text:
\begin{document}
\thispagestyle{empty}

\maketitle
\begin{abstract}
  We propose a method for computing approximations to the Hecke eigenvalues of a classical modular eigenform $f$, based on the analytic evaluation of $f$ at points in the upper half plane.
  Our approach works with arbitrary precision, allows for a strict control of the error in the approximation, and outperforms current exact computation methods.

  AMS 2010 classification: 11F25 (11F03).

  Keywords: Hecke eigenvalues, analytic evaluation.
\end{abstract}

\section{Introduction}\label{sect:intro}

Computing Fourier coefficients and Hecke eigenvalues of modular forms has long been of interest.  The relationship between the number of points modulo a prime $p$ on an elliptic curve $E$ and the coefficients of the modular form associated to $E$ is a sample motivation.  It is well-known that for modular forms that are eigenforms for the Hecke operators (and suitably normalized), the Fourier coefficients and Hecke eigenvalues agree.  In this paper we describe a new method to compute Hecke eigenvalues of eigenforms, as well as an implementation of the method.  An analogous version of this algorithm but for Maass forms was first described by Hejhal \cite{hejhal} and optimized by Booker, Str\"ombergsson, and Venkatesh \cite{booker}.

For example, consider the Ramanujan $\tau$-function given by
\[
\sum_{n \geq 1} \tau(n)q^n = q\prod_{n\geq 1} (1-q^n)^{24} = \Delta(z)
\]
where $q=2\pi i z$ and $\Im(z)>0$; $\Delta(z)$ is the unique modular cusp form of level 1 and weight 12.  A 1947 conjecture due to Lehmer \cite{lehmer}, asserts that the Fourier coefficients of $\Delta(z)$ never vanish.  Since Fourier coefficients of $\Delta(z)$ are also its Hecke eigenvalues, it can be said that one of the oldest unsolved conjectures about modular forms can be verified by computing the Hecke eigenvalues of a modular form.

The standard way to compute Hecke eigenvalues, as is done in Sage \cite{sage} and MAGMA \cite{magma}, is by the method of modular symbols (see \cite{stein} for an algorithmic introduction to modular symbols).  The advantage of this method is that it is quite general; in principle, it can be carried out for modular forms of any weight and level.  The issue is that, as the dimension of the space of modular forms increases (e.g., as the level or weight increases) the linear algebra required for the computation of Hecke eigenvalues becomes quite difficult and inefficient, making it hard to compute the $p$th eigenvalue for moderately large $p$.  Recently, Wuthrich \cite{wuthrich} proposed the computation of modular symbols which is faster for large level compared to traditional methods of computing modular symbols.  Also recently, PARI/GP \cite{pari} has started to include methods to compute Hecke eigenvalues based on trace formulas \cite{belabas}.

A significant breakthrough appeared in the second half of the last decade with the work of Couveignes and Edixhoven and their collaborators~\cite{couveignes}.
Their main result is an algorithm that computes the Galois representation over a finite field attached to a modular eigenform of level one in time polynomial in the logarithm of the cardinality of the finite field.
One can therefore compute the coefficients of such eigenforms (in characteristic zero) via a multimodular algorithm.
Unfortunately, we are not aware of any implementations of this algorithm that are available for public use.

We propose yet another way to compute Hecke eigenvalues.  The idea is relatively simple.  For $f$ a newform, we know that the action of the Hecke operator $T_p$ gives $T_pf = \lambda_pf$, where the equality is of functions.  In other words, if we could evaluate $T_pf$ and $f$ at some fixed point $z_0$ in the complex upper half-plane $\mathbb{H}$ at which $f$ does not vanish, then
\[
\lambda_p = \frac{T_pf(z_0)}{f(z_0)}.
\]
One clear drawback to our proposed method is that we only get a numerical approximation to $\lambda_p$.  This drawback is not too problematic.  For numerical experiments on $L(f,s)$, the L-function associated to $f$, only numerical approximations to the Dirichlet series coefficients (which are determined by the Hecke eigenvalues of $f$) are required.  If an exact representation of the Hecke eigenvalue is required (e.g., as an element of a number field), since the number field in which the Hecke eigenvalue lives is known, one can use LLL to find the Hecke eigenvalue exactly.  The advantages of our method are that it allows us to compute Hecke eigenvalues more quickly than using the traditional approach and that it can easily be made parallel.
As we indicate below, extensive benchmarks also show that our method exhibits performance that compares very favorably with modular symbol approaches.

In the special case of a level one eigenform $f$, we investigate a variant of the analytic evaluation method that writes $f$ as an explicit polynomial in the Eisenstein series $E_4$ and $E_6$, and evaluates these Eisenstein series at relevant points as described above.

\subsection*{Acknowledgments:}  We thank the Fulbright Commission of Ecuador for making the collaboration of Armend\'ariz, Ter\'an and Ryan possible.  We also thank David Farmer for some helpful conversations.

\section{Modular forms background}

We define the upper half plane $\HH=\{z\in\CC:\Im(z)>0\}$ and the group
\[
\SL_2(\RR)=\left\{ \begin{pmatrix}a&b\\c&d\end{pmatrix}:a,b,c,d\in\RR\text{ and } ad-bc = 1\right\}.
\]
The group $\SL_2(\RR)$ acts on $\HH$ via fractional linear transformations: if $\gamma=\left(\begin{smallmatrix} a&b\\c&d\end{smallmatrix}\right)$ then $\gamma z=\tfrac{az+b}{cz+d}$.

We also consider the discrete subgroup $\SL_2(\mathbb{Z})$, as well as the congruence subgroups $\Gamma_0(N)$ (for $N\geq 1$) of $\SL_2(\ZZ)$ given by
\[
  \Gamma_0(N) = \left\{\left(\begin{smallmatrix} a&b\\c&d\end{smallmatrix}\right)\in \SL_2(\ZZ)\colon c\equiv 0\pmod{N}\right\}.
\]

\begin{defn}\label{defn:MF}
A modular form of level $N$ and weight $k$ is a holomorphic function $f:\mathbb{H}\to\CC$ satisfying the following:
\begin{itemize}
\item The function $f$ has the symmetry
\[
f\left(\frac{az+b}{cz+d}\right)=(cz+d)^{k}f(z)
\]
for all elements
$\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)$ in
$\Gamma_0(N)$ and all $z\in\mathbb{H}$.
\item The function $f(z)$ is bounded as $z$ approaches the cusps of $\Gamma_0(N)$.
\end{itemize}
The space of all modular forms of weight $k$ and level $N$ is denoted $M_k(N)$.
\end{defn}

\begin{rem}
We define the slash notation:
\[
f\mid_{\gamma,k} (z) = f\left(\frac{az+b}{cz+d}\right)(cz+d)^{-k}\text{ for }\gamma=\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)
\]
which allows us to write the functional equation in Definition~\ref{defn:MF} as
\[
f\mid_{\gamma,k} (z)=f(z).
\]
We also observe that for $\gamma\in \GL_2(\RR)$, the group of invertible matrices with entries in $\RR$, we can define
\[
f\mid_{\gamma,k}(z) = \det(\gamma)^{k-1}f\left(\frac{az+b}{cz+d}\right)(cz+d)^{-k}\text{ for }\gamma=\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right) .
\]
\end{rem}
Since $\left(\begin{smallmatrix} 1& 1\\0&1\end{smallmatrix}\right)\in \Gamma_0(N)$ for all $N\geq 1$, we see that modular forms are periodic and, since they are holomophic on $\HH$ and at the cusps, they admit a Fourier expansion of the form
\[f(q)=\sum_{n\geq 0} a_n(f)q^n.\]

\begin{defn} A cusp form of weight $k$ and level $N$ is a modular form $f$ of weight $k$ and level $N$ that vanishes at the cusps.  In other words, $f$ is a cusp form if its Fourier expansion is of the form $\sum_{n\geq 1}a_n(f)q^n$.  We denote the space of all cusp forms of weight $k$ and level $N$ by $S_k(N)$.
\end{defn}

The arithmetically distinguished cuspforms that we are most interested in are those that are Hecke eigenforms.  Let $f$ be a modular form of weight $k$ and level $N$.  We now define the action of the Hecke operator $T_p$ on $f$ when $p$ and $N$ are relatively prime:
\begin{equation}\label{eq:Tp-a}
  T_p(f)(z) = p^{k-1} f(pz) + \frac{1}{p}\sum_{j\pmod{p}} f\left(\tfrac{z+j}{p}\right).
\end{equation}
Similarly, when $p\mid N$ we get
\begin{equation}\label{eq:Tp-b}
  T_p(f)(z) = \frac{1}{p}\sum_{j\pmod{p}} f\left(\tfrac{z+j}{p}\right).
\end{equation}

Among Hecke eigenforms, we are particularly interested in newforms:
\begin{defn}
If $d_1d_2=N$ and $f\in M_k(d_1)$, then we also have $f\in M_k(N)$ and also $g(z)=f(d_2z)\in M_k(N)$.  The subspace of $S_k(N)$ spanned by the forms obtained in one of these ways are the old forms and the orthogonal complement of the oldforms are the newforms.  The space of newforms is denoted $S_k^{\text{new}}(N)$.
\end{defn}

There is a basis for $S_k^{\text{new}}(N)$ consisting of eigenforms for the Hecke operators $T_p$.  These basis elements, called Hecke eigenforms, are also eigenforms for the Atkin-Lehner operator $W_N$ defined as follows:  let $w_N = \left(\begin{smallmatrix} 0 & -1\\N & 0\end{smallmatrix}\right)$.  Then
\begin{align*}
W_N(f)(z) &= N^{(2-k)/2} f\mid_{w_N,k}(z)\\
&=N^{(2-k)/2} f\left(\frac{-1}{Nz}\right)N^{k-1}(Nz)^{-k}.
\end{align*}
The map $W_N$ is an involution and so $W_N(f)(z)$ is also equal to $\pm f(z)$.  We will use this map to move points away from the real axis.

Two facts about modular forms to which we will refer later on are:
\begin{enumerate}
  \item Deligne's bound on Fourier coefficients: $|a_p(f)| \leq 2p^\frac{k-1}{2}$ and, consequently, $|a_n(f)|\leq d(n) n^{\frac{k-1}{2}}$, where $d(n)$ is the number of divisors of $n$.  We will use this bound when determining how many Fourier coefficients we need in order to be able to evaluate a given modular form $f$ at a given point $z$ in $\HH$.
  \item The Fourier coefficients of a normalized eigenform are real. This follows from Hecke's theorem relating Hecke eigenvalues to Fourier coefficients, since the Hecke operators are self-adjoint for the Petersson inner product and that the eigenvalues of a self-adjoint operator are real. This fact will be used throughout our computations.
\end{enumerate}

\section{Analytic preliminaries}\label{sec:analytic}

 We begin by stating some results related to bounding the error introduced when we evaluate a given classical modular form and its image under the Hecke operator $T_p$ at a point in the upper half-plane.

\subsection{Error}

We have a quantity defined as
\begin{equation*}
  z=\frac{x}{y}\quad\text{with }x,y\in\CC.
\end{equation*}
The numerator and denominator can be approximated to $x_A$, resp. $y_A$.  In particular, we think of $x_A$ as a numerical approximation to $T_p(f)(z_0)$, for some $z_0\in\HH$, and $y_A$ as $f(z_0)$.

Given $\eps>0$, we seek values of $\eps_x$ and $\eps_y$ ensure that
\begin{equation*}
  \text{if }|x-x_A|<\eps_x\text{ and }|y-y_A|<\eps_y\text{ then }|z-z_A|<\eps?
\end{equation*}

\begin{lem}
  With the above notation, let $e_x=x-x_A$ and $e_y=y-y_A$.
  Then
  \begin{equation*}
    z-z_A=\frac{e_x-e_yz_A}{y_A+e_y}.
  \end{equation*}
\end{lem}
\begin{proof}
 Straightforward calculation
\end{proof}

\begin{prop}
  \label{prop:quot}
  For any $h\in (0,1)$, if
  \begin{equation*}
    \eps_x<\frac{h\eps|y_A|}{2}
    \quad\text{and}\quad
    \eps_y<\min\left\{\frac{(1-h)\eps |y_A|}{2|z_A|}, \frac{|y_A|}{2}\right\},
  \end{equation*}
  then $|z-z_A|<\eps$.
\end{prop}
\begin{proof}
 Under the hypotheses, we have $|y_A+e_y|>|y_A|/2$ so
  \begin{equation*}
    |z-z_A|<\frac{2}{|y_A|}\left(|e_x|+|e_yz_A|\right)
    <h\eps+(1-h)\eps=\eps.
  \end{equation*}
\end{proof}

The value of the parameter $h$ can be chosen in such a way that the calculations of $x_A$ and of $y_A$ are roughly of the same level of difficulty.

In addition to the error in a quotient, we also point out that we need to take into account error in a sum.  In particular, to compute $T_p(f)(z_0)$ for some $z_0\in\HH$ we need to evaluate $f$ at $p+1$ (respectively, $p$) if $p$ does not divide (respectively, does divide) the level $N$.  In particular, suppose $x=\sum_{i=1}^N x^{(i)}$, $x_A = \sum_{i=1}^N x_A^{(i)}$ and that $|x^{(i)}-x^{(i)}_A|<\eps_i$.  We want an $\eps_i$ so that $|x^{(i)}-x^{(i)}_A|<\eps_i$ guarantees $|x-x_A|<\eps$.  By the triangle inequality, we can let $\eps_i = \tfrac{\eps}{N}$.

\subsection{Truncation error for normalised eigenforms}
Let $f\in S_k^\text{new}(\Gamma_0(N))$ be a classical newform of weight $k$ with Fourier expansion
\begin{equation*}
  f(z)=\sum_{n=0}^\infty a_n(f) e^{2\pi i nz}.
\end{equation*}
Given $T\in\NN$, we consider the truncated expansion
\begin{equation*}
  f_T(z)=\sum_{n=0}^T a_n(f) e^{2\pi i nz}.
\end{equation*}
Let
\begin{equation*}
  d=\begin{cases}
    \frac{k+1}{2} & \text{if $k$ is odd},\\
    \frac{k+2}{2} & \text{if $k$ is even}.
  \end{cases}
\end{equation*}

\begin{prop}
  \label{prop:truncate}
  Let $\eps>0$ and $y=\Im(z)$.
  If $T$ is such that
  \begin{equation*}
    T\geq \frac{d}{2\pi y}\qquad\text{and}\qquad
    \frac{d+1}{2\pi y}e^{-2\pi yT}T^d<\eps,
  \end{equation*}
  then $|f(z)-f_T(z)|<\eps$.
\end{prop}
\begin{proof}
  We have
  \begin{equation*}
    |f(z)-f_T(z)|
    =\left|\sum_{n=T+1}^\infty a_n(f) e^{2\pi i nz}\right|
    \leq\sum_{n=T+1}^\infty |a_n(f)| e^{-2\pi y n}.
  \end{equation*}
  Using Deligne's bound we have
  \begin{equation*}
    |a_n(f)|\leq d(n) n^{(k-1)/2}\leq n^{(k+1)/2}\leq n^d,
  \end{equation*}
  which gives us
  \begin{equation*}
    |f(z)-f_T(z)|\leq \sum_{n=T+1}^\infty n^d e^{-2\pi yn}.
  \end{equation*}
  Letting $g(x)=x^de^{-2\pi yx}$ we find easily that $g$ is decreasing for $x\geq d/(2\pi y)$.
  By hypothesis we have $T\geq d/(2\pi y)$, so we can compare the above infinite series with the improper integral
  \begin{equation*}
    |f(z)-f_T(z)|\leq \int_T^\infty x^de^{-2\pi yx} dx
    =\frac{e^{-2\pi yT}}{2\pi y}\sum_{j=0}^d\frac{d!}{j!(2\pi y)^{d-j}}T^j.
  \end{equation*}
  For any $j$ between $0$ and $d$, since $T\geq d/(2\pi y)$ we have
  \begin{equation*}
    \frac{(j+1)\cdots d}{(2\pi y)^{d-j}}\leq \left(\frac{d}{2\pi y}\right)^{d-j}<T^{d-j},
  \end{equation*}
  so
  \begin{equation*}
    |f(z)-f_T(z)|\leq \frac{e^{-2\pi yT}}{2\pi y}(d+1)T^d,
  \end{equation*}
  which is smaller than $\eps$ by hypothesis.
\end{proof}

The conditions on $T$ make it clear that larger values of $y$ allow for shorter truncations.

When our modular form is of level 1, we can use the modularity of $f$ to move the evaluation from the point $z=x+iy$ to another point $(az+b)/(cz+d)$ that has a larger imaginary part:
\begin{equation*}
  \Im\left(\frac{az+b}{cz+d}\right)=\frac{y}{(cx+d)^2+c^2y^2}.
\end{equation*}
The group $\SL_2(\ZZ)$ is generated by the matrices
\begin{equation*}
T = \begin{bmatrix} 1&1\\0&1 \end{bmatrix}
  \qquad\text{and}\qquad
S=\begin{bmatrix} 0 & 1\\-1&0 \end{bmatrix}.
\end{equation*}
The former does not change the imaginary part, so it is enough to consider the latter:
\begin{equation*}
  z=x+iy\longmapsto -\frac{1}{z}=-\frac{x}{x^2+y^2}+\frac{y}{x^2+y^2}\,i.
\end{equation*}
This leads us to a very simple approach: given $z=x+iy$, first use the periodicity of $f$ to ensure $|x|\leq 1/2$.
If $x^2+y^2\geq 1$, then evaluate $f$ at $z$; otherwise evaluate $f$ at $-1/z$ and use
\begin{equation*}
  f(z)=z^{-k}f\left(-\frac{1}{z}\right).
\end{equation*}

\begin{rem}\label{rem:inversion}
  In the case of modular forms of level $N>1$, it is harder to guarantee a large imaginary part.
  For example, consider $\Gamma_0(2)$.
  Let $\mathcal{F}$ be a fundamental domain for $\SL_2(\ZZ)$
\[
\mathcal{F}=\left\{z\in\HH: |\Re(z)| \leq \tfrac12 \text{ and } |z|\geq 1\right\},
\]
and $S$ and $T$ be the matrices defined above.  Then a fundamental domain for $\Gamma_0(2)$ is $\mathcal{F}\cup ST\mathcal{F}\cup S\mathcal{F}$.  Since $S\mathcal{F}$ extends down towards the origin, there is no obvious way to move a purely imaginary number with a small imaginary part (say less than 1) to a point with a larger imaginary part (say greater than 1).

We show how to overcome this difficulty in the cases $N=2,3$ by using Atkin-Lehner operators.
For $N=1$, the matrix $\left(\begin{smallmatrix}0 & -1\\1&0\end{smallmatrix}\right)$ sends a point inside the unit circle to one outside the unit circle.
This inversion along with translations (recall $f(z+m)=f(z)$ for all $m\in \ZZ$) allows us to assume an imaginary part bigger than $1$.
The matrix $w_N = \left(\begin{smallmatrix}0&-1\\N&0\end{smallmatrix}\right)$ is inversion across the circle centered at the origin and of radius $1/\sqrt{N}$.
Figure~\ref{fig:atkin} shows the outcome of applying this inversion for the cases $N=1,2,3,4$ in the computation of the Hecke operator $T_5$.
We can see that the imaginary part increases after applying $w_N$, but that for $N=4$ (and higher) this increase becomes too negligeable to be useful.
Related techniques (such as using $q$-expansions with respect to cups other than $i\infty$) may be appropriate for higher levels.
\end{rem}

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  \draw[fill,red] (i2) circle (0.02);
\end{tikzpicture}
\caption{The Atkin-Lehner operator $W_N$ performs inversion in the circle of radius $1/\sqrt{N}$.  These circles are illustrated above for $N=1,\dots,4$.  We see that the minimal guaranteed imaginary part decreases from $\sqrt{3}/2\approx 0.866$ in level $1$ to $0$ in level $4$ (and higher). We marked the points at which we need to evaluate $f$ in order to estimate $T_5(f)$ at $z=i$, as well as the corresponding points after optimizing the imaginary part via the Atkin-Lehner operator.}
\end{figure}

\section{Implementation details}

We now describe some of the details of our implementation, which can be found at~\cite{code}.
All inexact arithmetic, as well as elementary function evaluations, are done using ball arithmetic via Fredrik Johansson's library arb~\cite{arb}, as packaged in Sage.
This allows us to control the propagation of errors at the level of simple operations.

\subsection{Guaranteed error bounds}

In Section~\ref{sec:analytic} we stated and proved some error bounds.  Recall the notation of Proposition~\ref{prop:quot}.  In particular, we approximate $\lambda_p$ ($z_A$ in the notation of the Proposition) by taking the quotient $\tfrac{T_p(f)(z_0)}{z_0}$ (denoted $\tfrac{x_A}{y_A}$ in the Proposition).  We want to compute $\lambda_p$ to a chosen number of digits of accuracy.

In order to use the results of Proposition~\ref{prop:quot} in practice, we need a lower bound on $|y_A|$  and an upper bound on $|z_A|$ (which can be obtained from the lower bound on $|y_A|$ and an upper bound on $|x_A|$).

How do we bound $|x_A|$?  We compute a very coarse estimate $\tilde{x}$ to $x$, with $\tilde{\eps}_x$ just small enough that $|\tilde{x}|-2\tilde{\eps}_x>0$.  (We can start with $\tilde{\eps}_x=0.1$ and keep dividing by $10$ until the condition holds.)  Later we will make sure that $\eps_x$ is smaller than $\tilde{\eps}_x$.

Then we know that
\begin{equation*}
  |\tilde{x}-x|<\tilde{\eps}_x\qquad\text{and}\qquad
  |x_A-x|<\eps_x\leq\tilde{\eps}_x,
\end{equation*}
so
\begin{equation*}
\big||x_A|-|\tilde{x}|\big|\leq |x_A-\tilde{x}|<2\tilde{\eps}_x\qquad\Rightarrow\qquad 0< |\tilde{x}|-2\tilde{\eps}_x < |x_A| < |\tilde{x}|+2\tilde{\eps}_x,
\end{equation*}
giving us lower and upper bounds on $|x_A|$.  A similar argument works for $|y_A|$.

\subsection{Finding where to truncate}

Our code finds the $T$ described in Proposition~\ref{prop:truncate} as follows.   We increment $T$ by 1 until both conditions of Proposition~\ref{prop:truncate} are met.

The user is free to choose any value of $z_0$ at which to evaluate the modular form; when $z_0$ is chosen to have a moderately large imaginary part, then $T=1$ satisfies both conditions.  This is problematic because then the difference $|f(z)-f_T(z)|$ becomes zero and we are not really approximating anything. In this case, we arbitrarily increase $T$ to 100.

\subsection{Optimizing the choice of $z_0$}

Our code allows for the user to choose any value of $z_0$ at which to evaluate the modular form $f$.  When $z_0$ is closer to the real axis, our algorithm slows down considerably.  For such $z_0$ we need to compute a very large number of Fourier coefficients since in Proposition~\ref{prop:truncate} the value of $T$ becomes quite large.

As described in Remark~\ref{rem:inversion}, we use inversion across the unit circle or the circle centered at the origin and of radius $1/\sqrt{N}$ to make the imaginary part bigger.

\subsection{A mirror relation and the case for the imaginary axis}
\begin{prop}
  Let $N\in\NN$ and let $\{a_n\}$ be a collection of real numbers indexed by integers $n\in\ZZ$.
  Let
  \begin{equation*}
    f(z)=\sum_{n=-\infty}^{\infty} a_n e^{2\pi i nz/N},\qquad
    \text{for }z\in\HH.
  \end{equation*}
  Then
  \begin{equation*}
    f(-\overline{z})=\overline{f(z)}\qquad
    \text{for all }z\in\HH.
  \end{equation*}
\end{prop}

\begin{rem}  We point out that, even though the sum in Proposition goes from $-\infty$ to $+\infty$, since we are considering Hecke eigenforms, the sums we consider below go from $1$ to $+\infty$.
\end{rem}

\begin{proof}
  Write $z=x+iy\in\HH$, then $-\overline{z}=-x+iy\in\HH$.
  We have
  \begin{align*}
    e^{2\pi i nz/N} &= e^{-2\pi ny/N}
    \left(\cos(2\pi nx/N)+i\sin(2\pi nx/N)\right)\\
    e^{-2\pi i n\overline{z}/N} &= e^{-2\pi ny/N}
    \left(\cos(2\pi nx/N)-i\sin(2\pi nx/N)\right)
    =\overline{e^{2\pi i nz/N}},
  \end{align*}
  from which the claim follows because the coefficients $a_n$ are in $\RR$.
\end{proof}

\begin{cor}
  Given $f$ as in the Proposition:
  \begin{enumerate}[(a)]
    \item $f(iy)\in\RR$ for all $y\in\RR_{>0}$;
    \item if $y\in\RR_{>0}$, $p$ is prime and $b\in\{1,\dots,p-1\}$ then
      \begin{equation*}
        f\left(\frac{iy-b}{p}\right)=
        \overline{f\left(\frac{iy+b}{p}\right)}.
      \end{equation*}

\item Let $f\in M_k(\SL_2(\ZZ))$.  The summands in $T_pf(iy)$ (for $p>2$) come in pairs of conjugate complex numbers, which allows us to reduce the necessary computation in half.  In particular,
\begin{equation*}
  (T_p f)(iy) = p^{k-1} f(iyp) +\frac{1}{p} f\left(i\frac{y}{p}\right)
  + \frac{2}{p}\Bigg(\Re f\left(\frac{iy+1}{p}\right)+\dots+
  \Re f\left(\frac{iy+\frac{p-1}{2}}{p}\right)\Bigg).
\end{equation*}
  \end{enumerate}
\end{cor}
\begin{proof}
  Apply the Proposition with (a) $z=iy$ and (b) $z=(iy+b)/p$.  For (c) we apply parts (a) and (b) and use the fact that $f(z+1)=f(z)$.
\end{proof}

\begin{rem}
When $p=2$ there are no savings to be gained by using this symmetry.  Also, the corollary allows us to carry out computations over the real numbers.
\end{rem}

\subsection{Comparison of the two methods}

In this section, we compare the modular symbol method as implemented in Sage \cite{sage} to our method.  In Figure~\ref{fig:err} we see the results of the following computation:  for each $p$ we compute the $p$th Hecke eigenvalue in both ways (using modular symbols and using our analytic approach) and take $\log_{10}$ of the absolute value of the difference.  We see in Figure~\ref{fig:err} that for a fixed $F\in S_{24}(1)$ the two methods agree to at least $10^{-8}$.  In this section we use $\lambda_p$ to denote the $p$th Hecke eigenvalue computed via modular symbols and $\tilde{\lambda}_p$ to denote the $p$th Hecke eigenvalue approximated using our analytic approach.

\begin{figure}
\begin{center}
\includegraphics[width=3in]{f24-error-plot.png}
\end{center}
\caption{Let $\lambda_p$ be the $p$th Hecke eigenvalue computed using modular symbols and $\tilde{\lambda}_p$ be the $p$th Hecke eigenvalue approximated using our analytic approach.  The plot here is $\log_{10}(|\lambda_p-\tilde{\lambda}_p|)$.}\label{fig:err}
\end{figure}

In Table~\ref{tbl:classical-timings} we highlight some timings in which we computed Hecke eigenvalues using both methods.  The results of our method are computed via code like
\begin{verbatim}
sage: F12 = Newforms(1, 12, names='a')[0]
sage: p = next_prime(10000)
sage: err = 0.1
sage: e = eigenvalue_numerical(F12, p, err)
sage: e
-5.758585642481476962744000?e21
\end{verbatim}
and the results of the modular symbols method are computed via code like\footnote{We have chosen this somewhat cryptic way of computing the eigenvalue via modular symbols over more obvious and readable ones for two reasons: (a) Sage uses aggressive caching of various intermediate results, which makes it hard to obtain useful timings from repeated evaluations of the same eigenvalue; (b) it is the quickest way of getting the eigenvalue, since the more user-facing methods perform other bookkeeping that is not absolutely necessary for our purpose and can easily take up the bulk of the running time.  For instance, \texttt{e = F12[p]} for the example given in the text takes a little over 9 minutes, compared to about 1 second for the more verbose modular symbols method.}
\begin{verbatim}
sage: F12 = Newforms(1, 12, names='a')[0]
sage: p = next_prime(10000)
sage: ms = F12.modular_symbols(1)
sage: T = ms._eigen_nonzero_element(p)
sage: e = ms._element_eigenvalue(T, name='b')
sage: e
-5758585642481476962744
\end{verbatim}

  \begin{table}[h]
    \centering
    \begin{tabular}{rrr|rrr}
      \toprule
      level & weight & $p$ & modular symbols & analytic & Eisenstein\\
      \midrule
      $1$ & $12$ & $\numprint{   1009}$ &   $0.073$ &   $\mathbf{0.051}$ & $0.071$\\
        && $\numprint{  10007}$ &   $0.967$ &   $\mathbf{0.557}$ & $0.847$\\
        && $\numprint{ 100003}$ &  $13.710$ &  $\mathbf{7.270}$ & $9.274$\\
        && $\numprint{1000003}$ & $163.515$ & $\mathbf{79.494}$ & $100.782$\\
      \midrule
    $1$ & $24$ & $\numprint{   1009}$ &   $0.158$ &   $\mathbf{0.069}$ & $0.085$\\
        && $\numprint{  10007}$ &   $2.023$ &   $\mathbf{0.751}$ & $0.935$\\
        && $\numprint{ 100003}$ &  $27.190$ &  $\mathbf{8.742}$ & $9.941$\\
        && $\numprint{1000003}$ & $330.854$ & $\mathbf{95.078}$ & $108.923$\\
      \midrule
    $1$ & $100$ & $\numprint{101}$ & $0.057$ & $0.043$ & $\mathbf{0.013}$\\
        && $\numprint{   1009}$ &   $0.918$ &   $0.233$ & $\mathbf{0.146}$\\
        && $\numprint{  10007}$ &   $14.197$ &   $1.938$ & $\mathbf{1.913}$\\
        && $\numprint{ 100003}$ &   $186.873$ &  $24.117$ & $\mathbf{23.449}$\\
      \midrule
    $1$ & $200$ & $\numprint{101}$ &  $0.171$ & $1.160$ & $\mathbf{0.024}$\\
        && $\numprint{   1009}$ &   $2.443$ &   $3.101$ & $\mathbf{0.347}$\\
        && $\numprint{  10007}$ &   $36.195$ &   $9.965$ & $\mathbf{5.104}$\\
        && $\numprint{ 100003}$ &   $500.388$ &  $84.217$ &  $\mathbf{69.916}$\\
      \midrule
      $2$ & $8$ & $\numprint{   1009}$ &   $\mathbf{0.050}$ &   $0.055$\\
          && $\numprint{  10007}$ &   $0.681$ &   $\mathbf{0.596}$\\
          && $\numprint{ 100003}$ &   $8.452$ &   $\mathbf{6.627}$\\
          && $\numprint{1000003}$ & $108.695$ & $\mathbf{75.283}$\\
      \midrule
      $2$ & $48$ & $\numprint{   1009}$ &   $0.360$ &   $\mathbf{0.108}$\\
          && $\numprint{  10007}$ &   $4.927$ &   $\mathbf{1.326}$\\
          && $\numprint{ 100003}$ &   $67.787$ &   $\mathbf{15.784}$\\
      \midrule
      $3$ & $6$ & $\numprint{   1009}$ &   $\mathbf{0.041}$ &   $0.055$\\
          && $\numprint{  10007}$ &   $\mathbf{0.542}$ &   $0.615$\\
          && $\numprint{ 100003}$ &   $\mathbf{6.809}$ &   $6.881$\\
          && $\numprint{1000003}$ & $87.462$ & $\mathbf{78.406}$\\
      \midrule
      $3$ & $24$ & $\numprint{   1009}$ &   $0.206$ &   $\mathbf{0.097}$\\
          && $\numprint{  10007}$ &   $2.642$ &   $\mathbf{1.070}$\\
          && $\numprint{ 100003}$ &   $34.894$ &   $\mathbf{12.209}$\\
      \bottomrule
    \end{tabular}
    \caption{A summary of timings (in seconds; the \textbf{bold} figures identify the shortest time in each row) to compute Hecke eigenvalues in different ways.  The first uses the modular symbols method as implemented in Sage; the second way uses analytic evaluation of the eigenform itself; the third way, valid only in level $1$, uses analytic evaluation of the Eisenstein series $E_4$ and $E_6$.  The computations were done on one Intel i7-8550U core at 1.80GHz, on a machine with 16 GB RAM.  The benchmarks were repeated several times using the \texttt{pytest-benchmark} framework and the median timing is reported here.}\label{tbl:classical-timings}
  \end{table}

A clear difference between the two methods is the number of Fourier coefficients required for the computation of the $p$th Hecke eigenvalue.  In Figure~\ref{fig:T} we see how few coefficients are needed to compute $\tilde{\lambda}_p$.

\begin{figure}
\begin{center}
\includegraphics[width=3in]{f24-T-plot.png}
\end{center}
\caption{For each $p$ we calculate the number of coefficients of $F\in S_{24}(1)$ to compute $\tilde{\lambda}_p$, our approximation to the $p$th Hecke eigenvalue $\lambda_p$ using our analytic approach.  In principle, to compute $\lambda_p$ using the Fourier expansion of the modular form, one would need the first $p$ coefficients of the modular form.}\label{fig:T}
\end{figure}

\subsection{Improvements in Level 1} In the previous sections we described how to evaluate $f$ using its Fourier expansions and in Table~\ref{tbl:classical-timings} we provide some timings using this method.  In this section we focus on the specific case of level 1 and describe a faster way to evaluate $f$.  It is well known that the ring of modular forms of level 1 is generated by the Eisenstein series of levels 4 and 6.  What we do in this case is the following:  Let $z\in\HH$:
\begin{enumerate}
\item\label{step1} calculate roughly $k/12$ coefficients of the modular form $f$, $E_4$ and $E_6$;
\item\label{step2} identify a basis for $S_k(1)$ by finding $T = \{(a,b)\in\ZZ:4a+6b=k-12 \}$ and computing the first $k/12$ coefficients of $\Delta\cdot E_4^a E_6^b$ for all $(a,b)\in T$;
\item express $f$ as a linear combination of the basis elements found in Step \ref{step2} and replace $\Delta$ with $\tfrac{E_6^2-E_4^3}{1728}$;
\item\label{step4} calculate $x=E_4(z)$, $y=E_6(z)$ and the evaluate $f(z)$ by taking the appropriate algebraic combination of $x$ and $y$.
\end{enumerate}
\begin{rem}  Some comments on the steps are in order:
\begin{itemize}
\item In Step~\ref{step1}, the number of coefficients needed for large weight is much smaller than the number of coefficients needed to evaluate $f$ using its Forier expansion.
\item In Step~\ref{step4}, we use Fredrik Johansson's arb package \cite{arb}, available in Sage \cite{sage} to evaluate $E_4$ and $E_6$.
\end{itemize}
\end{rem}

Numerical experiments suggest that this approach via evaluation of Eisenstein series is particularly beneficial in the case of large weights (see Table~\ref{tbl:classical-timings}).

\bibliographystyle{plain}
\bibliography{hecke}

\end{document}


TITLE: What is the difference in conditions of a free electron in metal and a isolated free electron in space-time?
QUESTION [1 upvotes]: I am asking that what environment does the free electrons in metal have,what is net electric fields,force experienced by them.does they have transitions like valence electrons.can they be compared to an isolated electron?.

REPLY [0 votes]: The free electron model (FEM) was derived mostly by Sommerfeld in 1927. It does not yield band structures, so there's no energy bands, i.e. it cannot predict the existence of insulators and semiconductors. For such a description, one has to look at the nearly free electron model, which is able to describe energy bands.
In the FEM, the electrons have to satisfy Pauli's exclusion principle, so they cannot have any energy, unlike in the case of an isolated free electron in vacuum. Also, the electron-electron interaction is ignored, so there's no electric field involved in the Hamiltonian of the FEM.
One difference between the FEM and an isolated electron is the concept of temperature, which is well defined in the former case, but not in the latter. At absolute zero temperature, there is a maximum energy the electrons can possibly have in a metal, called the Fermi energy. But in the electron in vacuum case, the concept of temperature is ill defined.


TITLE: Why does the Born approximation for the scattering amplitude depend on the potential $V$ everywhere in space, unlike classical scattering?
QUESTION [2 upvotes]: The expression for the first-order Born approximation to the scattering amplitude goes like
$$
f(\theta) \propto \int d^3 r e^{-i ( k - k_i)r} V(r)
$$
and so in particular the value of $ V(r) $ everywhere is relevant. However, if we want to consider for example scatters at $ \theta \sim 0 $ it would seem that the only relevant parts of the potential are those far away from the origin. Classically, particles with high impact parameters scatter at $ \theta \sim 0, $ and it does not "see"  the potential at the origin at all during its orbit. Perhaps this is because the Born approximation is only approximate, but isn't it effective in the regime where $ V$ is small, which is certainly when $ r$ is big?
I am looking for a more physical/intuitive explanation of this difference between classical and quantum scattering rather than a mathematical rederivation of the Born approximation (unless of course there is a very helpful derivation).

REPLY [3 votes]: The answer, given by @user196574 is exshaustive, I only want to add, that integration over all space arises solely due to the choice of basis solution - a plane wave, with a fixed momentum, which exists everywhere in space. If you want to consider a scattering of localized object, for example, a Gaussian wavepacket, then performing first integral over momenta $k$:
$$
\int d^3 k \int d^3 r \ e^{-i (k - k_0) r} V(r) e^{-\alpha k^2 / 2} =
\ \int d^3 r \ V(r) e^{-(r-r_0)^2 / 2 \alpha}
$$
you have an expression, localized in space (decaying fast with the increase of distance).


TITLE: Stabilizer of the orthogonal group of the Minkowski form acting on the upper sheet of hyperboloid space
QUESTION [0 upvotes]: I was studying this paper and on page 8 the author defines the orthogonal group associated to the Minkowski linear form in the usual way ( $A \in O(1,n) \Longleftrightarrow \langle Ax | Ay \rangle =  \langle x|y \rangle \ \forall x, y \in \mathbb{R}^{1,n}$, where $\mathbb{R}^{1,n}$ is $\mathbb{R}^{n+1}$ with the Minkowski linear form). $\mathbb{H}^n$ is defined as the $x :  \langle x | x \rangle  = -1$ and $x_0 > 0$.
The part that I can't prove is that the stabilizer of $\mathbb{H}^n$ is equal to the subset of $O(1,n)$ such that the first column belongs to $\mathbb{H}^n$. I did prove that if $A \in Stab_{O(1,n)} \ \mathbb{H}^n$ then $A_0$ must belong to $\mathbb{H}^n$, but the reverse is escaping me.

REPLY [1 votes]: Intuitively speaking you have this: you know that any $A\in O(1,n)$ will map the point $x_0=(1,0,0,\dots,0)$ to an element of the two-sheeted hyperboloid. It could be the upper sheet or the lower sheet. With the extra condition of $A_{00}>0$ you know it will be mapped to the upper sheet. So one point from the upper sheet goes to a point on the upper sheet. From this you want to conclude that all points on the upper sheet stay on the upper sheet.
You will need some kind of continuity argument here. Something that says that if a point $x$ stays on the upper sheet, then a point $y$ close by will stay there, too. Assume this were not the case. Then you would have a continuous path from $x$ to $y$ made of points on the upper sheet, which gets mapped to something that lies partially on the upper sheet and partially on the lower sheet. This cannot be done without a discontinuity, and linear functions are continuous.
This is just a proof idea. You may need to flesh this out to make it rigorous enough for your formalism, but it should be enough to guide your approach.


text:
\begin{document}

{\let\thefootnote\relax\footnotetext{\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! MSC 2010:\,\,\,\,17B30, 17B55, 17B56, 05E15\\ Keywords: nilpotent k-ary Lie algebras, k-ary Lie algebra homology  \\
Contact: \href{mailto:emre-sen@uiowa.edu}{emre-sen@uiowa.edu} }}

\title{on the Homology of Nilpotent k-ary  Lie Algebras}

\begin{abstract} We introduce nilpotent k-ary Lie algebras including analogues of Heisenberg Lie algebras and free nilpotent Lie algebras.
We study homology of k-ary nilpotent Lie algebras by using a modification of Chevalley-Eilenberg complex. For some classes of nilpotent k-ary Lie algebras and in particular Heisenberg k-ary Lie algebras we give explicit formulas for Betti numbers. Representation stability of free nilpotent k-ary Lie algebras is proven and lower bounds for Betti numbers are described by Schur modules. We also verify that toral rank conjecture holds for the classes we studied. Moreover, for 2-step nilpotent k-ary Lie algebras, we give a refinement of it.

\end{abstract}

\author{Emre SEN}

\maketitle

\section{Introduction}

k-ary Lie algebra $\mathcal{L}$ is a vector space equipped with k-bracket
\begin{center}
$[.,\ldots,.]:\cL\times\ldots\times\cL\longrightarrow \cL$
\end{center}
satisfying:
\begin{enumerate}[label=\roman*)]
\item antisymmetry i.e. $[x_1,\ldots,x_k]=sgn(\sigma)[x_{\sigma(1)},\ldots,x_{\sigma(k)}]$ where $\sigma\in\mathfrak{S}_k$ and
\item \label{jacobi} generalized Jacobi identity:
\begin{center}
$[[x_1,x_2\ldots,x_k],x_{k+1},\ldots,x_{2k-1}]=\sum\limits^{k}_{i=1}[x_1,x_2,\ldots,x_{i-1},[x_i,x_{k+1},\ldots,x_{2k-1}],x_{i+1},\ldots,x_k]$
\end{center}
\end{enumerate}

It appears in the works of Filippov \cite{filippov26n} and Hanlon, Wachs \cite{hanlon1995lie}.
We can define \emph{lower central series} for k-ary Lie algebras as:
\begin{center}
$\cL\supset \underbrace{\left[\cL,\ldots,\cL\right]}_{k}\supset \underbrace{\left[\underbrace{\left[\cL,\ldots,\cL\right]}_{k},\cL,\ldots,\cL\right]}_{2k-1}\supset\ldots$
\end{center}
We call k-ary Lie algebra \emph{nilpotent} if lower central series terminates. In particular if the bracket of size greater than $k$ vanishes we call them \emph{ $2$-step nilpotent k-ary Lie algebras}. In other words, $\cL$ is 2-step nilpotent k-ary Lie algebra if and only if k-commutator $[\cL,\ldots,\cL]$ is contained in the center of $\cL$.
The aim of this paper is to introduce some classes of k-ary nilpotent Lie algebras and describe their homology by using a generalization of Chevalley-Eilenberg complex. Even in the classical set up $k=2$, computation of Betti numbers is a hard task, there are only limited cases that we know them explicitly. Most of them are dedicated for 2-step nilpotent (super) Lie algebras \cite{armstrong1997explicit}, \cite{sigg1996laplacian}, (\cite{getzler1999homology}), Heisenberg Lie algebras \cite{santrab} and some free nilpotent Lie algebras of small dimensions \cite{tirao2002homology}. Also, recently, it can be viewed as a problem of representation stability in the sense of Church and Farber \cite{church2013representation}. From this perspective homology of 2-step nilpotent Lie algebras, super Lie algebras and their analogues studied by S. Steven \cite{sam2013homology}.

Thus, we focus on k-ary generalizations of the above cases i.e. k-ary Heisenberg Lie algebras, free nilpotent k-ary Lie algebras etc. aiming to carry results on the homology of nilpotent Lie algebras into homology of nilpotent k-ary Lie algebras.

A real Heisenberg Lie algebra $H$ is $2m+1$ dimensional algebra with basis \\$\{x_1,\ldots,x_m,y_1,\ldots,y_m,z\}$ satisfying $[x_i,y_j]=\delta_{i,j}z$. Its homology is given in \cite{santrab}:
\begin{align*}
\vert H^i{(H)}\vert={2m\choose i}-{2m\choose i-2}
\end{align*}

We consider the following generalization of it: let $\hH$ be $km+1$ dimensional algebra with basis $\{x^1_{1},\ldots,x^1_{m},x^2_1,\ldots,x^2_m,\ldots,x^k_{m},z\}$ such that:
\begin{center}
$[x^1_{i},x^2_{i},\ldots, x^k_{i}]=z$
\end{center}
We prove that:
\begin{theorem}
Betti numbers are:
\begin{align}
\vert H^{ik-(i-1)}\left(\hH\right)\vert={km\choose i(k-1)+1}-{km\choose (i-1)(k-1)}
\end{align}
\end{theorem}

Another class of 2-step nilpotent Lie algebras was introduced by Armstrong, Cairns and Jessup in \cite{armstrong1997explicit}. We define k-ary analogue of it (see definition \ref{defACJ}), and study their homology in section \ref{homACJ}.\\

The following is a generalization of free Lie algebra for k-ary Lie algebras:
\begin{definition}\cite{friedmann2017generalization} The free k-ary Lie algebra on the set $X$ is a k-ary Lie algebra $\cL$ together with mapping $e:X\mapsto \cL$ with the following universal property: for each k-ary Lie algebra $\cK$ and for each mapping $f:X\mapsto \cK$ there is a unique k-ary Lie algebra homomorphism $F:\cL\mapsto\cK$ such that $f=F\circ e$
\end{definition}

The reason why we consider free k-ary Lie algebras is due to the well-known construction. Let $V$ be a vector space, $T(V)$ be its  tensor algebra. Let $F(V)=\{v\otimes u-u\otimes v\vert u,v\in T(V)\}$. $F(V)$ is free Lie algebra and it has a natural grading with respect to degree. Graded piece $F(V)_d$ of degree $d$ becomes both representation of symmetric group $\mathfrak{S}_d$ and $GL(V)$ which creates a rich combinatorics.

 The question of which irreducible representations appear in $F(V)_d$ solved by Klyachko \cite{klyachko1974lie}.
 The first few summands of $F(V)$ are:
\begin{gather}\label{freelie}
F(V)=V\oplus S_{11}V \oplus S_{21}V\oplus \left(S_{31}V\oplus S_{21^2}V\right)\oplus \cdots
\end{gather}
where $S_{\lambda}V$ is representation parametrized by Young diagram $\lambda$.
Free nilpotent Lie algebras are quotients of $F(V)$ i.e. taking quotient by elements with degrees greater then $d$ gives d-step nilpotent free Lie algebra. For example, 2-step free nilpotent Lie algebra is
\begin{align}
\mathfrak{n}_2=V\oplus \bigwedge^2 V,\hspace{1cm} [x,y]=x\wedge y
\end{align}

As an algebra, their structure is relatively simple but homologies are complicated, only 2-step case $\mathfrak{n}_2$ was solved, see \cite{jozefiak1988representation}, \cite{sigg1996laplacian}, \cite{sam2013homology}.

The decomposition of free k-ary Lie algebras into irreducible components is still an open problem, there are particular results    \cite{friedmann2017generalization}. For example, the second and the third graded components are parametrized by diagrams $\lambda=(1^k)$ and $\lambda=(2^{k-1},1)$ respectively.
We study homologies of free 2-step nilpotent k-ary Lie algebras:
\begin{align}
\mathfrak{n}^k_2=V\oplus\bigwedge^kV,\hspace{1cm} [x_1,x_2,\ldots,x_k]=x_1\wedge x_2\wedge\cdots\wedge x_k
\end{align}

We prove that each homology is a Schur module. However, we do not have general description of homologies, because to find irreducible summands of the complex $\left(\bigwedge\mathfrak{n}^k_2\right)$, the plethsym problem i.e. decomposition of $\bigwedge^j(\bigwedge^kV)$, $j,k\geq 3$ has to be solved. Nevertheless, we study lower bounds of Betti numbers in section \ref{sectionFree}.

For small dimensional case $\dim V=k$ we compute the homology of 3-step nilpotent k-ary Lie algebras. In section \ref{homfree3}, we show:
\begin{theorem} The ranks of homologies of free 3-step nilpotent k-ary Lie algebra $\mathfrak{n}^k_3$ of dimension $k+1$ are:
\begin{gather*}
\vert H^1(\mathfrak{n}^k_3)\vert=k\\
\vert H^k(\mathfrak{n}^k_3)\vert={2k+1\choose k}-\left(3k+2\right)\\
\vert H^{2k-1}(\mathfrak{n}^k_3)\vert=\left(2k+1\right)\left(k-1\right)
\end{gather*}

\end{theorem}

Beyond the explicit computation of Betti numbers, there are questions concerning the total homology of Lie algebras. For instance, the toral rank conjecture states that for any nilpotent Lie algebra $\mathfrak{n}$, the inequality
 \begin{align}
\sum\limits^{\dim\mathfrak{n}}_{i=0} H^i\left(\mathfrak{n}\right)\geq 2^{\dim\mathfrak{z}}
\end{align}
is valid where $\mathfrak{z}$ is the center of $\mathfrak{n}$. We verify that this holds for the algebras (definitions  \ref{defheis},\ref{defACJ},\ref{deffree2}, \ref{deffree3}) which are subject to this work In particular, for 2-step nilpotent k-ary Lie algebras, we improve the lower bound of total homology in theorem \ref{thmRefinement} by generalizing arguments of \cite{tirao2000refinement}.

Another open question about Lie algebra homology is the property M of Hanlon \cite{hanlon}, which relates total homology of current Lie algebra to homology of Lie algebra itself :

\begin{align}\label{defPropertyM}
H^*\left(\gG\otimes \faktor{\Cc[t]}{t^j}\right)=H^*\left(\gG\right)^{\otimes j}
\end{align}

For example, semisimple Lie algebras satisfy it. However, there are counterexamples to this conjecture, since then the purpose is to find algebras satisfying \ref{defPropertyM} above. Even the case of Heisenberg Lie algebras (k=2) of dimension greater or equal than $5$ remains unknown \cite{hanlon2002property}. We show that for k-ary Heisenberg Lie algebras with $k\geq 5$ Hanlon's conjecture \ref{defPropertyM} is not true.

Organization of the paper is : first we give definitions of some classes of k-ary Lie algebras and compare them with the classical (k=2) case in Section \ref{sectionExamples}. In the \ref{defResolution}rd section  we study their homologies. For k-ary Heisenberg Lie algebras, we give explicit Betti numbers. In section \ref{sectionFree}, we obtain lower bounds of Betti numbers and investigate which Schur modules can appear for some particular cases of free 2-step nilpotent k-ary Lie algebras. We discuss representation stability of some cases in \ref{sectionRep}. The last section \ref{sectionToral} is about toral rank conjecture  of nilpotent k-ary Lie algebras.

\section{Examples of Nilpotent k-ary Lie Algebras}\label{sectionExamples}
Here we present some examples of nilpotent k-ary Lie algebras:

\subsection{Generalization of Heisenberg Lie algebra}
The algebra H with basis $\{x,y,z\}$ and relation $[x,y]=z$ is called Heisenberg Lie algebra which is a central subject in many fields. We consider the following generalization of it:
\begin{definition}\label{defheis} A $km+1$ dimensional algebra $\hH$ with basis
\begin{center}
$\{x^1_{1},\ldots,x^1_{m},x^2_1,\ldots,x^2_m,\ldots,x^k_{m},z\}$
\end{center}
 and with the only nonzero k-bracket (up to permutation):
\begin{center}
$[x^1_{i},x^2_{i},\ldots x^k_{i}]=z$
\end{center}
is called \emph{k-ary Heisenberg Lie algebra}.
\end{definition}
It is clear that $[\hH,\ldots,\hH]=z$ which implies that $\hH$ is 2-step nilpotent k-ary Lie algebra.
\subsection{A class of nilpotent-ACJ}\label{defACJ} The following is generalization of the algebra studied in \cite{armstrong1997explicit}. Let $\gG$ be Lie algebra with basis $\{z,x^1_1,\ldots,x^1_m,x^2_1,\ldots x^2_m,\ldots,x^{k}_1,\ldots, x^k_m\}$ and with the bracket:
\begin{align}
[z,x^1_i,x^2_i,\ldots,x^{k-1}_i]=x^k_i
\end{align}
Center of the algebra is $\left\{x^k_1,\ldots,x^k_{m}\right\}$ which is equal to k-commutator of the algebra.

\subsection{Free 2-step nilpotent k-ary Lie algebras}\label{deffree2}
Let $V$ be vector space over $\Cc$ with $\dim V\geq k$. We consider $\mathfrak{g}=V\oplus\bigwedge^kV$ with respect to bracket:

\begin{center}
$\underbrace{\gG\times\cdots\times\gG}_{k-many}\mapsto \gG$\\
$[v_1,\ldots,v_k]=\begin{cases}
v_1\wedge\ldots\wedge v_k &\text{each}\quad v_i\in V \\
\quad\quad 0 &\text{otherwise}
\end{cases}$
\end{center}

Since the bracket is antisymmetric and satisfies generalized Jacobi identity identically, $\gG$ is free k-ary Lie algebra. Moreover, it is nilpotent, since $[\gG,\ldots,\gG]=\bigwedge^kV$.

\subsection{Free 3-step nilpotent Lie algebras}\label{deffree3}
To illustrate the main idea here, we choose $k=3$. Let $X$ be set, $X=\left\{x,y,z\right\}$. We use the Hall Basis $\left\{x,y,z,[xyz],[xy[xyz]],[yz[xyz]],[xz[xyz]]\right\}$ for the free 3-ary Lie algebra $\cL$ on the set $X$, where $[,,]$ is 3-bracket. It is nilpotent since the term $\left[\left[\left[\left[\cL,\cL,\cL\right],\cL,\cL\right],\cL,\cL\right],\cL,\cL\right]$ of lower central series vanishes.

\section{Chevalley-Eilenberg Complex and Homology}\label{defResolution}
Let $\gG$ be k-ary Lie algebra over $\Cc$. We use the below chain complex which mimics Chevalley-Eilenberg complex:

\begin{gather}
\begin{aligned}
\xymatrixcolsep{5pt}
\xymatrix{& \ar[rr]&&\bigwedge^{3k-2}\gG\ar[rrrr]^{\partial_{3k-2}}  &&&& \bigwedge^{2k-1}\gG\ar[rrrr]^{\partial_{2k-1}} &&&& \bigwedge^{k}\gG\ar[rrrr]^{\partial_{k}}  &&&&  \gG\ar[rr] &&\Cc
}
\end{aligned}
\label{defcomplex}
\end{gather}
with the differential

\begin{align}\label{defdiff}
\partial_t(x_1\wedge\ldots\wedge x_t)=\sum\limits_{\sigma\in D_{t,k}}sgn(\sigma)[x_{\sigma(1)},\ldots,x_{\sigma(k)}]\wedge x_{\sigma(k+1)}\wedge \cdots \wedge x_{\sigma(t)}
\end{align}

where
\begin{align}
D_{t,k}=\left\{\sigma\in\mathfrak{S}_t\,\,\vert\,\, \sigma(1)<\sigma(2)<\cdots<\sigma(k)\quad\text{and}\quad\sigma(k+1)<\cdots<\sigma(t)\right\}
\end{align}

This complex appears in \cite{hanlon1995lie}, and they compute homology of free k-ary Lie algebra $F(V)$. We just refer to their result:
\begin{proposition} $\partial^2=0$
\end{proposition}
\begin{proof}
See \cite{hanlon1995lie} theorem 1.6.
\end{proof}

Therefore, we can use chain complex \ref{defcomplex}, as generalization of Chevalley-Eilenberg complex, hence we define homology with trivial coefficients according to \ref{defdiff} i.e.
\begin{align}\label{defhomology}
H^{ik-i+1}(\gG)={\Large \cfrac{\ker\partial_{ik-i+1}}{\im\partial_{(i+1)k-i}}}
\end{align}

\subsection{Homology of k-ary Heisenberg Lie Algebras}\label{homHeis}

We prove the following:

\begin{theorem}\label{thmHeis} Betti numbers of k-ary Heisenberg Lie algebra $\hH$ are:

\begin{align}
\left|H^{ik-(i-1)}\left(\hH\right)\right|={km\choose i(k-1)+1}-{km\choose (i-1)(k-1)}
\end{align}
\end{theorem}
\begin{corollary}  In particular, when $k=2$ this recovers the well known result of \cite{santrab}.
\end{corollary}

Since $\left|H^{ik-(i-1)}\left(\hH\right)\right|=\left|\ker\partial_{ik-(i-1)}\right|-\left|\im\partial_{(i+1)k-i}\right|$ and $\dim\bigwedge^{ik-(i-1)}\left(\hH\right)=\left|\ker\partial_{ik-(i-1)}\right|+\left|\im\partial_{ik-(i-1)}\right|$ it suffices to compute the dimension of the images of the boundary maps. Answer is:
\begin{proposition}\label{thmHeisImage} The dimension of the image of $\partial_{ik-(i-1)}$ in $\bigwedge^{(i-1)k-(i-2)}\hH$ is
\begin{align}\label{heisenbergformula}
{km\choose (i-1)(k-1)}
\end{align}
\end{proposition}

\begin{proof}
We prove it by induction on the dimension of $\hH$. For simplicity, first we analyze the case $k=3$. Then we prove it for arbitrary $k$.\\
Let $k=3$, $m=1$. Hence $\hH$ has basis $\{x,y,z,w\}$ with bracket $[x,y,z]=w$. Therefore $\vert\im \partial_{3}\vert =1$ which is the value of \ref{heisenbergformula} at $i=1$.
Assume that the formula \ref{heisenbergformula} holds for $m$. Let $\hH'=\hH\oplus \Cc^3$ where $\Cc^3=<x,y,z>$. Now

\begin{align}
\bigwedge^{2i+1}\hH'=\bigwedge^{2i+1}\hH\oplus\left(\bigwedge^{2i}\hH\wedge \Cc^3\right)\oplus\left(\bigwedge^{2i-1}\hH\wedge\bigwedge^2\Cc^3\right)\oplus\left(\bigwedge^{2i-2}\hH\wedge\bigwedge^3\Cc^3\right)
\end{align}
The map $\partial^{'}_{2i+1}:\bigwedge^{2i+1}\hH'\mapsto \bigwedge^{2i-1}\hH'$ decomposes as:
\begin{align}
\partial^{'}_{2i+1}=\partial_{2i+1}\oplus\left(\partial_{2i}\otimes\Cc^3\right)\oplus\left(\partial_{2i-1}\otimes\bigwedge^2\Cc^3\right)\oplus\left(\partial_{2i-2}\otimes\bigwedge^3\Cc^3\right)
\end{align}
Therefore
\begin{gather}
\dim\!\im\partial^{'}_{2i+1}\!=\!\dim\!\im\partial_{2i+1}\!+\!\dim\!\im\left(\partial_{2i}\otimes\Cc^3\right)+\dim\!\im\left(\partial_{2i-1}\otimes\bigwedge^2\Cc^3\right)+\dim\!\im\left(\partial_{2i-2}\otimes\bigwedge^3\Cc^3\right)=\nonumber\\\dim\im\partial^{'}_{2i+1}=\dim\im\partial_{2i+1}+3\dim\im\partial_{2i}+3\dim\im\partial_{2i-1}+\dim\im\partial_{2i-2}\\
\dim\im\partial^{'}_{2i+1}={3m\choose 2i-2}+{3m\choose 2i-3}.3+{3m\choose 2i-4}.3+{3m\choose 2i-5}\\
\dim\im\partial^{'}_{2i+1}={3m+3\choose 2i-2}
\end{gather}
This finishes the proof for $k=3$

We follow similar arguments to prove the formula \ref{heisenbergformula} for arbitrary $k\geq 4$.
Let $m=1$, hence $\hH$ has basis $\{x^1,\ldots,x^k,w\}$ with bracket $[x^1,\ldots,x^k]=w$. The size of the image of $\partial_k$ is $1$ which coincides with \ref{heisenbergformula} at $i=1$. Assume that formula holds for $m$. Let $\hH'=\hH\oplus\Cc^k$ where $\Cc^k=\left\langle x^1,\ldots x^k \right\rangle$. We have the following decomposition of $\bigwedge^{ik-(i-1)}\hH'$ and the map $\partial_{ik-(i-1)}$:

\begin{align}
\bigwedge^{ik-(i-1)}\hH'=\bigoplus\limits^{k}_{j=0}\left(\bigwedge^{ik-(i-1)-j}\hH\wedge\bigwedge^j\Cc^k\right)
\end{align}

\begin{align}
\partial^{'}_{ik-(i-1)}=\bigoplus\limits^{k}_{j=0}\left(\partial_{ik-(i-1)-j}\otimes \bigwedge^{j}\Cc^{k}\right)
\end{align}

Therefore:
\begin{gather}
\dim\im\partial^{'}_{ik-(i-1)}=\sum\limits^{k}_{j=0}\dim\im\left(\partial_{ik-(i-1)-j}\otimes \bigwedge^{j}\Cc^{k}\right)=\\
\dim\im\partial^{'}_{ik-(i-1)}=\sum\limits^{k}_{j=0}\dim\im\left(\partial_{ik-(i-1)-j}\right){k\choose j}=\\
\dim\im\partial^{'}_{ik-(i-1)}=\sum\limits^{k}_{j=0}{km\choose ik-(i-1)-j}{k\choose j}=\\
\dim\im\partial^{'}_{ik-(i-1)}={km+k\choose ik-(i-1)}
\end{gather}
This finishes the proof.
\end{proof}

By using the proposition \ref{thmHeisImage}, we present the proof of the theorem \ref{thmHeis}:

\begin{proof}
Since $\dim\bigwedge^{ik-(i-1)}\left(\hH\right)=\left|\ker\partial_{ik-(i-1)}\right|+\left|\im\partial_{ik-(i-1)}\right|$, the size of the kernel is:
\begin{align}
{km+1\choose ik-(i-1)}-{km\choose (i-1)(k-1)}
\end{align}
Therefore by using $\left|H^{ik-(i-1)}\left(\hH\right)\right|=\left|\ker\partial_{ik-(i-1)}\right|-\left|\im\partial_{(i+1)k-i}\right|$ we get:

\begin{gather}
\left|H^{ik-(i-1)}\left(\hH\right)\right|={km+1\choose ik-(i-1)}-{km\choose (i-1)(k-1)}-{km\choose i(k-1)}=\\
{km\choose ik-(i-1)}+{km\choose ik-i}-{km\choose (i-1)(k-1)}-{km\choose i(k-1)}=\\
{km\choose ik-(i-1)}-{km\choose (i-1)(k-1)}
\end{gather}
or equivalently
\begin{align}
{km\choose i(k-1)+1}-{km\choose (i-1)(k-1)}
\end{align}
since $ik-(i-1)=i(k-1)+1$, which completes proof.
\end{proof}

\begin{remark} The formula is valid with constraints on the dimensions, there is no Poincare type duality theorem for the complex \ref{defcomplex}. Simply, we can choose
\begin{gather}
\alpha\leq \lfloor\cfrac{km+1}{2}\rfloor
\end{gather}
since we want $\bigwedge^{\alpha}\hH$ be the largest dimensional space, where $\alpha=ik-(i-1)$ and $\lfloor-\rfloor$ is floor function.
Substitution gives:
\begin{gather}
i\leq \cfrac{1}{k-1}\left(\lfloor\cfrac{km+1}{2}\rfloor-1\right)
\end{gather}
\end{remark}

\subsection{Homology of a k-ary ACJ Lie algebra}\label{homACJ}
Let $\gG$ be Lie algebra with basis\\ $\{z,x^1_1,\ldots,x^1_m,x^2_1,\ldots x^2_m,\ldots,x^{k}_1,\ldots, x^k_m\}$ and with the bracket:
\begin{align}
[z,x^1_i,x^2_i,\ldots,x^{k-1}_i]=x^k_i
\end{align}
The case  when $k=2$ were introduced by Armstrong, Cairns and Jessup in \cite{armstrong1997explicit}. $i$th Betti number is
\begin{align}
{m+1\choose \lfloor\frac{i+1}{2}\rfloor}{m\choose\lfloor\frac{i}{2}\rfloor}
\end{align}
where $\lfloor i\rfloor$ denotes the integer part of $i$. To compute it, they use derivation arising from adjoint map and then solve a recursion. We consider a similar reduction.

A k-ary analogue of adjoint map can be made in many ways, we choose:
\begin{gather}
ad(z):\bigwedge^{k-1}\gG\mapsto\gG\\
ad(z)(x_1\wedge\cdots\wedge x_{k-1})=[z,x_1,\ldots,x_k]
\end{gather}

The ACJ type k-ary Lie algebras given in \ref{defACJ} can be seen as a specific example of more general set up. Let $\gG$ be k-ary Lie algebra such that $\mathfrak{a}$ is codimension one abelian ideal in $\gG$ i.e.
\begin{gather}
\gG=\langle z\rangle\oplus\mathfrak{a}\\
[\gG,\ldots,\gG,\mathfrak{a}]\subseteq\mathfrak{a}
\end{gather}

Then:
\begin{proposition}
Homology of $\gG$ at $\alpha=ik-(i-1)$ is:
\begin{gather}\label{f1}
\vert H^{\alpha}(\gG)\vert=
{ km\choose \alpha}-{km\choose \alpha+k-2}+\dim\ker\theta_{\alpha-1}+\dim\ker\theta_{\alpha+k-2}
\end{gather}
where $\theta_{\alpha}:=ad(z)\left(\bigwedge^{\alpha}\mathfrak{a}\right)$.
\end{proposition}

\begin{proof}
We can express boundary maps in terms of $\theta$:
\begin{gather}
\vert H^{\alpha}(\gG)\vert=\vert\ker\partial_{\alpha}\vert-\vert\im\partial_{\alpha_{\alpha+k-1}}\vert=\\
\dim\bigwedge^{\alpha}\mathfrak{a}+\dim\ker\theta\left(\bigwedge^{\alpha-1}\mathfrak{a}\right)-\dim\theta\left(\bigwedge^{\alpha+k-2}\mathfrak{a}\right)=\\
\dim\bigwedge^{\alpha}\mathfrak{a}+\dim\ker\theta\left(\bigwedge^{\alpha-1}\mathfrak{a}\right)+\dim\ker\theta\left(\bigwedge^{\alpha+k-2}\mathfrak{a}\right)-\dim\bigwedge^{\alpha+k-2}\mathfrak{a}=\\
{\dim \mathfrak{a}\choose \alpha}-{\dim \mathfrak{a}\choose \alpha+k-2}+\dim\ker\theta_{\alpha-1}+\dim\ker\theta_{\alpha+k-2}
\end{gather}
The expression \ref{f1} holds.
\end{proof}

Therefore computation of the homology reduces to the computation of the dimension of the kernels of $\theta$.
The action of $\theta$ is :
\begin{gather}
\theta_{j}:\bigwedge^j\mathfrak{a}\mapsto\bigwedge^{j-k+2}\mathfrak{a}\\
\theta_j\left(x_1\wedge\cdots\wedge x_j\right)=\sum\limits_{i_1,\ldots, i_{j}} ad(z)(x_{i_1}\wedge \cdots\wedge x_{i_{j}})\wedge x_1\wedge \cdots\wedge\widehat{x_{i_a}}\wedge\cdots\wedge x_{j}=\\
\sum\limits_{i_1,\ldots, i_{j}}[z,x_{i_1},\ldots x_{i_{j}}]\wedge x_1\wedge \cdots\wedge\widehat{x_{i_a}}\wedge\cdots \wedge x_{j}
\end{gather}

We do not know how to find $\ker \theta_j$, since they are subject to recursive relations, and even the case $k=2$ requires work \cite{armstrong1997explicit},\cite{armstrong1996cohomology}. By straightforward computation, we only give the second homology:

\begin{proposition}
The second homology of ACJ type Lie algebra $\gG$ is
\begin{gather}
\left| H^k(\gG)\right|={km+1\choose k}-m{km-k\choose k-1}-{m+1\choose 2}
\end{gather}
\end{proposition}
\begin{proof}
Consider the sequence:
\begin{align}
\bigwedge^{2k-1}\gG\rightarrow\bigwedge^k\gG\rightarrow \gG
\end{align}
Then the image of $\partial_k$ is $\left\{x^k_1,\ldots,x^k_m\right\}$. Therefore the size of the kernel of $\partial_k$ is
\begin{align}
{km+1\choose k}-m
\end{align}
The possible contributors to image are the cycles of the forms:
\begin{enumerate}[label=\roman*)]
\item Cycles containing 2-pairs i.e. $z\wedge x^i_1\wedge\cdots\wedge x^i_{k-1}\wedge x^j_{1}\wedge\cdots\wedge x^j_{k-1}$, in total this gives ${m\choose 2}$ elements in $\bigwedge^k\gG$.
\item Cycles containing 1 pair i.e. $z\wedge x^i_1\wedge\cdots\wedge x^i_{k-1}\wedge *\wedge\cdots\wedge *$. The image is of the form $x^i_k\wedge *$, the total number is
\begin{align}
m\left({km-k+1\choose k-1}-{km-k\choose k-2}\right)
\end{align}
Therefore homology is
\begin{gather}
{km+1\choose k}-m-m\left({km-k+1\choose k-1}-{km-k\choose k-2}\right)-{m\choose 2}=\\
{km+1\choose k}-m{km-k\choose k-1}-{m+1\choose 2}
\end{gather}
\end{enumerate}
\end{proof}

\subsection{Homology of 3-step nilpotent k-ary Lie algebras of small rank}\label{homfree3}

Let $\gG$ be $3$-step nilpotent free k-ary Lie algebra on $V$ with $\dim V=k$ where $k\geq 4$. We fix Hall basis of $\gG$ as:\\
There are $k$ elements $x_1,\ldots,x_k$ of degree $1$\\
There is one element $y=[x_1\ldots x_k]$ of degree $k$\\
There are $k$ elements $[x_1,\ldots,\widehat{x_i},\ldots,y] $ of degree $2k-1$.

The complex below computes homology:
\begin{align}\label{f2}
\xymatrixcolsep{5pt}
\xymatrix{0\ar[rr]&&\bigwedge^{2k-1}\gG\ar[rrrr]^{\partial_{2k-1}} &&&& \bigwedge^{k}\gG\ar[rrrr]^{\partial_{k}}  &&&&  \gG \ar[rr] && \Cc
}
\end{align}

Image of $\partial_{k}$ is explicitly degree $k$ and $2k-1$ terms, hence $\vert\im\partial_k\vert=1+k$ and $\vert\ker\partial_k\vert={2k+1\choose k}-(k+1)$\\
\begin{lemma} The image of $\partial_{2k-1}$ is of size $k+1$.
\end{lemma}
\begin{proof}
Elements of $\bigwedge^{2k-1}\gG$ which does not contain at least $k-1$ degree one terms are in the kernel. In total we have four possibilities:
\begin{enumerate}[label=\roman*)]
\item All linear terms are in cocycle, no $y$, gives cocyles of $y$ and degree $2k-1$ elements, total size is ${k\choose k-1}$
\item All linear terms are in cocycle, with $y$, gives cycles of the form
\begin{center}
$\left( x_i\wedge z_i+x_j\wedge z_j\right)\wedge z_a\wedge z_b$
\end{center}
 $i,j$ different than $a,b$ and $z_i=[x_1,\ldots,\widehat{x_i},\ldots,x_k,y]$. There are only k many terms.
\item all but one linear terms, no y are in the kernel.
\item all but one linear terms with y, cycle of degree 2k-1 terms, there is only one element.
\end{enumerate}
Hence in total, $\vert\im\partial_{2k-1}\vert=k+1+k=2k+1$. Therefore kernel is ${2k+1\choose 2k-1}-(2k+1)$.
\end{proof}
By combining these results, we get:
\begin{theorem}
Homology of free 3-step nilpotent k-ary Lie algebra with $\dim V=k$ is:
\begin{gather*}
\vert H^1(\gG)\vert=k,\hspace{1cm}
\vert H^k(\gG)\vert={2k+1\choose k}-(3k+2),\hspace{1cm}
\vert H^{2k-1}(\gG)\vert=(2k+1)(k-1)
\end{gather*}
\end{theorem}

\paragraph{$k=3$}
Let $\gG$ be $3$-step nilpotent free 3-ary Lie algebra on $V$ with $\dim V=3$. The resolution is very similar to the previous case \ref{f2}, only the length of the complex is $4$, instead of $3$. We use the following Hall basis:\\
There are $3$ elements $x_1,x_2,x_3$ of degree 1\\
There is $1$ element $y=[x_1, x_2, x_3]$ of degree $3$\\
There are $3$ elements $[x_1, x_2,y], [x_1, x_3,y], [x_2,x_3,y]$ of degree $5$.

Chevalley-Eilenberg complex of $\gG$ is:
\[\xymatrixcolsep{5pt}
\xymatrix{0\ar[rr]&&\bigwedge^{7}\gG\ar[rrrr]^{\partial_7}&&&&\bigwedge^{5}\gG\ar[rrrr]^{\partial_{5}} &&&& \bigwedge^{3}\gG\ar[rrrr]^{\partial_{3}}  &&&&  \gG \ar[rr] && \Cc
}\]

Image of $\partial_3$ is of size $4$, hence kernel of $\partial_3$ is ${7\choose 3}-4=31$.
Elements of $\bigwedge^{5}\gG$ which does not contain at least $2$ degree one terms are in the kernel. Hence we have four cases for the image of $\partial_5$:
\begin{enumerate}[label=\roman*)]
\item All linear terms are in cycle, no $y$, gives cycles of $y$ and degree $5$ elements, total size is $3$
\item All linear terms are in cocycle, with $y$, gives elements of the form:
$\partial(x_1\wedge x_2\wedge x_3\wedge y\wedge z_1)=-x_3\wedge z_1\wedge z_3+x_2\wedge z_1\wedge z_2$\\
$\partial(x_1\wedge x_2\wedge x_3\wedge y\wedge z_2)=x_3\wedge z_2\wedge z_3+x_1\wedge z_1\wedge z_2$\\
$\partial(x_1\wedge x_2\wedge x_3\wedge y\wedge z_3)=-x_2\wedge z_2\wedge z_3-x_1\wedge z_1\wedge z_3$\\
They are linearly independent, hence image is $3$ dimensional for these cycles.
\item all but one linear terms, no y are in the kernel.
\item all but one linear terms with y: cycle of degree $5$ terms, there is only one element.
\end{enumerate}
Hence in total, $\vert\im\partial_{5}\vert=7$. Image of $\partial_7$ is zero which is obvious. By combining these results, we get:

\begin{theorem}\label{thm3}
Homology of free 3-step nilpotent 3-ary Lie algebra with $\dim V=3$ is:
\begin{gather*}
\vert H^1(\gG)\vert=3,\hspace{1cm}
\vert H^3(\gG)\vert=24,\hspace{1cm}
\vert H^{5}(\gG)\vert=14\hspace{1cm}
\vert H^{7}(\gG)\vert=1
\end{gather*}
\end{theorem}

\section{Free 2-step Nilpotent k-ary Lie Algebras}\label{sectionFree}
We divide this section into three parts. First we discuss homology of free 2-step nilpotent k-ary Lie algebras as $GL(V)$ representations, and state a result on the second homology when $k=3$. In the second part, we obtain lower bound of Betti numbers. We discuss representation stability of these algebras at the end of the section.

\subsection{Properties} We recall the definition \ref{deffree2} of free 2-step nilpotent k-ary Lie algebra: $\mathfrak{n}^k_2=V\oplus\bigwedge^kV$, $\dim_{\Cc} V=n\geq k$ with respect to bracket:
\begin{center}
$\mathfrak{n}^k_2\times\cdots\times\mathfrak{n}^k_2\rightarrow \mathfrak{n}^k_2$\\
$[v_1,\ldots,v_k]=\begin{cases}
v_1\wedge\cdots\wedge v_k &v_i\in V \quad \forall i\\
0 &\text{otherwise}
\end{cases}$
\end{center}
We also recall the complex \ref{defcomplex}:

\begin{align}\label{c2}
\xymatrixcolsep{5pt}
\xymatrix{& ...\bigwedge^{3k-2}\mathfrak{n}^k_2\ar[rrrr]^{\partial_{3k-2}}  &&&& \bigwedge^{2k-1}\mathfrak{n}^k_2\ar[rrrr]^{\partial_{2k-1}} &&&& \bigwedge^{k}\mathfrak{n}^k_2\ar[rrrr]^{\partial_{k}}  &&&&  \mathfrak{n}^k_2\ar[rr] &&\Cc
}
\end{align}
with the differential \ref{defdiff}:
\begin{align*}
\partial_t(x_1\wedge\ldots\wedge x_t)=\sum\limits_{\sigma\in D_{t,k}}sgn(\sigma)[x_{\sigma(1)},\ldots,x_{\sigma(k)}]\wedge x_{\sigma(k+1)}\wedge \cdots \wedge x_{\sigma(t)}
\end{align*}

\begin{lemma} Homology of the complex \ref{c2} is $GL(V)$ module.
\end{lemma}

\begin{proof}
Since $\mathfrak{n}^k_2=V\oplus\bigwedge^k V$ is representation of $GL(V)$, kernels and the images of boundary map $\partial$ are also representations of $GL(V)$, which makes each homology $GL(V)$ representation. In details, the action of $g\in GL(V)$ on $\mathfrak{n}^k_2$ is
\begin{align}
g[x_1,\ldots,x_k]=g\left(\sum_{\sigma\in\mathfrak{S}_k}\sign(\sigma)x_{\sigma(1)}\otimes\cdots\otimes x_{\sigma(k)}\right)=\sum_{\sigma}\sign(\sigma)gx_{\sigma(1)}\otimes\cdots\otimes gx_{\sigma(k)}=[gx_1,\ldots,gx_k]
\end{align}

and on cycles $\bigwedge^a\mathfrak{n}^k_2$ is $g.(v_1\wedge\ldots\wedge v_a)=gv_1\wedge\ldots\wedge gv_a$
\end{proof}

\begin{remark} There are many works for the case $k=2$ \cite{sigg1996laplacian}, \cite{jozefiak1988representation}. First of all, the plethsym $\bigwedge^j\left(\bigwedge^2V\right)$ is well-known \cite{landsberg}. Secondly, multiplicity of every irreducible summand in $\bigwedge\left(V\oplus\bigwedge^2V\right)$ is one. Combining these together with the result of Kostant \cite{kostant1963lie}, homology of free 2-step nilpotent Lie algebra is given by
\begin{align}\label{thmsigg}
H^p(\mathfrak{n}^2_2)=\bigoplus\limits_{\substack{\lambda=\lambda^{t}\\ \lambda\vdash p}} S_{\lambda}V
\end{align}
where $\lambda$ is self-conjugate partition of size $p$ \cite{sigg1996laplacian}, \cite{sam2013homology}.
\end{remark}

When $k\geq 3$, there is not any result relating homology of nilpotent k-ary algebras to eigenvalues of Lie algebra Laplacian  as in \cite{kostant1963lie}. Moreover, there is no formula to compute the plethsym $\bigwedge^j\left(\bigwedge^kV\right)$ $j,k\geq 3$. Therefore an analogous result to \ref{thmsigg} is a difficult one. Nevertheless we prove:

\begin{proposition} $H^{(3)}\left(\mathfrak{n}^3_2\right)=S_{221}V\oplus S_{2111}V\oplus S_{3211}V$
\end{proposition}
\begin{proof}
We will compute the kernel of $\partial_3$ and image of $\partial_5$ in the sequence \ref{defcomplex}. It is clear that every cycle of $\bigwedge^3\mathfrak{n}^3_2$ except $\bigwedge^3V$ is in the kernel of $\partial_3$. Therefore :

\begin{align}\label{k1}
\ker\partial_3=\bigwedge^2V\otimes\bigwedge^3V\bigoplus V\otimes\bigwedge^2\left(\bigwedge^3V\right)\bigoplus\bigwedge^3\left(\bigwedge^3V\right)
\end{align}
Before computing the image of $\partial_5$, we list some formulas which are consequences of Littlewood-Richardson and plethysm rules \cite{landsberg}:

\begin{gather}\label{k2}
\bigwedge^2V\otimes\bigwedge^3V\cong \bigwedge^5V\oplus S_{2111}V\oplus S_{221}V \\
\label{k3} V\otimes\bigwedge^2\left(\bigwedge^3V\right)\cong S_{3211}V\oplus S_{2221}V\oplus S_{22111}V\oplus S_{211111}V\oplus \bigwedge^7V \\
\label{k4} \bigwedge^4V\otimes\bigwedge^3V\cong S_{2221}V\oplus S_{22111}V\oplus S_{211111}V\oplus\bigwedge^7V
\end{gather}

Image of $\partial_5$ is
\begin{gather}
\im\partial_5=\bigwedge^5V\oplus\left(\bigwedge^4V\otimes\bigwedge^3V\bigcap V\otimes\bigwedge^2\left(\bigwedge^3V\right)\right)\oplus \bigwedge^3\left(\bigwedge^3V\right)\\
=S_{2221}V\oplus S_{22111}V\oplus S_{211111}V\oplus\bigwedge^7V \oplus \bigwedge^3\left(\bigwedge^3V\right)\label{k5}
\end{gather}
By formulas \ref{k1},\ref{k2},\ref{k3},\ref{k4} and \ref{k5}, homology is
\begin{gather}
H^{(3)}(\mathfrak{n}^3_2)=\cfrac{\ker\partial_3}{\im\partial_5}=S_{221}V\oplus S_{2111}V\oplus S_{3211}V
\end{gather}

\end{proof}

\subsection{Lower Bounds of Betti Numbers}
In this part, we study homology and lower bounds of Betti rank of free 2-step nilpotent k-ary Lie algebras. We start with an easy observation:

\begin{lemma} The decomposition of $\bigwedge^{ik-(i-1)}\mathfrak{n}^k_2$ is
\begin{gather}
\bigwedge^{ik-(i-1)}\mathfrak{n}^k_2=\bigoplus\limits^{ik-(i-1)}_{j=0}\left(\bigwedge^{ik-(i-1)-j}V\otimes\bigwedge^j\left(\bigwedge^kV\right)\right)
\end{gather}
\end{lemma}

\begin{proof}
It follows from Cauchy formula.
\end{proof}

\begin{lemma} Boundary maps are degree preserving, and their restrictions on each degree is:
\begin{gather}
\partial^j_{ik-(i-1)}:\bigwedge^{ik-(i-1)-j}V\otimes\bigwedge^j\left(\bigwedge^kV\right)\longrightarrow \bigwedge^{(i-1)k-i-j+1}V\otimes\bigwedge^{j+1}\left(\bigwedge^{k}V\right)
\end{gather}
\end{lemma}
\begin{proof}
Since center of $\mathfrak{n}^k_2$ is $\bigwedge^k V$, only nonzero bracket is of the form $[x_1\ldots,x_k]$, $x_i\in V$. Boundary map $\partial$ takes $k$ pairs from $\bigwedge^{ik-(i-1)-j}$, and adds $1$ element to $\bigwedge^j\left(\bigwedge^k V\right)$.
\end{proof}

\begin{proposition}\label{homologydecomposition} Homology of complex at $ik-(i-1)$ is
\begin{gather}
H^{ik-(i-1)}\left(\gG\right)=\cfrac{\ker\partial_{ik-(i-1)}}{im\partial_{(i+1)k-i}}=\bigoplus\limits^{ik-(i-1)}_{j=0}\cfrac{\ker\partial^j_{ik-(i-1)}}{\im\partial^j_{(i+1)k-i}}
\end{gather}
\end{proposition}
\begin{proof}
Cycles of the form $\bigwedge^mV\otimes\bigwedge^{(i+1)k-i-m}\left(\bigwedge^kV\right)$ maps to zero when $0\leq m<k$, so they are not in the image. The nonzero indices are same both boundary maps $\partial_{ik-(i-1)}$ and $\partial_{(i+1)k-i}$ in the homology.
\end{proof}

\begin{corollary} Betti rank is :
\begin{gather}
\left|H^{ik-(i-1)}\left(\mathfrak{n}^k_2\right)\right|=\sum\limits^{ik-(i-1)}_{j=0}\left|\ker\partial^j_{ik-(i-1)}\right|-\left|\im\partial^j_{(i+1)k-i}\right|
\end{gather}
\end{corollary}

\begin{proposition}
\begin{gather}
\left|H^{ik-(i-1)}\left(\mathfrak{n}^k_2\right)\right|=\sum\limits^{ik-(i-1)}_{j=0}\left|\ker\partial^j_{ik-(i-1)}\right|-\left|\im\partial^j_{(i+1)k-i}\right|\\
\geq \sum\limits^{(i-1)(k-1)}_{j=1}\left|\ker\partial^j_{ik-(i-1)}\right|-\left|\im\partial^j_{(i+1)k-i}\right|
\end{gather}
\end{proposition}
\begin{proof}
Cycles of the form
\begin{align}
\bigwedge^{ik-(i-1)-j}V\otimes\bigwedge^{j}\left(\bigwedge^{k}V\right)
\end{align} with $(i-1)(k-1)<j$ are in the kernel of $\partial^j_{ik-(i-1)}$. Hence, if we ignore their contribution to homology, we get smaller rank. \\
Also, $\bigwedge^{ik-(i-1)}V$ is always a submodule of $\bigwedge^{(i-1)k-(i-1)}V\otimes\bigwedge^kV$ by Pieri's rule. This gives the claimed lower bound.
\end{proof}

\begin{lemma}\label{pieri1}$\bigwedge^kV\otimes\bigwedge^{\alpha}\left(\bigwedge^kV\right)\cong \bigwedge^{\alpha+1}\left(\bigwedge^kV\right)\bigoplus S_{21^{\alpha-1}}\left(\bigwedge^kV\right)$
\end{lemma}
\begin{proof}
By Pieri's rule, $W\otimes\bigwedge^2W\cong \bigwedge^3W\oplus S_{21}W$. Replacing $W$ by $\bigwedge^kV$ proves the lemma.
\end{proof}

\begin{theorem}\label{thmLowerbound}
The rank of the homology at $ik-(i-1)$, $i\geq 2$, $H^{ik-(i-1)}\left(\mathfrak{n}^k_2\right)$ is bounded below by
\begin{gather}
\left|H^{ik-(i-1)}\left(\mathfrak{n}^k_2\right)\right|\geq{n\choose k}{x\choose (i-1)(k-1)}-{n\choose 2k}{x\choose (i-1)(k-1)-1}-{n\choose 0}{x\choose (i-1)(k-1)+1}\nonumber
\end{gather}
where $n$ is the dimension of $V$.
\end{theorem}
\begin{proof}
By proposition \ref{homologydecomposition}, homology of the sequence :

\begin{gather}
\label{h1}\bigwedge^{2k}V\otimes\bigwedge^{(i-1)(k-1)-1}\left(\bigwedge^kV\right)\longrightarrow \bigwedge^{k}V\otimes\bigwedge^{(i-1)(k-1)}\left(\bigwedge^k V\right)\longrightarrow\bigwedge^{(i-1)(k-1)+1}\left(\bigwedge^k V\right)
\end{gather}
contributes to the homology of the complex \ref{c2}.
By the lemma \ref{pieri1}, the middle term decomposes as:
\begin{gather}
\bigwedge^{2k}V\otimes\bigwedge^{(i-1)(k-1)-1}\left(\bigwedge^kV\right)\longrightarrow S_{\lambda}\left(\bigwedge^kV\right)\bigoplus \bigwedge^{(i-1)(k-1)+1}\left(\bigwedge^k V\right)\longrightarrow\bigwedge^{(i-1)(k-1)+1}\left(\bigwedge^k V\right)
\end{gather}
where $\lambda=\left(21^{(i-1)(k-1)-1}\right)$. Therefore homology $H^{ik-(i-1)}\mathfrak{n}^k_2$ contains
\begin{gather}
\coker\left(\bigwedge^{2k}V\otimes\bigwedge^{(i-1)k-(i)}V\longrightarrow S_{21^{(i-1)k-i}}V\right)
\end{gather}
as submodule. In particular, homology of the sequence \ref{h1} is bounded by
\begin{gather}
{n\choose k}{x\choose (i-1)(k-1)}-{n\choose 2k}{x\choose (i-1)(k-1)-1}-{n\choose 0}{x\choose (i-1)(k-1)+1}\nonumber
\end{gather}
Claim follows.
\end{proof}

\begin{proposition} The lower bound in theorem \ref{thmLowerbound} is asymptotically
\begin{align}
\cfrac{e^{2k}n^{2k}}{2k^{2k}}\left(\cfrac{1}{\pi k}-\cfrac{(\alpha+1)}{\sqrt{\pi k}}\cfrac{1}{2^{2k}}\right){x\choose \alpha}\cfrac{1}{(\alpha+1)(x-\alpha+1)}
\end{align}
with $\alpha=(i-1)(k-1)$
\end{proposition}
\begin{proof}
Assume that $n$ is large than $k$. Then we can use the approximation:
\begin{gather}
{n\choose k}\approx \cfrac{e^k}{\sqrt{2\pi k}}\left(\cfrac{n}{k}-\cfrac{1}{2}\right)^k
\end{gather}
We start with the bound in the Theorem \ref{thmLowerbound}, and simplify:
\begin{gather}
{x\choose \alpha}{n\choose k}-{x\choose \alpha-1}{n\choose 2k}-{x\choose \alpha+1}{n\choose 0}=\\
{x\choose \alpha}{n\choose k}-\cfrac{\alpha}{x-\alpha+1}{x\choose \alpha}{n\choose 2k}-\cfrac{x-\alpha}{\alpha+1}{x\choose \alpha}{n\choose 0}
\end{gather}
We approximate the coefficient:
\begin{gather}
\cfrac{\alpha}{\alpha+1}x+\cfrac{\alpha}{\alpha+1}-{n\choose 2k}\cfrac{\alpha}{x-\alpha+1}\implies\\
x^2+(2-\alpha)x+1-\alpha-(\alpha+1){n\choose 2k}\approx\\
\cfrac{e^{2k}}{2\pi k}\left(\cfrac{n}{k}-\cfrac{1}{2}\right)^{2k}+(2-\alpha)\cfrac{e^k}{\sqrt{2\pi k}}\left(\cfrac{n}{k}-\cfrac{1}{2}\right)^k+1-\alpha-(\alpha+1)\cfrac{e^{2k}}{\sqrt{4\pi k}}\left(\cfrac{n}{2k}-\cfrac{1}{2}\right)^{2k}
\end{gather}
\text{The leading term is}
\begin{gather}
\cfrac{e^{2k}n^{2k}}{2k^{2k}}\left(\cfrac{1}{\pi k}-\cfrac{(\alpha+1)}{\sqrt{\pi k}}\cfrac{1}{2^{2k}}\right)
\end{gather}
Therefore lower bound asymptotically is

\begin{align}
\cfrac{e^{2k}n^{2k}}{2k^{2k}}\left(\cfrac{1}{\pi k}-\cfrac{(\alpha+1)}{\sqrt{\pi k}}\cfrac{1}{2^{2k}}\right){x\choose \alpha}\cfrac{1}{(\alpha+1)(x-\alpha+1)}
\end{align}
In particular, positivity of the leading term is because of $2^{2k}\geq ki\sqrt{\pi k}$. This finishes the proof.
\end{proof}

For the second homology, we can track another approach.
\begin{proposition}\label{prop2}
 The second homology of free 2-step nilpotent k-ary Lie algebra always contains
\begin{gather}
 \bigoplus\limits^{k-1}_{j=1}S_{2^j1^{2k-2j+1}}V \subset H^{k}\left(\mathfrak{n}^k_2\right)
\end{gather}
\end{proposition}

\begin{proof}
By proposition \ref{homologydecomposition}, the second homology i.e. $i=1$ contains $\cfrac{\ker\partial^0_{ik-(i-1)}}{\im\partial^0_{(i+1)k-i}}$, which is :
\begin{gather}
 \cfrac{\bigwedge^{k-1}V\otimes\bigwedge^{k}V}{\bigwedge^{2k-1}V}= \bigoplus\limits^{k-1}_{j=1}S_{2^j1^{2k-2j-1}}V
\end{gather}
\end{proof}

\begin{corollary} The second homology of free 2-step nilpotent $k$-ary Lie algebra satisfies:
\begin{align}
\vert H^{(k)}\left(\mathfrak{n}^k_2\right)\vert\geq {n\choose k}{n\choose k-1}-{n\choose 2k-1}
\end{align}
\end{corollary}

\subsection{Representation stability}\label{sectionRep}

Representation stability is currently an active field and find wide range of applications in many branches including Lie algebra homology. For example, homologies of free nilpotent Lie algebras are representation stable \cite{church2013representation}, which follows representation stability of free Lie algebras. Multiplicity of irreducible representations in \ref{freelie} is given by the formula:
\begin{center}
$c_{\lambda}=\cfrac{1}{n}\sum\limits_{d\vert n}\mu(d)\chi_{\lambda}(\tau^{n/d})$
\end{center}
where $\chi_{\lambda}$ is the character of $\mathfrak{S}_n$-representation $\lambda$ and $\tau$ is the cycle $(12\ldots n)$ \cite{klyachko1974lie}.
For free k-ary Lie algebras, we do not have such a formula yet, since which representations occur in it is still an open problem \cite{friedmann2017generalization}.

\begin{theorem}
 Homologies of free nilpotent k-ary Lie algebras are representation stable.
\end{theorem}

\begin{proof}

Instead of representation stability techniques\footnotemark, we sketch direct arguments of \cite{tirao2002homology}\footnotetext{Representation stability of free k-ary Lie algebras will be a separate work}. Let $\Cc^n\hookrightarrow\Cc^{n+1}$ be an embedding. This induces the following inclusions:
\begin{enumerate}[label=\roman*)]
\item Free k-ary Lie algebras: $F^{(k)}(\Cc^n)\hookrightarrow F^{(k)}(\Cc^{n+1})$
\item Free s-step nilpotent k-ary Lie algebras $\mathfrak{n}^k_s(\Cc^n)\hookrightarrow\mathfrak{n}^k_s(\Cc^{n+1})$
\item Terms in the complex \ref{defcomplex}: $\bigwedge^{\alpha}\mathfrak{n}^k_s(\Cc^n)\hookrightarrow\bigwedge^{\alpha}\mathfrak{n}^k_s(\Cc^{n+1})$
\item In particular, $GL_n\mapsto GL_{n+1}$ via $A\rightarrow \begin{bmatrix}
A&0\\
0&1
\end{bmatrix}$
\end{enumerate}
We denote $\mathfrak{n}^k_s(\Cc^n)$ and $\mathfrak{n}^k_s(\Cc^{n+1})$ by $\mathfrak{n}_1$ and $\mathfrak{n}_2$ respectively. We compare homology $H^{\alpha}(n_1)$ and $H^{\alpha}(n_2)$.
If nonzero $[\lambda]\in H^{\alpha}(n_1)$, then $[\lambda]\in H^{\alpha}(n_2)$, since $[\lambda]$ is in the kernel of $\partial^2_{\alpha}$ via Young diagram $\left(\lambda,0\right)$. If $[\lambda]$ is in the image of $\partial^2_{\alpha+k-1}$ i.e. $[\lambda]=\partial^2_{\alpha+k-1}(\omega)$, then $\omega=[\left(\lambda,0\right)]$ by using $GL$ morphisms and in particular $\omega\in\bigwedge^{\alpha+k-1}n_2$. The restriction of $\omega$ gives either zero or $[\lambda]$. Therefore for large enough dimensions, $H^{\alpha}(n_1)$ is equal to $H^{\alpha}(n_2)$, which implies representation stability.
\end{proof}

\begin{example}
The second homology of $\mathfrak{n}^3_2$ is representation stable:
\begin{itemize}
\item $\dim V=3$, $H^{(3)}\left(\mathfrak{n}^3_2\right)= {\tiny\Yvcentermath1 \yng(2,2,1)}$ its dimension is $3$.
\item $\dim V\geq 4$ , $H^{(3)}\left(\mathfrak{n}^3_2\right)= {\tiny\Yvcentermath1 \yng(2,2,1)\oplus\yng(2,1,1,1)\oplus \yng(3,2,1,1)}$
\end{itemize}
\end{example}

It would be nice to get another stability in the direction of $k$, i.e. ith homologies of nilpotent k-ary Lie algebras while $k$ is increasing. However, except the first homology i.e. $H^1(\gG)=V$, this cannot not true, simply by the result of proposition \ref{prop2}:
\begin{gather}
 \cfrac{\bigwedge^{k-1}V\otimes\bigwedge^{k}V}{\bigwedge^{2k-1}V}= \bigoplus\limits^{k-1}_{j=1}S_{2^j1^{2k-2j-1}}V
\end{gather}
is always a submodule of the second homology of $\mathfrak{n}^k_2$, and not stabilizing.

\section{Toral Rank conjecture}\label{sectionToral}
Let $\mathfrak{n}$ be nilpotent Lie algebra with center $\mathfrak{z}$. Toral rank conjecture states that:
\begin{align}
\sum\limits^{\dim\mathfrak{n}}_{i=0} H^i\left(\mathfrak{n}\right)\geq 2^{\dim\mathfrak{z}}
\end{align}
We carry it into k-ary setup:
\begin{theorem} Toral rank conjecture  holds for  free 2-step nilpotent k-ary Lie algebras, k-ary Heisenberg Lie algebras, k-ary ACJ Lie algebras and free 3-step nilpotent k-ary Lie algebras   of small dimension.
\end{theorem}
\begin{proof}
In theorem \ref{thmRefinement}, we give a refinement of toral rank conjecture for 2-step nilpotent k-ary Lie algebras. Hence the only remaining cases are small dimensional 3-step nilpotent k-ary Lie algebras studied in \ref{homfree3}. The sum of the homologies when $k\geq 4$ is:
\begin{gather}
k+{2k+1\choose k}-(k+1)-(2k+1)+(2k+1)(k-1)=\\
{2k+1\choose k}+2k^2-3k-3
\end{gather}

It is clear that
\begin{align}
{2k+1\choose k}+2k^2-3k-3>{2k\choose k}\geq \cfrac{(2k)^k}{k^k}=2^k
\end{align}
Since the size of the center is $k$, toral rank conjecture holds for $k\geq 4$. If $k=3$, the total homology is $42$ according to the Theorem \ref{thm3} and greater than $8$.
\end{proof}

\subsection{Refinement of Toral rank Conjecture}

We prove the following:
\begin{theorem}\label{thmRefinement}
The total homology of any 2-step nilpotent $k$-ary Lie algebra $\gG$ satisfies

\begin{align}
\vert H^*(\gG)\vert\geq \sum\limits^{k-1}_{i=0}\left| \sum_{j=0}\left(-1\right)^{j}{\vert\mathfrak{v}\vert\choose kj+i}\right|2^{\vert\mathfrak{z}\vert}
\end{align}
where $\mathfrak{z}$ is the center of $\gG$ and $\mathfrak{v}$ is its vector space complement.
\end{theorem}

\begin{proof}
Let $\gG=\mathfrak{z}\oplus\mathfrak{v}$ as vector space, the complex \ref{defcomplex} becomes $\bigwedge\gG=\bigwedge\left(\mathfrak{z}\otimes\mathfrak{v}\right)$. Boundary maps are $\partial:\bigwedge^p\mathfrak{v}\otimes\bigwedge^q\mathfrak{z}\rightarrow
\bigwedge^{p-k}\mathfrak{v}\otimes\bigwedge^{q+1}\mathfrak{z}$. This implies that the complex $(\bigwedge\gG,\partial)$ is direct sum of complexes $(\bigwedge^{kj+i}\mathfrak{v}\otimes\bigwedge \mathfrak{z},\partial)$.

Homology of any finite complex satisfies:
\begin{gather}
\vert H^*(\gG)\vert\geq \left| \sum_{a=0}\left(-1\right)^{a}\dim\left(\bigwedge^{a}\gG\right)\right|
\end{gather}
Now, we replace $\gG$ by $\mathfrak{v}\oplus\mathfrak{z}$ and simplify the expression since the total homology is direct sum of all homologies in  mod k:
\begin{gather}
\vert H^*(\gG)\vert\geq \sum\limits^{k-1}_{i=0}\left| \sum_{j=0}\left(-1\right)^{j}\dim\left(\bigwedge^{kj+i}\mathfrak{v}\otimes\bigwedge\mathfrak{z}\right)\right|\\
\vert H^*(\gG)\vert\geq \sum\limits^{k-1}_{i=0}\left| \sum_{j=0}\left(-1\right)^{j}\dim\left(\bigwedge^{kj+i}\mathfrak{v}\right)\right|2^{\vert\mathfrak{z}\vert}\\
\vert H^*(\gG)\vert\geq \sum\limits^{k-1}_{i=0}\left| \sum_{j=0}\left(-1\right)^{j}{\vert\mathfrak{v}\vert\choose kj+i}\right|2^{\vert\mathfrak{z}\vert}
\end{gather}

\end{proof}

\begin{corollary} The refinement of the toral rank conjecture for 2-step nilpotent k-ary Lie algebras is
\begin{align*}
\vert H^*(\gG)\vert\geq \sum\limits^{k-1}_{i=0}\left| \sum_{j=0}\left(-1\right)^{j}{\vert\mathfrak{v}\vert\choose kj+i}\right|2^{\vert\mathfrak{z}\vert}>2^{\vert\mathfrak{z}\vert}
\end{align*}
When $k=2$, one get explicit formula obtained in \cite{tirao2000refinement}.

\end{corollary}

\begin{remark}When we apply this to 5-ary Heisenberg Lie algebra with $m=1$, complex is
\begin{gather}
0\longrightarrow \bigwedge^4\hH\longrightarrow\hH\longrightarrow 0
\end{gather}
and the total homology: $1+5+5=11$. The current 5-ary Lie algebra $\hH\otimes\cfrac{\Cc[t]}{t^2}$ is 2-step nilpotent, and its dimension is $12$. The lower bound of the total homology of any $12$ dimensional 2-step nilpotent 5-ary Lie algebra is $2900$ by the formula we obtained in the Theorem \ref{thmRefinement}.
Property M \ref{defPropertyM} asserts that
\begin{align}
\left|H\left(\hH\otimes \faktor{\Cc[t]}{t^j}\right)\right|=\left|H\left(\hH\right)\right|^{j}
\end{align}
However, at $j=2$, claimed total homology is $11^2=121<2900$
 which is impossible.\\
This suggests that for k-ary Lie algebras, exponent of the total homology might be $(k-1)j$ in \ref{defPropertyM}.
\end{remark}
Here we add comparison of the lower bounds according to Theorem \ref{thmRefinement} and exponents of $2$. The entries of the third column is due to the formula in \cite{tirao2000refinement}.

$\left[ \begin {array}{ccccccccc} \hline
n&k=2&\log_2&k=3&\log_2&k=4&\log_2&k=5&\log_2\\\hline
1&2& 1.0&2& 1.0&2& 1.0&2& 1.0\\\noalign{\medskip}2&2& 1.0&4& 2.0&4& 2.0&4& 2.0\\\noalign{\medskip}3
&4& 2.0&6& 2.584962500&8& 3.0&8& 3.0\\\noalign{\medskip}4&4& 2.0&12&
 3.584962501&14& 3.807354922&16& 4.0\\\noalign{\medskip}5&8& 3.0&18&
 4.169925001&28& 4.807354922&30& 4.906890596\\\noalign{\medskip}6&8&
 3.0&36& 5.169925000&48& 5.584962501&60& 5.906890595
\\\noalign{\medskip}7&16& 4.0&54& 5.754887502&96& 6.584962500&110&
 6.781359713\\\noalign{\medskip}8&16& 4.0&108& 6.754887502&164&
 7.357552004&220& 7.781359713\\\noalign{\medskip}9&32& 5.0&162&
 7.339850002&328& 8.357552004&400& 8.643856190\\\noalign{\medskip}10&
32& 5.0&324& 8.339850002&560& 9.129283017&800& 9.643856190
\\\noalign{\medskip}11&64& 6.0&486& 8.924812503&1120& 10.12928302&1450
& 10.50183718\\\noalign{\medskip}12&64& 6.0&972& 9.924812502&1912&
 10.90086681&2900& 11.50183718\\\noalign{\medskip}13&128& 7.0&1458&
 10.50977500&3824& 11.90086681&5250& 12.35810171\\\noalign{\medskip}14
&128& 7.0&2916& 11.50977500&6528& 12.67242534&10500& 13.35810171
\\\noalign{\medskip}15&256& 8.0&4374& 12.09473750&13056& 13.67242534&
19000& 14.21371180\\\noalign{\medskip}16&256& 8.0&8748& 13.09473750&
22288& 14.44397955&38000& 15.21371180\\\noalign{\medskip}17&512& 9.0&
13122& 13.67970001&44576& 15.44397955&68750& 16.06907210
\\\noalign{\medskip}18&512& 9.0&26244& 14.67970000&76096& 16.21553300&
137500& 17.06907210\\\noalign{\medskip}19&1024& 10.0&39366&
 15.26466251&152192& 17.21553300&248750& 17.92433701
\\\noalign{\medskip}20&1024& 10.0&78732& 16.26466251&259808&
 17.98708633&497500& 18.92433701\end {array} \right]$

\bibliographystyle{alpha}

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