centering
 \includegraphics[width=0.8\textwidth]{f2c.eps}
\end{minipage}}
\caption{(a) The 3-12-12 lattice with toroidal boundary condition;
(b) The triangular kagom$\acute{e}$ lattice with toroidal boundary
condition; (c) the $3^3.4^2$ lattice with toroidal boundary
condition.}
\end{figure}

Recently, the adjacency spectrum of 3.12.12 lattice has been
proposed in {\rm \cite{LY}.

\begin{theorem}\label{2-2} {\rm \cite{LY}}
 Let $J^{t}(m,n)$ be the 3.12.12 lattice with toroidal boundary condition. Then the adjacency spectrum is
\begin{eqnarray}
\nonumber Spec_A(J^{t}(m,n))&=&\textbf{\{}\underbrace{-2,-2,\dots,-2}_{(m+1)(n+1)},\underbrace{0,0,\dots,0}_{(m+1)(n+1)}\textbf{\}}\\
\nonumber &&  \bigcup\left\{ \frac{  1\pm \sqrt
{13\pm4\sqrt{3+2cos\alpha_i+2cos\beta_j+2cos\Big(\alpha_i+\beta_j\Big)}}
    }{2}:0\le i\le m, 0\le j\le n  \right\},\\ \nonumber
    \end{eqnarray}
    where $\alpha_i=\frac{2\pi i}{m+1}, \beta_j=\frac{2\pi
j}{n+1},  i=0,1,\dots,m;j=0,1,\dots,n.$
\end{theorem}

The following result has been obtained in {\rm \cite{NLB}, which
is an important relationship between $Spec_A(G)$ and $Spec_L(G)$.
Suppose that $G$ is an $r$-regular graph with $n$ vertices and
$Spec_A(G)=\{\lambda_1, \lambda_2, \dots, \lambda_n\}.$ Then
$$ Spec_L(G)=\Big\{r-\lambda_1, r-\lambda_2, \dots, r-\lambda_n\Big\}.$$

Note that $J^{t}(m,n)$ is the line graph of the subdivision of
$H^t (m,n)$ which is a $3$-regular graph with $2(m + 1)(n + 1)$
vertices and $J^{t}(m,n)$ has $6(m + 1)(n + 1)$ vertices. Hence,
we arrive to that

\begin{theorem}\label{4-3}
 Let $J^{t}(m,n)$ be the 3.12.12 lattice with toroidal boundary condition. Then the Laplacian spectrum is
\begin{eqnarray}
\nonumber Spec_L(J^{t}(m,n))&=&\textbf{\{}\underbrace{5,5,\dots,5}_{(m+1)(n+1)},\underbrace{3,3,\dots,3}_{(m+1)(n+1)}\textbf{\}}\\
\nonumber &&  \bigcup\left\{\frac{  5\pm \sqrt
{13\pm4\sqrt{3+2cos\alpha_i
+2cos\beta_j+2cos\Big(\alpha_i+\beta_j\Big)}}
    }{2}:0\le i\le m, 0\le j\le n \right\},\\ \nonumber
    \end{eqnarray}
    where $\alpha_i=\frac{2\pi i}{m+1}, \beta_j=\frac{2\pi
j}{n+1},  i=0,1,\dots,m;j=0,1,\dots,n.$
\end{theorem}

By the definition of the Laplacian-energy-like $LEL(G)$, one can
easily arrive to the following theorem.
\begin{theorem}\label{4-4}
For $J^{t}(m,n), J^c(m,n)$ and $J^f(m,n)$ with toroidal,
cylindrical, and free boundary conditions. Then
$$1.~~ \lim_{m\to\infty} \lim_{n\to
\infty}\frac{LEL\Big(J^{t}(m,n)\Big)}{6(m+1)(n+1)}=
 \lim_{m\to\infty}
\lim_{n\to\infty}\frac{LEL\Big(J^{c}(m,n)\Big)}{6(m+1)(n+1)}=
\lim_{m\to \infty}
\lim_{n\to\infty}\frac{LEL\Big(J^{f}(m,n)\Big)}{6(m+1)(n+1)}\approx
 1.3375 .$$\\
 $2.~~LEL\Big(J^t
(m,n)\Big)= LEL\Big(J^c (m,n)\Big)=LEL\Big(J^f (m,n)\Big)\approx
8.0250(m+1)(n+1).$
\end{theorem}
{\bf Proof.} By definitions of $H^t(m,n), H^c(m,n)$ and
$H^f(m,n)$, one can know that $H^c(m,n)$ and $H^f(m,n)$ are
spanning subgraphs of $H^t(m,n)$. Furthermore, the degree of
almost all vertices of $H^t(m,n), H^c(m,n)$ and $H^f(m,n)$ are 3.
Therefore, by Theorem 3.6 it is not difficult to arrive to that
$$ \lim_{m\to\infty} \lim_{n\to
\infty}\frac{LEL\Big(J^{t}(m,n)\Big)}{6(m+1)(n+1)}=
 \lim_{m\to \infty}
\lim_{n\to\infty}\frac{LEL\Big(J^{c}(m,n)\Big)}{6(m+1)(n+1)}=
\lim_{m\to \infty}
\lim_{n\to\infty}\frac{LEL\Big(J^{f}(m,n)\Big)}{6(m+1)(n+1)}.$$

It suffices to prove that
\begin{eqnarray}
\nonumber && \lim_{m\to\infty} \lim_{n\to
\infty}\frac{LEL\Big(J^{t}(m,n)\Big)}{6(m+1)(n+1)}\\
\nonumber &=&\lim_{m\to\infty} \lim_{n\to
\infty}\frac{1}{12(m+1)(n+1)}\sum_{i=0}^{m}\sum_{j=0}^{n}
\sqrt{5- \sqrt {13-4\sqrt{3+2cos\alpha_i
+2cos\beta_j+2cos\Big(\alpha_i+\beta_j\Big)}
    }}\\  \nonumber\\
     \nonumber  &&+\lim_{m\to\infty} \lim_{n\to
\infty}\frac{1}{12(m+1)(n+1)}\sum_{i=0}^{m}\sum_{j=0}^{n}
\sqrt{5- \sqrt {13+4\sqrt{3+2cos\alpha_i
+2cos\beta_j+2cos\Big(\alpha_i+\beta_j\Big)}
    }}\\  \nonumber
      \nonumber  &&+\lim_{m\to\infty} \lim_{n\to
\infty}\frac{1}{12(m+1)(n+1)}\sum_{i=0}^{m}\sum_{j=0}^{n}
\sqrt{5+ \sqrt {13-4\sqrt{3+2cos\alpha_i
+2cos\beta_j+2cos\Big(\alpha_i+\beta_j\Big)}
    }}\\  \nonumber
      \nonumber  &&+\lim_{m\to\infty} \lim_{n\to
\infty}\frac{1}{12(m+1)(n+1)}\sum_{i=0}^{m}\sum_{j=0}^{n}
\sqrt{5+ \sqrt {13+4\sqrt{3+2cos\alpha_i
+2cos\beta_j+2cos\Big(\alpha_i+\beta_j\Big)}
    }}\\ \nonumber
    &&+\frac{\sqrt{3}+\sqrt{5}}{6}\\  \nonumber
    \nonumber &=&\frac{1}{12}\int_{0}^{1}\int_{0}^{1}\sqrt{5-
\sqrt {13-4\sqrt{3+2cos 2\pi x +2cos 2\pi y+2cos 2\pi \Big(x+y\Big)}
    }}~dxdy\\  \nonumber\\
     \nonumber  &&+\frac{1}{12}\int_{0}^{1}\int_{0}^{1}\sqrt{5-
\sqrt {13+4\sqrt{3+2cos 2\pi x +2cos 2\pi y+2cos 2\pi \Big(x+y\Big)}
    }}~dxdy\\  \nonumber
      \nonumber  &&+\frac{1}{12}\int_{0}^{1}\int_{0}^{1}\sqrt{5+
\sqrt {13-4\sqrt{3+2cos 2\pi x +2cos 2\pi y+2cos 2\pi \Big(x+y\Big)}
    }}~dxdy\\  \nonumber
      \nonumber  &&+\frac{1}{12}\int_{0}^{1}\int_{0}^{1}\sqrt{5+
\sqrt {13+4\sqrt{3+2cos 2\pi x +2cos 2\pi y+2cos 2\pi \Big(x+y\Big)}
    }}~dxdy+\frac{\sqrt{3}+\sqrt{5}}{6} \\  \nonumber
   \nonumber &\approx & 1.3375 .
\end{eqnarray}

The above numerical integration value implies that
$J^{t}(m,n),J^{c}(m,n)$ and $J^{f}(m,n)$ have the same asymptotic
Laplacian-energy-like, i.e., $LEL\Big(J^t (m,n)\Big)= LEL\Big(J^c
(m,n)\Big)=LEL\Big(J^f (m,n)\Big)\\\approx 8.0250(m+1)(n+1)$ as
$m,n$ tends to infinity. \hfill $\blacksquare$

\subsection{The triangular kagom$\acute{e}$ lattice}

The triangular kagom$\acute{e}$ lattice with toroidal boundary
condition, denoted by $TKL^t(m,n)$, is depicted in Figure 2(b).
Ising spins and XXZ/Ising spins on the $TKL^t(m,n)$ have been
studied in {\rm \cite{YLC,SLM}}. In order to obtain the
Laplacian-energy-like of the The triangular kagom$\acute{e}$
lattice, we recall the spectrum and the Laplacian spectrum of
$TKL^{t}(m,n)$.

\begin{theorem}\label{2-3}{\rm
\cite{LY}}
 Let $\alpha_i=\frac{2\pi i}{m+1}, \beta_j=\frac{2\pi
j}{n+1},  i=0,1,\dots,m;j=0,1,\dots,n.$ Then the spectrum and the
Laplacian spectrum of $TKL^{t}(m,n)$ are
\begin{eqnarray}
\nonumber
Spec_A(TKL^{t}(m,n))&=&\textbf{\{}\underbrace{-2,-2,\dots,-2}_{3(m+1)(n+1)},
\underbrace{-1,-1,\dots,-1}_{(m+1)(n+1)},\underbrace{1,1,\dots,1}_{(m+1)(n+1)}\textbf{\}}\\
\nonumber &&  \bigcup \left\{ \frac{ 3\pm \sqrt
{13\pm4\sqrt{3+2cos\alpha_i
+2cos\beta_j+2cos\Big(\alpha_i+\beta_j\Big)}}
    }{2}:0\le i\le m, 0\le j\le n  \right\},\\ \nonumber
    \end{eqnarray}
    and
\begin{eqnarray}
\nonumber
Spec_L(TKL^{t}(m,n))&=&\textbf{\{}\underbrace{6,6,\dots,6}_{3(m+1)(n+1)},
\underbrace{3,3,\dots,3}_{(m+1)(n+1)},\underbrace{5,5,\dots,5}_{(m+1)(n+1)}\textbf{\}}\\
\nonumber &&  \bigcup\left\{ \frac{ 5\pm \sqrt
{13\pm4\sqrt{3+2cos\alpha_i
+2cos\beta_j+2cos\Big(\alpha_i+\beta_j\Big)}}
    }{2}:0\le i\le m, 0\le j\le n  \right\}.\\ \nonumber
\end{eqnarray}
\end{theorem}

 Note that the triangular $kagom\acute{e}$ lattice is the line graph
of the $3.12.12$ lattice and $TKL^{t}(m,n)$ is a $4$-regular graph
with $9(m + 1)(n + 1)$ vertices.

Consequently,  we can easily get the following Theorem.

\begin{theorem}\label{3-5}
 For $TKL^{t}(m,n), TKL^{c}(m,n)$ and $TKL^f(m,n)$ with toroidal,
cylindrical, and free boundary conditions. Then
$$1.\lim_{m\to\infty} \lim_{n\to
\infty}\frac{LEL(TKL^{t}(m,n))}{9(m+1)(n+1)}= \lim_{m\to\infty}
\lim_{n\to \infty}\frac{LEL(TKL^{c}(m,n))}{9(m+1)(n+1)}=
\lim_{m\to\infty} \lim_{n\to
\infty}\frac{LEL(TKL^{f}(m,n))}{9(m+1)(n+1)}\approx 1.7082.$$\\
$2.~LEL\Big(TKL^{t}(m,n)\Big)=LEL\Big(TKL^{c}(m,n)\Big)=LEL\Big(TKL^{f}(m,n)\Big)\approx
15.3738(m+1)(n+1).$
\end{theorem}

{\bf Proof.} By definitions of $TKL^{t}(m,n), TKL^{c}(m,n)$ and
$TKL^f(m,n)$, one can know that $TKL^{c}(m,n)$ and $TKL^f(m,n)$
are spanning subgraphs of $TKL^{t}(m,n)$. Furthermore, the degree
of almost all vertices of $TKL^{t}(m,n), TKL^{c}(m,n)$ and
$TKL^f(m,n)$ are 4. Therefore, by Theorem 3.6 it is not difficult
to arrive to that
$$\lim_{m\to\infty} \lim_{n\to
\infty}\frac{LEL\Big(TKL^{t}(m,n)\Big)}{9(m+1)(n+1)}=
\lim_{m\to\infty} \lim_{n\to
\infty}\frac{LEL\Big(TKL^{c}(m,n)\Big)}{9(m+1)(n+1)}=
\lim_{m\to\infty} \lim_{n\to
\infty}\frac{LEL\Big(TKL^{f}(m,n)\Big)}{9(m+1)(n+1)}.$$
 It suffices to prove that
\begin{eqnarray}
\nonumber && \lim_{m\to\infty} \lim_{n\to
\infty}\frac{LEL\Big(TKL^{t}(m,n)\Big)}{9(m+1)(n+1)}\\
\nonumber &=&\lim_{m\to\infty} \lim_{n\to
\infty}\frac{1}{18(m+1)(n+1)}\sum_{i=0}^{m}\sum_{j=0}^{n} \sqrt{5-
\sqrt {13-4\sqrt{3+2cos\alpha_i
+2cos\beta_j+2cos\Big(\alpha_i+\beta_j\Big)}
    }}\\  \nonumber\\
     \nonumber  &&+\lim_{m\to\infty} \lim_{n\to
\infty}\frac{1}{18(m+1)(n+1)}\sum_{i=0}^{m}\sum_{j=0}^{n}
\sqrt{5- \sqrt {13+4\sqrt{3+2cos\alpha_i
+2cos\beta_j+2cos\Big(\alpha_i+\beta_j\Big)}
    }}\\  \nonumber
      \nonumber  &&+\lim_{m\to\infty} \lim_{n\to
\infty}\frac{1}{18(m+1)(n+1)}\sum_{i=0}^{m}\sum_{j=0}^{n}
\sqrt{5+ \sqrt {13-4\sqrt{3+2cos\alpha_i
+2cos\beta_j+2cos\Big(\alpha_i+\beta_j\Big)}
    }}\\  \nonumber
      \nonumber  &&+\lim_{m\to\infty} \lim_{n\to
\infty}\frac{1}{18(m+1)(n+1)}\sum_{i=0}^{m}\sum_{j=0}^{n}
\sqrt{5+ \sqrt {13+4\sqrt{3+2cos\alpha_i
+2cos\beta_j+2cos\Big(\alpha_i+\beta_j\Big)}
    }}\\  \nonumber
    &&+\frac{\sqrt{3}+\sqrt{5}}{9}+\frac{\sqrt{6}}{3}\\  \nonumber
    \nonumber &=&\frac{1}{18}\int_{0}^{1}\int_{0}^{1}\sqrt{5-
\sqrt {13-4\sqrt{3+2cos 2\pi x +2cos 2\pi y+2cos 2\pi \Big(x+y\Big)}
    }}~dxdy\\  \nonumber\\
     \nonumber  &&+\frac{1}{18}\int_{0}^{1}\int_{0}^{1}\sqrt{5-
\sqrt {13+4\sqrt{3+2cos 2\pi x +2cos 2\pi y+2cos 2\pi \Big(x+y\Big)}
    }}~dxdy\\  \nonumber
      \nonumber  &&+\frac{1}{18}\int_{0}^{1}\int_{0}^{1}\sqrt{5+
\sqrt {13-4\sqrt{3+2cos 2\pi x +2cos 2\pi y+2cos 2\pi \Big(x+y\Big)}
    }}~dxdy\\  \nonumber
      \nonumber  &&+\frac{1}{18}\int_{0}^{1}\int_{0}^{1}\sqrt{5+
\sqrt {13+4\sqrt{3+2cos 2\pi x +2cos 2\pi y+2cos 2\pi \Big(x+y\Big)}
    }}~dxdy+\frac{\sqrt{3}+\sqrt{5}}{9}+\frac{\sqrt{6}}{3} \\  \nonumber
   \nonumber &\approx &  1.7082.
\end{eqnarray}

The above numerical integration value implies that $TKL^{t}(m,n)$,
$TKL^{c}(m,n)$ and $TKL^{f}(m,n)$ have the same asymptotic
Laplacian-energy-like, i.e.,
$LEL\Big(TKL^{t}(m,n)\Big)=LEL\Big(TKL^{c}(m,n)\Big)\\=LEL\Big(TKL^{f}(m,n)\Big)\approx
15.3738(m+1)(n+1)$  as $m,n$ tends to infinity. \hfill
$\blacksquare$

\subsection{The $3^3.4^2$ lattice}

The $3^3.4^2$ lattice with toroidal boundary condition, denoted by
$M^{t}(n,2m)$, can be constructed by starting with a $2m\times n$
square lattice and adding a diagonal edge connecting the vertices,
i.e., the upper left to the lower right corners of each square in
every other row as shown in Figure 2(c), where $a_1 = b_1, a_{2m}
= b_1^*, a_1^* = b_n, a_{2m}^* = b_n^*,$ and $(a_1, a_1^* ), (a_2,
a_2^*), \dots, (a_{2m}, a_{2m}^*), (b_1, b_1^*), (b_2, b_2^*),
\dots, (b_n, b_n^*),\\ (a_1, a_2^*), (a_3, a_4^* ),$ $ \dots,
(a_{2m-2}, a_{2m}^*)$ are edges in $M^t(n,2m)$.

Let $A(C_{2m})$ be the adjacency matrix of cycle $C_{2m}$, using
the result in {\rm \cite{YZ},the adjacency matrix $A(M^{t}(n,2m))$
of $M^{t}(n,2m)$ has the following form by a suitable labelling of
vertices of $M^{t}(n,2m)$:$$
A\Big(M^{t}(n,2m)\Big)=\left(
\begin{array}{cccccc}
  A(C_{2m}) & I_{2m}+F_{2m} & 0 &               \dots & 0 & I_{2m}+F_{2m}^T \\
  I_{2m}+F_{2m}^T & A(C_{2m}) & I_{2m}+F_{2m} & \dots & 0 & 0 \\
  0 & I_{2m}+F_{2m}^T & A(C_{2m}) &             \dots & 0 & 0 \\
  \vdots & \vdots & \ddots & \ddots & \ddots & \vdots \\
  0 & 0 & 0 & \dots & A(C_{2m}) & I_{2m}+F_{2m} \\
  I_{2m}+F_{2m} & 0 & 0 & \dots & I_{2m}+F_{2m}^T & A(C_{2m}) \\
\end{array}
\right)_{n\times n}$$

Notice that $M^{t}(n,2m)$ is an $r$-regular graph. Let
$L(M^{t}(n,2m))$ be the Laplacian matrix of $M^{t}(n,2m)$, it is
not difficult to obtain that $L(M^{t}(n,2m))$ is similar to the
block diagonal matrix whose diagonal blocks are
$$L_{ij}=\left(
\begin{array}{cc}
  5-\omega_n^i -\omega_n^{-i}  &  -1-\omega_n^{-i} -\omega_m^{j} \\
   -1-\omega_n^i -\omega_m^{-j} &  5-\omega_n^i -\omega_n^{-i} \\
\end{array}
\right)$$
 where $\omega_s=cos\frac{2\pi}{s}+i
sin\frac{2\pi}{s},i=0,1,\dots,m-1;j=0,1,\dots,n-1.$

Hence the Laplacian eigenvalues of $M^{t}(n,2m)$ are:
$$5-2cos\frac{2\pi i}{n} \pm\sqrt{3+2cos\frac{2\pi i}{m}
+2cos\frac{2\pi j}{n}+2cos\Big(\frac{2\pi i}{m}+\frac{2\pi
j}{n}\Big)}, \  \ i=0,1,\dots,m-1;j=0,1,\dots,n-1.$$

Similarly, it is not hard to derive the following theorem.
\begin{theorem}\label{4-7}
 For the $M^{t}(n,2m), M^{c}(n,2m)$ and $M^f(n,2m)$ with toroidal,
cylindrical, and free boundary conditions. Then
$$1.~\lim_{m\to\infty} \lim_{n\to
\infty}\frac{LEL\Big(M^t(n,2m)\Big)}{2mn}= \lim_{m\to\infty}
\lim_{n\to \infty}\frac{LEL\Big(M^{c}(n,2m)\Big)}{2mn}=
 \lim_{m\to\infty}
\lim_{n\to \infty}\frac{LEL\Big(M^{f}(n,2m)\Big)}{2mn}\approx
2.1525 .$$\\
$ 2.~LEL \Big(M^t(n,2m)\Big)=LEL \Big(M^c(n,2m)\Big)=LEL
\Big(M^f(n,2m)\Big)\approx 4.3050mn.$
\end{theorem}

{\bf Proof.} By the definition of the Laplacian-energy-like, we
can easily get that
\begin{eqnarray}
\nonumber  LEL\Big(M^t(n,2m)\Big)&=&
\sum_{i=0}^{m-1}\sum_{j=0}^{n-1}
\sqrt{5-2cos\frac{2\pi i}{m} -\sqrt{3+2cos\frac{2\pi i}{m} +2cos\frac{2\pi j}{n}+2cos(\frac{2\pi i}{m}+\frac{2\pi j}{n})}}\\
\nonumber     &&+  \sum_{i=0}^{m-1}\sum_{j=0}^{n-1}
\sqrt{5-2cos\frac{2\pi i}{m} +\sqrt{3+2cos\frac{2\pi i}{m}
+2cos\frac{2\pi j}{n}+2cos(\frac{2\pi i}{m}+\frac{2\pi j}{n})} }.\\
\nonumber
\end{eqnarray}

As an analogue to preceding proof, based on Theorem 3.6, one can
get that
$$\lim_{m\to\infty} \lim_{n\to
\infty}\frac{LEL\Big(M^t(n,2m)\Big)}{2mn}= \lim_{m\to\infty}
\lim_{n\to \infty}\frac{LEL\Big(M^{c}(n,2m)\Big)}{2mn}=
 \lim_{m\to\infty}
\lim_{n\to \infty}\frac{LEL\Big(M^{f}(n,2m)\Big)}{2mn}.$$

It suffices to prove that
\begin{eqnarray}
\nonumber &&\lim_{m\to\infty} \lim_{n\to
\infty}\frac{LEL\Big(M^t(n,2m)\Big)}{2mn}\\ \nonumber &=&
\lim_{m\to\infty} \lim_{n\to
\infty}\frac{1}{2mn}\sum_{i=0}^{m-1}\sum_{j=0}^{n-1} \sqrt{
5-2cos\frac{2\pi i}{n} -\sqrt{3+2cos\frac{2\pi i}{m}
+2cos\frac{2\pi j}{n}+2cos\Big(\frac{2\pi i}{m}+\frac{2\pi
j}{n}\Big)} } \\
 \nonumber & &+\lim_{m\to\infty}
\lim_{n\to
\infty}\frac{1}{2mn}\sum_{i=0}^{m-1}\sum_{j=0}^{n-1}\sqrt{
5-2cos\frac{2\pi i}{n} +\sqrt{3+2cos\frac{2\pi i}{m} +2cos\frac{2\pi
j}{n}+2cos\Big(\frac{2\pi i}{m}+\frac{2\pi
j}{n}\Big)} }\\
   \nonumber &=&\frac{1}{2}\int_{0}^{1}\int_{0}^{1}
\sqrt{ 5-2cos2\pi x -\sqrt{3+2cos 2\pi x
+2cos2\pi y+2cos 2\pi(x+y) }} \ ~dxdy \\
\nonumber & &+\frac{1}{2}\int_{0}^{1}\int_{0}^{1} \sqrt{ 5-2cos2\pi
x +\sqrt{3+2cos 2\pi x
+2cos2\pi y+2cos 2\pi(x+y) }} \ ~dxdy \\
  \nonumber &\approx & 2.1525 .
\end{eqnarray}
The above numerical integration value implies that $M^t(n,2m)$,
$M^{c}(n,2m)$ and $M^{f}(n,2m)$ have the same asymptotic
Laplacian-energy-like, i.e., $LEL \Big(M^t(n,2m)\Big) =LEL
\Big(M^c(n,2m)\Big)=LEL \Big(M^f(n,2m)\Big)\\ \approx 4.3050mn$
 as
$m,n$ tends to infinity. Summing up, we complete the proof. \hfill
$\blacksquare$

\vspace{5pt} \noindent {\bf Remark 4.8}\ In the Figure 2 each kind
of graphs of three kinds of boundary conditions of graphs have the
same the asymptotic Laplacian-energy as $m,n$ tends to infinity.

\section{Concluding remarks}
The calculations of some topological indexes in terms of various
lattices have attracted the attention of many physicists as well
as mathematicians. In this paper, we have deduced the explicit
formulae expressing the Laplacian-energy-like of some lattices
with toroidal, cylindrical, and free boundary conditions, the
explicit asymptotic values of Laplacian-energy-like in these
lattices are obtained via the applications of analysis approach
with the help of calculational software.

 Let $\{G_n\}$ be a
sequence of finite simple graphs with bounded average degree, it
is difficult to calculate its asymptotic Laplacian-energy-like
directly, however, we can find a sequence of graphs $H_n$ with
bounded average degree, which satisfies $\left|V(G_n)\right|$ and
$\left|V(H_n)\right|$ and almost all vertices of $G_n$ and $H_n$
have the same degrees. If we can formulate the asymptotic
Laplacian-energy-like of $H_n$ immediately, then by Theorem 3.6,
$G_n$ and $H_n$ have the same asymptotic Laplacian-energy-like.
Therefore, Theorem 3.6 provides a very effective approach to
handle the asymptotic Laplacian-energy-like of a graph with
bounded average degree. For instance, dealing with the problem of
the asymptotic Laplacian-energy-like of the hexagonal lattice with
the free boundary is not an easy work but we deduced it in a
simple approach.

We can convert some harder problems to easy ones and
simultaneously obtain many results  by utilizing the approach.
Moreover, we showed that the Laplacian-energy-like per vertex of
the many types of lattices is independent of the three boundary
conditions. It is no difficulty to see that the conclusion is true
in general. Actually, the approach can be used widely to formulate
the other topological indexes of various lattices.

 \vspace{5pt} \noindent
{\bf Acknowledgments}\

The work of J. B. Liu is partly supported by the Natural Science
Foundation of Anhui Province of China under Grant No.KJ2013B105;
The work of X. F. Pan is partly supported by the National Science
Foundation of China under Grant Nos.10901001, 11171097, and
11371028.

config: arxiv


TITLE: Resolution of Gorenstein rational singularities on a surface
QUESTION [6 upvotes]: I am reading Artin's notes "Lipman's Proof of Resolution of Singularities for Surfaces" from the book "Arithmetic Geometry". I am very confused by the proof of Lemma $6.5.$ (I am formulating it below in a little bit different way than it appears in the text)

Lemma 6.5: Let $(A,\mathfrak m, k)$ be a normal complete excellent ring of dimension $2$ that defines a rational double point (rational Gorenstein singularity). Denote by $X$ the blow-up of the unique closed point in Spec $A$. Assume that the exceptional divisor $E$ is equal to $2C$, where $C$ is a line in $\mathbb P^2_k$. And let $X' \to X$ be a sequence of blow-ups in closed points $p_1, \dots, p_n$, s.t. $X'$ is regular at evert point of the strict transform $C'$, then $\Sigma_{i} [k(p_i):k]=3$.

The key step is to compute $\deg_C \mathcal O_X(-C)|_C$ (Note that $\mathcal O_X(-C)$ isn't locally free since $C$ isn't a Cartier divisor, but it is always reflexive, in particular torsion-free. Hence, $\mathcal O_C(-C)$ is always an invertible sheaf). Artin claims that it is equal to $-1$, but I don't understand his argument.

In our case, since $2C$ is isomorphic to a double line in $\mathbb P^2$, the degree is the same as for such a line, i.e., $[-C,C]=-1$.

How could one put this into a rigorous argument? It is not clear how to relate $\deg_C \mathcal O_C(-C)$ with this immersion since $\mathcal O_C(-C)^{\otimes 2} \neq \mathcal O_C(-E)$.
P.S. By a rational singularity I mean that for any normal modification $f:X \to Spec A$ we have $H^1(X,\mathcal O_X)=0$. If $A$ is also Gorenstein, it is called rational double point. The latter condition is equivalent to $\dim_k \mathfrak m/\mathfrak m^2 \leq 3$.
P.S.2. In the formulation of Lemma $E$ should be equal to a double line $2C$ with respect to the natural immersion $X \to \mathbb P^2_A$ defined by the sheaf $\mathcal O_X(-E)$.
UPD: Jason Starr mentioned in the comments that if $A$ is defined over a field $k$, then $A\cong k[[x,y,z]]/(F(x,y,z)-G(x,y,z))$, where $F$ is homogeneous quadratic polynomial and $G$ is of degree at least $3$. We can do almost the same without assuming that $A$ is defined over a field. Namely, since $A$ is a rational double point $\dim_k \mathfrak m^n/\mathfrak m^{n+1}=2n+1$. Then we have $3$ generators for $\mathfrak m$ and there is precisely one relation in degree $2$ between them in $gr_{\mathfrak m} A$. Let this relation be $F(x,y,z)=G(x,y,z)$, where $F,G\in k[T_1,T_2,T_3]$ are polynomials of degree $2$ and $3$ respectively ($F$ is also homogeneous). Since $E\cong Proj(gr_{\mathfrak m} A)$ we conclude that $E\cong V(F) \subset \mathbb P^2_k$. Taking into account that $E=2C$ we can actually choose (after a suitable linear change of coordinates) $F(T_1,T_2,T_3)=T_1^2$.
But I still don't understand what is the connection between $\deg \mathcal O_C(-C)$ and the intersection number $[-C,C]$ inside $\mathbb P^2_k$.

REPLY [4 votes]: The explanation given in the text indeed seems to be too terse.  What follows is taken from an insert I have in my copy of that book (since the argument I came up with when I read the article many years ago did not fit in the margin as other clarifications did). I hope it is helpful, and that I haven't made some blunder.
We'll use the notation in Artin's article (which I won't explain here). The idea is to give a description of $O_C(-C)$ that is intrinsic to the infinitesimal thickening $C \hookrightarrow Z$, without direct reference to the ambient $X_1$. That will allow us to make a switch to transfer the degree calculation to a more convenient choice of $X_1$!
By definition the $O_C$-module $O_C(-C)$ on $C \simeq \mathbf{P}^1_k$ is
$$(O_{X_1}(-C) \otimes_{O_{X_1}} O_C)/(O_C\mbox{-}{\rm{torsion}}) = (O_{X_1}(-C) \otimes_{O_{X_1}} (O_{X_1}/O_{X_1}(-C)))/(O_C\mbox{-}{\rm{torsion}})$$
that is torsion-free with generic rank 1 by design, so it is invertible.  Moreover, we claim that the canonical surjection of $O_{X_1}$-modules
$$q:O_{X_1}(-C) \twoheadrightarrow O_C(-C)$$
has kernel exactly $O_{X_1}(-2C)$. The quotient $O_{X_1}(-C)/O_{X_1}(-2C)$ is easily checked to be torsion-free as an $O_C$-module and has generic rank 1 as such, so it is invertible as such. Thus, since a surjection between invertible sheaves is an isomorphism, to identify $\ker q$ as claimed it suffices to show that the $O_{X_1}$-linear restriction $q: O_{X_1}(-2C) \to O_C(-C)$ vanishes.  But the target
of this restriction is a torsion-free $O_C$-module, so it suffices to check the vanishing near the generic point of $C$, hence over the regular locus $X_1^{\rm{reg}}$, where it is clear.  The upshot is that we have
$$O_C(-C) \simeq O_{X_1}(-C)/O_{X_1}(-2C)$$
as $O_C$-modules.
Let $I$ denote the coherent ideal sheaf of $C$ in $Z$, so this is a square-zero ideal by definition of $Z$ (as the ideal sheaf $O_{X_1}(-C)$ inside $O_{X_1}$ has square contained in $O_{X_1}(-2C)$) and hence $I$ is naturally an $O_C$-module.  As such, we have
$$I := {\rm{image}}(O_{X_1}(-C) \to O_Z = O_{X_1}/O_{X_1}(-2C)) =
O_{X_1}(-C)/O_{X_1}(-2C) \simeq O_C(-C),$$
the final isomorphism being what we established above.
Thus, we have an intrinsic description of the $O_C$-module $O_C(-C)$ as the coherent ideal sheaf of $C$ inside the square-zero thickening $Z$ of $C$. The conclusion is that $O_C(-C)$ as an $O_C$-module depends
only in the data of the infinitesimal closed immersion $C \hookrightarrow Z$ and not on the ambient $X_1$.
But we can identify $Z$ scheme-theoretically as a doubled-line in $\mathbf{P}^2_k$ with $C = Z_{\rm{red}}$ a straight line in this projective plane. Hence, to compute the degree of the $O_C$-module $O_C(-C)$ it suffices to do the calculation in an ambient $\mathbf{P}^2_k$. Now we can run the calculations in reverse, with $C$ inside $\mathbf{P}^2_k$ having ideal sheaf $O_{\mathbf{P}^2}(-1)$:
$$O_C(-C) \simeq O_{\mathbf{P}^2}(-C)/O_{\mathbf{P}^2}(-2C) \simeq
O_{\mathbf{P}^2}(-1)/O_{\mathbf{P}^2}(-2).$$
Since $C = \mathbf{P}^1_k$ has structure sheaf $O_C$ with Euler characteristic equal to 1, so ${\rm{deg}}_C(L) = \chi(L) - 1$ for any line bundle $L$ on $C$, to show ${\rm{deg}}_C(O_C(-C)) = -1$ is the same as showing $\chi(O_C(-C))=0$. From the above displayed expression for $O_C(-C)$ in terms of $O_{\mathbf{P}^2}(-r)$'s, it is the same to show
that $\chi(O_{\mathbf{P}^2}(-1))=\chi(O_{\mathbf{P}^2}(-2))$.  Both of
these latter Euler characteristics are equal to 0.