# Drazin and group invertibility in algebras spanned by two idempotents

Rounak Biswas  Falguni Roy

###### Abstract

For two given idempotents $p\text{ and }q$ from an associative algebra
$\mathcal{A},$ in this paper, we offer a comprehensive classification of
algebras spanned by the idempotents $p\text{ and }q$. This classification is
based on the condition that $p\text{ and }q$ are not tightly coupled and
satisfies $(pq)^{m-1}=(pq)^{m}$ but $(pq)^{m-2}p\neq(pq)^{m-1}p$ for some
$m(\geq 2)\in\mathbb{N}.$ Subsequently, we categorized all the group
invertible elements and established an upper bound for Drazin index of any
elements in these algebras spanned by $p,q$. Moreover, we formulate a new
representation for the Drazin inverse of $(\alpha p+q)$ under two different
assumptions, $(pq)^{m-1}=(pq)^{m}$ and $\lambda(pq)^{m-1}=(pq)^{m},$ here
$\alpha$ is a non-zero and $\lambda$ is a non-unit real or complex number.

###### keywords:

Finite dimensional algebra , group inversion , Drazin inversion , idempotent

###### MSC:

15A30 , 15A09 , 16S15 , 17C27

[inst1]organization=Department of Mathematical and Computational Science,
National Institute of Technology Karnataka, city=Surathkal, postcode=575025,
state=Karnataka, country=India

Research highlight 1

Research highlight 2

##  1 Introduction

Throughout this paper, $\mathcal{A}$ always represent an associative algebra
over the field of scalars $\mathbb{C}$ or $\mathbb{R},$ with an identity $I$.
An element $a\in\mathcal{A}$ is said to be idempotent if $a^{2}=a$. In this
paper $p$ and $q$ will consistently refer to two idempotent elements in
$\mathcal{A}$. For given $p,q$; alg$(p,q)$ denotes the subalgebra of
$\mathcal{A}$ spanned by $p\text{ and }q$, i.e. alg$(p,q)$ is composed of all
possible finite linear combinations of elements from the list

$\displaystyle p,pq,pqp,(pq)^{2}\cdots(pq)^{m}\cdots$ (1) $\displaystyle
q,qp,qpq,(qp)^{2}\cdots(qp)^{m}\cdots.$

Here, the order of an element from (1) is defined by the number of factors
present. The notations $(pq)_{i},\text{ }(qp)_{i}$ denotes the upper and lower
element of the $i$-th column in the list (1). For given $p,q$ if two members
from the list (1) with order difference $\leq 1$ are equal, then following
[9], $p,q$ termed as tightly coupled idempotents. In simpler terms, we say
that $p$ and $q$ are tightly coupled if, for some $i\in\mathbb{N}$, either the
product $(pq)_{i}$ or $(qp)_{i}$ matches one of the five neighbouring elements
from the list $(\ref{list})$. Specifically, this means that $(pq)_{i}$ (or
$(qp)_{i}$) coincides with any of the five adjacent elements( called
neighbourhood): ${(pq)_{i-1},(pq)_{i+1},(qp)_{i-1},(qp)_{i},(qp)_{i+1}}$ (or
${(qp)_{i-1},(qp)_{i+1},(pq)_{i-1},(pq)_{i},(pq)_{i+1}}$), for some
$i\in\mathbb{N}$. Böttcher and Spitkovsky [9, 5] studied the algebra spanned
by $p$ and $q,$ when they are tightly coupled. Depending on the relation
between $p$ and $q,$ they classified alg$(p,q)$ into four different types
$Z_{m},U_{m},D_{m}$, $D_{m}^{*}$ and established

$\displaystyle D_{m}^{*}$ $\displaystyle\cong Z_{m-2}\oplus D_{2}^{*}\text{
}(m\geq 2),$ $\displaystyle D_{m}$ $\displaystyle\cong Z_{m-2}\oplus
D_{2}\text{ }(m\geq 2),$ (2) $\displaystyle U_{m}$ $\displaystyle\cong
Z_{m-1}\oplus U_{1}\text{ }(m\geq 1),$

where $m\in\mathbb{N}.$ Here $Z_{m}=\mathrm{alg}(p,q),$ when $p,q$ satisfy one
of the following conditions depending on $m$

$\displaystyle\begin{cases}(qp)_{k}=(pq)_{k}=0,\text{ if }m=2k-2\\\
(qp)_{k}=0\text{ or }(pq)_{k}=0,\text{if }m=2k-1\end{cases},$

and for construction of the algebras $U_{m},D_{m}^{*},D_{m}$ interested
readers can refer to [9]. However, when attempting to extend this
classification beyond tightly coupled scenarios, Böttcher and Spitkovsky [9,
5] acknowledged the challenges and complexities involved in categorizing the
algebra alg$(p,q)$. In this article, we establish that starting from one of
the two elements in any column of the list (1), there is only one way to go
beyond tightly coupled. Specifically, if one initiates with $(pq)^{m}$(or
$(qp)^{m}$) from the $2m$-th column of the list (1), then the only way go
beyond tightly coupled is when $(pq)^{m}=(pq)^{m-k}$(or$(qp)^{m}=(qp)^{m-k}$),
for some $k\in\mathbb{N}$ and $k<m.$ Furthermore, we provide a complete
classification of the algebra alg$(p,q)$, when $p\text{ and }q$ are not
tightly coupled and satisfies $(pq)^{m-1}=(pq)^{m},$
$(pq)^{m-2}p\neq(pq)^{m-1}p.$ The investigation of these algebraic structures
is motivated by the goal of comprehending Drazin and group inevitability
within this context.

For $a\in\mathcal{A},$ the Drazin inverse of $a$ is the unique element
$b\in\mathcal{A}$(denoted by $a^{D}$), satisfying

$ab=ba,\text{ }ab^{2}=b\text{ and }a^{k+1}b=a^{k},$ (3)

for some $k\in\mathbb{N}\cup\\{0\\}$. The least $k$ satisfying (3), is known
as the Drazin index $i(a)$ of $a$. In the particular case $i(a)=1,$ the Drazin
inverse is termed as group inverse( symbolized by $a^{g}$). As demonstrated by
Drazin [4], the Drazin invertibility of $a$ is equivalent to existence of two
elements $x,y\in\mathcal{A}$ satisfying

$a^{k_{1}+1}x=a^{k_{1}}\text{ and }ya^{k_{2}+1}=a^{k_{2}},$ (4)

for some positive integer $k_{1},k_{2}$. The smallest $k_{1},k_{2}$ satisfying
(4) is referred to as the left and right index of $a$, respectively. Moreover
if $k_{1},k_{2}$ is finite and the least integer satisfying (4) then
$k_{1}=k_{2}=i(a)$ with $a^{D}=a^{k_{1}}x^{k_{1}+1}=y^{k_{1}+1}a^{k_{1}}.$ One
of the interesting problems in Drazin inverse theory is to investigate Drazin
invertibility of $\alpha p+\beta q,$ where $\alpha,\beta$ are scalars. This
specific problem has garnered great interest from various researchers in
recent years. Deng [2] considered this problem for idempotent operators on a
Hilbert space, when $\alpha,\beta\in\\{1,-1\\}$ which is extended in the
setting of a Banach algebra by Zhang, Wu [3], where they provided an
expression of $(\alpha p+\beta q)^{D}$ assuming $pqp=0$ and $pqp=pq$, for non
zero scalars $\alpha,\beta$. More results on this problem can be found in [6,
12, 13, 14]. In particular, for tightly coupled idempotents, $p,q$; alg$(p,q)$
becomes finite-dimensional, and hence, every element in this algebra becomes
Drazin invertible. However, the same statement is not true in the case of
group inverse. Böttcher and Spitkovsky [8] solved this problem of group
invertibility in $Z_{m}$, and that is enough to settle the case for other
algebras due to (1). But we observe that the Theorem 2.2. [8], which is
essential for investigating group invertibility in $Z_{m}$, lacks
consideration of the case when $Z_{m}$ and $\mathcal{A}$ possess different
identity elements. An example (Example 3.2) narrating this observation will be
discussed in Section 3. This paper addresses this gap by presenting a
comprehensive version of Theorem 2.2. [8], and utilizing this, we solve the
problem of group invertibility in alg$(p,q),$ when $p,q$ are not tightly
coupled and satisfies $(pq)^{m-1}=(pq)^{m},$ $(pq)^{m-2}p\neq(pq)^{m-1}p$ for
some $m\in\mathbb{N}.$ Providing a representation of the Drazin inverse (or
group inverse) for any element in associative algebra poses a significant
challenge once it is established that the element possesses such an inverse.
In this paper, we will provide a new representation of $(\alpha p+q)^{D}$
under two different conditions; the first one is when $(pq)^{m-1}=(pq)^{m}$
and another one is when $\lambda(pq)^{m-1}=(pq)^{m},$ here $\alpha$ is a non-
zero and $\lambda$ is a non-unit scalar.

The paper is organized in the following manner. Section 2, presents the
classification of algebras generated by $p\text{ and }q$, when
$(pq)^{m}=(pq)^{m-1}$. Section 3, deals with the Darzin index and the group
invertible elements in the algebras discussed in section 2. Representations of
the Drazin inverse of $(\alpha p+q)$ under two different assumption
$(pq)^{m-1}=(pq)^{m}$ and $\lambda(pq)^{m-1}=(pq)^{m}$ are in Section 4.

##  2 Algebras spanned by non-tightly coupled idempotents

Due to the complexity in extending the neighbourhood beyond tightly coupled,
Böttcher and Spitkovsky [9] settled the case for the algebra alg$(p,q)$, when
$pqp=p$. But it turns out that there is only one way to extend the
neighborship beyond tightly coupled; our next lemma establishes this.

######  Lemma 2.1.

For $m,k\in\mathbb{N}$, where $k<m$, if $p$ and $q$ satisfy one of the
following conditions, then they become tightly coupled:

  1. (i)

$(pq)^{m-k}p=(pq)^{m};$

  2. (ii)

$(qp)^{m-k}q=(pq)^{m};$

  3. (iii)

$(qp)^{m-k}=(pq)^{m}.$

###### Proof.

  1. (i)

Since $(pq)^{m-k}p=(pq)^{m}$, multiplying both sides by $q$ from the right
yields $(pq)^{m-k+1}=(pq)^{m}$. Consequently, we get
$(pq)^{m-k}p=(pq)^{m-k+1}$, indicating that $p$ and $q$ are tightly coupled.

  2. (ii)

Multiplying $(pq)^{m}=(qp)^{m-k}q$ by $p$ from left we obtain
$(pq)^{m}=(pq)^{m-k+1}$, therefore $(qp)^{m-k}q=(pq)^{m-k+1}.$

  3. (iii)

Similarly multiplying $(pq)^{m}=(qp)^{m-k}$ by $p$ from left we obtain
$(qp)^{m-k}=p(qp)^{m-k}.$

∎

Therefore, starting from $(pq)^{m}$ for some $m\in\mathbb{N}$, there is only
one way to extend the neighbourship beyond tightly coupled, i.e. when
$(pq)^{m}=(pq)^{m-k}$ for some integer $k<m.$ For this paper, our focus is on
the particular case when $k=1$. In the following lemma, we provide some
properties of the idempotents $p,q$ when they satisfy $(pq)^{m}=(pq)^{m-1}$
for some $m\in\mathbb{N}$.

######  Lemma 2.2.

Let $p,q$ satisfies $(pq)^{m-1}=(pq)^{m}$ for $n\in\mathbb{N}$,

  1. (i)

$(pq)^{m-1}=(pq)^{m+k}$, for $k\in\mathbb{N}$;

  2. (ii)

if $(pq)^{n_{1}}=(pq)^{n_{2}}$, where $n_{1},n_{2}\in\mathbb{N}$ such as
$1\leq n_{2}<n_{1}\leq m$ then $(pq)^{m}=(pq)^{m-i}$ for some
$i\in\mathbb{N}$;

  3. (iii)

if $(pq)^{m}=(pq)^{m-k}$ for some $k\in\mathbb{N}$, where $1<k<m$, then
${pq}^{m-2}=(pq)^{m-1}.$

###### Proof.

  1. (i)

One can verify this easily.

  2. (ii)

If $n_{1}=m$ then $i=m-n_{2}$. Now if $n_{1}<m$ then multiplying both side of
$(pq)^{n_{1}}=(pq)^{n_{2}}$ with $(pq)^{m-n_{1}}$ we get

$(pq)^{m}=(pq)^{n_{2}+m-n_{1}}.$

Hence $(pq)^{m}=(pq)^{m-i}$ where $i=n_{1}-n_{2}.$

  3. (iii)

Since $(pq)^{m}=(pq)^{m-k}$ then multiplying both side by $(pq)^{k-2}$ we
obtain

$\displaystyle(pq)^{m-2}=(pq)^{m+k-2}{}=(pq)^{m-1}.$

∎

######  Remark 1.

According to Lemma 2.2, if $m$ is the least positive integer satisfying
$(pq)^{m}=(pq)^{m-1}$ then $(pq)^{j}\neq(pq)^{l}$ for $j,l\in\mathbb{N}$ where
$j<l<m$.

In the remaining part of this section, we will provide a complete
classification of all algebras spanned by the idempotents $p,q;$ where $p,q$
are not tightly coupled and satisfies $(pq)^{m-1}=(pq)^{m}$ but
$(pq)^{m-2}p\neq(pq)^{m-1}p$ for some $m\in\mathbb{N}.$ Because of these
assumptions on $p,q$, the infinite list $(\ref{list})$ terminate to a finite
list; hence, alg$(p,q)$ becomes finite dimensional. Before proceeding further,
let’s revisit the following lemma from [9]; it provides the complete
classification of alg$(p,q)$ when $pqp=p$. Note that in the context of any
algebra $\mathcal{X}$ and for $A\in\mathcal{X}$; the notations
$\mathcal{N}(\mathcal{X})$ and $\sigma(A)$, represent the collection of
nilpotent elements in $\mathcal{X}$ and the spectrum of $A$ within
$\mathcal{X}$, respectively.

######  Lemma 2.3.

[9] Upto isomorphism, there are precisely four algebras, namely
$W_{3},W_{3}\oplus Z_{1},W_{4},$ $W_{4}\oplus Z_{1},$ spanned by $p,q$; where
$pqp=p$ and $q,p,qp,pq$ are pairwise distinct.

  1. (i)

For the case when $q=qpq$, $q+p=qp+pq,$ we have $\mathrm{alg}(p,q)\cong W_{3}$
where $\mathrm{dim}\text{ }W_{3}=3$, $\mathrm{dim}\text{
}\mathcal{N}(W_{3})=2.$

  2. (ii)

If $q\neq qpq$ with $qp+pq=qpq+p$, we have $\mathrm{alg}(p,q)\cong W_{3}\oplus
Z_{1}$ where $\mathrm{dim}\text{ }(W_{3}\oplus Z_{1})=4$ and
$\mathrm{dim}\text{ }\mathcal{N}(W_{3}\oplus Z_{1})=2.$

  3. (iii)

For the case when $q=qpq$ but $q+p\neq qp+pq$, we have $\mathrm{alg}(p,q)\cong
W_{4},$ with $\mathrm{dim}\text{ }W_{4}=4$, $\mathrm{dim}\text{
}\mathcal{N}(W_{4})=3.$

  4. (iv)

If $q\neq qpq$ with $qp+pq\neq qpq+p$, then $\mathrm{alg}(p,q)\cong
W_{4}\oplus Z_{1}$ where $\mathrm{dim}\text{ }(W_{4}\oplus Z_{1})=5$,
$\mathrm{dim}\text{ }\mathcal{N}(W_{4}\oplus Z_{1})=3$.

######  Lemma 2.4.

Let $p\text{, }q$ are not tightly coupled and satisfy

$(pq)^{m-1}=(pq)^{m}\text{ but }(pq)^{m-2}p\neq(pq)^{m-1}p,$ (5)

for some $m(\geq 2)\in\mathbb{N},$ where $m$ is the least positive integer
satisfying $(\ref{least})$. Then the set

$\\{p,q,pq,qp,\cdots,(pq)^{m-2}p,(qp)^{m-2}q\\}$ (6)

is linearly independent.

###### Proof.

Since $m$ is the least positive integer satisfying (5), therefore by Lemma
2.2, $(pq)^{m-1}\neq(pq)^{m-k}$ for $k\in\mathbb{N}$ where $2\leq k<m.$ Now
consider an element from the linear span of the set $(\ref{collection})$ which
is claimed to be zero,

$\sum_{i}x_{i}(pq)_{i}+\sum_{i}y_{i}(qp)_{i},$ (7)

where $x_{i},y_{i}$ are scalars. Let $l$ denote the smallest order of the
products that appear in $(\ref{linear_com})$ with a nonzero coefficient. First
let assume $l$ to be an even integer, then if $x_{l}\neq 0$, multiplying (7)
by $p$ in left side and with $q$ in right side, respectively, we obtain,

$x_{l}(pq)^{\frac{l}{2}}+\sum_{i\in\\{l+1,l+3,\cdots\\}}(x_{i}+x_{i+1})(pq)^{\frac{i+1}{2}}+\sum_{i\in\\{l,l+2,\cdots\\}}(y_{i}+y_{i+1})(pq)^{\frac{i+2}{2}}=0.$
(8)

Next multiplying (8) by $(pq)^{m-\frac{l}{2}-2}$ we get

$x_{l}(pq)^{m-2}+\left(\sum_{i>l}x_{i}+\sum_{i\geq
l}y_{i}\right)(pq)^{m-1}=0.$ (9)

Now again multiplying (9) by $pq$ and subtracting from (9) we get

$x_{l}\left((pq)^{m-2}-(pq)^{m-1}\right)=0,$

hence we obtain $x_{l}=0$ as $(pq)^{m-2}-(pq)^{m-1}\neq 0,$ a contradiction.
Similarly, let $y_{l}\neq 0,$ then multiply (7) by $pq$ from left and from
right by $p(qp)^{m-\frac{l}{2}-2}$, hence we have

$y_{l}p(qp)^{m-2}-\sum_{i\neq l}(x_{i}+y_{i})p(qp)^{m-1}=0.$ (10)

Since $p(qp)^{m-2}\neq p(qp)^{m-1},$ from (10) we obtain $y_{l}=0$, a
contradiction.

In the case when $l$ is odd, in a similar way, one can verify that
$x_{l}=y_{l}=0$. Hence, the collection $(\ref{collection})$ consists of
linearly independent elements. ∎

If the assumptions outlined in Lemma 2.4 are satisfied by $p$ and $q$, then in
an analogous way like [9, Lemma 5.1], using Lemma 2.4 one can verify the
subsequent lemma.

######  Lemma 2.5.

Let $p\text{ and }q$ satisfy the assumptions of Lemma 2.4.

  1. (i)

If $(qp)^{m-1}=(qp)^{m}$ and $(qp)^{m-1}+(pq)^{m-1}=(qp)^{m-1}q+(pq)^{m-1}p$
then

$p,q,pq,qp,\cdots,(pq)^{m-1},(qp)^{m-1}q$

forms a linearly independent set where $\mathrm{dim}\text{
}\mathrm{alg}(p,q)=4m-3.$

  2. (ii)

If $(qp)^{m-1}\neq(qp)^{m}$ and $(qp)^{m}+(pq)^{m-1}=(qp)^{m-1}q+(pq)^{m-1}p$
then

$p,q,pq,qp,\cdots,(pq)^{m-1}p,(qp)^{m-1}q$

forms a linearly independent set where $\mathrm{dim}\text{
}\mathrm{alg}(p,q)=4m-2.$

  3. (iii)

If $(qp)^{m-1}=(qp)^{m}$ and
$(qp)^{m-1}+(pq)^{m-1}\neq(qp)^{m-1}q+(pq)^{m-1}p$ then

$p,q,pq,qp,\cdots,(pq)^{m-1}p,(qp)^{m-1}q$

forms a linearly independent set where $\mathrm{dim}\text{
}\mathrm{alg}(p,q)=4m-2.$

  4. (iv)

If $(qp)^{m-1}\neq(qp)^{m}$ and
$(qp)^{m}+(pq)^{m-1}\neq(qp)^{m-1}q+(pq)^{m-1}p$ then

$p,q,pq,qp,\cdots,(pq)^{m-1}p,(qp)^{m-1}q,(qp)^{m}$

forms a linearly independent set with $\mathrm{dim}\text{
}\mathrm{alg}(p,q)=4m-1.$

Now, we are ready to establish the main result of this section.

######  Theorem 2.6.

Let $p,q$ satisfy the conditions of Lemma 2.4.

  1. (i)

If $(qp)^{m-1}=(qp)^{m}$ and $(qp)^{m-1}+(pq)^{m-1}=(qp)^{m-1}q+(pq)^{m-1}p$
then here $\mathrm{alg}(p,q)\cong Z_{4m-6}\oplus W_{3}$ with
$\mathrm{dim}\text{ }\mathcal{N}\left(\mathrm{alg}(p,q)\right)=4m-6.$

  2. (ii)

In the case when $(qp)^{m-1}\neq(qp)^{m}$ and
$(qp)^{m}+(pq)^{m-1}=(qp)^{m-1}q+(pq)^{m-1}p$, then $\mathrm{alg}(p,q)\cong
Z_{4m-5}\oplus W_{3}$ and $\mathrm{dim}\text{
}\mathcal{N}\left(\mathrm{alg}(p,q)\right)=4m-5.$

  3. (iii)

If $(qp)^{m-1}=(qp)^{m}$ and
$(qp)^{m-1}+(pq)^{m-1}\neq(qp)^{m-1}q+(pq)^{m-1}p$ then here
$\mathrm{alg}(p,q)\cong Z_{4m-6}\oplus W_{4}$ and $\mathrm{dim}\text{
}\mathcal{N}\left(\mathrm{alg}(p,q)\right)=4m-5.$

  4. (iv)

In the case when $(qp)^{m-1}\neq(qp)^{m}$ and
$(qp)^{m}+(pq)^{m-1}\neq(qp)^{m-1}q+(pq)^{m-1}p$, then here
$\mathrm{alg}(p,q)\cong Z_{4m-5}\oplus W_{4}$ and $\mathrm{dim}\text{
}\mathcal{N}\left(\mathrm{alg}(p,q)\right)=4m-4.$

###### Proof.

Since $p\text{ and }q$ satisfies the assumptions of Lemma 2.4; therefore they
fit precisely into one of the four scenarios described in Lemma 2.5. The
complete characterization of the algebra spanned by $p$ and $q$ can be
achieved by assessing their defining relations. Therefore, two algebras
falling under the same category of Lemma 2.5 are isomorphic.

  1. (i)

Choose $p_{0},q_{0}$ are to be idempotents of the type $Z_{4m-6}$, then
$(q_{0}p_{0})^{m-1}=(p_{0}q_{0})^{m-1}=0$. Furthermore let $p,q$ are
idempotents of type $W_{3}.$ Put

$\bf{p}=\begin{bmatrix}p_{0}&\\\ &p\\\ \end{bmatrix}\text{ and
}\bf{q}=\begin{bmatrix}q_{0}&\\\ &q\\\ \end{bmatrix},$

then $\textbf{p},\textbf{q}\in Z_{4m-6}\oplus W_{3}$ and

$\displaystyle(\textbf{qp})^{m-1}=\begin{bmatrix}0&\\\
&(qp)^{m-1}\end{bmatrix}\text{ and }(\textbf{pq})^{m-1}=\begin{bmatrix}0&\\\
&(pq)^{m-1}\end{bmatrix}.$

Now since $p,q$ are idempotents of type $W_{3}$, therefore $qpq=q,pqp=p$ and
$q+p=qp+pq.$ By using these properties of $p,q$ we get
$(\textbf{pq})^{m}=(\textbf{pq})^{m-1}$,
$(\textbf{qp})^{m}=(\textbf{qp})^{m-1}$ and

$(\textbf{qp})^{m-1}+(\textbf{pq})^{m-1}=(\textbf{pq})^{m-1}\textbf{p}+(\textbf{qp})^{m-1}\textbf{q}.$

Clearly by the construction of p and q they are not tightly coupled and $m$ is
least positive integer satisfying $(\textbf{pq})^{m-1}=(\textbf{pq})^{m}$ and
$(\textbf{pq})^{m-2}\textbf{p}\neq(\textbf{pq})^{m-1}\textbf{p}$. Hence in
this case alg$(\textbf{p,q})$$\cong Z_{4m-6}\oplus W_{3}$ and

$\text{dim }\mathcal{N}\left(\text{alg}(\textbf{p,q})\right)=\text{dim
}\mathcal{N}(Z_{4m-6})+\text{dim }\mathcal{N}(W_{3})=4m-8+2=4m-6.$

Similarly, like ((i)), one can prove the other three case ((iii)), ((iv)) and
((ii)). ∎

##  3 Group and Drazin invertibility

One of the main reasons behind the study of the finite-dimensional algebra
generated by two tightly coupled idempotents is to understand group and Drazin
invertibility in these types of algebras. Since every element in a finite-
dimensional algebra is Drazin invertible, group invertibility, in particular
properly group invertibility (i.e. when an element is group invertible without
being invertible), is the main concern here. The following theorem of [8]
addresses this problem of properly group invertibility in $Z_{m}.$

######  Theorem 3.1.

[8] For $m\geq 1$, $I\in Z_{m}$. Let $A\in Z_{m}$, such that

$A=x_{1}p+y_{1}q+x_{2}pq+y_{2}qp+\cdots,$

where $\\{x_{1},x_{2}\cdots,y_{1},y_{2},\cdots\\}$ are scalars and $p,q$ are
idempotents of type $Z_{m}.$ Then $A$ is properly group invertible if and only
if

  1. (i)

$A=0;$
or

  2. (ii)

either $x_{1}\neq 0,y_{1}=0$ or $x_{1}=0,y_{1}\neq 0$ and $0$ is a root of
$\psi(t)$ with multiplicity at least $l(m,y_{1});$ where

$l(m,y_{1})=\begin{cases}\lceil\frac{m}{4}\rceil-1\text{ if }m=1\text{
}\mathrm{mod}\text{ }4\text{ and }y_{1}=0\\\ \lceil\frac{m}{4}\rceil\text{
otherwise }\end{cases},$

and $\psi(t)$ is defined in $(\ref{hom})$. Moreover
$\sigma(A)=\\{x_{1},y_{1}\\}$.

Here in Theorem 3.1, the claim that the identity element $I\in Z_{m}$ is not
true in general, and due to this, $A$ can be properly group invertible mean
while $x_{1}y_{1}\neq 0.$ This can be seen in the following example.

######  Example 3.2.

Let

$p_{1}=\begin{bmatrix}1&1&0\\\ 0&0&0\\\ 0&0&0\\\ \end{bmatrix}\text{ and
}q_{1}=\begin{bmatrix}0&0&0\\\ 0&1&0\\\ 0&0&0\\\ \end{bmatrix}\in
M_{3}(\mathbb{C}).$

It is easy to check that $p_{1}$ and $q_{1}$ are two idempotents satisfying
$q_{1}p_{1}=0$. Therefore $p_{1},q_{1}$ are idempotents of the type $Z_{3}.$
It is easily seen that rank $(p_{1}+q_{1})=$ rank $(p_{1}+q_{1})^{2},$ hence
$p_{1}+q_{1}$ is properly group invertible. Clearly, here the identity
$I\notin\mathrm{alg}(p_{1},q_{1})$ and the unit element in
$\mathrm{alg}(p_{1},q_{1})$ is the element

$\begin{bmatrix}1&0&0\\\ 0&1&0\\\ 0&0&0\\\ \end{bmatrix}.$

By [9, Lemma 4.2]

$p+q-pq-qp+pqp+qpq-\cdots$ (11)

is the unit in $Z_{m},$ we will denote it by $I_{Z_{m}}$. Due to the
preservation of group invertibility under isomorphism, Böttcher and Spitkovsky
[8, Theorem 2.2] considered only the case when algebra $Z_{m}$ is generated by

$p=\begin{bmatrix}I&B\\\ 0&0\\\ \end{bmatrix}\text{ and
}q=\begin{bmatrix}0&0\\\ C&I\\\ \end{bmatrix},$ (12)

for some suitable $B$ and $C$. For the $p,q$ given in (12), $I=I_{Z_{m}}$. Now
if we consider the idempotents

$p^{\prime}=\begin{bmatrix}I&B_{1}&B_{2}\\\ 0&0&0\\\ 0&0&0\\\
\end{bmatrix}\text{ and }q^{\prime}=\begin{bmatrix}0&0&0\\\ C_{1}&I&C_{2}\\\
0&0&0\\\ \end{bmatrix},$ (13)

then for suitable $B_{1},C_{1}$, alg$(p^{\prime},q^{\prime})=Z_{m}$, but here
$I_{Z_{m}}\neq I$. Note that similarly as given in [8], depending on $m$, it
is possible to choose $B_{1},C_{1}$ such that $p^{\prime},q^{\prime}$ are of
type $Z_{m}$, and no additional condition is required on $B_{2},C_{2}$.
Now if $Z_{m}=\text{alg}(p,q)$ then any elements $A\in Z_{m}$ has the
representation

$A=x_{1}p+y_{1}q+x_{2}pq+y_{2}qp+\cdots,$ (14)

where $x_{i},y_{i}$ are scalars. For further use, we recall the following
functions introduced by [8]

$\displaystyle\upvarphi_{00}(t)$
$\displaystyle=x_{1}+(x_{2}+x_{3})t+(x_{4}+x_{5})t^{2}+\cdots$
$\displaystyle\upvarphi_{11}(t)$
$\displaystyle=y_{1}+(y_{2}+y_{3})t+(y_{4}+y_{5})t^{2}+\cdots$
$\displaystyle\upvarphi_{01}(t)$
$\displaystyle=(x_{1}+x_{2})+(x_{3}+x_{4})t+\cdots$ (15)
$\displaystyle\upvarphi_{10}(t)$
$\displaystyle=(y_{1}+y_{2})+(y_{3}+y_{4})t+\cdots$ $\displaystyle\psi(t)$
$\displaystyle=\upvarphi_{00}(t)\upvarphi_{11}(t)-t\upvarphi_{01}(t)\upvarphi_{10}(t),$

and we also define

$\displaystyle\upvarphi_{02}(t)$
$\displaystyle=x_{1}+x_{3}t+x_{5}t^{2}+\cdots$
$\displaystyle\upvarphi_{12}(t)$
$\displaystyle=y_{1}+y_{3}t+y_{5}t^{2}+\cdots$
$\displaystyle\upvarphi_{02^{\prime}}(t)$
$\displaystyle=x_{2}+x_{4}t+x_{6}t^{2}+\cdots$ (16)
$\displaystyle\upvarphi_{12^{\prime}}(t)$
$\displaystyle=y_{2}+y_{4}t+y_{6}t^{2}+\cdots$ $\displaystyle\psi_{1}(t)$
$\displaystyle=\upvarphi_{00}(t)\upvarphi_{12}(t)-t\upvarphi_{10}(t)\upvarphi_{02^{\prime}(t)}$
$\displaystyle\psi_{2}(t)$
$\displaystyle=\upvarphi_{00}(t)\upvarphi_{12^{\prime}}(t)-\upvarphi_{10}(t)\upvarphi_{02}(t)$

Our next lemma establishes a relation between zeros of the polynomials
$\psi,\psi_{1}$ and $\psi_{2}$ when $y_{1}=0$.

######  Lemma 3.3.

Whenever $y_{1}=0,$ in $(\ref{hom})$ and $(\ref{hom2})$, if $0$ is a root of
the polynomial $\psi$ with multiplicity $n$, then $0$ is also a root of the
polynomials $\psi_{1}$ and $\psi_{2}$ with at least the same multiplicity.

###### Proof.

Since $y_{1}=0$ then we obtain

$\psi(0)=\psi_{1}(0)=\psi_{2}(0)=0.$

Thus, the statement holds for $n=1$ case. Let assume it is true for $n=n-1$
case, i.e. if $0$ is a root of the polynomial $\psi$ with multiplicity $n-1$,
then $0$ is also a root of $\psi_{1}$ and $\psi_{2}$ with multiplicity $n-1.$
Now assume $0$ as a root of $\psi$ with multiplicity $n$, then

$\psi(0)=\diff*{\psi}{t}{(0)}=\cdots=\diff*[n]{\psi}{t}{(0)}=0.$

Again from $\diff*[n]{\psi}{t}{(0)}=0$, choosing $x_{k},y_{k}=0$ for $k<1$ we
obtain

$\displaystyle\diff*[n]{\psi}{t}{(0)}=$
$\displaystyle\sum_{k=0}^{n}\left((x_{2n-2k}+x_{2n-2k+1})(y_{2k}+y_{2k+1})\right.$
$\displaystyle-\left.(x_{2n-2k-1}+x_{2n-2k})(y_{2k+1})(y_{2k+2})\right)$
$\displaystyle=$ $\displaystyle-
x_{2n}y_{2}+x_{2n-1}y_{3}+x_{2n-2}(y_{2}-y_{4})+x_{2n-3}(y_{5}-b_{3})$
$\displaystyle+\cdots+x_{2}(y_{2n-2}-y_{2n})+x_{1}(y_{2n+1}-y_{2n-1})$
$\displaystyle=$ $\displaystyle{}0.$

By previous assumption $0$ is also a root of $\psi_{1}$ with multiplicity
$n-1$, therefore

$\displaystyle\diff*[n-1]{\psi_{1}}{t}{(0)}=$
$\displaystyle{}\sum_{k=0}^{n-1}\left((x_{2n-2k-2}+x_{2n-2k-1})y_{2k+1}\right.$
$\displaystyle-\left.(y_{2n-2k-3}+y_{2n-2k-2})x_{2k+2}\right)$
$\displaystyle=$
$\displaystyle{}-x_{2n-2}y_{2}+x_{2n-3}y_{3}-x_{2n-4}y_{4}+\cdots-
x_{2}y_{2n-2}+x_{1}y_{2n-1}$ $\displaystyle{}=$ $\displaystyle 0.$

Now

$\displaystyle\diff*[n]{\psi_{1}}{t}{(0)}=$
$\displaystyle{}\sum_{k=0}^{n}\left((x_{2n-2k}+x_{2n-2k+1})y_{2k+1}\right.$
$\displaystyle-\left.(y_{2n-2k-1}+y_{2n-2k})x_{2k+2}\right)$ $\displaystyle=$
$\displaystyle-x_{2n}y_{2}+x_{2n-1}y_{3}-x_{2n-2}y_{4}+\cdots-
x_{2}y_{2n}+x_{1}y_{2n+1}$ $\displaystyle=$
$\displaystyle\diff*[n]{\psi}{t}{(0)}+\diff*[n-1]{\psi_{1}}{t}{(0)}.$

Therefore by $\diff*[n]{\psi}{t}{(0)}=0\text{ and
}\diff*[n-1]{\psi_{1}}{t}{(0)}=0$ it follows that
$\diff*[n]{\psi_{1}}{t}{(0)}=0.$ Hence $0$ is a root of the function
$\diff*[n]{\psi_{1}}{t}{}$, it implies $0$ is a root of $\psi_{1}$ with
multiplicity at least $n.$ Similarly one can proof for $\psi_{2}$ also. ∎

Now we are ready to present the version of [8, Theorem 2.2], when $I\notin
Z_{m}.$

######  Theorem 3.4.

Let $n\geq 1$, $I\notin Z_{m}$ and $p,q$ are idempotent of type $Z_{m}.$ If
$A\in Z_{m}$ is in the form $(\ref{represe})$, then $A$ is properly group
invertible if and only if one of the following is true

  1. (i)

$A=0;$

  2. (ii)

$x_{1}y_{1}\neq 0$;

  3. (iii)

either $x_{1}\neq 0,y_{1}=0\text{ or }x_{1}=0,y_{1}\neq 0$ and $0$ is a root
of $\psi(t)$ with multiplicity at least $\lceil\frac{m}{4}\rceil.$

Moreover $\sigma(A)=\\{x_{1},y_{1},0\\}.$

###### Proof.

The case when $m=1$ or $2$ is trivial, so let $m\geq 3.$ It is straightforward
to verify that if $p,q$ is of the form (13), then $I\notin Z_{m}$ and

$A=\begin{bmatrix}\upvarphi_{00}(B_{1}C_{1})&\upvarphi_{01}(B_{1}C_{1})B_{1}&\upvarphi_{02}(B_{1}C_{1})B_{2}+\upvarphi_{02^{\prime}}(B_{1}C_{1})B_{1}C_{2}\\\
C_{1}\upvarphi_{10}(B_{1}C_{1})&\upvarphi_{11}(C_{1}B_{1})&\upvarphi_{12}(C_{1}B_{1})C_{2}+\upvarphi_{12^{\prime}}(C_{1}B_{1})C_{1}B_{2}\\\
0&0&0\\\ \end{bmatrix}.$

As mentioned in [8] here the matrix $B_{1}C_{1}$ is nilpotent with degree
$\lceil\frac{m}{4}\rceil-1$ when $m=1\text{ mod }4$ and $B_{1}C_{1},\text{
}C_{1}B_{1}$ are nilpotent with degree $\lceil\frac{m}{4}\rceil$ in other
cases. Now by [11, Theorem 3.1], if the $2\times 2$ block

$A_{1}=\begin{bmatrix}\upvarphi_{00}(B_{1}C_{1})&\upvarphi_{01}(B_{1}C_{1})B_{1}\\\
C_{1}\upvarphi_{10}(B_{1}C_{1})&\upvarphi_{11}(C_{1}B_{1})\\\ \end{bmatrix}$
(17)

is invertible, then $A$ becomes properly group invertible. According to
Theorem 3.1, if $x_{1}y_{1}\neq 0$, then $A_{1}$ is invertible; hence for the
case $x_{1}y_{1}\neq 0,$ $A$ is properly group invertible. Next, if
$x_{1}=y_{1}=0$, then $A$ is a nilpotent matrix, which is group invertible if
and only if $A=0.$ Thus $(\ref{11k})$ and $(\ref{13k})$ follows.
Now assume $x_{1}\neq 0,y_{1}=0.$ Since $\upvarphi_{00}(0)=x_{1}\neq 0$ and
$B_{1}C_{1}$ is nilpotent, therefore $\upvarphi_{00}(B_{1}C_{1})$ is
invertible, and the Schur complement of $\upvarphi_{00}(B_{1}C_{1})$ in $A$ is

$\frac{1}{\upvarphi_{00}(t)}\begin{bmatrix}{\psi(t)}(C_{1}B_{1})&{\psi_{1}(t)}(C_{1}B_{1})C_{2}+{\psi_{2}(t)}(C_{1}B_{1})C_{1}B_{2}\\\
0&0\\\ \end{bmatrix}.$ (18)

But $\psi(0)=0$, so (18) is a nilpotent matrix. Again the Schur complement of
$\upvarphi_{00}^{2}(t)-t\upvarphi_{01}(t)\upvarphi_{10}(t)(B_{1}C_{1})$ in
$A^{2}$ is

$\frac{1}{\upvarphi_{00}^{2}(t)-t\upvarphi_{01}(t)\upvarphi_{10}(t)}\begin{bmatrix}{\psi(t)}(C_{1}B_{1})&{\psi_{1}(t)}(C_{1}B_{1})C_{2}+{\psi_{2}(t)}(C_{1}B_{1})C_{1}B_{2}\\\
0&0\\\ \end{bmatrix}^{2}.$

Thus rank $(A)$=rank $(A^{2})$ if and only if Schur complement of
$\upvarphi_{00}(t)(B_{1}C_{1})$ in $A$ and Schur complement of
$\upvarphi_{00}^{2}(t)-t\upvarphi_{01}(t)\upvarphi_{10}(t)(B_{1}C_{1})$ in
$A^{2}$ have the same rank. Since $(\ref{schur})$ is a nilpotent matrix;
therefore, this is possible only when $(\ref{schur})$ is the zero matrix. Now
for $Z_{m}$, $B_{1}C_{1}$ is nilpotent matrix of degree
$\lceil\frac{m}{4}\rceil-1$ when $m=1$ mod $4$ and $B_{1}C_{1},C_{1}B_{1}$ are
nipotent with degree $\lceil\frac{m}{4}\rceil$ otherwise. Hence
${\psi(t)}(C_{1}B_{1})=0$ if and only if $0$ is a root of $\psi(t)$ with
multiplicity atleast $\lceil\frac{m}{4}\rceil$. According to Lemma 3.3, if $0$
is a root of $\psi(t)$ with multiplicity $\lceil\frac{m}{4}\rceil$, then $0$
is also a root of $\psi_{1}(t)$ and $\psi_{2}(t)$ with at least the same
multiplicity. Hence we get
${\psi_{1}(t)}(C_{1}B_{1})C_{2}+{\psi_{2}(t)}(C_{1}B_{1})C_{1}B_{2}=0$. This
proves $(\ref{12k})$ when $x_{1}\neq 0$ and $y_{1}=0.$ Likewise, the situation
when $x_{1}=0$ but $y_{1}\neq 0$ can be treated by considering the Schur
complement of $\upvarphi_{11}(C_{1}B_{1}).$ Moreover since
$\sigma(A)=\sigma(A_{1})\cup\\{0\\},$ therefore from Theorem 3.1 it also
follows that $\sigma(A)=\\{0,x_{1},y_{1}\\}$. ∎

With the help of the previous theorem and [8, Theorem 3.5], it is possible to
classify all the properly group invertible elements in the algebra
$Z_{m}\oplus W_{3}$ and $Z_{m}\oplus W_{4}.$ Our next two results describe
that.

######  Theorem 3.5.

If $A\in Z_{m}\oplus W_{3}$ is of the form $(\ref{represe}),$ then $A$ is
properly group invertible if and only if one of the following is true

  1. (i)

$A=0$;

  2. (ii)

$\displaystyle\sum_{i}(x_{2i-1}+y_{2i})=\sum_{i}(x_{2i}-y_{2i})=\sum_{i}y_{i}=0$
and

    1. (a)

$x_{1}y_{1}\neq 0$;
or

    2. (b)

$x_{1},y_{1}$ satisfies condition ((iii)) of Theorem 3.4;

  3. (iii)

$\sum_{i}(x_{i}+y_{i})\neq 0$ and

    1. (a)

all the coefficients of $A$ in $(\ref{represe})$ are $0$, except for the last
three;
or

    2. (b)

$x_{1}y_{1}\neq 0$;
or

    3. (c)

$x_{1},y_{1}$ satisfies condition ((iii)) of Theorem 3.4.

Moreover $\sigma(A)=\\{0,x_{1},y_{1},x+y\\}$ where $x=\sum_{i}x_{i}$ and
$y=\sum_{i}y_{i}.$

###### Proof.

Let $\textbf{p},\textbf{q}$ are idempotents of type $Z_{m}\oplus W_{3}$ then

$\bf{p}=\begin{bmatrix}p_{0}&\\\ &p\end{bmatrix}\text{ and
}\bf{q}=\begin{bmatrix}q_{0}&\\\ &q\end{bmatrix}$ (19)

where $p_{0},q_{0}$ are type $Z_{m}$ idempotents and $p,q$ are idempotents of
type $W_{3}.$ Denote $A_{0}=x_{1}p_{0}+y_{1}q_{0}+x_{2}p_{0}q_{0}\cdots$ and
$A_{1}=x_{1}p+y_{1}q+x_{2}pq\cdots.$ Then by Theorem 3.4 and [8, Theorem 3.5]

$\sigma(A)=\sigma(A_{0})\cup\sigma(A_{1})=\\{0,x_{1},y_{1},x+y\\}.$

According to [8, Theorem 3.5] $A_{1}$ is not invertible; therefore $A$ is
properly group invertible if and only if $A_{0}$ and $A_{1}$ are group
invertible. Since in $W_{3}$

$\displaystyle A_{1}$ $\displaystyle=x_{1}p+y_{1}q+x_{2}pq\cdots$
$\displaystyle=\sum_{i}(x_{2i-1}+y_{2i})p+\sum_{i}(x_{2i}-y_{2i})pq+\sum_{i}(y_{2i-1}+y_{2i})q,$

hence by [8, Theorem 3.5] $A_{1}$ is group invertible if and only if one of
the following holds

  1. 1.

$x_{i},y_{i}=0$ for all $i$;

  2. 2.

$\displaystyle\sum_{i}(x_{2i-1}+y_{2i})=\sum_{i}(x_{2i}-y_{2i})=\sum_{i}(y_{2i-1}+y_{2i})=0$;

  3. 3.

$\displaystyle\sum_{i}x_{i}+y_{i}\neq 0$.

From here, the desired result can be confirmed with the help of Theorem 3.4. ∎

######  Theorem 3.6.

If $A\in Z_{m}\oplus W_{4}$ is of the form $(\ref{represe})$ then $A$ is
properly group invertible if one of the followings holds

  1. (i)

$A=0$;

  2. (ii)

$\displaystyle\sum_{i}x_{2i-1}=\sum_{i}x_{2i}=\sum_{i}y_{2i-1}=\sum_{i}y_{2i}=0$
and

    1. (a)

$x_{1}y_{1}\neq 0$;
or

    2. (b)

$x_{1},y_{1}$ satisfies condition ((iii)) of Theorem 3.4;;

  3. (iii)

$\displaystyle\sum_{i}(x_{i}+y_{i})\neq 0$,
$\displaystyle\left(\sum_{i}x_{2i-1}\right)\left(\sum_{i}y_{2i-1}\right)=\left(\sum_{i}x_{2i}\right)\left(\sum_{i}y_{2i}\right)$
and

    1. (a)

all the coefficients of $A$ in $(\ref{represe})$ are $0$, except for the last
four;
or

    2. (b)

$x_{1}y_{1}\neq 0$;
or

    3. (c)

$x_{1},y_{1}$ satisfies condition ((iii)) of Theorem 3.4.

Moreover $\sigma(A)=\\{0,x_{1},y_{1},x+y\\}$ where $x=\sum_{i}x_{i}$ and
$y=\sum_{i}y_{i}.$

###### Proof.

This follows from Theorem 3.4 and [8, Theorem 3.5]. ∎

As already mentioned, being a finite-dimensional algebra, every element in
$Z_{m}$ is Drazin invertible, and it is possible to provide an upper bound for
the Drazin index of any element $A\in Z_{m}.$ The following theorem describes
this.

######  Theorem 3.7.

Let $m\geq 1$, and $A\in Z_{m}$ is not group invertible. If $A$ is nilpotent
then

$i(A)\leq\begin{cases}2\lfloor\frac{m}{4}\rfloor\text{ if }m=0\text{
}\mathrm{mod}\text{ }4\\\ 2\lfloor\frac{m}{4}\rfloor+1\text{ if }m=1\text{ or
}2\text{ }\mathrm{mod}\text{ }4\\\ 2\lfloor\frac{m}{4}\rfloor+2\text{ if
}m=3\text{ }\mathrm{mod}\text{ }4\end{cases},$

and if $A$ is not nilpotent then $i(A)\leq\lceil\frac{m}{4}\rceil.$

###### Proof.

Since $A\in Z_{m}$, therefore $A$ is of the form (14). Now if $x_{1}y_{1}\neq
0,$ then by Theorem 3.4, $A$ is group invertible, and if exactly one of
$x_{1},y_{1}$ is $0$, then by Theorem 3.4, $A$ is Drazin invertible with
Drazin index $\leq\lceil\frac{m}{4}\rceil$. Finally if $x_{1}=0=y_{1}$, then

$A=x_{2}pq+y_{2}qp+x_{3}pqp+y_{3}qpq+\cdots,$

which is a nilpotent element in alg$(p,q)$. Here using induction, it becomes
evident that $(pq)_{k+1}$ and $(qp)_{k+1}$ are the elements with the lowest
order in the representation of $A^{k}.$ If $m=0$ mod $4$, then any element in
$Z_{m}$ with order $\geq 2\lfloor\frac{m}{4}\rfloor+1$ is equals to $0,$ hence
$A^{2\lfloor\frac{m}{4}\rfloor}=0.$ Therefore $i(A)\leq
2\lfloor\frac{m}{4}\rfloor.$ Likewise, one can confirm the validity of other
cases as well. ∎

######  Remark 2.

Using Theorem 3.7, one can find an upper for the Drazin index of elements in
algebra $U_{m},D_{m},D_{m}^{*}$.

##  4 Representation of Drazin and group inverse

Given that $p,q$ satisfies $(pq)^{m-1}=(pq)^{m}$, then the finite
dimensionality of the algebra alg$(p,q)$ ensures that the Drazin inverse of
$(\alpha p+q)$ always exists and lies in alg$(p,q)$. However, expressing the
representation of $(\alpha p+q)^{D}$ using $p,q$ is still a challenging task.
The representation of $(\alpha p+\beta q)$ under the assumption
$(pq)^{n}=(pq)^{n-1}$ is already provided by Shi and Guolin [6] for
idempotents in a Banach algebra, but they proved it using the integral
representation of the Drazin inverse in a Banach algebra. But here, we provide
a proof which is also suitable for an associative algebra.

######  Theorem 4.1.

Let $\alpha$ be a non-zero scalar. If $(pq)^{m-1}=(pq)^{m}$ then $(\alpha
p+q)$ is Drazin invertible and $i(\alpha p+q)\leq\begin{cases}2,\text{ if
}\alpha\neq-1\\\ 3,\text{ if }\alpha=-1\end{cases}.$ Moreover

$(\alpha p+q)^{D}=\begin{cases}\textbf{A}^{3}(\alpha p+q)^{2},\text{ if
}\alpha\neq-1\\\ \textbf{B}^{4}(\alpha p+q)^{3},\text{ if
}\alpha=-1\end{cases},$ (20)

where

$\displaystyle\textbf{A}=$
$\displaystyle\displaystyle\sum_{i=1}^{2m-3}(-1)^{i-1}\left(\left(\lfloor\frac{i}{2}\rfloor+\frac{\lfloor\frac{i}{2}\rfloor+\phi(i)}{\alpha}\right)(pq)^{\lfloor\frac{i}{2}\rfloor}p^{\phi(i)}\right.$
$\displaystyle+\left.\left(\lfloor\frac{i}{2}\rfloor+\phi(i)+\frac{\lfloor\frac{i}{2}\rfloor}{\alpha}\right)(qp)^{\lfloor\frac{i}{2}\rfloor}q^{\phi(i)}\right)$
$\displaystyle-\left(m-1+\frac{m-1}{\alpha}\right)(qp)^{m-1}-\left(\frac{m-1}{\alpha}-\frac{\alpha}{(1+\alpha)^{2}}\right)(pq)^{m-1}$
$\displaystyle+\left(\frac{m-1}{\alpha}+\frac{1}{(1+\alpha)^{2}}\right)(qp)^{m-1}q,$

and

$\textbf{B}=\displaystyle\sum_{i=1}^{2m-3}\phi(i)\left((qp)^{\lfloor\frac{i}{2}\rfloor}q^{\phi(i)}-(pq)^{\lfloor\frac{i}{2}\rfloor}p^{\phi(i)}\right)$

here the function $\phi$ is defined as $\phi(i)=\begin{cases}0,\text{ if
}i\text{ is even,}\\\ 1,\text{ if }i\text{ is odd}\end{cases}$.

###### Proof.

First let assume that $\alpha\neq-1$, then by [10, Lemma] to prove (20), it is
enough to prove that $\textbf{A}(\alpha p+q)^{3}=(\alpha p+q)^{2}$ and
$(\alpha p+q)^{3}\textbf{A}^{\prime}=(\alpha p+q)^{2},$ for some
$\textbf{A}^{\prime}\in\mathcal{A}.$ Now we compute the coefficient of
$(pq)^{\lfloor\frac{i}{2}\rfloor}p^{\phi(i)}$ in the expansion
$\textbf{A}(\alpha p+q)^{3}$ for $2<i\leq 2m-3$. Let assume $i$ to be even,
then $i=2k$ for some $k\in\mathbb{N}$, and the coefficient of
$(pq)^{\lfloor\frac{i}{2}\rfloor}p^{\phi(i)}\left(=(pq)^{k}\right)$ is

$\displaystyle\left(\frac{k-1}{\alpha}+k-2\right)$
$\displaystyle\alpha-\left(\frac{k-1}{\alpha}+k-1\right)(2\alpha+\alpha^{2})$
$\displaystyle+\left(\frac{k}{\alpha}+k-1\right)(1+\alpha+\alpha^{2})-\left(\frac{k}{\alpha}+k\right)$
$\displaystyle{}=$ $\displaystyle 0.$

Next if $i$ is odd then $i=2k+1$ for some $k\in\mathbb{N},$ then the
coefficient of
$(pq)^{\lfloor\frac{i}{2}\rfloor}p^{\phi(i)}\left(=(pq)^{k}p\right)$ is

$\displaystyle-\left(\frac{k-1}{\alpha}+k-1\right)$
$\displaystyle\alpha^{2}+\left(\frac{k}{\alpha}+k-1\right)(2\alpha^{2}+\alpha)$
$\displaystyle-\left(\frac{k}{\alpha}+k\right)(\alpha+\alpha^{2}+\alpha^{3})+\left(\frac{k+1}{\alpha}+k\right)\alpha^{3}$
$\displaystyle{}=$ $\displaystyle 0.$

In particular, the coefficient of $p\text{ and }pq$ in $\textbf{A}(\alpha
p+q)^{3}$ are $\alpha^{2}$ and $\alpha$, respectively. Hence, we obtain that
in the expansion of $\textbf{A}(\alpha p+q)^{3}$, the coefficient of
$(pq)^{\lfloor\frac{i}{2}\rfloor}p^{\phi(i)}$ is

$\begin{cases}\alpha^{2},\text{ if }i=1\\\ \alpha,\text{ if }i=2\\\ 0,\text{
if }3\leq i\leq 2m-3\end{cases}.$

Likewise, it can be verified that the coefficient of
$(qp)^{\lfloor\frac{i}{2}\rfloor}q^{\phi(i)}$ is

$\begin{cases}1,\text{ if }i=1\\\ \alpha,\text{ if }i=2\\\ 0,\text{ if }3\leq
i\leq 2m-2\end{cases}.$

Now, if $i=2m-2$, then the coefficient of
$(pq)^{\lfloor\frac{i}{2}\rfloor}p^{\phi(i)}\left(=(pq)^{m-1}\right)$ is

$\displaystyle\left(\frac{m-2}{\alpha}+m-3\right)$
$\displaystyle\alpha-\left(\frac{m-2}{\alpha}+m-2\right)(2\alpha+\alpha^{2})$
$\displaystyle+\left(\frac{m-1}{\alpha}+m-2\right)(1+2\alpha+\alpha^{2})$
$\displaystyle-\left(\frac{m-1}{\alpha}-\frac{\alpha}{(1+\alpha)^{2}}\right)(1+\alpha)^{2}$
$\displaystyle=$ $\displaystyle 0$

In a similar manner, it can be confirmed that the coefficient of
$(pa)^{m-1}p,$ $(qp)^{m-1}q$ and $(qp)^{m}$ is $0$ in the expansion of
$\textbf{A}(\alpha p+q)^{3}.$ Hence, through evaluating the coefficients of
each term in the expansion of $\textbf{A}(\alpha p+q)^{3}$, we obtain that

$\textbf{A}(\alpha p+q)^{3}=(\alpha p+q)^{2}.$

Certainly, by choosing

$\displaystyle\textbf{A}^{\prime}=$
$\displaystyle\displaystyle\sum_{i=1}^{2m-3}(-1)^{i-1}\left(\left(\lfloor\frac{i}{2}\rfloor+\frac{\lfloor\frac{i}{2}\rfloor+\phi(i)}{\alpha}\right)(pq)^{\lfloor\frac{i}{2}\rfloor}p^{\phi(i)}\right.$
$\displaystyle\left.+\left(\lfloor\frac{i}{2}\rfloor+\phi(i)+\frac{\lfloor\frac{i}{2}\rfloor}{\alpha}\right)(qp)^{\lfloor\frac{i}{2}\rfloor}q^{\phi(i)}\right)+\left(m-1+\frac{m-1}{\alpha}\right)(qp)^{m-1}$
$\displaystyle+\left(\frac{1}{(\alpha+1)^{2}}+1-m\right)(pq)^{m-1}+\left(m-1+\frac{\alpha}{(1+\alpha)^{2}}\right)(pq)^{m-1}p,$

one can proof $(\alpha p+q)^{3}\textbf{A}^{\prime}=(\alpha p+q)^{2}.$
Therefore if $\alpha\neq-1$ then $i(\alpha p+q)\leq 2$ and

$(\alpha p+q)^{D}=\textbf{A}^{3}(\alpha p+q)^{2}.$

Now if $\alpha=-1$ then

$\displaystyle\textbf{B}(q-p)^{4}=$
$\displaystyle\left(\displaystyle\sum_{i=1}^{2m-3}\phi(i)\left((qp)^{\lfloor\frac{i}{2}\rfloor}q^{\phi(i)}-(pq)^{\lfloor\frac{i}{2}\rfloor}p^{\phi(i)}\right)\right)(q-p)^{4}$
$\displaystyle{}=$
$\displaystyle\left((q-p)+(pq)^{m-1}p-(qp)^{m-1}q\right)(q-p)^{2}$
$\displaystyle{}=$ $\displaystyle(q-p)^{3}+\left((qp)^{m}q-(qp)^{m-1}q\right.$
$\displaystyle\left.+(pq)^{m-1}q-(pq)^{m}p\right)$ $\displaystyle{}=$
$\displaystyle{(q-p)^{3}}$ $\displaystyle{}=$
$\displaystyle(q-p)^{4}\textbf{B}.$

Therefore $i(q-p)\leq 3$ and $(q-p)^{D}=\textbf{B}^{4}(\alpha p+q)^{3}.$ ∎

In this situation, the next question is if $\lambda(pq)^{m-1}=(pq)^{m}$ for
some non-unit scalar $\lambda$, then what conclusions can be drawn regarding
the Drazin invertibility of $(\alpha p+q).$ Chen et al. [14] recently
investigated a specific instance of this problem, specifically when
$pqp=\lambda p.$ In their study, they established the group invertibility
$c_{1}p+c_{2}q+c_{3}pq+c_{4}qp+c_{5}pqp,$ where $c_{1},\cdots,c_{5}$ are
scalar. In the subsequent theorem, we establish the group invertibility of
$(\alpha p+q),$ when $(pq)^{m}=\lambda(pq)^{m-1}.$

######  Theorem 4.2.

Let $\alpha$ be a non-zero scalar. If $\lambda(pq)^{m-1}=(pq)^{m}$ for some
non-unit scalar $\lambda,$ then $(\alpha p+q)$ is group invertible and

$(\alpha p+q)^{g}=\begin{cases}\textbf{A}^{2}(\alpha p+q),\text{ if
}\alpha\neq-1\\\ \textbf{B}^{2}(\alpha p+q),\text{ if }\alpha=-1\end{cases},$
(21)

where

$\displaystyle\textbf{A}=$
$\displaystyle\displaystyle\sum_{i=1}^{2m-3}(-1)^{i-1}\left(\left(\lfloor\frac{i}{2}\rfloor+\frac{\lfloor\frac{i}{2}\rfloor+\phi(i)}{\alpha}\right)(pq)^{\lfloor\frac{i}{2}\rfloor}p^{\phi(i)}\right.$
$\displaystyle\left.+\left(\lfloor\frac{i}{2}\rfloor+\phi(i)+\frac{\lfloor\frac{i}{2}\rfloor}{\alpha}\right)(qp)^{\lfloor\frac{i}{2}\rfloor}q^{\phi(i)}\right)-\left(m-1+\frac{m-1}{\alpha}\right)(qp)^{m-1}$
$\displaystyle+a_{1}(pq)^{m-1}+a_{2}(pq)^{m-1}p+b_{1}(qp)^{m-1}q+b_{2}(qp)^{m},$

and

$\textbf{B}=\displaystyle\sum_{i=1}^{2m-3}\phi(i)\left((qp)^{\lfloor\frac{i}{2}\rfloor}q^{\phi(i)}-(pq)^{\lfloor\frac{i}{2}\rfloor}p^{\phi(i)}\right)+\frac{(qp)^{m-1}q-(pq)^{m-1}p}{1-\lambda}$

here $\phi$ is the same function as defined in Theorem 4.1, and

$\displaystyle a_{1}={}$
$\displaystyle\frac{(\alpha+1)\left(m(\lambda-1)+1-2\lambda\right)}{\alpha(\lambda-1)^{2}},$
$\displaystyle a_{2}={}$
$\displaystyle\frac{m(1+\alpha-\lambda-\alpha\lambda)+(\lambda-\alpha+2\alpha\lambda)}{\alpha(\lambda-1)^{2}},$
$\displaystyle b_{1}={}$
$\displaystyle\frac{m(\alpha+1)(1-\lambda)+\alpha\lambda+2\lambda-1}{\alpha(\lambda-1)^{2}},$
$\displaystyle b_{2}={}$
$\displaystyle-\frac{m(1+\alpha-\alpha\lambda-\lambda)+\lambda(\alpha+1)}{\alpha(\lambda-1)^{2}}.$

###### Proof.

First, consider the case when $\alpha\neq-1$. Similar to the proof of Theorem
4.1, to prove (21), it is enough if we prove that $\textbf{A}(\alpha
p+q)^{2}=\alpha p+q$ and $(\alpha p+q)^{2}\textbf{A}=\alpha p+q.$ Now, one can
verify that in the expansion of $\textbf{A}(\alpha p+q)^{2}$, the coefficient
of $(pq)^{\lfloor\frac{i}{2}\rfloor}p^{\phi(i)}$ is

$\begin{cases}\alpha,\text{ if }i=1\\\ 0,\text{ if }2\leq i\leq
2m-3\end{cases}$

and the coefficient of $(qp)^{\lfloor\frac{i}{2}\rfloor}q^{\phi(i)}$ is

$\begin{cases}1,\text{ if }i=1\\\ 0,\text{ if }2\leq i\leq 2m-2\end{cases}.$

Next the coefficient of $(pq)^{m-1}$ in the expansion of $\textbf{A}(\alpha
p+q)^{2}$ is

$\displaystyle-\left(\frac{m-2}{\alpha}+m-2\right)\alpha+$
$\displaystyle\left(\frac{m-1}{\alpha}+m-2\right)(1+\alpha)+a_{1}(1+\lambda\alpha)$
$\displaystyle+a_{2}(\lambda+\lambda\alpha)$ $\displaystyle{}=$
$\displaystyle(m-1)\left(1+\frac{1}{\alpha}\right)+a_{1}(1+\alpha\lambda)+a_{2}(\lambda+\lambda\alpha).$
(22)

Identically we obtain the coefficient of $(pq)^{m-1}p$ in $\textbf{A}(\alpha
p+q)^{2}$ is

$\displaystyle\left(\frac{m-1}{\alpha}+m-2\right)\alpha+a_{1}(\alpha^{2}+\alpha)+a_{2}(\alpha^{2}+\lambda\alpha).$
(23)

Now solving the system of linear equations obtained from (4.2) and (23) i.e.

$\displaystyle(m-1)\left(1+\frac{1}{\alpha}\right)+a_{1}(1+\alpha\lambda)+a_{2}(\lambda+\lambda\alpha)=$
$\displaystyle 0\text{ and }$
$\displaystyle\left(\frac{m-1}{\alpha}+m-2\right)+a_{1}(\alpha+1)+a_{2}(\alpha+\lambda)=$
$\displaystyle 0$

we get

$\displaystyle a_{1}={}$
$\displaystyle\frac{(\alpha+1)\left(m(\lambda-1)+1-2\lambda\right)}{\alpha(\lambda-1)^{2}},$
$\displaystyle a_{2}={}$
$\displaystyle\frac{m(1+\alpha-\lambda-\alpha\lambda)+(\lambda-\alpha+2\alpha\lambda)}{\alpha(\lambda-1)^{2}}.$

Analogously solving the system of linear equations obtained from the
coefficient of $(qp)^{m-1}q$ and $(qp)^{m}$ in $\textbf{A}(\alpha p+q)^{2}$,
the values of $b_{1},\text{ }b_{2}$ can be confirmed. Hence we get

$\textbf{A}(\alpha p+q)^{2}=\alpha p+q.$

In the same way, by comparing the coefficient of each term in the expansion of
$(\alpha p+q)^{2}\textbf{A}$ one can verify

$(\alpha p+q)^{2}\textbf{A}=\alpha p+q.$

Hence $(\alpha p+q)$ is group invertible and

$(\alpha p+q)^{g}=\textbf{A}^{2}(\alpha p+q).$

Now if $\alpha=-1$ then

$\displaystyle\textbf{B}(q-p)^{2}=$
$\displaystyle(q-p)+(pq)^{m-1}p-(qp)^{m-1}q+\frac{(qp)^{m-1}q-(pq)^{m-1}p}{1-\lambda}(q-p)^{2}$
$\displaystyle=$ $\displaystyle(q-p)$ $\displaystyle=$
$\displaystyle(q-p)^{2}\textbf{B}.$

Therefore $(q-p)$ is group invertible and

$(q-p)^{g}=\textbf{B}^{2}(q-p).$

∎

## References

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# Reimagining an autonomous vehicle

Jeffrey Hawke, Haibo E, Vijay Badrinarayanan, Alex Kendall
Wayve Technologies Ltd., London, United Kingdom

###### Abstract

The self driving challenge in 2021 is this century’s technological equivalent
of the space race, and is now entering the second major decade of development.
Solving the technology will create social change which parallels the invention
of the automobile itself. Today’s autonomous driving technology is laudable,
though rooted in decisions made a decade ago. We argue that a rethink is
required, reconsidering the autonomous vehicle (AV) problem in the light of
the body of knowledge that has been gained since the DARPA challenges which
seeded the industry. What does AV2.0 look like? We present an alternative
vision: a recipe for driving with machine learning, and grand challenges for
research in driving.

> Keywords: autonomous driving, policy learning, deep learning

##  1 Introduction

The modern autonomous driving industry draws its roots from the DARPA Urban
Challenge in 2007 [1]. Since then, we have seen rapid growth of research in
both academic and industrial settings. The majority of this has been in
industry, as many technical problems are beyond the resources of academic
laboratories. The challenge has been harder than anticipated, with a widely
held view that the industry has been humbled by this experience. Nevertheless,
the technology continues to advance and it is likely that we are beyond
Gartner’s _trough of disillusionment_ [2], supported by increasing public
milestones and further investment.

A key question this provokes is simply: are autonomous vehicles (AVs) solved?
All indications point to ‘no’. AV technology has not easily scaled thus far,
hence there must be some gap. A challenge in assessing this question is that
the industry is tight-lipped, due to the competition and the economic
opportunity on offer. Our assessment in this paper is through our own
experience developing autonomous driving systems, with additional insight from
industry publicity, research, and conversations.

We assess the primary factors for this lack of delivery to be: 1) technical
scalability, 2) safety-critical engineering effort, 3) unit economics
(profitability of an AV based on utilisation and costs), 4) regulation. Of
these, we argue that the primary factor is _technical scalability_. By this,
we mean the ability of the decision-making software systems to generalize to
new situations quickly with sufficient performance for deployment. First, let
us discuss the other factors.

Making a safety-critical decision-making system of any type is no small
engineering feat, but it is one that may be solved with time and investment.
Similarly, we do not consider regulation to be the sole barrier. Regulators
worldwide have shown a willingness to support autonomous driving [3, 4, 5],
and we consider this simply a matter of time, pending technological
performance. Finally, we do not believe that unit economics are a dominant
factor. While retrofit AVs are costly [6], these costs have reduced with
maturity and volume such that it could be addressed if sufficient utilisation
and scale was technically possible. This scale problem is that one that
venture capital models are very familiar with: invest until scale is achieved,
bring costs down with scale, then extract profit. Given the availability of
capital, we can discount this as the primary factor.

This leaves us with technical scalability as the underlying problem, which
suggests that we have yet to really solve autonomous driving. Why might this
be? We propose that we have been solving the problem in the wrong direction,
and there is ample opportunity for research to further the field.

The technology that brought us from the DARPA era to today can be described as
solving specialized general intelligence by combining components of even
narrower intelligence. We refer to this as AV1.0. This was necessary given the
constraints and goals of 2007, but arguably limits generalization to complex
situations due to brittle decision-making. The alternative lies in AV2.0: a
complete rethink of what it means to architect a decision-making system for
driving with machine learning.

##  2 AV1.0: The limitations of a classical autonomous driver

AVs today are designed around the same deliberative robotics architecture,
which is arguably an expansion of the sense-plan-act paradigm as in [7]. For a
detailed assessment, we refer the reader to surveys such as [8, 9]. We argue
that the key gap is decision-making intelligence. To assess this, let’s
consider the problem addressed by classical AV components, and limitations on
decision-making.

  1. 1.

Sensing. _Problem_ : Can we observe sufficient information about the
environment to make the correct driving decision? _Limitation_ : Sensor
development has focused on increasing range and fidelity, arguably due to the
industry shift towards autonomous trucking business models on highways.
Improved fidelity is certainly helpful, but this is not a limitation for non-
highway applications as there is sufficient raw information present to make a
decision.

  2. 2.

Scene Representation: building a hand-crafted representation of the world.

    1. (a)

Localization & mapping. _Problem_ : Where is my robot in a known map? This
simplifies real-time decisions by transferring part of the representation
offline with clean, curated data [10]. _Limitation_ : 1) maintenance
complexity as the world changes, and 2) there is an unclear need if downstream
decision-making were sufficiently robust.

    2. (b)

Perception. _Problem_ : Can we extract sufficient context from the raw data
for decision making? _Limitation_ : Modern perception algorithms are extremely
good, and a supervised approach is largely sufficient given data, time, and
resources. We can perceive almost anything desired. Rather, the limitation is
the hand-crafted representation itself. Do we have the information necessary
for the decision? Furthermore, reducing sensor data to symbolic data may not
generalize well, for example, failing to correctly interpret a pedestrian
walking a bicycle given a hand-crafted taxonomy.

    3. (c)

Behavior prediction. _Problem_ : Many decisions require an estimate of future
state, which is complex due to the dynamic scene. _Limitation_ : Prediction is
sensitive to upstream error and has a dependency with planning. This means
that an isolated prediction system will always have some representational
error.

  3. 3.

Planning: decision-making, given a world representation.

    1. (a)

Behavior planning _Problem_ : how should an autonomous vehicle achieve its
goal (e.g., route) given the current representation. _Limitation_ : 1) it is
extremely difficult to determine whether the input representation is necessary
and sufficient for all decisions; 2) it is difficult to separate this from
behavioral prediction, as decisions made by the planner will influence other
actors and thus the prediction; 3) behavioral planners can be described as a
highly engineered _expert system_ [11], which are well known for being
brittle. Despite this immense human effort, we have reached the same
conclusion as thirty years earlier: symbolic expert systems are inherently
limiting [12].

    2. (b)

Motion planning. _Problem_ : generate a metric trajectory for a short horizon,
given local constraints [13]. _Limitation_ : This works well, but cannot
overcome limitations of upstream decisions or hand-crafted constraints.

  4. 4.

Control. _Problem_ : How does a vehicle execute some trajectory? _Limitation_
: Vehicle dynamics and control are relatively well understood with remaining
challenges in low friction.

With the exception of behavior prediction and planning, we consider the
majority of these to be sufficiently mature for driving based on the success
of respective benchmarks. Incrementally increasing perception performance does
not enable a human observer to now make an effective judgement of how a car
should drive given this increase in information. While further gains may be
had, we do not believe any of these areas will offer a step change to unlock
scalable driving.

What evidence exists for behavior prediction and planning as the limiting
factor? We refer to the increasing focus of the AV industry on research
associated with motion prediction. Since 2018, we have seen a flurry of
published research and datasets. Examples include Nutonomy’s NuScenes dataset
[14], Waymo Open Motion Dataset [15], the Lyft Prediction Dataset [16], and
Argo’s Argoverse [17]. A substantial focus of these is prediction, evidenced
by [18, 19, 20, 21].

Given this landscape, there are two possible conclusions. Either, solving
_behavior prediction & planning_ as defined by these boundaries will enable
self-driving, or we need a rethink of this decomposition to achieve an
autonomous future. We think the former is unlikely to be sufficient alone.

##  3 AV2.0: The solution of a data-driven learned driver

Figure 1: How should we reimagine an autonomous vehicle, given the progress
the scientific and engineering academy has made since 2007? We propose that
the classical deliberative architecture needs a rethink. We reflect on the
long-held ‘sense-plan-act’ paradigm (substantially simpler than the robot
architectures used in modern autonomous vehicles), and pose the joint sensing
and planning problem as one that may be solved by data.

###  3.1 Solving driving with data

With the challenges of the classical approach in mind, how should we best
solve driving? It is hard to combine hand-crafted abstraction layers in a
complex decision-making system without brittleness.

We have seen good progress in similar fields by posing a complex problem as
one that is able to modeled end-to-end by data. Examples include natural
language processing with GPT-3 [22], and in games with MuZero [23] and
AlphaStar [24]. In these problems, the solution to the task was sufficiently
complex that hand-crafted abstraction layers and features were unable to
adequately model the problem. Driving is similarly complex, hence we claim
that it requires a similar solution.

The solution we pursue is a holistic learned driver, where the driving policy
may be thought of as learning to estimate the motion the vehicle should
conduct given some conditioning goals. This is different to simply applying
increasing amounts of learning to the components of a classical architecture,
where the hand-crafted interfaces limit the effectiveness of data [25].

At a high level, the route to a learned driver may be expressed very simply as
the following:

  1. 1.

Frame the driving problem as one that may be solved by data.

  2. 2.

Build a source of data, for sampling and curation of data at sufficient scale
and diversity.

  3. 3.

Build a data engine, which is able to train and iterate effectively on this
shifting corpus.

  4. 4.

Build an experimental environment to explore the problem, in simulation and
reality.

  5. 5.

Iteratively improve driving, through changes in modeling, data, and problem
formulation.

The key shift is _reframing the driving problem as one that may be solved by
data_. This means removing abstraction layers used in this architecture and
bundling much of the classical architecture into a neural network, as outlined
in Figure 1. This is frequently depicted as the ‘end-to-end’ approach to
driving [26]. However, this does not mean we believe the driving problem is
simply solved by attaching a neural network directly to vehicle actuators. For
instance, classical control methods together with learned representation and
abstraction layers are still very effective at trajectory following.

###  3.2 Grand challenges for learned driving

With this move away from modularity, we arrive at a deep learning model that
encapsulates much of the function of a classical robotics architecture. We
suggest posing driving as a model-based policy learning problem such as [27]:
one where we learn both a predictive model of the world (conditioned on ego-
vehicle actions) and a model-based policy. We may also retain many of the key
inductive biases for driving [28, 29, 30] as part of this framework. However,
this means more than learning a representation independently of control [31].
To build this next-generation autonomy architecture, we suggest grand
challenges for driving research.

Work on many of these challenges exists in current research across robotics
and machine learning. However, many of these wider research threads have yet
to focus on autonomous driving.

  1. 1.

Vehicle adaptability: deployment with different sensors, vehicle platforms,
and use cases. A learned driver needs to be agnostic to sensor types, sensor
rigs, and robot dynamics. It should adapt the collective experience of all
driving to its vehicle, e.g., [32, 33]. This enables using a policy or
adaptation modules across heterogeneous fleets and cultures.

  2. 2.

Modeling real-world complexity: multi-agent, dynamic environments where future
state is conditioned on current action [34]. To model the true scene dynamics,
a learned driver needs to model prediction and planning jointly, where
prediction is conditioned on action [27]. This enables learning from imagined
states which increases sample efficiency and policy robustness.

  3. 3.

Learning from accessible off-policy data: train and evaluate from scalable
data sources. Currently, much research is focused on online and on-policy
learning, which is unlikely to have impact beyond toy reasoning problems. We
need to learn in a way that is feasible [35], as on-policy scaling is
extremely slow and costly [36]. A particular need is effective off-policy
evaluation to compare driving decisions, e.g., [37].

  4. 4.

Safety under uncertainty: know when and what we don’t know. In addition to
common learning failures [38], a learned driver needs to identify when
possible decisions are uncertain [39, 34, 40]. This enables robust performance
and a safety case, by revoking control to simpler but robust systems to safely
halt the vehicle when no clear decision exists.

  5. 5.

Interpretability of failures: disentangle the causal factors in decisions. A
learned driver needs to be able to identify causal factors [41] in decision-
making where the encoding is sufficiently disentangled [42]. E.g., failing to
stop for a traffic light by incorrectly associating it to our vehicle is
distinct from failures to see it due to weather. This interpretability enables
development and will enable verification [43].

  6. 6.

Generalization to new situations: a generalizable policy requires complete but
lean generalizable representations. A learned driver needs to generalize to a
new distribution every time it goes on the road [44], requiring repurposable
driving knowledge [45]. For example, our driver needs to reason about a
multitude of different bicycle types: road bikes, cargo bikes, and commuter
e-bicycles share many similarities (e.g., use of dedicated cycle lanes), but
have large dissimilarities in visual appearance and speed. Additionally, a
learning-based AV stack is transferable to a new geographic environment with
relative ease, if the representation generalizes. This is a revolution: no
classical AV stack can do this.

  7. 7.

Driving reward: optimization criteria for society’s changing driving needs. A
learned driver may start with supervision, but it may be difficult to go
beyond human performance. Additionally, driving requirements are unlikely to
be static, thus we must adapt to changing driving regulations as AVs become
the dominant road user. For example, the following distance may be decreased
to enable increased vehicle density on a largely automated road. We won’t have
the opportunity to regather expert supervisory data, hence we need a learning
signal to shape driving behavior accordingly. This is largely unexplored [46,
47].

In our opinion, truly solving learned autonomous driving requires solving each
of these challenges. There are wider questions of using deep learning for
decision-making which apply more broadly than driving, including ethics. We
consider these beyond the scope of this paper.

##  4 Learned driving is not uniquely challenged by safety-critical
requirements

Do safety-critical requirements mean that learned driving is impossible? This
medium is insufficient for a comprehensive analysis, but in short, no. There
is no inherent safety reason why we should not pursue a data-driven driver.
This hinges on a number of key theses.

  1. 1.

Safety assurance of an AV depends on (1) the design of an architecture which
includes, but is not limited to, the neural net, plus (2) an engineering
effort in verification and validation. Core safety tools should apply with
some thoughtful adaptation [48].

  2. 2.

Within the broader architecture of the AV, redundant safety may be achieved
with interpretable methods designed to identify and resolve specific failure
modes. These methods cannot offer the generalized decision-making that a
neural net can, but they are able to ensure that a vehicle will not cause harm
in a very specific way [49].

  3. 3.

Scalable safety arguments will use a large non-stationary corpus of empirical
evidence. This is not unique to learned driving, and there is precedent in
other domains such as in medicine [50]. We anticipate that AVs may benefit
from similar methods to quickly revalidate the learned driver as part of a
wider system, however, this is an open question.

##  5 What’s required to address these grand challenges?

To see significant progress in the twenty-first century’s space race, we
suggest the following.

For the academic community, we encourage exploring the full _embodied
intelligence_ problem space [51] beyond just modeling, including data
curriculum, sensor configuration, and robot geometry. Within modeling
research, we encourage even more focus on off-policy learning and evaluation,
to make better use of available data with increased research impact.

For the industry research community, we encourage sharing of data of rare
events and data curricula for the full driving problem for benchmarks. Today
most datasets describe sub-problems under the assumption that the set is
sufficient for solving the whole [14, 15, 16, 17], which we consider
insufficient for driving research. We also encourage continued open
collaboration with academia by sharing progress and the insights on the
problems gathered from real world testing at scale.

Beyond this, we see a need for increased availability of holistic simulators.
Tools such as CARLA [52] are still nascent, and research benchmarking
standards require more efforts to mature.

In solving the driving problem together, we have the potential to unlock both
great societal value and also to discover and create embodied intelligence in
the open world. We encourage research efforts on these grand challenges, to
tackle one of the most complex problems of technical scalability.

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