Math Overflow Recent Questions


Determinant as a Hamiltonian

Sat, 07/14/2018 - 13:40

Are there two symplectic structures $\omega_1, \omega_2$ on $M_{2n}(\mathbb{R})$ such that the function $Det:M_{2n}(\mathbb{R})\to \mathbb{R}$ is completely integrable with respect to $\omega_{1}$ but is not completely integrable with respect to $\omega_2$

Note that we do not limit the symplectic structures to structures with constant coefficients.

I had already asked a weaker version of this question in the following link but I confess that I did not understand the details(How does the action of orthonormal group guarantee's the integrability?)

(In)stability of a two-dimensional dynamical system

Sat, 07/14/2018 - 13:04

Consider the following system of coupled differential equations $$ \dot{x}_1(t) = -x_1(t) - \cos(\omega t)x_1(t) + \cos(\omega t)x_2(t), \ x_1(0)\in\mathbb{R},\\ \dot{x}_2(t) = -\gamma x_2(t) - \cos(\omega t)x_2(t) + \cos(\omega t)x_1(t), \ x_2(0)\in\mathbb{R}, $$ where $\omega$ and $\gamma$ are positive real constants. Observe that $\bar{x}=(\bar{x}_1,\bar{x}_2)=(0,0)$ is an equilibrium of the above system.

It is almost trivial to see that if $\gamma=1$ then $\bar x$ is attractive. Indeed, in this case, we have that $x(t)=[x_1(t), x_2(t)]^\top$ can be explicitly computed as $$ x(t) = \exp\left(\begin{bmatrix}-t &0\\ 0 & -t\end{bmatrix} + \frac{1}{\omega}\sin(\omega t)\begin{bmatrix}-1 &1\\ 1 & -1\end{bmatrix}\right)x(0), $$ so that $x(t)\to 0$ for $t\to \infty$.

However in case $\gamma\ne 1$ proving the attractiveness of the origin is not obvious (and perhaps not even true!).

In particular, numerical simulations seem to suggest that for $\gamma$ and $\omega$ sufficiently small (e.g. $\gamma=0.001$ and $\omega=10$) the equilibrium $\bar{x}$ is not attractive.

I've struggled a lot to find a way of formally proving this, with no luck. So I decided to post the problem here hoping that some of you will provide some useful suggestions or tips. Thank you!

I post here the Mathematica code that I've used in my simulations:

(* nominal values for simulation *) values = {gamma -> 0.001, w -> 10}; equations = { {x1'[t], x2'[t]} == {-x1[t] - Cos[w*t]*x1[t] + Cos[w*t]*x2[t], -gamma*x2[t] - Cos[w*t]*x2[t] + Cos[w*t]*x1[t]}, {x1[0], x2[0]} == {0.1, 0.1}}; {x1t, x2t} = NDSolveValue[equations /. values, {x1[t], x2[t]}, {t, 0, 1000}]; Plot[x1t, {t, 0, 1000}, PlotRange -> {-0.2, 0.2}] Plot[x2t, {t, 0, 1000}, PlotRange -> {-0.2, 0.2}]

Existence of non-null-homotopic map from $M^n$ to $S^{n-1}$

Sat, 07/14/2018 - 09:27

Let $M^n$ be compact, connected, oriented $n$-dimensional smooth manifold without boundary, the Hopf degree theorem states that the homotopy class of continuous maps from $M^n$ to $S^n$ is classified by its degree. What about continuous map from $M^n$ to $S^{n-1}$ (assume $n>0$) ? I am mostly interested in the case $n>2$.

More specially, when does there exist a non null-homotopic map in the case $n=3$? If $H^2(M,\Bbb Z) \not=0$ then it's true as $[M,\Bbb{C}P^{\infty}]=H^2(M,\Bbb Z)$ and $\Bbb CP^1 \cong S^2$. This is not necessary as $M=S^3$ shows.

What is the minimal dimension of a complex realising a group representation?

Sat, 07/14/2018 - 02:42

This question is inspired by this one, which was about representations that can be realised homologically by an action on a graph (i.e., a 1-dimensional complex).

Many interesting integral representations of groups arise from a group acting on a simplicial complex that is homotopy equivalent to a wedge of spheres, by applying homology. A classical example is the action of groups of Lie type on spherical buildings. On homology this gives an integral form of the Steinberg representation.

One may ask if there exists a complex of lower dimension than the Tits building that realises the (integral) Steinberg representation in this way. I expect that the answer is No, but how to prove it?

More generally, given an integral $G$-representation that can be realised as the homology of a spherical complex with an action of $G$, is there an effective lower bound on the dimension of such a complex? One obvious lower bound is given by the minimal length of a resolution by permutation representations. Is this something that has been studied?

When every open cover admits a $\sigma$-disjoint subcover?

Sat, 07/14/2018 - 02:41

We say that a sequence $(\mathcal X_n)$ of families of subsets of a topological space $X$ is a $\sigma$-disjoint cover of $X$ if every family $\mathcal X_n$ consists of mutually disjoint sets and $\bigcup\limits_n\bigcup\mathcal X_n=X$.

Let us say that a space $X$ is weakly Lindelof, if every open cover of $X$ admits a $\sigma$-disjoint subcover. Clearly, every Lindelof space is weakly Lindelof.

Question 1. Is there any well-known in the literature name for the class of "weakly Lindelof" spaces?

The following question concerns weaker property than "weak Lindeloffness".

Question 2. Does there exist a $\sigma$-disjoint cover of a Banach space $X$ by open balls of diameters $\le 1$?

"Flat links", a reference request

Sat, 07/14/2018 - 01:21

A hyperbolic link is one whose complement admits a hyperbolic metric. Hyperbolic links, and especially hyperbolic knots, are quite popular these days. However, I am currently interested in links whose complement admits a flat (i.e. locally euclidean) metric. If I got it right, the major difference from the hyperbolic case is the existence (at least, I want for it to exist) of a natural compactification. This compactification makes the 3-sphere into an Alexandrov space with conical singularities along the components of the link, or something like this. Is there such a construction somewhere in the literature?

[EDIT] I think I need to clarify this a bit. What I want is a metric on ${\mathbb S}^3$ such that

1) The metric is flat outside of the link.

2) Each component of the link has a neighborhood isometric to a product of a conical point by ${\mathbb S}^1$.

(So, if it is a ``negative'' conical point, then it won't be an Alexandrov space.)

Moments of area of random triangle inscribed in a circle

Fri, 07/13/2018 - 14:47

The $2m$th moment of the (random) area of the triangle whose vertices are three independent, uniformly distributed random points on the unit circle appears to be $((3m)!/(m!)^3)/16^m$. Can anyone prove this? Better yet, can anyone give a conceptual explanation for why this moment should be rational? (If this observation is not new, references would be appreciated.)

This question was inspired by John Baez's posts and

A geometric proof of Krull's Principal ideal theorem

Fri, 07/13/2018 - 12:34

Krull's height theorem states that in a Noetherian, local ring $(A,\mathfrak m)$, for any $f \in \mathfrak m$, the minimal prime ideal containing $(f)$ is at most height $1$.

This is a very geometric statement and is essentially saying that hypersurfaces can cut down the dimension by at most one. However, I have never seen a geometrically motivated proof. All the proofs I have seen essentially muck around with symbolic powers of prime ideals and this seems very ad-hoc/unmotivated to me.

Surely a geometrical statement should have a geometric proof!

Does someone have a geometric way of seeing why this theorem should be true or what is going on? Or what is going on geometrically with the standard proof and symbolic powers?

Bezout theorem for germs of holomorphic functions

Fri, 07/13/2018 - 11:05


It was pointed out by @Dmitri that two smooth curves given by $f=y$ and $g=y+x^k$ in $\mathbb C^2$ provide a simple counterexample.

Let $f_1, \ldots, f_p, g_1, \ldots, g_q$ be germs of holomorphic functions at $0$ in $\mathbb C^{p+q}$. Assume that the multiplicity ${\rm mult}(f_1, \ldots, f_p)$ is finite let $$ V:= \{ f_1 = \ldots = f_p =0\} $$ be (germ of) the zero variety of $f_1, \ldots, f_p$, and denote by ${\rm mult}_V(g_1, \ldots, g_q)$ the maximum number for points in the fiber of the restriction $g|_V$, where $g:=(g_1, \ldots, g_q)$.

Is the following true?

$$ {\rm mult}(f_1, \ldots, f_p, g_1, \ldots, g_q) \le {\rm mult}(f_1, \ldots, f_p) \cdot {\rm mult}_V(g_1, \ldots, g_q) $$

For example, if $f=y$ and $g=y+x^k$, we have ${\rm mult} (f) = {\rm mult} (g) = 1$ but ${\rm mult}(f, g) = k$. On the other hand, for $V=\{f=0\}$, we have ${\rm mult}_V (g) = k$, so the above relation is correct.

OLD QUESTION (the answer is no).

Is the following local holomorphic variant of Bezout theorem true:

Let $f_1, \ldots, f_p, g_1, \ldots, g_q$ be germs of holomorphic functions at $0$ in $\mathbb C^{p+q}$. Then $$ {\rm mult}(f_1, \ldots, f_p, g_1, \ldots, g_q) = {\rm mult}(f_1, \ldots, f_p) \cdot {\rm mult}(g_1, \ldots, g_q), \quad (*) $$ provided the first multiplicity is finite.

Here for $m$ holomorphic germs $(h_1, \dots, h_m)$ in $\mathbb C^n$, $m\le n$, the multiplicity ${\rm mult}(h_1, \ldots, h_m)$ is defined as the minimum dimension of the quotient $$ \cal O/\langle h_1, \ldots, h_m, l_1, \ldots, l_{n-m} \rangle $$ of the ring $\cal O = \cal O_{\mathbb C^n,0}$ of germs at $0$ of holomorphic functions in $\mathbb C^n$ modulo the ideal generated by the germs $h_1, \ldots, h_m, l_1, \ldots, l_{n-m}$, with $l_j$ being linear functions, and the minimum is taken over all choices of the $l_j$.

Equivalently, the multiplicity can be defined as the minimum number of points (over all choices of the linear functions $l_j$) in the preimage of a generic point in $\mathbb C^n$ near $0$ under the map $$ z\mapsto (h_1(z), \ldots, h_m(z), l_1(z), \ldots, l_{n-m}(z)), $$ where the same notation is used for the germ representative functions. (Instead of taking generic point, one can take the maximal number of points in a preimage.)

Orthogonal Grassmanians: cases where $\text{OG}( \mathbb{P}^1 , Q) \not \simeq \mathbb{P}^3$

Fri, 07/13/2018 - 08:26

Let $Q = \{ q(x_0, \dots, x_4) = 0 \}$ be a quadric-threefold over a field $k$. Are there cases where the orthogonal Grassmanian $\text{OG}( \mathbb{P}^1 , Q)$ is not a copy of $\mathbb{P}^3$?

Here's a Lemma (Kollar) that says it is a $\mathbb{P}^3$ iff it contains a line:

Lemma 61

(1) $\text{OG}(\mathbb{P}^1,Q)$ is a Severi-Brauer variety

(3) $\text{OG}(\mathbb{P}^1,Q) \simeq \mathbb{P}^3 \longleftrightarrow \text{OG}(\mathbb{P}^1,Q) \text{ contains a line }\longleftrightarrow q \sim y_0y_1 + y_2y_3 + a y_4^2$

but isn't this variety the Grassmanian of lines?

Here I believe $\simeq$ means "isomorphism of varieties" (instead of birational equivalence $\stackrel{bir}{\sim}$) and the symbol $\sim$ is "rational equivalence of quadratic forms".

I am still parsing the definition based on the comments I am wondering:

  • Are there other 3-dimensional Severi-Brauer varieties over $\mathbb{Q}$ other than $\{ 0\}$ and $\mathbb{P}^3$ ?
    I believe the answer is no because quadric sections themselves are varieties of this type.

However Wikipedia says this:

In mathematics, a Severi–Brauer variety over a field K is an algebraic variety V which becomes isomorphic to a projective space over an algebraic closure of K.

In dimension one, the Severi–Brauer varieties are conics. The corresponding central simple algebras are the quaternion algebras. The algebra $(a,b)_K$ corresponds to the conic C(a,b) with equation $$ C(a,b) = \big\{ z^{2} -ax^{2}-by^{2} = 0 \big\} $$ and the algebra $(a,b)_K$ splits, that is, $(a,b)_K$ is isomorphic to a matrix algebra over $K$, if and only if $C(a,b)$ has a point defined over $K$: this is in turn equivalent to $C(a,b)$ being isomorphic to the projective line over $K$.

So we get that quaternions play a role in SB varieties and that conics (and one could imagine quadrics) are Severi-Brauer varieties. Now I am confused again, that if the conic has a single point over $K$, then it is a projective line $\mathbb{P}^1(K)$.

Does this correspond to the trivial element of the Brauer group? There are even expositions of Severi-Brauer varieties that are more geometric and avoid the use of Galois cohomology.

In Neukirch's Cohomology of Number Fields, they give a defintion of the Brauer Group:

The Brauer group of a field $K$ is the set of similarity classes $[A]$ of central $K$-simple algebras $A$ endowed with the multiplication $$[A][B] = [A \otimes_K B] $$ Two central simple $K$-algebras $A$ and $B$ are similar if $$ A \otimes_K M_r(K) \simeq B \otimes_K M_s(K) $$ for some $r,s$. (Chapter 6.3)

The group $\mathbb{Q}/\mathbb{Z}$ keeps being mentions in regards to various exact sequences and the Hasse principle. It seems the challenge is to stay grounded and not to float away with these definitions, for what is a very concrete object.

Density on Hölder spaces whose elements vanish on the boundary

Fri, 07/13/2018 - 08:13

I would like to ask the following problem.

Let $\Omega$ be a $C^{r+1,\alpha}$ domain, $r\in \mathbb{N}, 0<\alpha<1.$ We denote $$C^{r,\alpha}_{0}(\overline{\Omega})=\{f\in C^{r,\alpha}(\overline{\Omega}): f=0 \mbox{ on }\partial \Omega\},$$ here $C^{r,\alpha}(\overline{\Omega})$ is Holder spaces. Is $C^{r+1,\alpha}_0(\overline{\Omega})$ dense in $C^{r,\alpha}_0(\overline{\Omega})?$

We see that when $r\geq 2$, the answer is positive. For any $u\in C^{r,\alpha}_{0}(\overline{\Omega}),$ we have $$\Delta u := f\in C^{r-2,\alpha}_{0}(\overline{\Omega}).$$ Then there exists a sequence $f_n\in C^{r-1,\alpha}_{0}(\overline{\Omega})$ such that $f_n\rightarrow f$ in $C^{r-2,\alpha}_{0}(\overline{\Omega}).$ With each $f_n,$ there exists unique $u_n\in C^{r+1,\alpha}_{0}(\overline{\Omega})$ such that

$$\left\{\begin{array}{ll}\Delta u_n=f_n &\mbox{ in }\Omega\\ u_n =0 &\mbox{ on }\partial \Omega. \end{array}\right.$$ Therefore, by eliptic regularity, we obtain that $||u_n-u||_{C^{2,\alpha}}\leq C||f_n-f||_{C^{r-2,\alpha}}.$ It implies the conclution. It seems that we can not apply the above method for the case $r=1.$

Examples of "miraculous" proofs

Fri, 07/13/2018 - 07:24

Concerning the proof that $\zeta(3)$ is irrational, Van der Poorten famously noted that

"Apéry's incredible proof appears to be a mixture of miracles and mysteries".

Indeed, many ideas introduced in Apéry's proof such as $\zeta(3)=\frac{5}{2} \sum_{k=1}^{\infty}{\frac{ (-1)^{k-1}} {\binom {2k}{k}k^{3}}}$ and the recurrence $n^3u_n + (n-1)^3 u_{n-2} = (34n^3-51n^2+27n-5)u_{n-1}, n\geq2$ amazed contemporary mathematicians (although the fast-converging series was already derived several years earlier by Hjortnaes).

What are other examples of "miraculous" proofs, whose ingredients amazed mathematicians of their time? In particular, proofs such that:

  • a large portion of the Theorems involved in the proof represent entirely new ideas,
  • these Theorems are applicable to a wide area of mathematics,
  • the general atmosphere among contemporary mathematicians was a mixture of surprise and awe ("where did this come from?").

Integral structures via lattices

Thu, 07/12/2018 - 18:52

I am looking at the paper "p-adic Groups" by Bruhat (in the Boulder Proceedings, 1965). I have a question about one of the statements. Let $k$ be the quotient field of a complete discrete valuation ring $\mathcal{O}$. Let $V$ be a vector space over $k$. Let $L$ be a lattice in $V$. Choose a basis for $L$.

Bruhat states the following in pp. 63-64:

(1) The algebra $\mathcal{O}[GL]:=\mathcal{O}[g_{ij},(det(g_{ij}))^{-1}]$ is an $\mathcal{O}$-structure for $GL(V)$.

(2) More generally, if $G$ is a linear algebraic group over $k$, then given a faithful rational representation $\rho:G \rightarrow GL(V)$, the image of $\mathcal{O}[GL(V)]$ in $k[G]$ is an $\mathcal{O}$-structure for $G$.

(3) Any $\mathcal{O}$-structure for $G$ may be obtained in this way.

How is (3) proven?

intersection of sets

Thu, 07/12/2018 - 03:48

Let $A_n=\{d:d|n\}\cup \{kn|D(n)\leq k \leq n-1\}$ where $D(n)$ is the number of divisors of $n$.

Is it correct that
$$|A_i \cap A_j |=(i,j),$$ for each $i,j \in N$?

A commuting pair of functions with no common bijective part

Thu, 07/12/2018 - 03:44

Let $X$ be set. Let $f$ be a function from $X$ into $X$. For a given set $E\subseteq X$, we say $E$ determines a $U$-part of $f$ if $f(E)\subseteq E$ and the restriction $f:E\to E$ is a bijection.

I am looking for a commuting pair of injective functions $f:X\to X$ and $g:X\to X$ (I mean $fg=gf$) such that both $f$ and $g$ have $U$-parts but not common $U$-part.

Is there a function from a Suslin tree to itself which send compatible elements to incompatible elements?

Thu, 07/12/2018 - 03:43

Given a Suslin tree $S$, I would like to know wether it is possible or impossible to define a function $f:S\to S$ such that for every $x,y\in S$ with $x<y$ we have $f(x)\perp f(y)$ and a proof or reference to a proof.

Notice $S$ is not required to have a single root.

Would the same result holds for a $S$ non-special Aronszajn trees which is not necessarily Suslin?

Matrix and Sparse Matrix Addition

Thu, 07/12/2018 - 03:31

I am a student of computer science. I have problem of some basic mathematics computation. I want to clarify it. I have a project which based on addition of a normal matrix and a Sparse matrix. I just want to know that if this addition problem. My normal matrix have a dimension 4*4 and my sparse matrix have only two non zero element in cell(0,0) and (1,1). How can I performed addition? Is that necessary to mention the dimension of sparse matrix?

It is very helpful to me if someone suggest some tutorial link of sparse matrix consist of sparse matrix operations.

Looking for positive reply.

Thank you in advance

Small modules over finite group with large cohomology

Thu, 07/12/2018 - 03:15

Looking at this Example of group cohomology not annihilated by exponent of $G$? I stumbled upon one question I couldn't solve (probably because it's hard), so I post it here.

Using Lyndon resolvent, we can always find $G$-module $M$ of rank equal to number of relators in some presentation such that $H^2(G, M)$ has maximal exponent (and is equal to $\Bbb Z/|G|$). Because one-relator finite group is cyclic, I asked myself

  1. Is it true that if there's ideal $R$ with $H^2(G, \Bbb Z[G]/R)$ of exponent $|G|$, then $G$ cyclic?

  2. If there's $2$-generated $G$-module $M$ with $H^2$ of exponent $|G|$, what can we say about $G$?

Of course, we can look not only on second cohomology, and obtain series of filtrations on category of finite groups: $G \in BC_{n, k}$ (BC for "big cohomology") if exists a $k$-generated module s. t. $exp(H^n(G, M) = |G|$. (or maybe $H^{\leq n}$, or $H^{\geq n}$,..). Looks interesting.

Example of a Non-double commuting pair of isometric operators with no common shift part

Thu, 07/12/2018 - 02:54

Before to state the problem, let us mention the Wold decomposition theorem.

Theorem. Every isometric operator is a direct sum of a unitary and shift.

Let $(T_1,T_2)$ be a commuting pair (I mean $T_1T_2=T_2T_1$) of isometric operators on the Hilbert space $H$.

I am looking for such a pair satisfying the followings properties:

1) Neither $T_1$ nor $T_1$ is unitary.

2) $T_1T_2^*$ is not the same as $T_2^*T_1$.

3) If $P$ is a projection commuting with both $T_1$ and $T_2$ such that the restrictions $T_n:PH\to PH$ are both shifts, then $P=0$.

Remark. By a shift, I mean a direct sum of unilateral shifts.

Argmax of weighted sum of binomials

Thu, 07/12/2018 - 02:32

(Writing my thesis, I encountered the following problem. It is secondary to the topic of the thesis and I have the solution that is enough for the purposes of the thesis—but inner perfectionist is still not happy.)

Fix two positive integers $n$ and $2 \leq \ell \leq n$ and define a discrete function $F : \{ 1, 2, \dotsc, n-\ell+1\} \rightarrow \mathbb N$ in the following way: $$ F(w) = F_{n,\ell}(w) = w \sum_{i=0}^{\ell-1} \binom{n-w}{i} \,. $$

What is $\underset{x}{\operatorname{argmax}} F(w)$? I can prove (see below) that $$ \underset{x}{\operatorname{argmax}} F(w) \in \left\{ \left\lfloor \frac{n+1}{\ell} \right\rfloor, \left\lceil \frac{n}{\ell} \right\rceil \right\}. $$ And though this result is efficient and easy to calculate, I somehow still want to see in some sense closed formula. Can you help me?

Proof. First of all, it is easy to see that $$ \left\lfloor \frac{n+1}{\ell} \right\rfloor = \left\lceil \frac{n}{\ell} \right\rceil \quad\text{or}\quad \left\lfloor \frac{n+1}{\ell} \right\rfloor + 1 = \left\lceil \frac{n}{\ell} \right\rceil. $$ Then, to prove the statement, it is sufficient to show that $F(w)$ increases for $w < \left\lfloor \frac{n+1}{\ell} \right\rfloor$ and decreases for $w \geq \left\lceil \frac{n}{\ell} \right\rceil$.

Consider a finite difference: $$ \Delta F(w) = F(w+1) - F(w) \,. $$

It can be expanded as follows: \begin{split} \Delta F(w) &= F(w+1) - F(w) \\ &= (w+1)\sum_{i=0}^{\ell-1}\binom{n-w-1}{i} - w\sum_{i=0}^{\ell-1}\binom{n-w}{i} \\ &= (w+1)\sum_{i=0}^{\ell-1}\binom{n-w-1}{i} - w\sum_{i=0}^{\ell-1}\left(\binom{n-w-1}{i} + \binom{n-w-1}{i-1}\right) \\ &= \sum_{i=0}^{\ell-1}\binom{n-w-1}{i} - w\sum_{i=0}^{\ell-1}\binom{n-w-1}{i-1} \,. \end{split}

We have: \begin{split} \Delta F(w) &= \sum_{i=0}^{\ell-1}\left( \binom{n-w-1}{i} - w\binom{n-w-1}{i-1} \right) \\ &= \sum_{i=0}^{\ell-1} \frac{(n-w-1)!}{i!(n-w-i)!}\left( n-i-w(i+1) \right) \,. \end{split} If we require that $$ w \leq \frac{n-\ell+1}{\ell} = \frac{n+1}{\ell} - 1 \,, $$ then it follows also that $$ w < \frac{n-i}{i+1} \quad \text{ for all } i < \ell-1 \,; $$ hence, each of the terms $(n-i-w(i+1))$ is positive for $i < \ell-1$ and $(n-\ell+1 - w\ell) \geq 0$. Therefore, $$ F(1) < F(2) < \dotsb < F\left(\left\lfloor \frac{n+1}{\ell} - 1 \right\rfloor\right) < F\left(\left\lfloor \frac{n+1}{\ell} \right\rfloor\right) \,. $$

On the other hand, we can write: \begin{split} \Delta F(w) &= \sum_{i=0}^{\ell-1}\binom{n-w-1}{i} - w\sum_{i=0}^{\ell-1}\binom{n-w-1}{i-1} \\ &= \sum_{i=0}^{\ell-1}\binom{n-w-1}{i} - w\sum_{i=0}^{\ell-2}\binom{n-w-1}{i} \\ &= \binom{n-w-1}{\ell-1} + (1-w)\sum_{i=0}^{\ell-2}\binom{n-w-1}{i} \,. \end{split} And, if $w > 1$, we have: \begin{split} \Delta F(w) &< \binom{n-w-1}{\ell-1} + (1-w)\binom{n-w-1}{\ell-2} \\ &= \frac{(n-w-1)!}{(\ell-1)!(n-\ell-w+1)!} (n-w\ell) \,. \end{split}

If we further require $w \ge \frac{n}{\ell}$, then $\Delta F(w) < 0$ and $$ F\left( \left\lceil \frac{n}{\ell} \right\rceil \right) > F\left( \left\lceil \frac{n}{\ell} \right\rceil +1 \right) > \dots > F(n-\ell+1) \,. $$