# An indirect method for solving combinatorial problems

M. Shevtsova  Margarita Shevtsova: Department of Discrete Mathematics, Moscow
institute of physics and technology, Dolgoprudny 141700, Russia.
<EMAIL_ADDRESS>,  A. Kanel-Belov  Alexei Kanel-Belov: Department of
Discrete Mathematics, Moscow institute of physics and technology, Dolgoprudny
141700, Russia<EMAIL_ADDRESS>and  M. Golafshan  Mehdi Golafshan:
Department of Discrete Mathematics, Moscow institute of physics and
technology, Dolgoprudny 141700, Russia<EMAIL_ADDRESS>
###### Abstract.

This article introduces a pedagogical method for solving combinatorial
problems that frequently involve structures that are unfamiliar or less
familiar. Indeed, an indirect method has been proposed in order to evade any
possible questions that might obstruct our path to the correct answer and
final solution. With this method, the desired configuration can be formulated.
This method is effective for solving problems in Olympiads and mathematical
competitions as well as for teaching combinations and counting science in the
elementary school mathematics curriculum.

Keywords: Problem solving in combinatorics$\cdot$ Mathematical
olympiads$\cdot$ Combinatorial pedagogy$\cdot$ Teaching combinatorics$\cdot$
Combinatorial thinking stages

MSC 2020: 97D30; 97D50; 97K10; 97K20

##  1\. Introduction

Combinatorics is a significant subfield of discrete mathematics, which is
defined as “a vibrant branch of contemporary mathematics that is widely
applied in business and industry” (National Council of Teachers of Mathematics
[NCTM], 2000, p. 31). Combinatorics provides tools for dealing with both
everyday life and professional practice, and it is connected to numerous
branches of mathematics and other disciplines (for example, computer science,
communication, genetics, and statistics). Thus, it is critical to integrate
combinatorics into the mathematics curriculum beginning in the early
elementary grades and continuing through senior high school (English, 1993;
NCTM, 2000).

Although, combinatorial problems from mathematical olympiads require a high
level of wit and creativity to solve. There are difficult problems that can be
solved almost entirely by a single brilliant idea. In order to approach these
problems with confidence, it is necessary to have previously dealt with
problems of similar difficulty. A large number of combinatorial problems have
a quick and/or simple solution. However, this does not mean the problem at
hand is not difficult. The difficulty of a problem in combinatorics is
frequently due to the fact that the solution is well “hidden”. In recent
years, good and coherent books have been written in this field, among which we
can mention [1]. Moreover, articles such as [2] have been written on this
subject.

In addition, solving combinatorial problems is related to the nature of
combinatorial thinking. It is discussed in greater detail in article [3].

###  1.1.  Purpose of the study

Combinations, or discrete mathematics in general, are a type of science,
though some of the problems they solve have ancient mathematical roots.
Problem solving and teaching are both critical components of learning this
branch of mathematics. We attempt to introduce a method that, to some extent,
lifts the veil of secrecy surrounding combinatorial problems.

Therefore, this research has two general and serious goals:

  1. 1)

Developing and implementing a pedagogical and methodical approach to teaching
combinatorial concepts and solving combinatorial problems for teachers;

  2. 2)

Development of methods for solving combinatorial problems in Olympiads.

###  1.2. Mathematical backgrounds

Here, establishes the terminology and concepts that will be used throughout
the paper. Though the majority of you probably already know it, yet their
summary is not without charm.

Notations: We are solely concerned with finite objects. We employ the
conventional set-theoretic notation. Let $A,B$ are two finite sets. $|A|$
denotes the size (the cardinality) of a set $A$ Moreover, $A\times
B=\\{(a,b):a\in A,b\in B\\}$ is Cartesian product of $A$ times $B$. In set-
builder notation, $A\cup B=\\{x:x\in A\text{ or }x\in B\\}$.

A permutation of $A$ is a bijection $f:A\to A$.

Terminologies: We will discuss briefly some concepts in enumerative
combinatorics.

Let $A,B$ be finite sets that are pairwise disjoint. Then

  * •

Addition principle: $|A\cup B|=|A|+|B|.$

  * •

Product principle: $|A\times B|=|A|\cdot|B|.$

We propose a slightly more complicated definition. Then, the number of
permutations of $n$ objects with $n_{1}$ identical objects of type $1$,
$n_{2}$ identical objects of type $2$, . . . , and $n_{k}$ identical objects
of type $k$.

  * •

Permutations with repetition: $\frac{n!}{n_{1}!n_{2}!\cdots n_{k}!}.$

Refer to references [4] or [5] for additional information and study.

###  1.3. Methodology

We speculated as mathematics teachers and researchers about a claim that
learning combinatorial concepts necessitates a unique way of thinking, and a
review of the related literature revealed that some researchers acknowledged
this speculation and labeled it combinatorial thinking. Investigating certain
situations, systematically producing all cases, converting the problem into
another combinatorial problem, and understanding combinatorial reasoning were
the four levels of understanding combinatorial thinking we utilized.

The assessment of relevant literature in the combinatorial problem convinced
us that much more research is needed to illuminate the teaching and learning
of combinatorics as an engaging aspect of both school and university
mathematics curricula.

The authors drew on their previous teaching experiences to conduct their
research. It is a kind of reverse engineering that we are trying to do here.
We ask the following general question at each step of solving the relevant
problem:

What is the source of my blunder and my confusion in solving this problem?

###  1.4. Strategy

When we are faced with a problem involving combinations that are not the same
as the previous ones we have solved, we employ this strategy and ask the this
question. At the end, in each step to prevent mistakes, we will try to design
a problem for the answer and see if the answer corresponds to the main problem
one by one or not.

To improve our proficiency with this approach and to make our discussion more
coherent, we begin by examining two problems in this manner. Then, we apply
the same strategy to two additional problems and summarize our findings in the
discussion section.

##  2\. Finding

We start by gradually expressing the desired approach through a few examples,
and then we look at each one separately.

Sample 1. Look at the points in Figure 1\. How many squares can be formed by
connecting points with lines that are parallel to the square’s sides? (With
the exception of the outer square.)

Figure 1. Set of integer points

Our main goal is not to solve this problem and find an answer; rather, we want
to establish a methodical and purposeful approach that we can apply to other
problems.

Step i) This can be a kind of wording problem. In fact, we only have one
shape, and we are not sure what to count as part of it. What we do know is
counting the number of squares. So, the first question that may cause us to
become perplexed is: 111The word configuration is used to describe a finite
collection $p_{1}\ldots p_{n}$ of points in $\mathbb{R}^{2}$ and some lines.

What is the configuration of the squares in the figure?

To answer this question, one must act as if it were a wording problem,
establishing a logical link between the problem form and a mathematical
concept embedded in the figure.

Roughly speaking, a square is made up of four equal line segments with the
facing segments parallel to each other, as we know from basic geometry (of
course, considering the inside, which we did not mention here because it has
no effect on the answer to the problem). So regardless of the size of the
squares, we are looking for something like Figure 2:

Figure 2. Square in different geometric structure

Here are three different colored shapes in purple, red, and green (And maybe
black points!). Each of these is considered a square in different categories,
yet what is meant is a red square (why?) 222The discussion of the purple
square goes back to concepts of Integer Geometry that are beyond the scope of
basic mathematics. You can look at [6] for further details.. Another simple,
short, and crucial thing is required in order to more precisely determine your
desired configuration. We must once again determine what caused our mistake
and respond accordingly.

Step ii) This section could have been a subsection of the previous section,
but because it is so important, we will look at it separately. What we will do
now is specify the configuration type. However, this necessitates a thorough
examination of the problem statement. This is what confuses us:

Which of the following squares should be counted according to the problem (red
or green or both)?

Let us go over the problem once more, and then bold the section where the
distinction between green and red squares is made.

The points marked in the following Cartesian coordinate system are integer.

How many squares with sides that are parallel to the coordinate axes and
vertices that are these points can be built?

Therefore, due to the bold part, we only have to consider the red squares. But
from a pedagogical point of view, it is good to ask the question, what would
the problem be if we were to count the green squares as well? Ask students to
discuss this with each other in class. Students can learn and consolidate what
they have learned by creating a problem based on various data and answers. But
the problem that can be raised here is a dual problem:

Sample $1^{\prime}$. Look at the points in Figure 3\. How many squares with
vertices that are these points can be built? ( To see the solution to this
problem, you can refer to [7].)

Figure 3. Set of integer points

So far, we have discovered that regardless of the square size, we are looking
for the configuration in Figure 4.

Figure 4. The configuration structure that is considered.

Now that we have a clear understanding of the configuration structure,
regardless of the size of the squares, we can move on to this.

Step iii) In this section, we will review the allowed sizes of squares. We now
have a very large setup of objects and conditions. So, in order to make
counting easier, we must break it down into smaller structures. These smaller
structures are squares and can come in a variety of sizes. However, we are
once again perplexed by the following question:

What is the allowable size for squares?

The simplest criterion that determines the size of a square is its side. As a
result, we must respond to the following question: What are the allowed
segments that can be sides of this square? If we call each row and each column
a line, then we can say that the points on all the lines are equal. As a
result, the type of square is determined by the line segment that is the
square’s floor side( See Figure 5).

Figure 5. Different sizes of squares in the configuration

According to the above and the figure, we find that the minimum distance
between two different horizontal points that can be the sides of a square is
equal to $1$, and the maximum distance between two horizontal points that can
be the sides of a square is equal to $4$. The desired configuration is now
fully revealed to us after three steps.

In the final discussion, fourth, we count the final number of squares.

Step iv) First, for the convenience of work, we show a square with sides
$1,2,3$ and $4$ with the symbols of $A_{1},A_{2},A_{3}$ and $A_{4}$,
respectively. So the question in mathematical form is as follows: Find the
$|A_{i}|$ for all $i=1,2,3,4$. What is the fundamental and ambiguous question
that arises for each of these sets? After counting, we ask ourselves whether
we have counted too much or too little? Therefore, the following puzzling
question arises for us:

How do we make sure that it is true?

If we are not experts at solving combinatorial problems and are unfamiliar
with the process, we may count each square in $A_{i}$ individually. As an
example, let’s say we did this for $A_{2}$. In this process, the number of
squares of size $2$ may be more or less than counted. But how do we know if
this number is less than or greater than the correct number? Attempting to
create a problem for the answer we provide is a good guideline. For example,
if you found $8$ squares of size $2$, ask the question: If we had $8$ squares
of size $2$, would this be our configuration? We will find a one-to-one
correspondence between the answer and the desired problem if we proceed in
this manner. However, for the sake of convenience and to avoid mistakes, we
can ask the following questions:

How do you begin counting and how do you count according to specific rules?

To do this, we do the following: to count the number of elements $A_{2}$: In
the figure, consider virtual horizontal rails at a distance of two units ( See
Figure 6).

Figure 6. Structures of squares with a side of two units

According to the figures from left to right, if we move the rails one unit
down, we have $3$ pairs of these virtual rails. The number of squares between
the blue, green, and brown rails is indicated by the symbols $A_{21},A_{22}$
and $A_{23}$, respectively. Hence, $|A_{2}|=|A_{21}|+|A_{22}|+|A_{23}|.$ That
is, we use the principle of addition. Due to the symmetry of the shape in
terms of the number of points as well as the rails, we count only the number
of squares created by the blue rail ( See Figure 7).

Figure 7. Counting squares with sides of two units in the first structure

According to the figure, $|A_{21}|=3$, and so on $|A_{22}|=|A_{23}|$. Then
$|A_{2}|=9$. If we repeat the process for virtual rails at distances of $1$,
$3$, and $4$, we get the following results. The number of pairs of rails at a
distance of one is $4$, and between each of these rails we have $4$ squares on
one side; So according to the principle of multiplication we have:
$|A_{1}|=4\cdot 4$. Also, $|A_{3}|=2\cdot 2$, and $|A_{4}|=1\cdot 1$.

Finally, based on the principle of addition, the total number of squares
equals: $|A|=|A_{1}|+|A_{2}|+|A_{3}|+|A_{4}|$ i.e., $|A|=30$.

To help you better understand and practice this method, we will look at
another combinatorial problem together. Naturally, we chose a less well-known
problem this time. As a result, the following problem will be examined in this
manner:

Sample 2. What is the number of all possible ways to read “Open!” in Figure 8,
according to these conditions:

  1. a)

the square with next letter must share a side with the previous square;

  2. b)

the sequence of letters in the read squares must make up the word “Open!”.

Figure 8. Word game table

We attempt to replicate the method used to solve the previous problem and
reach a general conclusion about this one in particular.

Step i) This is an intriguing problem that also serves as an excellent
educational example. While the symmetrical structure of this configuration may
be perplexing at first, it ultimately aids in our solution. In this problem,
we have been asked to come up with alternative ways to read the word “Open!”
in the hypothetical form given the above conditions. The first step is to
establish a meaningful and palatable link between the form and the problem. As
a result, the first question that may cause us to become perplexed is:

What is the configuration of the word “Open!” in the figure?

We first examine the word structure, then do the same for the shape, and
finally make a connection between them.

This word contains five symbols, and more importantly, it begins with the “O”
symbol and concludes with the “!” symbol (See Figure  9).

Figure 9. Main configuration

A circle indicates the location of the symbols. For the time being, we can say
that this structure exists independently of the figure’s main configuration,
and that the general structure reveals its nature when it interacts with the
shape.

In Figure 10, we attempt to specify one or more of the possible readings of
the word “Open!” by specifying the initial and final letters:

Figure 10. Types of configuration

Thus far, the word has been configured in part.

Step ii) What may cause confusion here (though hopefully not too much!) is
whether this is the correct configuration in the form that corresponds to the
word in question. Thus, the issue that concerns us here is as follows:

Which cells can be configured in the desired manner and which cells cannot?

As previously stated, it is critical to pay close attention to the problem’s
assumptions and conditions. The cells whose adjacent symbols should be
adjacent are mentioned in this case. What may cause students to become
confused in this instance is the term “adjacent.” Adjacent here, the existing
cells of two consecutive symbols must have a common side. That is, structures
as follows are not allowed ( Figure 11).

Figure 11. Unacceptable configurations

Solicit from students the creation of a problem for this structure that is
unrelated to our configuration. One of the problems that can be raised is the
following:

Sample $2^{\prime}$. What is the number of all possible ways to read “Open!”
in the picture such that, the sequence of letters in the read squares must
make up the word “Open!”.

The configuration has been precisely defined. That is, it is critical that the
signs are adjacent conditionally and have no bearing on their proximity from
the left, top, or right. That is, the word can be read both left to right and
right to left.

Step iii) We organize the authorized structures in this section according to
the legal system in order to familiarize ourselves with all the configuration
structures. Thus, the question that prompts our skepticism in this instance is
as follows:

What general structure should these cells have in order to represent this
word?

If we pay attention to the following figure, there are only “!” symbols in the
four corners, and we have only one “O” in the center. So the center square
must be present in our structures (See Figure 12).

Figure 12. Common configuration features

This means that the olive cell must appear as the first symbol in each of the
configurations, and one of the four orange cells must also appear in our
structure. Now, let us count.

Step iv) However, this question perplexes us:

How and where to begin counting?

We begin by selecting “O”. We fix it because it is required in all structures.
Additionally, the selection of “!”’s is straightforward. We will use one of
them as an illustration. There are four ways we can do this.

If we number the “!”’s in this manner “$\text{!}_{1}$”, “$\text{!}_{2}$”,
“$\text{!}_{3}$”, and “$\text{!}_{4}$”, the $A_{k}$ collection is a set of
“Open!” word configurations in which “$\text{!}_{k}$” is the final symbol. So
if we show the total number of allowed words with $|A|$, according to addition
principle we have

$|A|=|A_{1}|+|A_{2}|+|A_{3}|+|A_{4}|.$

On the other hand, we consider only one of the structures due to the symmetry
of the symbol distribution. They are denoted by a dash in Figure 13.

Figure 13. Components of configurations

As a result of the multiplication and symmetry principles, we have:
$|A|=4\cdot|A_{1}|.$ As a result, we only need to count one of them. For
instance, you could do this for $A_{1}$. However, how is $|A_{1}|$ to be
calculated? According to the figure, in order to get from “O” to “!” using the
desired structure, we must ascend two times and turn right two times. Of
course, the order in which the movements occur is irrelevant (why?).

If we depict the upward movement with the $\bf{U}$ and the right movement with
the $\bf{R}$, we must count the structure of the $\bf{UURR}$, which equals the
repetition: $|A_{1}|={4!}/{2!2!}$, i.e., $|A_{1}|=6$. Thus, the total number
equals: $|A|=4\cdot 6$, i.e., $|A|=24$.

##  3\. Discussion

The following questions could arise if we try to solve the problem using the
two examples given:

  1. 1)

Identify the objects that need to be counted and look at the configuration of
the objects;

  2. 2)

Indicate the configuration’s attributes;

  3. 3)

Look for a structure in the set up (for example, divide objects into classes);

  4. 4)

Try to use the product or addition principle, and then verify your conclusion
by establishing a one-to-one relationship between the items you enumerated and
the configuration’s component pieces.

We produce a response for each step in order to try to make a problem out of
the bad answers and figure out why they were wrong. That’s why we call it an
indirect method ( We have placed these questions in grey boxes.).

Finally, we try to go over each step once with the following query to make
sure we have covered everything:

How do we ensure that the counting is carried out properly?

The steps will undoubtedly be time-consuming and unpleasant, but they will be
helpful. Besides, after we become professionals, we are not always required to
record every action we conduct.

##  4\. Conclusion Remarks

Combinatorics is recognised as one of the more difficult mathematical topics
to teach and learn, as we have previously discussed and encountered. Since
most issues lack simple fixes, there is a great deal of confusion over how to
approach them and which method to employ. There are many instances where two
different approaches that result in two different answers to the same problem
can seem equally compelling. While there are other ways to solve combinatorial
puzzles besides the one described in this article, it might be helpful when
these puzzles are challenging. At the start of teachers’ combinatorics
instruction, it can also be employed as a teaching strategy.

Finally, the authors want to investigate more issues relating to teaching
discrete and combinatorial mathematics in the future.

### Acknowledgement

Alexei Kanel-Belov is supported by the Russian Science Foundation 22-11-00177.

## References

  * [1] P. Soberón. Problem-Solving Methods in Combinatorics-An Approach to Olympiad Problems. Birkhäuser (2013). DOI: 10.1007/978-3-0348-0597-1.
  * [2] M. M. Eizenberg and O. Zaslavsky. Students’ Verification Strategies for Combinatorial Problems. Math. Think. Learn., vol. 6 (2004): 15-36. DOI: 10.1207/s15327833mtl0601-2.
  * [3] Y. M. Hidayati, C. Sa’dijah and S. A. Qohar. Combinatorial Thinking to Solve the Problems of Combinatorics in Selection Type. Int. J. Learn. Teach. Educ. Res., vol. 18 (2019): 65-75. DOI: 10.26803/IJLTER.18.2.5.
  * [4] M. Bona. A Walk Through Combinatorics: An Introduction to Enumeration and Graph Theory.  World Scientific Publishing (2006). DOI: 10.1142/10258.
  * [5] M. Keller and W. Trotter. Applied Combinatorics.  American Institute of Mathematics (2017). ISBN: 13: 9781973702719.
  * [6] O. N. Karpenkov. Geometry of Continued Fractions.  AACIM, vol. 26, Springer, Berlin, Heidelberg (2022). DOI: 10.1007/978-3-662-65277-0.
  * [7] T. Andreescu and Z. Feng. A Path to Combinatorics for Undergraduates-Counting Strategies.  Birkhäuser (2004). DOI: 10.1007/978-0-8176-8154-8.


# A NOTE ON THE NON-TRIVIAL ELEMENTS IN THE COHOMOLOGY GROUPS OF THE STEENROD
ALGEBRA

Đặng Võ Phúc  Faculty of Education Studies
University of Khanh Hoa, 01 Nguyen Chanh, Khanh Hoa, Viet Nam
<EMAIL_ADDRESS>
###### Abstract.

Let $F_{2}$ be the prime field of two elements and let $GL_{s}:=GL(s,F_{2})$
be the general linear group of rank $s.$ Denote by $\mathscr{A}$ the Steenrod
algebra over $F_{2}.$ The (mod-2) Lambda algebra, $\Lambda,$ is one of the
tools to describe those mysterious "Ext-groups". In addition, the $s$-th
algebraic transfer of William Singer [26] is also expected to be a useful tool
in the study of them. This transfer is a homomorphism
$Tr_{s}:F_{2}\otimes_{GL_{s}}P_{\mathscr{A}}(H_{*}(B\mathbb{V}_{s},F_{2}))\to{\rm
Ext}_{\mathscr{A}}^{s,s+*}(F_{2},F_{2}),$ where $\mathbb{V}_{s}$ denotes the
elementary abelian $2$-group of rank $s$, and $H_{*}(B\mathbb{V}_{s})$ is the
homology group of the classifying space of $\mathbb{V}_{s},$ while
$P_{\mathscr{A}}(H_{*}(B\mathbb{V}_{s},F_{2}))$ means the primitive part of
$H_{*}(B\mathbb{V}_{s},F_{2})$ under the action of $\mathscr{A}.$ It has been
shown that $Tr_{s}$ is highly non-trivial and, more precisely, that $Tr_{s}$
is an isomorphism for $s\leq 3.$ In addition, Singer proved that $Tr_{4}$ is
an isomorphism in some internal degrees. He was also investigated the image of
the fifth transfer by using invariant theory.

In this note, we use another method to study the image of $Tr_{5}.$ More
precisely, by direct computations using a representation of $Tr_{5}$ over the
algebra $\Lambda,$ we show that $Tr_{5}$ detects the non-zero elements
$h_{0}d_{0}\in{\rm Ext}_{\mathscr{A}}^{5,5+14}(F_{2},F_{2}),\
h_{2}e_{0}\in{\rm Ext}_{\mathscr{A}}^{5,5+20}(F_{2},F_{2})$ and
$h_{1}h_{4}c_{0}\in{\rm Ext}_{\mathscr{A}}^{5,5+24}(F_{2},F_{2}).$ The same
argument can be used for homological degrees $s\geq 6$ under certain
conditions.

###### Key words and phrases:

Steenrod algebra, Lambda algebra, Algebraic transfer, Cohomology of Steenrod
algebra

###### 2010 Mathematics Subject Classification:

55S10, 55S05, 55T15

##  1\. The Singer algebraic transfer and the (mod-2) Lambda algebra

Let $\mathbb{V}_{s}$ be a rank $s$ elementary abelian 2-group and let
$B\mathbb{V}_{s}$ denote the classifying space of $\mathbb{V}_{s}.$ We may
equally well view $\mathbb{V}_{s}$ as an $s$-dimensional $F_{2}$-vector space.
As well-known, $H^{*}(\mathbb{V}_{s},F_{2})\cong S(\mathbb{V}_{s}^{*}),$ the
symmetric algebra over the dual space
$\mathbb{V}_{s}^{*}=H^{1}(\mathbb{V}_{s},F_{2}).$ Pick $x_{1},\ldots,x_{s}$ to
be a basis of $H^{1}(\mathbb{V}_{s},F_{2}).$ Then, we have
$H^{*}(B\mathbb{V}_{s},F_{2})=H^{*}(\mathbb{V}_{s},F_{2})\cong\mathcal{P}_{s}:=F_{2}[x_{1},\ldots,x_{s}],$
the polynomial algebra on $s$ generators of degree one. It is considered as an
unstable $\mathscr{A}$-module. The canonical $\mathscr{A}$-action on
$\mathcal{P}_{1}$ is extended to an $\mathscr{A}$-action on
$F_{2}[x_{1},x_{1}^{-1}],$ the ring of finite Laurent series (see [2], [29]).
Then, $\overline{\mathcal{P}}=\langle\\{x_{1}^{t}|\ t\geq-1\\}\rangle$ is
$\mathscr{A}$-submodule of $F_{2}[x_{1},x_{1}^{-1}].$ One has a short-exact
sequence:

(1.1)
$0\to\mathcal{P}_{1}\xrightarrow{q}\overline{\mathcal{P}}\xrightarrow{\pi}\Sigma^{-1}F_{2},$

where $q$ is the inclusion and $\pi$ is given by $\pi(x_{1}^{t})=0$ if
$t\neq-1$ and $\pi(x_{1}^{-1})=1.$ Denote by $e_{1}$ the element in ${\rm
Ext}_{\mathscr{A}}^{1}(\Sigma^{-1}F_{2},\mathcal{P}_{1}),$ which is
represented by the cocyle associated to (1.1). For each $i\geq 1,$ the short-
exact sequence

(1.2)
$0\to\mathcal{P}_{i+1}\cong\mathcal{P}_{i}\otimes_{F_{2}}\mathcal{P}_{1}\xrightarrow{1\otimes_{F_{2}}q}\mathcal{P}_{i}\otimes_{F_{2}}\overline{\mathcal{P}}\xrightarrow{1\otimes_{F_{2}}\pi}\Sigma^{-1}\mathcal{P}_{i},$

determines a class $(e_{1}\times\mathcal{P}_{i})\in{\rm
Ext}_{\mathscr{A}}^{1}(\Sigma^{-1}\mathcal{P}_{i},\mathcal{P}_{i+1}).$ Then,
using the cross product and Yoneda product, we have the element

$e_{s}=(e_{1}\times\mathcal{P}_{s-1})\circ(e_{1}\times\mathcal{P}_{s-2})\circ\cdots\circ(e_{1}\times\mathcal{P}_{1})\circ
e_{1}\in{\rm Ext}_{\mathscr{A}}^{s}(\Sigma^{-s}F_{2},\mathcal{P}_{s}).$

Let $\Delta(e_{1}\times\mathcal{P}_{i}):{\rm
Tor}_{s-i}^{\mathscr{A}}(F_{2},\Sigma^{-s}\mathcal{P}_{i})\to{\rm
Tor}_{s-i-1}^{\mathscr{A}}(F_{2},\mathcal{P}_{i+1})$ be the connecting
homomorphism associated to (1.2). Then, we have a composition of the
connecting homomorphisms

$\overline{\varphi}_{s}=\Delta(e_{1}\times\mathcal{P}_{s-1})\circ\Delta(e_{1}\times\mathcal{P}_{s-2})\circ\cdots\circ\Delta(e_{1}\times\mathcal{P}_{1})\circ\Delta(e_{1})$

from ${\rm Tor}_{s}^{\mathscr{A}}(F_{2},\Sigma^{-s}F_{2})$ to ${\rm
Tor}_{0}^{\mathscr{A}}(F_{2},\mathcal{P}_{s})=F_{2}\otimes_{\mathscr{A}}\mathcal{P}_{s}$,
determined by $\overline{\varphi}_{s}(z)=e_{s}\cap z$ for any $z\in{\rm
Tor}_{s}^{\mathscr{A}}(F_{2},\Sigma^{-s}F_{2}).$ Here $\cap$ denotes the cap
product in homology with $F_{2}$-coefficients. The image of
$\overline{\varphi}_{s}$ is a submodule of the variant
$(F_{2}\otimes_{\mathscr{A}}\mathcal{P}_{s})^{GL_{s}},$ where $GL_{s}$ denotes
the general linear group of rank $s$ over $F_{2}.$ Hence,
$\overline{\varphi}_{s}$ induces homomorphism

$\varphi_{s}:{\rm
Tor}_{s}^{\mathscr{A}}(F_{2},\Sigma^{-s}F_{2})\to(F_{2}\otimes_{\mathscr{A}}\mathcal{P}_{s})^{GL_{s}}.$

Since the supension $\Sigma^{-s}$ induces an isomorphism

${\rm Tor}_{s,*}^{\mathscr{A}}(F_{2},\Sigma^{-s}F_{2})\cong{\rm
Tor}_{s,s+*}^{\mathscr{A}}(F_{2},F_{2}),$

we have the homomorphism

$\varphi_{s}:{\rm
Tor}_{s,s+*}^{\mathscr{A}}(F_{2},F_{2})\to(F_{2}\otimes_{\mathscr{A}}\mathcal{P}_{s})_{*}^{GL_{s}}.$

Let $P_{\mathscr{A}}(H_{*}(B\mathbb{V}_{s},F_{2}))$ be the primitive part
consisting of all elements in $H_{*}(B\mathbb{V}_{s},F_{2}),$ which are
annihilated by every positive degree operation in $\mathscr{A}.$ Then, the
dual homomorphism

$Tr_{s}:F_{2}\otimes_{GL_{s}}P_{\mathscr{A}}(H_{*}(B\mathbb{V}_{s},F_{2}))\to{\rm
Ext}_{\mathscr{A}}^{s,s+*}(F_{2},F_{2})$

of $\varphi_{s}$ is called the $s$-th Singer algebraic transfer. It is
expected to be a useful tool in describing the cohomology groups of
$\mathscr{A}$ by means of invariant theory and the Peterson "hit problem" of
determining a minimal set of $\mathscr{A}$-generatos for $\mathcal{P}_{s}$
(see [3], [8], [12], [13, 14, 15, 16, 17, 18, 20, 21, 22, 23], [28]). It was
then investigated by many authors (see Boardman [3], Chơn and Hà [5], Crossley
[6], Hà [7], Minami [11], the present author [15, 16, 18, 19, 20, 22, 23, 24,
25], and others). By Boardman [3] and Singer [26], $Tr_{s}$ is an isomorphism
for $s\leq 3.$ In [26], Singer also gave computations to show that $Tr_{4}$ is
an isomorphism in a range of internal degrees.

As it is known, the homological algebra $\\{H_{d}(B\mathbb{V}_{s},F_{2})|\
d\geq 0\\}$ is dual to $\mathcal{P}_{s}.$ Moreover, it is isomorphic to
$\Gamma(a_{1},\ldots,a_{s}),$ the divided power algebra generated by
$a_{1},\ldots,a_{s},$ each of degree one, where $a_{j}=a_{j}^{(1)}$ is dual to
$x_{j}.$ Here the duality is taken with respect to the basis of
$\mathcal{P}_{s}$ consisting of all monomials in $x_{1},\ldots,x_{s}.$ The
algebra $\\{H_{d}(B\mathbb{V}_{s},F_{2})|\ d\geq 0\\}$ is a right
$\mathscr{A}$-module. The right $\mathscr{A}$-action on this algebra is given
by $(a^{(j)}_{t})Sq^{s}=\binom{j-s}{s}a^{(j-s)}_{t}=Sq^{s}_{*}(a^{(j)}_{t})$
and Cartan’s formula. Note that $Sq^{s}_{*}$ is the dual Steenrod operation.

The (mod 2) Lambda algebra $\Lambda$ (see Bousfield et al. [4]) is also one of
the tools to compute the cohomology groups of $\mathscr{A}.$ Recall that
$\Lambda$ is defined as a differential, bigraded, associative algebra with
unit over $F_{2}$ is generated by $\lambda_{k}\in\Lambda^{1,k},$ satisfying
the Adem relations

(1.3) $\lambda_{k}\lambda_{2k+s+1}=\sum_{j\geq
0}\binom{s-j-1}{j}\lambda_{k+s-j}\lambda_{2k+1+j}\ \ (k\geq 0,\ s\geq 0)$

and differential

(1.4) $\partial(\lambda_{s-1})=\sum_{j\geq
1}\binom{s-j-1}{j}\lambda_{s-j-1}\lambda_{j-1}\ \ (s\geq 1),$

where $\binom{s-j-1}{j}$ denotes the modulo $2$ value of the binomial
coefficient, with the usual convention $\binom{s-j-1}{j}=0$ for $j>s-j-1.$ In
addition,

$H^{s,*}(\Lambda,F_{2})={\rm Ext}_{\mathscr{A}}^{s,s+*}(F_{2},F_{2}).$

Now, for non-negative integers $t_{1},\ldots,t_{s},$ a monomial
$\lambda_{t_{1}}\ldots\lambda_{t_{s}}\in\Lambda$ is called the monomial of the
length $s$. We shall write $\lambda_{T},\ T=(t_{1},\ldots,t_{s})$ for
$\prod_{1\leq k\leq s}\lambda_{t_{k}}$ and refer to $\ell(T)=s$ as the length
of $T.$ Note that the algebra $\Lambda$ is not commutative and that the
bigrading of a monomial indexed by $T$ may be written $(s,d),$ where the
homological degree $s,$ as above, is the length of $T,$ and $d=\sum_{1\leq
k\leq s}t_{k}.$ A monomial $\lambda_{T}$ is called admissible if $t_{k}\leq
2t_{k+1}$ for all $1\leq k\leq s-1.$ By the relations (1.3), the
$F_{2}$-vector subspace

$\Lambda^{s,*}=\langle\\{\lambda_{T}|\ T=(t_{1},\ldots,t_{s}),\ t_{k}\geq
0,1\leq k\leq s\\}\rangle$

of $\Lambda$ has an additive basis consisting of all admissible monomials of
length $s.$ In [5], Chơn and Hà defined a homomorphism
$\psi_{s}:H_{*}(B\mathbb{V}_{s},F_{2})\longrightarrow\Lambda^{s,*},$ which is
considered as a representation in the algebra $\Lambda$ of the Singer transfer
$Tr_{s}$ and determined by the following inductive formula:

$\psi_{s}(a^{T})=\left\\{\begin{array}[]{ll}\lambda_{t_{1}}&\text{if}\;\ell(T)=1,\\\
\sum_{j\geq t_{s}}\psi_{s-1}(\prod_{1\leq k\leq
s-1}a_{k}^{(t_{k})}Sq^{j-t_{s}})\lambda_{j}&\text{if}\;\ell(T)>1,\end{array}\right.$

for any $a^{T}:=\prod_{1\leq k\leq s}a_{k}^{(t_{k})}\in
H_{*}(B\mathbb{V}_{s},F_{2})$ and $T=(t_{1},t_{2},\ldots,t_{s}).$ Note that
$\psi_{s}$ is not an $\mathscr{A}$-homomorphism. The authors showed in [5]
that if $\theta\in P_{\mathscr{A}}(H_{*}(B\mathbb{V}_{s},F_{2})),$ then
$\psi_{s}(\theta)$ is a cycle in $\Lambda^{s,*}$ and
$Tr_{s}([\theta])=[\psi_{s}(\theta)]$. Therefrom, we have a homomorphism of
algebras

$\psi=\\{\psi_{s}:\ s\geqslant 0\\}:\\{H_{*}(B\mathbb{V}_{s},F_{2}):\
s\geqslant 0\\}\to\\{\Lambda^{s,*}:\ s\geqslant 0\\}=\Lambda,$

which induces the Singer transfer.

By using invariant theory, Singer showed in [26] that the non-zero element
$Ph_{1}\in{\rm Ext}_{\mathscr{A}}^{5,5+9}(F_{2},F_{2})$ was not detected by
$Tr_{5}.$ The aim of the present note is also to study the image of $Tr_{5}.$

##  2\. Main result

The main result of this note is the following.

######  Main Theorem.

The non-zero elements

  1. (i)

$h_{0}d_{0}\in{\rm Ext}_{\mathscr{A}}^{5,5+14}(F_{2},F_{2}),$

  2. (ii)

$h_{2}e_{0}=h_{0}g_{1}\in{\rm Ext}_{\mathscr{A}}^{5,5+20}(F_{2},F_{2}),$

  3. (iii)

$h_{1}h_{4}c_{0}\in{\rm Ext}_{\mathscr{A}}^{5,5+24}(F_{2},F_{2})$

are in the image of the fifth algebraic transfer.

We prove the theorem by direct computations using the representation in the
Lambda algebra of the fifth algebraic transfer.

###### Proof of main theorem.

It is well-known that there exists an endomorphism $Sq^{0}$ of the lambda
algebra $\Lambda,$ determined by $Sq^{0}(\lambda_{T}=\prod_{1\leq k\leq
s}\lambda_{t_{k}})=\prod_{1\leq k\leq s}\lambda_{2t_{k}+1},$ where
$\lambda_{T}$ is not necessarily admissible. It respects the Adem relations in
(1.3) and commutes with the differential $\partial$ in (1.4) above. Then,
$Sq^{0}$ induces the classical squaring operation in the Ext groups

$Sq^{0}:H^{s,*}(\Lambda,F_{2})={\rm Ext}^{s,s+*}_{\mathscr{A}}(F_{2},F_{2})\to
H^{s,s+2*}(\Lambda,F_{2})={\rm Ext}^{s,2(s+*)}_{\mathscr{A}}(F_{2},F_{2}).$

According to Liulevicius [10], $Sq^{0}$ is not the identity map. Moreover, it
commutes with Kameko’s squaring operation (see [8])

$\widetilde{Sq^{0}}:F_{2}\otimes_{GL_{s}}P_{\mathscr{A}}(H_{*}(B\mathbb{V}_{s},F_{2}))\to
F_{2}\otimes_{GL_{s}}P_{\mathscr{A}}(H_{s+2*}(B\mathbb{V}_{s},F_{2}))$

through the $s$-th Singer transfer. In what follows,

$(Sq^{0})^{t}:{\rm Ext}^{*,*}_{\mathscr{A}}(F_{2},F_{2})\to{\rm
Ext}^{*,*}_{\mathscr{A}}(F_{2},F_{2})$

denotes the composite $Sq^{0}\ldots Sq^{0}$ ($t$ times of $Sq^{0}$) if $t>1,$
is $Sq^{0}$ if $t=1,$ and is the identity map if $t=0.$ A family
$\\{a_{t}:\,t\geq 0\\}\subset{\rm Ext}_{\mathscr{A}}^{s,s+*}(F_{2},F_{2})$ is
called a $Sq^{0}$-family if $a_{t}=(Sq^{0})^{t}(a_{0})$ for $t\geq 0.$
According to Lin [9], the algebra $\\{{\rm
Ext}_{\mathscr{A}}^{s,s+*}(F_{2},F_{2}):\ s\geqslant 0\\}$ for $s\leqslant 4$
is generated by $h_{t},\ c_{t},\ d_{t},\ e_{t},\ f_{t},\ g_{t+1},\ p_{t},\
D_{3}(t),\ p^{\prime}_{t}$ for $i\geqslant 0$ and subject to the relations
$h_{i}h_{i+1}=0,$ $h_{i}h^{2}_{i+2}=0,$ and $h_{i}^{3}=h^{2}_{i-1}h_{i+1}$
together with the relations $h^{2}_{t}h^{2}_{t+3}=0,h_{j}c_{t}=0$ for
$j\in\\{t-1,\ t,\ t+2,\,t+3\\}.$ Furthermore, the set of the elements $d_{t},\
e_{t},\ f_{t},\ g_{t+1},\ p_{t},\ D_{3}(t)$ and $p^{\prime}_{t}$ for
$t\geqslant 0,$ is an $F_{2}$-basis for the indecomposable elements in ${\rm
Ext}_{\mathscr{A}}^{4,4+*}(F_{2},F_{2}).$ By Lin [9], we have

$\begin{array}[]{ll}\vskip 6.0pt plus 2.0pt minus
2.0pth_{t}&=\\{\lambda_{2^{t}-1}=(Sq^{0})^{t}(\lambda_{0})\\}\in{\rm
Ext}_{\mathscr{A}}^{1,2^{t}}(F_{2},F_{2})\ \mbox{for $t\geq 0$},\\\ \vskip
6.0pt plus 2.0pt minus
2.0ptc_{t}&=\\{(Sq^{0})^{t}(\lambda_{3}^{2}\lambda_{2})\\}\in{\rm
Ext}_{\mathscr{A}}^{3,2^{t+3}+2^{t+1}+2^{t}}(F_{2},F_{2})\ \mbox{for $t\geq
0$},\\\ \vskip 6.0pt plus 2.0pt minus
2.0ptd_{t}&=\\{(Sq^{0})^{t}(\lambda_{3}^{2}\lambda_{2}\lambda_{6}+\lambda_{3}^{2}\lambda_{4}^{2}+\lambda_{3}\lambda_{5}\lambda_{4}\lambda_{2}+\lambda_{7}\lambda_{1}\lambda_{5}\lambda_{1})\\}\\\
\vskip 6.0pt plus 2.0pt minus 2.0pt&\quad\in{\rm
Ext}_{\mathscr{A}}^{4,2^{t+4}+2^{t+1}}(F_{2},F_{2})\ \mbox{for $t\geq 0$},\\\
e_{t}&=\\{(Sq^{0})^{t}(\lambda_{3}^{3}\lambda_{8}+(\lambda_{3}\lambda_{5}^{2}+\lambda_{3}^{2}\lambda_{7})\lambda_{4}+(\lambda_{3}^{2}\lambda_{9}+\lambda_{9}\lambda_{3}^{2})\lambda_{2})\\}\\\
\vskip 6.0pt plus 2.0pt minus 2.0pt&\quad\in{\rm
Ext}_{\mathscr{A}}^{4,2^{t+4}+2^{t+2}+2^{t}}(F_{2},F_{2})\ \mbox{for $t\geq
0$},\\\
g_{t+1}&=\\{(Sq^{0})^{t}(\lambda_{7}^{2}\lambda_{0}\lambda_{6}+(\lambda_{3}^{2}\lambda_{9}+\lambda_{9}\lambda_{3}^{2})\lambda_{5}+(\lambda_{3}^{2}\lambda_{11}+\lambda_{3}\lambda_{9}\lambda_{5})\lambda_{3})\\}\\\
&\quad\in{\rm Ext}_{\mathscr{A}}^{4,2^{t+4}+2^{t+3}}(F_{2},F_{2})\ \mbox{for
$t\geq 0$}.\end{array}$

Combining this and the previous results by Tangora [27], we get

${\rm
Ext}_{\mathscr{A}}^{5,5+n}(F_{2},F_{2})=\left\\{\begin{array}[]{ll}\langle
h_{0}d_{0}\rangle&\mbox{if $n=14$},\\\ \langle h_{2}e_{0}\rangle&\mbox{if
$n=20$},\\\ \langle h_{1}h_{4}c_{0}\rangle&\mbox{if
$n=24$},\end{array}\right.$

and $h_{0}d_{0}\neq 0,\ h_{2}e_{0}=h_{0}g_{1}\neq 0,\
h_{1}h_{4}c_{0}=h_{3}e_{0}\neq 0.$ Here, $h_{t}=[\lambda_{2^{t}-1}]$ is the
Adams element (see [1]) in ${\rm
Ext}_{\mathcal{A}_{2}}^{1,2^{t}}(F_{2},F_{2}),$ for $0\leq t\leq 4.$

The case (i). Obviously, $\lambda_{0}\in\Lambda^{1,0},$ and

$\overline{d}_{0}=\lambda_{3}^{2}\lambda_{2}\lambda_{6}+\lambda_{3}^{2}\lambda_{4}^{2}+\lambda_{3}\lambda_{5}\lambda_{4}\lambda_{2}+\lambda_{7}\lambda_{1}\lambda_{5}\lambda_{1}\in\Lambda^{4,14}$

are the cycles in the algebra $\Lambda.$ Recall that

$h_{0}=[\lambda_{0}]\in{\rm Ext}_{\mathscr{A}}^{1,1}(F_{2},F_{2}),\ \
d_{0}=[\overline{d}_{0}]\in{\rm Ext}_{\mathscr{A}}^{4,18}(F_{2},F_{2}).$

We consider the element $u_{14}$ in $H_{14}(B\mathbb{V}_{5},F_{2}),$ which is
the following sum:

$\begin{array}[]{ll}&a_{1}^{(0)}a_{2}^{(6)}a_{3}^{(5)}a_{4}^{(2)}a_{5}^{(1)}+a_{1}^{(0)}a_{2}^{(6)}a_{3}^{(4)}a_{4}^{(3)}a_{5}^{(1)}+a_{1}^{(0)}a_{2}^{(6)}a_{3}^{(3)}a_{4}^{(4)}a_{5}^{(1)}+\vskip
6.0pt plus 2.0pt minus
2.0pta_{1}^{(0)}a_{2}^{(6)}a_{3}^{(2)}a_{4}^{(5)}a_{5}^{(1)}\\\
&\quad+a_{1}^{(0)}a_{2}^{(6)}a_{3}^{(1)}a_{4}^{(6)}a_{5}^{(1)}+a_{1}^{(0)}a_{2}^{(5)}a_{3}^{(6)}a_{4}^{(1)}a_{5}^{(2)}+a_{1}^{(0)}a_{2}^{(5)}a_{3}^{(5)}a_{4}^{(2)}a_{5}^{(2)}+\vskip
6.0pt plus 2.0pt minus
2.0pta_{1}^{(0)}a_{2}^{(5)}a_{3}^{(4)}a_{4}^{(3)}a_{5}^{(2)}\\\
&\quad+a_{1}^{(0)}a_{2}^{(5)}a_{3}^{(3)}a_{4}^{(4)}a_{5}^{(2)}+a_{1}^{(0)}a_{2}^{(5)}a_{3}^{(2)}a_{4}^{(5)}a_{5}^{(2)}+a_{1}^{(0)}a_{2}^{(5)}a_{3}^{(1)}a_{4}^{(6)}a_{5}^{(2)}+\vskip
6.0pt plus 2.0pt minus
2.0pta_{1}^{(0)}a_{2}^{(5)}a_{3}^{(5)}a_{4}^{(1)}a_{5}^{(3)}\\\
&\quad+a_{1}^{(0)}a_{2}^{(3)}a_{3}^{(6)}a_{4}^{(2)}a_{5}^{(3)}+a_{1}^{(0)}a_{2}^{(6)}a_{3}^{(2)}a_{4}^{(3)}a_{5}^{(3)}+a_{1}^{(0)}a_{2}^{(6)}a_{3}^{(1)}a_{4}^{(4)}a_{5}^{(3)}+\vskip
6.0pt plus 2.0pt minus
2.0pta_{1}^{(0)}a_{2}^{(5)}a_{3}^{(2)}a_{4}^{(4)}a_{5}^{(3)}\\\
&\quad+a_{1}^{(0)}a_{2}^{(3)}a_{3}^{(4)}a_{4}^{(4)}a_{5}^{(3)}+a_{1}^{(0)}a_{2}^{(3)}a_{3}^{(2)}a_{4}^{(6)}a_{5}^{(3)}+a_{1}^{(0)}a_{2}^{(3)}a_{3}^{(6)}a_{4}^{(1)}a_{5}^{(4)}+\vskip
6.0pt plus 2.0pt minus
2.0pta_{1}^{(0)}a_{2}^{(3)}a_{3}^{(5)}a_{4}^{(2)}a_{5}^{(4)}\\\
&\quad+a_{1}^{(0)}a_{2}^{(3)}a_{3}^{(4)}a_{4}^{(3)}a_{5}^{(4)}+a_{1}^{(0)}a_{2}^{(3)}a_{3}^{(3)}a_{4}^{(4)}a_{5}^{(4)}+a_{1}^{(0)}a_{2}^{(3)}a_{3}^{(2)}a_{4}^{(5)}a_{5}^{(4)}+\vskip
6.0pt plus 2.0pt minus
2.0pta_{1}^{(0)}a_{2}^{(3)}a_{3}^{(1)}a_{4}^{(6)}a_{5}^{(4)}\\\
&\quad+a_{1}^{(0)}a_{2}^{(5)}a_{3}^{(3)}a_{4}^{(1)}a_{5}^{(5)}+a_{1}^{(0)}a_{2}^{(6)}a_{3}^{(1)}a_{4}^{(2)}a_{5}^{(5)}+a_{1}^{(0)}a_{2}^{(5)}a_{3}^{(2)}a_{4}^{(2)}a_{5}^{(5)}+\vskip
6.0pt plus 2.0pt minus
2.0pta_{1}^{(0)}a_{2}^{(3)}a_{3}^{(4)}a_{4}^{(2)}a_{5}^{(5)}\\\
&\quad+a_{1}^{(0)}a_{2}^{(5)}a_{3}^{(1)}a_{4}^{(3)}a_{5}^{(5)}+a_{1}^{(0)}a_{2}^{(3)}a_{3}^{(3)}a_{4}^{(3)}a_{5}^{(5)}+a_{1}^{(0)}a_{2}^{(3)}a_{3}^{(1)}a_{4}^{(5)}a_{5}^{(5)}+\vskip
6.0pt plus 2.0pt minus
2.0pta_{1}^{(0)}a_{2}^{(6)}a_{3}^{(1)}a_{4}^{(1)}a_{5}^{(6)}\\\
&\quad+a_{1}^{(0)}a_{2}^{(5)}a_{3}^{(2)}a_{4}^{(1)}a_{5}^{(6)}+a_{1}^{(0)}a_{2}^{(3)}a_{3}^{(4)}a_{4}^{(1)}a_{5}^{(6)}+a_{1}^{(0)}a_{2}^{(3)}a_{3}^{(3)}a_{4}^{(2)}a_{5}^{(6)}+a_{1}^{(0)}a_{2}^{(6)}a_{3}^{(6)}a_{4}^{(1)}a_{5}^{(1)}.\end{array}.$

Then, it is $\mathscr{A}^{+}$-annihilated. Indeed, by the unstable condition,
we need to compute the actions of the Steenrod squares $Sq^{1},Sq^{2}$ and
$Sq^{4}.$ It is not difficult to check that $(u_{14})Sq^{i}=0$ for $i=1,2.$ By
direct calculation, $(u_{14})Sq^{4}$ is the following sum:

$\begin{array}[]{ll}&a_{1}^{(0)}a_{2}^{(5)}a_{3}^{(3)}a_{4}^{(1)}a_{5}^{(1)}+a_{1}^{(0)}a_{2}^{(3)}a_{3}^{(5)}a_{4}^{(1)}a_{5}^{(1)}+a_{1}^{(0)}a_{2}^{(5)}a_{3}^{(3)}a_{4}^{(1)}a_{5}^{(1)}+\vskip
6.0pt plus 2.0pt minus
2.0pta_{1}^{(0)}a_{2}^{(3)}a_{3}^{(5)}a_{4}^{(1)}a_{5}^{(1)}\\\
&\quad+a_{1}^{(0)}a_{2}^{(3)}a_{3}^{(3)}a_{4}^{(3)}a_{5}^{(1)}+a_{1}^{(0)}a_{2}^{(3)}a_{3}^{(3)}a_{4}^{(3)}a_{5}^{(1)}+a_{1}^{(0)}a_{2}^{(5)}a_{3}^{(1)}a_{4}^{(3)}a_{5}^{(1)}+\vskip
6.0pt plus 2.0pt minus
2.0pta_{1}^{(0)}a_{2}^{(3)}a_{3}^{(1)}a_{4}^{(5)}a_{5}^{(1)}\\\
&\quad+a_{1}^{(0)}a_{2}^{(5)}a_{3}^{(1)}a_{4}^{(3)}a_{5}^{(1)}+a_{1}^{(0)}a_{2}^{(3)}a_{3}^{(1)}a_{4}^{(5)}a_{5}^{(1)}+a_{1}^{(0)}a_{2}^{(3)}a_{3}^{(5)}a_{4}^{(1)}a_{5}^{(1)}+\vskip
6.0pt plus 2.0pt minus
2.0pta_{1}^{(0)}a_{2}^{(5)}a_{3}^{(3)}a_{4}^{(1)}a_{5}^{(1)}\\\
&\quad+a_{1}^{(0)}a_{2}^{(3)}a_{3}^{(3)}a_{4}^{(2)}a_{5}^{(2)}+a_{1}^{(0)}a_{2}^{(5)}a_{3}^{(3)}a_{4}^{(1)}a_{5}^{(1)}+a_{1}^{(0)}a_{2}^{(3)}a_{3}^{(5)}a_{4}^{(1)}a_{5}^{(1)}+\vskip
6.0pt plus 2.0pt minus
2.0pta_{1}^{(0)}a_{2}^{(3)}a_{3}^{(2)}a_{4}^{(3)}a_{5}^{(2)}\\\
&\quad+a_{1}^{(0)}a_{2}^{(3)}a_{3}^{(3)}a_{4}^{(3)}a_{5}^{(1)}+a_{1}^{(0)}a_{2}^{(3)}a_{3}^{(3)}a_{4}^{(2)}a_{5}^{(2)}+a_{1}^{(0)}a_{2}^{(3)}a_{3}^{(3)}a_{4}^{(3)}a_{5}^{(1)}+\vskip
6.0pt plus 2.0pt minus
2.0pta_{1}^{(0)}a_{2}^{(3)}a_{3}^{(2)}a_{4}^{(3)}a_{5}^{(2)}\\\
&\quad+a_{1}^{(0)}a_{2}^{(5)}a_{3}^{(1)}a_{4}^{(3)}a_{5}^{(1)}+a_{1}^{(0)}a_{2}^{(3)}a_{3}^{(1)}a_{4}^{(5)}a_{5}^{(1)}+a_{1}^{(0)}a_{2}^{(3)}a_{3}^{(1)}a_{4}^{(5)}a_{5}^{(1)}+\vskip
6.0pt plus 2.0pt minus
2.0pta_{1}^{(0)}a_{2}^{(5)}a_{3}^{(1)}a_{4}^{(3)}a_{5}^{(1)}\\\
&\quad+a_{1}^{(0)}a_{2}^{(3)}a_{3}^{(3)}a_{4}^{(1)}a_{5}^{(3)}+a_{1}^{(0)}a_{2}^{(3)}a_{3}^{(3)}a_{4}^{(1)}a_{5}^{(3)}+a_{1}^{(0)}a_{2}^{(3)}a_{3}^{(1)}a_{4}^{(3)}a_{5}^{(3)}+\vskip
6.0pt plus 2.0pt minus
2.0pta_{1}^{(0)}a_{2}^{(3)}a_{3}^{(1)}a_{4}^{(3)}a_{5}^{(3)}\\\
&\quad+a_{1}^{(0)}a_{2}^{(3)}a_{3}^{(2)}a_{4}^{(2)}a_{5}^{(3)}+a_{1}^{(0)}a_{2}^{(3)}a_{3}^{(1)}a_{4}^{(3)}a_{5}^{(3)}+a_{1}^{(0)}a_{2}^{(3)}a_{3}^{(2)}a_{4}^{(2)}a_{5}^{(3)}+\vskip
6.0pt plus 2.0pt minus
2.0pta_{1}^{(0)}a_{2}^{(3)}a_{3}^{(1)}a_{4}^{(3)}a_{5}^{(3)}\\\
&\quad+a_{1}^{(0)}a_{2}^{(3)}a_{3}^{(3)}a_{4}^{(1)}a_{5}^{(3)}+a_{1}^{(0)}a_{2}^{(3)}a_{3}^{(3)}a_{4}^{(2)}a_{5}^{(2)}+a_{1}^{(0)}a_{2}^{(3)}a_{3}^{(3)}a_{4}^{(1)}a_{5}^{(3)}+\vskip
6.0pt plus 2.0pt minus
2.0pta_{1}^{(0)}a_{2}^{(3)}a_{3}^{(2)}a_{4}^{(3)}a_{5}^{(2)}\\\
&\quad+a_{1}^{(0)}a_{2}^{(5)}a_{3}^{(1)}a_{4}^{(3)}a_{5}^{(1)}+a_{1}^{(0)}a_{2}^{(3)}a_{3}^{(1)}a_{4}^{(5)}a_{5}^{(1)}+a_{1}^{(0)}a_{2}^{(3)}a_{3}^{(2)}a_{4}^{(3)}a_{5}^{(2)}+\vskip
6.0pt plus 2.0pt minus
2.0pta_{1}^{(0)}a_{2}^{(3)}a_{3}^{(1)}a_{4}^{(3)}a_{5}^{(3)}\\\
&\quad+a_{1}^{(0)}a_{2}^{(3)}a_{3}^{(1)}a_{4}^{(3)}a_{5}^{(3)}+a_{1}^{(0)}a_{2}^{(3)}a_{3}^{(3)}a_{4}^{(1)}a_{5}^{(3)}+a_{1}^{(0)}a_{2}^{(5)}a_{3}^{(1)}a_{4}^{(1)}a_{5}^{(3)}+\vskip
6.0pt plus 2.0pt minus
2.0pta_{1}^{(0)}a_{2}^{(3)}a_{3}^{(1)}a_{4}^{(1)}a_{5}^{(5)}\\\
&\quad+a_{1}^{(0)}a_{2}^{(3)}a_{3}^{(2)}a_{4}^{(2)}a_{5}^{(3)}+a_{1}^{(0)}a_{2}^{(5)}a_{3}^{(1)}a_{4}^{(1)}a_{5}^{(3)}+a_{1}^{(0)}a_{2}^{(3)}a_{3}^{(1)}a_{4}^{(1)}a_{5}^{(5)}+\vskip
6.0pt plus 2.0pt minus
2.0pta_{1}^{(0)}a_{2}^{(3)}a_{3}^{(2)}a_{4}^{(2)}a_{5}^{(3)}\\\
&\quad+a_{1}^{(0)}a_{2}^{(3)}a_{3}^{(3)}a_{4}^{(1)}a_{5}^{(3)}+a_{1}^{(0)}a_{2}^{(3)}a_{3}^{(1)}a_{4}^{(3)}a_{5}^{(3)}+a_{1}^{(0)}a_{2}^{(3)}a_{3}^{(1)}a_{4}^{(3)}a_{5}^{(3)}+\vskip
6.0pt plus 2.0pt minus
2.0pta_{1}^{(0)}a_{2}^{(5)}a_{3}^{(1)}a_{4}^{(1)}a_{5}^{(3)}\\\
&\quad+a_{1}^{(0)}a_{2}^{(3)}a_{3}^{(1)}a_{4}^{(1)}a_{5}^{(5)}+a_{1}^{(0)}a_{2}^{(3)}a_{3}^{(1)}a_{4}^{(1)}a_{5}^{(5)}+a_{1}^{(0)}a_{2}^{(5)}a_{3}^{(1)}a_{4}^{(1)}a_{5}^{(3)}+\vskip
6.0pt plus 2.0pt minus
2.0pta_{1}^{(0)}a_{2}^{(3)}a_{3}^{(3)}a_{4}^{(1)}a_{5}^{(3)}\\\
&\quad+a_{1}^{(0)}a_{2}^{(3)}a_{3}^{(3)}a_{4}^{(1)}a_{5}^{(3)}.\end{array}$

Thus, $(u_{14})Sq^{4}=0.$ Then, using the differential (1.4) and the
representation of $Tr_{5}$ over $\Lambda,$ we see that
$\begin{array}[]{ll}\psi_{5}(u_{14})&=\lambda_{0}\lambda_{3}^{2}\lambda_{2}\lambda_{6}+\lambda_{0}\lambda_{3}^{2}\lambda_{4}^{2}+\lambda_{0}\lambda_{3}\lambda_{5}\lambda_{4}\lambda_{2}+\lambda_{0}\lambda_{7}\lambda_{1}\lambda_{5}\lambda_{1}+\partial(\lambda_{0}\lambda_{3}^{2}\lambda_{9}+\lambda_{0}\lambda_{3}\lambda_{9}\lambda_{3})\\\
&=\lambda_{0}\overline{d}_{0}+\partial(\lambda_{0}\lambda_{3}^{2}\lambda_{9}+\lambda_{0}\lambda_{3}\lambda_{9}\lambda_{3}).\end{array}$
is a cycle in $\Lambda^{5,14}$ and is a representative of $h_{0}d_{0}\in{\rm
Ext}_{\mathscr{A}}^{5,5+14}(F_{2},F_{2}).$ Therefore, $h_{0}d_{0}\in{\rm
Im}(Tr_{5}).$

The case (ii). Obviously, $\lambda_{3}\in\Lambda^{1,3}$ and

$\overline{e}_{0}=\lambda_{3}^{3}\lambda_{8}+(\lambda_{3}\lambda_{5}^{2}+\lambda_{3}^{2}\lambda_{7})\lambda_{4}+\lambda_{7}\lambda_{5}\lambda_{3}\lambda_{2}+\lambda_{3}^{2}\lambda_{5}\lambda_{6}\in\Lambda^{4,17}$

are the cycles in the algebra $\Lambda$ and they are representative of the
non-zero elements $h_{2}\in{\rm Ext}_{\mathscr{A}}^{1,4}(F_{2},F_{2})$ and
$e_{0}\in{\rm Ext}_{\mathscr{A}}^{4,21}(F_{2},F_{2})$ respectively. We
consider the element $u_{20}\in H_{20}(B\mathbb{V}_{5},F_{2})$, which is the
following sum:

$\begin{array}[]{ll}&a_{1}^{(3)}a_{2}^{(5)}a_{3}^{(5)}a_{4}^{(5)}a_{5}^{(2)}+a_{1}^{(3)}a_{2}^{(5)}a_{3}^{(5)}a_{4}^{(6)}a_{5}^{(1)}+a_{1}^{(3)}a_{2}^{(3)}a_{3}^{(5)}a_{4}^{(8)}a_{5}^{(1)}+\vskip
6.0pt plus 2.0pt minus
2.0pta_{1}^{(3)}a_{2}^{(5)}a_{3}^{(3)}a_{4}^{(8)}a_{5}^{(1)}\\\
&\quad+a_{1}^{(3)}a_{2}^{(3)}a_{3}^{(6)}a_{4}^{(7)}a_{5}^{(1)}+a_{1}^{(3)}a_{2}^{(5)}a_{3}^{(7)}a_{4}^{(4)}a_{5}^{(1)}+a_{1}^{(3)}a_{2}^{(7)}a_{3}^{(5)}a_{4}^{(4)}a_{5}^{(1)}+\vskip
6.0pt plus 2.0pt minus
2.0pta_{1}^{(3)}a_{2}^{(3)}a_{3}^{(9)}a_{4}^{(4)}a_{5}^{(1)}\\\
&\quad+a_{1}^{(3)}a_{2}^{(9)}a_{3}^{(3)}a_{4}^{(4)}a_{5}^{(1)}+a_{1}^{(3)}a_{2}^{(3)}a_{3}^{(9)}a_{4}^{(3)}a_{5}^{(2)}+a_{1}^{(3)}a_{2}^{(9)}a_{3}^{(3)}a_{4}^{(3)}a_{5}^{(2)}+\vskip
6.0pt plus 2.0pt minus
2.0pta_{1}^{(3)}a_{2}^{(5)}a_{3}^{(9)}a_{4}^{(2)}a_{5}^{(1)}\\\
&\quad+a_{1}^{(3)}a_{2}^{(9)}a_{3}^{(5)}a_{4}^{(2)}a_{5}^{(1)}+a_{1}^{(3)}a_{2}^{(5)}a_{3}^{(10)}a_{4}^{(1)}a_{5}^{(1)}+a_{1}^{(3)}a_{2}^{(9)}a_{3}^{(6)}a_{4}^{(1)}a_{5}^{(1)}+\vskip
6.0pt plus 2.0pt minus
2.0pta_{1}^{(3)}a_{2}^{(3)}a_{3}^{(11)}a_{4}^{(2)}a_{5}^{(1)}\\\
&\quad+a_{1}^{(3)}a_{2}^{(11)}a_{3}^{(3)}a_{4}^{(2)}a_{5}^{(1)}+a_{1}^{(3)}a_{2}^{(5)}a_{3}^{(5)}a_{4}^{(3)}a_{5}^{(4)}+a_{1}^{(3)}a_{2}^{(5)}a_{3}^{(3)}a_{4}^{(5)}a_{5}^{(4)}+\vskip
6.0pt plus 2.0pt minus
2.0pta_{1}^{(3)}a_{2}^{(3)}a_{3}^{(5)}a_{4}^{(5)}a_{5}^{(4)}\\\
&\quad+a_{1}^{(3)}a_{2}^{(3)}a_{3}^{(12)}a_{4}^{(1)}a_{5}^{(1)}+a_{1}^{(3)}a_{2}^{(11)}a_{3}^{(4)}a_{4}^{(1)}a_{5}^{(1)}+a_{1}^{(3)}a_{2}^{(7)}a_{3}^{(8)}a_{4}^{(1)}a_{5}^{(1)}+\vskip
6.0pt plus 2.0pt minus
2.0pta_{1}^{(3)}a_{2}^{(7)}a_{3}^{(7)}a_{4}^{(1)}a_{5}^{(2)}\\\
&\quad+a_{1}^{(3)}a_{2}^{(13)}a_{3}^{(2)}a_{4}^{(1)}a_{5}^{(1)}+a_{1}^{(3)}a_{2}^{(14)}a_{3}^{(1)}a_{4}^{(1)}a_{5}^{(1)}+a_{1}^{(3)}a_{2}^{(6)}a_{3}^{(5)}a_{4}^{(3)}a_{5}^{(3)}+\vskip
6.0pt plus 2.0pt minus
2.0pta_{1}^{(3)}a_{2}^{(5)}a_{3}^{(3)}a_{4}^{(6)}a_{5}^{(3)}\\\
&\quad+a_{1}^{(3)}a_{2}^{(3)}a_{3}^{(6)}a_{4}^{(5)}a_{5}^{(3)}+a_{1}^{(3)}a_{2}^{(6)}a_{3}^{(3)}a_{4}^{(3)}a_{5}^{(5)}+a_{1}^{(3)}a_{2}^{(3)}a_{3}^{(3)}a_{4}^{(6)}a_{5}^{(5)}+\vskip
6.0pt plus 2.0pt minus
2.0pta_{1}^{(3)}a_{2}^{(3)}a_{3}^{(6)}a_{4}^{(3)}a_{5}^{(5)}\\\
&\quad+a_{1}^{(3)}a_{2}^{(5)}a_{3}^{(3)}a_{4}^{(3)}a_{5}^{(6)}+a_{1}^{(3)}a_{2}^{(3)}a_{3}^{(5)}a_{4}^{(3)}a_{5}^{(6)}+a_{1}^{(3)}a_{2}^{(3)}a_{3}^{(3)}a_{4}^{(5)}a_{5}^{(6)}+\vskip
6.0pt plus 2.0pt minus
2.0pta_{1}^{(3)}a_{2}^{(3)}a_{3}^{(3)}a_{4}^{(3)}a_{5}^{(8)}\\\
&\quad+a_{1}^{(3)}a_{2}^{(3)}a_{3}^{(3)}a_{4}^{(4)}a_{5}^{(7)}+a_{1}^{(3)}a_{2}^{(3)}a_{3}^{(5)}a_{4}^{(2)}a_{5}^{(7)}+a_{1}^{(3)}a_{2}^{(3)}a_{3}^{(6)}a_{4}^{(1)}a_{5}^{(7)}+\vskip
6.0pt plus 2.0pt minus
2.0pta_{1}^{(3)}a_{2}^{(3)}a_{3}^{(3)}a_{4}^{(9)}a_{5}^{(2)}\\\
&\quad+a_{1}^{(3)}a_{2}^{(3)}a_{3}^{(3)}a_{4}^{(10)}a_{5}^{(1)}+a_{1}^{(3)}a_{2}^{(5)}a_{3}^{(3)}a_{4}^{(7)}a_{5}^{(2)}+a_{1}^{(3)}a_{2}^{(5)}a_{3}^{(7)}a_{4}^{(3)}a_{5}^{(2)}+a_{1}^{(3)}a_{2}^{(7)}a_{3}^{(5)}a_{4}^{(3)}a_{5}^{(2)}.\end{array}$

Then it is $\mathscr{A}^{+}$-annihilated. Indeed, it is not difficult to
verify that $(u_{20})Sq^{i}=0$ for $i=1,2.$ A direct calculation shows that
$(u_{20})Sq^{4}$ is the following sum:
$\begin{array}[]{ll}&a_{1}^{(3)}a_{2}^{(3)}a_{3}^{(5)}a_{4}^{(3)}a_{5}^{(2)}+a_{1}^{(3)}a_{2}^{(5)}a_{3}^{(3)}a_{4}^{(3)}a_{5}^{(2)}+a_{1}^{(3)}a_{2}^{(3)}a_{3}^{(3)}a_{4}^{(6)}a_{5}^{(1)}+\vskip
6.0pt plus 2.0pt minus
2.0pta_{1}^{(3)}a_{2}^{(3)}a_{3}^{(5)}a_{4}^{(4)}a_{5}^{(1)}\\\
&\quad+a_{1}^{(3)}a_{2}^{(3)}a_{3}^{(3)}a_{4}^{(6)}a_{5}^{(1)}+a_{1}^{(3)}a_{2}^{(5)}a_{3}^{(3)}a_{4}^{(4)}a_{5}^{(1)}+a_{1}^{(3)}a_{2}^{(3)}a_{3}^{(3)}a_{4}^{(6)}a_{5}^{(1)}+\vskip
6.0pt plus 2.0pt minus
2.0pta_{1}^{(3)}a_{2}^{(3)}a_{3}^{(7)}a_{4}^{(2)}a_{5}^{(1)}\\\
&\quad+a_{1}^{(3)}a_{2}^{(7)}a_{3}^{(3)}a_{4}^{(2)}a_{5}^{(1)}+a_{1}^{(3)}a_{2}^{(3)}a_{3}^{(5)}a_{4}^{(4)}a_{5}^{(1)}+a_{1}^{(3)}a_{2}^{(3)}a_{3}^{(7)}a_{4}^{(2)}a_{5}^{(1)}+\vskip
6.0pt plus 2.0pt minus
2.0pta_{1}^{(3)}a_{2}^{(5)}a_{3}^{(3)}a_{4}^{(4)}a_{5}^{(1)}\\\
&\quad+a_{1}^{(3)}a_{2}^{(7)}a_{3}^{(3)}a_{4}^{(2)}a_{5}^{(1)}+a_{1}^{(3)}a_{2}^{(3)}a_{3}^{(5)}a_{4}^{(3)}a_{5}^{(2)}+a_{1}^{(3)}a_{2}^{(5)}a_{3}^{(3)}a_{4}^{(3)}a_{5}^{(2)}+\vskip
6.0pt plus 2.0pt minus
2.0pta_{1}^{(3)}a_{2}^{(5)}a_{3}^{(5)}a_{4}^{(2)}a_{5}^{(1)}\\\
&\quad+a_{1}^{(3)}a_{2}^{(3)}a_{3}^{(7)}a_{4}^{(2)}a_{5}^{(1)}+a_{1}^{(3)}a_{2}^{(5)}a_{3}^{(5)}a_{4}^{(2)}a_{5}^{(1)}+a_{1}^{(3)}a_{2}^{(7)}a_{3}^{(3)}a_{4}^{(2)}a_{5}^{(1)}+\vskip
6.0pt plus 2.0pt minus
2.0pta_{1}^{(3)}a_{2}^{(5)}a_{3}^{(6)}a_{4}^{(1)}a_{5}^{(1)}\\\
&\quad+a_{1}^{(3)}a_{2}^{(5)}a_{3}^{(6)}a_{4}^{(1)}a_{5}^{(1)}+a_{1}^{(3)}a_{2}^{(3)}a_{3}^{(7)}a_{4}^{(2)}a_{5}^{(1)}+a_{1}^{(3)}a_{2}^{(7)}a_{3}^{(3)}a_{4}^{(2)}a_{5}^{(1)}+\vskip
6.0pt plus 2.0pt minus
2.0pta_{1}^{(3)}a_{2}^{(3)}a_{3}^{(3)}a_{4}^{(3)}a_{5}^{(4)}\\\
&\quad+a_{1}^{(3)}a_{2}^{(3)}a_{3}^{(5)}a_{4}^{(3)}a_{5}^{(2)}+a_{1}^{(3)}a_{2}^{(5)}a_{3}^{(3)}a_{4}^{(3)}a_{5}^{(2)}+a_{1}^{(3)}a_{2}^{(3)}a_{3}^{(3)}a_{4}^{(3)}a_{5}^{(4)}+\vskip
6.0pt plus 2.0pt minus
2.0pta_{1}^{(3)}a_{2}^{(3)}a_{3}^{(3)}a_{4}^{(5)}a_{5}^{(2)}\\\
&\quad+a_{1}^{(3)}a_{2}^{(5)}a_{3}^{(3)}a_{4}^{(3)}a_{5}^{(2)}+a_{1}^{(3)}a_{2}^{(3)}a_{3}^{(3)}a_{4}^{(3)}a_{5}^{(4)}+a_{1}^{(3)}a_{2}^{(3)}a_{3}^{(3)}a_{4}^{(5)}a_{5}^{(2)}+\vskip
6.0pt plus 2.0pt minus
2.0pta_{1}^{(3)}a_{2}^{(3)}a_{3}^{(5)}a_{4}^{(3)}a_{5}^{(2)}\\\
&\quad+a_{1}^{(3)}a_{2}^{(7)}a_{3}^{(4)}a_{4}^{(1)}a_{5}^{(1)}+a_{1}^{(3)}a_{2}^{(7)}a_{3}^{(4)}a_{4}^{(1)}a_{5}^{(1)}+a_{1}^{(3)}a_{2}^{(3)}a_{3}^{(3)}a_{4}^{(3)}a_{5}^{(4)}+\vskip
6.0pt plus 2.0pt minus
2.0pta_{1}^{(3)}a_{2}^{(3)}a_{3}^{(3)}a_{4}^{(5)}a_{5}^{(2)}\\\
&\quad+a_{1}^{(3)}a_{2}^{(3)}a_{3}^{(3)}a_{4}^{(6)}a_{5}^{(1)}+a_{1}^{(3)}a_{2}^{(3)}a_{3}^{(3)}a_{4}^{(5)}a_{5}^{(2)},\end{array}$
and therefore by the unstable condition, $u_{20}\in
P_{\mathscr{A}}(H_{20}(B\mathbb{V}_{5},F_{2})).$ Then, using the differential
(1.4) the representation of the rank 5 transfer over the lambda algebra we see
that
$\begin{array}[]{ll}\psi_{5}(u_{20})&=\lambda_{3}(\lambda_{3}^{3}\lambda_{8}+(\lambda_{3}\lambda_{5}^{2}+\lambda_{3}^{2}\lambda_{7})\lambda_{4}+\lambda_{7}\lambda_{5}\lambda_{3}\lambda_{2}+\lambda_{3}^{2}\lambda_{5}\lambda_{6})\\\
&\quad+\lambda_{3}^{2}\lambda_{5}\lambda_{4}\lambda_{5}+\lambda_{3}^{2}\lambda_{5}\lambda_{3}\lambda_{6}+\lambda_{3}^{2}\lambda_{5}\lambda_{6}\lambda_{3}\\\
&=\lambda_{3}\overline{e}_{0}+\partial(\lambda_{3}^{2}\lambda_{5}\lambda_{10}+\lambda_{3}^{2}\lambda_{12}\lambda_{3}+\lambda_{3}\lambda_{4}\lambda_{7}^{2}+\lambda_{3}\lambda_{0}\lambda_{11}\lambda_{7}).\end{array}$
is a cycle in $\Lambda^{5,20}$ and is a representative of the non-zero
$h_{2}e_{0}\in{\rm Ext}_{\mathscr{A}}^{5,5+20}(F_{2},F_{2}).$ Thus,
$h_{2}e_{0}$ is in the image of $Tr_{5}.$

The case (iii). We see that
$\lambda_{1}\in\Lambda^{1,1},\,\lambda_{15}\in\Lambda^{1,15}$ and
$\lambda_{3}^{2}\lambda_{2}\in\Lambda^{3,8}$ are cycles in the algebra
$\Lambda$ and that $h_{1}=[\lambda_{1}]\in{\rm
Ext}_{\mathscr{A}}^{1,2}(F_{2},F_{2}),\,h_{4}=[\lambda_{15}]\in{\rm
Ext}_{\mathscr{A}}^{1,16}(F_{2},F_{2})$ and
$c_{0}=[\lambda_{3}^{2}\lambda_{2}]\in{\rm
Ext}_{\mathscr{A}}^{3,11}(F_{2},F_{2}).$ We consider the following element in
$H_{24}(B\mathbb{V}_{5},F_{2})$:

$u_{24}=a_{1}^{(1)}a_{2}^{(15)}a_{3}^{(3)}a_{4}^{(3)}a_{5}^{(2)}+a_{1}^{(1)}a_{2}^{(15)}a_{3}^{(3)}a_{4}^{(4)}a_{5}^{(1)}+a_{1}^{(1)}a_{2}^{(15)}a_{3}^{(5)}a_{4}^{(2)}a_{5}^{(1)}+a_{1}^{(1)}a_{2}^{(15)}a_{3}^{(6)}a_{4}^{(1)}a_{5}^{(1)}.$

Then, it is $\mathscr{A}^{+}$-annihilated. Indeed, by the unstable condition,
we need only to consider the effects of the Steenrod squares $Sq^{1}$ and
$Sq^{2}.$ A direct computations shows that

$\begin{array}[]{ll}(u_{24})Sq^{1}&=a_{1}^{(1)}a_{2}^{(15)}a_{3}^{(3)}a_{4}^{(3)}a_{5}^{(1)}+a_{1}^{(1)}a_{2}^{(15)}a_{3}^{(3)}a_{4}^{(3)}a_{5}^{(1)}\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt&\quad+a_{1}^{(1)}a_{2}^{(15)}a_{3}^{(5)}a_{4}^{(1)}a_{5}^{(1)}+a_{1}^{(1)}a_{2}^{(15)}a_{3}^{(5)}a_{4}^{(1)}a_{5}^{(1)}=0,\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt(u_{24})Sq^{2}&=a_{1}^{(1)}a_{2}^{(15)}a_{3}^{(3)}a_{4}^{(2)}a_{5}^{(1)}+a_{1}^{(1)}a_{2}^{(15)}a_{3}^{(3)}a_{4}^{(2)}a_{5}^{(1)}=0.\end{array}$

Now, using the representation in $\Lambda$ of the fifth algebraic transfer, we
get

$\begin{array}[]{ll}\vskip 6.0pt plus 2.0pt minus
2.0pt\psi_{5}(a_{1}^{(1)}a_{2}^{(15)}a_{3}^{(3)}a_{4}^{(3)}a_{5}^{(2)})&=\lambda_{1}\lambda_{15}(\lambda_{3}^{2}\lambda_{2}+\lambda_{3}\lambda_{4}\lambda_{1}+\lambda_{4}\lambda_{3}\lambda_{1}),\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\psi_{5}(a_{1}^{(1)}a_{2}^{(15)}a_{3}^{(3)}a_{4}^{(4)}a_{5}^{(1)})&=\lambda_{1}\lambda_{15}(\lambda_{3}\lambda_{4}\lambda_{1}+\lambda_{4}\lambda_{3}\lambda_{1}+\lambda_{5}\lambda_{2}\lambda_{1}),\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\psi_{5}(a_{1}^{(1)}a_{2}^{(15)}a_{3}^{(5)}a_{4}^{(2)}a_{5}^{(1)})&=\lambda_{1}\lambda_{15}(\lambda_{5}\lambda_{2}\lambda_{1}+\lambda_{6}\lambda_{1}^{2}),\\\
\vskip 6.0pt plus 2.0pt minus
2.0pt\psi_{5}(a_{1}^{(1)}a_{2}^{(15)}a_{3}^{(6)}a_{4}^{(1)}a_{5}^{(1)})&=\lambda_{1}\lambda_{15}\lambda_{1}\lambda_{6}\lambda_{1}^{2},\end{array}$

Then, since $u_{24}\in P_{\mathscr{A}}(H_{24}(B\mathbb{V}_{5},F_{2})),$
$\psi_{5}(u_{24})=\lambda_{1}\lambda_{15}\lambda_{3}^{2}\lambda_{2}$ is a
cycle in $\Lambda^{5,24}$ and is a representative of the non-zero element
$h_{1}h_{4}c_{0}\in{\rm Ext}_{\mathscr{A}}^{5,29}(F_{2},F_{2}).$ This implies
that $h_{1}h_{4}c_{0}\in{\rm Im}(Tr_{5}).$ The proof of the main theorem is
completed.

∎

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# The Potential of Neural Speech Synthesis-Based Data Augmentation
for Personalized Speech Enhancement

###### Abstract

With the advances in deep learning, speech enhancement systems benefited from
large neural network architectures and achieved state-of-the-art quality.
However, speaker-agnostic methods are not always desirable, both in terms of
quality and their complexity, when they are to be used in a resource-
constrained environment. One promising way is personalized speech enhancement
(PSE), which is a smaller and easier speech enhancement problem for small
models to solve, because it focuses on a particular test-time user. To achieve
the personalization goal, while dealing with the typical lack of personal
data, we investigate the effect of data augmentation based on neural speech
synthesis (NSS). In the proposed method, we show that the quality of the NSS
system’s synthetic data matters, and if they are good enough the augmented
dataset can be used to improve the PSE system that outperforms the speaker-
agnostic baseline. The proposed PSE systems show significant complexity
reduction while preserving the enhancement quality.

Index Terms—  personalized speech enhancement, data augmentation, speech
synthesis

##  1 Introduction

Designing algorithms for speech enhancement (SE) is a long-standing research
problem in which the current state-of-the-art methods use deep neural networks
(DNNs) [1, 3, 4, 5, 6]. DNN-based noise suppression algorithms typically
utilize a training set prepared by artificially mixing arbitrary noise sounds
with clean speech signals from many different speakers. As a result, fully-
trained SE systems attempt to enhance any speech within a given input mixture.
These models can generalize to the unseen test speakers if the model’s
computational capacity is large enough to encompass the variations found in
thousands of speakers and noise types. Therefore, generalist models come at
the cost of increased test-time complexity.

Recent studies have explored methods for developing target speaker extraction
models, which employ a conditioning mechanism to focus on the target speaker
out of a mixture of multiple talkers and noise sources. A straightforward
strategy for targeting on a particular speaker is to condition the model with
a speaker embedding [7], inferred from around 5 to 10 seconds of enrollment
data (i.e., clean speech) of the target speaker [8]. Likewise, embedding-based
conditioning frameworks effectively merge the tasks of speaker extraction with
speech enhancement [9, 6].

A certain type of personalized speech enhancement (PSE) method seeks the
potential benefits of personalizing the model beyond just better performance.
Its goal is to reduce the model complexity while preserving the enhancement
quality. It is possible because on the one hand, learning to enhance a single
speaker is simpler than enhancing many speakers, allowing for less complex
models. Therefore, PSE may be seen as a model compression mechanism, as more-
efficient PSE models may be deployed in place of larger speaker-agnostic SE
models without sacrificing performance [10, 11, 12, 13]. On the other hand,
personalization is a challenging optimization task unless it gets help from
the enrollment procedure that can provide the target speaker’s information.
However, the reference signals acquired in this way come at the cost of
untrustworthy recording quality and privacy concerns. Consequently, reducing
the amount of the target speaker’s clean speech is an additional constraint
for PSE systems toward data efficiency.

In this work, we investigate a novel data augmentation strategy for PSE
systems using neural speech synthesis (NSS) data. We consider the scenario in
which a PSE model has access to a very limited amount of speech data from the
target speaker, e.g., 5 seconds of clean reference speech. We investigate
whether this amount is good enough for some off-the-shelf NSS systems to
generate clean speech signals, while preserving the speaker’s identity. In
this constrained condition, their synthesis quality must vary depending on the
model’s performance, and there are various ways to evaluate the perceptual
quality of synthesized speech, such as naturalness, intelligibility,
personality, emotions, etc. Hence, our goal is to analyze the correlation
between the varying quality of synthesized speech and the PSE performance
using this augmented dataset. We compare two different NSS-augmented datasets
and their usability on PSE models. Furthermore, we also compare the proposed
approach against non-personalized SE models as well as self-supervised PSE
models [13]. Our results show that NSS dataset augmentation is useful for PSE,
especially in cases where the model is too small to generalize well to the
test speaker. We also observe that the quality of synthesized speech impacts
PSE performance. Our findings are consistent across different model sizes,
where NSS-augmented PSE models outperform speaker-agnostic models. This
suggests that sub-optimal NSS synthesis is still advantageous in the context
of personalizing speech enhancement systems with a simpler training framework
compared to existing methods.

Set | Subset | Duration | Quantity | Description | Corpus
---|---|---|---|---|---
$\mathbb{G}$ | $\mathbb{G}_{\text{tr}}$ | 443 h | 1132 spkrs | Clean speech from many anonymous speakers | LibriSpeech [14]
$\mathbb{G}_{\text{vl}}$ | 8h | 20 spkrs
$\mathbb{S}^{(1:20)}$ | $\mathbb{S}^{(1:20)}_{\text{tr+vl}}$ |  $5\times 30$ sec/spkr | 20 spkrs | Target speakers set used for synthesis | LibriSpeech [14]
$\mathbb{S}^{(1:20)}_{\text{te}}$ | 30 sec/spk | 20 spkrs | Clean speech target speaker set only used for evaluation
$\mathbb{S}^{(21:26)}$ | $\mathbb{S}^{(21:26)}_{\text{tr+vl}}$ | 3 sec/spkr | 6 spkrs | Target speaker set used for synthesis | LibriSpeech [14]
$\mathbb{S}^{(21:26)}_{\text{te}}$ |  $\sim$8 sec/spk | 6 spkrs | Clean speech target speaker set only used for evaluation
$\tilde{\mathbb{S}}^{(1:20)}$ | $\tilde{\mathbb{S}}^{(1:20)}_{\text{tr}}$ | 60 sec/spkr | 20 spkrs | Synthesized target speaker utterances (YourTTS) |
$\tilde{\mathbb{S}}^{(1:20)}_{\text{vl}}$ | 30 sec/spkr | 20 spkrs |
$\tilde{\mathbb{S}}^{(21:26)}$ | $\tilde{\mathbb{S}}^{(21:26)}_{\text{tr}}$ | 21 sec/spkr | 6 spkrs |  Synthesized target speaker utterances (YourTTS, AudioLM) |
$\tilde{\mathbb{S}}^{(21:26)}_{\text{vl}}$ | 7 sec/spkr | 6 spkrs |
$\mathbb{N}$ | $\mathbb{N}_{\text{tr}}$ | 5h | 616 noises | Injection noises used during SE and PSE training | MUSAN [15]
$\mathbb{N}_{\text{vl}}$ | 0.5h | 60 noises
$\mathbb{N}_{\text{te}}$ | 0.5h | 60 noises |  Injection noises not seen during training, used to prepare speaker-specific test sets
$\mathbb{T}$ |  |  | 88156 sentences |  Sentences used for synthesis in both {tr, vl} partitions of $\tilde{\mathbb{S}}^{(1:20)}$ and $\tilde{\mathbb{S}}^{(21:26)}$. |  VCTK [16]

Table 1: Description of the datasets used in experiments.

##  2 Proposed Method

The generalists: A fully-supervised framework for training SE models defines a
large set $\mathbb{G}$ of clean utterances from many anonymous speakers. They
are mixed with various noise signals sampled from noise corpus $\mathbb{N}$ at
random signal-to-noise ratios (SNRs) to simulate arbitrary contaminated
speech, i.e., $x=s+n$ where $s\in\mathbb{G}$ and $n\in\mathbb{N}$. The SE
model is a mapping function $f(\cdot)$ with trainable parameters
$\mathcal{W}_{\text{SE}}$ that aims to recover $s$ from $x$, i.e.,
$f(x;\mathcal{W}_{\text{SE}})=y\approx s$. Our experiments use negative
signal-to-distortion ratio (SDR) [17] as the loss function for the SE system:

$\mathcal{L}_{\text{Neg-
SDR}}(\hat{v}\|v)=-10\log_{10}\left[\frac{\sum_{t}(v_{t})^{2}}{\sum_{t}(v_{t}-\hat{v}_{t})^{2}}\right],$
(1)

where $v$ is the clean signal and $\hat{v}$ is the estimated signal. We refer
to such models as generalists and use them as a baseline.

PSE using NSS: The transfer learning approach can introduce a bias towards a
particular speaker. As opposed to random initialization, transfer learning
deems to be highly beneficial for specialists if the speaker-specific clean
speech is available for finetuning, which we denote by $\mathbb{S}^{(i)}$ with
a speaker index $i$. We assume $|\mathbb{S}^{(i)}|\ll|\mathbb{G}|$ due to any
technical challenges in acquiring clean recordings or privacy concerns of the
user, e.g., a few seconds. Therefore, we propose the data augmentation
technique for PSE using neural speech synthesis (NSS) systems. Here, we assume
that the NSS system is a text-to-speech (TTS) system that can generate any
utterances that sound similar to the target speaker.

Data augmentation via NSS: Our goal in using the NSS systems is to generate as
many new utterances per speaker as needed using an NSS model $g(\cdot)$ with
pretrained parameters $\mathcal{W}_{\text{NSS}}$. A sentence is sampled from a
large set $t\in\mathbb{T}$ fed to the NSS model as
$\tilde{s}=g(t;\mathcal{W}_{\text{NSS}}|s)$, where $s$ works as a condition to
preserve target speaker’s personality as $s\in\mathbb{S}^{(i)}$. Thus, we get
a synthesized set of target speaker utterances $\tilde{\mathbb{S}}^{(i)}$ of
any predefined size. In training phase $\tilde{s}$ is used the same way as $s$
to construct a noisy mixture $\tilde{x}=\tilde{s}+n$. We define the task of
data augmented PSE as
$f(\tilde{x};\mathcal{W}_{\text{PSE}})=\tilde{y}\approx\tilde{s}$. We
hypothesize that the quality of the speech synthesis system affects the
performance of the PSE, where two main factors are involved in defining the
NSS systems’ performance: the general quality of the speech signal and
personality. While these two concepts are not straightforward to quantify, we
empirically show that the two NSS systems, in comparison, are with different
performances, and they are correlated to their usefulness in the PSE task. Our
experiments address the following research questions: (a) does the quality of
an NSS system impact its usefulness towards PSE? (b) for each of the tested
NSS systems, how much generated data is needed in order for a transfer
learning-based PSE system to perform comparably to large generalist models?
(c) how much lossless compression can we achieve with NSS-augmented PSE models
in comparison with generalist SE models?

##  3 Experimental Setup

In our experiments, we take the baseline speaker-agnostic generalist model
proposed in [13] and finetune it with utterances synthesized by two off-the-
shelf NSS models that exhibit varying performances. We repeat the finetuning
process for multiple target speakers to comprehensively assess the proposed
method. The speech enhancement model complexity between four preset sizes in
order to investigate the merit of PSE in terms of model compression.

###  3.1 NSS Models

Our first NSS system is YourTTS [18], a multi-lingual multi-speaker TTS model
composed of a transformer-based encoder, a normalizing flow decoder, and a
HiFi-GAN vocoder. Second, we choose AudioLM [19], which is a novel auto-
regressive speech synthesis system based on a language modeling (LM) approach.
As opposed to YourTTS, AudioLM does not require textual data to synthesize
speech, because the LM generates a word sequence on the fly. Instead of
attempting to reproduce this model, whose pretrained version is unavailable,
we use the small number of examples published in the authors’
website111https://google-research.github.io/seanet/audiolm/examples/. Although
these examples are a small dataset, their relative higher quality provides an
interesting comparison point to the large quantity of YourTTS results.

###  3.2 Datasets

Table 1 describes all the datasets used in our experiments. The subscripts
‘tr’, ‘vl’, and ‘te’ denote training, validation, and test subsets,
respectively. We chose to work with two different subsets of speakers
$\mathbb{S}^{(1:20)}$ and $\mathbb{S}^{(21:26)}$ for better comparison.
$\mathbb{S}^{(1:20)}$ contains 20 speakers from LibriSpeech’s [14] train-
clean-100 subset to match the training setup in [13]. As for the second set
$\mathbb{S}^{(21:26)}$, we are based on the high-quality audio samples
generated by AudioLM available online. Since the identity of those speakers is
not fully known, we work with only 6 identifiable speakers from test-clean,
$\mathbb{S}^{(21:26)}$.

We generate the augmented version
$\tilde{\mathbb{S}}^{(1:20)}_{\text{YourTTS}}$ using YourTTS. For a given
speaker ID $i$, the system takes a 5-second-long clean reference audio of the
target speaker from $s\sim\mathbb{S}^{(i)}$ and a random sentence
$t\sim\mathbb{T}$, which are the input pair $(s,t)$ to YourTTS that generates
$\tilde{s}$. We repeat the process until we collect a pre-defined duration of
speech for the target speaker.

Data augmentation for the second subset ${\mathbb{S}}^{(21:26)}$ is conducted
using both NSS systems. First, as for AudioLM, we conveniently repurpose their
publicly available synthesis results. The 7-second-long AudioLM utterances in
$\tilde{\mathbb{S}}_{\text{AudioLM}}^{(21:26)}$ are generated from a 3 seconds
of audio prompt; four such synthesized examples are available per speaker,
making 28 seconds of synthesized audio in total. We generate another augmented
set $\tilde{\mathbb{S}}_{\text{YourTTS}}^{(21:26)}$ using the same 3-second
long prompt, but this time we can synthesize as long utterances as we want
because we have access to the pretrained model. All the utterances are
resampled to 16 kHz.

For noise subsets we use sound-bible partition of MUSAN [15] only for test-
time mixtures $\mathbb{N}_{\text{te}}$, 60 noise files from free-sound folder
are set aside for validation mixtures $\mathbb{N}_{\text{vl}}$ and the rest of
the signals from free-sound are used for training time noisy mixtures
$\mathbb{N}_{\text{tr}}$. Mixturre SNR is chosen from $[-5,5]$ dB at random.

###  3.3 Implementation

All models in our experiment are based on the well-known monoaural time-domain
source separation DNN, ConvTasNet (CTN) [20]. Following [13], we define four
different sizes of the model: large, medium, small and tiny. With each size
variant the number of channels in the bottleneck module and convolutional
blocks is reduced by factor of 2. Consequently, the number of trainable
parameters in each model is 138.8K, 224.1K, 437.8K, and 1M, from the tiny to
the large models, respectively. We assume smaller models are more suitable for
on-device speech enhancement.

The generalist models are pretrained using the Asteroid implementation of
ConvTasNet [21] as in [13] and then finetuned using the proposed augmented
datasets. We use Adam optimizer [22] with a learning rate of $1\mathrm{e}{-5}$
and batch size of 8. After seeing 500 mixtures the model is validated on a
fixed set of mixtures depending on the size of the training set: for
$\tilde{\mathbb{S}}^{(1:20)}$ the validation set consists of 100 mixtures, for
$\tilde{\mathbb{S}}_{\text{AudioLM}}^{(21:26)}$ and
$\tilde{\mathbb{S}}_{\text{YourTTS}}^{(21:26)}$ the validation set consisted
of 30 mixtures. The training continues until there is no validation SDR
improvement for 5000 mixtures. To show the improvement of finetuned models we
report SDR, PESQ [23] and eSTOI [24] for both test and validation sets.

Subset | MOS (est.) | Cosine Similarity
---|---|---
$\mathbb{S}^{(1:20)}_{\text{YourTTS}}$; 60 sec. | 3.78 | 0.80
$\mathbb{S}^{(1:20)}_{\text{YourTTS}}$; 120 sec. | 3.81 | 0.80
$\mathbb{S}^{(21:26)}_{\text{AudioLM}}$; 21 sec. | 4.35 | 0.96
$\mathbb{S}^{(21:26)}_{\text{YourTTS}}$; 21 sec. | 4.18 | 0.87
$\mathbb{S}^{(21:26)}_{\text{YourTTS}}$; 60 sec. | 4.12 | 0.88

Table 2: Quality evaluation of synthesized speech for YourTTS and AudioLM
generated sets.

Experiment #1 | Size | SDRI (te) | SDR (te) | eSTOI (te) | PESQ (te) | SDRI (vl) | SDR (vl) | eSTOI (vl) | PESQ (vl)
---|---|---|---|---|---|---|---|---|---
Baseline Generalist, trained from $\mathbb{G}$, tested on $\mathbb{S}^{(i)}$ $(i\in\\{1,\ldots,20\\})$ | L | 9.84 | 10.38 | 0.70 | 1.68 | 11.76 | 11.25 | 0.81 | 2.29
M | 9.74 | 10.25 | 0.69 | 1.59 | 11.34 | 10.83 | 0.79 | 2.19
S | 8.75 | 9.29 | 0.66 | 1.50 | 10.45 | 9.94 | 0.76 | 2.01
T | 7.89 | 8.43 | 0.64 | 1.42 | 9.70 | 9.19 | 0.74 | 1.89
PSE Model, | L | 10.13 | 10.65 | 0.69 | 1.65 | 12.12 | 11.61 | 0.81 | 2.29
finetuned from $\tilde{\mathbb{S}}^{(i)}_{\text{YourTTS}}$, | M | 9.53 | 10.04 | 0.68 | 1.59 | 11.77 | 11.27 | 0.80 | 2.22
$|\tilde{\mathbb{S}}^{(i)}_{\text{YourTTS}}|=$60 sec. | S | 8.69 | 9.20 | 0.64 | 1.47 | 10.86 | 10.35 | 0.77 | 2.01
$(i\in\\{1,\ldots,20\\})$ | T | 7.84 | 8.35 | 0.62 | 1.38 | 10.83 | 9.61 | 0.74 | 1.85
PSE Model, | L | 9.98 | 10.49 | 0.68 | 1.61 | 12.12 | 11.57 | 0.82 | 2.27
finetuned from $\tilde{\mathbb{S}}^{(i)}_{\text{YourTTS}}$ | M | 9.39 | 9.90 | 0.67 | 1.56 | 11.87 | 11.32 | 0.80 | 2.19
$|\tilde{\mathbb{S}}^{(i)}_{\text{YourTTS}}|=$120 sec. | S | 8.56 | 9.07 | 0.64 | 1.45 | 10.93 | 10.38 | 0.77 | 1.97
$(i\in\\{1,\ldots,20\\})$ | T | 7.47 | 7.98 | 0.61 | 1.34 | 10.10 | 9.55 | 0.74 | 1.80
PseudoSE +DP [13], | L | 10.40 | 10.91 | 0.72 | 1.62 |  |  |  |
trained via SSL, | M | 10.19 | 10.70 | 0.71 | 1.58 |  |  |  |
tested on $\mathbb{S}^{(i)}$ | S | 9.88 | 10.39 | 0.70 | 1.55 |  |  |  |
$(i\in\\{1,\ldots,20\\})$ | T | 9.40 | 9.91 | 0.68 | 1.49 |  |  |  |

Table 3: Experimental results for speaker set 1. Best results for each size
and model are indicated in bold.

Experiment #2 | Size | SDRI (te) | SDR (te) | eSTOI (te) | PESQ (te) | SDRI (vl) | SDR (vl) | eSTOI (vl) | PESQ (vl)
---|---|---|---|---|---|---|---|---|---
Baseline Generalist, trained from $\mathbb{G}$, tested on $\mathbb{S}^{(i)}$ $(i\in\\{21,\ldots,26\\})$ | L | 11.26 | 10.75 | 0.69 | 1.38 | 14.23 | 13.20 | 0.75 | 1.71
M | 10.12 | 9.61 | 0.67 | 1.32 | 12.43 | 11.40 | 0.73 | 1.60
S | 10.31 | 9.80 | 0.65 | 1.28 | 12.87 | 11.83 | 0.72 | 1.55
T | 8.88 | 8.38 | 0.64 | 1.24 | 11.29 | 10.26 | 0.71 | 1.46
PSE Model, | L | 11.81 | 11.31 | 0.69 | 1.43 | 14.69 | 13.66 | 0.75 | 1.75
finetuned from $\tilde{\mathbb{S}}^{(i)}_{\text{AudioLM}}$ | M | 11.36 | 10.86 | 0.69 | 1.38 | 14.08 | 13.05 | 0.75 | 1.69
$|\tilde{\mathbb{S}}^{(i)}_{\text{AudioLM}}|=$21 sec. | S | 10.73 | 10.22 | 0.66 | 1.31 | 13.37 | 12.34 | 0.72 | 1.57
$(i\in\\{21,\ldots,26\\})$ | T | 9.98 | 9.47 | 0.65 | 1.26 | 12.48 | 11.45 | 0.71 | 1.48
PSE Model, | L | 10.99 | 10.48 | 0.64 | 1.37 | 14.30 | 13.81 | 0.74 | 1.68
finetuned from $\tilde{\mathbb{S}}^{(i)}_{\text{YourTTS}}$ | M | 10.51 | 10.00 | 0.67 | 1.33 | 13.87 | 13.37 | 0.73 | 1.63
$|\tilde{\mathbb{S}}^{(i)}_{\text{YourTTS}}|=$21 sec. | S | 9.89 | 9.38 | 0.64 | 1.27 | 13.03 | 12.54 | 0.70 | 1.51
$(i\in\\{21,\ldots,26\\})$ | T | 9.25 | 8.75 | 0.64 | 1.24 | 11.88 | 11.39 | 0.70 | 1.44
PSE Model, | L | 11.40 | 10.89 | 0.69 | 1.39 | 12.89 | 12.45 | 0.74 | 1.71
finetuned from $\tilde{\mathbb{S}}^{(i)}_{\text{YourTTS}}$ | M | 10.78 | 10.27 | 0.67 | 1.35 | 12.54 | 12.10 | 0.73 | 1.66
$|\tilde{\mathbb{S}}^{(i)}_{\text{YourTTS}}|=$60 sec. | S | 10.22 | 9.72 | 0.63 | 1.30 | 11.69 | 11.25 | 0.70 | 1.55
$(i\in\\{21,\ldots,26\\})$ | T | 9.42 | 8.91 | 0.62 | 1.23 | 10.83 | 10.39 | 0.68 | 1.40

Table 4: Experimental results for speaker set 2. Best results for each size
and model are indicated in bold

###  3.4 Experiment #1: Comparison with other PSE methods

The goal of the experiments with $\tilde{\mathbb{S}}^{(1:20)}$ is to show the
benefit of synthetic data augmentation compared to speaker-agnostic models. We
generate two sets using YourTTS, with length 60 or 120 seconds, in order to
determine the minimal amount of synthesized data needed to achieve enhancement
performance comparable to the generalist’s. These experiments use the same
test set as generalist and specialist models in [13], allowing for a direct
comparison.

###  3.5 Experiment #2: Comparison between NSS schemes

Experiments with $\tilde{\mathbb{S}}^{(21:26)}$ aim to determine the impact of
the quality of the synthesized speech as well as the amount of high-quality
data needed for fine-tuning as opposed to lower-quality counterparts. To this
end, we finetune the generalist with the two synthesized sets
$\tilde{\mathbb{S}}^{(21:26)}_{\text{YourTTS}}$ and
$\tilde{\mathbb{S}}^{(21:26)}_{\text{AudioLM}}$, resulting in $6\times 2$ PSE
models.

##  4 Results and Discussion

###  4.1 NSS Systems Performance

Evaluating the quality of the synthesized utterances $\tilde{\mathbb{S}}$ is
infeasible due to the subjective nature of assessment and the difficulty in
conducting a listening test on a large-scale synthesized dataset.
Additionally, reliable objective metrics, such as PESQ and STOI, cannot be
applied due to the lack of clean references.

To verify our assumption about the quality difference between the two NSS
models, we use non-intrusive objective quality evaluation metrics. Instead of
the mean opinion scores (MOS), we use an open-source neural network MOS
estimator [25], with which we can indirectly compare the speech quality of the
synthesized results. In addition, to compare the personality-preservation
performance of the differently synthesized speech, we extract x-vectors with
the neural speaker encoder developed by the SpeechBrain project [26]. Then, we
compute the average cosine similarity between the synthesized utterances’
x-vector and that of the ground-truth target speaker. As can be seen in Table
2, the estimated MOS scores of YourTTS samples are lower than the quality of
AudioLM samples. Furthermore, the high cosine similarity indicates that
speaker personality is better preserved using AudioLM.

Note that these results do not necessarily imply the overall performance of
the two NSS systems in comparison, as their direct comparison is unfair due to
various reasons. For example, the AudioLM’s demo examples we collected from
the authors’ website might not correctly represent the model’s overall
quality. Meanwhile, the test speakers in $\mathbb{S}^{(1:20)}$ were already
seen by YourTTS during its pretraining, so the results might be better than
its actual performance on unseen speakers. However, if we limit the comparison
to examples used in this paper, it is convincing that AudioLM’s examples are
better than YourTTS’s. Next, we will see their influence on PSE.

###  4.2 Experiment #1

The results of Experiment #1 are summarized in Table 3. First, on the test
set, we can see that the proposed PSE models, in general, underperform the
baseline generalist model except for a few large model cases. Meanwhile, their
performance on the validation set is indeed consistently better than the
baseline. Considering that the validation set was also built based on
YourTTS’s synthesized utterances, the noticeable performance improvement on
the validation set appears to signify an overfitting case. In other words, the
finetuning-based personalization was done on the wrong speaker due to the
mismatch between the target speaker and the synthesized speech in terms of
personality. This trend does not change even if we double the size of
synthesized speech to 120 seconds—the larger augmented set actually worsens
the situation. For comparison, we reproduce the results of one of the self-
supervised learning (SSL) methods from [13], namely pseudo speech enhancement
(PseudoSE) and data purification (DP), which shows the best results even
though they do not use any clean speech of the target speaker. Experiment #1
results suggest that a well-designed SSL can outperform the finetuning-based
PSE if the data augmentation does not maintain the personality of the target
speaker.

###  4.3 Experiment #2

Experiment #2’s results are shown in Table 4. We see that the NSS-based data
augmentation, if better NSS results from AudioLM are used, improves the PSE
performance in every way with a significant margin (more than 1 dB in terms of
SDR). An interesting observation we can make here is that there is a clear gap
between the PSE models depending on which augmented set they are finetuned
from, i.e., $\tilde{\mathbb{S}}^{(i)}_{\text{AudioLM}}$ vs.
$\tilde{\mathbb{S}}^{(i)}_{\text{YourTTS}}$. This result resonates with Table
2, where the AudioLM examples showed better speech quality and personalization
performance in all metrics. In this smaller subset with six speakers, it
appears that YourTTS adapts to the target speakers better than it does to the
first speaker set in Experiment #1. Yet, there is a clear gap between the two
NSS systems’ impact on PSE. Note that the direct comparison to the PseudoSE+DP
results shown in Table 3 is impossible due to the mismatch of the two subsets.

##  5 Conclusion

Our work investigated the potential of neural speech synthesis methods for
adapting speech enhancement models towards a particular speaker. We employed
two off-the-shelf neural synthesis systems (YourTTS and AudioLM) to synthesize
new speech utterances as if they were spoken by the target speaker. Because
YourTTS and AudioLM vary in performance depending on the conditioning
mechanism and their own model capacity, we assessed the output speech quality
and personality-preservation of both systems using non-intrusive metrics. Our
experiments demonstrate that speech synthesis quality does correlate with
usefulness towards PSE. Using the best-quality synthesis dataset, we show that
it is possible to implement efficient PSE systems via a simple finetuning
approach.

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