# DePS: An improved deep learning model for de novo peptide sequencing

Cheng Ge
Institute of Bioinformatics and Medical Engineering
School of Electrical and Information Engineering
Jiangsu University of Technology
China, Changzhou 213001
<EMAIL_ADDRESS>
Yi Lu
Institute of Bioinformatics and Medical Engineering
School of Electrical and Information Engineering
Jiangsu University of Technology
China, Changzhou 213001
Jia Qu
School of Computer Science and Artificial Intelligence & Aliyun School of Big
Data
Changzhou University
China, Changzhou 213164
Liangxu Xie
Institute of Bioinformatics and Medical Engineering
School of Electrical and Information Engineering
Jiangsu University of Technology
China, Changzhou 213001
Feng Wang
School of Computer Science and Artificial Intelligence & Aliyun School of Big
Data
Changzhou University
China, Changzhou 213164
Hong Zhang
School of Mathematics
China University of Mining and Technology
China, Xuzhou 221116
Ren Kong
Institute of Bioinformatics and Medical Engineering
School of Electrical and Information Engineering
Jiangsu University of Technology
China, Changzhou 213001
<EMAIL_ADDRESS>
Shan Chang
Institute of Bioinformatics and Medical Engineering
School of Electrical and Information Engineering
Jiangsu University of Technology
China, Changzhou 213001
<EMAIL_ADDRESS>

###### Abstract

De novo peptide sequencing from mass spectrometry data is an important method
for protein identification. Recently, various deep learning approaches were
applied for de novo peptide sequencing and DeepNovoV2 is one of the
represetative models. In this study, we proposed an enhanced model, DePS,
which can improve the accuracy of de novo peptide sequencing even with missing
signal peaks or large number of noisy peaks in tandem mass spectrometry data.
It is showed that, for the same test set of DeepNovoV2, the DePS model
achieved excellent results of 74.22%, 74.21% and 41.68% for amino acid recall,
amino acid precision and peptide recall respectively. Furthermore, the results
suggested that DePS outperforms DeepNovoV2 on the cross-species dataset.

##  1 Introduction

Protein identification is an essential step in proteomics research, and tandem
mass spectrometry (MS/MS) is one of the principal methods to resolve this
problem. Shotgun proteomics is a method to identify complex protein mixtures
using a combination of high-performance liquid chromatography and mass
spectrometry [1]. During the identification process, proteins are
enzymatically digested firstly, and the resulting peptide mixtures are ionized
and then scanned by MS/MS to obtain a set of spectra data. Finally, peptide
identification and sequence assembly steps are performed to detect the amino
acid sequence of the protein [2]. In particular, mass spectrometry-based
peptide identification is an important part of protein identification.
Currently, there are two major categories for mass spectrometry-based peptide
identification methods: (1) database searching, (2) de novo sequencing.
Database searching methods have been the main methods of peptide
identification for decades. It compares the mass spectrometry data of the
unknown peptide with the database data, looks for sequences that match the
known peptide and selects the corresponding peptide with the highest match
score. Common database search algorithms include Mascot [3], X!Tandem [4],
pFind [5] and PEAKS DB [6]. However, the database search is only suitable for
peptides already collected in the database. Once the query peptide does not
exist in the database, then the method will no longer be suitable. Therefore,
De novo peptide sequencing is a good solution as an alternative to database
search methods. For de novo peptide sequencing, the amino acid of peptide
could be obtained by using the mass difference between two fragment ions
without database search. There are several algorithms for de novo peptide
sequencing. Traditional algorithms include method of exhaustion and graph-
based method. The exhaustive method [7, 8] lists all possible candidate
peptides based on the mass of the parent ion, and then compares the candidate
peptides with the experimental profile to find the best matching candidate
peptide. This method is capable of identifying shorter peptide sequences, but
is not suitable for longer peptide sequences or for data produced by low
precision mass spectrometers. The graph-based method [9, 10, 11, 12, 13, 14],
which is more commonly used, constructs a mass spectrum peak linkage graph. If
the mass difference between two peaks is equal to the mass of an amino acid
residue, the two mass spectrum peaks are added to the (V, E) graph as two
vertices and an edge. Once the linkage map has been constructed, all possible
pathways are searched to obtain candidate peptides. Finally, the candidate
peptides are sorted and output by means of a scoring function. The
disadvantage of this method is that it is not applicable for the data with
lost ion peaks, which is occasionally produced by low accuracy mass
spectrometer. Meanwhile, higher precision means that less nodes are merged and
the generated spectrum graph has more vertices, which directly leads to a
higher computational complexity. Muth et al. [15] provided a detailed overview
of traditional algorithms and evaluated their performance on selected
experimental and simulated MS/MS datasets. Similarly, Mohammed et al. [16]
presented a theoretical framework and a data processing workflow for
visualizing and comparing the results of these different types of algorithms.
Deep learning has evolved rapidly over the past several decades. It has
excelled in feature extraction in large and complex datasets and is widely
used in proteomics research [17, 18]. Deep learning based de novo peptide
sequencing algorithms have been developed in recent years, including DeepNovo
[19], pDeep [20] and DeepNovoV2 [21]. Among them, DeepNovoV2 is a state-of-
the-art model of deep learning-based approaches for de novo peptide sequencing
due to its excellent feature extraction network and unique feature
representation. Its model structure consists of a Long Short Term Memory
(LSTM) network and a T-net model from the Point Net [22]. However, de novo
peptide sequencing methods are strongly influenced by the quality of the mass
spectrometry data. To improve the accuracy of de novo sequencing in the low
accuracy tandem mass spectrometry data, we proposed the DePS model.
Specifically, for the problem of missing signal peaks, DePS model expands the
feature dimension based on DeepNovoV2. Meanwhile, to address the problem of a
large number of noisy peaks in tandem mass spectrometry data, we were inspired
by the Deep Residual Shrinkage Networks [23], which performs well on signal
data with noise, and constructed a CNN network with an attention mechanism.

Figure 1: Tandem mass spectrometry spectrum in different situations. Take the
peptide sequence ”ANELLL” as an example, and only b-ions are shown here. Under
the ideal conditions, the tandem mass spectrometry spectrum contains complete
b-ion peaks. Each neighbouring b-ion peak differs by one amino acid mass from
each other. In the case of missing b-ion peaks, the difference between
neighbouring b-ion peaks is the mass of two or more amino acids. In the case
of including noise peaks, the b-ion peak is difficult to identify. In
realistic situations, they may even occur at the same time.

##  2 Methods

###  2.1 Mass spectrometry data

Mass spectrometry data includes a mass spectrometry and tandem mass
spectrometry. The tandem mass spectrometry is obtained by fragmenting the
peptides in a mass spectrometry. The tandem mass spectrometry mainly includes
three complementary types of fragment ions, (a_ion, x_ion), (b_ion, y_ion) and
(c_ion, z_ion), depending on different fragmentation methods and locations of
the peptide chain break [24]. The most commonly used fragmentation methods are
collision-induced dissociation (CID) and electron transfer dissociation (ETD).
CID can produce b and y ions, and ETD mainly produces c and z ions. There are
other types of ions that could be generated because of the incomplete
dissociation of the peptide fragment, such as b and y ions that have lost
ammonia (b-$\rm NH_{3}$, y-$\rm NH_{3}$) or water (b-$\rm H_{2}O$, y-$\rm
H_{2}O$). We can plot the tandem mass spectra using the mass/charge (m/z) and
intensity values of the ions. Normally, tandem mass spectrometry data may have
some signal peaks lost or contain a large number of noise peaks, as shown in
Figure 1, which adds difficulties to deep learning based de novo peptide
sequencing algorithms.

Figure 2: The model structure of DePS. Firstly, different feature data is
obtained from the mass spectrometry and these are passed into the CNN and LSTM
respectively. After that, the output features of the two networks are
concentrated and finally the peptide sequences are obtained through the output
layer.

###  2.2 Model construction

The DEPS model mainly consists of feature extraction, CNN and LSTM module, as
shown in Figure 2. It obtains feature representation of tandem mass
spectrometry by using a feature extraction module and extracts features using
a CNN module. The LSTM module is used to extract the language model
information of peptides. It includes sequence features after embedding and
hidden state that encoding by position embedding methods [25].

###  2.3 Feature Extraction

We use a set of [m/z, intensity] pairs to represent the tandem mass
spectrometry spectrum and only select the top n most intense peaks (n = 500).
A mass list, called mass_AA_list, is created. It contains 26 elements (20
amino acid residues, 3 post-translational modifications, and 3 special tokens:
”go”, ”end” and ”pad”). As shown in Figure 3, we used a feature extraction
framework similar to DeepNovoV2. First, we get the theoretical ion masses in
an iterative fashion based on the peptide sequence. It is a matrix of shape
(v,k), which is denoted Mtheoretical. After that, the observed ion masses are
obtained from the spectrum, and matched to the theoretical ion masses.
Finally, the obtained result is concatenated with the intensity values of the
peaks to obtaining the feature matrix. Different from DeepNovoV2, we take into
account the condition of signal peak loss to obtain the theoretical ion
masses.

Figure 3: The process of feature extraction. Where prefix_max is an iterative
variable with an initial value of 0, mass[E] represents the mass of amino acid
E, mass_AA_list is a list of the masses of 26 tokens, and candidate_b
represents the mass of the candidate b-ion. T denotes the length of the
sequence.

Take the peptide sequence "SLETENR" as an example. The sequence is preceded by
an initial token (’go’, mass=0). After that, we use the mass of ’go’ as the
mass of the N-terminal, and add the resulting mass to each element of
mass_AA_list. This step is used to obtain the theoretical mass of the b-ion
(candidate b_ion). Once the b-ion mass is obtained, the masses of y-ion and
a-ion can be calculated using the mass relationship between the ions.
Similarly, we can get candidate_b_H2O, candidate_b_NH3, candidate_b(2+),
candidate_y_H2O, candidate_y(2+), candidate_y_NH3, candida te_a_H2O,
candidate_a(2+) and candidate_a_NH3 by using the mass relationship between the
ions. We add 23 amino acid masses to each candidate ion mass individually,
because, without considering the masses of the 3 special tokens masses, there
are 23 possible arrangements of amino acids when the difference in mass
between the signal peaks is two amino acid masses. Then, the feature of each
token becomes (26,58) dimensions by dimensional transformation. This is a
significant improvement compared to the feature dimension of DeepNovoV2
(26,12). This method is a good solution to the problem of missing key signal
peaks. Further, in order to match with Mobserved, the feature matrix dimension
needs to be reconstructed as (1,26*58). Now, we have a matrix of Mtheoretical.
Next, the Mtheoretical matrix is matched with the Mobserved matrix obtained
from the mass spectra to get the feature matrix by doing the operation of
Equation (1).

$\sigma(D)=\mathrm{e}^{\left(-\left|M_{observed}-M_{theoretical}\right|*c\right)}$
(1)

where c=100. Then, the peak intensities are concatenated with the feature
matrix, as shown in Equation (2).

$F=\sigma(D)\oplus I$ (2)

In this way, the final feature matrix is obtained.

###  2.4 CNN model

Changes in the arrangement of amino acids in the peptide sequence will lead to
the feature change of the tandem mass spectra. This constantly changing
feature order makes it more difficult for CNNs to extract features. To solve
this problem, DeepNovoV2 uses the T-net network [22], which performs well in
dealing with the problem of changing order of point cloud features. The T-net
network consists of three 1D convolutions, one global-max-pooling, and two
fully connected layers. The global-max-pooling operation which achieves order
invariance makes the T-net network perform excellent in tasks with changing
features.

Figure 4: DePS-net. It consists mainly of three one-dimensional convolutional
layers, a deep shrinkage network module, a global max-pooling layer and three
fully connected layers.

We propose a novel CNN model, which we call DePS-net, retains the global-max-
pooling layer and adds a deep shrinkage network module, as shown in Figure 4.
DePS-net consists mainly of three one-dimensional convolutional layers, a deep
shrinkage network module [23], a global max-pooling layer and three fully
connected layers. The deep shrinkage network module consists mainly of
attention mechanism and soft thresholding. The model adaptively thresholds
through an attention mechanism, where each channel of the feature corresponds
to a threshold. It can set separate thresholds for each feature on a feature-
by-feature basis, and is therefore suitable for situations where the noise
content within each feature varies. Soft thresholding is a function that
shrinks the input data towards zero and is often used in signal noise
reduction algorithms. It is formulated as follows:

$y=\left\\{\begin{array}[]{ll}x-\tau&x>\tau\\\ 0&-\tau\leq x\leq\tau\\\
x+\tau&x<-\tau\end{array}\right.$ (3)

where x denotes the input feature, y denotes the output feature and $\tau$
denotes the threshold value. The derivative of the soft thresholding function
is formulated as follows:

$\frac{\partial y}{\partial x}=\left\\{\begin{array}[]{ll}1&x>\tau\\\
0&-\tau\leq x\leq\tau\\\ 1&x<-\tau\end{array}\right.$ (4)

We can see that the derivative of the soft thresholding function is either
zero or one. Thus, unimportant features are eliminated through a combination
of attention mechanism and soft thresholding.

In DePS-net model, features are first convolved by three 1D convolution layers
(1D convolutional convolution, batch normalization (BN), ReLU activation).
Then they are fed into a deep shrinkage network module. The specific steps of
deep shrinkage network module include: (1) Changing the features into absolute
values and performing global-max-pooling; (2) Passing through a fully
connected layer, a batch normalization layer, a ReLU activation layer, a fully
connected layer, and a Sigmoid activation function layer in order; (3)
Weighting the result with the result after the global-max-pooling layer; (4)
Shorting the features with the results after 1D convolution layers and
performing a soft thresholding operation. After that, the features are
subjected to global-max-pooling. Finally, the features are passed through
three fully connected layers.

##  3 Experiments And Results

###  3.1 Dataset

The same dataset used in DeepNovoV2 are used here, the DDA dataset of Hela
samples111PRG 2018: Evaluation of Data-Independent Acquisition (DIA) for
Protein Quantification in Academic and Core Facility Settings.
https://abrf.org/research-group/proteomics-research-group-prg (denoted as ABRF
dataset). It is divided into training set, validation set and test set in the
same way as in DeepNovoV2. As the ABRF dataset consists of only a single
species, we also test DePS with cross species data published by Tran et
al.[19] (MSV000081382), which contains 9 different species data.

###  3.2 Implementation Details

The implementation is based on Pytorch using an NVIDIA Tesla V100 32GB GPU.
The model is trained using the Adam optimizer and a batch size of 32 with a
learning rate of 0.001. We use focal loss as suggested by Tran et al. [19]
when training the model.

To compare the performance of DePS with DeepNovoV2, we set up two tasks. Task
1 contains two subtasks. The CNN model is used in subtask 1, which is used to
compare the performance of T-net with DePS-net in different feature
dimensions. A CNN+LSTM structure is used in subtask 2, which is used to
compare the performance of DeepNovoV2 with DePS in different feature
dimensions. Task2 is used to test the generalisation ability of DeepNovoV2 and
DePS on a cross species dataset.

###  3.3 Evaluation metrics

In order to evaluate the performance of the model, we use the same evaluation
metrics as those proposed by DeepNovoV2. The evaluation metrics include Amino
Acid Recall (AAR), Amino Acid Precision (AAP), and Peptide Recall (PR).

$AAR=\frac{recall\\_AA\\_total}{target\\_len\\_total}$ (5)
$AAP=\frac{recall\\_AA\\_total}{predicted\\_len\\_total}$ (6)

$PR=\frac{recall\\_peptide\\_total}{predicted\\_count}$ (7)

where recall_AA_total represents the total number of amino acids that were
correctly predicted, target_len_total represents the total number of target
amino acids, predicted_len_total represents the total number of predicted
amino acids, recall_peptide_total represents the total number of peptides that
were correctly predicted, and predicted_count represents the total number of
predicted peptides.

###  3.4 Experiment Results

Results of feature dimension expansion
The experimental results of Task1 are shown in Table 1. Comparing Model1 and
Model3, Model2 and Model4, it can be seen that, with the model structure
unchanged, it achieves a better result in all three evaluation metrics by
extending the feature dimensions whether T-net or DePS-net. As the dimension
expands from 12 to 58, DePS-net has increased its value in the AAR by 1.79
percent. In terms of AAP, this value is 1.95 percent. In terms of PR, it is
1.64 percent. Similarly, T-net has increased its value in the AAR by 1.77
percent. In terms of AAP, this value is 1.96 percent. In terms of PR, it is
1.26 percent. Hereby, we construct expanded features that include masses of
original ions and the masses of original ions plus one extra amino acid. The
expanded features allow CNN models to extract more useful information and
could alleviate the problems of signal peaks missing.
Results of different feature extraction models
Comparing Model1 and Model2, Model3 and Model4, it can be seen that DePS-net
outperforms T-net on all three evaluation metrics with the same feature
dimensions. When the feature dimensions are 12, DePS-net is 0.19 percent
higher than T-net in AAR. In terms of AAP, it is 0.12 percent higher than
T-net. In terms of PR, it is 0.66 percent higher than T-net. When the feature
dimensions are 58, DePS-net is 0.21 percent higher than T-net in AAR. In terms
of AAP, it is 0.11 percent higher than T-net. In terms of PR, it is 1.04
percent higher than T-net. This suggests that the DePS-network model can
slightly improve the accuracy of de novo peptide sequencing in the presence of
a large number of noise peaks.
Results of CNN+LSTM structure
Comparing Model1 and Model5, Model2 and Model6, Model3 and Model7, Model4 and
Model8, it can be seen that the overall performance of the models has been
improved with the LSTM model added in the model.
DePS vs DeepNovoV2
Comparing the loss curves of DePS and DeepNovoV2 on the training set and
validation set, as shown in Figure 5, it can be seen that DePS can converge
faster and outperforms DeepNovoV2 as increasing itration. Comparing Model5 and
Model8 in Table 1, it can be seen that the results of DePS model show a
significant improvement over DeepNovoV2. DePS is 1.49 percent higher than
DeepNovoV2 in AAR. In terms of AAP, it is 1.47 percent higher than DeepNovoV2.
In terms of PR, it is 1.87 percent higher than DeepNovoV2. Further, we
analysed the precision and recall of each peptide, as shown in Figure 6, which
clearly shows the performance of the different models in the test set.
Comparing model 5 with model 8, it can be seen that the mean and lower bound
of precision and recall for the peptides of DePS are higher than those of
DeepNovoV2.

Table 1: Experimental results of Task 1. Task | Model | Model structure | Num of features | AAR (%) | AAP (%) | PR (%)
---|---|---|---|---|---|---
subtask1 | Model1 | T-net | 12 | 71.83 | 71.91 | 38.12
Model2 | DePS-net | 12 | 72.02 | 72.03 | 38.78
Model3 | T-net | 58 | 73.60 | 73.87 | 39.38
Model4 | DePS-net | 58 | 73.81 | 73.98 | 40.42
subtask2 | Model5 (DeepNovoV2) | T-net+LSTM | 12 | 72.73 | 72.74 | 39.81
Model6 | DePS-net+LSTM | 12 | 73.00 | 72.80 | 40.52
Model7 | T-net+LSTM | 58 | 74.16 | 74.18 | 40.59
Model8 (DePS) | DePS-net+LSTM | 58 | 74.22 | 74.21 | 41.68
Figure 5: The loss curves of DePS and DeepNovoV2 on the training set and
validation set.  Figure 6: Box plots of test results for the eight models. It
shows the precision and recall of each peptide in the results of each model.

(a)

(b)

(c)

Figure 7: Bar Chart of experimental results for Task 2. Amino acid recall,
amino acid precision and peptide recall of DeepNovoV2 and DePS.

To validate the generalization ability of DePS, we conduct experiments on
cross species datasets. The experimental results of Task 2 are shown in figure
7, and it can be seen that DePS still performs well on cross species datasets.
DePS is 2.3 to 5.3 percent higher than DeepNovoV2 in Amino Acid Recall. In
terms of Amino Acid Precision, it is 2.4 to 5.7 percent higher than
DeepNovoV2. In terms of Peptide Recall, it is 3.4 to 5 percent higher than
DeepNovoV2. Moreover, DePS outperforms DeepNovoV2 in all three evaluation
metrics on cross species datasets. In particular, on datasets where DeepNovoV2
predicts poorly, DePS has a large improvement. This situation occurs because
the loss of signal peaks and a large number of noisy peaks can lead to poor
predictions from DeepNovoV2, whereas DePS is better able to solve these
problems.

##  4 Conclusion and Future Research

This paper proposed a deep learning-based model for de novo peptide
sequencing, called DePS. It achieves better performance than DeepNovoV2 by
increasing the feature dimension and constructing a novel CNN network. By
expanding the dimensionality of the feature matrix, DePS solves the problem of
missing key signal peaks in tandem mass spectra very well. By proposing a
DePS-net model with global-max-pooling, an attention mechanism and soft
thresholding, DePS effectively improves the accuracy of de novo peptide
sequencing in the presence of a large number of noise peaks. On ABRF dataset,
the DePS model achieved excellent results of 74.22%, 74.21% and 41.68% for
amino acid recall, amino acid precision and peptide recall respectively. This
result is significantly better than DeepNovoV2. On additional cross species
data, the DePS model outperformed DeepNovoV2, which also shows that our model
has great generalization ability. In the future, we will optimise the feature
representation and improve the feature extraction network.

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# Depth and minimal number of generators of square free monomial ideals

Dorin Popescu  Dorin Popescu, Institute of Mathematics ”Simion Stoilow”,
Research unit 5, University of Bucharest, P.O.Box 1-764, Bucharest 014700,
Romania<EMAIL_ADDRESS>
###### Abstract.

Let $I$ be an ideal of a polynomial algebra over a field generated by square
free monomials of degree $\geq d$. If $I$ contains more monomials of degree
$d$ than $(n-d)/(n-d+1)$ multiplied with the number of square free monomials
of $S$ of degree $d$ then $\operatorname{depth}_{S}I\leq d$, in particular the
Stanley’s Conjecture holds in this case.
Key words  : Monomial Ideals, Depth, Stanley depth.
2000 Mathematics Subject Classification: Primary 13C15, Secondary 13F20,
13F55, 13P10.

The support from the CNCSIS grant PN II-542/2009 of Romanian Ministry of
Education, Research and Inovation is gratefully acknowledged.

Let $S=K[x_{1},\ldots,x_{n}]$ be the polynomial algebra in $n$-variables over
a field $K$ and $I\subset S$ a square free monomial ideal. Let $d$ be a
positive integer and $\rho_{d}(I)$ be the number of all square free monomials
of degree $d$ of $I$.

The proposition below was repaired using an idea of Y. Shen to whom we owe
thanks.

######  Proposition 1.

If $I$ is generated by square free monomials of degree $\geq d$ and
$\rho_{d}(I)>((n-d)/(n-d+1)){n\choose d}$ then $\operatorname{depth}_{S}I\leq
d$.

###### Proof.

Apply induction on $n$. If $n=d$ then there exists nothing to show. Suppose
that $n>d$. Let $\nu_{i}$ be the number of the square free monomials of degree
$d$ from $I\cap(x_{i})$. We may consider two cases renumbering the variables
if necessary.

Case 1 $\nu_{1}>((n-d)/(n-d+1)){n-1\choose d-1}$.

Let $S^{\prime}:=K[x_{2},\ldots,x_{n}]$ and
$x_{1}c_{1},\ldots,x_{1}c_{\nu_{1}}$, $c_{i}\in S^{\prime}$ be the square free
monomials of degree $d$ from $I\cap(x_{1})$. Then $J=(I:x_{1})\cap S^{\prime}$
contains $(c_{1},\ldots,c_{\nu_{1}})$ and so
$\rho_{d-1}(J)\geq\nu_{1}>((n-d)/(n-d+1)){n-1\choose d-1}.$ By induction
hypothesis, we get $\operatorname{depth}_{S^{\prime}}J\leq d-1$. It follows
$\operatorname{depth}_{S}JS\leq d$ and so $\operatorname{depth}_{S}I\leq d$ by
[7, Proposition 1.2].

Case 2 $\nu_{i}\leq((n-d)/(n-d+1)){n-1\choose d-1}$ for all $i\in[n]$.

We get $\sum_{i=1}^{n}\nu_{i}\leq n((n-d)/(n-d+1)){n-1\choose d-1}$. Let
$A_{i}$ be the set of the square free monomials of degree $d$ from
$I\cap(x_{i})$. A square free monomial from $I$ of degree $d$ will be present
in $d$-sets $A_{i}$ and it follows

$\rho_{d}(I)=|\cup_{i=1}^{n}A_{i}|\leq(n/d)((n-d)/(n-d+1)){n-1\choose
d-1}=((n-d)/(n-d+1)){n\choose d}$

if $n\geq d+1$. Contradiction!

######  Remark 2.

If $I$ is generated by square free monomials of degree $\geq d$, then
$\operatorname{depth}_{S}I\geq d$. Indeed, since $I$ has a square free
resolution the last shift in the resolution of $I$ is at most $n$. Thus if $I$
is generated in degree $\geq d$, then the resolution can have length at most
$n-d$, which means that the depth of $I$ is greater than or equal to $d$ (this
argument belongs to J. Herzog). Hence in the setting of the above proposition
we get $\operatorname{depth}_{S}I=d$.

######  Corollary 3.

Let $I$ be an ideal generated by $\mu(I)$ square free monomials of degree $d$.
If $\mu(I)>((n-d)/(n-d+1)){n\choose d}$ then $\operatorname{depth}_{S}I=d$.

######  Example 4.

Let $I=(x_{1}x_{2},x_{2}x_{3})\subset S:=K[x_{1},x_{2},x_{3}].$ Then $d=2$ and
$\mu(I)=2>(1/2){3\choose 2}$. It follows that $\operatorname{depth}_{S}I=2$ by
the above corollary.

######  Example 5.

Let
$I=(x_{1}x_{2},x_{1}x_{3},x_{1}x_{4},x_{2}x_{3},x_{2}x_{5},x_{3}x_{4},x_{3}x_{5},x_{4}x_{5})\subset\\\
S:=K[x_{1},\ldots,x_{5}]$. Then $d=2$ and $\mu(I)=8>(3/4){5\choose 2}$ and so
$\operatorname{depth}_{S}I=2$.

Next lemma presents a nice class of square free monomial ideals $I$ with
$\mu(I)={n\choose d+1}\leq((n-d)/(n-d+1)){n\choose d}$ but
$\operatorname{depth}_{S}I=d$. We suppose that $n\geq 3$. Let $w$ be the only
square free monomial of degree $n$ of $S$, that is $w=\Pi_{j=1}^{n}x_{i}$. Set
$f_{i}=w/(x_{i}x_{i+1})$ for $1\leq i<n$, $f_{n}=w/(x_{1}x_{n})$ and let
$L_{n}:=(f_{1},\ldots,f_{n-1})$, $I_{n}:=(L,f_{n})$ be ideals of $S$ generated
in degree $d=n-2$. We will see that $\operatorname{depth}_{S}I_{n}=n-2$ even
$\mu(I_{n})=n={n\choose d+1}$.

######  Lemma 6.

Then $\operatorname{depth}_{S}L_{n}=n-1$ and
$\operatorname{depth}_{S}I_{n}=n-2$.

###### Proof.

Apply induction on $n\geq 3$. If $n=3$ then $L_{3}=(x_{3},x_{1})$,
$I_{3}=(x_{1},x_{2},x_{3})$ and the result is trivial. Assume that $n>3$. Note
that $(L_{n}:x_{n})=L_{n-1}S=(I_{n}:x_{n})$ because
$f_{n},f_{n-1}\in(L_{n}:x_{n})$. We have

$L_{n}=(L_{n}:x_{n})\cap(x_{n},L_{n})=(L_{n-1}S)\cap(x_{n},f_{n-1}),$
$I_{n}=(I_{n}:x_{n})\cap(x_{n},I_{n})=(L_{n-1}S)\cap(x_{n},f_{n-1},f_{n})=(L_{n-1}S)\cap(x_{n},u)\cap(x_{1},x_{n-1},x_{n}),$

where $u=w/(x_{1}x_{n-1}x_{n})$. But $(x_{1},x_{n-1})$ is a minimal prime
ideal of $L_{n-1}S$ and so we may remove $(x_{1},x_{n-1},x_{n})$ above, that
is $I_{n}=(L_{n-1}S)\cap(x_{n},u)$. On the other hand,
$(L_{n-1}S)+(x_{n},u)=(x_{n},I_{n-1})$ and
$(L_{n-1}S)+(x_{n},f_{n-1})=(x_{n},L_{n-1})S$ because $f_{n-1}\in L_{n-1}S$.
We have the following exact sequences

$0\rightarrow S/L_{n}\rightarrow S/L_{n-1}S\oplus S/(x_{n},f_{n-1})\rightarrow
S/(x_{n},L_{n-1}S)\rightarrow 0,$ $0\rightarrow S/I_{n}\rightarrow
S/L_{n-1}S\oplus S/(x_{n},u)\rightarrow S/(x_{n},I_{n-1}S)\rightarrow 0.$

By induction hypothesis $\operatorname{depth}L_{n-1}=n-2$ and
$\operatorname{depth}I_{n-1}=n-3$ and so
$\operatorname{depth}_{S}S/(x_{n},L_{n-1}S)=n-3$,
$\operatorname{depth}_{S}S/(x_{n},I_{n-1}S)=n-4$. As
$\operatorname{depth}_{S}S/(x_{n},f_{n-1})=\operatorname{depth}_{S}S/(x_{n},u)=n-2$,
it follows $\operatorname{depth}_{S}S/L_{n}=n-2$,
$\operatorname{depth}_{S}S/I_{n}=n-3$ by the Depth Lemma applied to the above
exact sequences.

Now, let $I$ be an arbitrary square free monomial ideal and $P_{I}$ the poset
given by all square free monomials of $I$ (a finite set) with the order given
by the divisibility. Let ${\mathcal{P}}$ be a partition of $P_{I}$ in
intervals $[u,v]=\\{w\in P_{I}:u|w,w|v\\}$, let us say
$P_{I}=\cup_{i}[u_{i},v_{i}]$, the union being disjoint. Define
$\operatorname{sdepth}{\mathcal{P}}=\operatorname{min}_{i}\operatorname{deg}v_{i}$
and
$\operatorname{sdepth}_{S}I=\operatorname{max}_{\mathcal{P}}\operatorname{sdepth}{\mathcal{P}}$,
where ${\mathcal{P}}$ runs in the set of all partitions of $P_{I}$. This is
the so called the Stanley depth of $I$, in fact this is an equivalent
definition given in a general form by [1].

For instance, in Example 4, we have
$P_{I}=\\{x_{1}x_{2},x_{2}x_{3},x_{1}x_{2}x_{3}\\}$ and we may take
${\mathcal{P}}:\ \
P_{I}=[x_{1}x_{2},x_{1}x_{2}x_{3}]\cup[x_{2}x_{3},x_{2}x_{3}]$ with
$\operatorname{sdepth}_{S}{\mathcal{P}}=2$. Moreover, it is clear that
$\operatorname{sdepth}_{S}I=2$.

######  Remark 7.

If $I$ is generated by $\mu(I)>{n\choose d+1}$ square free monomials of degree
$d$ then $\operatorname{sdepth}_{S}I=d$. Since $((n-d)/(n-d+1)){n\choose
d}\geq{n\choose d+1}$, the Proposition 1 says that in a weaker case case
$\operatorname{depth}_{S}I\leq\operatorname{sdepth}_{S}I$, which was in
general conjectured by Stanley [8]. Stanley’s Conjecture holds for
intersections of four monomial prime ideals of $S$ by [2] and [4] and for
square free monomial ideals of $K[x_{1},\ldots,x_{5}]$ by [3] (a short
exposition on this subject is given in [5]). It is worth to mention that
Proposition 1 holds in the stronger case when $\mu(I)>{n\choose d+1}$ (see
[6]), but the proof is much more complicated and the easy proof given in the
present case has its importance.

In the Example 5 we have
$P_{I}=[x_{1}x_{2},x_{1}x_{2}x_{4}]\cup[x_{1}x_{3},x_{1}x_{3}x_{5}]\cup[x_{1}x_{4},x_{1}x_{4}x_{5}]\cup[x_{2}x_{3},x_{1}x_{2}x_{3}]\cup[x_{3}x_{4},x_{1}x_{3}x_{4}]\cup[x_{3}x_{5},x_{3}x_{4}x_{5}]\cup[x_{4}x_{5},x_{2}x_{4}x_{5}]\cup[x_{2}x_{3}x_{4},x_{2}x_{3}x_{4}]\cap[x_{2}x_{3}x_{5},x_{2}x_{3}x_{5}]\cup(\cup_{\alpha}[\alpha,\alpha])$,
where $\alpha$ runs in the set of square free monomials of $I$ of degree
$4,5$. It follows that $\operatorname{sdepth}_{S}I=3$. But as we know
$\operatorname{depth}_{S}I=2$.

## References

  * [1] J. Herzog, M. Vladoiu, X. Zheng, How to compute the Stanley depth of a monomial ideal, J. Algebra, 322 (2009), 3151-3169.
  * [2] A. Popescu, Special Stanley Decompositions, Bull. Math. Soc. Sc. Math. Roumanie, 53(101), no 4 (2010), arXiv:AC/1008.3680.
  * [3] D. Popescu, An inequality between depth and Stanley depth, Bull. Math. Soc. Sc. Math. Roumanie 52(100), (2009), 377-382, arXiv:AC/0905.4597v2.
  * [4] D. Popescu, Stanley conjecture on intersections of four monomial prime ideals, arXiv.AC/1009.5646.
  * [5] D. Popescu, Bounds of Stanley depth, An. St. Univ. Ovidius. Constanta, 19(2),(2011), 187-194.
  * [6] D. Popescu, Depth of factors of square free monomial ideals, Preprint, 2011.
  * [7] A. Rauf, Depth and Stanley depth of multigraded modules, Comm. Algebra, 38 (2010),773-784.
  * [8] R. P. Stanley, Linear Diophantine equations and local cohomology, Invent. Math. 68 (1982) 175-193.


Square root problem and Subnormal Aluthge transforms]Square root problem and Subnormal Aluthge transforms
H. El Azhar, A. Hanine, K. Idrissi and E. H. Zerouali]Hamza El Azhar $^{1}$, Abdelouahab Hanine$^{1}$, Kaissar Idrissi and El Hassan Zerouali $^{1}$
$^{1}$Center of mathematical research of Rabat, P.O. box 1014, Department of Mathematics, Faculty of sciences, Mohammed V university in Rabat, Rabat, Morocco.
<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>
[2020]Primary 47B20, 47B37.

For a non negative measure $\mu$ with $p$ atoms,  we study the  relation between the Square Root Problem of $\mu$ and  the problem of subnormality of ${\tilde W_\mu}$ the Aluthge transform of the associated unilateral weighted shift. We
use an approach based on   uniquely represented elements in the support of $\mu*\mu$. We first show that if ${\tilde W_\mu}$ is subnormal, then $2p-1\le card(supp(\mu*\mu))\le [\frac{(p-1)^2+6}{2}]$. We rewrite several results known for finitely atomic measure having  at most five atoms and  give a
complete solution for measures  six atoms.

The  authors are supported by the CeReMaR and the  Hassan II Academy of sciences. The last author is supported by African university of Sciences and technology-Abuja. Nigeria.

§ INTRODUCTION

Let us denote  $\mathcal{H}$ an infinite dimensional  Hilbert space and   let $\mathcal{L(H)}$ be  the space of all bounded linear operators on $\mathcal{H}$. An operator  $T \in \mathcal{L(H)}$ is normal if $TT^*=T^*T$,  is subnormal if it is the restriction of some normal operators and is hyponormal if $T^*T-TT^*\ge 0$. Here $T^*$ stands for the usual adjoint operator of $T$. The polar decomposition of an operator is given by the unique representation $T= U|T|$, where $|T| = (T^*T)^{\frac{1}{2}}$ and $U$ is a partial isometry satisfying $ker U = ker T$ and $ker U^*=ker T^*$. The Aluthge transform is then given by the expression

$$\tilde T= |T|^{\frac{1}{2}}U|T |^{\frac{1}{2}}.$$

The Aluthge  transform was introduced in  [1] by Aluthge, in
order to extend several  inequalities valid for hyponormal operators, and has received deep attention in the recent years.

We consider below $\mathcal{H}=l^2(\mathbb{Z}_+)$ endowed by some  orthonormal basis $\{e_n\}_{n\in\mathbb{Z}_+}$. The forward  shift operator  $W_{\alpha}$  is defined on the basis  by $W_\alpha e_n=\alpha_ne_{n+1}$, where   $\alpha=\{\alpha_n\}_{n\geq 0}$ is a given  a  sequence of positive real numbers (called weights). We associate with $W_\alpha$ the moments sequence  obtained  by
$$\gamma_0=1  \mbox{  and } \gamma_k\equiv\gamma_k(\alpha):=\alpha_0^2\alpha_1^2\cdots\alpha_{k-1}^2 \mbox{ for } k\geq 1.$$

We will say that a sequence $\gamma$ admits a representing signed measure (called also a charge) supported in $K \subset {\mathbb R}$,  if

\begin{equation}\label{cmp}
\gamma_n = \int_Kt^nd\mu(t)\: \: \mbox{ for every } \: n\ge 0 \:\: \: \mbox{and} \: \: supp(\mu) \subset K.
\end{equation}

The weighted shift operator is bounded if and only if  $\|W_{\alpha}\|=sup_{n\geq0}\alpha_n<+\infty$.  It is clear  that $W_{\alpha}$ is never  normal  and that  $W_\alpha$ is hyponormal precisely when $\alpha_n$ is non decreasing. On the other hand,   Berger's Theorem says, $W_{\alpha}$  is a subnormal operator, if and only if, there exists a nonnegative Borelean measure $\mu$ (called Berger measure), which is a representing measure of $\{\gamma_n\}_{n\geq0}$ and such that $supp(\mu)\subset[0,\|W_{\alpha}\|^2]$.  In this case we will also write $W_{\alpha}=W_{\mu}$

The Aluthge transform $\tilde W_\alpha$ of a weighted shift  $W_\alpha$  is also a weighted shift, denoted below   $W_{\tilde \alpha}$ . Indeed, it is easy  to check that  $|W_\alpha| e_n=\alpha_ne_n$ and that $Ue_n=e_{n+1}$. It   follows then that

$$\tilde W_\alpha e_n= \sqrt{\alpha_n\alpha_{n+1}}e_{n+1}= {\tilde \alpha_n}e_{n+1}= W_{\tilde \alpha}e_n.$$

Notice also that ${\tilde \gamma}^2_n= \frac{1}{\alpha_0}\gamma_n \gamma_{n+1}$. In several recent  papers, the problem of subnormatiy of the Aluthge transform of  weighted shifts  was considered. See [5], for example.
The next question has been considered in several recent papers

(SP)  Under what conditions is subnormality  of a weighted shift preserved under the Aluthge

For given two measures, let   $*$ denote the multiplicative convolution defined as follows

[\nu*\mu](A)=\int_{\mathbb{R}} \chi_A(xy) d\nu(x)d\mu(y)

The square root problem is usually stated as follows

(SRP): Given a non negative measure $\mu$. Under what conditions, there exist a nonnegative measure $\nu$ such that $\nu*\nu= \mu$.

In [4],  it has been observed that

Let $W_\mu$ be subnormal  weighted shift with  associated representing moment measure  $\mu$. Then  ${\tilde W}_\mu$  is subnormal  if and only if  there exists a $\mathbb{R}^+$-supported probability measure $\nu$ such that $\nu*\nu=\mu*t\mu$.

It follows from the previous proposition that

$$ \mu \mbox{ has a square root } \Rightarrow {\tilde W}_\mu \mbox{ is subnormal} $$

The question if the reverses implication holds is stated in some recent papers. More precisely, the next question is    suggested by <cit.>  and stated  in [4].

Problem 1.   If $W_\mu$ is  a subnormal weighted shift with Berger measure $\mu$,  are the following statements equivalent?

(i) $\mu$ as a square root;

(ii) The Aluthge transform ${\tilde W}_\mu$ is subnormal.

The case of finite atomic measures with at most 5 atoms has been completely solved in [4]. The general case of finite atomic measure remains as an  open conjecture in <cit.>.

Conjecture 4.6.  If $W_\mu$   subnormal with a finitely atomic Berger measure $\mu$ then the following statements are equivalent:

(i) $\mu$ has a square root;

(ii) $ {\tilde W}_\mu$ is subnormal.

The main goal of this note is to investigate Conjecture 4.6. We focus our approach on uniquely represented atoms in $supp(\mu*t\mu)$. This allows to rewrite several  proofs from [4] in a short manner. We also provide an affirmative answer in the case of finite atomic Berger measures with six atoms.

§  SUPPORT OF FINITE ATOMIC MEASURES ASSOCIATED WITH WEIGHTED SHIFTS HAVING  SUBNORMAL ALUTHGE TRANSFORM.

Before starting our study let us mention some useful remarks

* First, for  $\mu$ a non negative measure such that $\mu= a\delta_0+\mu_1$ for some $a>0$ and $0\notin supp(\mu_1)$, here $\delta_0$ is the Dirac measure at zero, we have

* $\mu*\mu=(a^2+2a\|\mu_1\|)\delta_0 +\mu_1*\mu_1$, where $\|\mu_1\|$ is the total variation of $\mu_1$.

* $\mu*t\mu=  (a\delta_0+\mu_1)*(t\mu_1)=  a\|t\mu_1\|\delta_0+\mu_1*t\mu_1. $

It follows that $\mu$,  (resp. $\mu*t\mu$) admits a square root if and only if $\mu_1$, (resp. $\mu_1*t\mu_1$) admits a square root. Thus in  Conjecture 4.6, we can assume without any loss of generality that $0\notin supp(\mu)$.

* Second, let $\mu$ be a given non negative measure and $x$ be   non negative number. Denote $\mu x(t)= \mu(xt)$ and $\mu_x(t)=\mu(t^x)$, the image measures by the mappings $t\to xt$ and $t\to t^x$,respectively. We  have the following,

* $\mu x$ has a square root, if and only if $\mu$ has a square root. Thus, without any loss of generality, we can suppose that $\lambda_1 =1$.

* It is not difficult to see that  $\mu_x$ has a square root, if and only if $\mu$ has a square root. Thus, if we put $r=\frac{\lambda_2}{\lambda_1}$, we get a root exists for one $r$ if and only if it  exists for all $ r$. This fact has been noticed in  remark 2.8 in [3], for $ \mu:=a_1\delta_{\lambda_1}+\cdots + a_p\delta_{\lambda_p}$ when $\lambda_i=r^i$ is a geometric sequence.

In the sequel   $
\mu:=a_1\delta_{\lambda_1}+\cdots + a_p\delta_{\lambda_p}
$ is  a  $p$ atomic measures associated with a weighted shifts   $W_\mu$ and  write further
$0< \lambda_1=1 <\lambda_2 <\cdots <\lambda_p.$
It is clear that

$$supp(\mu*t\mu) = \{\lambda_1^2 <\lambda_1\lambda_2<\lambda_2^2\le \cdots \le \lambda_{p-1}^2< \lambda_{p-1}\lambda_p<\lambda_p^2\}$$
and if $\nu$ solves  $\nu*\nu=\mu*t\mu$, we get $supp(\nu)=supp(\mu)$. In particular  $\mu:=b_1\delta_{\lambda_1}+\cdots + b_p\delta_{\lambda_p}$ for some positive numbers $b_1, \cdots, b_p$.
We also

\begin{equation}\label{nu*nu}
\nu*\nu= \sum_{i=1}^pb_i^2\delta_{\lambda_i^2} + 2\sum_{i\le j}^pb_ib_j\delta_{\lambda_i\lambda_j} \mbox{and } \mu*t\mu =  \sum_{i=1}^pa_i^2\lambda_i\delta_{\lambda_i^2} + \sum_{i\le j}^pa_ia_j(\lambda_i+\lambda_j)\delta_{\lambda_i\lambda_j}.
\end{equation}

The general picture of $\nu*\nu$ is given by the next diagram

\begin{array}{ccccccc}
\lambda_1^2&< \lambda_1\lambda_2& < \lambda_1\lambda_3&<\cdots&<\cdots&<\cdots&< \lambda_1\lambda_p\\
& \wedge   &\wedge  & \wedge & \cdots  &\cdots  & \wedge \\
& \lambda_2^2&<\lambda_2\lambda_3&<\cdots&<\cdots& <\cdots&<\lambda_2\lambda_p\\
&   &\wedge  & \wedge & \cdots  &\cdots  & \wedge \\
& & \lambda_3^2& \ddots& \vdots& \vdots& \lambda_3\lambda_p\\
& & & \ddots& \vdots& \vdots& \vdots\\
& & & & &  \lambda_{p-1}^2&<\lambda_{p-1}\lambda_p\\
& & & & &  &\wedge \\
& & & & & &<\lambda^2_p
\end{array}

In particular $2p-1 \le card(supp(\mu*\mu))\le \frac{p(p+1)}{2}$. In the case  $\lambda_1 = 0$ we have $2p-2 \le card(supp(\mu*\mu))\le \frac{p(p-1)}{2}+1$. We will show below that the inequality $card(supp(\mu*t\mu))\le \frac{p(p+1)}{2}$ can be improved when $\mu*t\mu$ admits a square root.

§.§ Uniquely represented atoms and subnormality of Aluthge transform.

Let $\mu:=a_1\delta_{\lambda_1}+\cdots + a_p\delta_{\lambda_p}$ be a finitely atomic measure. An atom $\lambda_i\lambda_j$ in $supp(\mu *\mu)$  will be said to be uniquely represented if, for every $k,l $ satisfying
$\lambda_i\lambda_j= \lambda_k\lambda_l$, we have $\{i,j\}=\{k,l\}$. The subset of uniquely represented atoms will be denoted ${\cal U}r_\mu$ and the subset of non uniquely represented atoms will be denoted ${\cal NU}r_\mu=supp(\mu*\mu)\setminus {\cal U}r_\mu$.

Clearly ${\cal U}r_\mu\ne \emptyset$ since  $\{\lambda_1^2, \lambda_1\lambda_2, \lambda_{p-1}\lambda_p,\lambda_p^2\}\subset {\cal U}r_\mu$.
We assemble in the next proposition some additional  properties of ${\cal U}r_\mu$ to be used in our context.
Suppose the exists a non negative measure $\nu$ such that  $\nu*\nu=\mu*t\mu$. Then

* For every set $\{i,j, k,l \}$, containing at least 2 numbers,  we have
$$  \{\lambda_i\lambda_j, \lambda_j\lambda_k, \lambda_k\lambda_l,\lambda_l\lambda_i \}  \cap{\cal NU}r_\mu\ne \emptyset.$$

* For every $i\ne j$ and $i\ne k $, we have $  \{\lambda_i^2, \lambda_i\lambda_j, \lambda_j\lambda_k, \lambda_i\lambda_k\} \cap{\cal NU}r_\mu \ne \emptyset.$

* For every $i\ne j$, we have $  \{\lambda_i^2,\lambda_i\lambda_j,\lambda_j^2\} \cap{\cal NU}r_\mu\ne \emptyset.$

* If $ \{\lambda_i^2, \lambda_i\lambda_j, \lambda_j\lambda_k, \lambda_k^2\} \subset {\cal U}r_\mu$ for some  $i\ne j\ne k $, then $  \lambda_j^2=\lambda_i\lambda_k.$

* $\{ \lambda_2^2, \lambda_{p-1}^2, \lambda_1\lambda_p, \lambda_1\lambda_{p-1}, \lambda_2\lambda_p\} \subset {\cal NU}r_\mu.$

We have
$$\nu*\nu= \sum_{i=1}^pb_i^2\delta_{\lambda_i^2} + 2\sum_{i\le j}^pb_ib_j\delta_{\lambda_i\lambda_j} \mbox{ and }  \mu*t\mu =  \sum_{i=1}^pa_i^2\lambda_i\delta_{\lambda_i^2} + \sum_{i\le j}^pa_ia_j(\lambda_i+\lambda_j)\delta_{\lambda_i\lambda_j}.$$

Then by matching the coefficients  of the adequate  masses in both sides, we derive that  for  $\lambda_i\lambda_j \in  {\cal U}r_\mu,  $ we have  $ 2b_ib_j= a_ia_j(\lambda_i+\lambda_j),$ ($ b_i^2=a_i^2\lambda_i,$ if $i=j$).

* Suppose $i\ne k$ and $j\ne l$ are such that $  \{ \lambda_i\lambda_j, \lambda_j\lambda_k, \lambda_k\lambda_l,\lambda_l\lambda_i, \}  \subset {\cal U}r_\mu $. We get

$$ \left\{\begin{array}{rcl}  2b_ib_j=a_ia_j(\lambda_i+\lambda_j), & & 2b_jb_k=a_ja_k(\lambda_j+\lambda_k)\\
2b_kb_l=a_ka_l(\lambda_k+\lambda_l), & & 2b_lb_i=a_la_i(\lambda_l+\lambda_i).
\end{array}
\right.

Multiplying both  sides gives $(\lambda_i+\lambda_j)(\lambda_k+\lambda_l)  =  (\lambda_j+\lambda_k)(\lambda_l+\lambda_i)$ and expanding, we obtain

$$(\lambda_i-\lambda_k)(\lambda_l-\lambda_j) =0. $$

* Derives from 1. with $i= l$.

* Derives from 2. with $j= k$.

* If $ \{\lambda_i^2, \lambda_i\lambda_j, \lambda_j\lambda_k, \lambda_k^2\} \subset {\cal U}r_\mu$, we will derive that

\left\{\begin{array}{lcl}
b_i= a_i\sqrt{\lambda_i}, & &
2b_ib_j= a_ia_j(\lambda_i+\lambda_j)\\
2b_kb_j= a_ka_j(\lambda_k+\lambda_j),& & b_k=  a_k\sqrt{\lambda_k}, \\
\end{array}
\right.

which gives  by multiplying both sides

(\lambda_j+\lambda_i)\sqrt{\lambda_k} = (\lambda_j+\lambda_k)\sqrt{\lambda_i}.

And then
$$(\sqrt{\lambda_k} -\sqrt{\lambda_i})(\lambda_j-\sqrt{\lambda_i\lambda_k})=0.$$

Now, since $\lambda_i\ne \lambda_k$; we get the desired result.

* $\{ \lambda_2^2, \lambda_{p-1}^2, \lambda_1\lambda_p, \lambda_1\lambda_{p-1}, \lambda_2\lambda_p\} \cap {\cal U}r_\mu =\emptyset.$ The case $p=3$ is trivial, we  suppose that $p\ge 4$.

* For $\{ \lambda_2^2, \lambda_{p-1}^2, \lambda_1\lambda_p\} \cap {\cal U}r_\mu =\emptyset$, we use $4.$ with $(i,j)=(1,2)$, $(i,j)=(p-1,p)$ and   $(i,j)=(1,p)$, respectively.

* The cases $ \lambda_2\lambda_{p}$ and   $\lambda_1\lambda_{p-1} $   are symmetric. We only show $ \lambda_2\lambda_{p}\notin {\cal U}r_\mu$.
If $ \lambda_2\lambda_{p}\in {\cal U}r_\mu, $ we will get  $\{\lambda_1\lambda_2,  \lambda_2\lambda_p\} \subset {\cal U}r_\mu $,  and hence
$ \lambda_2^2= \lambda_1\lambda_p$. In particular, it follows that   $\lambda_1\lambda_{p-1}\in {\cal U}r_\mu $ and as before
$ \lambda_{p-1}^2= \lambda_1\lambda_p$. Contradiction.

We deduce  the next immediate corollaries

For every $1< k<p$, $\{\lambda_1\lambda_k,\lambda_2\lambda_k\}\cap{\cal NU}r_\mu \ne \emptyset $ and $\{ \lambda_{p-1}\lambda_k,\lambda_p\lambda_k \}\cap{\cal NU}r_\mu \ne \emptyset.$ Furthermore, there is at most one $k$ such that $\{\lambda_1\lambda_k,\lambda_p\lambda_k \}\subset {\cal U}r_\mu$ and  if such $k$ exists, we will have $\lambda_k^2=\lambda_1\lambda_p$.

We use 1. with  $\{\lambda_1^2,  \lambda_1\lambda_2, \lambda_1\lambda_k,\lambda_2\lambda_k\}\cap{\cal NU}r_\mu \ne \emptyset $ and $\{ \lambda_{p-1}\lambda_k,\lambda_p\lambda_k, \lambda_{p-1}\lambda_p, \lambda_p^2 \}\cap{\cal NU}r_\mu\ne \emptyset$ respectively. For the additional assumption, we use 4. in Proposition <ref>.

For every $2< k<p-1$, any of the sets $\{\lambda_1\lambda_k,\lambda_2\lambda_k, \lambda_k^2\}\cap{\cal NU}r_\mu $ and $\{\lambda_p\lambda_k,\lambda_{p-1}\lambda_k, \lambda_k^2\}\cap{\cal NU}r_\mu $ contains at least two elements.

Derives from  $\{\lambda_1\lambda_k, \lambda_2\lambda_k\}\cap{\cal NU}r_\mu \ne \emptyset, $ $\{\lambda_1\lambda_k, \lambda_k^2\}\cap{\cal NU}r_\mu\ne \emptyset $ and $\{\lambda_2\lambda_k, \lambda_k^2\}\cap{\cal NU}r_\mu \ne \emptyset.$

§.§ Number of atoms of $p$ atomic measure such that $ {\tilde W}_\mu$ is subnormal

We start with the next auxiliary immediate lemma of independent interest.

Let $\mu$ be a non negative measure and let ${\cal NU}r_\mu$ the set of non uniquely represented elements in $\mu*t\mu$. Then

$$card(supp(\mu*t\mu)) \le \left[ \frac{p(p+1)-card({\cal NU}r_\mu)}{2}\right]$$

\begin{array}{cc|ccccc|cc}
{\lambda_1^2}&< \lambda_1\lambda_2& < \lambda_1\lambda_3&<\cdots&<\fbox{$\lambda_1\lambda_k$}&<\cdots&<\lambda_1\lambda_{p-2}&<\bf{\lambda_1\lambda_{p-1}}&< \bf{ \lambda_1\lambda_p}\\
& \wedge   &\wedge  & \wedge & \wedge &\wedge &\wedge &\wedge & \wedge \\
& \bf{\lambda_2^2}&<\lambda_2\lambda_3&<\cdots&<\fbox{$\lambda_2\lambda_k$}&<\cdots &<\lambda_2\lambda_{p-2}&<\lambda_2\lambda_{p-1}&<\bf{\lambda_2\lambda_p}\\\hline
&   &\wedge  & \wedge &\cdots& \cdots  &\cdots & \cdots  & \wedge \\
& & \lambda_3^2& \ddots& \ddots&\cdots& &\vdots <\lambda_3\lambda_{p-1}&< \lambda_3\lambda_p\\
&   &  &  &\cdots& \cdots &\vdots &\cdots  & \wedge \\
& & & \ddots& \vdots& &\vdots &\vdots& \vdots\\
&   &  & &\wedge& &  &\wedge  & \wedge \\
& & && \fbox{$ \lambda_{k}^2$} &\dots&\vdots & <\fbox{$\lambda_{k}\lambda_{p-1}$}& <\fbox{$\lambda_{k}\lambda_p$}\\
& & & &&\ddots &\vdots &\vdots& \vdots\\
& & &&& & \lambda_{p-2}^2& \lambda_{p-2}\lambda_{p-1}& \lambda_{p-2}\lambda_p\\
\hline
& & & &&& & \bf{\lambda_{p-1}^2}&<\lambda_{p-1}\lambda_p\\
& & & && &&  &\wedge \\
& & & && && &<\lambda^2_p
\end{array}

Bold ${\cal NU}r_\mu$  elements and $3(p-4)$ ${\cal NU}r_\mu$ elements from $\{\lambda_1\lambda_k,\lambda_2\lambda_k,\lambda_k^2,\lambda_k\lambda_{p-1},\lambda_k\lambda_p\}\; \; 3\le k\le p-2.$

Noticing that if $\mu*t\mu$ admits a square root, then for $p\ge 4$, we have ${\cal NU}r_\mu$  contains at least $5$ elements given  by Proposition <ref>  (in bold in Figure <ref>), $2(p-4)$ elements coming from the $p-4$ central columns by corollary <ref>, and  additional $p-4$ elements from the two last columns by the second part of the corollary <ref>   (
boxed in Figure <ref>), we derive that $ card({\cal NU}r_\mu)\ge 3p-7.$ From the previous discussion and Proposition <ref>, we deduce
Let $p\ge 4$. If $\mu$ is a $p$ atomic measure such that $ {\tilde W}_\mu$ is subnormal, then
$$2p-1\le card(supp(\mu*t\mu)) \le  \left[\frac{(p-1)^2+6}{2}\right].$$

The two examples below show that  both inequalities in Proposition <ref> are sharp,

* Take $\mu=\frac{1}{8}\delta_1+\frac{\sqrt{2}-1}{2}\delta_2+\frac{7-4\sqrt{2}}{4}\delta_4+\frac{\sqrt{2}-1}{2}\delta_8+\frac{1}{8}\delta_{16}$. Then, $\nu=\frac{\sqrt{2}}{4}\delta_{1}+\frac{2-\sqrt{2}}{2}\delta_{2}+\frac{\sqrt{2}}{4}\delta_{4}$ is a square root of $\mu$ and then  ${\tilde W}_\mu$ is subnormal, a direct calculation gives :

card( supp(\mu*t\mu))=card(\left\{1, 2, 4, 8, 16, 32, 64, 128, 256\right\})=9.

* Take $\mu=\frac{1}{4}\delta_1+\frac{1}{3}\delta_3+\frac{1}{6}\delta_6+\frac{1}{9}\delta_9+\frac{1}{9}\delta_{18}+\frac{1}{36}\delta_{36}$. Then, $\nu=\frac{1}{2}\delta_{1}+\frac{1}{3}\delta_{3}+\frac{1}{6}\delta_{6}$ is a square root of $\mu$ and then  ${\tilde W}_\mu$ is subnormal, a direct calculation gives :

card(supp(\mu*t\mu))=card(\left\{1, 3, 6, 9, 18, 27, 36, 54, 81, 108, 162, 216, 324, 648, 1296 \right\})=15.

Since $\{\lambda_2^2,\lambda_{p-1}^2\}\subset  {\cal NU}r_\mu,$ there exists $i\le p-2$ and $j \ne 3$ such that $\lambda_2^2= \lambda_1\lambda_j$ and
$\lambda_{p-1}^2= \lambda_i\lambda_p$ for some $i\le p-2$. In particular, if $p=3$, we get necessarily  $\lambda_2^2=  \lambda_1\lambda_p=\lambda_1\lambda_3.$ The reverse result is given by the next corollary

Under the notations above with $p\ge 3$. If the identities $\lambda_2^2= \lambda_1\lambda_p$ or   $\lambda_{p-1}^2= \lambda_1\lambda_p$, then $p=3$.

If we assume that $\lambda_2^2=\lambda_1\lambda_p$, then $\{\lambda_1\lambda_{p-1}, \lambda_{p-1}\lambda_p\}\subset{\cal U}r_\mu$. By Corollary <ref>, we get $\lambda_{p-1}^2=\lambda_1\lambda_p=\lambda_2^2$, and thus $p=3$.

For $p\ge 4$, the above result is improved as follows.
Under the notations above, if $ (\lambda_2^2,\lambda_{p-1}^2)= (\lambda_1\lambda_{p-1}, \lambda_2\lambda_{p})$ then $p = 4$.

Let $p\ge 6$ and  assume  $\lambda_2^2= \lambda_1\lambda_{p-1}$ and  $\lambda_{p-1}^2= \lambda_2\lambda_{p}$.   From the inequalities

$$\lambda_1^2 < \lambda_1\lambda_2 < \lambda_1\lambda_3<\cdots \lambda_1\lambda_{p-1}=  \lambda_2^2 \, \mbox{  and } \,
\lambda_{p-1}^2 =\lambda_2\lambda_p < \lambda_3\lambda_p<\cdots <  \lambda_p^2,  $$

it comes that $\lambda_1\lambda_2 < \lambda_1\lambda_3<\cdots \lambda_1\lambda_{p-2}$
and $ \lambda_3\lambda_p < \lambda_4\lambda_p<\cdots <  \lambda_p^2 $
are   uniquely represented. In particular, $\{\lambda_1^2 , \lambda_1\lambda_3 , \lambda_3\lambda_p ,\lambda_p^2 \} \subset {\cal U}r_\mu$ and  $\{\lambda_1^2 , \lambda_1\lambda_4, \lambda_4\lambda_p, \lambda_p^2  \} \subset {\cal U}r_\mu$. We use Proposition <ref> again to get  $\lambda_3^2 =\lambda_1\lambda_p= \lambda_4^2$. Thus $\lambda_3=\lambda_4.$  Contradiction.

For $p=5$, suppose $\lambda_2^2= \lambda_1\lambda_{4}$ and  $\lambda_{4}^2= \lambda_2\lambda_{5}$. We derive in particular that $\{\lambda_1\lambda_3 ,  \lambda_3\lambda_5\} \subset  {\cal NU}r_\mu $ and then $\lambda_3^2=\lambda_1\lambda_5. $ This last fact forces $\{\lambda_1\lambda_3 ,  \lambda_2\lambda_3\} \subset  {\cal U}r_\mu $ and contradicts Proposition <ref>.

§ FINITE ATOMIC MEASURES ASSOCIATED WITH WEIGHTED SHIFTS HAVING  SUBNORMAL ALUTHGE TRANSFORM.

In the sequel, we will say that  $p$ atomic measure $\mu$ has a geometric support if it is of the next form

$$ supp(\mu ) = \{a,ar,\cdots,ar^{p-1}\} \: \: \: \mbox{ for some } a, r  \in {\mathbb R}^*.$$

We start with the next useful lemma
Let $\mu$ be $p-$atomic such that  $ {\tilde W}_\mu$ is subnormal. Then  $\mu$ has a geometric support if and only if  $card(supp(\mu*t\mu)) =2p-1$.

Proof. For the direct implication, suppose $supp(\mu ) = \{a,ar,\cdots,ar^{p-1}\}$. Direct computations, give
$supp(\mu*t\mu)= \{a^2, a^2r^{}, \cdots,a^2r^{2p-2}\} $ and hence  $card(supp(\mu*t\mu)) =2p-1$.

Conversely, write $supp(\mu)= \{\lambda_1,\cdots, \lambda_p\}$ and suppose that  $card(supp(\mu*t\mu)) =2p-1$. It follow that
$$supp(\mu*t\mu)=\{\lambda_1^2<\lambda_1\lambda_2<\lambda_2^2<\cdots< \lambda_{p-1}^2<\lambda_{p-1}\lambda_p<\lambda_p^2\}.$$

\lambda_i\lambda_{i+2}\in ]\lambda_i\lambda_{i+1},\lambda_{i+1}\lambda_{i+2}[\cap supp(\mu*t\mu)=\{\lambda_{i+1}^2\},
$ for every $1\le i \le p-2$,
we derive that   $\lambda_i\lambda_{i+2}=\lambda_{i+1}^2$, and then $\frac{\lambda_{i+1}}{\lambda_i}=\frac{\lambda_{i+2}}{\lambda_{i+1}}$. Thus $r=:\frac{\lambda_2}{\lambda_1}= \frac{\lambda_3}{\lambda_2}=\cdots=\frac{\lambda_{p}}{\lambda_{p-1}}$. Finally $\mu$ has a geometric support with $a=\lambda_1$ and

§.§ The case   $card(supp(\mu))= 3$.

Under the notations above, if $p=3$ then  the following are equivalent

* $ {\tilde W}_\mu$ is subnormal;

* $\mu = a_1\delta_{\lambda_1}+a_2\delta_{\lambda_1r} +a_3\delta_{\lambda_1r^2}$  with $ r=\frac{\lambda_2}{\lambda_1}
>1$ and where  $a_1, a_2, a_3$ are non negative numbers
satisfying  $a_1+a_2+a_3=1$ and $a_2^2=4a_1a_3$.

* $\mu $ admits a square root.

$(1.\Rightarrow 2.)$ Suppose that $ {\tilde W}_\mu$ is subnormal.
Since $supp(\mu*t\mu)=\{\lambda_1^2,\lambda_1\lambda_2, \lambda_{2}^2, \lambda_{1}\lambda_3, \lambda_{2}\lambda_3,\lambda_3^2\}$ and ${\cal NU}r_\mu = \{\lambda_2^2, \lambda_{1}\lambda_3\}$ it follows that $\lambda_{2}^2= \lambda_{1}\lambda_3$ and $card(supp(\mu))=5=2\times3-1,$ and  hence  $\mu$ has a geometric support and then  $\mu = a_1\delta_{\lambda_1}+a_2\delta_{\lambda_1r} +a_3\delta_{\lambda_1r^2}$.

Now writing   $\mu*t\mu=\nu*\nu$ for some $\nu=b_1\delta_{\lambda_1}+b_2\delta_{\lambda_1r} +b_3\delta_{\lambda_1r^2}$, we get

\mu*t\mu= & a_1^2\lambda_1\delta_{\lambda_1^2}+a_1a_2\lambda_1(1+r)\delta_{\lambda_1^2 r}+\lambda_1(a_1a_3(1+r^2)+a_2^2r)\delta_{\lambda_1^2 r^2} \\ & +a_2a_3\lambda_1(r+r^2)\delta_{\lambda_1^2 r^3}+   a_3^2\lambda_1 r^2\delta_{\lambda_1^2 r^4}\\ \nu *\nu = & b_1^2\delta_{\lambda_1^2}+2b_1b_2\delta_{\lambda_1^2 r}+(2b_1b_3+b_2^2)\delta_{\lambda_1^2r^2} +2b_2b_3\delta_{\lambda_1^2 r^3}+   b_3^2\delta_{\lambda_1^2 r^4}.\end{array}$$

Identifying the coefficient of the  atoms, we obtain $b_1= a_1\sqrt{\lambda_1}$, $b_2 = a_2\sqrt{\lambda_1}\frac{1+r}{2}$ and
$b_3= a_3r\sqrt{\lambda_1}$. Using the identity $2b_1b_3+b_2^2=\lambda_1(a_1a_3(1+r^2)+a_2^2r) $ we get  $ a_2^2=4a_1a_3$.

$(2.\Rightarrow 3.)$ Let  $\mu = a_1\delta_{\lambda_1}+a_2\delta_{\lambda_1r} +a_3\delta_{\lambda_1 r^2}$, where   $a_1, a_2, a_3$ are non negative numbers
satisfying  $a_1+a_2+a_3=1$ and $a_2^2=4a_1a_3$. Then  $\mu_0=\sqrt{a_1}\delta_{\sqrt{\lambda_1}}+\sqrt{a_3}\delta_{r\sqrt{\lambda_1}}$ satisfies $\mu = \mu_0*\mu_0$.

$(3.\Rightarrow 1.)$ Derives from $\mu = \mu_0*\mu_0\Rightarrow \mu*t\mu= \mu_0*t\mu_0* \mu_0*t\mu_0$ and Proposition <ref>.

§.§ The case   $card(supp(\mu))= 4$.

Under the notations above, if $p=4$ then $ {\tilde W}_\mu$ is not subnormal.
Assume $ {\tilde W}_\mu$ is  subnormal. It will follow that
$$7\le card(supp(\mu*t\mu) )\le \left[\frac{(4-1)^2+6}{2}\right] = 7,$$
and hence
$supp(\mu)$ is geometric. Assume $\lambda_1=1$. We put $\lambda_i= r^{i-1}$ for $1\le i\le 4$ and write
$\mu=\sum\limits_{i=0}^{i=3}a_i\delta_{r^i}$, we get

a_0^2\delta_{1}+a_0a_1(1+r)\delta_{r}+(a_1^2r+a_0a_2(1+r^2))\delta_{r^2}+(a_1a_2(r+r^2)+a_0a_3(1+r^3))\delta_{r^3}+(a_2^2r^2+a_1a_3(r+r^3))\delta_{r^4} +a_2a_3(r^2+r^3)\delta_{r^5}+a_3^2r^3\delta_{r^6}.

Also, we set $\nu=\sum\limits_{i=0}^{i=3}b_i\delta_{r^i}$.
for $\nu$ such that $\mu*t\mu=\nu*\nu$. Matching the coefficients of ${\cal U}r_\mu$ elements, we derive the identities

$$ b_0^2 = a_0^2, \;
2b_0b_1 = a_0a_1(1+r), \;
2b_2b_3= a_2a_3(r^2+r^3) \;
\mbox{ and } \;  b_3^2=r^3a_3^2.

From which, we deduce $
b_0 = a_0 , \;  b_1= a_1\frac{(1+r)}{2}, \;
b_2= a_2\frac{(1+r)\sqrt{r}}{2}, \; \mbox{ and }

Plugging in the additional equation

2b_0b_2+b_1^2 =  a_0a_2(1+r^2)+a_1^2r,

we obtain

-  a_0a_2(1+r^2)-a_1^2r.$$

The previous equality holds for arbitrary $r$, as noticed  in the beginning of  the second section,  and hence all  coefficients in the polynomial equation (in $\sqrt{r}$) vanish. In particular, with $r\sqrt{r}$,  we deduce that $a_0a_2=0$. Contradiction

Since every measure with a square root satisfies $ {\tilde W}_\mu$ is  subnormal,  we deduce the next corollary,

Under the notations above, if $p=4$ then $\mu$ has no square root.

Recal that  we are assuming here that $0\not\in supp(\mu)$. The case where $0$ is in the support the result derives from   3 atomic  support measure satisfying proposition <ref>.

§.§ The case   $card(supp(\mu))= 5$.

The main result in this subsection  stated next is form [4]. We give a simpler proof based on uniquely represented elements.
Let $\mu=\sum_{i=1}^5a_i\delta_{\lambda_i}$ a probability measure with $0<\lambda_1<\cdots <\lambda_5$. The following are equivalent

* $\mu*t\mu$ has a square  root

* $\mu$ has geometric support and $\frac{a_2^2}{a_1}  = \frac{a_4^2}{a_5} $ and  $a_3= \frac{a_2^2}{4a_1} +2\sqrt{a_1 a_5}.$

* $\mu$ has a square root.

Proof. Since $\mu*t\mu$ has a square  root, we get
$$9\le card(supp(\mu*t\mu) )\le \left[\frac{(5-1)^2+6}{2}\right] = 11.$$

We first prove that $card(supp(\mu*t\mu) )\ne 11$. Counting non uniquely represented elements, it suffices to show that $\lambda_2\lambda_4 \in {\mathcal NU}r_\mu $. Indeed,  if $\lambda_2^2  = \lambda_1\lambda_4 $ we get $  \lambda_2\lambda_3=\lambda_1\lambda_5 $ and then $\lambda_3\lambda_4= \lambda_2\lambda_5. $ Which forces $\lambda_3^2  = \lambda_2\lambda_4.$  If $\lambda_2^2  = \lambda_1\lambda_3 $, we distinguish two subcases,

* $     \lambda_1\lambda_4= \lambda_2\lambda_3\Rightarrow \lambda_3^2=\lambda_2\lambda_4.$

* $   \lambda_1\lambda_4= \lambda_3^2.$ Then since, $ \lambda_4^2=\lambda_3\lambda_5$, we get $ \lambda_1\lambda_5= \lambda_3\lambda_4$ and hence $\lambda_2\lambda_5\in {\mathcal U}r_\mu $.  Contradiction.

We claim that $card(supp(\mu*t\mu) )=9$ and hence that   $\mu$ has a geometric support. To this aim let us first suppose that $card(supp(\mu*t\mu) )=10$ and show that $\lambda_2^2=\lambda_1\lambda_3 $ and   $\lambda_4^2  = \lambda_3\lambda_5$.
If $\lambda_2^2\ne \lambda_1\lambda_3 $, we get $\lambda_1\lambda_3\in {\mathcal U}r_\mu$ and $\lambda_4^2  = \lambda_3\lambda_5$. Thus

$$\begin{array}{lll} supp(\mu*t\mu)&=&\{\lambda_1^2<\lambda_1\lambda_2<\lambda_1\lambda_3 <\lambda_2^2<\lambda_2\lambda_3<\, \lambda_3^2\, <\lambda_3\lambda_4<\,  \lambda_4^2\, <\lambda_4\lambda_5< \lambda_5^2  \} \\ &=&\{\lambda_1^2<\lambda_1\lambda_2<\lambda_1\lambda_3 <\lambda_2^2< \lambda_2\lambda_3<\lambda_2\lambda_4<\lambda_2\lambda_5<\lambda_3\lambda_5<\lambda_4\lambda_5< \lambda_5^2 \}.
\end{array} $$

We deduce that $ \; \lambda_3^2=\lambda_2\lambda_4,$
from which we get
$$\frac{\lambda_3}{\lambda_2} = \frac{\lambda_4}{\lambda_3} = \frac{\lambda_5}{\lambda_4} (=r), $$

and together with  the  additional equality $\lambda_2^2=\lambda_1\lambda_4$, we obtain
$$ \lambda_2=\lambda_1r^2, \lambda_3=\lambda_1r^3, \lambda_4=\lambda_1r^4 \mbox{ and } \lambda_5=\lambda_1r^5.$$

We put now  $\mu=a_1\delta_{\lambda_1}+ \sum_{i=2}^5 a_i \delta_{\lambda_1 r^i}$, and $\nu=b_1\delta_{\lambda_1}+ \sum_{i=2}^5 b_i \delta_{\lambda_1 r^i}$. Matching the uniquely represented elements in $\mu*t\mu=\nu*\nu$ we get:
$  b_1^2 = a_1^2\lambda_1, \;  2b_1b_2 = a_1a_2\lambda_1(1+r^2), \;
2b_1b_3= a_1a_3{\lambda_1}(1+r^3), \; 2b_4b_5=a_4a_5{\lambda_1}(r^4+r^5)$ and $
b_5^2=a_5^2{\lambda_1}r^5. $

$$ b_1 = a_1\sqrt{\lambda_1}, \; b_2 = a_2\sqrt{\lambda_1}\frac{(1+r^2)}{2}, \;  b_3= a_3\sqrt{\lambda_1}\frac{(1+r^3)}{2}, \:
b_4=a_4\sqrt{\lambda_1}r^{\frac32}\frac{(1+r)}{2}, \;

Using the additional equation

$$ b_3^2+2b_2b_4=\lambda_1a_3^2r^3+\lambda_1a_2a_4(r^2+r^4),$$
we get

and then $a_3= 0$ as the coefficient in $r^6$. Contradiction.

So $\lambda_2^2=\lambda_1\lambda_3$, and similarly $\lambda_4^2=\lambda_3\lambda_5$. We have  $\lambda_1\lambda_5 \in {\cal NU}r_\mu$ and  $\lambda_1\lambda_5 \in \{\lambda_3^2, \lambda_2\lambda_4\}$, (otherwise, either $\lambda_1\lambda_4$, or $\lambda_2\lambda_5$ is uniquely represented, which is not true). Now in both cases, we get $\mu$ has geometric support.

It remains to characterise measures $\mu$ with   geometric support such that  $\mu*t\mu$ has a square  root. We have the following
Under the notations above, suppose $\lambda_1=1$ and let $\mu=\sum_{i=1}^5a_i \delta_{r^i}$. Then $\mu*t\mu$ has a square root if and only if  $\frac{a_2^2}{a_1}  = \frac{a_4^2}{a_5} $ and  $a_3= \frac{a_2^2}{4a_1} +2\sqrt{a_1 a_5}.$

To this aim, suppose $\lambda_i= r^{i-1}$ for $1\le i\le 5$. The identity $ \nu*\nu=\mu*t\mu$ provides 9 non linear equations. We  use  the  4  equations deriving from uniquely represented elements,

$$ b_1^2 = a_1^2,  \;  \;
2b_1b_2 = a_1a_2(1+r),  \; \;
2b_4b_5 = a_4a_5(r^3+r^4), \; \;
\mbox{ and } b_5^2=a_5^2r^4,

to get
$$  b_1 = a_1, \;
b_2 = a_2\frac{(1+r)}{2}, \;
b_4=a_4\frac{r+r^2}{2} \; \text{ and } \;

Comparing the expression of  $b_3$ obtained from  the next two  equations.

\left\{
\begin{array}{lll}
2b_1b_3+ b_2^2 &=& a_1a_3(1+r^2)+a_2^2r\\
2b_3b_5+ b_4^2 &=& a_3a_5(r^2+r^4)+a_4^2r^3\\
\end{array}
\right.

we derive
$$b_3=a_3\frac{(1+r^2)}{2}-\frac{a_2^2}{a_1}\frac{(1-r)^2}{8} =a_3\frac{(1+r^2)}{2}-\frac{a_4^2}{a_5}\frac{(1-r)^2}{8} .$$

\begin{equation}
\end{equation}

Because of the additional compatibility equation

$$ 2b_1b_5+ b_3^2+ 2b_2b_4 = a_1a_5(1+r^4)+a_3^2r^2+ a_2a_4(r+r^3)$$
we obtain

$$a_3= \frac{a_2^2}{4a_1} +2\sqrt{a_1 a_5}.$$

To end the proof of Lemma <ref> and Proposition <ref>, we
use , as in  <cit.>, the measure $\nu =\sqrt{a_1}\delta_{r}+\frac{a_2}{2\sqrt{a_1}}\delta_{r^3}+\sqrt{a_5}\delta_{r^5}$  to provide a square root of $\mu$.

§ THE CASE OF ATOMIC MEASURES WITH  $6$ ATOMS.

In this case, when $ {\tilde W}_\mu$ is  subnormal, we get
$$11\le card(supp(\mu*t\mu) )\le \left[\frac{(6-1)^2+6}{2}\right] = 15.$$

We will see in this section that the above inequality is optimal.

We start with the next lemma which improves Lemma <ref>.
Let $\mu=\sum_{i=1}^6a_i\delta_{\lambda_i}$  as above such that
$\mu*t\mu$ has a square  root. Then $\lambda_2^2\ne \lambda_1\lambda_{5}$ and  $\lambda_{5}^2\ne  \lambda_2\lambda_{6}$.

Because of symmetry, we only show that $\lambda_2^2\ne \lambda_1\lambda_5$. If it is not the case, we will get $r_1=\frac{\lambda_2}{\lambda_1} = \frac{\lambda_5}{\lambda_2} $ and
$\{ \lambda_1^2,\lambda_1\lambda_2,\lambda_1\lambda_3,\lambda_1\lambda_4\}\subset{\cal U}r_\mu$, so by  Proposition <ref> $\{ \lambda_2\lambda_3,\lambda_2\lambda_4\}\subset {\cal NU}r_{\mu} $

We will have  necessarily,

\begin{equation}\label{(2,3)=(1,6)}
\lambda_2\lambda_3=\lambda_1\lambda_6 \mbox{ and } \lambda_3^2=\lambda_2\lambda_4.
\end{equation}

Then   $\frac{\lambda_2}{\lambda_1} =\frac{\lambda_6}{\lambda_3} = \frac{\lambda_5}{\lambda_2}  $ and $\frac{\lambda_3}{\lambda_2}=\frac{\lambda_4}{\lambda_3}$. From Proposition <ref>  again $\{\lambda_3\lambda_6, \lambda_4\lambda_6, \lambda_5^2\} \subset {\cal NU}r_{\mu}. $ Thus
$\lambda_5^2=\lambda_4\lambda_6$ (or equivalently $\frac{\lambda_5}{\lambda_4}=\frac{\lambda_6}{\lambda_5}$). Otherwise $\lambda_5^2=\lambda_3\lambda_6 $ and then $\lambda_4\lambda_6\in {\cal U}r_{\mu}$. Contradiction.

Writing    $\lambda_2\lambda_4\lambda_5=\lambda_3^2\lambda_5=\lambda_2\lambda_6\lambda_3$, we derive that
$ \lambda_5\lambda_4=\lambda_3\lambda_6,  $
which leads to $\frac{\lambda_3}{\lambda_2}=\frac{\lambda_4}{\lambda_3}=\frac{\lambda_5}{\lambda_4}=\frac{\lambda_6}{\lambda_5}=r_2.$ We deduce that  $r_1=\frac{\lambda_6}{\lambda_3} =r_2^3$ and then
$\lambda_k= r^{k+1}\lambda_1$ for $1 <k\le 6$, with   $r=r_2$.
Since $\{ \lambda_1^2,\lambda_1\lambda_2,\lambda_1\lambda_3,\lambda_1\lambda_4\}\subset{\cal U}r_\mu$, we get:

b_1^2=a_1^2\lambda_1,\; \;
2b_1b_2=a_1a_2\lambda_1(1+r^3), \; \;
2b_1b_3=a_1a_3\lambda_1(1+r^4), \;  \mbox{  and  }
2b_1b_4=a_1a_4\lambda_1(1+r^5). $$

That gives $b_1=a_1\sqrt{\lambda_1}, \; b_2=a_2\sqrt{\lambda_1}\frac{1+r^3}{2}, \; b_3=a_3\sqrt{\lambda_1}\frac{1+r^4}{2}$ and $\; b_4=a_4\sqrt{\lambda_1}\frac{1+r^5}{2}.$

Replacing in the additional equation $\, \, b_3^2+2b_2b_4=a_3^2\lambda_1r^4+a_2a_4\lambda_1r^3(1+r),$ we obtain

$$a_3^2\frac{(1+r^4)^2}{4}+a_2a_4\frac{(1+r^3)(1+r^5)}{2}-a_3^2r^4-a_2a_4r^3(1+r)=0 $$

and then $\frac12a_2a_4= 0$ as it is the coefficient of $r^5$.
Let $\mu=\sum_{i=1}^6a_i\delta_{\lambda_i}$  as above such that
$\mu*t\mu$ has a square  root. Then either $\lambda_2^2\ne \lambda_1\lambda_{4}$ or   $\lambda_{5}^2\ne  \lambda_3\lambda_{6}$.

Suppose that   $(\lambda_2^2,\lambda_5^2)= (\lambda_1\lambda_4, \lambda_6\lambda_3)$  (or equivalently
$\frac{\lambda_2}{\lambda_1}= \frac{\lambda_4}{\lambda_2}$ and $\frac{\lambda_6}{\lambda_5}= \frac{\lambda_5}{\lambda_3})$. It follows that  $\{\lambda_1\lambda_3, \lambda_6\lambda_4\}\subset{\cal U}r_\mu$. Using  Proposition <ref> (1.), it follows that  $\lambda_2\lambda_3$ and  $\lambda_4\lambda_5$ are
not uniquely represented. Then $\lambda_2\lambda_3\in \{ \lambda_1\lambda_5,\lambda_1\lambda_6\}$ and $\lambda_4\lambda_5\in \{ \lambda_1\lambda_6,\lambda_2\lambda_6\}$ .
If we assume that  $\lambda_2\lambda_3=\lambda_1\lambda_6$ (or symmetrically  $\lambda_4\lambda_5=\lambda_1\lambda_6$), we will get
$\lambda_1\lambda_5\in{\cal U}r_\mu$, which contradicts Proposition <ref> (5.).
We derive that

$$    \lambda_2\lambda_3=\lambda_1\lambda_5 \mbox{ and } \lambda_4\lambda_5=\lambda_2\lambda_6 $$

and then
$(r_1=r_2^2)=\frac{\lambda_2}{\lambda_1}= \frac{\lambda_4}{\lambda_2}=\frac{\lambda_5}{\lambda_3}= \frac{\lambda_6}{\lambda_5}.$

Since $\{\lambda_1^2,\lambda_1\lambda_3\}\in{\cal U}r_\mu$ and $\{\lambda_6^2,\lambda_4\lambda_6\}\in{\cal U}r_\mu$ we get $\{\lambda_3^2\}\not\in{\cal U}r_\mu$, $\{\lambda_4^2\}\not\in{\cal U}r_\mu$.

We clearly  have  $\frac{\lambda_3}{\lambda_2}= \frac{\lambda_4}{\lambda_3}= \frac{\lambda_5}{\lambda_4}=(r_2)$ and hence    $\lambda_i=r^i\lambda_1$ for $1<i<6$, and $\lambda_6=r^7\lambda_1$.
From the description  above, we obtain:

$$  b_1=a_1\sqrt{\lambda_1}, \;
b_2=\frac12 a_2\sqrt{\lambda_1}(1+r^2), \;
b_6=a_6\sqrt{r\lambda_1}r^3, \:
b_4=\frac12 a_4\sqrt{r\lambda_1}(1+r^3), $$

b^2_2+2b_1b_4= a_2^2 r^2\lambda_1+a_1a_4\lambda_1(1+r^4).

This leads to the next identity,

$$   a_2^2 r^2+a_1a_4(1+r^4)-\frac14 a_2^2(1+r^2)^2-a_1a_4\sqrt{r}(1+r^3)=0.

and  the coefficient  $a_1a_4$ of $\sqrt{r}$ vanishes as before. Contradiction.

It follows  that only one of the 3 next cases may occur $(\lambda_2^2,\lambda_5^2)= (\lambda_1\lambda_3, \lambda_3\lambda_6)$, $(\lambda_2^2,\lambda_5^2)= (\lambda_1\lambda_4, \lambda_4\lambda_6)$ and
$(\lambda_2^2,\lambda_5^2)= (\lambda_1\lambda_3, \lambda_4\lambda_6)$. From  the first two symmetric cases, we have

Let $\mu=\sum_{i=1}^6a_i\delta_{\lambda_i}$ be with $6$ atoms.
I) If $(\lambda_2^2,\lambda_5^2)= (\lambda_1\lambda_4, \lambda_4\lambda_6)$,
then the following are equivalent

* $\mu*t\mu$ has a square  root.

* $\lambda_3^2=\lambda_1\lambda_6$ and
$(a_2^2, a_3^2,a_5^2) =(4a_1a_4, 4a_1a_6,4a_4a_6)$.

* $\mu$ has a square  root.

II) If $(\lambda_2^2,\lambda_5^2)= (\lambda_1\lambda_3, \lambda_3\lambda_6)$,
then the following are equivalent

* $\mu*t\mu$ has a square  root.

* $\lambda_4^2=\lambda_1\lambda_6$ and
$(a_2^2, a_4^2,a_5^2) =(4a_1a_3, 4a_1a_6,4a_3a_6)$.

* $\mu$ has a square  root.

Proof. Because of symmetry, we only show $I)$. Since the implication  $3. \Rightarrow 2.$ is well known, it remains to show  $1. \Rightarrow 2.$  and  $2. \Rightarrow 3.$

$1. \Rightarrow 2.$ From $\lambda_2^2=\lambda_1\lambda_4$, as in the proof of Lemma <ref> we obtain
$   \lambda_2\lambda_3=\lambda_1\lambda_5$. Using $\lambda_5^2= \lambda_4\lambda_6, $ we get $ \lambda_3\lambda_5=\lambda_2\lambda_6$.
$$ \frac{\lambda_2}{\lambda_1}=\frac{\lambda_4}{\lambda_2}= \frac{\lambda_5}{\lambda_3} =R \mbox{ and }  \frac{\lambda_3}{\lambda_2}=\frac{\lambda_5}{\lambda_4}=  \frac{\lambda_6}{\lambda_5}=r

we deduce that,

\begin{array}{ccccc}
\lambda_2=R\lambda_1& \lambda_3=rR\lambda_1 & \lambda_4=R^2\lambda_1
& \lambda_5=rR^2\lambda_1 &  \lambda_6=(rR)^2\lambda_1
\end{array}

If  $\lambda_3\lambda_6 \in {\cal NU}r_\mu $ then  clearly $\lambda_3\lambda_6 \in \{\lambda_2^4,\lambda_4\lambda_5\}$. If $\lambda_3\lambda_6 = \lambda^2_4$, we get $R=r^3$ and if $\lambda_3\lambda_6 = \lambda_4\lambda_5$, we have $R=r^2$ . Hence arises the  next 3 cases

* $R=r^2.$ We obtain  the next spectral picture of $supp(\mu*t\mu)$:

\begin{array}{ccccccccccccc}
& &&&&\lambda_2\lambda_4&&  \lambda_4^2 &\lambda_4\lambda_5&&&\\
& && && \|&& \|&\|&&&\\
\lambda_1^2&<\lambda_1\lambda_2& <\lambda_1\lambda_3&<\lambda_1\lambda_4&<\lambda_2\lambda_3&<\lambda_1\lambda_6 &<\lambda_2\lambda_5&<\lambda_2\lambda_6&<\lambda_3\lambda_6&<\lambda_4\lambda_6&<\lambda_5\lambda_6&<\lambda_6^2 \\
& &&\|&\|&\|&\|&\|&& \|&&\\
& && \lambda_2^2&\lambda_1\lambda_5&\lambda_3^2 &\lambda_3\lambda_4& \lambda_3\lambda_5& &\lambda_5^2&
\end{array}

* $R=r^3.$  We get the next spectral picture of $supp(\mu*t\mu)$:

$$ \begin{array}{cccccccccccccc}
\lambda_1^2&<\lambda_1\lambda_2& <\lambda_1\lambda_3&<\lambda_1\lambda_4&<\lambda_2\lambda_3&<\lambda_1\lambda_6&<\lambda_2\lambda_4 &<\lambda_2\lambda_5&<\lambda_2\lambda_6&<\lambda_4^2&<\lambda_4\lambda_5&<\lambda_4\lambda_6&<\lambda_5\lambda_6&<\lambda_6^2 \\
&&&\|&\|&\|&&\|&\|&\|&& \|&&\\
&&&\lambda_2^2&\lambda_1\lambda_5&\lambda_3^2& &\lambda_3\lambda_4& \lambda_3\lambda_5&\lambda_3\lambda_6& &\lambda_5^2&
\end{array}

* $R\notin \{r^2, r^3\}.$  We have a "Maximal cardinal support" ( 15 atoms) and we get  the next spectral picture of $supp(\mu*t\mu)$:

$$ \hspace{-2cm}\begin{array}{ccccccccccccccc}
\lambda_1^2&<\lambda_1\lambda_2& <\lambda_1\lambda_3&<\lambda_1\lambda_4&<\lambda_2\lambda_3&<\lambda_2\lambda_4&<\lambda_1\lambda_6 &<\lambda_2\lambda_5&<\lambda_4^2&<\lambda_2\lambda_6&<\lambda_4\lambda_5&<\lambda_3\lambda_6&<\lambda_4\lambda_6&<\lambda_5\lambda_6&<\lambda_6^2 \\
&&&\|&\|&&\|&\|&&\|&&& \|&&\\
&&&\lambda_2^2&\lambda_1\lambda_5&&\lambda_3^2 &\lambda_3\lambda_4&& \lambda_3\lambda_5&& &\lambda_5^2&
\end{array}
From the associated diagrams, in the  3 cases above, We use the next common   equations to determine, $b_1, b_2, b_3,b_4, b_5$ and $  b_6,$

\left\{
\begin{array}{lll}
b_1^2 = a_1^2\lambda_1  & 2b_1b_2 = a_1a_2\lambda_1(1+R)\\
2b_1b_3 = a_1a_3{\lambda_1}(1+rR)&
b_2^2+2b_1b_4 =a_2^2 \lambda_1R+ a_1a_4{\lambda_1}(1+R^2)\\
2b_5b_6 = a_5a_6\lambda_1rR^2(1+r) &
\end{array}
\right.

so that
\left\{
\begin{array}{ccc}
b_1 = a_1\sqrt{\lambda_1},  &  b_2 = a_2\sqrt{\lambda_1}\frac{(1+R)}{2}, & b_3 = a_3\sqrt{\lambda_1}\frac{(1+rR)}{2}\\
b_4=\frac{\sqrt{\lambda_1}}{2}\left[a_4(1+R^2)-\frac{a_2^2}{a_1}\frac{(1-R)^2}{4}\right] & b_5=a_5\sqrt{\lambda_1}\frac{R(1+r)}{2} &b_6=a_6\sqrt{\lambda_1}rR
\end{array}
\right.

Plugging in the additional 3 common equations,

\left\{
\begin{array}{lll}
2b_1b_5+ 2b_2b_3 &=& a_1a_5{\lambda_1}(1+rR^2)+ a_2a_3{\lambda_1}R(1+r)\\
2b_2b_5+ 2b_3b_4 &=& a_2a_5{\lambda_1}R(1+rR)+a_3a_4{\lambda_1}R(r+R)\\
2b_4b_6+ b_5^2 &=& a_4a_6{\lambda_1}R^2(r^2+1)+a_5^2{\lambda_1}rR^2, \\
\end{array}
\right.

we obtain the next necessary compatibility conditions:

$$ (2a_1a_5 -a_2a_3)(1-R)(1-rR) = 0,$$

$$ (2a_3a_4 -a_2a_5)(1-R^2)(1-rR) = 0,$$

and hence

$$  2a_5a_1=a_2a_3, \; \;
2a_3a_4=a_2a_5 \; \mbox{ and } a_5^2=4a_4a_6$$

It is clear that the  tree obtained relations  are equivalent to $a_2^2=4a_1a_4, \, a_3^2=4a_1a_6$ and $a_5^2=4a_4a_6$.

$2. \Rightarrow 3.$  Direct computations show that the 3 atomic measure
is  a square root of  $\mu$.
For $p=6$, in the previous situations, the number of equations arising from the diagrams are either 12, 14 or 15. We only used the 9 common ones. As shown by Proposition <ref>, the remaining   compatibility   equations are automatically fulfilled.

Let $\mu=\sum_{i=1}^6a_i\delta_{\lambda_i}$  be  such that $(\lambda_2^2,\lambda_5^2)= (\lambda_1\lambda_3, \lambda_4\lambda_6)$. Then ${\tilde W}_\mu$ is not subnormal.

Let us start with the geometric case. Identifying the coefficients associated with $\nu*\nu$ and $\mu*t\mu$, we obtain a system of 11 nonlinear equations. We use the next six one to get expression of $b_i, i \le 6$.

b_1^2- a_1^2&=0\\
(2b_1b_3+b_2^2)- (a_1a_3(1+r^2)+a_2^2r)&=0 \\
(2b_4b_6+b_5^2)- (a_4a_6(r^3+r^5)+a_5^2r^4)&=0\\
+2b_5b_6 -a_5a_6(r^4+r^5)&=0\\
b_6^2 -a_6^2r^5 &=0.
\end{array}$$

We have,

\begin{array}{ccc}
b_1= a_1, & b_2= \frac{1}{2}a_2(1+r),& b_3=  (\frac{a_3}{2}(1+r^2)-\frac{a_2^2}{8a_1}(1-r)^2)\\
b_6= a_6r^2\sqrt{r},& b_5= \frac{1}{2}a_5(1+r)r\sqrt{r}, & b_4= ( \frac{a_4}{2}(1+r^2) -\frac{a_5^2}{8a_6}(1-r)^2)\sqrt{r}).
\end{array}

Plugging in the $ 7^{th} $  equation,  $(2b_1b_4+2b_2b_3)-(a_1a_4(1+r^3)+a_2a_3(r+r^2))=0$, we derive

a_1a_4[(1+r^2)\sqrt{r}-1-r^3] - \frac{a_5^2a_1}{4a_6}(1-r)^2\sqrt{r}+ \frac{1}{2}a_2a_3(1+r)(1-r)^2 -\frac{a_2^3}{8a_1}(1+r)(1-r)^2=0.
$ And because of the coefficient of $r\sqrt{r}$ that should vanish, we obtain $\frac{a_5^2a_1}{4a_6}=0$. Contradiction.

Suppose now that $supp(\mu)$ is not geometric and write  $\frac{\lambda_2}{\lambda_1} = \frac{\lambda_3}{\lambda_2}=r_1$ and
$ \frac{\lambda_5}{\lambda_4} = \frac{\lambda_6}{\lambda_5} =r_2$.

Aiming to describe the spectral picture, we distinguish 3 cases

$ a)$ $\{\lambda_1\lambda_4,\lambda_3\lambda_6\} \subset \mathcal{NU}r_\mu$

$ b)$ $\lambda_1\lambda_4 \in \mathcal{U}r_\mu $ and $\lambda_3\lambda_6 \notin \mathcal{U}r_\mu$ (or by symmetry $\lambda_1\lambda_4 \notin \mathcal{U}r_\mu $ and $\lambda_3\lambda_6 \in \mathcal{U}r_\mu$.

$ c)$ $\{\lambda_1\lambda_4,\lambda_3\lambda_6\} \subset \mathcal{U}r_\mu$.

$ a)$  If  $\{\lambda_1\lambda_4,\lambda_3\lambda_6\} \subset \mathcal{NU}r_\mu$, we get the
additional  information $\lambda_1\lambda_4\in \{\lambda_2\lambda_3, \lambda_3^2\}$ and $\lambda_3\lambda_6 \in \{\lambda_4^2,\lambda_4\lambda_5\} $. Thus

* $(\lambda_1\lambda_4,\lambda_3\lambda_6 )= (\lambda_2\lambda_3,\lambda_4\lambda_5 )$. This implies
$\frac{\lambda_2}{\lambda_1} = \frac{\lambda_3}{\lambda_2}=\frac{\lambda_4}{\lambda_3}=\frac{\lambda_5}{\lambda_4} = \frac{\lambda_6}{\lambda_5}$  and then $\mu$ has a geometric support. Contradiction.

* $(\lambda_1\lambda_4,\lambda_3\lambda_6 )= (\lambda_2\lambda_3,\lambda_4^2 )$, ( or symmetrically $(\lambda_1\lambda_4,\lambda_3\lambda_6 )= (\lambda_3^2,\lambda_4\lambda_5 )$).

It will follow that $\frac{\lambda_2}{\lambda_1} =\frac{\lambda_4}{\lambda_3} =\frac{\lambda_6}{\lambda_4}$ and then $r_1=r_2^2(=r^2).$  This gives $
\lambda_2=\lambda_1r^2 \; \lambda_3=\lambda_1r^4 \; \lambda_4=\lambda_1r^6 \;$   $\lambda_5=\lambda_1r^7\; \lambda_6=\lambda_1r^8. $
From the above expressions it  comes that
$\lambda_1\lambda_{5} \in  {\cal U}r_\mu$, which contradicts Proposition <ref>.

* $(\lambda_1\lambda_4,\lambda_3\lambda_6 )= (\lambda_3^2,\lambda_4^2)$. We get $r_1^2=\frac{\lambda_3}{\lambda_1} =\frac{\lambda_4}{\lambda_3} =\frac{\lambda_6}{\lambda_4} =r_2^2$ and then $r_1=r_2 (=r)$. We deduce that $   \lambda_2=r\lambda_1 \; \lambda_3=r^2\lambda_1 \; \lambda_4=r^4\lambda_1 \;   \lambda_5=r^5\lambda_1\; \lambda_6=r^6\lambda_1. $

This  leads to the equations:

\left\{
\begin{array}{lllllll}
b_1^2 &=& a_1^2\lambda_1, & b_6^2 &=& a_6^2{\lambda_1}r^6, \\
2b_1b_2 &=& a_1a_2\lambda_1(1+r), &  2b_5b_6 &=& a_5a_6\lambda_1r^5(1+r),\\
2b_2b_3 &=&a_2a_3{\lambda_1}(r+r^2), &  2b_4b_5 &=& a_4a_5{\lambda_1}r^4(1+r).
\end{array}
\right.

Hence    $b_1, b_2, b_3,b_4,b_5$ and $b_6$ are given by
b_1 = a_1\sqrt{\lambda_1}, \;   b_2 = a_2\sqrt{\lambda_1}\frac{(1+r)}{2}, \; b_3 = a_3\sqrt{\lambda_1}r, \;
b_4 = a_4 \sqrt{\lambda_1}r^2,  \; b_5 =a_5 \sqrt{\lambda_1}r^2\frac{(1+r)}{2} $ and $ b_6=a_6 \sqrt{\lambda_1} r^3.$

Replacing in the additional  equation

$$   2b_1b_4+ b_3^2 = a_1a_4{\lambda_1}(r^4+1)+a_3^2{\lambda_1}r^2, $$
we get :

a_1a_4(1-r^2)^2=0.     $$
Which is impossible, and then $ {\tilde W}_\mu$  is not subnormal.

$b)$  $\lambda_1\lambda_4 \in \mathcal{U}r_\mu $ and $\lambda_3\lambda_6 \notin \mathcal{U}r_\mu$. In this case, we have $\lambda_3\lambda_6\in \{ \lambda_4\lambda_5, \lambda_4^2\}$ and $\lambda_2\lambda_4\in{\cal NU}r_\mu$. Moreover $\lambda_2\lambda_4\in \{\lambda_1\lambda_5,\lambda_3^2,\lambda_1\lambda_6\}$.

* $\lambda_3\lambda_6 = \lambda_4\lambda_5$, gives $\frac{\lambda_2}{\lambda_1} =\frac{\lambda_3}{\lambda_2}$ and $ \frac{\lambda_4}{\lambda_3}=\frac{\lambda_5}{\lambda_4}=\frac{\lambda_6}{\lambda_5}$ and if moreover  $\lambda_2\lambda_4\in \{\lambda_1\lambda_5,\lambda_3^2\}$, we obtain  $\frac{\lambda_2}{\lambda_1}=\frac{\lambda_3}{\lambda_2} =\frac{\lambda_4}{\lambda_3} =\frac{\lambda_5}{\lambda_4}=\frac{\lambda_6}{\lambda_5}$ that contradicts $\lambda_1\lambda_4 \in \mathcal{U}r_\mu.$ It follows that $\lambda_2\lambda_4=\lambda_1\lambda_6$ and  then $r_1=r_2^2.$ This gives  $
\lambda_2=r^2\lambda_1 \; \lambda_3=r^4\lambda_1 \; \lambda_4=r^5\lambda_1 \;   \lambda_5=r^6\lambda_1\; \lambda_6=r^7\lambda_1. $ It results that

\left\{
\begin{array}{l}
b_1^2 = a_1^2\lambda_1,  \quad \quad \quad\quad\quad \quad \quad b_6^2=a_6^2{\lambda_1}r^7\\
2b_1b_4= a_1a_4\lambda_1(1+r^5),  \quad 2b_5b_6= a_5a_6r^6\lambda_1(1+r).\\
\end{array}
\right.

\begin{array}{cccc}
b_1 = a_1\sqrt{\lambda_1},& b_4 =a_4 \sqrt{\lambda_1} \frac{(1+r^5)}{2}, &    b_5 = a_5\sqrt{\lambda_1}r^{\frac52}\frac{(1+r)}{2}, & b_6=a_6 \sqrt{\lambda_1} r^{\frac72}
\end{array}

Replacing in the additional equation:
$2b_4b_6 + b_5^2=a_4a_6{\lambda_1}r^5(1+r^2)+a_5^2\lambda_1r^6 $, we find

a_5^2 r^{\frac32}\frac{(1-r)^2}{4}+a_4a_6(r^{\frac32}-1)(r^{\frac72}-1)=0.

Which is again impossible and hence  $ {\tilde W}_\mu$ is not subnormal.

* $\lambda_3\lambda_6 = \lambda_4^2.$ As before  $\lambda_2\lambda_4\ne \lambda_3^2$. Here we have two subcases:

$i)$  $\lambda_2\lambda_4 = \lambda_1\lambda_5$.  Which gives
$ \lambda_2=r\lambda_1 \; \lambda_3=r^2\lambda_1 \; \lambda_4=r^4\lambda_1 \;   \lambda_5=r^5\lambda_1\; \lambda_6=r^6\lambda_1. $ Thus $\lambda_1\lambda_4=\lambda_3^2$, contradiction.

$ii)$   $\lambda_2\lambda_4= \lambda_1\lambda_6$, gives   $\frac{\lambda_6}{\lambda_4}=\frac{\lambda_2}{\lambda_1}$ and we have
$\frac{\lambda_4}{\lambda_3} =\frac{\lambda_6}{\lambda_4}$, we obtain $\frac{\lambda_2}{\lambda_1}=\frac{\lambda_4}{\lambda_3}$  that contradicts $\lambda_1\lambda_4 \in \mathcal{U}r_\mu $.

$ c)$ $\{\lambda_1\lambda_4,\lambda_3\lambda_6\} \subset \mathcal{U}r_\mu$. We deduce that
$\{\lambda_2\lambda_4,\lambda_3\lambda_5\} \subset \mathcal{NU}r_\mu$.
Since $\lambda_2\lambda_4=\lambda_3^2$ will imply $\lambda_1\lambda_4=\lambda_2\lambda_3$, we have
$\lambda_2\lambda_4\ne\lambda_3^2$ and similarly $\lambda_3\lambda_5\ne\lambda_4^2$. Thus $\lambda_2\lambda_4\in \{\lambda_1\lambda_5, \lambda_1\lambda_6\}$ and $\lambda_3\lambda_6\in \{\lambda_2\lambda_6, \lambda_1\lambda_6\}$.

$i)$  If we suppose now that $\lambda_2\lambda_4=\lambda_1\lambda_5$ and $\lambda_3\lambda_5= \lambda_2\lambda_6$, we will have $r_1=r_2$ and hence
$\frac{\lambda_2}{\lambda_1} =\frac{\lambda_3}{\lambda_2}=\frac{\lambda_5}{\lambda_4}=\frac{\lambda_6}{\lambda_5}(=r)$. We deduce that
$\lambda_2=r\lambda_1,\lambda_3=r^2\lambda_1,  \lambda_5=r\lambda_4$ and $\lambda_6=r^2\lambda_4$.
If we  show that $\lambda_4^2 \in{\cal U}r_\mu$, we will get a contradiction since $\{\lambda_1^2, \lambda_1\lambda_4\} \subset  {\cal U}r_\mu$.

Suppose $\lambda_4^2=\lambda_i\lambda_j$ for $i\le 3$ and $j\ge 5$. We will have $\lambda_4^2=r^{i-1}r^{j-4}\lambda_1\lambda_4$ and thus
$\lambda_4=r^{i+j-5}\lambda_1$. Now since $i+j-5 \le 3$ and $\lambda_4 \notin\{\lambda_2, \lambda_3\}$, it follows that  $\lambda_4= \lambda_1r^3$ and then
$ \lambda_1\lambda_4=\lambda_2\lambda_2$. Impossible

$ii)$ $(\lambda_2\lambda_4, \lambda_3\lambda_5) =(\lambda_1\lambda_6, \lambda_1\lambda_6), $ that gives $\lambda_2\lambda_4= \lambda_3\lambda_5$ and hence $\lambda_1\lambda_4= \lambda_2\lambda_5$, impossible.

$iii)$ $(\lambda_2\lambda_4, \lambda_3\lambda_5) =(\lambda_1\lambda_5, \lambda_1\lambda_6)$. We will obtain $r_1=r_2$ and $r_2=r_1^2$. Contradiction again.

$(iv)$ $(\lambda_2\lambda_4, \lambda_3\lambda_5) =(\lambda_1\lambda_6\lambda_2\lambda_6)$. Is the symmetric case of $(iii)$, which run in the same manner.

§ CONCLUDING REMARKS

As shown in the case $p=6$, the number of atoms of $\mu*t\mu$ in the case where $\mu$ admits a square root may be $12, 14$ or $15$. It is intriguing to determine the set of integers $q$ such that there exist $\mu$ for which $\mu*t\mu$ admits a square root and such that $card(\mu*t\mu)= q$.

It is not difficult to see that the method used in this paper provides an algorithm to solve the problem for arbitrary number of atoms. If is not too long to treat this problem for small number of atoms (say for $p\le 10$), it is more complicated to run in all computations for big numbers of atoms.

Recall that  a weighted shift is said to be  recursively generated   if its moment sequence $(\gamma_n)_{n\ge 0}$satisfies

$$\gamma_{n+r}= a_{r-1}\gamma_{n+r-1}+ a_{r-2}\gamma_{n+r-2}+\cdots +a_1\gamma_{n+1}+a_{0}\gamma_{n},$$

where $a_1, \cdots, a_{r-1}$ are given numbers and $r\ge 1$ is an integer.
It is known that a subnormal weighted shift is recursively generated if and only if the associated Berger measure in finitely atomic.
In connection with the investigations of this paper, we mention the next partial version of a question  stated in [5] as question 4.2.

"We do not know of a recursively generated subnormal weighted shift (with
the trivial exceptions of scalar multiples of the unweighted shift)  for which any of the Aluthge transform,
or root shifts are subnormal."

The approach used  of this paper provides a promising way to solve this problem. For recursively generated shift of order $ r\le 6$, a complete description is given.
[1]  A. Aluthge,   On $p$-polynomial operators for ${0<p<1}$,
Integral Equations Operator Theory 13 (1990), 307-315.
[2] R. Ben Taher, M. Rachidi and H. Zerouali,   On the Aluthge transforms of weighted shifts with moments of Fibonacci type. Application to subnormality,  Integral Equations and Operator Theory  $\mathbf{82}$  (2015), 287-299.
[3]   R. Curto and J. Exner Berger Measure for Some Transformations
of Subnormal Weighted Shifts,  Integr. Equ. Oper. Theory 84,  (2016) 429–450.
[4] R. Curto, J. Kim and J. Yoon,  The Aluthge transform of unilateral weighted shifts and the Square Root Problem for finitely atomic measures, Math. Nachr. 292 (2019) 2352–2368.[5] G.R. Exner, Aluthge transforms and $n-$contractivity of weighted shifts, J. Operator Theory 61 (2009)  419–438.
[6] G.R. Exner, J.Y. Jin, I.B. Jung and M.R. Lee,   Weighted shifts induced by Hamburger moment sequences ,
J. Math. Anal. Appl. 427 (2015) 581–599.
[7] S. H. Lee, W.  Y.  Lee and J. Yoon
Subnormality of Aluthge transforms of weighted shifts, Integral Equations and Operator Theory $\mathbf{72}$ (2012),