# Deep Reinforcement Learning based Online Scheduling Policy for Deep Neural
Network Multi-Tenant Multi-Accelerator Systems

Francesco G. Blanco, Enrico Russo, Maurizio Palesi, Davide Patti, Giuseppe
Ascia, Vincenzo Catania<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>of
CataniaCataniaItaly

(2024)

###### Abstract.

Currently, there is a growing trend of outsourcing the execution of DNNs to
cloud services. For service providers, managing multi-tenancy and ensuring
high-quality service delivery, particularly in meeting stringent execution
time constraints, assumes paramount importance, all while endeavoring to
maintain cost-effectiveness. In this context, the utilization of heterogeneous
multi-accelerator systems becomes increasingly relevant. This paper presents
RELMAS, a low-overhead deep reinforcement learning algorithm designed for the
online scheduling of DNNs in multi-tenant environments, taking into account
the dataflow heterogeneity of accelerators and memory bandwidths contentions.
By doing so, service providers can employ the most efficient scheduling policy
for user requests, optimizing Service-Level-Agreement (SLA) satisfaction rates
and enhancing hardware utilization. The application of RELMAS to a
heterogeneous multi-accelerator system composed of various instances of Simba
and Eyeriss sub-accelerators resulted in up to a 173% improvement in SLA
satisfaction rate compared to state-of-the-art scheduling techniques across
different workload scenarios, with less than a 1.5% energy overhead.

††journalyear: 2024††copyright: rightsretained††conference: 61st ACM/IEEE
Design Automation Conference; June 23–27, 2024; San Francisco, CA,
USA††booktitle: 61st ACM/IEEE Design Automation Conference (DAC ’24), June
23–27, 2024, San Francisco, CA, USA††doi: 10.1145/3649329.3657319††isbn:
979-8-4007-0601-1/24/06

##  1\. Introduction

Modern real-world applications present increasingly sophisticated challenges
that demand equally complex solutions. Among various approaches, Deep Learning
(DL), particularly Deep Neural Networks (DNNs), has demonstrated unparalleled
capability in addressing these complex problems across diverse applications
such as image recognition, object detection, natural language processing, IoT,
social networks, medical diagnosis, drug discovery, autonomous driving, and
service robotics. As a result, the demand for and reliance on DNNs have
significantly grown in various domains.

However, executing DNNs is neither straightforward nor inexpensive due to
extensive hardware requirements. With the increasing number of DL models
developed daily, incorporating a rising number of parameters, greater memory,
and computational resources become necessary (xu_nature18, ). Although
computational requirements continue to increase, the same cannot be said for
computational power, due to the slowdown of Moore’s law and Dennard scaling. A
common solution nowadays is to use Domain-Specific Accelerators (DSAs) for
executing DNNs.

On the other hand, applications seeking fast, easy, and cost-effective
solutions are increasingly delegating DNN execution to cloud services, which
benefit from the computational power of large accelerators available in data
centers. In the field of Engineering, there is a trend towards simplifying
complex tasks, making them accessible to users and developers without
extensive technical knowledge. This trend is evident in paradigms like FaaS
(Function-as-a-Service, i.e., Serverless Computing) (baldini_racc17, ), which
is now extended to Machine Learning tasks through the INFaaS (Inference-as-a-
Service) paradigm (romero_usenix21, ). Software developers are delegating such
tasks to cloud services, enabling them to focus on core application aspects.
As a result, efficient hardware resource management becomes crucial for
service providers.

A promising near-term solution is to develop scheduling algorithms that
efficiently map different neural networks from different tenants onto the same
hardware resource, which typically is a multi-accelerator system (MAS),
composed of cores called sub-accelerators (SAs). These algorithms must
guarantee user-provided constraints, particularly ensuring that DNN model
instances are executed within specified maximum deadlines.

This paper presents RELMAS, an approach to address the challenge of concurrent
scheduling of various DNN models on a MAS with heterogeneous SAs. The main
objective is to ensure diverse Quality of Service (QoS) levels, particularly
in meeting maximum deadline constraints. RELMAS is an online scheduling method
that creates a model of the environment, consisting of a MAS with a fixed
number of SAs, and a varying number of instances of different DNN models that
arrive online and need to be scheduled and completed within specified maximum
deadlines. The proposed solution employs Deep Reinforcement Learning,
specifically the Deep Deterministic Policy Gradient (DDPG) algorithm. We
decided to unconventionally coordinate DDPG algorithm with Long-Short-Term-
Memory (LSTM) networks, thus providing a technique to automatically understand
deadline constraints among different DNN models instances and provide a
schedule that effectively balances the workload through space and time. The
algorithm effectively converges, resulting in a substantial reduction in
deadline misses and a significant increase in deadline hits.

##  2\. Related Work

While the topic of multi-tenancy DNN scheduling in conventional general-
purpose cores has been widely studied in both software and hardware since the
introduction of chip-multi-processors, addressing multi-tenancy execution in
DSAs is a more recent endeavor. Diverse approaches have emerged from academic
research that can be classified into two categories (kim_hpca23, ): static and
dynamic. Static mechanisms involve configuring software or hardware statically
(i.e., no decisions are taken at runtime) to adjust the memory access rates of
contentious applications. Dynamic mechanisms, on the other hand, utilize
runtime information to adaptively adjust the contentious nature of an
application based on the actual amount of memory traffic in the system.

Prior works like LayerWeaver (oh_hpca21, ), PREMA (choi2020prema, ), and AI-MT
(baek_isca20, ) propose time-multiplexing DNN execution, either statically or
dynamically. However, they often suffer from low hardware utilization,
especially when layers cannot fully utilize all available resources. To
enhance hardware utilization, static partition techniques such as HDA
(kwon_hpca21, ) and MAGMA (kao_hpca22, ) propose spatial partitioning of
compute or memory resources. However, their static nature hinders adaptability
to different dynamic scenarios. While AI-MT employs multi-accelerators with
homogeneous cores (specifically, systolic array architecture), MAGMA also
supports multi-accelerators with heterogeneous cores. Additionally, the multi-
tenancy support in AI-MT relies on heuristics, while MAGMA uses a research
process based on a genetic algorithm. AI-MT considers potential inter-layer
dependencies, unlike MAGMA. MAGMA is an offline scheduling strategy, requiring
all job characteristics and arrival times to be known beforehand, which does
not align with the requirements of on-demand services. Planaria
(ghodrati2020planaria, ) and Veltair (liu_asplos22, ) propose dynamic
allocation of compute resources for co-running workloads. Despite their
ability to adaptively allocate compute resources, such as processing elements
for Planaria or CPU cores for Veltair, they lack the capability to dynamically
manage memory resources.In contrast, MoCA (kim_hpca23, ) highlights the
importance of memory-centric resource management, proposing a dynamic
partition of both compute and memory resources throughout execution.

Although homogeneous-core accelerators have been thoroughly explored in the
literature, heterogeneous architectures, often incorporating varied dataflows,
enable more efficient processing of a diverse range of DNN layers, serving as
both flexible and cost-effective solutions (symons2022towards, ). However, in
a data center environment that provides on-demand inference services – with
unpredictable job arrivals and Service Level Agreements primarily focused on
execution within set deadlines – the necessity arises for RELMAS, a sample-
efficient, online DNN scheduling algorithm, that mitigates the risk of
resource underutilization and QoS violations while maintaining bandwidth
constraints.

##  3\. Problem formulation

Figure 1. Diagram of the reference Multi-Accelerator Heterogeneous
Architecture used in evaluations.

We consider a fixed hardware accelerator architecture, which consists of $M$
heterogeneous sub-accelerators (SAs), such as the one depicted in Fig.1. In
the context of this multi-accelerator system (MAS), individual SAs exhibit
distinct specifications encompassing dataflow capabilities, processing unit
counts, memory hierarchy configurations, and buffer sizes. All the SAs share
the off-chip memory bandwidth. Such a system, which can ideally be deployed in
both data center and edge scenarios, operates in a multi-tenant fashion,
meaning it can receive and fulfill inference requests from multiple users. The
resulting workload within the MAS can be parallelized at the layer
granularity, allowing for concurrent processing of different layers on
different SAs.

An user request comprises the task of executing a DNN model inference while
adhering to latency constraints specified in a Service Level Agreement (SLA).
We refer to each request as a job. Each job consists of several sub-jobs
(SJs): one for each layer of the requested DNN model. At runtime, all the sub-
jobs to be executed are collected in a virtually infinite ready queue (RQ).
SJs are added to the RQ when a new request arrives and are removed when they
start their execution on a SA or their deadline (stemming from the SLA)
passes.

The objective of this work is to design a scheduling algorithm with the
primary goal of maximizing the SLA satisfaction rate, namely the fraction of
jobs completed while meeting the specified latency requirements, or conversely
minimizing the deadline misses.

We make the reasonable assumption that all potential DNN models that may be
requested are known in advance. This is in line with prior works
(art:microsoft_soifer, ). Layer dependencies and parameters of all these
models are known. Thus, it is possible to compile a table containing: the
latencies $c^{m}_{i,s}$ and the required bandwidths $b^{m}_{i,s}$ to execute
the $s$-th SJ of the $i$-th potential request on the $m$-th SA. Due to the
heterogeneity of the MAS, each layer executed on different SAs exhibits
varying latency, energy, and bandwidth requirements. However, in an online
context, during periods of low system activity, any unfamiliar model could
undergo a “registration phase”, involving compiling latency tables for the new
model. Once registered, these models are integrated into the system.

On the other hand, arrival times of each request $a_{j}$ are not known in
advance. Thus, an online scheduling strategy is needed. Such a scheduler
should determine the start time $s_{j}$ for each SJ and the SA on which the SJ
will be executed. A deadline miss will happen if the finish time
$f_{j}>s_{j}+q_{j}$, where $q_{j}$ is the latency requirement of the request.
The finish time cannot be simply calculated summing the computational costs of
each layer on the assigned SA because of the limited memory bandwidth shared
by all the SAs. When the sum of the bandwidth required by multiple SJs
executing in parallel on multiple SAs exceeds the available memory bandwidth,
all the SJs experience a slowdown proportional to the respective required
bandwidth. This translates to the same amount of stall cycles for all the
overlapping sub-jobs. Hence, it is important for the online scheduler to
wisely orchestrate the execution of compute-intensive and memory-bound sub-
jobs in the RQ to minimize bandwidth contentions.

The described problem can be considered as a generalization of the Flexible
Job Shop Scheduling (FJSP) problem and, in Graham’s notation
(graham1979optimization, ), it can be expressed as $FJSP\ |\ prec\ |\ U$,
where $prec$ refers to the presence of precedence constraints among the SJs
and $U$ is the number of tardy jobs, which is the objective to be minimized.
It has been proved that the FJSP problem is NP-hard (dauzere2023flexible, ).

##  4\. RELMAS Online Scheduler

(a) High-level visualization of the RELMAS policy in action during deployment.

(b) RELMAS policy learning process.

Figure 2. Overview of the proposed online scheduler in production and learning
phases.

In this section we introduce RELMAS, the proposed approach to online
scheduling in a multi-accelerator heterogeneous system based on Deep
Reinforcement Learning (DRL). For this section, we adopt a top-down approach:
we first depict the working of RELMAS in a deployment setting, then we deepen
the RELMAS agent learning process. For an introduction to the fundamentals of
deep reinforcement learning, please refer to (arulkumaran_spm17, ).

###  4.1. Scheduling

In the target scenario, the environment consists in the MAS executing DNN
layer inferences and receiving new requests. The latter represent the
stochastic component of the environment. The state of the environment at any
given time includes the current system load information, i.e., the number of
cycles for which each SA will be busy executing a previously assigned SJ, and
the ready queue (RQ), i.e., the collection of SJs ready to be executed. The
action, which is the decision taken by the RELMAS policy agent for each SJ,
can be articulated in a priority, defining temporal scheduling, and a SA
choice, defining spatial allocation of the SJ in the MAS. Fig. 2(a) shows an
overview of RELMAS in a deployment setting. It can be noted that the scheduler
is triggered periodically with a period $T_{s}$. During a period, the MAS
execute the latest elaborated schedule, which is represented graphically in
the space-time graph in Fig. 2.1. At the end of the period, SJs remained
unexecuted are collected in the so called residual RQ (Fig. 2.3). These SJs
are merged with the new SJs related to job requests arrived and queued in the
meantime (Fig. 2.4).

This overall RQ (Fig. 2.5) is sorted in ascending order of absolute deadline
and submitted for scheduling by the RELMAS policy. We decided to implement the
RL policy as a Long-Short-Term-Memory (LSTM) recurrent neural architecture.
The rationale is that it helps dealing with a state of variable size (variable
number of SJs in RQ) and LSTMs are good in detecting temporal relationships
and patterns of SJs sequences. The LSTM cell with a hidden size of $h$ is
succeeded by two fully connected layers: the first projects to $h/2$ units and
applies a Rectified Linear Unit (ReLU) activation function, while the second
projects to $G$ values and applies Hyperbolic Tangent (Tanh) activation
function, encoding the decision, i.e., the action, for the corresponding SJ.

As shown in Fig. 2.6, the LSTM sequentially ingests the encoded representation
of each SJ in the RQ. Hence, if $R$ sub-jobs reside in the RQ, $R$ timesteps
of the LSTM policy are executed. The encoded representation (Fig. 2.7) of each
SJ is an array of features including all the information that could be useful
to derive an optimal scheduling: an identifier of the model requested of which
the SJ represent a layer, an identifier of the layer, the deadline
representing the latency constraint111The deadline for each job request is
explicitly stated and reiterated within the representation of each SJ. This
redundancy enables the policy to differentiate between two identical SJs (same
layer, same model), each associated with distinct request latency
constraints., the waiting time informing about the distance in time from the
arrival time of the request, the computational times and bandwidth
requirements on the SAs. Thus, the length $F$ of each encoding is equal to
$4+2M$.

In Fig. 2.5 it is also possible to spot an extra element prepended to the
encoded representation of the RQ. As no preemption is assumed, it is important
to instruct the policy about the duration for which each SA will be busy, we
do this by using a virtual SJ as a primer for the LSTM. All its features are
zero except for computational times that instead represent the number of
cycles before each SA will be available.

As output of the LSTM policy, after discarding the first timestep
(corresponding to the primer) we get a decision encoding (Fig. 2.8) for each
SJ in the ready queue. Due to the Tanh activation function, all the output
values relies in the $[-1,1]$ range that for each SJ are: the temporal
priority and one value $u_{m}$ for each SA. The index associated with the
highest $u_{m}$ determines the spatial allocation of the SJ in the MAS. Thus,
the length $G$ of each encoding is equal to $1+M$.

###  4.2. Learning

As shown in Fig. 2(b), we employ Deep Deterministic Policy Gradient (DDPG)
which has emerged as a prominent algorithm in the field of reinforcement
learning, particularly for solving high-dimensional continuous action space
problems. For the sake of brevity, we will not describe all the mathematical
details of DDPG, please refer to (lillicrap2015continuous, ) for a detailed
explanation. In this section, we instead focus on the choices made to adapt
DDPG to the problem addressed by this paper.

As in other RL algorithms, the goal in DDPG is to learn an optimal policy that
maximizes a cumulative reward signal over time. To accomplish this, DDPG
employs two essential components: the actor and the critic networks. Both are
aimed to construct parameterized functions that iteratively converge to the
corresponding optimal functions. Due to the exceptional capability of DNNs to
approximate non-linear functions, two neural networks are used for these
components. The actor network represent the parameterized deterministic policy
function $\mu_{\theta}$ and provides a continuous action output given an input
state representation. On the other hand, the critic network estimates the
Q-function $Q_{\phi}$, which quantifies the expected cumulative reward of
following a given policy for a specific state and action pair.

We already discussed the policy as an LSTM-based architecture. Likewise, we
propose to adopt the same architecture for the critic network (Fig. 2.9). The
state and action pairs are concatenated horizontally as shown in Fig. 2.10 and
provided as input to the critic for Q-value estimation. Thus, the input length
at each timestep is equal to $F+G$, while only one output value
$Q_{\phi}(s,a)$ per timestep is projected from the last hidden state.

As in the baseline DDPG algorithm, experiences are stored in a replay buffer
(Fig. 2.11) and sampled from it during the training phase for weight updates.
Each experience is a tuple $(s_{t},a_{t},r_{t},s_{t+1})$: $s_{t}$ is the
environment state encoding at time $t$, $a_{t}$ is action representation of
the decision taken by the policy, $r_{t}$ is the reward value obtained as an
effect and $s_{t+1}$ is the state representation after executing the action.
As we assume RELMAS to work in a periodic fashion and due to the stochastic
request arrivals, $s_{t+1}$ can also include stochastic request arrivals,
therefore breaking a deterministic causality chain from $s_{t}$ to $s_{t+1}$.
Indeed, we found out that learning performance increases when $s_{t+1}$ only
encodes the residual RQ instead of the overall RQ, eliminating the stochastic
component from the experiences and allowing the algorithm to effectively infer
causality chains.

Regarding the reward function, we defined it as follows:

$\displaystyle r_{t}$
$\displaystyle=\sum_{l\in{RQ}_{t}}\Delta_{t,l}\left(A_{t,l}+\gamma\frac{(a_{l}+q_{l})-f_{l}}{q_{l}}\right)$

$\Delta_{t,l}=\begin{cases}1,&f_{l}<t+T_{s}\\\
\delta,&\text{otherwise}\end{cases}$ $A_{t,l}=\begin{cases}\alpha,&f_{l}\leq
a_{l}+q_{l}\\\ -\beta,&\text{otherwise}\end{cases}$

where $\delta\in[0,1]$ quantifies the relevance of events beyond the next
scheduling period, $\alpha$ is the deadline hit reward, $\beta$ is the
deadline miss penalty and $\gamma$ weights the significance of SJs normalized
slacks. The introduction of normalized slacks, implemented as a scaled sum
term, effectively mitigates resource underutilization. While compliance with
SLA requirements is the foremost priority, this strategy additionally
incentivizes the reduction of resource usage duration. This alignment serves
the dual objective of not only fulfilling service commitments but also
boosting operational efficiency.

##  5\. Evaluation

In this section, we present the evaluation of our novel online scheduling
technique. To conduct our evaluation, we considered the hardware
specifications detailed in Table 1. Our chosen hardware configuration is
rooted in a heterogeneous accelerator architecture following the principles
described in (kwon_hpca21, ), boasting an equivalent off-chip memory bandwidth
and total number of MAC units as in other works focusing on systolic arrays
(kim_hpca23, ). We are aiming for a multi-chip-module-based architecture in
which sub-accelerators take the form of chiplets interconnected by means of a
network-on-package (NoP). The on-package links utilize a silicon interposer
and incorporate efficient intra-package signaling circuits, as in
(shao_micro19, ). The selected architectural configuration comprises six sub-
accelerators, which are instances of two widely recognized state-of-the-art
DNN accelerators, as visualized in Fig.1. Specifically, half of these sub-
accelerators adopt a row-stationary dataflow and an internal organization
closely resembling Eyeriss(chen2019eyeriss, ), while the remaining half employ
a weight-stationary dataflow and an architecture akin to a Simba
(shao_micro19, ) chiplet. We have chosen to integrate both small and large
instances from these two categories of sub-accelerators (see Table 1). This
strategic decision enables more effective orchestration of memory-bound and
compute-intensive workloads.

Table 1. Parameters of the sub-accelerators classes considered in evaluations. | Eyeriss | Simba
---|---|---
Parameter | Small | Large | Small | Large
Frequency | 1 GHz
DRAM Bandwidth | 16 GB/s
NoP Bandwidth, Energy | 100 GB/s, 1.3 pJ/bit
Dataflow | Row Stationary | Weight Stationary
Num. PEs | 256 | 512 | 16 | 32
MACs per PE | 1 | 1 | 16 | 16
Global Buffer | 64 KiB | 64 KiB | 32 KiB | 64 KiB
PE Buffers | 220 B | 220 B | 24 KiB | 24 KiB
Table 2. Benchmark workloads used in evaluations. Workload | Domain | DNN Models
---|---|---
Light | Image Classification | SqueezeNet
| Object Detection | YOLO-Lite
| NLP | Keyword Spotting
Heavy | Image Classification | AlexNet, InceptionV3, ResNet50
| Object Detection | YOLO-V2
Mixed |  Light $\cup$ Heavy

In our evaluations we consider seven different state-of-the-art DNN models
from image classification, object detection and natural language processing
(NLP) domains. These DNN models features different model sizes, computational,
memory and bandwidths requirements rendering them highly representative of a
wide spectrum of DNN workloads. Similarly to (ghodrati2020planaria, ) and
(kim_hpca23, ), we organized these models in three workload sets classified by
size as summarized in Table 2. To generate multi-tenant scenarios from each of
these three workload sets, we consider a time interval during which inter-
arrival times of inference requests are drawn from a Pareto distribution,
emulating task dispatching in data centers according to insights from
(art:googleworkload, ).

Each request is associated with a SLA requirement defined by the QoS level.
Adopting the PREMA (choi2020prema, ) approach, the latency requirement is
calculated multiplying a coefficient, that we name QoS factor, by the minimum
execution latency of the requested job (i.e., without interference of other
jobs). Specifically, we identify a QoS factor for the baseline QoS level (QoS-
Medium), then we derive other two targets: QoS-Low which indicates relaxed
latency constraints corresponding to $1.2\times$ QoS-Medium and QoS-High
denoting $0.8\times$ QoS-Medium, similarly to (ghodrati2020planaria, ) and
(kim_hpca23, ).

We assess the energy, latency and memory bandwidth requirements figures of
each DNN layers on each sub-accelerator using the validated Timeloop/Accelergy
(parashar2019timeloop, ) simulation infrastructure. Additionally, we develop a
multi-accelerator multi-tenant simulation platform that takes into account
layer latencies, layer dependencies and runtime memory bandwidth contentions
to accurately determine the start and finish times of each sub-job when
executing according to a specific scheduling algorithm. This methodology
allows us to effectively evaluate the target metrics. Across all the
evaluations, the following reward coefficients were used during RELMAS
learning: $\alpha=0.10$, $\beta=0.11$, $\gamma=0.05$ and $\delta=0.01$.

###  5.1. SLA Satisfaction Rate

(a) Light Workload

(b) Heavy Workload

(c) Mixed Workload

Figure 3. SLA Satisfaction Rate comparison against other baselines for
different workload sets.

In Fig. 3 we show the results of the evaluation of the three workload sets in
Table 2 with three different QoS targets. We adopt the SLA Satisfaction Rate
as our primary comparative metric. To evaluate the efficacy of the proposed
approach, we conduct comparative analyses against robust baselines, which draw
inspiration from and are adapted from previous works in the field. The
following baseline algorithms are considered: (1) FCFS-H, which is the First-
Come-First-Served scheduling algorithm for priority assignment to each sub-job
in the ready queue, paired with an heuristic for the selection of the sub-
accelerator. The heuristic consists in choosing the SA which minimizes the
finish time of the sub-job. (2) PREMA-H, which is a scheduling algorithm based
on (choi2020prema, ), combining a token mechanism based on waiting time and
Shortest-Job-First priority assignment. As the original PREMA work targets a
monolithic architecture, for a fair comparison we combine it with the
aforementioned heuristic. (3) Herald (kwon_hpca21, ), which is a layer
scheduling algorithm designed for heterogenous architectures, focusing on load
balancing of the SAs. (4) MAGMA (kao_hpca22, ), consisting in a genetic
algorithm approach to scheduling optimization. Our implementation adopts the
same custom genetic operators of the original work. For a fair comparison, we
used the same evaluation platform as RELMAS and adapted its fitness function
to take into account the SLA satisfaction rate.

All the evaluated algorithms are supposed to be triggered for each scheduling
period at runtime. For each tested scenario we trained RELMAS considering a
hidden size of 256 for both the actor and the critic. As shown in Fig. 3, the
trained policy is able to match or outperform the other methods across all the
scenarios in terms of SLA satisfaction rate.

With respect to Herald, RELMAS achieves a geometric mean improvement of
$59.4\%$, while with respect to PREMA-H it achieves a geometric mean
improvement of $109\%$. The gratest improvements with respect to these two
baselines are achieved for the Mixed workload. We believe, this effect should
be attributed to high variability of model sizes in the Mixed workload, that
cause the scheduling process to be more complex. In particular, the Shortest-
Job-First strategy of PREMA, allocates more priority to the Light models in
the Mixed workload starving the Heavy models.

RELMAS achieves the same SLA Satisfaction Rate of MAGMA in the Light and Heavy
workloads. However, it is worth noting that MAGMA is executed with same
parameters of the original paper, i.e., 100 generations of 100 individuals,
for each scheduling period to optimally scheduling a ready queue. For the
evaluation of individual fitness, we employ the identical simulation platform,
previously delineated and utilized for RELMAS training. These choices resulted
in MAGMA scheduling runtimes of several minutes on a standard workstation for
each simulated scheduling period, which is reasonable as MAGMA, unlike RELMAS,
was born to be an offline scheduling policy, not an online one. While this
places our MAGMA implementation in the realm of limited practicality within
real-world contexts, it serves as a strong benchmark for evaluating the
proposed approach. Furthermore, RELMAS is capable to outperform MAGMA in a
Mixed workload workload scenario with up to $22.5\%$. SLA Satisfaction Rate
improvement. We will analyze the overhead of the proposed solution in Sec.
5.3.

###  5.2. Bandwidth Sensitivity

Figure 4. Impact of memory bandwidth reduction on SLA Satisfaction Rate for
different scheduling strategies.

RELMAS is designed to be aware of the bandwidth required by each DNN layer to
be executed without stalls. To this end, this information is embedded as a
feature in encoded LSTM policy input. This makes the proposed technique more
robust in memory-constrained contexts. We test a Light workload scenario and
in Fig. 4 we show the SLA Satisfaction Rate of different scheduling strategies
normalized to their respective best result achieved. It can be noted that,
independently from their absolute performance, some scheduling approaches tend
to experience an abrupt degradation when decreasing the memory bandwidth, with
respect to their performance in a bottleneck-free situation. On the other
hand, RELMAS degradation is slower, especially in the low-bandwidth region,
indicating an higher capability of the proposed approach to better orchestrate
layers on the heterogeneous hardware optimizing memory bandwidth utilization.
We found that the same conclusions remain valid for the other considered
workloads; therefore, we are reporting the results for the Light workload
only.

###  5.3. Overhead Assessment

We now analyze the overhead of the proposed solution for the online scheduling
in a multi-accelerator system. As already described, RELMAS consists in a DNN
model composed of an LSTM cell followed by two projection fully connected
layers. The policy works recurrently ingesting one input state encoding at
each timestep, one for each layer in the ready queue. To gain insight into the
computational complexity, it is informative to note that for a policy with an
hidden size of 256, the number of Multiply-Accumulate (MAC) operations per
layer amounts to 316 288, representing about a mere $0.04\%$ and $0.007\%$ of
the MAC operations required by the AlexNet and ResNet50 models respectively.
We opted to implement the scheduling policy within one of the accelerators
designated for the policy’s execution. This was achieved by communicating the
accelerator’s busy status to the scheduler, which was done by adding the
overhead time to the first element in the State encoding, as shown in Fig.
2.5. Our experiments revealed that this implementation incurred no overhead
impact on the scheduling process itself. We conducted an accurate evaluation
of the energy overhead assuming that the scheduler is executed on the MAS
itself. In particular, we assumed that one of the Simba Small SAs (Fig. 1) is
responsible for running each timestep of the RELMAS policy. We evaluated the
energy required for this setup using Timeloop (parashar2019timeloop, ). We
then assessed the overall energy overhead to the workload execution
considering different hidden sizes for the LSTM cell and the Mixed workload.
The results are shown in Fig. 5. We found that, for the considered workload,
there is no significant SLA Satisfaction Rate improvement for an hidden size
bigger than 128. In any case, the energy overhead does not exceed $1.3\%$. We
also noticed that decreasing the scheduling period can increase the
performance of the scheduler, because it enables shorter reaction times. This
comes to the cost of a slight increase in energy overhead due to the
increasing average length of residual ready queues. In other words, a layer
will be scheduled multiple times on average before being actually executed,
because it is often forwarded to the next scheduling period.

Figure 5. Energy overhead of the proposed scheduling algorithm varying the
hidden size of the LSTM policy and the scheduling period.

##  6\. Conclusion

This paper presents RELMAS, a low-overhead deep reinforcement learning based
online scheduling policy for DNN multi-tenant heterogeneous multi-accelerator
systems. We believe that reinforcement learning could be a viable and low-
overhead solution for online scheduling and dispatching in multi-accelerator
multi-tenant heterogenous systems. For the proposed methodology, we included
computation times and memory bandwidths requirements as features for the RL
policy. We think many more could be explored to enable other functionalities.
For instance, tenant-wise SLA satisfaction and fairness awareness could be
enabled by exploring new state encoding and online tenant-wise deadline misses
closed-loop control with dynamic reward shaping. Furthermore, we
unconventionally considered an LSTM architecture for the policy but other
trending architectures could be explored and their overhead assessed such as
transformer-based models.

###### Acknowledgements.

This work has been (partially) supported by MUR project ARS01_00592 reCITY.
The work of M. Palesi, who has contributed to the development of Secs. 3 and
4, has been supported by the Spoke 1 ”FutureHPC & BigData” of the Italian
Research Center on High-Performance Computing, Big Data and Quantum Computing
(ICSC). The work of V. Catania, who has contributed to Sec. 5, has been
supported by PNRR MUR project PE0000013-FAIR.

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Transposed Poisson structures on
generalized Witt algebras and Block Lie algebras111 The work was supported by
FCT UIDB/MAT/00212/2020, UIDP/MAT/00212/2020, 2022.02474.PTDC and by CMUP,
member of LASI, which is financed by national funds through FCT — Fundação
para a Ciência e a Tecnologia, I.P., under the project with reference
UIDB/00144/2020. 222Corresponding author<EMAIL_ADDRESS>

Ivan Kaygorodov333CMA-UBI, Universidade da Beira Interior, Covilhã, Portugal;
<EMAIL_ADDRESS>& Mykola Khrypchenko444 Departamento de Matemática,
Universidade Federal de Santa Catarina, Brazil; and CMUP, Departamento de
Matemática, Faculdade de Ciências, Universidade do Porto, Rua do Campo Alegre
s/n, 4169–007 Porto, Portugal<EMAIL_ADDRESS>

Abstract: We describe transposed Poisson structures on generalized Witt
algebras $W(A,V,\langle\cdot,\cdot\rangle)$ and Block Lie algebras $L(A,g,f)$
over a field $F$ of characteristic zero, where $\langle\cdot,\cdot\rangle$ and
$f$ are non-degenerate. More specifically, if $\dim(V)>1$, then all the
transposed Poisson algebra structures on $W(A,V,\langle\cdot,\cdot\rangle)$
are trivial; and if $\dim(V)=1$, then such structures are, up to isomorphism,
mutations of the group algebra structure on $FA$. The transposed Poisson
algebra structures on $L(A,g,f)$ are in a one-to-one correspondence with
commutative and associative multiplications defined on a complement of the
square of $L(A,g,f)$ with values in the center of $L(A,g,f)$. In particular,
all of them are usual Poisson structures on $L(A,g,f)$. This generalizes
earlier results about transposed Poisson structures on Block Lie algebras
$\mathcal{B}(q)$.

Keywords: Transposed Poisson algebra, generalized Witt algebra, Block Lie
algebra, $\delta$-derivation.

MSC2020: primary 17A30; secondary 17B40, 17B61, 17B63.

## Introduction

Poisson algebras originated from the Poisson geometry in the 1970s and have
shown their importance in several areas of mathematics and physics, such as
Poisson manifolds, algebraic geometry, operads, quantization theory, quantum
groups, and classical and quantum mechanics. One of the popular topics in the
theory of Poisson algebras is the study of all possible Poisson algebra
structures with fixed Lie or associative part [1, 17, 9, 10]. Recently, Bai,
Bai, Guo, and Wu [2] have introduced a dual notion of the Poisson algebra,
called transposed Poisson algebra, by exchanging the roles of the two binary
operations in the Leibniz rule defining the Poisson algebra. They have shown
that a transposed Poisson algebra defined this way not only shares common
properties of a Poisson algebra, including the closedness under tensor
products and the Koszul self-duality as an operad, but also admits a rich
class of identities. More significantly, a transposed Poisson algebra
naturally arises from a Novikov-Poisson algebra by taking the commutator Lie
algebra of the Novikov algebra. Thanks to [3], any unital transposed Poisson
algebra is a particular case of a “contact bracket” algebra and a quasi-
Poisson algebra. Later, in a recent paper by Ferreira, Kaygorodov, and
Lopatkin a relation between $\frac{1}{2}$-derivations of Lie algebras and
transposed Poisson algebras has been established [7]. These ideas were used to
describe all transposed Poisson structures on Witt and Virasoro algebras in
[7]; on twisted Heisenberg-Virasoro, Schrödinger-Virasoro and extended
Schrödinger-Virasoro algebras in [19]; on oscillator algebras in [3]; Witt
type Lie algebras in [12]. It was proved that each complex finite-dimensional
solvable Lie algebra has a non-trivial transposed Poisson structure [13]. The
${\rm Hom}$\- and ${\rm BiHom}$-versions of transposed Poisson algebras and
transposed Poisson bialgebras have been considered in [14, 15]. The algebraic
and geometric classification of $3$-dimensional transposed Poisson algebras is
given in [4]. For the list of actual open questions on transposed Poisson
algebras see [3].

The first non-trivial example of a transposed Poisson algebra was constructed
on the Witt algebra with the multiplication law $[e_{i},e_{j}]=(i-j)e_{i+j}$
for $i,j\in\mathbb{Z}$ (see, [7]). This attracted certain interest to the
description of transposed Poisson structures on Lie algebras related to the
Witt algebra. Thus, all transposed Poisson structures on the Virasoro algebra
[7], Block type Lie algebras and Block type Lie superalgebras [11], Witt type
Lie algebras [12] have been described. In the last years, the concept of Witt
type and Block type Lie algebra has been enlarged and generalized by various
authors, such as Kawamoto, Osborn, Đoković, Zhao, Xu, Passman, Jordan, etc.
(see, for example, [11, 12] and references therein). In the present paper, we
study transposed Poisson structures on the class of generalized Witt algebras
defined by Đoković and Zhao in [6] and Block algebras defined by Block in [5].
We use the standard method of characterization of transposed Poisson algebra
structures on a fixed Lie algebra $\mathfrak{L}$ based on the description of
the space of $\frac{1}{2}$-derivations of $\mathfrak{L}$.

Our work consists of two main parts. Section 2 is devoted to a description of
$\frac{1}{2}$-derivations and transposed Poisson structures on generalized
Witt algebras $W(A,V,\langle\cdot,\cdot\rangle)$, which result in the
following theorem.

######  Theorem A (Theorem 13).

Let $W(A,V,\langle\cdot,\cdot\rangle)$ be a generalized Witt algebra with non-
degenerate $\langle\cdot,\cdot\rangle$ and $\mathrm{char}(F)=0.$

  1. (i)

If $\dim(V)>1$, then all the transposed Poisson algebra structures on
$W(A,V,\langle\cdot,\cdot\rangle)$ are trivial.

  2. (ii)

If $\dim(V)=1$, say, $V=\operatorname{span}_{F}\\{v\\}$, then the transposed
Poisson algebra structures on $W(A,V,\langle\cdot,\cdot\rangle)$ are exactly
mutations of the product $(a\otimes v)\cdot(b\otimes v)=(a+b)\otimes v$.

In Section 3 we study the same questions on Block Lie algebras $L(A,g,f)$ and
obtain the following result.

######  Theorem B (Theorems 22 and 29).

Let $L(A,g,f)$ be a Block Lie algebra with non-degenerate $f$ and
$\mathrm{char}(F)=0.$

  1. (i)

If $g=0,$ then there is only one non-trivial transposed Poisson algebra
structure $\cdot$ on $L(A,0,f)$. It is given by $u_{0}\cdot u_{0}=u_{0}.$

  2. (ii)

If $g\neq 0$ and $(g(a),h(a))\neq(0,-1)$ for all $a\in A$, then all the
transposed Poisson algebra structures on $L(A,g,f)$ are trivial.

  3. (iii)

If $g\neq 0$ and there is $a\in A$, such that $(g(a),h(a))\neq(0,-1)$, then
the transposed Poisson algebra structures on $L(A,g,f)$ are usual Poisson
algebra structures that are extensions by zero of commutative associative
products $*$ on the complement $V=\operatorname{span}_{F}\\{u_{a}\mid
g(a)=h(a)+2=0\\}$ of $[L,L]$ with values in
$Z(L)=\operatorname{span}_{F}\\{u_{a}\mid g(a)=h(a)+1=0\\}$.

##  1\. Definitions and preliminaries

All the algebras below will be over a field $F$ of characteristic zero and all
the linear maps will be $F$-linear, unless otherwise stated.

######  Definition 1.

Let ${\mathfrak{L}}$ be a vector space equipped with two nonzero bilinear
operations $\cdot$ and $[\cdot,\cdot].$ The triple
$({\mathfrak{L}},\cdot,[\cdot,\cdot])$ is called a transposed Poisson algebra
if $({\mathfrak{L}},\cdot)$ is a commutative associative algebra and
$({\mathfrak{L}},[\cdot,\cdot])$ is a Lie algebra that satisfies the following
compatibility condition

$\displaystyle 2z\cdot[x,y]=[z\cdot x,y]+[x,z\cdot y].$ (1)

Transposed Poisson algebras were first introduced in a paper by Bai, Bai, Guo,
and Wu [2].

######  Definition 2.

Let $({\mathfrak{L}},[\cdot,\cdot])$ be a Lie algebra. A transposed Poisson
algebra structure on $({\mathfrak{L}},[\cdot,\cdot])$ is a commutative
associative multiplication $\cdot$ on $\mathfrak{L}$ which makes
$({\mathfrak{L}},\cdot,[\cdot,\cdot])$ a transposed Poisson algebra.

######  Definition 3.

Let $({\mathfrak{L}},[\cdot,\cdot])$ be an algebra and
$\varphi:\mathfrak{L}\to\mathfrak{L}$ a linear map. Then $\varphi$ is a
$\frac{1}{2}$-derivation if it satisfies

$\displaystyle\varphi\big{(}[x,y]\big{)}=\frac{1}{2}\big{(}[\varphi(x),y]+[x,\varphi(y)]\big{)}.$
(2)

Observe that $\frac{1}{2}$-derivations are a particular case of
$\delta$-derivations introduced by Filippov in [8] (see also [20] and
references therein). The space of all $\frac{1}{2}$-derivations of an algebra
$\mathfrak{L}$ will be denoted by $\Delta(\mathfrak{L}).$

Definitions 1 and 3 immediately imply the following key Lemma.

######  Lemma 4.

Let $({\mathfrak{L}},[\cdot,\cdot])$ be a Lie algebra and $\cdot$ a new binary
(bilinear) operation on ${\mathfrak{L}}$. Then
$({\mathfrak{L}},\cdot,[\cdot,\cdot])$ is a transposed Poisson algebra if and
only if $\cdot$ is commutative and associative and for every
$z\in{\mathfrak{L}}$ the multiplication by $z$ in $({\mathfrak{L}},\cdot)$ is
a $\frac{1}{2}$-derivation of $({\mathfrak{L}},[\cdot,\cdot]).$

The basic example of a $\frac{1}{2}$-derivation is the multiplication by a
field element. Such $\frac{1}{2}$-derivations will be called trivial.

######  Theorem 5.

Let ${\mathfrak{L}}$ be a Lie algebra without non-trivial
$\frac{1}{2}$-derivations. Then all transposed Poisson algebra structures on
${\mathfrak{L}}$ are trivial.

Given a Lie algebra $(\mathfrak{L},[\cdot,\cdot])$ denote by $Z(\mathfrak{L})$
its center, i.e. $Z(\mathfrak{L})=\\{a\in\mathfrak{L}\mid[a,b]=0,\ \forall
b\in\mathfrak{L}\\}$, and by $[\mathfrak{L},\mathfrak{L}]$ its square, i.e.
$[\mathfrak{L},\mathfrak{L}]=\operatorname{span}_{F}\\{[a,b]\mid
a,b\in\mathfrak{L}\\}$. Fix a complement $V$ of $[\mathfrak{L},\mathfrak{L}]$
in $\mathfrak{L}$. Then any commutative associative product $*:V\times V\to
Z(\mathfrak{L})$ defines a transposed Poisson algebra structure $\cdot$ on
$\mathfrak{L}$ by means of

$\displaystyle(a_{1}+a_{2})\cdot(b_{1}+b_{2})=a_{1}*b_{1},$ (3)

where $a_{1},b_{1}\in V$ and $a_{2},b_{2}\in[\mathfrak{L},\mathfrak{L}]$.
Indeed, the right-hand side of 1 is zero, because $z\cdot x,z\cdot y\in
Z(\mathfrak{L})$, and the left-hand side of 1 is zero by 3, because
$[x,y]\in[\mathfrak{L},\mathfrak{L}]$. We say that $\cdot$ is the extension by
zero of $*$. Observe that $\cdot$ is at the same time a usual Poisson
structure on $(\mathfrak{L},[\cdot,\cdot])$.

##  2\. Transposed Poisson structures on generalized Witt algebras

###  2.1. Generalized Witt algebras

Đoković and Zhao [6] introduced the following generalization of the classical
Witt algebra.

######  Definition 6.

Let $F$ be a field, $(A,+)$ a non-trivial abelian group, $V\neq\\{0\\}$ a
vector space and $\langle\cdot,\cdot\rangle:V\times A\to F$ a map linear in
the first variable and additive in the second one. Denote $W:=FA\otimes_{F}V$
and define the product $[\cdot,\cdot]$ on $W$ by means of

$\displaystyle[a\otimes v,b\otimes w]=(a+b)\otimes(\langle v,b\rangle
w-\langle w,a\rangle v).$ (4)

Then $(W,[\cdot,\cdot])$ is a Lie algebra called a generalized Witt algebra.

When it is necessary to specify $A$, $V$ and $\langle\cdot,\cdot\rangle$, one
writes $W=W(A,V,\langle\cdot,\cdot\rangle)$. We assume that
$\langle\cdot,\cdot\rangle$ is non-degenerate, i.e.

$\displaystyle\langle V,a\rangle=\\{0\\}$ $\displaystyle\Leftrightarrow a=0.$
(5)

We also assume that $\mathrm{char}(F)=0$. Then it follows from 5 that $A$ is
torsion-free.

The algebra $W(A,V,\langle\cdot,\cdot\rangle)$ is a generalization of the so-
called Witt type Lie algebra $V(f)$ (corresponding to an additive map $f$)
introduced555Notice that ‘$V$’ in $V(f)$ is not the same space $V$ from
Definition 6. by Yu in [18]. We recall its definition using the notation from
the present paper. Given an abelian group $A$, a field $F$ and a function
$f:A\to F$, define $V(f)$ to be a vector space with basis $\\{e_{a}\\}_{a\in
A}$ and multiplication

$\displaystyle[e_{a},e_{b}]=(f(b)-f(a))e_{a+b}.$ (6)

Without loss of generality, one assumes that $f(0)=0$. Then $V(f)$ is a Lie
algebra if and only if

$\displaystyle\big{(}f(a+b)-f(a)-f(b)\big{)}\big{(}f(a)-f(b)\big{)}=0$

for all $a,b\in A$. Observe that in general one does not require that $f$ be
additive. However, it turns out to be so if $|f(A)|\geq 4$ by [18, Lemma 4.6].

######  Lemma 7.

Let $\dim(V)=1$ and $\langle\cdot,\cdot\rangle$ be non-degenerate. Then
$W(A,V,\langle\cdot,\cdot\rangle)$ is isomorphic to the Witt type Lie algebra
$V(f)$ for some additive injective $f:A\to F$ with $|f(A)|=\infty$.

###### Proof.

Choose $v\in V\setminus\\{0\\}$. Then $\\{a\otimes v\mid a\in A\\}$ is a basis
of $W$. Define an additive map $f:A\to F$ by $f(a)=\langle v,a\rangle$ and a
bijective linear map $\varphi:W\to V(f)$ by $\varphi(a\otimes v)=e_{a}$. Then
by 6 and 4

$\displaystyle[\varphi(a\otimes v),\varphi(b\otimes v)]$
$\displaystyle=[e_{a},e_{b}]=(f(b)-f(a))e_{a+b}=(\langle v,b\rangle-\langle
v,a\rangle)e_{a+b}$ $\displaystyle=(\langle v,b\rangle-\langle
v,a\rangle)\varphi((a+b)\otimes v)=\varphi([a\otimes v,b\otimes v]).$

Observe that $f$ is injective by 5, because $\langle
v,a\rangle=0\Leftrightarrow\langle V,a\rangle=\\{0\\}$. Since $A$ is torsion-
free, then $|A|=\infty$, whence $|f(A)|=\infty$ as well. ∎

Since transposed Poisson structures on $V(f)$ were described in [12], we only
need to deal with the case $\dim(V)>1$.

######  Lemma 8.

Let $\dim(V)>1$. If $a\neq 0$, then there exist two linearly independent
$v^{\prime},v^{\prime\prime}\in V$ such that $\langle v^{\prime},a\rangle\neq
0\neq\langle v^{\prime\prime},a\rangle$.

###### Proof.

Denote $V_{0}=\\{v\in V\mid\langle v,a\rangle=0\\}$. If $V_{0}=\\{0\\}$, there
is nothing to prove. Otherwise, choose $0\neq v_{0}\in V_{0}$. Since $a\neq
0$, by 5 there is $v^{\prime}\in V$ such that $\langle v^{\prime},a\rangle\neq
0$. Observe that $v^{\prime}$ and $v_{0}$ are linearly independent, since
otherwise $v^{\prime}=kv_{0}$ and $\langle v^{\prime},a\rangle=k\langle
v_{0},a\rangle=0$. Then $v^{\prime}$ and $v^{\prime}+v_{0}$ are also linearly
independent and $\langle v^{\prime}+v_{0},a\rangle=\langle
v^{\prime},a\rangle\neq 0$. So, we may choose
$v^{\prime\prime}=v^{\prime}+v_{0}$. ∎

###  2.2. $\frac{1}{2}$-derivations of generalized Witt algebras

Observe that $W$ is an $A$-graded algebra, namely

$\displaystyle W=\bigoplus_{a\in A}W_{a},\text{ where }W_{a}=a\otimes
V=\\{a\otimes v\mid v\in V\\}.$

For all $a\in A$ and $v\in V$ denote, for simplicity,

$\displaystyle v_{a}:=a\otimes v.$ (7)

Any linear map $\varphi:W\to W$ decomposes as

$\displaystyle\varphi=\sum_{a\in A}\varphi_{a},$

where $\varphi_{a}:W\to W$ is a linear map such that
$\varphi_{a}(W_{b})\subseteq W_{a+b}$ for all $b\in A$. In particular,
$\varphi\in\Delta(W)$ if and only if $\varphi_{a}\in\Delta(W)$ for all $a\in
A$. We write

$\displaystyle\varphi_{a}(v_{b})=d_{a}(v_{b})_{a+b},$ (8)

where $d_{a}:W\to V$.

######  Lemma 9.

Let $\varphi_{a}:W\to W$ be a linear map satisfying 8. Then
$\varphi_{a}\in\Delta(W)$ if and only if for all $x,y\in A$ and $v,w\in V$

$\displaystyle 2d_{a}(\langle v,y\rangle w_{x+y}-\langle w,x\rangle v_{x+y})=$
$\displaystyle\quad\langle d_{a}(v_{x}),y\rangle w-\langle w,a+x\rangle
d_{a}(v_{x})+\langle v,a+y\rangle d_{a}(w_{y})-\langle d_{a}(w_{y}),x\rangle
v.$ (9)

###### Proof.

By 4, 8 and 7 we have

$\displaystyle 2\varphi_{a}([v_{x},w_{y}])$
$\displaystyle=2\varphi_{a}(\langle v,y\rangle w_{x+y}-\langle w,x\rangle
v_{x+y})=2d_{a}(\langle v,y\rangle w_{x+y}-\langle w,x\rangle
v_{x+y})_{a+x+y}$

and

$\displaystyle[\varphi_{a}(v_{x}),w_{y}]+[v_{x},\varphi_{a}(w_{y})]$
$\displaystyle=[d_{a}(v_{x})_{a+x},w_{y}]+[v_{x},d_{a}(w_{y})_{a+y}]$
$\displaystyle=\langle d_{a}(v_{x}),y\rangle w_{a+x+y}-\langle w,a+x\rangle
d_{a}(v_{x})_{a+x+y}$ $\displaystyle\quad+\langle v,a+y\rangle
d_{a}(w_{y})_{a+x+y}-\langle d_{a}(w_{y}),x\rangle v_{a+x+y}.$

∎

######  Lemma 10.

Let $\dim(V)>1$, $a\neq 0$ and $\varphi_{a}\in\Delta(W)$ satisfying 8. Then
$\varphi_{a}=0$.

###### Proof.

Substitute $y=0$ into 9:

$\displaystyle-2\langle w,x\rangle d_{a}(v_{x})=-\langle w,a+x\rangle
d_{a}(v_{x})+\langle v,a\rangle d_{a}(w_{0})-\langle d_{a}(w_{0}),x\rangle v,$

that is

$\displaystyle\langle w,a-x\rangle d_{a}(v_{x})=\langle v,a\rangle
d_{a}(w_{0})-\langle d_{a}(w_{0}),x\rangle v.$ (10)

Then setting $x=a$ in 10 we obtain

$\displaystyle\langle v,a\rangle d_{a}(w_{0})=\langle d_{a}(w_{0}),a\rangle
v.$ (11)

By Lemma 8 there are two linearly independent $v^{\prime},v^{\prime\prime}\in
V$ such that $\langle v^{\prime},a\rangle,\langle
v^{\prime\prime},a\rangle\neq 0$. Choosing consecutively $v=v^{\prime}$ and
$v=v^{\prime\prime}$ in 11 we have

$\displaystyle d_{a}(w_{0})=\frac{\langle d_{a}(w_{0}),a\rangle}{\langle
v^{\prime},a\rangle}v^{\prime}=\frac{\langle d_{a}(w_{0}),a\rangle}{\langle
v^{\prime\prime},a\rangle}v^{\prime\prime}.$

By the linear independence of $v^{\prime}$ and $v^{\prime\prime}$,

$\displaystyle d_{a}(w_{0})=0$ (12)

for all $w\in V$. It follows from 10 that

$\displaystyle\langle w,a-x\rangle d_{a}(v_{x})=0.$

If $x\neq a$, then $\langle w,a-x\rangle\neq 0$ for some $w\in V$ by 5. Thus,

$\displaystyle d_{a}(v_{x})=0,\text{ if }x\neq a.$ (13)

Now substitute $x=a$ and $y=-a$ into 9 and use 12:

$\displaystyle 0=\langle d_{a}(v_{a}),a\rangle w+2\langle w,a\rangle
d_{a}(v_{a})+\langle d_{a}(w_{-a}),a\rangle v.$

Since $A$ is torsion-free, then $a\neq-a$, so $d_{a}(w_{-a})=0$ by 13. Taking
consecutively $w=v^{\prime}$ and $w=v^{\prime\prime}$ we have

$\displaystyle d_{a}(v_{a})=-\frac{\langle d_{a}(v_{a}),a\rangle}{2\langle
v^{\prime},a\rangle}v^{\prime}=-\frac{\langle d_{a}(v_{a}),a\rangle}{2\langle
v^{\prime\prime},a\rangle}v^{\prime\prime},$

whence

$\displaystyle d_{a}(v_{a})=0$ (14)

by the linear independence of $v^{\prime}$ and $v^{\prime\prime}$. Combining
13 and 14, we conclude that $\varphi_{a}=0$. ∎

######  Lemma 11.

Let $\varphi_{0}\in\Delta(W)$ satisfying 8 with $a=0$. Then
$\varphi_{0}\in\operatorname{span}_{F}\\{\mathrm{id}\\}$.

###### Proof.

For $a=0$, equality 9 takes the form

$\displaystyle 2d_{0}(\langle v,y\rangle w_{x+y}-\langle w,x\rangle
v_{x+y})=\langle d_{0}(v_{x}),y\rangle w-\langle w,x\rangle
d_{0}(v_{x})+\langle v,y\rangle d_{0}(w_{y})-\langle d_{0}(w_{y}),x\rangle v.$
(15)

Then $y=0$ in 15 gives

$\displaystyle\langle w,x\rangle d_{0}(v_{x})=\langle d_{0}(w_{0}),x\rangle
v.$

If $x\neq 0$, then choosing $w\in V$ with $\langle w,x\rangle\neq 0$, we
obtain

$\displaystyle d_{0}(v_{x})=\frac{\langle d_{0}(w_{0}),x\rangle}{\langle
w,x\rangle}v=:k_{x}v,\text{ if }x\neq 0.$ (16)

In particular, $d_{0}(v_{x})=d_{0}(v_{-x})$ for all $x\neq 0$. On the other
hand, taking $y=-x\neq 0$ in 15, we have

$\displaystyle 2d_{0}(\langle v,x\rangle w_{0}+\langle w,x\rangle
v_{0})=\langle d_{0}(v_{x}),x\rangle w+\langle w,x\rangle d_{0}(v_{x})+\langle
v,x\rangle d_{0}(w_{x})+\langle d_{0}(w_{x}),x\rangle v,$ (17)

which for $w=v$ gives

$\displaystyle 2\langle v,x\rangle d_{0}(v_{0})=\langle d_{0}(v_{x}),x\rangle
v+\langle v,x\rangle d_{0}(v_{x}).$

Choosing $v\in V$ with $\langle v,x\rangle\neq 0$ and applying 16, we conclude
that

$\displaystyle d_{0}(v_{0})=k_{x}v=d_{0}(v_{x}),\text{ if }\langle
v,x\rangle\neq 0.$ (18)

If $x\neq 0$ and $\langle v,x\rangle=0$, then, thanks to 16, equality 17
becomes

$\displaystyle 2\langle w,x\rangle d_{0}(v_{0})=\langle w,x\rangle
d_{0}(v_{x})+\langle d_{0}(w_{x}),x\rangle v=\langle w,x\rangle k_{x}v+\langle
k_{x}w,x\rangle v=2k_{x}\langle w,x\rangle v.$

Choosing an arbitrary $w\in V$ with $\langle w,x\rangle\neq 0$, we arrive at

$\displaystyle d_{0}(v_{0})=k_{x}v=d_{0}(v_{x}),\text{ if }x\neq 0\text{ and
}\langle v,x\rangle=0.$ (19)

Combining 18, 19 and 8, we finally prove the desired fact. ∎

######  Proposition 12.

If $\dim(V)>1$, then $\Delta(W)=\operatorname{span}_{F}\\{\mathrm{id}\\}$.

###### Proof.

The inclusion $\Delta(W)\subseteq\operatorname{span}_{F}\\{\mathrm{id}\\}$ is
Lemmas 10 and 11. The converse inclusion is trivial. ∎

######  Theorem 13.

Let $\mathrm{char}(F)=0$ and $\langle\cdot,\cdot\rangle$ be non-degenerate.

  1. (i)

If $\dim(V)>1$, then all the transposed Poisson algebra structures on
$W(A,V,\langle\cdot,\cdot\rangle)$ are trivial.

  2. (ii)

If $\dim(V)=1$, say, $V=\operatorname{span}_{F}\\{v\\}$, then the transposed
Poisson algebra structures on $W(A,V,\langle\cdot,\cdot\rangle)$ are exactly
mutations of the product $(a\otimes v)\cdot(b\otimes v)=(a+b)\otimes v$.

###### Proof.

(i) is an immediate consequence of Proposition 12 and [7, Theorem 8], while
(ii) follows from Lemma 7 and [12, Proposition 26]. ∎

##  3\. Transposed Poisson structures on Block Lie algebras

###  3.1. Block Lie algebras

Another generalization of the Witt algebra is the class of Lie algebras
studied by Block in [5].

######  Definition 14.

Let $F$ be a field, $(A,+)$ a non-trivial abelian group, $g:A\to F$ an
additive map and $f:A\times A\to F$ an anti-symmetric biadditive map. The
Block algebra $L(A,g,f)$ is the $F$-vector space with basis $\\{u_{a}\\}_{a\in
A}$ and product

$\displaystyle[u_{a},u_{b}]=(f(a,b)+g(a-b))u_{a+b}.$ (20)

It is known [5] (and, in fact, easy to see) that $L(A,g,f)$ is a Lie algebra
if and only if either $g=0$ or there exists an additive map $h:A\to F$ such
that for all $a,b\in A$:

$\displaystyle f(a,b)=g(a)h(b)-g(b)h(a).$ (21)

We will write $L=L(A,g,f)$ for the simplicity of notation. We will also assume
that $\mathrm{char}(F)=0$ and $f$ is non-degenerate in the sense that

$\displaystyle f(a,A)=\\{0\\}\Leftrightarrow a=0.$ (22)

Then, as in Section 2.1, this implies that $A$ is torsion-free.

Observe that $L(A,g,f)$ is a generalization of the Block Lie algebra
$\mathcal{B}(q)$ studied in [11] (it had been introduced in [16] under
slightly different assumptions on $q$ and on the basis). Recall that
$\mathcal{B}(q)$, where $q\in\mathbb{C}$, is the complex Lie algebra with a
basis $\\{L_{m,i}\mid m,i\in\mathbb{Z}\\}$, where

$\displaystyle[L_{m,i},L_{n,j}]=(n(i+q)-m(j+q))L_{m+n,i+j}$

for all $i,j,m,n\in\mathbb{Z}$. It is immediately seen that
$\mathcal{B}(q)=L(A,g,f)$, where $F=\mathbb{C}$,
$A=\mathbb{Z}\times\mathbb{Z}$,

$\displaystyle g(m,i)=-qm\text{ and }f((m,i),(n,j))=ni-mj$ (23)

for $(m,i),(n,j)\in\mathbb{Z}\times\mathbb{Z}$. If $q\neq 0$, then the
corresponding map $h$ from 21 can be chosen to be

$\displaystyle h(m,i)=i/q.$ (24)

Observe that $f$ is non-degenerate, because $f((m,i),(0,-1))=m$ and
$f((m,i),(1,0))=i$.

We will need descriptions of $Z(L)$ and $[L,L]$ in the general case.

######  Lemma 15.

Let $f$ be non-degenerate.

  1. (i)

If $g=0$, then $Z(L)=\operatorname{span}_{F}\\{u_{0}\\}$. Otherwise,
$Z(L)=\operatorname{span}_{F}\\{u_{a}\mid g(a)=h(a)+1=0\\}$.

  2. (ii)

If $g=0$, then $[L,L]=\operatorname{span}_{F}\\{u_{a}\mid a\neq 0\\}$.
Otherwise, $[L,L]=\operatorname{span}_{F}\\{u_{a}\mid g(a)\neq 0\text{ or
}h(a)+2\neq 0\\}$.

###### Proof.

(i). Let $g=0$. The inclusion $\operatorname{span}_{F}\\{u_{0}\\}\subseteq
Z(L)$ is trivial. Conversely, if $x=\sum x_{a}u_{a}\in Z(L)$ and $x_{a}\neq 0$
for some $a\neq 0$, then choose $b\in A$ such that $f(a,b)\neq 0$ (it exists
due to non-degeneracy of $f$) and calculate

$\displaystyle[x,u_{b}]=\sum_{c\neq
a}x_{c}f(c,b)u_{c+b}+x_{a}f(a,b)u_{a+b}\neq 0.$

Let $g\neq 0$. If $g(a)=h(a)+1=0$, then for all $b\in A$

$\displaystyle f(a,b)+g(a-b)=g(a)h(b)-g(b)h(a)+g(a)-g(b)=g(b)-g(b)=0,$

so $[u_{a},u_{b}]=0$. This proves the inclusion
$\operatorname{span}_{F}\\{u_{a}\mid g(a)=h(a)+1=0\\}\subseteq Z(L)$.
Conversely, assume that $x=\sum x_{a}u_{a}\in Z(L)$. Then $[x,u_{0}]=0$
implies $g(a)=0$ for all $a$ with $x_{a}\neq 0$. Consequently,
$[x,u_{b}]=-g(b)\sum x_{a}(h(a)+1)u_{a+b}$. Choosing $b\in A$ with $g(b)\neq
0$, we conclude that $h(a)+1=0$ whenever $x_{a}\neq 0$. Thus,
$Z(L)\subseteq\operatorname{span}_{F}\\{u_{a}\mid g(a)=h(a)+1=0\\}$.

(ii). Let $g=0$. If $[u_{a},u_{b}]\neq 0$, then $b\neq-a$, since otherwise
$f(a,b)=f(a,-a)=0$. Hence,
$[u_{a},u_{b}]=f(a,b)u_{a+b}\in\operatorname{span}_{F}\\{u_{a}\mid a\neq
0\\}$. This proves $[L,L]\subseteq\operatorname{span}_{F}\\{u_{a}\mid a\neq
0\\}$. Conversely, for any $a\neq 0$ there exists $b\in A$ such that
$f(a,b)\neq 0$. Then $[u_{a-b},u_{b}]=f(a-b,b)u_{a}=f(a,b)u_{a}\neq 0$, whence
$u_{a}\in[L,L]$.

Let $g\neq 0$. If $[u_{a},u_{b}]\neq 0$, then either $g(a+b)\neq 0$ or
$h(a+b)+2\neq 0$, since otherwise

$\displaystyle f(a,b)+g(a-b)$ $\displaystyle=g(a)h(b)-g(b)h(a)+g(a)-g(b)$
$\displaystyle=g(a)(-h(a)-2)+g(a)h(a)+g(a)+g(a)=0.$

Hence, $[L,L]\subseteq\operatorname{span}_{F}\\{u_{a}\mid g(a)\neq 0\text{ or
}h(a)+2\neq 0\\}$. Conversely, take $a\in A$ with $g(a)\neq 0$ or $h(a)+2\neq
0$. If $g(a)\neq 0$, then $[u_{a},u_{0}]=g(a)u_{a}\neq 0$, so $u_{a}\in[L,L]$.
Otherwise, $h(a)+2\neq 0$ and

$\displaystyle f(a-b,b)+g(a-b-b)$
$\displaystyle=f(a,b)+g(a-2b)=g(a)h(b)-g(b)h(a)+g(a)-2g(b)$
$\displaystyle=-g(b)h(a)-2g(b)=-g(b)(h(a)+2),$

so choosing $b\in A$ with $g(b)\neq 0$ we have
$[u_{a-b},u_{b}]=-g(b)(h(a)+2)u_{a}\neq 0$, whence $u_{a}\in[L,L]$. ∎

We will also need the following technical lemma.

######  Lemma 16.

Let $\alpha,\beta:A\to F$ two non-zero additive functions. Then there exists
$a\in A$ such that $\alpha(a)\neq 0\neq\beta(a)$.

###### Proof.

Assume that for any $a\in A$ either $\alpha(a)=0$ or $\beta(a)=0$. Then
$A=\ker\alpha\cup\ker\beta$. Since $\ker\alpha$ and $\ker\beta$ are subgroups
of $A$, then either $\ker\alpha\subseteq\ker\beta$, in which case
$A=\ker\beta$, or $\ker\beta\subseteq\ker\alpha$, in which case
$A=\ker\alpha$. Hence, either $\alpha=0$ or $\beta=0$, a contradiction. ∎

###  3.2. $\frac{1}{2}$-derivations of Block Lie algebras

It follows from 20 that $L=\bigoplus_{a\in A}Fu_{a}$ is an $A$-grading, so any
linear map $\varphi:L\to L$ decomposes into the direct sum of linear maps

$\displaystyle\varphi=\sum_{a\in A}\varphi_{a},$

where $\varphi_{a}(u_{b})\in Fu_{a+b}$ for all $b\in A$. Moreover,
$\varphi\in\Delta(L)$ if and only if $\varphi_{a}\in\Delta(L)$ for all $a\in
A$. As usual, we write

$\displaystyle\varphi_{a}(u_{b})=d_{a}(u_{b})u_{a+b},$ (25)

where $d_{a}:L\to F$.

######  Lemma 17.

Let $\varphi_{a}:L\to L$ be a linear map satisfying 25. Then
$\varphi_{a}\in\Delta(L)$ if and only if for all $x,y\in A$

$\displaystyle 2(f(x,y)+g(x-y))d_{a}(x+y)=$
$\displaystyle\quad(f(a+x,y)+g(a+x-y))d_{a}(x)+(f(x,a+y)+g(x-a-y))d_{a}(y).$
(26)

###### Proof.

By 20 and 25 we have

$\displaystyle 2\varphi_{a}([u_{x},u_{y}])$
$\displaystyle=2\varphi_{a}((f(x,y)+g(x-y))u_{x+y})=2(f(x,y)+g(x-y))d_{a}(x+y)u_{a+x+y}$

and

$\displaystyle[\varphi_{a}(u_{x}),u_{y}]+[u_{x},\varphi_{a}(u_{y})]$
$\displaystyle=[d_{a}(x)u_{a+x},u_{y}]+[u_{x},d_{a}(y)u_{a+y}]$
$\displaystyle=(f(a+x,y)+g(a+x-y))d_{a}(x)u_{a+x+y}$
$\displaystyle\quad+(f(x,a+y)+g(x-a-y))d_{a}(y)u_{a+x+y}.$

∎

####  3.2.1. The case $g=0$

Assume first that $g=0$.

######  Lemma 18.

Let $a\neq 0$ and $\varphi_{a}\in\Delta(L)$ satisfying 25. Then
$\varphi_{a}=0$.

###### Proof.

Taking $y=-x$ in 26 and using anti-symmetry of $f$, we have

$\displaystyle
0=f(a+x,-x)d_{a}(x)+f(x,a-x)d_{a}(-x)=-f(a,x)(d_{a}(x)+d_{a}(-x)).$

Hence,

$\displaystyle d_{a}(-x)=-d_{a}(x),\text{ if }f(a,x)\neq 0.$ (27)

Now, substitute $y=-a$ into 26:

$\displaystyle 2f(x,-a)d_{a}(x-a)=f(a+x,-a)d_{a}(x)=f(x,-a)d_{a}(x),$

whence

$\displaystyle d_{a}(x)=2d_{a}(x-a),\text{ if }f(a,x)\neq 0.$

Since $f(a,a+x)=f(a,x)$, the latter is equivalent to

$\displaystyle d_{a}(a+x)=2d_{a}(x),\text{ if }f(a,x)\neq 0.$ (28)

On the other hand, $y=a$ in 26 gives

$\displaystyle
2f(x,a)d_{a}(x+a)=f(a+x,a)d_{a}(x)+f(x,2a)d_{a}(a)=f(x,a)d_{a}(x)+2f(x,a)d_{a}(a).$

If $f(a,x)\neq 0$, then using 28, we come to $4d_{a}(x)=d_{a}(x)+2d_{a}(a)$.
Consequently,

$\displaystyle 3d_{a}(x)=2d_{a}(a),\text{ if }f(a,x)\neq 0.$ (29)

However, $f(a,x)\neq 0\Leftrightarrow f(a,-x)\neq 0$, so replacing $x$ by $-x$
in 29 and taking into account $\mathrm{char}(F)=0$, we have

$\displaystyle d_{a}(-x)=d_{a}(x),\text{ if }f(a,x)\neq 0.$ (30)

Combining 27 and 30, we conclude that

$\displaystyle d_{a}(x)=0,\text{ if }f(a,x)\neq 0.$ (31)

Now assume that $f(a,x)=0$. Since $a\neq 0$, by 22 there exists $y\in A$ such
that $f(a,y)\neq 0$. Observe that $f(a,x+y)=f(a,y)\neq 0$. Then
$d_{a}(y)=d_{a}(x+y)=0$ thanks to 31, so 26 takes the form

$\displaystyle 0=f(a+x,y)d_{a}(x).$

By Lemma 16 applied to $f(a+x,-)$ and $f(a,-)$, whenever $x\neq-a$, the
element $y$ can be chosen in a way that $f(a+x,y)\neq 0\neq f(a,y)$. Thus, we
have proved

$\displaystyle d_{a}(x)=0,\text{ if }f(a,x)=0\text{ and }x\neq-a.$ (32)

Finally, taking $y=-a-x$ in 26 we see that the right-hand side is zero, while
the left-hand side equals $2f(a,x)d_{a}(-a)$. Choosing $x\in A$ such that
$f(a,x)\neq 0$, we show that $d_{a}(-a)=0$. Combining this with 31 and 32, we
get the desired fact. ∎

######  Lemma 19.

Let $\varphi_{0}\in\Delta(L)$ satisfying 25 with $a=0$. Then
$\varphi_{0}(x)=\varphi_{0}(y)$ for all $x,y\neq 0$.

###### Proof.

Write 26 with $a=0$:

$\displaystyle 2f(x,y)d_{0}(x+y)=f(x,y)d_{0}(x)+f(x,y)d_{0}(y).$

Consequently,

$\displaystyle 2d_{0}(x+y)=d_{0}(x)+d_{0}(y),\text{ if }f(x,y)\neq 0.$ (33)

Observe that $f(x,y)=f(x+y,-y)$, so applying 33 with $(x,y)$ replaced by
$(x+y,-y)$, we have

$\displaystyle 2d_{0}(x)=d_{0}(x+y)+d_{0}(-y),\text{ if }f(x,y)\neq 0.$ (34)

Combining 33 and 34, we come to

$\displaystyle 3d_{0}(x)=d_{0}(y)+2d_{0}(-y),\text{ if }f(x,y)\neq 0.$ (35)

However, $f(x,-y)=-f(x,y)$, so replacing $y$ by $-y$ in 35, we obtain

$\displaystyle 3d_{0}(x)=d_{0}(-y)+2d_{0}(y),\text{ if }f(x,y)\neq 0.$ (36)

It follows from 35 and 36 that $d_{0}(y)=d_{0}(-y)$, so

$\displaystyle d_{0}(x)=d_{0}(y),\text{ if }f(x,y)\neq 0,$ (37)

because $\mathrm{char}(F)=0$.

Now let $x,y\neq 0$. By Lemma 16 applied to $f(x,-)$ and $f(y,-)$ there exists
$z\in A$ such that $f(x,z)\neq 0\neq f(y,z)$. Then 37 gives

$\displaystyle d_{0}(x)=d_{0}(z)=d_{0}(y),\text{ if }x,y\neq 0,$

as needed. ∎

######  Lemma 20.

The linear map $\alpha:L\to L$ given by

$\displaystyle\alpha(u_{a})=\begin{cases}u_{0},&a=0,\\\ 0,&a\neq
0,\end{cases}$

is a $\frac{1}{2}$-derivation of $L$.

###### Proof.

Observe by Lemma 15(i) that $\alpha(L)\subseteq Z(L)$, so the right-hand side
of 2 is always zero for $\varphi=\alpha$. Now, $\alpha([L,L])=\\{0\\}$ by
Lemma 15(ii). Thus, the left-hand side of 2 is always zero as well. ∎

######  Proposition 21.

We have $\Delta(L)=\operatorname{span}_{F}\\{\mathrm{id},\alpha\\}$.

###### Proof.

The inclusion
$\Delta(L)\subseteq\operatorname{span}_{F}\\{\mathrm{id},\alpha\\}$ is Lemmas
18 and 19. The converse inclusion is Lemma 20. ∎

######  Theorem 22.

Let $\mathrm{char}(F)=0$ and $f$ be non-degenerate. Then there is only one
non-trivial transposed Poisson algebra structure $\cdot$ on $L(A,0,f)$. It is
given by

$\displaystyle u_{0}\cdot u_{0}=u_{0}.$ (38)

###### Proof.

Let $\cdot$ be a non-trivial transposed Poisson algebra structure on
$L(A,0,f)$. By Propositions 21 and 4 for any $a\in A$ there are
$k_{a},l_{a}\in F$ such that

$\displaystyle u_{a}\cdot
u_{b}=k_{a}u_{b}+l_{a}\alpha(u_{b})=\begin{cases}(k_{a}+l_{a})u_{0},&b=0,\\\
k_{a}u_{b},&b\neq 0.\end{cases}$ (39)

Since $|A|>2$ ($A$ is torsion-free), for any $a\neq 0$ there exists
$b\not\in\\{0,a\\}$. Then by 39 and commutativity of $\cdot$ we have
$k_{a}u_{b}=u_{a}\cdot u_{b}=u_{b}\cdot u_{a}=k_{b}u_{a}$. Consequently,
$k_{a}=0$ for $a\neq 0$. Similarly, $(k_{a}+l_{a})u_{0}=u_{a}\cdot
u_{0}=u_{0}\cdot u_{a}=k_{0}u_{a}$ gives $k_{0}=l_{a}=0$ for $a\neq 0$. Thus,
the only non-zero product $u_{a}\cdot u_{b}$ is $u_{0}\cdot u_{0}=l_{0}u_{0}$.
So, up to isomorphism, $\cdot$ is of the form 38.

Conversely, in view of Lemma 15 the product 38 is of the form 3, so
$(L,\cdot,[\cdot,\cdot])$ is a transposed Poisson (and usual Poisson) algebra.
∎

######  Remark 23.

Consider $\mathcal{B}(0)$ as the complex Block algebra
$L(\mathbb{Z}\times\mathbb{Z},0,f)$, where $f$ is given by 23. Then we obtain
the description of transposed Poisson algebra structures on $\mathcal{B}(0)$
given in [11, Theorem 2.14] as a particular case of Theorem 22.

####  3.2.2. The case $g\neq 0$

In this case, as it was commented above, there exists an additive map $h:A\to
F$ such that 21 holds.

######  Lemma 24.

Let $a\neq 0$ and $\varphi_{a}\in\Delta(L)$ satisfying 25. If $g(x)f(a,x)\neq
0$, then $\varphi_{a}(x)=0$.

###### Proof.

Consider first $y=0$ in 26

$\displaystyle 2g(x)d_{a}(x)=g(a+x)d_{a}(x)+(f(x,a)+g(x-a))d_{a}(0).$

Then

$\displaystyle g(a-x)d_{a}(x)=(f(a,x)+g(a-x))d_{a}(0).$ (40)

Replacing $x$ by $-x$, we obtain

$\displaystyle g(a+x)d_{a}(-x)=(-f(a,x)+g(a+x))d_{a}(0).$ (41)

On the other hand, $y=-x$ in 26 gives

$\displaystyle
4g(x)d_{a}(0)=(-f(a,x)+g(a+2x))d_{a}(x)+(-f(a,x)+g(2x-a))d_{a}(-x).$ (42)

Multiplying both sides of 42 by $g(a-x)g(a+x)=g(a)^{2}-g(x)^{2}$ and using 41
and 40, we get

$\displaystyle 4g(x)(g(a)^{2}-g(x)^{2})d_{a}(0)$
$\displaystyle=(-f(a,x)+g(a+2x))g(a+x)(f(a,x)+g(a-x))d_{a}(0)$
$\displaystyle\quad+(-f(a,x)+g(2x-a))g(a-x)(-f(a,x)+g(a+x))d_{a}(0).$ (43)

We have

$\displaystyle(-f(a,x)+g(a+2x))(f(a,x)+g(a-x))$
$\displaystyle=(g(a)+g(x)/2)^{2}-(f(a,x)-3g(x)/2)^{2},$
$\displaystyle(-f(a,x)+g(2x-a))(-f(a,x)+g(a+x))$
$\displaystyle=(f(a,x)-3g(x)/2)^{2}-(g(a)-g(x)/2)^{2}.$

Since

$\displaystyle(g(a)+g(x)/2)^{2}-(g(a)-g(x)/2)^{2}$ $\displaystyle=2g(a)g(x),$
$\displaystyle(g(a)+g(x)/2)^{2}+(g(a)-g(x)/2)^{2}$
$\displaystyle=2g(a)^{2}+g(x)^{2}/2,$

the coefficient of $d_{a}(0)$ on the right-hand side of 43 equals

$\displaystyle g(a)\cdot
2g(a)g(x)+g(x)(2g(a)^{2}+g(x)^{2}/2)-2g(x)(f(a,x)-3g(x)/2)^{2}$
$\displaystyle=g(x)(4g(a)^{2}+g(x)^{2}/2-2(f(a,x)-3g(x)/2)^{2}).$

Subtracting the coefficient of $d_{a}(0)$ on the left-hand side of 43, we
obtain

$\displaystyle
g(x)(9g(x)^{2}/2-2(f(a,x)-3g(x)/2)^{2})=2g(x)f(a,x)(3g(x)-f(a,x)).$

Thus, under the assumption $g(x)f(a,x)\neq 0$, 43 is equivalent to

$\displaystyle(3g(x)-f(a,x))d_{a}(0)=0.$ (44)

Case 1. $f(a,x)\neq 3g(x)$. Then 44 gives

$\displaystyle d_{a}(0)=0.$ (45)

Case 1.1. $g(a)\neq g(x)$. It follows from 45 and 40 that $d_{a}(x)=0$.

Case 1.2. $g(a)=g(x)$. Then $g(a+x)=2g(x)\neq 0$, so $d_{g}(-x)=0$ by 41.
Moreover, $-f(a,x)+g(a+2x)=-f(a,x)+3g(x)\neq 0$, so 45 and 42 yield
$d_{a}(x)=0$.

Case 2. $f(a,x)=3g(x)\neq 0$. Then 40 becomes

$\displaystyle g(a-x)d_{a}(x)=(3g(x)+g(a-x))d_{a}(0)=(2g(x)+g(a))d_{a}(0).$
(46)

Since $f(a,x)=3g(x)$ is invariant under the replacement of $x$ by $kx$, then
46 implies

$\displaystyle g(a-kx)d_{a}(kx)=(2kg(x)+g(a))d_{a}(0).$ (47)

On the other hand, $y=2x$ in 26 gives

$\displaystyle-2g(x)d_{a}(3x)=(5g(x)+g(a))d_{a}(x)-(4g(x)+g(a))d_{a}(2x).$
(48)

Multiplying both sides of this equality by $g(a-x)g(a-2x)g(a-3x)$ and using 47
we get

$\displaystyle-2g(x)g(a-x)g(a-2x)(6g(x)+g(a))d_{a}(0)$
$\displaystyle=(5g(x)+g(a))g(a-2x)g(a-3x)(2g(x)+g(a))d_{a}(0)$
$\displaystyle\quad-(4g(x)+g(a))g(a-x)g(a-3x)(4g(x)+g(a))d_{a}(0).$ (49)

Comparing the coefficients of $g(a)^{i}g(x)^{j}d_{a}(0)$, $0\leq i+j\leq 4$,
in 49, we see that 49 is equivalent to $36g(x)^{4}d_{a}(0)=0$. Hence, we again
have 45.

Case 2.1. $g(a)\neq g(x)$. Then 45 and 40 yield $d_{a}(x)=0$.

Case 2.2. $g(a)=g(x)$. Then $g(a-kx)=(1-k)g(x)\neq 0$ for $k\neq 1$, so
$d_{a}(2x)=d_{a}(3x)=0$ by 47. Moreover, $5g(x)+g(a)=6g(x)\neq 0$, so 45 and
48 imply $d_{a}(x)=0$. ∎

######  Lemma 25.

Let $a\neq 0$ and $\varphi_{a}\in\Delta(L)$ satisfying 25. If $g(x)f(a,x)=0$,
then $d_{a}(x)=0$, unless $g(a)=g(x)=0$, $h(a)=1$ and $h(x)=-2$.

###### Proof.

Case 1. $g(a+x)\neq 0$. Since $a\neq 0$, by Lemma 16 applied to $f(a,-)$ and
$g$ there exists $y\in A$ such that $f(a,y)\neq 0\neq g(y)$. Observe that for
any $k\neq 0$ we have

$\displaystyle f(a,ky)\neq 0\neq g(ky),$ (50)

because $\mathrm{char}(F)=0$. We affirm that $k$ can be chosen in a way that

$\displaystyle f(a,x+ky)\neq 0\neq f(a+x,ky)+g(a+x-ky).$ (51)

Indeed, $f(a,y)\neq 0$, so there exists at most one $k$ such that
$f(a,x)+kf(a,y)=0$. If $f(a+x,y)-g(y)=0$, then $f(a+x,ky)+g(a+x-ky)=g(a+x)\neq
0$ for all $k$. Otherwise, there exists at most one $k$ such that
$k(f(a+x,y)-g(y))+g(a+x)=0$. Thus, 51 and 50 hold for infinitely many integer
$k\neq 0$ (recall that $\mathrm{char}(F)=0$). Then $d_{a}(ky)=d_{a}(x+ky)=0$
for any such $k$ by Lemma 24. Consequently, applying 26 with $y$ replaced by
$ky$ and using 51 we prove $d_{a}(x)=0$.

Case 2. $g(a+x)=0$. Then

$\displaystyle f(a+x,y)+g(a+x-y)=-g(y)(h(a+x)+1).$

Case 2.1. $h(a+x)+1\neq 0$. Then the same argument as in Case 1 gives
$d_{a}(x)=0$.

Case 2.2. $h(a+x)+1=0$. Then $0=g(x)f(a,x)=g(a)^{2}=g(x)^{2}$, so 26
simplifies to

$\displaystyle 2g(y)h(a)d_{a}(x+y)=g(y)h(a)d_{a}(y).$

Observe that $h(a)\neq 0$, since otherwise $f(a,y)=0$ for all $y\in A$
contradicting 22. Hence,

$\displaystyle 2d_{a}(x+y)=d_{a}(y),\text{ if }g(y)\neq 0.$ (52)

On the other hand, since $g(a)=0$, then substituting $x=0$ in 26 we get

$\displaystyle d_{a}(y)=(h(a)+1)d_{a}(0),\text{ if }g(y)\neq 0.$ (53)

Observe that $g(x+y)=g(y)$ whenever $g(x)=0$, so 53 and 52 yield

$\displaystyle 2(h(a)+1)d_{a}(0)=(h(a)+1)d_{a}(0).$

It follows that $d_{a}(0)=0$, unless $h(a)+1=0$. In any case, 53 implies

$\displaystyle d_{a}(y)=0,\text{ if }g(y)\neq 0.$ (54)

Write 26 replacing $x$ by $x-y$, where $g(y)\neq 0$. Since
$d_{a}(y)=d_{a}(x-y)=0$ by 54, we come to

$\displaystyle(h(x)+2)d_{a}(x)=0.$

Thus, $d_{a}(x)=0$, unless $h(x)=-2$ (in which case $h(a)=1$). ∎

######  Lemma 26.

Let $\varphi_{0}\in\Delta(L)$ satisfying 25 with $a=0$. Then
$\varphi_{0}\in\operatorname{span}_{F}\\{\mathrm{id}\\}$.

###### Proof.

Writing 26 with $a=0$, we get

$\displaystyle 2(f(x,y)+g(x-y))d_{0}(x+y)=(f(x,y)+g(x-y))(d_{0}(x)+d_{0}(y)).$
(55)

Case 1. $g(x)\neq 0$. Then put $y=0$ in 55 to get:

$\displaystyle d_{0}(x)=d_{0}(0),\text{ if }g(x)\neq 0.$ (56)

Case 2. $g(x)=0$ and $h(x)\neq-1$. Then $f(x,y)=-g(y)h(x)$, and 55 is
equivalent to

$\displaystyle 2g(y)d_{0}(x+y)=g(y)(d_{0}(x)+d_{0}(y)).$ (57)

Choose $y\in A$ such that $g(y)\neq 0$. Then $d_{0}(y)=d_{0}(x+y)=d_{0}(0)$ by
56. Hence, 57 yields $d_{0}(x)=d_{0}(0)$.

Case 3. $g(x)=0$ and $h(x)=-1$. Let us use 55 with $(x,y)$ replaced by
$(x+y,-y)$. Observe that $f(x+y,-y)=-f(x,y)=-g(y)$, so we get

$\displaystyle 6g(y)d_{0}(x)=3g(y)(d_{0}(x+y)+d_{0}(-y)).$ (58)

Choosing $y\in A$ with $g(y)\neq 0$ we have $d_{0}(x+y)=d_{0}(-y)=d_{0}(0)$ by
56. Thus, $d_{0}(x)=d_{0}(0)$ due to 58. ∎

Given $\lambda,\mu\in F$, we introduce the following notation:

$\displaystyle A_{(\lambda,\mu)}:=\\{a\in A\mid g(a)=\lambda\text{ and
}h(a)=\mu\\}.$

######  Lemma 27.

Let $a\in A_{(0,-2)}$ and $b\in A_{(0,-1)}$. Then the linear map
$\alpha_{(a,b)}:L\to L$ given by

$\displaystyle\alpha_{(a,b)}(u_{c})=\begin{cases}u_{b},&c=a,\\\
0,&\text{otherwise},\end{cases}$

is a $\frac{1}{2}$-derivation of $L$.

###### Proof.

Observe by Lemma 15(i) that $\alpha_{(a,b)}(L)\subseteq Z(L)$, showing that
the right-hand side of 2 is always zero for $\varphi=\alpha_{(a,b)}$.
Furthermore, $\alpha_{(a,b)}([L,L])=\\{0\\}$ by Lemma 15(ii), and the left-
hand side of 2 is always zero as well. ∎

######  Proposition 28.

We have
$\Delta(L)=\operatorname{span}_{F}(\\{\mathrm{id}\\}\cup\\{\alpha_{(a,b)}\mid
a\in A_{(0,-2)}\text{ and }b\in A_{(0,-1)}\\})$.

###### Proof.

The fact that any $\varphi\in\Delta(L)$ is a linear combination of
$\mathrm{id}$ and $\alpha_{(a,b)}$ follows from Lemmas 24, 25 and 26.
Conversely, the inclusion $\mathrm{id}\in\Delta(L)$ is trivial, and
$\alpha_{(a,b)}\in\Delta(L)$ is Lemma 27. ∎

######  Theorem 29.

Let $\mathrm{char}(F)=0$, $g\neq 0$ and $f$ be non-degenerate. If
$(g(a),h(a))\neq(0,-1)$ for all $a\in A$, then all the transposed Poisson
algebra structures on $L(A,g,f)$ are trivial. Otherwise, the transposed
Poisson algebra structures on $L(A,g,f)$ are exactly extensions by zero of
commutative associative products $*$ on the complement
$V=\operatorname{span}_{F}\\{u_{a}\mid g(a)=h(a)+2=0\\}$ of $[L,L]$ with
values in $Z(L)=\operatorname{span}_{F}\\{u_{a}\mid g(a)=h(a)+1=0\\}$.

###### Proof.

Let $\cdot$ be a transposed Poisson algebra structure on $L(A,g,f)$. If
$(g(a),h(a))\neq(0,-1)$ for all $a\in A$, then $A_{(0,-1)}=\emptyset$, so by
Proposition 28 we have $\Delta(L)=\\{\mathrm{id}\\}$. It follows from [7,
Theorem 8] that $\cdot$ is trivial.

Assume that $(g(a),h(a))=(0,-1)$ for some $a\in A$. Then
$(g(2a),h(2a))=(0,-2)$, so both $A_{(0,-1)}$ and $A_{(0,-2)}$ are non-empty.
By Propositions 28 and 4 for any $a\in A$ there are $k_{a}\in F$ and
$\\{l_{a}^{(x,y)}\\}_{x\in A_{(0,-2)},y\in A_{(0,-1)}}\subseteq F$, such that

$\displaystyle u_{a}\cdot u_{b}=k_{a}u_{b}+\sum_{x\in A_{(0,-2)},y\in
A_{(0,-1)}}l_{a}^{(x,y)}\alpha_{(x,y)}(u_{b})=\begin{cases}k_{a}u_{b}+\sum_{y\in
A_{(0,-1)}}l_{a}^{(b,y)}u_{y},&b\in A_{(0,-2)},\\\ k_{a}u_{b},&b\not\in
A_{(0,-2)}.\end{cases}$ (59)

Since $\mathrm{char}(F)=0$ and $A$ is torsion-free, then $A\setminus
A_{(0,-1)}$ and $A\setminus A_{(0,-2)}$ are infinite (indeed, if $a\in
A_{(\lambda,\mu)}$ with $(\lambda,\mu)\neq(0,0)$, then $ka\not\in
A_{(\lambda,\mu)}$ for all $k\neq 1$). So, for any $a\not\in A_{(0,-2)}$ there
exists $b\not\in A_{(0,-2)}$, $b\neq a$. Then by 59 and commutativity of
$\cdot$ we have $k_{a}u_{b}=u_{a}\cdot u_{b}=u_{b}\cdot u_{a}=k_{b}u_{a}$.
Consequently,

$\displaystyle k_{a}=0\text{ for }a\not\in A_{(0,-2)}.$ (60)

Now let $a\in A_{(0,-2)}$ and $b\not\in A_{(0,-1)}\cup A_{(0,-2)}$. Then
$k_{b}u_{a}+\sum_{y\in A_{(0,-1)}}l_{b}^{(a,y)}u_{y}=u_{b}\cdot
u_{a}=u_{a}\cdot u_{b}=k_{a}u_{b}$ gives

$\displaystyle k_{a}=0\text{ for }a\in A_{(0,-2)}.$ (61)

It follows from 60 and 61 and the commutativity of $\cdot$ that

$\displaystyle u_{a}\cdot u_{b}=\begin{cases}\sum_{y\in
A_{(0,-1)}}l_{a}^{(b,y)}u_{y},&a,b\in A_{(0,-2)},\\\
0,&\text{otherwise}.\end{cases}$

Thus, $\cdot$ is of the form 3 for the commutative associative product
$u_{a}*u_{b}=\sum_{y\in A_{(0,-1)}}l_{a}^{(b,y)}u_{y}\in Z(L)$ on
$V=\operatorname{span}_{F}\\{u_{a}\mid a\in A_{(0,-2)}\\}$.

Conversely, in view of Lemma 15 the product 38 is of the form 3, so
$(L,\cdot,[\cdot,\cdot])$ is a transposed Poisson (and usual Poisson) algebra.
∎

######  Remark 30.

Consider $\mathcal{B}(q)$ with $q\neq 0$ as the complex Block algebra
$L(\mathbb{Z}\times\mathbb{Z},g,f)$, where $g$, $f$ and $h$ are given by 23
and 24. Then $(g(m,i),h(m,i))=(0,-2)\Leftrightarrow(m,i)=(0,-2q)$ and
$(g(m,i),h(m,i))=(0,-1)\Leftrightarrow(m,i)=(0,-q)$, so we again obtain the
description of transposed Poisson algebra structures on $\mathcal{B}(q)$ given
in [11, Theorem 2.14] as a particular case of Theorem 29.

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