TITLE: Why do some Lagrangians have both a rotational and a non-rotational part of their kinetic energy?
QUESTION [1 upvotes]: I was going through a derivation and I came across a term in the kinetic energy of the Lagrangian that had both a velocity term and a rotational term. So I looked it up and I found these two links
https://hepweb.ucsd.edu/ph110b/110b_notes/node19.html
https://hepweb.ucsd.edu/ph110b/110b_notes/node20.html
In the first link, they say

For a rigid body, we will find in the equations that the motion can be separated into the motion of the center of mass and the rotation around the center of mass.
[...]
For the purposes of calculation, we will assume that the body is made up of a set of discrete masses, labeled by an index $\alpha$. In any inertial frame, the velocity of one of those masses is
$$\vec{v}_\alpha = \vec{V} + \vec{\omega} \times \vec{r}_\alpha$$
In this equation, the $\vec{v}_\alpha$ is given in the inertial frame, $\vec{V}$ is the velocity of the center of mass of the object, and $\vec{r}_\alpha$ is the position of the mass in the body frame in which it is at rest.

In the second link, they then show the following

From Lagrangian dynamics, we know that we can extract the physics if we know the kinetic and potential energy, $T$ and $U$. For now we will not be applying any potential so we only have the kinetic energy.
\begin{align}
T&=\sum\limits_\alpha T_\alpha = {1\over 2}\sum_\alpha m_\alpha v_\alpha^2 = {1\over 2}\sum_\alpha m_\alpha \left(\vec{V} + \vec{\omega}\times\vec{r}_\alpha\right)^2\\
T &= {1\over 2}\sum\limits_\alpha m_\alpha \left(V^2 + 2\vec{V}\cdot(\vec{\omega}\times\vec{r}_\alpha)+(\vec{\omega}\times\vec{r}_\alpha)^2\right)\\
T &= V^2{1\over 2}\sum\limits_\alpha m_\alpha + \vec{V}\cdot\left(\vec{\omega}\times\sum_\alpha m_\alpha \vec{r}_\alpha\right)+{1\over 2}\sum\limits_\alpha m_\alpha(\vec{\omega}\times\vec{r}_\alpha)^2
\end{align}
[Note that]
$\sum_\alpha m_\alpha \vec{r}_\alpha = 0$, [so then]
\begin{align}
T &= {1\over 2}M V^2 + {1\over 2}\sum\limits_\alpha m_\alpha(\vec{\omega}\times\vec{r}_\alpha)^2\\
T& = T_{CM} + T_{rot}\\
\end{align}

I'm confused why the kinetic energy has to be broken up this way. For example, consider the example with a pendulum.

Intuitively looking at it, I would say there is a torque from gravity on the pendulum causing it to move. So the force should be accounted for in this "rotational" part of the kinetic energy. However, I understand the standard derivation of a pendulum to be as follows:
The only force acting on the pendulum is gravity. So the potential energy is
$$V = mgr_y = -mg\ell\cos(\theta)$$
The kinetic energy is
$$
\begin{align}
T &= \frac{1}{2} m \lVert \mathbf{v}\rVert^2\\
&= \frac{1}{2} m (v_x^2 +v_y^2)\\
\end{align}
$$
Note,
$$
\begin{align}
v_x &= \frac{\partial r_x}{\partial \theta} \frac{\partial \theta}{\partial t}\\
&=  \frac{\partial}{\partial \theta}(\ell\sin(\theta)) \dot{\theta}\\
&=  \ell\cos(\theta) \dot{\theta}\\
v_x^2&=  \ell^2\cos^2(\theta) \dot{\theta}^2
\end{align}
$$
and
$$
\begin{align}
v_y &= \frac{\partial r_y}{\partial \theta} \frac{\partial \theta}{\partial t}\\
&=  \frac{\partial}{\partial \theta}(-\ell\cos(\theta)) \dot{\theta}\\
&=  \ell\sin(\theta) \dot{\theta}\\
v_y^2&=  \ell^2\sin^2(\theta) \dot{\theta}^2
\end{align}
$$
Thus,
\begin{align}
v_x^2+v_y^2&=  \ell^2\cos^2(\theta) \dot{\theta}^2+ \ell^2\sin^2(\theta) \dot{\theta}^2\\
&= \ell^2 \dot{\theta}^2 (\cos^2(\theta)+\sin^2(\theta))\\
&= \ell^2 \dot{\theta}^2\\
\end{align}
So then, the kinetic energy is
$$
T =\frac{1}{2} m (v_x^2 +v_y^2) = \frac{1}{2} m \ell^2 \dot{\theta}^2
$$
So then the Lagrangian is given by
$$\begin{align}
L & = T - V\\
&= \frac{1}{2} m \ell^2 \dot{\theta}^2 - (-mg\ell\cos(\theta)) \\
&= \frac{1}{2} m \ell^2 \dot{\theta}^2 + mg\ell\cos(\theta) \\
\end{align}$$
which yields
$$\ell\ddot{\theta}+g\sin(\theta)=0$$
Notice in my derivation however, the kinetic energy term for the pendulum was stated using $(v_x^2 +v_y^2)$ which seems analogous to the $T_{CM}$ term from earlier and not $T_{rot}$.
My Question
Why is velocity broken up into $\vec{v}_\alpha = \vec{V} + \vec{\omega} \times \vec{r}_\alpha$? Why doesn't the derivation of a pendulum use $\vec{\omega} \times \vec{r}_\alpha$ given that the mass rotates around a point?

REPLY [0 votes]: It is convenient to decompose the motion of a particle on a rotating frame (rigid body) to the translational velocity of the center of mass, plus a rotation about the center of mass. The reason the center of mass is important is because momentum of a body relates to the motion of the center of mass, and hence forces (time derivative of momentum) relate to the motion of the center of mass also.
Additionally, the kinetic energy of a moving rigid body is an invariant (constant) scalar value that does not depend on the location it is resolved at.
You can arrive at the same kinetic energy $T$ value, regardless of how motion is decomposed, and the choice of the center of mass is purely for convenience.
Consider a body pinned at point A

Kinetic energy is calculated at various points, depending on the definition of the mass moment of inertia at those points.
$$ \require{cancel} \begin{aligned}
 T & =\vec{\omega} \cdot I_A \vec{\omega} + m \cancel{(\vec{v}_A \cdot \vec{v}_A)} \\
  & = \vec{\omega} \cdot I_B \vec{\omega} + m( \vec{v}_B \cdot \vec{v}_B)\\
  & = \vec{\omega} \cdot I_C \vec{\omega} + m( \vec{v}_C \cdot \vec{v}_C)\\
  & = \vec{\omega} \cdot I_G \vec{\omega} + m( \vec{v}_G \cdot \vec{v}_G) \\
\end{aligned}$$
All of the above return the same exact value.


TITLE: Pushout exists in category of pointed topological spaces
QUESTION [2 upvotes]: It is not mentioned anywhere but do push out always exists in pointed spaces?  (I will omit the points below)  We have something like
$\require{AMScd}$
\begin{CD}
    X @>f>> Y\\
    @V g V V @VV  V\\
    Z @>>> Y\vee Z/\sim
\end{CD}
I claim the push out is $Y \vee Z / \sim$.
where $Y\vee Z$ is the pointed space, $Y \sqcup Z / *_Y \cup *_Z $.
$\sim$ is the relation generated by $i_Yf(x)\sim i_Zg(x)$ for $x \in X$. $i_X,i_Z$ being inclusions.
Proof: A cocone of the push out diagarm, with maps $g':Z \rightarrow A, f':Y \rightarrow A$ induces a map $Y \vee Z \rightarrow A$. But the conditions that it is a cocone induces a map $Y \vee Z /\sim \rightarrow A$.

REPLY [0 votes]: Most categories of "all" mathematical objects of a given type have all (small) limits and colimits. Examples are categories of algebraic obects, in which limits are constructed as in sets, or a more general underlying category, and colimits are constructed out of the free algebras and categories of topological objects, as you have here. Unpointed topological spaces have all limits and colimits constructed as in sets, with an appropriately maximal or minimal topology. Pointed topological spaces then inherit all limits and colimits: limits and connected colimits, such as pushouts, are constructed as in spaces, while for non-connected colimits we also have to identify the basepoints, as in the wedge product. This can be seen as a special case of the story about algebraic objects mentioned above: a pointed object is an algebra for the very boring theory that says "there's a chosen element and no operations."


TITLE: Find the Principal Value of the integral $\int_{-\infty}^\infty \frac{x sin(x)}{x^2+2x+2}dx$
QUESTION [6 upvotes]: This problem comes from a preliminary exam from 2009
"Find the Principal Value of the integral
$$\int_{-\infty}^\infty \frac{x \sin(x)}{x^2+2x+2}dx"$$
My attempt at solution: Letting $f(z)=\frac{z \sin(z)}{z^2+2z+2}$,  the residue theorem implies that $ \int_{-\infty}^\infty f(x)dx=2\pi i \text{Res}(f, -1+i)$, but I get that Res$(f, -1+i)=(\frac{1}{2}+\frac{i}{2})\sin(-1+i)$, and I don't know how to compute $\sin(-1+i)$.
Also, I was hoping someone could help me to understand why the problem asks for the "Principal Value" rather than the value because the only singularities of the function $f(z)=\frac{z \sin(z)}{z^2+2z+2}$ are at $-1\pm i$, neither of which are on the real axis.

REPLY [3 votes]: Note that the integral of interest can be written
$$\int_{-\infty}^{\infty}\frac{x\sin x}{x^2+2x+1}dx=\text{Im}\left(\int_{-\infty}^{\infty}\frac{xe^{ix}}{x^2+2x+1}dx\right) \tag 1$$
Let's examine the integral
$$\oint_C\frac{ze^{iz}}{z^2+2z+1}dz$$
where $C$ is the closed contour comprised of
$(i)$ the real-line segment from $(-R,0)$ to $(R,0)$
$(ii)$ the semicircle $C_R$ in the upper-half plane, centered at the origin with radius $R$.
Thus, we can write
$$\begin{align}
\oint_C\frac{ze^{iz}}{z^2+2z+1}dz&=\int_{-R}^{R}\frac{xe^{ix}}{x^2+2x+1}dx+
\int_{C_R}\frac{ze^{iz}}{z^2+2z+1}dz \tag 2\\\\
\end{align}$$
Note that in the limit as $R\to \infty$, the imaginary part of first integral on the right-hand side of $(2)$ equals the Cauchy Principal Value of the integral of interest in $(1)$.  The second integral on the right-hand side of $(2)$ can be shown using Jordan's Lemma to go to zero in the limit as $R\to \infty$.  Thus, from the residue theorem we have
$$\int_{-\infty}^{\infty}\frac{x\sin x}{x^2+2x+1}dx=\text{Im}\left(2\pi i \,\text{Res}\left(\frac{ze^{iz}}{z^2+2z+2},z=-1+i\right)\right)\tag 3$$
The residue in $(3)$ can be evaluated as
$$\begin{align}
\text{Res}\left(\frac{ze^{iz}}{z^2+2z+2},z=-1+i\right)&=
\lim_{z\to -1+i}\frac{(z+1-i)ze^{iz}}{(z+1-i)(z+1+i)}\\\\
&= \frac{(-1+i)e^{i(-1+i)}}{2i} \\\\
&=\frac{e^{-1}}{2i}\sqrt{2}e^{i(3\pi/4-1)}\tag 4
\end{align}$$
Substituting $(4)$ into $(3)$ reveals that
$$\bbox[5px,border:2px solid #C0A000]{\int_{-\infty}^{\infty}\frac{x\sin x}{x^2+2x+1}dx=\pi e^{-1}(\cos 1 -\sin 1)}$$


TITLE: An unfair die problem. Finding the probability that $4$ of the outcomes are either one or two, $4$ are either three or four, etc.
QUESTION [0 upvotes]: This is Exercise 6.1 from Ross' A first course in Probability.
Suppose that an unfair die lands on each of the odd numbers $1,3,5$ with probability $1/9$ and on each of the even numbers with probability $2/9$.
Suppose now that $12$ independent tosses of the die are made. Find the probability that $4$ of the outcomes are either one or two, $4$ are either three or four, and $4$ are either five or six.
The answer states $\frac{(12)!}{(4!)^3} (1/3)^{12}$.
I can see that we should have $(1/3)^{12}$ as the probability of each instance, but I am not sure why the count of all these outcomes should be $\frac{(12)!}{(4!)}^3$. If we simply group either one or two, either three or four, and either five or six as $A,B,C$ events, then $\frac{(12)!}{(4!)}^3$ gives the total number of distinct sequences. However, each $A$ can be $1,1,1,1$ or $1,1,1,2$ and so on. So how are we able to disregard the possible combinations within each of these groups in counting the total number of outcomes?

REPLY [0 votes]: $\left.{12!}\middle/{4!^3}\right.$ counts the arrangements of 'places' for each group of results.   We multiply this by the probability that each place has that result.
$\left. 1\middle/{3^4}\right.$ is the probability that each of the four 'places' we 'select' for 1 or 2 do each have one from those results.   We don't care which of those results we have, or what order they come among themselves, just that we have four occurances of that group.
And likewise the others, so that $\left.{12!}\middle/({4!^4}\,{3^{12}})\right.$ is the probability for obtaining some arrangement of four occurances from each three result groups.


TITLE: How to solve these recurrence relations by using generating function
QUESTION [6 upvotes]: First of all, I want to make an apology for my English for I'm not an English native speaker.
I'm reading Discrete Mathematics and its Applications recent days, and I am stumped by these three problems.

$a_k=a_{k-1}+2a_{k-2}+2^k, a_0=4, a_1=12$
$a_k=4a_{k-1}-4a_{k-2}+k^2, a_0=2, a_1=5$
$a_k=2a_{k-1}-3a_{k-2}+4^k+6, a_0=20, a_1=60$

Anyone has any idea? Your replies will be highly appreciated.

Problem solved. Thanks to you all. And here are my solutions.

Problem 1.
$$\begin{cases}a_k=a_{k-1}+2a_{k-2}+2^k\\ a_0=4\\ a_1=12\end{cases}$$
Let $f(x)$ denote the generating function for the sequence $a_k$, then we get $f(x)=\sum\limits_{k\ge0}a_kx^k$
Take the first equation, then multiply each term by $x^k$
$$a_kx^k=a_{k-1}x^k+2a_{k-2}+2^kx^k$$
And sum each term from 2 since it's a 2-order recurrence relation.
$$\begin{align*}\sum\limits_{k\ge2}a_kx^k&=\sum\limits_{k\ge2}a_{k-1}x^k + \sum\limits_{k\ge2}2a_{k-2}x^k+\sum\limits_{k\ge2}2^kx^k\\&=\sum\limits_{k\ge2}a_{k-1}x^k + 2\sum\limits_{k\ge2}a_{k-2}x^k+\sum\limits_{k\ge2}2^kx^k\end{align*}$$

The idea is to manipulate each term so that you can write them as expressions in terms of the generating function $f(x)$ and known series representations.

$\begin{align*}\sum\limits_{k\ge2}a_kx^k&=\left(\sum\limits_{k\ge2}a_kx^k+a_1x+a_0\right)-a_1x-a_0\\&=\sum\limits_{k\ge0}a_kx^k-a_1x-a_0\\&=f(x)-4x-12\end{align*}$

$\begin{align*}\sum\limits_{k\ge2}a_{k-1}x^k&=x\sum\limits_{k\ge2}a_{k-1}x^{k-1}\\&=x\sum\limits_{k\ge1}a_kx^k\\&=x\left(\sum\limits_{k\ge0}a_kx^k-a_0\right)\\&=x(f(x)-4)\end{align*}$

$\begin{align*}\sum\limits_{k\ge2}a_{k-2}x^k&=x^2\sum\limits_{k\ge2}a_{k-2}x^{k-2}\\&=x\sum\limits_{k\ge0}a_kx^k\\&=x^2f(x)\end{align*}$

$\begin{align*}\sum\limits_{k\ge2}2^kx^k&=\sum\limits_{k\ge0}2^kx^k-2x-1\\&=\frac{1}{1-2x}-2x-1\end{align*}$

Right now, we have
$$f(x)-4x-12=x(f(x)-4)+2x^2f(x)+\frac{1}{1-2x}-2x-1$$
Rewrite the equation to solve $f(x)$, and we have
$$f(x)=\frac{2}{3}\frac{1}{(1-2x)^2}+\frac{38}{9}\frac{1}{1-2x}-\frac{8}{9}\frac{1}{1+x}$$
Finally,
$$\begin{align*}a_k&=\frac{2}{3}C(2+k-1, 2-1)2^k+\frac{38}{9}·2^k-\frac{8}{9}(-1)^k\\&=\frac{2}{3}(1+k)2^k+\frac{38}{9}·2^k-\frac{8}{9}(-1)^k\\&=\frac{44}{9}·2^k+\frac{2}{3}k·\frac{k}{3}-\frac{8}{9}(-1)^k\end{align*}$$
Problem 2.
Most steps are the same as they were in problem 1, so I'll only write the steps of conjugating $\sum\limits_{k\ge2}k^2x^k$
$$\begin{align*}\sum\limits_{k\ge2}k^2x^k&=\sum\limits_{k\ge0}k^2x^k-x\\&=x\sum\limits_{k\ge0}k^2x^{k-1}-x\\&=x\sum\limits_{k\ge0}k·kx^{k-1}-x\\&=x\left(\sum\limits_{k\ge0}k·x^k\right)'-x\\&=x\left(x\sum\limits_{k\ge0}k·x^{k-1}\right)'-x\\&=x\left(x\left(\sum\limits_{k\ge0}x^k\right)'\right)'-x\\&=x\left(x\left(\frac{1}{1-x}\right)'\right)'-x\\&=x\left(\frac{x}{(1-x)^2}\right)'-x\\&=\frac{x(1+x)}{(1-x)^3}-x\end{align*}$$

My notes, using generating functions to solve recurrence relations

REPLY [3 votes]: Here's a (partial) solution for the second recurrence relation. There's no particular reason for me having chosen this particular recurrence among the other two; this is just used as an example to give you an idea of how to solve the others. I intentionally left out most of the details to let this serve more as a reference for you to check your solution at different checkpoints rather than as an answer key.
$$\begin{cases}a_k=4a_{k-1}-4a_{k-2}+k^2\\ a_0=2\\ a_1=5\end{cases}$$
Let $f(x)$ denote the generating function for the sequence $a_k$; that is, $f(x)=\sum\limits_{k\ge0}a_kx^k$. Taking the first equation, multiply each term by $x^k$ and sum each term over all positive $k\ge2$. (Why?)
$$\color{red}{\sum_{k\ge2}a_kx^k}=4\color{blue}{\sum_{k\ge2}a_{k-1}x^k}-4\color{green}{\sum_{k\ge2}a_{k-2}x^k}+\color{orange}{\sum_{k\ge2}k^2x^k}$$
The idea is to manipulate each term so that you can write them as expressions in terms of the generating function $f(x)$ and known series representations. For instance, consider the LHS:
$$\begin{align*}\sum_{k\ge2}a_kx^k&=\sum_{k\ge2}a_kx^k+a_1x+a_0-a_1x-a_0\\&=\sum_{k\ge0}a_kx^k-a_1x-a_0\\&=f(x)-5x-2\end{align*}$$
Doing the same throughout, you end up with a rewritten equation that can be solved for $f(x)$.
$$\begin{align*}\color{red}{f(x)-5x-2}&=4\color{blue}{x(f(x)-2)}-4\color{green}{x^2f(x)}+\color{orange}{\frac{x(1+x)}{(1-x)^3}-x}\\[1ex]
f(x)&=\frac{2}{(1-x)^3}+\frac{5}{(1-x)^2}+\frac{13}{1-x}+\frac{6}{(1-2x)^2}-\frac{24}{1-2x}
\end{align*}$$
From here, you need to determine an appropriate series representation for $f(x)$. All the information you need happens to be contained in the geometric series,
$$\sum_{k\ge0}x^k=\frac{1}{1-x}\quad\text{for }|x|<1$$
and its derivatives.
You should arrive at the final solution
$$a_k=20-9(2^{1+k})+8k+3k(2^{1+k})+k^2$$


TITLE: Integral on interval $[-\infty,W_t]$, $W_t$ is Brownian motion
QUESTION [1 upvotes]: Basicaly I have an expectation of an integral on the interval which contains Brownian motion and it look like this.
$$
E\left[e^{W_t}\cdot\int_{-\infty}^{W_t} e^{-z^2}dz\right]
$$
$W_t$ is Brownian motion, $t$ is fixed. My question is: Can I write this is any other form or, can calculate this?
Any help would be appreciated. Thanks!

REPLY [1 votes]: Hint::
For the function
$$
f(x):=\int\limits^{x}_{-\infty}e^{-r^2}\mathrm d r\,,
$$
using Leibniz's rule and Ito's lemma tells us
$$
\mathrm df(W_t) = e^{-W_t^2}\mathrm dW_t-W_t~e^{-W_t^2}\mathrm dt\,.
$$
Therefore,

$$
\int\limits^{W_t}_{-\infty}e^{-r^2}\mathrm d r = \dfrac{\sqrt{\pi}}{2}{}{}+{}\int\limits_{0}^{t}e^{-W_u^2}\mathrm dW_u-\int\limits_{0}^{t}W_u~e^{-W_u^2}\mathrm du\,.
$$

This puts the argument of the expectation in a form that is "more appealing". Can you proceed from here?
Edit:
This approach leads to a successive evaluation of expectations by successive use of Ito's lemma. The result is a series representation of the desired expectation. So, for instance,
expansions like
$$
\mathbb{E}\left[e^{W_t}\int\limits_{0}^{t}e^{-W_u^2}\mathrm dW_u\right] = \mathbb{E}\left[\int\limits_{0}^{t}e^{-W_u^2}\mathrm dW_u\right] + \mathbb{E}\left[\int\limits_{0}^{t}e^{W_u-W_u^2}\mathrm du\right] + \frac{1}{2}\mathbb{E}\left[\int\limits_{0}^{t}e^{W_u}\mathrm du\int\limits_{0}^{t}e^{-W^2_u}\mathrm dW_u\right]
$$
follow from observing
$$
e^{W_t}=1 + \int\limits_{0}^{t}e^{W_u}\mathrm dW_u + \frac{1}{2}\int\limits_{0}^{t}e^{W_u}\mathrm du\,.
$$
This is akin to situations in ordinary calculus where integrals are solved by obtaining a Taylor expansion for the solution. However, this is more involved than it needs to be. Certainly, a more direct approach would be to use the properties of the standard normal CDF, $\Phi$, and $W_t\sim\mathcal{N}\left(0,t\right)$ to justify the following:
$$
\begin{eqnarray*}
\mathbb{E}\left[e^{W_t}\int\limits^{W_t}_{-\infty}e^{-u^2}\mathrm du\right] = \mathbb{E}\left[\sqrt{\pi}e^{W_t}\Phi(\sqrt{2}W_t)\right] = \sqrt{\pi}\frac{1}{\sqrt{2\pi t}}\int\limits_{-\infty}^{\infty}\Phi(\sqrt{2}x)e^{-\frac{1}{2}\frac{x^2}{t}+x}\mathrm dx = \sqrt{\pi}\frac{e^{\frac{1}{2}t}}{\sqrt{2\pi(2t)}}\int\limits^{\infty}_{-\infty}\Phi(y)e^{-\frac{1}{4t}\left(y-\sqrt{2}t\right)^2}\mathrm dy = \sqrt{\pi}e^{\frac{1}{2}t}\Phi\left(\frac{\sqrt{2}t}{\sqrt{1+2t}}\right),
\end{eqnarray*}
$$
resulting in a final form for the solution originally suggested by @Did.


text:
\begin{document}

\title{Vanishing Flats: A Combinatorial Viewpoint on the Planarity of Functions and Their Application}
\author{Shuxing Li, Wilfried Meidl, Alexandr Polujan, Alexander Pott, Constanza Riera, and Pantelimon St\u anic\u a
\thanks{S. Li is with Institute of Algebra and Geometry, Faculty of Mathematics, Otto von Guericke University Magdeburg, Universit\"atsplatz 2, 39106 Magdeburg, Germany (email: shuxing\_li@sfu.ca).}
\thanks{W. Meidl is with Johann Radon Institute for Computational and Applied Mathematics, Austrian Academy of Sciences, Altenbergerstrasse 69, 4040 Linz, Austria (email: meidlwilfried@gmail.com).}
\thanks{A. Polujan is with Institute of Algebra and Geometry, Faculty of Mathematics, Otto von Guericke University Magdeburg, Universit\"atsplatz 2, 39106 Magdeburg, Germany (email: alexandr.polujan@ovgu.de).}
\thanks{A. Pott is with Institute of Algebra and Geometry, Faculty of Mathematics, Otto von Guericke University Magdeburg, Universit\"atsplatz 2, 39106 Magdeburg, Germany (email: alexander.pott@ovgu.de).}
\thanks{C. Riera is with Department of Computer Science, Electrical Engineering and Mathematical Sciences, Western Norway University of Applied Sciences, 5020 Bergen, Norway (email: csr@hvl.no).}
\thanks{P. St\u anic\u a is with Department of Applied Mathematics, Naval Postgraduate School, Monterey, CA 93943--5216 (email: pstanica@nps.edu).}
}

\maketitle

\begin{abstract}
For a function $f$ from $\F_2^n$ to $\F_2^n$, the planarity of $f$ is usually measured by its differential uniformity and differential spectrum. In this paper, we propose the concept of vanishing flats, which supplies a combinatorial viewpoint on the planarity. First, the number of vanishing flats of $f$ can be regarded as a measure of the distance between $f$ and the set of almost perfect nonlinear functions. In some cases, the number of vanishing flats serves as an ``intermediate'' concept between differential uniformity and differential spectrum, which contains more information than differential uniformity, however less than the differential spectrum. Secondly, the set of vanishing flats forms a combinatorial configuration called partial quadruple system, since it conveys a detailed structural information about $f$. We initiate this study by considering the number of vanishing flats and the partial quadruple systems associated with monomials and Dembowski-Ostrom polynomials. In addition, we present an application of vanishing flats to the partition of a vector space into disjoint equidimensional affine spaces. We conclude the paper with several further questions and challenges.
\end{abstract}

\begin{IEEEkeywords}
Almost perfect nonlinear function, Dembowski-Ostrom polynomial, differential uniformity, differential spectrum, equivalence, finite field, monomial, partial quadruple system, planarity, vanishing flat, vector space partition.
\end{IEEEkeywords}

\section{Introduction}

In this paper, we propose a combinatorial viewpoint to assess the planarity of functions from $\F_2^n$ to $\F_2^n$.  A function $f: \F_2^n \rightarrow \F_2^n$ achieves the highest planarity if and only if for each nonzero $a \in \F_2^n$, the function $f(x+a)-f(x)$ is as balanced as possible. A function from $\F_2^n$ to $\F_2^n$ with the highest planarity is called almost perfect nonlinear and will be formally defined later.  Keeping in mind that planarity is defined for functions over vector spaces, we always adopt the finite field model $\Ftn$ to represent the vector space $\F_2^n$, as each mapping defined over $\Ftn$ can be described by a polynomial. In fact, the univariate polynomial representation of functions over $\F_{2^n}$ may better serve our purpose and will be used throughout.

We first recall some basic concepts and notation. For a polynomial $f$ over $\Ftn$ and $a \in \Ftn^*$, the \emph{first-order derivative of $f$ at $a$} is the polynomial $f(x+a)+f(x)$. Given a polynomial $f$ and field elements $a \in \Ftn^*$, $b \in \Ftn$, we define
$$
\de_f(a,b)=|\{ x \in \Ftn \mid f(x+a)+f(x)=b \}|.
$$
Clearly, $\de_f(a,b)$ must be even (since, if $x_0$ is a solution of the above equation, so is $x_0+a$). The \emph{differential uniformity} of $f$ is defined as
$$
\de_f=\max_{a \in \Ftn^*, b \in \Ftn} \de_f(a,b)
$$
and the \emph{differential spectrum} of $f$ is the sequence $(\ell_{f,0},\ell_{f,2},\ldots,\ell_{f,\de_f})$, where $\ell_{f,2i}$ is the frequency of $2i$ in the multiset $[\de_f(a,b) \mid a \in \Ftn^*, b \in \Ftn]$. For the sake of convenience, we use $\ell_{2i}$ to represent $\ell_{f,2i}$, whenever the function $f$ is clear from the context. The polynomial $f$ is called {\em almost perfect nonlinear} (APN), if its differential uniformity is $\de_f=2$, which is the smallest possible. Equivalently, $f$ is APN if and only if the first-order derivative $f(x+a)+f(x)$ is a $2$-to-$1$ mapping for each $a \in \Ftn^*$. For $a \in \Ftn^*$, we define the \emph{differential uniformity of $f$ along the direction $a$} as
\[
\de_f(a)=\max_{b \in \Ftn} \de_f(a,b).
\]
Moreover, for $a \in \Ftn^*$, we define
\[
E_f(a)=\left\{ f(x+a)+f(x) \mid x \in \Ftn \right\}
\]
to be the image set of the first-order derivative of $f$ along the direction $a$. Note that $\de_f(a) \ge 2$ for each $a \in \Ftn^*$ and $f$ is APN if and only if $\de_f(a)=2$ for each $a \in \Ftn^*$. The recent paper~\cite{BKKRS} calls a function $f$ \emph{partially $a$-APN}, if $\de_f(a)=2$ at some $a \in \Ftn^*$. Surely, $f$ is partially $a$-APN if and only if $|E_f(a)|=2^{n-1}$.

There has been intensive research about APN functions, and  we refer to~\cite{BN15} and \cite{P}, for comprehensive surveys and extensive references on the subject. The fact that most known APN polynomials are quadratic and there is only one known class of APN permutations over $\Ftn$ with $n=6$, further motivates our research into the ``second best'' polynomials, which are not APN, but are closest to APN functions in some sense (this was one of the reasons for the introduction of the previously mentioned partially APN functions). One natural research venue  is  finding polynomials with differential uniformity $4$, which is the second smallest possible. Along this direction, a considerable amount of research dealt with polynomials having differential uniformity $4$, also satisfying some other  properties (see~\cite{BCC,BL,BTT,N,QTTL},  for example). The main purpose of this paper is to propose yet another research venue, namely the concept of vanishing flats, which offers a combinatorial viewpoint to understand the planarity of functions. The number of vanishing flats serves as a measure of the planarity of functions, or equivalently, measure the distance from a function to the set of APN functions. The set of vanishing flats forms a configuration called partial quadruple system, which is closely related to the classical Steiner quadruple system. The partial quadruple system contains detailed structural information of $f$. For instance, one can easily determine the directions along which a function $f$ is partially APN, by reading through the partial quadruple system of $f$.

The rest of the paper is organized as follows. In Section~\ref{sec2}, we introduce the vanishing flats and the partial quadruple system. Both the number of vanishing flats and the partial quadruple system are invariants under the CCZ-equivalence~\cite{CCZ}, as we shall show. Some relations among the differential uniformity, the number of vanishing flats and differential spectrum are discussed. With respect to the partial quadruple systems, some combinatorial implications of differential uniformity and differential spectrum are revealed. In Section~\ref{sec3}, we consider the partial quadruple system and the number of vanishing flats associated with monomials and Dembowski-Ostrom polynomials. We determine the partial quadruple systems of the inverse and the Gold functions. For general Dembowski-Ostrom polynomials, a formula for the number of vanishing flats is presented. In addition, the number of vanishing flats of a few more classes of monomials follows directly from their differential spectra. Section~\ref{sec4} presents an application of vanishing flats to a problem of partitioning vector spaces over finite fields into a collection of disjoint equidimensional affine subspaces, in the spirit of Baum and Neuwirth~\cite{BN75}. By extending the notion of vanishing flats, we propose a new approach to generate this type of partition. In Section~\ref{sec5}, we give some concluding remarks and raise some future research problems.

\section{Vanishing flats and partial quadruple systems associated with Boolean functions}\label{sec2}

In this section, we propose the concept of vanishing flats and a configuration named partial quadruple system associated with non-APN functions.
For a polynomial $f: \Ftn \rightarrow \Ftn$, we define the set of \emph{critical directions} of $f$ as
\begin{align*}
D_f&=\{ a \in \Ftn^* \mid \mbox{$f(x+a)+f(x)$ is not $2$-to-$1$} \}\\
   &=\{a \in \Ftn^* \mid \de_f(a) \ge 4\}.
\end{align*}
Clearly, $f$ is APN if and only if $D_f=\es$, namely, there exists no critical direction of $f$, or, the first-order derivative is as much balanced as possible along each direction. Indeed, in order to verify the APN property, elegant results in \cite{BD,CK} say that checking about half of all the directions would suffice. More precisely, if $\Ftn^* \sm D_f$ contains an $(n-1)$-dimensional vector space over $\F_2$ minus the zero vector, then this much weaker condition ensures that $f$ is APN \cite[Theorem 2]{CK} (this result was extended in~\cite[Theorem 6.2]{CP} to show that if $\Ftn^* \sm D_f$ contains a $t$-dimensional vector space over $\F_2$ minus the zero vector then the differential uniformity is $\leq 2^{n-t}$). Surely, for a non-APN function $f$, the set $D_f$ indicates the critical directions where the $2$-to-$1$ property fails. For instance, for $a \in \Ftn^*$, a function $f$ is partially $a$-APN if and only if $a \not\in D_f$. Consequently, we consider the set $D_f$ to be a crucial ingredient conveying much information related to the planarity of $f$. In order to understand $D_f$, we trace back to its source and arrive at a well-known alternative definition of APN functions.

For $n \ge 2$, define the set of all $2$-dimensional flats in $\Ftn$ as
\begin{align*}
\cB_n=\{ \{x_1,x_2,x_3, & x_4\}  \mid \mbox{$x_1+x_2+x_3+x_4=0$ } \\
                             & \mbox{and $x_1,x_2,x_3,x_4 \in\Ftn$ are distinct} \}.
\end{align*}
Conventionally, each subset in $\cB_n$ is called a \emph{block}. The classical \emph{Steiner quadruple system} is a pair $(\Ftn,\cB_n)$, so that each $3$-subset of $\Ftn$ is contained in exactly one block of $\cB_n$. A function $f: \Ftn \rightarrow \Ftn$ is APN if and only if for each $\{x_1,x_2,x_3,x_4\} \in \cB_n$,
$$
f(x_1)+f(x_2)+f(x_3)+f(x_4) \ne 0.
$$
Namely, the summation of $f$ over each $2$-dimensional flat is non-vanishing. For a function $f: \Ftn \rightarrow \Ftn$, define the set of \emph{vanishing flats} with respect to $f$ as
\begin{align*}
\VB_{n,f}=\{ \{x_1,x_2,& x_3, x_4\} \in \cB_n \mid \\
                           &  f(x_1)+f(x_2)+f(x_3)+f(x_4)=0 \}.
\end{align*}
Note that each vanishing flat $\{x_1,x_2,x_3,x_4\}$ in $\VB_{n,f}$, gives rise to three distinct critical directions $a_1=x_1+x_2$, $a_2=x_1+x_3$ and $a_3=x_1+x_4$, such that $f(x+a_i)+f(x)$ is not $2$-to-$1$ for each $1 \le i \le 3$. Therefore, the set of critical directions $D_f$ follows immediately from $\VB_{n,f}$. Since $\VB_{n,f} \subset \cB_n$, we call the set system $(\Ftn,\VB_{n,f})$ a \emph{partial quadruple system}. Indeed, a partial quadruple system is an instance of the much more general class of configurations named \emph{packings}. For comprehensive surveys about general Steiner systems and packings, we refer to \cite{CM} and \cite{SWY}, respectively.

For an APN function $f$, the associated partial quadruple system is degenerate as the block set $\VB_{n,f}$ is empty. Now, we proceed to justify that the partial quadruple system is a proper measure of the planarity of non-APN functions.

Let $f$ and $g$ be two polynomials from $\F_{2^n}$ to $\F_{2^n}$. They are {\em CCZ-equivalent} \cite{CCZ} if there exists an affine permutation $A$ on $\F_{2}^n \times \F_{2}^n$ such that
\begin{equation}\label{eqn-CCZ}
\left\{\colvec{x}{g(x)} \bigm| x\in\F_{2}^n \right\}=\left\{A \colvec{x}{f(x)} \bigm| x\in\F_{2}^n \right\},
\end{equation}
where by choosing a basis of $\F_{2^n}$ over $\F_2$, the polynomials $f$ and $g$ are regarded as two mappings from $\F_2^n$ to $\F_2^n$. As it is well-known, the CCZ-equivalence preserves the differential uniformity, and therefore the APN property~\cite[Proposition 2]{BCP}. Recall that two partial quadruple systems are \emph{isomorphic} if there exists a bijective mapping between their point sets, which also induces a bijective mapping between their block sets. Next, we will show that the CCZ-equivalence also preserves partial quadruple systems up to isomorphism. Note that we will use the notation $\cong$ to represent two isomorphic partial quadruple systems.

\begin{theorem}
\label{thm-CCZinvariant}
Let $f$ and $g$ be two CCZ-equivalent functions from $\F_{2^n}$ to $\F_{2^n}$, in which
\begin{align*}
&\left\{\colvec{x}{g(x)} \bigm|  x\in\F_{2}^n \right\} \\
=&\left\{ \begin{pmatrix}
A_{11} & A_{12}\\
A_{21} & A_{22}
\end{pmatrix} \colvec{x}{f(x)}+\colvec{u}{v} \bigm|  x\in\F_{2}^n \right\}.
\end{align*}
Then, $\{x_1,x_2,x_3,x_4\} \in \VB_{n,f}$ if and only if $\{y_1,y_2,y_3,y_4\} \in \VB_{n,g}$, where $y_i=A_{11} x_i+A_{12} f(x_i)+u$.
Consequently, $(\Ftn,\VB_{n,f}) \cong (\Ftn,\VB_{n,g})$ and the number of vanishing flats is an invariant under the CCZ-equivalence.
\end{theorem}
\begin{proof}
Suppose $\{x_1,x_2,x_3,x_4\} \in \VB_{n,f}$. Then
\begin{align*}
x_1+x_2+x_3+x_4&=0, \\
f(x_1)+f(x_2)+f(x_3)+f(x_4)&=0.
\end{align*}
Let $T$ be a mapping from $\F_2^n$ to $\F_2^n$, such that $T(x)=A_{11}x+A_{12}f(x)+u$. By the defining relation of the CCZ-equivalence, $T$ induces a permutation on $\Ftn$. Moreover, we let $y_i=T(x_i)$ and $g(y_i)=A_{21} x_i+A_{22} f(x_i)+v$. A direct computation shows that
\begin{align*}
y_1+y_2+y_3+y_4&=0, \\
g(y_1)+g(y_2)+g(y_3)+g(y_4)&=0.
\end{align*}
Therefore, $\{y_1,y_2,y_3,y_4\} \in \VB_{n,g}$ and $T$ induces an injective mapping from $\VB_{n,f}$ to $\VB_{n,g}$. Reversing the above argument, one can show that by applying the inverse of $T$ to a vanishing flat $\{y_1,y_2,y_3,y_4\} \in \VB_{n,g}$, we obtain a vanishing flat $\{x_1,x_2,x_3,x_4\} \in \VB_{n,f}$. Hence, $T$ induces a bijection between $\VB_{n,f}$ and $\VB_{n,g}$, and therefore $(\Ftn,\VB_{n,f}) \cong (\Ftn,\VB_{n,g})$.
\end{proof}

For a polynomial $f$ over $\Ftn$ and $\sig \in \Gal(\Ftn/\F_2)$ (the Galois group of $\Ftn$ over $\F_2$), we define  $\sig f$ to be a function satisfying  $\sig f(a)=\sig(f(a))$, for each $a \in \Ftn$. It is easy to see that both applying a Galois automorphism of the field extension $\Ftn/\F_2$ and adding an affine function keep the partial quadruple systems invariant. In addition, if $f$ is a permutation, then the partial quadruple systems associated with $f$ and $f^{-1}$ are isomorphic.

\begin{remark}\label{rem-transformation}
Let $f$ be a polynomial over $\Ftn$. Then we have the following$:$
\begin{itemize}
\item[$(1)$] For each $\sig \in \Gal(\Ftn/\F_2)$, we have $\VB_{n,f}=\VB_{n,\sig f}$.
\item[$(2)$] For each affine polynomial $A$ over $\Ftn$, we have $\VB_{n,f}=\VB_{n,f+A}$.
\item[$(3)$] If $f$ is a permutation, then $(\Ftn,\VB_{n,f}) \cong (\Ftn,\VB_{n,f^{-1}})$.
\end{itemize}
\end{remark}

A particularly interesting quantity associated with $(\Ftn,\VB_{n,f})$ is the size of the block set $\VB_{n,f}$, or equivalently, the number of vanishing flats of $f$. Note that the number of vanishing flats is $0$ if and only if $f$ is APN. In this sense, the number of vanishing flats measures the distance between $f$ and the set of APN functions. In fact, the size of $\VB_{n,f}$ follows from the differential spectrum of $f$.

\begin{theorem}
\label{thm-diffspecVB}
For a function $f: \Ftn \rightarrow \Ftn$, we have $\displaystyle \left|\VB_{n,f}\right|=\frac{1}{3}\sum_{a \in \Ftn^*, b \in \Ftn} \binom{\de_f(a,b)/2}{2}$.
\end{theorem}
\begin{proof}
By definition, each pair $(a,b)$ with $\de_f(a,b) \le 2$ is irrelevant to vanishing flats. Thus, we only need to consider the pairs $(a,b) \in \Ftn^* \times \Ftn$ with $\de_f(a,b) \ge 4$. For each such pair, there exist distinct $x_1, x_2 \in \Ftn$, such that $f(x_1+a)+f(x_1)=f(x_2+a)+f(x_2)=b$ and therefore $\{x_1,x_1+a,x_2,x_2+a\}$ forms a vanishing flat. In total, the number of vanishing flats derived from these $(a,b)$ pairs is equal to $\sum_{a \in \Ftn^*, b \in \Ftn} \binom{\de_f(a,b)/2}{2}$. On the other hand, we claim that each vanishing flat can be derived exactly three times by considering the $(a,b)$ pairs. Indeed, for distinct $a_1$, $a_2$, $a_3$ with $a_1+a_2+a_3=0$, a vanishing flat $\{x,x+a_1,x+a_2,x+a_3\}$ satisfying
\begin{align*}
f(x+a_1)+f(x)&=f(x+a_2)+f(x+a_3)=b_1, \\
f(x+a_2)+f(x)&=f(x+a_1)+f(x+a_3)=b_2, \\
f(x+a_3)+f(x)&=f(x+a_1)+f(x+a_2)=b_3,
\end{align*}
occurs exactly three times when we consider the pairs $(a_i,b_i)$, $1 \le i \le 3$. Consequently, $|\VB_{n,f}|=\frac{1}{3}\sum_{a \in \Ftn^*, b \in \Ftn} \binom{\de_f(a,b)/2}{2}$.
\end{proof}

Observe that both the differential uniformity and the number of vanishing flats carry partial information of the differential spectrum. In general, they offer different viewpoints to assess the planarity of functions. On the other hand, when a function has two-valued differential spectrum, these three notions are equivalent. We recall that a function $f$ over $\Ftn$ with two-valued differential spectrum has the property that $\{ \de_f(a,b) \mid a \in \Ftn^*, b \in \Ftn \}=\{0,2^s\}$ for some positive integer $1 \le s \le n$ \cite[Lemma 3]{BCC}. In the following result,  for the sake of simplicity, we only concentrate on the nonzero entries in the differential spectrum.

\begin{proposition}\label{prop-equivalence}
Let $f$ be a polynomial over $\Ftn$ with two-valued differential spectrum $\{0,2^s\}$, where $1 \le s \le n$. Then the following are equivalent$:$
\begin{itemize}
\item[$(1)$] $\de_f=2^s$.
\item[$(2)$] $|\VB_{n,f}|=\frac{2^{n-2}(2^n-1)(2^{s-1}-1)}{3}$.
\item[$(3)$] $\ell_{f,0}=(2^n-2^{n-s})(2^n-1)$ and $\ell_{f,2^s}=2^{n-s}(2^n-1)$.
\end{itemize}
\end{proposition}
\begin{proof}
The proof follows easily from the definitions and Theorem~\ref{thm-diffspecVB}.
\end{proof}

For a function $f$ over $\Ftn$, the number of its vanishing flats tells the number of blocks in the partial quadruple system $(\Ftn,\VB_{n,f})$. As one may expect, the differential spectrum of $f$ supplies more detailed structural information of $(\Ftn,\VB_{n,f})$. Indeed, for $x \in \Ftn$ and $a \in \Ftn^*$ such that $f(x+a)+f(x)=b$, the value of $\de_f(a,b)$ renders the number of vanishing flats containing the pair $x$ and $x+a$.

\begin{theorem}\label{thm-diffspecPQS}
Let $f$ be a function over $\Ftn$.
\begin{itemize}
\item[$(1)$] For $x \in \Ftn$ and distinct $a_1, a_2 \in \Ftn^*$, if the three elements $x$, $x+a_1$ and $x+a_2$ belong to a vanishing flat, then $\de_f(a) \ge 4$ for each $a \in \{ a_1, a_2, a_1+a_2 \}$.
\item[$(2)$] For $x \in \Ftn$ and $a \in \Ftn^*$, such that $f(x+a)+f(x)=b$, the two elements $x$ and $x+a$ belong to $\frac{\de_f(a,b)}{2}-1$ vanishing flats.
\end{itemize}
\end{theorem}
\begin{proof}
The proof of (1) is easy and hence omitted. For (2), note that there are $\frac{\de_f(a,b)}{2}$ pairs of $x_i$ and $x_i+a$, where $1 \le i \le \frac{\de_f(a,b)}{2}$ and $x_1=x$, such that $f(x_i+a)+f(x_i)=b$. Consequently, there are exactly $\frac{\de_f(a,b)}{2}-1$ vanishing flats containing $x$ and $x+a$.
\end{proof}

\begin{remark}
We note that the converse of Theorem~\ref{thm-diffspecPQS}(1) is not true. For instance, consider the inverse function $f(x)=x^{-1}$ over $\F_{2^n}$ with $n$ being even. According to Proposition~\ref{prop-mono} below, $\de_f(a) \ge 4$ for each $a \in \Ftn^*$. On the other hand, by Theorem~\ref{thm-PQSinv}, for distinct $a_1,a_2 \in \Ftn$, the three elements $0$, $a_1$ and $a_2$ belong to the same vanishing flat if and only if $\frac{a_1}{a_2}$ is a third root of unity in $\F_{2^n}$.
\end{remark}

\section{Vanishing Flats and Partial Quadruple Systems of Monomials and Dembowski-Ostrom Polynomials}\label{sec3}

In this section, we investigate the vanishing flats and the partial quadruple systems associated with monomials and Dembowski-Ostrom polynomials.

\subsection{Vanishing Flats and Partial Quadruple Systems of Monomials}

We first consider the vanishing flats and the partial quadruple systems derived from monomials. For the sake of convenience, we write $\de_d:=\de_{x^d}$, $\de_d(a,b):=\de_{x^d}(a,b)$, $\de_d(a):=\de_{x^d}(a)$, $D_d:=D_{x^d}$ and $\VB_{n,d}:=\VB_{n,x^d}$. A polynomial $f$ being monomial ensures that the first-order derivative $f(x+a)+f(x)$ behaves uniformly at each $a \in \Ftn^*$.

\begin{proposition}
\label{prop-mono}
Let $x^d$ be a non-APN monomial over $\Ftn$. Then the multiset $[\de_d(a,b) \mid b \in \Ftn]$ is the same for each $a \in \Ftn^*$. Consequently, $D_d=\Ftn \sm \{0\}$ and
\begin{equation}\label{eqn-VBf}
|\VB_{n,d}| \ge \begin{cases}
              \frac{2^n+1}{3} & \mbox{if $n$ odd} \\
              \frac{2^n-1}{3} & \mbox{if $n$ even}.
            \end{cases}
\end{equation}
\end{proposition}
\begin{proof}
For distinct $a_1,a_2 \in \Ftn^*$ and $b \in \Ftn$, $x$ is a solution to $(x+a_1)^d+x^d=b$ if and only if $\frac{a_2}{a_1}x$ is a solution to $(x+a_2)^d+x^d=\left(\frac{a_2}{a_1}\right)^db$. Hence, $\de_d(a_1,b)=\de_d\left(a_2,\left(\frac{a_2}{a_1}\right)^db\right)$ and consequently, the multiset $[\de_d(a,b) \mid b \in \Ftn]$, and therefore $\de_d(a)$ is the same for each $a \in \Ftn^*$. Since $x^d$ is not APN, we have $\de_d(a) \ge 4$ for each $a \in \Ftn^*$ and thus, $D_d=\Ftn \sm \{0\}$. Namely, $x^d$ has $2^n-1$ critical directions. Note that along each critical direction, there exists at least one vanishing flat. Also, each vanishing flat gives three distinct critical directions. To cover all $2^n-1$ critical directions, one needs at least $\frac{2^n-1}{3}$ vanishing flats. Hence, we have $|\VB_{n,d}| \ge \frac{2^n-1}{3}$, which leads to~\eqref{eqn-VBf}.
\end{proof}

\begin{remark}\label{rem-upperbound}
In addition to \eqref{eqn-VBf}, since $\VB_{n,f} \subset \cB_n$, we have a natural upper bound
$$
|\VB_{n,f}| \le |\cB_n|=\frac{2^{n-2}(2^{n-1}-1)(2^n-1)}{3}.
$$
The upper bound is tight as $\VB_{n,1}=\cB_n$ and $|\VB_{n,1}|=\frac{2^{n-2}(2^{n-1}-1)(2^n-1)}{3}$.
\end{remark}

Equation \eqref{eqn-VBf} gives a general lower bound on the number of vanishing flats for monomials. In fact, when $n$ is even, the lower bound is tight.

\begin{theorem}
\label{thm-PQSinv}
Let $n$ be even and $3 \nmid d$. Let $\al$ be a primitive element of $\Ftn$ and $\ze=\al^{\frac{2^n-1}{3}}$. Then $\VB_{n,d} \supseteq \left\{ \left\{0,\al^i,\al^i\ze,\al^i\ze^2\right\} \mid 0 \le i \le \frac{2^n-4}{3} \right\}$. In particular, for the inverse function $x\mapsto x^{-1}:=x^{2^n-2}$, we have $\VB_{n,-1}=\left\{ \left\{0,\al^i,\al^i\ze,\al^i\ze^2\right\} \mid 0 \le i \le \frac{2^n-4}{3} \right\}$.
\end{theorem}
\begin{proof}
If $3 \nmid d$, then by definition, $\VB_{n,d} \supseteq \left\{ \left\{0,\al^i,\al^i\ze,\al^i\ze^2\right\} \mid 0 \le i \le \frac{2^n-4}{3} \right\}$. Now, we proceed to show that each vanishing flat in $\VB_{n,-1}$ contains $0$, by reduction to absurdity. Assume that $\{x_1,x_2,x_3,x_4\}$ is a vanishing flat with $x_1x_2x_3x_4 \ne 0$. Combining $x_1+x_2=x_3+x_4$ and $x_1^{-1}+x_2^{-1}=x_3^{-1}+x_4^{-1}$, we have $x_1+x_2=x_3+x_4=a$ and $x_1x_2=x_3x_4=b$ for some nonzero $a$ and $b$. Hence, $x_i$, $1 \le i \le 4$, are four distinct solutions  to the quadratic equation $x^2+ax+b=0$, which is impossible.

Without loss of generality, we assume $x_4=0$ and $x_3=\be$ for some nonzero $\be$. Hence, we have $x_1+x_2=\be$ and $x_1^{-1}+x_2^{-1}=\be^{-1}$. It is easy to see that $\{x_1,x_2\}=\{\be\ze,\be\ze^2\}$. Therefore, two elements $0$ and $\be$ determine the whole vanishing flat $\{x_1,x_2,x_3,x_4\}=\left\{0,\be,\be\ze,\be\ze^2\right\}$. In conclusion, $\VB_{n,-1}=\left\{ \left\{0,\al^i,\al^i\ze,\al^i\ze^2\right\} \mid 0 \le i \le \frac{2^n-4}{3} \right\}$.
\end{proof}

The partial quadruple system can also be determined for the non-APN Gold function.  We shall use $(a,b)$ to denote the greatest common divisor of $a$ and $b$.

\begin{theorem}\label{thm-PQSGold}
Let $f(x)=x^{2^t+1}$ be a function over $\Ftn$ with $(n,t)=s>1$. For $a \in \Fts \sm \{0,1\}$ and $x \in \Ftn^*$, we define a $2$-dimensional vector space $V_{a,x}=\{0,x,ax,(1+a)x\}$ and
\begin{align*}
U_{a,x}=\{ & \left\{c,x+c,ax+c,(1+a)x+c\right\} \mid \mbox{$c$ ranges} \\
& \mbox{over all coset representatives of $V_{a,x}$ in $\Ftn$} \}.
\end{align*}
Then $\displaystyle \VB_{n,f}=\bigcup_{\substack{a \in \Fts \sm \{0,1\} \\ x \in \Ftn^*}} U_{a,x}$ and $\displaystyle |\VB_{n,f}|=\frac{2^{n-2}(2^{s-1}-1)(2^n-1)}{3}$.
\end{theorem}
\begin{proof}
Note that if $\{x_1,x_2,x_3,x_4\}$ is a vanishing flat of $f$, then so is $\{x_1+c,x_2+c,x_3+c,x_4+c\}$ for each $c \in \Ftn$. Hence, it suffices to analyze the $2$-dimensional vanishing subspaces, which have the form $\{0,x_1,x_2,x_3\}$, where the nonzero $x_1$, $x_2$, $x_3$ are distinct, $x_1+x_2+x_3=0$ and $f(0)+f(x_1)+f(x_2)+f(x_3)=x_1^{2^t+1}+x_2^{2^t+1}+x_3^{2^t+1}=0$.  Note that $x_1^{2^t+1}+x_2^{2^t+1}+x_3^{2^t+1}=x_1^{2^t+1}+x_2^{2^t+1}+(x_1+x_2)^{2^t+1}=0$. Simplifying the last equation, we have $(\frac{x_1}{x_2})^{2^t}=\frac{x_1}{x_2}$. Thus, $\frac{x_1}{x_2} \in \Fts$. Since $x_1 \ne 0$ and $x_1 \ne x_2$, we have $\frac{x_1}{x_2} \in \Fts \sm \{0,1\}$. Indeed, we have shown for each $1 \le j<\ell\le 3$, $\frac{x_j}{x_\ell} \in \Fts \sm \{0,1\}$. Note that each $2$-dimensional vanishing subspace $\{0,x_1,x_2,x_3\}$ can be expressed in exactly six different ways: $V_{\frac{x_2}{x_1},x_1}$, $V_{\frac{x_3}{x_1},x_1}$, $V_{\frac{x_1}{x_2},x_2}$, $V_{\frac{x_3}{x_2},x_2}$, $V_{\frac{x_1}{x_3},x_3}$ and $V_{\frac{x_2}{x_3},x_3}$. Consequently, we have $\VB_{n,f}=\bigcup_{\substack{a \in \Fts \sm \{0,1\} \\ x \in \Ftn^*}} U_{a,x}$. Moreover, $|\VB_{n,f}|=\frac{2^{n-2}(2^s-2)(2^n-1)}{6}=\frac{2^{n-2}(2^{s-1}-1)(2^n-1)}{3}$.
\end{proof}

So far, we have determined the partial quadruple systems associated with the inverse function (Theorem~\ref{thm-PQSinv}) and the Gold  functions (Theorem~\ref{thm-PQSGold}). It is worthy to note that there are a few power functions whose differential spectra are known. Therefore, employing Theorem~\ref{thm-diffspecVB}, the number of vanishing flats follows immediately. We summarize these results in Table~\ref{tab-VF}, where $s=(n,t)$ and $K$ is the Kloosterman sum with the following explicit expression \cite[Remark 3]{BCC11}:
$$
K=1+\frac{(-1)^{n-1}}{2^{n-1}}\sum_{i=0}^{\lf \frac{n}{2} \rf} (-1)^i \binom{n}{2i}7^i.
$$
For two positive integers $a$ and $b$, we define
$$
\De(a,b)=\begin{cases}
  1 & \mbox{if $a \mid b$,} \\
  0 & \mbox{if $a \nmid b$.}
\end{cases}
$$
By Proposition~\ref{prop-mono}, for every monomial, each entry among the differential spectrum is divisible by $2^n-1$. For simplicity, we shall use $w_i=\ell_i/(2^n-1)$ in Table~\ref{tab-VF}. The column ``Ref'' indicates the references where the differential spectra are calculated.

\begin{table*}\renewcommand{\arraystretch}{1.5}

\begin{center}
\caption{Power functions $x^d$ over $\Ftn$ with known differential spectra and number of vanishing flats}

\begin{tabular}{|c|c|c|c|c|}

\hline
$n$ & $d$ & $|\VB_{n,d}|$ & Differential Spectrum &  Reference \\ \hline
\multirow{2}{*}{$n \ge 2$} & $2^t+1$ & \multirow{2}{*}{$\dfrac{2^{n-2}(2^{s-1}-1)(2^n-1)}{3}$} & $w_0=2^n-2^{n-s}$ & \multirow{2}{*}{\cite[Section 5.2]{BCC}} \\
& $1 \le t \le n/2$ & & $w_{2^s}=2^{n-s}$ & \\ \hline
$n \ne 3t$ & $2^{2t}-2^t+1$ & \multirow{2}{*}{$\dfrac{2^{n-2}(2^{s-1}-1)(2^n-1)}{3}$} & $w_0=2^n-2^{n-s}$ & \multirow{2}{*}{\cite[Theorem 2]{BCC}} \\
$n/s$ odd & $2 \le t \le n/2$ & & $w_{2^s}=2^{n-s}$ & \\ \hline
\multirow{3}{*}{$n$ even}& \multirow{3}{*}{$2^n-2$} & \multirow{3}{*}{$\dfrac{2^n-1}{3}$} & $w_0=2^{n-1}+1$ & \multirow{3}{*}{\cite[Proposition 6]{N}} \\
 & & & $w_{2}=2^{n-1}-2$ & \\
 & & & $w_{4}=1$ & \\ \hline
\multirow{3}{*}{$n=4t$}& \multirow{3}{*}{$2^{2t}+2^t+1$} & \multirow{3}{*}{$\dfrac{\left(2^{n-3}-2^{3t-3}\right)(2^n-1)}{3}$} & $w_0=5\cdot2^{n-3}-2^{3t-3}$ & \multirow{2}{*}{\cite[Example 4]{BCC}} \\
 & & & $w_{2}=2^{n-2}+2^{3t-2}$ & \\
 & & & $w_{4}=2^{n-3}-2^{3t-3}$ & \cite[Theorem 1]{XY}\\ \hline
\multirow{4}{*}{$n \ge 6$}& \multirow{4}{*}{$7$} & & $w_0=2^{n-1}+2w_6+w_4$ & \multirow{4}{*}{\cite[Theorem 5]{BCC11}} \\
& & $\left(\frac{2^{n-2}+1-3w_4}{6}+(-1)^n\frac{K}{8}\right)$ & $w_{2}=2^{n-1}-3w_6-2w_4$ & \\
& & $\cdot (2^n-1)$ & $w_{4}=\De(2,n)$ & \\
& & & $w_{6}=\frac{2^{n-2}+1-5w_4}{6}+(-1)^n\frac{K}{8}$ & \\ \hline
\multirow{5}{*}{$n \ge 6$}& $2^{n-2}-1$ &  & $w_0=2^{n-1}+2w_6+3w_8$ & \\
& or & $\Big(\frac{2^{n-1}-3-(-1)^n5}{12}+(-1)^n\frac{K}{8}$ & $w_{2}=2^{n-1}-3w_6-4w_8$ & \cite[Corollary 5]{BCC11} \\
& $2^{\frac{n-1}{2}}-1$ & $+w_8\Big)\cdot(2^n-1)$ & $w_{6}=\frac{2^{n-1}-3-(-1)^n5}{12}$ & \cite[Theorem 5]{BP} \\
& $n$ odd & & $+(-1)^n\frac{K}{8}-w_8$ & \\
&  & & $w_{8}=\De(3,n)$ & \\ \hline
& \multirow{4}{*}{$2^{\frac{n}{2}}-1$} &  & $w_0=2^{n-1}+2^{\frac{n}{2}-1}-2+w_4$ & \multirow{4}{*}{\cite[Theorem 7]{BCC11}} \\
$n \ge 6$ & & $\left(\left(2^{\frac{n}{2}-1}-1\right)\Big(2^{\frac{n}{2}-2}-1\Big)\right.$ & $w_{2}=2^{n-1}-2^{\frac{n}{2}-1}+1-2w_4$ & \\
$n$ even & & $+w_4\Big)\cdot\frac{2^n-1}{3}$ & $w_{4}=1-\De(4,n)$ & \\
& & & $w_{2^{\frac{n}{2}}-2}=1$ & \\ \hline
& \multirow{3}{*}{$2^{\frac{n}{2}+1}-1$} &  & $w_0=2^{n-1}+2^{\frac{n}{2}-1}-1$ & \multirow{3}{*}{\cite[Theorem 8]{BCC11}} \\
$n \ge 6$  & & $\dfrac{2^{\frac{n}{2}-2}\left(2^{\frac{n}{2}-1}-1\right)(2^n-1)}{3}$ & $w_{2}=2^{n-1}-2^{\frac{n}{2}-1}$ & \\
$n$ even & & & $w_{2^{\frac{n}{2}}}=1$ & \\
\hline
& \multirow{3}{*}{$2^{\frac{n+3}{2}}-1$} & & $w_0=2^{n-1}+2w_6+2\De(3,n)$ & \multirow{3}{*}{\cite[Theorems 1,5]{BP}} \\
$n \ge 7$  & & $\left(\frac{2^{n-2}+1}{6}-\frac{K}{8}\right)(2^n-1)$ & $w_{2}=2^{n-1}-3w_6-3\De(3,n)$ & \\
$n$ odd & & & $w_6=\frac{2^{n-2}+1}{6}-\frac{K}{8}$ & \\ \hline
& & & $w_0=89 \cdot 2^{n-7}+7\cdot 2^{t-7}(4-K)$ & \multirow{5}{*}{\cite[Theorem 1.4]{XYY}} \\
$n=2t$ & $2^{t+1}+2^{\frac{t+1}{2}}+1$ & & $w_{2}=5 \cdot 2^{n-5}-5 \cdot 2^{t-5}(4-K)$ & \\
$t \ge 5$ & or & $\dfrac{2^{n-2}(2^n-1)}{3}$ & $w_{4}=7 \cdot 2^{n-6}+9 \cdot 2^{t-6}(4-K)$ & \\
$t$ odd & $2^{t+1}+3$ &  & $w_{6}=2^{n-5}-2^{t-5}(4-K)$ & \\
& & & $w_{8}=2^{n-7}-2^{t-7}(4-K)$ & \\ \hline
\end{tabular}
\label{tab-VF}
\end{center}
\end{table*}

\begin{remark}
The following are some observations based on Table~\textup{\ref{tab-VF}}, regarding the relationship between the differential uniformity, the number of vanishing flats and the differential spectrum.
\begin{itemize}
\item[$(1)$] By Proposition~\textup{\ref{prop-equivalence}}, for the first two classes, the differential uniformity, the number of vanishing flats and the differential spectrum carry the same information.
\item[$(2)$] Comparing with the differential uniformity, there are some cases where the number of vanishing flats is a finer criterion to evaluate the planarity. For example, the third and fourth classes have the same differential uniformity but distinct number of vanishing flats. The same happens to the fifth and sixth classes when $3 \nmid n$.
\item[$(3)$] By Theorem~\textup{\ref{thm-diffspecVB}}, the number of vanishing flats follows from the differential spectrum. The converse is not true as the number of vanishing flats does not contain the information of $w_0$ and $w_2$. For example, whether $3\, |\, n$ or not, the differential spectra of the ninth class differ at $w_0$ and $w_2$. However, the number of vanishing flats remains the same.
\end{itemize}
\end{remark}

In Table~\ref{tab-VFexample}, we list the number of vanishing flats of all monomials over $\F_{2^n}$, where $2 \le n \le 8$. We use the superscript $\bigstar$ to mark the monomials whose number of vanishing flats is not obtained from Table~\ref{tab-VF} or Remark~\ref{rem-upperbound}. Note that by Remark~\ref{rem-transformation}, the number of vanishing flats of $x^d$ is the same as $x^{2^id}$, for each $0 \le i \le n-1$, as well as, $x^{2^id^{-1}}$ for each $0 \le i \le n-1$, if $(d,2^n-1)=1$. Among these monomials which necessarily have the same number of vanishing flats, we only list one of them.

\begin{table*}\renewcommand{\arraystretch}{1.5}

\begin{center}
\caption{The number of vanishing flats of $x^d$ over $\Ftn$, for $2 \le n \le 8$}
\begin{tabular}{|c|c|}
\hline
$n$ & $\left(d,|\VB_{n,d}|\right)$ \\ \hline
$2$ & $(1,1)$ \\ \hline
$3$ & $(1,14)$, $(3,0)$ \\ \hline
$4$ & $(1,140)$, $(3,0)$, $(5,20)$, $(7,5)$ \\ \hline
$5$ & $(1,1240)$, $(3,0)$, $(5,0)$, $(15,0)$ \\ \hline
\multirow{2}{*}{$6$} & $(1,10416)$, $(3,0)$, $(5,336)$, $(7,84)$, $(9,1008)$, $(11,336)^\bigstar$,  \\
                     & $(15,126)$, $(21,2520)^\bigstar$, $(27,1260)^\bigstar$, $(31,21)$  \\ \hline
$7$ & $(1,85344)$, $(3,0)$, $(5,0)$, $(7,889)$, $(9,0)$, $(11,0)$, $(19,889)^\bigstar$, $(21,889)$, $(23,0)$, $(63,0)$ \\ \hline
\multirow{4}{*}{$8$} & $(1,690880)$, $(3,0)$, $(5,5440)$, $(7,3655)$, $(9,0)$, $(11,5185)^\bigstar$, $(13,5185)^\bigstar$, $(15,1785)$,\\
                     & $(17,38080)$, $(19,4420)^\bigstar$, $(21,2040)$, $(23,4930)^\bigstar$, $(25,4420)^\bigstar$, $(27,15810)^\bigstar$, \\
                     & $(31,2380)$, $(39,0)$, $(43,27625)^\bigstar$, $(45,1785)^\bigstar$, $(51,66300)^\bigstar$, $(53,7480)^\bigstar$, $(55,5440)^\bigstar$,  \\
                     &  $(63,3570)$, $(85,174760)^\bigstar$, $(87,24480)^\bigstar$, $(95,2380)^\bigstar$, $(111,1020)^\bigstar$, $(119,41905)^\bigstar$, $(127,85)$
                 \\ \hline
\end{tabular}
\label{tab-VFexample}
\end{center}
\end{table*}

Finally, we mention a connection between the vanishing flats of monomials and the binary cyclic codes defined by the same monomials. For a more detailed account on such cyclic codes, we refer the reader to~\cite{CCZ,CTZ}.

\begin{proposition}\label{prop-vfcdoe}
Let $\al$ be a primitive element of $\Ftn$. Let $C_d$ be a binary cyclic code of length $2^n-1$ having two zeroes $\al$ and $\al^d$, in which each nonzero element of $\Ftn$ indexes a coordinate of the codewords in $C_d$. Each vanishing flat of $x^d$ must be one of the following$:$
\begin{itemize}
\item[$(1)$] For each weight three codeword, the zero element plus the three nonzero elements indexing three nonzero coordinates form a vanishing flat.
\item[$(2)$] For each weight four codeword, the four nonzero elements indexing four nonzero coordinates form a vanishing flat.
\end{itemize}
Consequently, the number of weight three codewords of $C_d$ equals the number of vanishing flats in $\VB_{n,d}$ containing $0$, and the number of weight four codewords of $C_d$ equals the  number of vanishing flats in $\VB_{n,d}$ not containing $0$.
\end{proposition}
\begin{proof}
First, we note that the binary cyclic code $C_d$ has a parity-check matrix
$$
\begin{pmatrix}
1 & \al & \al^2 & \cdots & \al^{2^n-2} \\
1 & \al^d & \al^{2d} & \cdots & \al^{(2^n-2)d}
\end{pmatrix}.
$$
The rest follows easily from the definition of vanishing flats.
\end{proof}

\begin{remark}
Proposition~\ref{prop-vfcdoe} can be further extended. Let $f$ be a polynomial over $\Ftn$. Using the same notation as in Proposition~\ref{prop-vfcdoe}, there is a one-to-one correspondence between the vanishing flats of $f$ and the indices of the weight four codewords in the binary linear code having a parity-check matrix
$$
\begin{pmatrix}
0 & 1 & \al & \al^2 & \cdots & \al^{2^n-2} \\
f(0) & f(1) & f(\al) & f(\al^{2}) & \cdots & f(\al^{2^n-2})
\end{pmatrix}.
$$
\end{remark}

\subsection{Vanishing Flats and Partial Quadruple Systems of Dembowski-Ostrom Polynomials}

A polynomial $f$ over $\Ftn$ is of \emph{Dembowski-Ostrom} (DO) type if
$$
f=\sum_{0 \le i<j<n} c_{ij}x^{2^i+2^j},
$$
where $c_{ij} \in \Ftn$. For DO polynomials, we have the following result on the number of vanishing flats.

\begin{theorem}
\label{thm-cosetstrucDO}
Let $\displaystyle f(x)=\sum_{0 \le i<j <n}c_{i,j}x^{2^i+2^j}$ be a non-APN DO polynomial over $\Ftn$, $n \ge 2$. The following hold$:$
\begin{itemize}
\item[$(1)$] If $\{x_1,x_2,x_3,x_4\} \in \VB_{n,f}$, then $\{\{x_1+a,x_2+a,x_3+a,x_4+a\} \mid a \in \Ftn\} \subset \VB_{n,f}$ for each $a \in \Ftn$. Consequently, $2^{n-2} \bigm| |\VB_{n,f}|$ and $|\VB_{n,f}| \ge 2^{n-2}$.
\item[$(2)$] For each $a \in \Ftn$, the subset $\{a,x_1+a,x_2+a,x_1+x_2+a\} \in \VB_{n,f}$ if and only if
$$
\sum_{0 \le i<j <n}c_{i,j}\left(x_1^{2^i}x_2^{2^j} + x_1^{2^j}x_2^{2^i}\right) = 0.
$$
\end{itemize}
\end{theorem}
\begin{proof}
(1) Let $\displaystyle f(x)=\sum_{0 \le i<j <n} c_{ij}x^{2^i+2^j}$. Since $\{x_1,x_2,x_3,x_4\} \in \VB_{n,f}$, we have $x_1+x_2+x_3+x_4=0$ and $f(x_1)+f(x_2)+f(x_3)+f(x_4)=0$. For $a \in \Ftn$ and $1 \le \ell \le 4$, we have
\begin{align*}
f(x_\ell+a)=&\sum_{0 \le i<j <n} c_{ij}(x_\ell+a)^{2^i+2^j}\\
=&\sum_{0 \le i<j <n} c_{ij} \left(x_\ell^{2^i+2^j}+a^{2^i}x_\ell^{2^j}+a^{2^j}x_\ell^{2^i}+a^{2^i+2^j}\right)\\
=& f(x_{\ell})+f(a)+\sum_{0 \le i<j <n} c_{ij}\left(a^{2^i}x_\ell^{2^j}+a^{2^j}x_\ell^{2^i}\right)
\end{align*}
Note that $\displaystyle \sum_{\ell=1}^4 c_{ij}a^{2^i}x_\ell^{2^j}=c_{ij}a^{2^i}\left(\sum_{\ell=1}^4 x_\ell\right)^{2^j}=0$ and $\displaystyle \sum_{\ell=1}^4 c_{ij}a^{2^j} x_\ell^{2^i}=c_{ij}a^{2^j}\left(\sum_{\ell=1}^4 x_\ell\right)^{2^i}=0$. Therefore, we have $f(x_1+a)+f(x_2+a)+f(x_3+a)+f(x_4+a)=0$. Hence, for each $a \in \Ftn$, $\{x_1+a,x_2+a,x_3+a,x_4+a\} \in \VB_{n,f}$. Namely, $\VB_{n,f}$ contains all $2^{n-2}$ cosets of the flat $\{x_1,x_2,x_3,x_4\}$. Indeed, $\VB_{n,f}$ consists of a collection of $2$-dimensional vector spaces and all their cosets, which implies our claim.

(2) By the first part, $\{a,x_1+a,x_2+a,x_1+x_2+a\} \in \VB_{n,f}$ if and only if $\{0,x_1,x_2,x_1+x_2\} \in \VB_{n,f}$. Note that
$$
f(x_1+x_2)=f(x_1)+f(x_2)+\sum_{0 \le i<j <n} c_{ij}\left(x_1^{2^i}x_2^{2^j}+x_1^{2^j}x_2^{2^i}\right).
$$
Hence, $\{0,x_1,x_2,x_1+x_2\} \in \VB_{n,f}$ if and only if $\sum_{0 \le i<j <n} c_{ij}\left(x_1^{2^i}x_2^{2^j}+x_1^{2^j}x_2^{2^i}\right)=0$.
\end{proof}

Let $f(x)=\sum_{0 \le i<j<n}c_{ij}x^{2^i+2^j}$ be a DO polynomial over $\Ftn$. For $a \in \Ftn^*$, we define a linearized polynomial
$$
L_{f,a}(x)=\sum_{0 \le i<j \le n-1}c_{ij} \left(a^{2^i}x^{2^j}+a^{2^j}x^{2^i}\right).
$$
For such a DO polynomial $f$, we next define the multiset
\begin{equation}
\label{eqn-Rf}
R_f=\left[ \rank(L_{f,a}) \mid a \in \Ftn^* \right].
\end{equation}
The following theorem describes how to determine the number of vanishing flats for a DO polynomial.

\begin{theorem}
\label{thm-vanishingnumDO}
Let $f$ be a DO polynomial. Then $\displaystyle |\VB_{n,f}|=\frac{2^{n-2}}{3}\sum_{h \in R_f}\left(2^{n-h-1}-1\right)$.
\end{theorem}
\begin{proof}
For $a \in \Ftn^*$, we have $f(x+a)+f(x)=L_{f,a}(x)+f(a)$. Hence, to determine the differential spectrum of $f$, it suffices to consider the multiset $[L_{f,a}(x)+f(a) \mid x \in \Ftn]$. Define $h_a=\rank(L_{f,a})$, then the multiset $\left[ L_{f,a}(x)+f(a) \mid x \in \Ftn \right]$ contains $2^{h_a}$ elements each with multiplicity $2^{n-h_a}$. Thus, $\sum_{b \in \Ftn} \binom{\de_f(a,b)/2}{2}=2^{h_a}\binom{2^{n-h_a}/2}{2}=2^{n-2}\left(2^{n-h_a-1}-1\right)$. Hence, by Theorem~\ref{thm-diffspecVB}, we have $|\VB_{n,f}|=\frac{1}{3}\sum_{{\substack a \in \Ftn^* \\ b \in \Ftn}} \binom{\de_f(a,b)/2}{2}=\frac{2^{n-2}}{3}\sum_{h \in R_f}\left(2^{n-h-1}-1\right)$.
\end{proof}

\section{Partitioning vector spaces into disjoint equidimensional affine subspaces via vanishing flats}
\label{sec4}

Given a vector space over a finite field, a long-standing problem is to find a partition of that vector space into a collection of equidimensional vector subspaces, see for instance~\cite{B,HKK}. As an example, for a vector space $V=\Fq^{2n}$ over $\Fq$, a collection of $q^n+1$ subspaces of $V$ with dimension $n$, which contains each nonzero element of $V$ exactly once, is the well-known configuration named \emph{spread}, which has been intensively studied in finite geometry (see~\cite{LP}, for example). A perhaps less known partition was proposed by Baum and Neuwirth~\cite{BN75}, which aims to partition a vector space over a finite field into a collection of disjoint affine subspaces with equal dimension. In this section, we present an application of vanishing flats which generates this kind of partition in an elegant way. We call an $r$-dimensional affine subspace an $r$-flat. The following is a definition of~\cite{BN75}.

\begin{definition}
Let $n \ge 2$ and $1 \le d \le n-1$. A {\em cover} of $\F_{2}^n$ with dimension $d$ is a partition of $\F_{2}^n$ into $2^{n-d}$ disjoint affine subspaces each having dimension $d$. A $d$-dimensional linear subspace of $\F_2^n$ and all its cosets forms a {\em trivial cover}.
\end{definition}

Let $\{A_i\}_{i=1}^{2^{n-d}}$ and $\{B_i\}_{i=1}^{2^{n-d}}$ be two covers of $\F_2^n$ with dimension $d$. They are \emph{equivalent} if there exists an affine permutation $T$, such that $T(A_i)=B_i$, where the order of the affine subspaces in $\{B_i\}_{i=1}^{2^{n-d}}$ can be adjusted if necessary. Clearly, two trivial covers with the same dimension are equivalent. A major question is, are there any covers inequivalent to the trivial ones? In~\cite{BN75}, Baum and Neuwirth proposed to classify all covers up to equivalence, but that seems to be elusive, so far. On the other hand, they made considerable progress along this direction and in particular, they provided constructions of nontrivial covers. To describe their results, we need a few more concepts.

For an affine subspace $A_i$, we use $A_i^*$ to denote the associated linear subspace. A cover $\{A_i\}_{i=1}^{2^{n-d}}$ of $\F_2^n$ is called \emph{nonparallel} if $A_i^* \ne A_j^*$ for every distinct $i$ and $j$. A cover $\{A_i\}_{i=1}^{2^{n-d}}$ of $\F_2^n$ is called \emph{totally skew} if $A_i^* \cap A_j^*=\{0\}$ for every distinct $i$ and $j$. In \cite{BN75}, the following nontrivial covers were obtained.

\begin{result}\label{res-nonpara}
For each $n \ge 3$, there exists a nonparallel cover of $\F_2^n$ with dimension $d$, where $1 \le d \le n-2$. Note that each cover of $\F_2^n$ with dimension $(n-1)$ necessarily consists of two parallel $(n-1)$-flats. Hence, there exists a nonparallel cover of $\F_2^n$ with dimension $d$ if and only if $1 \le d \le n-2$.
\end{result}

\begin{result}\label{res-skew}
For each $n \ge 3$, there exists a totally skew cover of $\F_2^n$ with dimension $d$, where $1 \le d \le \frac{n-1}{2}$.
\end{result}

In this section, we employ DO monomials to supply new constructions of totally skew covers. Again, we use the finite field $\Ftn$ to represent the vector space $\F_2^n$. Given a polynomial $f$ over $\Ftn$ and $A=\{x_1,x_2,x_3,x_4\} \in \VB_{n,f}$, define $f(A)=\{ f(x_1), f(x_2), f(x_3), f(x_4)\}$. Moreover, for a permutation polynomial $f$ and a subset $\cA \subset \VB_{n,f}$, we define
\begin{align*}
f(\cA)=\{ \{ f(x_1), f(x_2),  f(x_3), & f(x_4) \} \mid \\
                                      &\{ x_1, x_2, x_3, x_4\} \in \cA\}.
\end{align*}
The next theorem outlines the strategy of generating covers by using permutation polynomials and their vanishing flats.

\begin{theorem}
\label{thm-covergene}
Let $f$ be a permutation polynomial over $\Ftn$. Suppose $\cA \subset \VB_{n,f}$ is a cover of $\Ftn$ with dimension $2$. Then, so is $f(\cA)$.
\end{theorem}
\begin{proof}
The property of vanishing flats implies that $f(\cA)$ is a collection of $2$-dimensional affine subspaces. The permutation property ensures that $f(\cA)$ is a cover.
\end{proof}

The above theorem indicates that given a cover with dimension $2$ contained in the vanishing flats of a permutation polynomial, a new cover can be derived by applying the permutation. Naturally, the most interesting permutations are the DO permutations, whose vanishing flats are formed by a union of trivial covers with dimension $2$:

\begin{theorem}
\label{thm-coverDO}
Let $f$ be a DO permutation over $\Ftn$, which is not APN. Then $\VB_{n,f}$ is a disjoint union of trivial covers $\cA_i$, $\displaystyle 1 \le i \le \frac{1}{3}\sum_{h \in R_f}\left(2^{n-h-1}-1\right)$, with dimension $2$,
where
$$
\cA_i=\{ \{ c,x_i+c,y_i+c,x_i+y_i+c \} \mid c \in \Ftn \}
$$
and $R_f$ is defined in \eqref{eqn-Rf}. The set $f(\cA_i)$ is a cover with dimension $2$ for each $i$. In addition, $f(\cA_i)$ is totally skew if and only if the following hold$:$
\begin{itemize}
\item[$(1)$] $\de_f(x_i)=\de_f(y_i)=\de_f(x_i+y_i)=4$,
\item[$(2)$] $E_f(x_i)$, $E_f(y_i)$ and $E_f(x_i+y_i)$ are pairwise disjoint.
\end{itemize}
\end{theorem}
\begin{proof}
The first part follows from Theorems~\ref{thm-cosetstrucDO}(1), \ref{thm-vanishingnumDO} and \ref{thm-covergene}. For distinct $c, d \in \Ftn$, write
\begin{align*}
A_i(c)&=\{c, x_i+c, y_i+c, x_i+y_i+c\} \in \cA_i,  \\
A_i(d)&=\{d, x_i+d, y_i+d, x_i+y_i+d\} \in \cA_i.
\end{align*}
Note that $f(\cA_i)$ is totally skew if and only if for every distinct $c,d \in \Ftn$, $f(A_i(c))^* \cap f(A_i(d))^*=\{0\}$. Namely,
$\{0, f(x_i+c)+f(c), f(y_i+c)+f(c), f(x_i+y_i+c)+f(c)\}$ and $\{0, f(x_i+d)+f(d), f(y_i+d)+f(d), f(x_i+y_i+d)+f(d)\}$ intersect trivially at $0$, for every distinct $c,d \in \Ftn$.  This is equivalent to the two conditions on the top of the next page.

\begin{figure*}[t]
{\flushleft (a) For each $c \in \Ftn$ and $d \in \Ftn \sm \{ c, x_i+c, y_i+c, x_i+y_i+c \}$,}
\begin{equation*}
\begin{aligned}
f(x_i+c)+f(c) &\ne f(x_i+d)+f(d), \\
f(y_i+c)+f(c) &\ne f(y_i+d)+f(d), \\
f(x_i+y_i+c)+f(c) &\ne f(x_i+y_i+d)+f(d).
\end{aligned}
\end{equation*}

{\flushleft (b) For each $c \in \Ftn$,}
\begin{equation*}
\begin{aligned}
&f(x_i+c)+f(c) \notin \{ f(y_i+d)+f(d), f(x_i+y_i+d)+f(d) \mid d \in \Ftn \sm \{ c, x_i+c, y_i+c, x_i+y_i+c \} \}, \\
&f(y_i+c)+f(c) \notin \{ f(x_i+d)+f(d), f(x_i+y_i+d)+f(d) \mid d \in \Ftn \sm \{ c, x_i+c, y_i+c, x_i+y_i+c \} \}, \\
&f(x_i+y_i+c)+f(c) \notin \{ f(x_i+d)+f(d), f(y_i+d)+f(d) \mid d \in \Ftn \sm \{ c, x_i+c, y_i+c, x_i+y_i+c \} \}.
\end{aligned}
\end{equation*}
\hrulefill
\end{figure*}

Noting that $f(c)+f(x_i+c)+f(y_i+c)+f(x_i+y_i+c)=0$, we observe that Condition (a) is equivalent to $\de_f(x_i)=\de_f(y_i)=\de_f(x_i+y_i)=4$ and Condition (b) is equivalent to $E_f(x_i)$, $E_f(y_i)$ and $E_f(x_i+y_i)$ being pairwise disjoint.
\end{proof}

By restricting to DO monomials, we expect more specific information about the associated covers.  We note that each DO monomial is CCZ-equivalent to a Gold function. In view of Theorem~\ref{thm-CCZinvariant}, we only need to consider covers derived from Gold functions. For $f(x)=x^{2^t+1}$ over $\Ftn$, $f$ is a permutation if and only if $(2^t+1,2^n-1)=1$ and is not APN if and only if $(n,t)>1$. We note that $(2^t+1,2^n-1)=1$ is equivalent to $\frac{n}{(n,t)}$ being odd. As a preparation, we have the following crucial lemma which follows from \cite[Theorem 3]{CH}.

\begin{lemma}
\label{lem-numsol}
Let $x \in \Ftn$. Suppose $s=(t,n)$ and $z \in \Fts^*$. Then the equation
$$
x^{2^t}+x=z
$$
has no solution in $\Ftn$ if and only if $\frac{n}{(n,t)}$ is odd. Moreover, suppose that $\frac{n}{(n,t)}$ is odd, then for $x,y \in \Ftn$ and $z_1, z_2 \in \Fts$, the equation
\begin{equation}
\label{eqn-numsol}
x^{2^t}+x+z_1=y^{2^t}+y+z_2
\end{equation}
holds if and only if $z_1=z_2$ and $x,y$ belong to the same additive coset of $\Fts$ in $\Ftn$.
\end{lemma}
\begin{proof}
 According to \cite[Theorem 3]{CH}, the equation $x^{2^t}+x=z$ has no solution in $\Ftn$ if and only if $\sum_{i=0}^{\frac{n}{(n,t)}-1} z^{2^{ti}} \ne 0$. Since $z \in \Fts^*$, $s=(t,n)$ and $\frac{n}{(n,t)}$ odd, we have $\sum_{i=0}^{\frac{n}{(n,t)}-1} z^{2^{ti}}=z \ne 0$ and the first part follows. For the second part, the sufficiency is clear. Note that $x^{2^t}+x+z_1=y^{2^t}+y+z_2$ is equivalent to $(x+y)^{2^t}+(x+y)=z_1+z_2$. Since $\frac{n}{(n,t)}$ is odd, then~\eqref{eqn-numsol} holds only if $z_1=z_2$, which forces $x,y$ being in the same additive coset of $\Fts$ in $\Ftn$.
\end{proof}

Now we proceed to consider DO monomials.

\begin{corollary}\label{cor-coverGold}
Let $f(x)=x^{2^t+1}$ be over $\Ftn$, where $1 \le t \le n-1$, $(2^t+1,2^n-1)=1$ and $s=(n,t)>1$. Then we have
\begin{itemize}
\item[$(1)$] $\VB_{n,2^t+1}$ is a disjoint union of trivial covers $\left\{\cA_i \mid 1 \le i \le \frac{(2^{s-1}-1)(2^n-1)}{3} \right\}$ with dimension $2$, where
    $$
      \cA_i=\left\{ \{ c,x_i+c,y_i+c,x_i+y_i+c \} \mid c \in \Ftn \right\},
    $$
    for $x_i, y_i \in \Ftn^*$ and $\frac{x_i}{y_i} \in \Fts \sm \{0,1\}$. For each $i$, $f(\cA_i)$ is a cover with dimension $2$.
\item[$(2)$] For $c \in \Ftn$, write
    $$
      A_i(c)=\{c, x_i+c, y_i+c, x_i+y_i+c\} \in \cA_i,
    $$
    where $x_i, y_i \in \Ftn^*$ and $\frac{x_i}{y_i} \in \Fts \sm \{0,1\}$. Write $x_i=\al z_i$ and $y_i=\al w_i$ for some $\al \in \Ftn^*$ and distinct $z_i, w_i \in \Fts^*$. Let $b_{\al,j}+\al\Fts$, $1 \le j \le 2^{n-s}$, be $2^{n-s}$ additive cosets of $\al\Fts$ in $\Ftn$.
    Then $f(A_i(c))$ is parallel to $f(A_i(d))$ if and only if $c,d \in b_{\al,j}+\al\Fts$, for some $j$. Otherwise, $f(A_i(c))^* \cap f(A_i(d))^*=\{0\}$. In particular, $f(\cA_i)$ can be partitioned into $2^{n-s}$ subsets $\bigcup_{c \in b_{\al,j}+\al\Fts}f(A_i(c))$, $1 \le j \le 2^{n-s}$, where each subset contains $2^{s-2}$ parallel $2$-flats and for every pair of $2$-flats from distinct subsets, their associated linear subspace intersect trivially. Consequently, for each $i$, $f(\cA_i)$ is a totally skew cover with dimension $2$ if $s=2$ and is a nontrivial cover with dimension $2$ having parallel $2$-flats, if $s>2$.
\end{itemize}
\end{corollary}
\begin{proof}
We only need to show Part (2), as Part (1) follows from Theorems~\ref{thm-PQSGold}, \ref{thm-cosetstrucDO}, and \ref{thm-coverDO}. Note that $x_i=\al z_i$, $y_i=\al w_i$, where $z_i,w_i \in \Fts^*$ are distinct. Since $z_i^{2^t}=z_i$ and $w_i^{2^t}=w_i$, we can compute $f(A_i(c))^*$ and $f(A_i(d))^*$ in the middle of the next page.

\begin{figure*}[t]
\begin{align*}
f(A_i(c))^*&=\{0, f(x_i+c)+f(c), f(y_i+c)+f(c), f(x_i+y_i+c)+f(c)\} \\
           &=\left\{0, x_ic^{2^t}+x_i^{2^t}c+x_i^{2^t+1}, y_ic^{2^t}+y_i^{2^t}c+y_i^{2^t+1}, (x_i+y_i)c^{2^t}+(x_i+y_i)^{2^t}c+(x_i+y_i)^{2^t+1} \right \} \\
           &=\left\{0, \al z_i^{2^t} c^{2^t}+\al^{2^t}z_ic+\al^{2^t+1}z_i^2, \al w_i^{2^t} c^{2^t}+\al^{2^t}w_ic+\al^{2^t+1}w_i^2\right., \\
           & \quad \quad \quad \quad \quad \quad \quad \quad \quad\left. \al (z_i+w_i)^{2^t} c^{2^t}+\al^{2^t}(z_i+w_i)c+\al^{2^t+1}(z_i+w_i)^2 \right\} \\
           &=\Bigg\{0, \al^{2^t+1}\left(\left(\frac{z_ic}{\al}\right)^{2^t}+\frac{z_ic}{\al}+z_i^2\right), \al^{2^t+1}\left(\left(\frac{w_ic}{\al}\right)^{2^t}+\frac{w_ic}{\al}+w_i^2\right), \\
         & \quad \quad \quad \quad \quad \quad \quad \quad \quad \al^{2^t+1}\left(\left(\frac{(z_i+w_i)c}{\al}\right)^{2^t}+\frac{(z_i+w_i)c}{\al}+(z_i+w_i)^2\right) \Bigg\}
\end{align*}
and similarly,
\begin{align*}
f(A_i(d))^*&=\Bigg\{0, \al^{2^t+1}\left(\left(\frac{z_i d}{\al}\right)^{2^t}+\frac{z_i d}{\al}+z_i^2\right), \al^{2^t+1}\left(\left(\frac{w_i d}{\al}\right)^{2^t}+\frac{w_i d}{\al}+w_i^2\right) \\
         & \quad \quad \quad \quad \quad \quad \quad \quad \quad \al^{2^t+1}\left(\left(\frac{(z_i+w_i)d }{\al}\right)^{2^t}+\frac{(z_i+w_i) d}{\al}+(z_i+w_i)^2\right) \Bigg\}.
\end{align*}
\hrulefill
\end{figure*}

Since $z_i, w_i \in \Fts^*$ are distinct, by Lemma~\ref{lem-numsol}, $|f(A_i(c))^* \cap f(A_i(d))^*| \ge 2$ if and only if $\left(\frac{\be c}{\al}\right)^{2^t}+\frac{\be c}{\al}+\be^2=\left(\frac{\be d}{\al}\right)^{2^t}+\frac{\be d}{\al}+\be^2$ for some $\be \in \{ z_i, w_i, z_i+w_i \}$. Each of these three cases is equivalent to $\left(\frac{c+d}{\al}\right)^{2^t}=\frac{c+d}{\al}$. Hence, $f(A_i(c))^*=f(A_i(d))^*$ if and only if $c,d \in b_{\al,j}+\al\Fts$, for some $j$. Otherwise, $f(A_i(c))^* \cap f(A_i(d))^*=\{0\}$. Consequently, $f(\cA_i)$ can be partitioned into $2^{n-s}$ subsets $\bigcup_{1 \le j \le 2^{n-s}} (\bigcup_{c \in b_j+\al\Fts} f(A_i(c)))$, where for every pair of $2$-flats from distinct subsets, their associated linear spaces intersect trivially. Note that $f(A_i(c))=f(A_i(d))$ if and only if $c+d \in \{ 0,x_i,y_i,x_i+y_i \}$. Thus, each subset $\bigcup_{c \in b_j+\al\Fts} f(A_i(c))$ contains $\frac{2^s}{4}=2^{s-2}$ parallel $2$-flats.
\end{proof}

Consequently, we have the following necessary and sufficient condition determining which DO monomial leads to totally skew covers.

\begin{corollary}\label{cor-skewcoverGold}
Let $f(x) = x^{2^t+1}$ be over $\Ftn$, where $(2^t+1,2^n-1)=1$ and  $s=(n,t)>1$. For each trivial cover
$$
\cA_i=\{ \{ c,x_i+c,y_i+c,x_i+y_i+c \} \mid c \in \Ftn \},
$$
where $x_i, y_i \in \Ftn^*$ and $\frac{x_i}{y_i} \in \Fts \sm \{0,1\}$, the cover $f(\cA_i)$ is totally skew if and only if $n \equiv 2 \pmod 4$ and $s=2$. More precisely, let $\ga$ be a primitive element of $\Ftn$. Then each trivial cover $\cA_j=\{ \{ c+\ga^j\F_4 \} \mid c \in \Ftn\}$, where $0 \le j \le \frac{2^n-4}{3}$, gives a totally skew cover $f(\cA_j)$.
\end{corollary}
\begin{proof}
We only need to prove the necessary and sufficient condition, as the rest follows immediately. The sufficiency follows from Corollary~\ref{cor-coverGold}(2). Conversely, if $f(\cA_i)$ is totally skew, by Corollary~\ref{cor-coverGold}(2), $s=(n,t)=2$. Assume $n \equiv 0 \pmod 4$. Then we have $t \equiv 2 \pmod 4$. In this case, $(2^t+1,2^n-1)=5$, which gives a contradiction. Hence, we must have $n \equiv 2 \pmod 4$.
\end{proof}

We observe that the concept of vanishing flats can be extended to higher dimensions. Let $f$ be over $\Ftn$, an $s$-flat in $\F_{2^n}$ is called a \emph{vanishing $s$-flat} of $f$, if its image under $f$ is again an $s$-flat. Clearly, a vanishing flat of $f$ is a vanishing $2$-flat. The following theorem generalizes Corollary~\ref{cor-skewcoverGold} and produces totally skew covers with higher dimensions.
For a subset $A \subset \Ftn$ and a function $f$ over $\Ftn$, we write $f(A)=\{f(a) \mid a \in A\}$.

\begin{theorem}\label{thm-coverhigh}
Let $f(x)=x^{2^t+1}$ be over $\Ftn$, with $(2^t+1,2^n-1)=1$ and $s=(n,t)>1$. For every $ \al \in \Ftn^*$, the set $\{ c+\al\Fts \mid c \in \Ftn \}$ forms a trivial cover of $\Ftn$ with dimension $s$, which is a vanishing $s$-flat of $f$. Moreover, $\{ f(c+\al\Fts) \mid c \in \Ftn \}$ is a totally skew cover of $\Ftn$ with dimension $s$.
\end{theorem}
\begin{proof}
Note that

\begin{align*}
 &f(c+\al\Fts) \\
=&\left\{ c^{2^t+1}+c^{2^t}\al z+c\al^{2^t}z^{2^t}+\al^{2^t+1}z^{2^t+1} \mid z \in \Fts \right\} \\
=&c^{2^t+1}+S_c,
\end{align*}
where
$$S_c=\left\{(c^{2^t}\al+c\al^{2^t})z+\al^{2^t+1}z^2 \mid z \in \Fts\right\}.$$
Clearly, $S_c$ is a linear subspace. Since $f$ is a permutation, the dimension of $S_c$ is $s$. Consequently, $f(c+\al\Fts)$ is an $s$-flat and hence, $c+\al\Fts$ is a vanishing $s$-flat, for each $c \in \Ftn$. To prove that $\{ f(c+\al\Fts) \mid c \in \Ftn \}$ is a totally skew cover, it suffices to show that $S_c=S_d$ if and only if $c+d \in \al\Fts$ and $S_c \cap S_d= \{0\}$, otherwise.

 Assume $S_c \cap S_d$ contains a nonzero element. Then there exist $z_1, z_2 \in \Fts^*$, such that
\begin{equation}\label{eqn-S}
(c^{2^t}\alpha+c\alpha^{2^t})z_1+\alpha^{2^t+1}z_1^2=(d^{2^t}\alpha+d\alpha^{2^t})z_2+\alpha^{2^t+1}z_2^2.
\end{equation}
Rewrite \eqref{eqn-S} as
$$
(c^{2^t}z_1+d^{2^t}z_2)\alpha+(cz_1+dz_2)\alpha^{2^t}+(z_1+z_2)^2\alpha^{2^t+1}=0.
$$
Noting that $z_1^{2^t}=z_1$ and $z_2^{2^t}=z_2$, we have
$$
(cz_1+dz_2)^{2^t}\alpha+(cz_1+dz_2)\alpha^{2^t}+(z_1+z_2)^2\alpha^{2^t+1}=0,
$$
and therefore,
$$
(\frac{cz_1+dz_2}{\alpha})^{2^t}+\frac{cz_1+dz_2}{\alpha}=(z_1+z_2)^2.
$$
Since $\frac{cz_1+dz_2}{\alpha} \in \Ftn$ and $z_1, z_2 \in \Fts^*$, by Lemma~\ref{lem-numsol}, the above equation holds only if $z_1=z_2$. By Equation~\eqref{eqn-S}, $z_1=z_2$ implies $(\frac{c+d}{\alpha})^{2^t}=\frac{c+d}{\alpha}$, and therefore, $c+d\in\al\Fts$. Conversely, if $c+d\in\al\Fts$, a direct computation shows $S_c=S_d$.
\end{proof}

\begin{remark}
Assume the conditions of Theorem~\textup{\ref{thm-coverhigh}} hold. Let $A$ be a $2$-dimensional subspace of $\al\Fts$. Suppose $\cA=\{A_i \mid 1 \le i \le 2^{n-2} \}$ is a trivial cover of $\Ftn$ formed by $A$ and all its cosets. By Corollary~\textup{\ref{cor-coverGold}(2)}, as a cover of $\Ftn$, $f(\cA)$ can be partitioned into $2^{n-s}$ subsets, where each subset contains $2^{s-2}$ parallel $2$-flats. By taking a union of all the parallel $2$-flats in each subset, we exactly recover the totally skew cover of $\Ftn$ with dimension $s$ derived in Theorem~\textup{\ref{thm-coverhigh}}.
\end{remark}

\section{Concluding remarks}\label{sec5}

In this paper, we investigated the planarity of a function from the viewpoint of vanishing flats. We showed that in some cases, the number of vanishing flats is a criterion that is finer than the differential uniformity and coarser than the differential spectrum. Moreover, the partial quadruple system formed by all the vanishing flats contains detailed information about $f$. For monomials and DO polynomials, we studied their number of vanishing flats and partial quadruple systems in more detail. As an application, we extended the concept of vanishing flats and proposed a new construction of totally skew covers using DO monomials.

We believe that the vanishing flats and the partial quadruple systems deserve further investigation. We propose next some further questions:
\begin{itemize}
\item[$(1)$] The lower bound in \eqref{eqn-VBf} is tight when $n$ is even, which is achieved by the inverse function. Are there any other monomials achieving this lower bound? Another natural question is to consider is whether~\eqref{eqn-VBf} is tight when $n$ is odd. If so, characterize the monomials achieving this lower bound.
\item[$(2)$] By Theorem~\ref{thm-cosetstrucDO}(1), the number of vanishing flats of each non-APN DO polynomial over $\Ftn$ is lower bounded by $2^{n-2}$. According to Theorem~\ref{thm-PQSGold}, DO monomials cannot achieve this lower bound. On the other hand, is there any DO polynomial having exactly $2^{n-2}$ vanishing flats?
\item[$(3)$] So far, we only know the partial quadruple systems associated with the inverse and Gold functions. It is interesting to determine the partial quadruple systems associated with other polynomials.
\item[$(4)$] Instead of the partial quadruple systems, an easier problem is to determine the number of vanishing flats associated with other polynomials. For instance, the open cases in Table~\ref{tab-VFexample} may serve as a starting point.
\item[$(5)$] In Corollary~\ref{cor-skewcoverGold}, the $\frac{2^n-1}{3}$ trivial covers correspond to $\frac{2^n-1}{3}$ totally skew covers. It is interesting to consider the equivalence problem of these totally skew covers. Moreover, we ask whether the totally skew covers derived from DO monomials are equivalent to those in \cite[Theorem 2]{BN75}.
\item[$(6)$] In Corollary~\ref{cor-coverGold}, the totally skew covers were derived from DO monomials whenever $n \equiv 2 \pmod 4$. On the other hand, there are some known DO permutations other than monomials \cite{BCHO}. Thus, we ask if any totally skew covers can be derived from these DO permutations, especially when $n \not\equiv 2 \pmod 4$.
\end{itemize}

\section*{Acknowledgement}

Shuxing Li is supported by the Alexander von Humboldt Foundation. Wilfried Meidl is supported by the FWF Project P 30966. This paper was started while the second, fifth and  sixth-named authors visited the fourth-named author at Otto von Guericke University Magdeburg in the spring of 2019. These authors would like to thank the host institution for the excellent working conditions.

config: arxiv


TITLE: Exponential equation for three known points
QUESTION [4 upvotes]: This answer (https://math.stackexchange.com/a/680695/438808) was helpful to me in finding the quadratic equation for three known points.
Is there a similar solution for finding the exponential equation (y = ab^x + c) for three known points?
Thanks in advance!

REPLY [0 votes]: I've sat down with pen and paper for a bit to try and figure this out and I've come with a solution (not very elegent but it works):
First instead of a variable base, let your equation be y=ae^bx+c. You don't have to do this but I prefer it:
Let m^x = e^bx
e^ln(m)x = e^bx
ln(m) = b
Substituting your 3 points gives 3 equations:

y1 = ae^bx1 + c
y2 = ae^bx2 + c
y3 = ae^bx3 + c

Take equations [ 1) - 2) ] / [ 1) - 3) ] and simplify to get:
(y1-y2)/(y1-y3) = (e^bx1-e^bx2)/(e^bx1-e^bx3)
Now you can solve for b (just shove it into symbolab or whatever). If an answer can't be found, then it is impossible to represent the set of points as an exponential
Now that you have b, sub this back into equations 1) - 2) and solve for a. Now that you have a, solve for c and done
I know this post is pretty old, but posting because it's one of the first links on google when you search how to fit an exponential


TITLE: A confusion about notation in Goldstein
QUESTION [7 upvotes]: On treating systems of particles, Goldstein starts with the consideration that whenever there are $k$ particles on a system, the $i$-th one obeys the relation
$$\dfrac{d}{dt}{\bf p}_i = {\bf F}_i^{(e)}+\sum_{j=1}^k {\bf F}_{ji}$$
Where ${\bf F}_{i}^{(e)}$ is the external force on the $i$-th particle and ${\bf F}_{ji}$ is the force that the $j$-th particle applies on the $i$-th one. Now if we compute the work on moving the system along a curve $\gamma$ we have:
$$W=\sum_{i=1}^k\int_\gamma {\bf F}_i = \sum_{i=1}^k \int_{\gamma}{\bf F}_i^{(e)}+\sum_{i,j=1}^k\int_{\gamma}{\bf F}_{ji}$$
Now he says the following (on p. 10)

In the special case that the external forces are derivable in terms of the gradient of a potential, the first term can be written as
  $$\sum_{i=1}^k \int_{\gamma} {\bf F}_i^{(e)} = -\sum_{i=1}^k \int_{\gamma} {\bf \nabla}_iV_i=-\sum_{i=1}^k V_i\bigg|_{a}^b$$
  where the $i$ subscript on the del symbol indicates that the derivatives are with respect to the coordinates of $\mathbf{r}_i$.

Now I can't understand why this $i$ on the del symbol. Basically we are considering the configuration manifold to be $\mathbb{R}^3$ so the natural coordinate functions are $x,y,z: \mathbb{R}^3\to \mathbb{R}$ which are just projections onto the axis and with those coordinates, there's just one del operator, namely:
$${\bf \nabla} = \dfrac{\partial}{\partial x} e_1 + \dfrac{\partial}{\partial y}e_2+\dfrac{\partial}{\partial z}e_3$$
So what really means "differentiate with respect to the coordinates of $\mathbf{r}_i$? And why do we need it? Does it have any relationship with the idea of taking the configuration space for a system of $k$ particles as $\mathbb{R}^{3k}$?

REPLY [9 votes]: This is typical to see in situations where the potential is a function of the coordinates of more than one particle:
$$
V=V(\mathbf r_1,\ldots,\mathbf r_N)=V(x_1,y_1,z_1,\ldots,x_N,y_N,z_N).
$$
The force produced by such a potential on the $i$th particle is the gradient of this function with respect to that particle's coordinates, while holding all the other particle coordinates constant. That's the meaning behind the notation
$$\nabla_i = \dfrac{\partial}{\partial x_i} e_x + \dfrac{\partial}{\partial y_i}e_y+\dfrac{\partial}{\partial z_i}e_z.$$
It is still a vector-valued quantity, but it takes different values for the different particles.
If you want to systematize the use of subscripts, then the coordinate space of your system is described by coordinate functions $x_{ki}$, where $k=1,2,3$ describes the three spatial dimensions, and $i=1,\ldots,N$ labels the particles. In that notation,
$$
\nabla_i=\dfrac{\partial}{\partial x_{ik}}e_k
$$
with Einstein summation understood.

REPLY [8 votes]: What Goldstein means by $\nabla_iV_i$ is
$$
\nabla_iV_i=\left(\frac{\partial}{\partial x_{1,i}}\hat{x}_{1,i}+\frac{\partial}{\partial x_{2,i}}\hat{x}_{2,i}+\frac{\partial}{\partial x_{3,i}}\hat{x}_{3,i}\right)V_i
$$
which is indeed a vector. Here, $\mathbf r_i=(x_{1,i},\,x_{2,i},\,x_{3,i})$ is the position of the $i$th particle (with respect to the origin), so $\nabla_i$ is the gradient of the potential from the particle.
Thinking of $\nabla_iV_i$ as some sort of "dot product" (to make it a scalar quantity) is erroneous because a dot product needs two vectors and $V_i$ is clearly a scalar quantity.


TITLE: Oscillation of non-uniform plank on parallel springs
QUESTION [1 upvotes]: A plank of length $l$ and mass $m$ is placed on two parallel springs, each with spring constant $k$ and equidistant from the plank's horizontal center of gravity. When the plank is displaced from it's equilibrium position, it's expected to oscillate at this frequency:
$$ f = \frac{1}{2\pi}\sqrt{\frac{2k}{m}} $$
Now, if the plank isn't of uniform density, such that at equilibrium, the force acting on spring 1 is $amg$ and that on spring 2 is $bmg$ where $a + b = 1$. At what frequency will the plank oscillate at when it's displaced from its equilibrium position?
Is it possible to obtain an estimate of its horizontal center of gravity at equilibrium (or rather, constants $a$, $b$) by observing the oscillations, say by taking a fourier transform of the plank's oscillation waveform?

REPLY [1 votes]: First you decided on the independent degrees of freedom. I choose the center of the plank and I track the translation $x$ and rotation $\theta$ from the equilibrium conditions.
The displacements of each spring are:
$$\begin{align}
  x_1 & = x - \frac{\ell}{2} \theta \\
  x_2 & = x + \frac{\ell}{2} \theta
\end{align} $$
Each spring force is:
$$\left. \begin{align}
  F_1 & = a m g - k x_1 \\
  F_2 & = b m g - k x_2
\end{align} \right\}
\begin{aligned}
  F_1 & = a m g + \frac{\ell}{2} k \theta - k x \\
  F_2 & = b m g - \frac{\ell}{2} k \theta - k x
\end{aligned}$$
The displacement of the center of mass is $x_C = x + \frac{\ell}{2} (b-a) \theta$ and hence the center of mass acceleration (needed for the equations of motion) is $ \ddot{x}_C = \ddot{x} + \frac{\ell}{2} (b-a) \ddot{ \theta}$.
The EOM are:
$$\begin{align}
  m \ddot{x}_C & = F_1 + F_2 - m g \\
  I_C \ddot{\theta} & = a \ell F_1 - b \ell F_2
\end{align} $$
where $I_C$ is the mass moment of inertia about the center of mass. The above is solved by $x(t) = X \sin \omega t$ and $\theta(t) = \Theta \sin \omega t$. This produces the system of equations of
$$ \begin{align}
  2 k X & = m \omega^2 \left( X + \Theta \frac{\ell}{2} (b-a) \right) \\
  k X \ell (a-b) + k \frac{\ell^2}{2} \Theta (a+b) & = I_C \Theta \omega^2
\end{align} $$
This has two solutions for frequency $\omega_T$ and $\omega_R$ for translational and rotational modes of vibration.
The two degrees of freedom are coupled with
$$ -\frac{X}{\Theta} = \frac{\frac{\ell}{2} m \omega^2 (a-b)}{ (2k-m \omega^2)}$$
The left hand side of this equation is the center of rotation position (distance) from the center of the plank. Pure translation occurs when $\omega^2 = 2 \frac{k}{m}$ and pure rotation when $a=b$.
$$\begin{align}
  \omega^2_T & = \frac{k}{m} \left(1+ \frac{m \ell^2 (1-2 a b)}{2 I_C} + \sqrt{ 1 + \left( \frac{m \ell^2 (1-2 a b)}{2 I_C} \right)^2 - \frac{2 a b m \ell^2}{I_C} } \right) \\
  \omega^2_R & = \frac{k}{m} \left(1+ \frac{m \ell^2 (1-2 a b)}{2 I_C} - \sqrt{ 1 + \left( \frac{m \ell^2 (1-2 a b)}{2 I_C} \right)^2 - \frac{2 a b m \ell^2}{I_C} } \right)
\end{align} $$
Edit 1
To estimate $a$ $b$ from the resulting motion, maybe you can solve the equations of motion using the normalized frequency $n^2 = \frac{\omega^2}{2 k/m}$ and center of rotation location $r=-\frac{X}{\Theta}$.
$$ \begin{align}
  a &= \frac{2 I_C n^2}{m \ell^2} + \frac{ r (1-n^2) (2 r+\ell)}{n^2 \ell^2} \\
  b &= \frac{2 I_C n^2}{m \ell^2} + \frac{ r (1-n^2) (2 r-\ell)}{n^2 \ell^2}
\end{align} $$


TITLE: General solution of ODE: why are there numbers as coefficients in the solution?
QUESTION [1 upvotes]: I'm asked to solve the following ODE:
$y^{(4)}-2y^{(3)}+5y^{(2)}=0$
I solve as usual, and got
$$c_1+c_2x+e^x(c_3cos(2x)+c_4sin(2x))$$
But in the solution and in WolframAlpha, they have:
$$c_1+c_2x+\frac{1}{25}(c_3(4cos(2x)+3sin(2x))+c_4(3cos(2x)-4sin(2x)))e^x$$
I don't see where this $\frac{1}{25}$ and those 3 and 4 come from. I thought that 25 is maybe the norm squared of $3cos(2x)-4sin(2x)$ but then, why do we have to normalize and where do the 3 and 4 as coefficients come from ? I thought that it's maybe because there are two sines and cosines and thus they are from the same family, but in this case, shouldn't we multiply by x instead of just a number ? Because in this case, they remain in the same family ($3cos(2x)$ and $4cos(2x)$ still belong to the same family)
Thanks for your help.

REPLY [2 votes]: Ultimately, it should be clear that they are equivalent. Let me rewrite yours as $$y=a_1+a_2x+e^x\bigl(a_3\cos(2x)+a_4\sin(2x)\bigr).$$ Setting $c_1=a_1,$ $c_2=a_2,$ $c_3=4a_3+3a_4,$ and $c_4=3a_3-4a_4,$ we then have $$y=c_1+c_2x+\frac1{25}\Bigl(c_3\bigl(4\cos(2x)+3\sin(2x)\bigr)+c_4\bigl(3\cos(2x)-4\sin(2x)\bigr)\Bigr)e^x.$$ On the other hand, starting with WA's solution, we set $a_1=c_1,$ $a_2=c_2,$ $a_3=\frac{4c_3+3c_4}{25},$ and $a_4=\frac{3c_3-4c_4}{25},$ yielding your solution.

I can't say for sure how WA obtained that solution or why they left it as they did, but here's an idea of an approach similar to that which it may have taken. Consider the ODE $$z''-2z'+5z=0\tag{$\star$}.$$ If $z(x)$ is a solution to $(\star),$ then any $y(x)$ satisfying $y''=z$ will satisfy the given ODE. On the other hand, if $y(x)$ is a solution to the given ODE, then $z=y''$ will be a solution to $(\star).$ Thus, it may be that WA began by considering $(\star),$ instead. The characteristic equation of $(\star)$ is $$\lambda^2-2\lambda+5=0,$$ which readily has solutions $\lambda=1\pm2i.$ Thus, the general solution to $(\star)$ is of one of the following equivalent forms: $$z(x)=b_1e^{(1+2i)x}+b_2e^{(1-2i)x}\\ z(x)=b_1e^{x+2xi}+b_2e^{x-2xi}\\ z(x)=\left(b_1e^{2xi}+b_2e^{-2xi}\right)e^x\\ z(x)=\left(b_1\bigl(\cos(2x)+i\sin(2x)\bigr)+b_2\bigl(\cos(2x)-i\sin(2x)\bigr)\right)e^x\\ z(x)=\left(\bigl(b_1+b_2)\cos(2x)+(b_1-b_2)i\sin(2x)\right)e^x.$$ There's no real reason to leave it this way, of course, so at this point, WA may have replaced it with $$z(x)=\bigl(c\cos(2x)+d\sin(2x))e^x,$$ or may have simply skipped right to this form.
At this point, having found $z(x),$ it may have found the general antiderivative (which would correspond to $y'$), then found the general antiderivative yet again. As I discuss in this answer, iterated integration by parts gives us the general formula $$\int e^{ax}\sin(bx)\,dx=e^{ax}\left(\frac{a}{a^2+b^2}\sin(bx)-\frac{b}{a^2+b^2}\cos(bx)\right)+C.$$ A similar approach yields $$\int e^{ax}\cos(bx)\,dx=e^{ax}\left(\frac{a}{a^2+b^2}\cos(bx)+\frac{b}{a^2+b^2}\sin(bx)\right)+C.$$ In our particular case, this means that $$\int e^x\cos(2x)\,dx=e^x\left(\frac15\cos(2x)+\frac25\sin(2x)\right)+C=\frac15\left(\cos(2x)+2\sin(2x)\right)e^x+C,$$ and similarly, $$\int e^x\sin(2x)\,dx=\frac15\left(\sin(2x)-2\cos(2x)\right)e^x+C.$$
Thus, $$y'(x)=\frac15\Bigl(c\bigl(\cos(2x)+2\sin(2x)\bigr)+d\cdot\bigl(\sin(2x)-2\cos(2x)\bigr)\Bigr)e^x+c_2,$$ or $$y'(x)=\frac15\Bigl((c-2d\bigr)\cos(2x)+\bigl(2c+d\bigr)\sin(2x)\Bigr)e^x+c_2.$$ Repeating this process again yields $$y(x)=\frac1{25}\Bigl(\bigl(c-2d\bigr)\bigl(\cos(2x)+2\sin(2x)\bigr)+\bigl(2c+d\bigr)\cdot\bigl(\sin(2x)-2\cos(2x)\bigr)\Bigr)e^x+c_2x+c_1,$$ or $$y(x)=c_1+c_2x+\frac1{25}\Bigl(c\bigl(4\sin(2x)-3\cos(2x)\bigr)-d\bigl(4\cos(2x)+3\sin(2x)\bigr)\Bigr)e^x.$$ At this point, we can let $c_3=-d$ and $c_4=-c,$ whence $$y(x)=c_1+c_2x+\frac1{25}\Bigl(c_3\bigl(4\cos(2x)+3\sin(2x)\bigr)+c_4\bigl(3\cos(2x)-4\sin(2x)\bigr)\Bigr)e^x.$$ I can't really see any reason for this final substitution, though, so I suspect WA didn't go about it in exactly this way.


TITLE: What determines the maximal dimension of the irreps of a (finite) group?
QUESTION [28 upvotes]: I am chemist and ask for apologies for all my mathematical inabilities when asking this question in advance, but after quite a bit of searching I found that this problem could be "open" or at least hard enough to find addressed in the literature and also advanced enough that it's possibly suitable to be asked here.
I work over a subset of the finite groups called point groups these are all the ((essentially) finite) subgroups of $O(3)$. The "degeneracies" arising in those groups are of importance in Chemistry so I started to investigate them. With "degeneracy" the dimensions of the irreps over $\mathbb R$, (this is of essential importance) occurring for the group under consideration is meant. Whenever a group is represented by an irrep of dimension $n>1$ we speak about $n$-fold degeneracy.
The term degeneracy in this context relates to the fact that the quantum mechanical eigenstates of (symmetric) systems/molecules form such $n$-dimensional sub spaces of the Hilbert space. Since the Hamilton operator is self-adjunct it makes sense to regard representations over $\mathbb R$ instead of over the usually more elegant $\mathbb C$.
My first question is, under which conditions in terms of group elements degeneracy can occur in a group? And the second question is how is the maximal dimension $n_{\max}$ of real irreps, over all irreps of the group, determined by the structure of the group?
My primary observation is that groups that contain exactly one generator $y$ of order $m=3$, like a group $\langle x,y,...| x^2=y^m = 1 = ...\rangle $ have $n_{\max}=2$
Then there are just a very few point groups with $n_{\max} > 2$. We call them "high symmetry groups". Basically its the symmetry groups of the tetrahedron, the octahedron and the icosahedron (with 2 or 3 certain subgroups of theirs), where the former two have $n_{\max}=3$, the latter $n_{\max}=5$.
These high symmetry groups all have two generators, the tetrahedral groups one of order $m=3$, the octahedral groups one of order $m=4$ and the icosahedral groups one of order $5$.
So I would assume that there is a connection between the order of the generators and the degeneracies that can occur in a group. What it is exactly remains very obscure to me. So I would be very grateful about any hints also to the literature.

Edit
Since the audience is so fantastically knowledgeable, I can't resist to make a small comment on the motivation of my research in the hope it might ring some bell and give rise to more inspiring comments:
The motivation of my question is, that you can see certain interesting physical properties in the states (of physical systems) if they are degenerate. "States" are some manifestations of irreps where we have direct numerical access to and a good intuition about their visual representations. It happens that there are certain, it seems, deep connections between the angular momentum operator, which is essentially an infinitesimal rotation in the (physical space) $\mathbb R^3$, and the occurrence of degeneracy (at least if it's 2-fold). At the same time we see that these states which are instances of degenerate representations are transformed into each others by (finite) rotations. Such that the question was arising if all such degeneracies are related to rotations, or if there is at least something in the structure of rotations that is general in $d>1$ dimensional representations.
There was recently a result that suggested that there is a "hidden" anti-unitary symmetry (state transforming operator) at the basis of any $2$-fold degeneracy of the form
$$ \mathcal{O} = i \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \mathcal{K}$$ (with $\mathcal{K}$ as complex conjugation operator) that transforms between the two degenerate irreps. Such an operator can be easily constructed from the angular momentum operator, for example, but not only, constructions of operators like {\mathcal{O}} based on spin or time reversal and others are possible if one considers more general "parametrizations" of physical systems then only $\mathbb R^3$. Anyway, as far as symmetries of the real space $\mathbb R^3$ are concerned it seems that rotation (and infintesimal rotation) are crucial and I aim to  understand what "crucial" here exactly means.
What I find in particular intriguing in the light of the answer from @QiaochuYuan is the connection with "non-Abelianess", because the defining relation of $J$, the angular momentum or also spin operator is the so called "angular momentum algebra", that is a commutator relation between its components
$$ [J_i,J_j] = i \varepsilon_{ijk} J_k$$
(with the Levi-Civita symbol $\varepsilon$). So this seems to suggest that angular momentum is an essential source of non-commutativity somehow. I like then to understand what types of sources else there might be, for this non-commutativity in quantum mechanical systems, especially if only representations of states in $\mathbf{R}^3$ are considered. One thing which complicates the question is that in most systems rotational symmetry "is broken" but degeneracy can nevertheless occur. Then my question would be, what is the explicit form of the $\mathcal{O}$ operator and if it can be continuously related to angular momentum if one views the "symmetry breaking" as a continuous process.
(Sorry for the post-answer edit, I hope it complies with the MO rules!)

REPLY [21 votes]: A simple bound on the largest dimension of a complex irreducible representation (which is either equal to or half of the largest dimension of a real irreducible representation) is the following: we know that

$|G| = \sum d_i^2$ where $d_i$ are the dimensions of the irreducibles,
the number of (complex) irreducibles is the number $c(G)$ of conjugacy classes, and
the size $a(G) = |G^{ab}|$ of the abelianization is the number of $1$-dimensional irreducibles (so the number of $d_i$ terms equal to $1$).

It follows that the largest dimension $d_{max}$ satisfies $a(G) + d_{max}^2 \le |G| \le a(G) + (c(G) - a(G)) d_{max}^2$, and rearranging these gives
$$\sqrt{ \frac{|G| - a(G)}{c(G) - a(G)} } \le d_{max} \le \sqrt{|G| - a(G)}.$$
$c(G)$ is a measure of "how abelian" $G$ is; it's a nice exercise to show that  $\frac{c(G)}{|G|}$ is the probability that two random elements of $G$ commute. Roughly speaking this means that $d_{max}$ is a measure of "how nonabelian" $G$ is. For example, if $G = A_5$ is the icosahedral group then $|G| = 60, a(G) = 1, c(G) = 5$ gives
$$ \sqrt{ \frac{59}{4} } \approx 3.84 \dots \le d_{max} \le \sqrt{59} \approx 7.68 \dots $$
so $4 \le d_{max} \le 7$, and since we also know that the dimensions $d_i$ divide $|G|$ we have $4 \le d_{max} \le 6$, and the true value $d_{max} = 5$ is right in the middle. Loosely speaking this says that $A_5$ is "more nonabelian" than, say, a dihedral group, which satisfies $d_{max} = 2$.
This bound is most useful when the abelianization is large. A different bound useful when the center $Z$ is large is the following: we know that

by Schur's lemma every irreducible representation has a central character, and if $\lambda : Z \to \mathbb{C}^{\times}$ is a central character, then the irreducibles with central character $\lambda$ can be identified with simple modules over the twisted group algebra obtained by quotienting $\mathbb{C}[G]$ by the relations $z = \lambda(z)$ for $z \in Z(G)$,
every twisted group algebra as above has dimension $|G/Z|$, so the dimensions $d_i(\lambda)$ of the irreducibles with central character $\lambda$ satisfy $|G/Z| = \sum d_i(\lambda)^2$,
the number of irreducibles with a fixed central character is the number of conjugacy classes of $G/Z$ satisfying a certain condition, and in particular is at most the number of conjugacy classes of $G/Z$.

Now it follows that the largest dimension $d_{max}$ satisfies $d_{max}^2 \le |G/Z| \le c(G/Z) d_{max}^2$, which gives
$$\sqrt{ \frac{|G/Z|}{c(G/Z)} } \le d_{max} \le \sqrt{|G/Z|}.$$
For example, the upper bound is tight for a finite Heisenberg group $H_3(\mathbb{F}_p)$, which satisfies $|G/Z| = p^2$ and has $p^2$ one-dimensional characters and $p - 1$ irreducibles of dimension $p$. The lower bound actually produces $1$ here which shows that it can be worse than the previous lower bound (which applied here gives $\sqrt{ \frac{p^3 - p^2}{p^2 + p - 1} } \approx \sqrt{p}$). The size of the center is another measure of "how abelian" $G$ is so this gives another sense in which $d_{max}$ measures "how nonabelian" $G$ is.