The set of idempotents of an algebraic semigroup is closed, since every idempotent is given by the polynomial equation $e^2 = e$. For an algebraic monoid, the group of invertible elements is open and is an algebraic group.

Definition of Algebra and Associative Multiplication

The talk is based on the results of two papers by Michel Brion, each of which consists of two parts.

An algebraic semigroup is an algebraic variety $S$ with a multiplication that is a morphism and is associative. The product of elements $x$ and $y$ is written without a sign: $xy$.

An algebraic monoid is an algebraic semigroup in which there exists a neutral element with respect to multiplication. If a neutral element exists, it is unique.

Algebraic Groups and Equations

Algebraic Group and the Inverse Map

An algebraic group is a group in which every element is invertible. The map that takes an element to its inverse is a morphism. From the speaker's recollection, Bingley's book contains an exercise claiming that the inversion morphism holds automatically. If all elements of a group are invertible, the proof uses the fact that inversion is a morphism, together with the property of preimages.

Idempotents in a Semigroup

An element $e$ of a semigroup $S$ is called an idempotent if $e^2 = e$. The set of all idempotents in $S$ is denoted by $E(S)$.

The set $E(S)$ is closed in $S$. It is defined by the equations $e^2 = e$, whose left and right sides are polynomial functions of $e$.

Example: Projectors

Let $S$ be the semigroup of all matrices under multiplication. Then the idempotents in $S$ are the matrices $P$ for which $P^2 = P$. Such matrices are called projectors. They are the matrices of linear maps that project the space onto some subspace along another subspace.

Remarks on Idempotents

Statements about idempotents are often complicated, so from here on they are presented as remarks. The first remark concerns semigroup homomorphisms.

Idempotents and Subsemigroups

Let φ: A → B be a homomorphism of semigroups. This is a map compatible with multiplication: φ(ab) = φ(a)φ(b) for all a, b ∈ A.

Under a homomorphism, the image of an idempotent is an idempotent. Let e be an idempotent in A, that is, e·e = e. Then φ(e) = φ(e·e) = φ(e)·φ(e), which implies that φ(e) is an idempotent in B.

The preimage of any element x ∈ B is a subsemigroup of A if and only if x is an idempotent. Indeed, if a and b lie in φ⁻¹(x), then φ(ab) = φ(a)φ(b) = x·x. For ab to also belong to φ⁻¹(x), we need x·x = x, that is, x must be an idempotent.

Fixed Elements and Examples

Closure Condition and Subsemigroup

Let us fix the set X′ and require that it be a subsemigroup. This means the following: the product of any two elements X and Y that map into X′ also maps into X′. The mapping φ is a homomorphism by definition, so φ(XY) = φ(X)·φ(Y). Therefore, the equality linking X³ with X′, which concerns the triple product of elements, is required to hold.

Idempotents and Their Preimages

Idempotents are distinguished by the fact that their preimages do not coincide with the sets A and B, where A is a semigroup.

Definition of an Ideal

A subset of a semigroup is called an ideal if its multiplication by elements of the semigroup S on one side remains within this subset. If the condition holds for both sides, the ideal is called two-sided.

Difficulty with Factorization of an Ideal

Factorization of an ideal presents the greatest difficulty when working with this material.

One-Sided Ideals and the Identity Element

Semigroup theory is closer to algebraic geometry than group theory is. The one-sided ideals associated with an idempotent are divided into left and right ones. Which is which is not specified in the notation.

Let $e$ be an idempotent of a semigroup $S$. Two one-sided ideals are considered: $Se$ and $eS$. The set $Se$ consists of the products $xe$ over all $x \in S$, and $eS$ consists of the products $ex$ over all $x \in S$. The set $eSe$ consists of the products $exe$. Subscripts are used for notation: $S_e$, ${}_eS$, and ${}_eS_e$ correspond to these sets. One-sided multiplication by $S$ yields an ideal, whereas one-sided multiplication by $e$ does not have this property.

The elements of these sets are distinguished by the fact that $e$ is an identity for them. If $x = xe$, then $x \in Se$. If $x = ex$, then $x \in eS$. An element for which $e$ is an identity on both sides satisfies the equality $x = ex = xe$. Such an element belongs to $eSe$. Conversely, every element of $eSe$ has this property, since $e$ is idempotent.

The set $eSe$ is a subsemigroup with identity $e$. If $T$ is any submonoid of the semigroup $S$ with identity element $e$, then $T \subseteq eSe$. The identity element of a submonoid is itself an idempotent, since $e \cdot e = e$. Therefore, $eSe$ is the maximal submonoid with identity element $e$. For the one-sided sets $Se$ and $eS$, the identity property of $e$ holds only on one side.

As an example, consider a block-form idempotent matrix: the diagonal contains identity and zero blocks. Its square equals itself. If an arbitrary matrix is multiplied on the left or on the right by such a matrix, only the block corresponding to the identity blocks remains. This part is a monoid with identity element $e$. It is asserted that any submonoid of matrices is contained in $eSe$ for some idempotent $e$, not necessarily of the indicated block form.

For a subset of $eSe$ with zero or trivial multiplication structure, the following holds. If we take two elements $x$ and $y$ from this subset, their product equals $y$. Thus, the subset is a subsemigroup with trivial right multiplication.

There is also mention of a statement about linear systems. According to the author, it is considered proven, and the proof is carried out rigorously, without reliance on intuition. The formulation of the statement is not fully reconstructed in the record.

The sets $Se$, $eS$, and $eSe$ are used to understand the structure of $S$. The map $x \mapsto xe$, that is, right multiplication by $e$, has image $Se$. The author gave this image a special term, but the record breaks off before naming it.

Retractions and Projections

A retraction maps a set onto its subset. The elements of the subset remain in place, while all other elements are glued together into the subset.

Formally, a retraction of a set X onto a subset S is defined by the condition r(x) = x for every x in S.

A retraction can be compared with a projection. A retraction is a projection without a linear component.

A retraction defined on a semigroup is not required to preserve its structure. Such a map is a retraction of sets, but not of semigroups. If the map preserves multiplication, it is a retraction of semigroups.

If an element already has the form xe, where e is an idempotent, then multiplication by e does not change it: (xe)e = xe.

A retraction is a left inverse of the inclusion of the subset S into X. Calling it an inverse map is incorrect, since the inclusion is, as a rule, not invertible.

Ideals and Mappings

One-Sided Inverse Element and the Set eS

We consider the set eS, where e is an idempotent of the semigroup S. Multiplication is defined in the maximal submonoid with identity element. The condition formulated for e determines the structure of this set.

eS as an Ideal and a Mapping

The set eS is an ideal in S. Next, we consider a mapping that defines a semigroup structure on eS.

Closedness of eS under Multiplication

Take two elements ex and ey from eS. Their product equals exey. This result is multiplied on the right by e. Since e is an idempotent, it can be replaced by e in this expression.

Гомоморфизмы полугрупп

Homomorphisms of Semigroups

Отображение является гомоморфизмом полугрупп. Этот вывод следует из предыдущих рассуждений. Далее в пересказываемом фрагменте речь переходит к примеру, однако изложение примера в этом отрезке не приведено.

The mapping is a homomorphism of semigroups. This conclusion follows from the preceding reasoning. Further, in the fragment being retold, the discussion moves on to an example; however, the presentation of the example is not given in this segment.

Example with an Arbitrary Matrix

Consider an arbitrary matrix written in block form (A B; C D). Let S denote the set of such matrices. If a matrix from S is multiplied by the idempotent of the form e₀₀, the result is a matrix of the form (a 0), where a depends on the original matrix.

The mapping φ constructed in this way is not a semigroup homomorphism. The product of two matrices from the image of φ does not preserve the required form, so the homomorphism condition is not satisfied.

The situation changes if we restrict ourselves to the subset of matrices of the form (a 0). On this subset, the mapping φ is a semigroup homomorphism.

The layer over the idempotent E consists of matrices of the form e₀₀. It is precisely such matrices that, when multiplied on the left by the chosen idempotent, leave the original matrix A unchanged. If one such matrix is multiplied by another of the same kind, the result is again a matrix of the same form, that is, the product equals the first factor.

For this set, the equalities that define a right zero and a left identity element hold. The proof of the corresponding theorem follows.

Existence of idempotents in semigroups

The speaker claims that every semigroup has idempotents, but immediately clarifies that in the general case this is false. Example: the natural numbers under addition form a semigroup in which the sum n + n is never equal to n. Therefore, it has no idempotents.

For algebraic groups, there exists a proof of the existence of idempotents. According to the speaker, it is very similar to the problem for algebraic semigroups.

In the proof, it is important not to write the equality of powers of an element A of the form A^k = A^n, and not to multiply both sides by the inverse element.

Outside the algebraic case, idempotents may be absent.

The speaker intends to prove the result only over a field of odd characteristic. In his opinion, this restriction will not affect the exposition.

Closedness under multiplication

Field and characteristic

The statement is formulated over an algebraically closed field. According to the speaker, it holds over any algebraically closed field, regardless of characteristic. He notes that he does not remember exactly whether the characteristic needs to be taken into account. Possibly the arbitrary case reduces to odd characteristic, but this requires complicated arguments, which he does not go through.

Reduction to a semigroup without proper closed subsemigroups

Let $S$ be a semigroup in which an idempotent is sought. The speaker claims that one may assume $S$ contains no proper closed subsemigroups.

The justification proceeds in steps. If $S$ contains a closed subsemigroup $S_1$, we pass to it. If $S_1$ contains a closed subsemigroup $S_2$, we pass to $S_2$. The process continues as long as there are proper closed subsemigroups. It is finite, since algebraic varieties satisfy the descending chain condition. At the last step, one obtains a subsemigroup containing no proper closed subsemigroups. If it contains an idempotent, then that idempotent also belongs to the original set.

Density of the powers of an element

Let $x$ be an arbitrary element of $S$. Consider the subsemigroup generated by the powers of $x$. Its closure is a closed subsemigroup of $S$. Since there are no proper closed subsemigroups, the closure coincides with $S$. Hence the set of powers of $x$ is dense in $S$.

The same reasoning applies to the closure of any set that is closed under multiplication: it is also closed under multiplication.

Density of the shift $yS$

Let $y$ be an arbitrary element of $S$. The set $yS$ is closed under multiplication: the product $ys_1 \cdot ys_2$ equals $y\,(s_1 y s_2)$, which again lies in $yS$. Therefore the closure of $yS$ is a closed subsemigroup. Since there are no proper such subsemigroups, $yS$ is dense in $S$.

Constructibility of $yS$ and intersection with dense sets

The set $yS$ is the image of the morphism of left multiplication by $y$. The image of a morphism is constructible, so it contains a nonempty open subset. The speaker points out that this scheme does not work for groups. In a group, multiplication by $y$ is an automorphism, and its image coincides with the whole group. In a semigroup there is no such invertibility.

Thus, $yS$ contains a nonempty open subset and is also dense in $S$. A dense set intersects every nonempty open set. Consequently, these two sets have a common point.

Next, the speaker formulates a statement about the existence, for any $X$ and $Y$, of a number $n$ such that $X$ belongs to the corresponding set. In the notation, this statement is not completed.

Constructibility and Density of Sets

Decomposition of S into Layers

The set S is represented as a union of layers S_N, where N runs over the natural numbers. The layer S_N consists of all elements x such that the power x^N belongs to the given set. The union of all layers coincides with S.

Constructibility of the Layers

Each layer S_N is constructible. The justification is as follows. Consider the direct product containing the set of all pairs (x, y) satisfying x^N = y. This set forms the required part of the product. Its projection onto the first component is S_N. The image under a morphism is constructible, so S_N is constructible.

Lemma on Countable Unions

Next, we use a lemma whose proof is given later. Suppose a variety is represented as a countable union of constructible sets. Then one of these sets contains a nonempty open subset. Applied to this situation, this gives the following: there exists N such that S_N contains a nonempty open subset U.

Nesting of the Layers

The layers are nested within one another. If an element belongs to S_N, then it also belongs to S_{N+1}. Thus, S_N ⊆ S_{N+1}. The speaker notes that this simplifies the problem, but nesting by itself does not resolve the question.

A Countable Field and a Counterexample

For a countable field, the lemma in the form described is false, even under the assumption that the layers are nested. Example: if the line over a countable field is represented as a countable union of points, then at each finite step we obtain a finite set (one point, two points, three points, and so on), while the union of all steps gives the whole line. Therefore, for a countable field, the conclusion about a nonempty open subset cannot be relied upon in this form.

Density of Powers and Intersection with an Open Set

For an open subset U from a nonempty open set, another dense set is taken. It has the form {x^{Nm} : m ∈ ℕ}. This set is dense in S. Since U is open and nonempty, and the set is dense, they intersect. There exists m such that x^{Nm} lies in U, and hence also in S_N. Further in the argument, it is required to clarify what it means for x^{Nm} to belong to the layer S_N; however, this part of the proof is not given in this fragment.

Commutativity and Powers

To find an idempotent, the element s raised to the power m is denoted by the letter S. Then S^(m−1)·Z = S^(2m−2)·Z for some Z. From this equality it follows that the element in question is an idempotent, since it coincides with its own square.

The justification relies on commutativity. Powers of a single element commute with one another. If the set under consideration is dense, then its closure is commutative. Therefore, one may assume from the outset that the group is commutative.

The notion of the commutator is not used in this context.

Next, the right-hand side of the equality is transformed. The middle factor is taken from it and replaced by the expression S^m. Then the last factor S is replaced by what stands on the right. The purpose of the transformation is to obtain an element equal to its own square.

The lemma on which this reasoning is based is proved by induction on the dimension. The inequality m ≥ 1 requires a separate explanation, which is not given here.

Locally closed parts and dimension

Reduction to locally closed parts

A constructible set is by definition a union of locally closed sets. Therefore, every constructible set can be replaced by the set of its locally closed parts.

Passage to irreducible components

Every locally closed set can be replaced by its irreducible components. Thus, it suffices to consider locally closed irreducible sets.

Passage to closed sets

Each such set can be replaced by its closure. According to the speaker, this does not change the argument.

Passage to an irreducible set

Next, the set X may be assumed to be irreducible. To do this, one passes from X to an open subset that contains only one irreducible component. As a result, it suffices to prove that X coincides with one of the sets under consideration.

Induction on dimension

The proof proceeds by induction on the dimension of X.

• Dimension 1. The closed subsets of X in this case are either finite or coincide with X. Since the set under consideration is neither finite nor equal to X, such sets cannot be represented in the indicated way.
• Dimension greater than 1. The set X is intersected with a hypersurface Y that does not contain all the components of X. The intersection gives a representation of Y in a lower dimension. By the induction hypothesis, this representation is impossible, which yields a contradiction.

The speaker noted that there is a second proof of this result, which is given elsewhere.

Transfer of Idempotents under a Homomorphism

Condition on the Subsets

For the formulation, the subsets under consideration must not coincide with Y. In other words, Y must not be contained in X.

Corollary for Surjective Homomorphisms

The corollary strengthens the statement of the first remark. It concerns the case where the homomorphism φ is surjective. In this case, idempotents map to idempotents.

The corollary follows from combining the proven theorem with the two statements of the first remark.

Let e be an idempotent of the subsemigroup B. By the theorem, the subsemigroup A has an idempotent that maps to e. Thus, for every idempotent of the image, there exists an idempotent preimage.

This reasoning relies on the algebraic nature of the semigroups. For arbitrary semigroups, the statement may not hold.

Examples

The embedding of the natural numbers into the integers does not give a surjective mapping. The zero element of the integers has no preimage among the natural numbers, so the idempotent 0 is not covered.

The mapping from the natural numbers to Z_p that assigns each number its remainder modulo p is a homomorphism. It is surjective onto Z_p. This example is regarded as convincing.

Remark on Components

The discussion now turns to components. The formulation of the remark is not given in this fragment.

Connectedness and algebraic semigroups

For connected components, the picture in algebraic semigroups is not as simple as in algebraic groups. However, a similar construction works.

Any finite set is an algebraic set. Any abstract finite semigroup is algebraic. The converse also holds: for any algebraic semigroup, one can consider the set of its connected components.

A natural semigroup structure can be introduced on this set. To multiply two connected components, take their representatives and multiply them. The image of a connected set is connected, so the product lies entirely within a single connected component. That component is then declared to be the image.

Thus, a natural semigroup structure arises on the set of connected components. There is also a canonical map: each element is assigned its connected component. This map is a homomorphism.

This construction does not work for irreducible components. To illustrate, consider a semigroup that is a monoid.

Examples with Matrices and Zero Entries

The Set of Matrices with a Bounded Number of Nonzero Entries

We consider matrices in which each row and each column contains at most one nonzero entry. Such matrices resemble permutation matrices, but singular matrices are allowed, and the nonzero entries can be arbitrary, not necessarily equal to 0 or 1. The product of matrices is governed as follows: the rows of the result are determined by the left factor, and the columns are determined by the right factor.

The set of such matrices is closed under multiplication and forms a semigroup. Setting nonzero entries to zero is also allowed.

Irreducible Components

According to the speaker, this set has n! irreducible components. They correspond to the permutations from the symmetric group S_n and to the diagonal matrices D_n, including singular ones.

Each component is specified by a set of positions at which a nonzero entry may stand. All other positions contain zeros. Each irreducible component is isomorphic to an affine space of the corresponding dimension. For example, if three positions are chosen, then any three numbers are written in them. The component is isomorphic to an affine space as a variety.

The maximal components correspond to fillings of positions in which each row and each column contains exactly one entry. A component with fewer positions is obtained from these subsets, and these subsets are included in the maximal ones.

Connectedness

All irreducible components intersect at the zero matrix. Therefore the set forms one connected component, even though the number of irreducible components is n!.

Homomorphisms

A homomorphism that maps components to components according to a certain rule is not uniquely determined. The reason is that the zero element belongs to all components, and it is unclear into which connected component the zero should map.

The image of an irreducible set under a homomorphism remains irreducible. It lies in exactly one irreducible component. Therefore the image of the semigroup remains a semigroup.

A Proposition Without Proof

The speaker was going to state a proposition without proof. It is related to the group S_m. The statement is not completed in the transcript.

Irreducible Components and Homomorphisms

The set of irreducible components may be a group $S_N$, that is, the symmetric group of degree $N$. A homomorphism with such properties does not exist, because for some elements the number of values under the action of the operation is greater than one.

The irreducible component containing the identity is uniquely determined. For an algebraic monoid, it is the only irreducible component.

Statement of the Open Group Theorem

Let $G_M$ be the group of invertible elements of the monoid $M$. The theorem states that $G_M$ is open in $M$ and is an algebraic group. The proof is not presented, since going through it would take too much time. According to the speaker, this fact is not difficult to prove.

The theorem had already been formulated in other talks by the speaker. In the irreducible case, it was proved by Rittatore. For the reducible case, the result, as the speaker understands it, belongs to Brn. Brn notes that his argument almost completely coincides with Rittatore's. The speaker points out that Brn does not state directly that the arguments are identical, but confirms that they are close.

As an example, the set of all matrices is considered. The fragment devoted to this example breaks off in the source material.

Preimage of the Unit and the Reducible Component

Preimage of the Unit under Multiplication

Consider the set of pairs $(x, y)$ such that $xy = 1$. The projection of this set onto the first component coincides with the set of invertible elements. The speaker noted that the image of the map in this case is precisely the required set, and questioned whether it would be open.

The statement that the unit always lies in a single irreducible component remains an open question. The speaker formulated it as a matter to be verified. For the irreducible component containing the unit, the openness of the set of invertible elements within it was discussed. The passage to other components and the multiplication of invertible elements by them did not receive a complete proof: the justification that the product of an element from one component with invertible elements remains invertible in another component is not completed in the record.

Left and Right Inverses

The first statement says: if an element $x$ has a left inverse $y$ ($yx = 1$), then $x$ is invertible. The proof proceeds as follows. Let $xz = 1$ be a right inverse. Then $y = y(xz) = (yx)z = z$, whence $x$ is invertible.

It follows that left multiplication by $x$ is an injective map. The speaker referred to Borel's theorem: an injective endomorphism of an algebraic variety into itself is surjective. She noted that for normal varieties, an injective morphism is an isomorphism. The speaker did not verify the normality condition in the formulation of Borel's theorem. The endomorphism in this case is surjective, but not necessarily an isomorphism, since the inverse map is not constructed.

Non-invertible Elements as a Two-sided Ideal

The second statement says: the set $M \setminus G$, where $G$ is the group of invertible elements, is a two-sided ideal. Let $x$ be non-invertible and $y$ an arbitrary element. Suppose that $xy$ is invertible, with inverse $z$. Then $x(yz) = 1$, that is, $x$ has a right inverse, and by the first statement $x$ is invertible. Contradiction. Similarly for $yx$.

The speaker called $M \setminus G$ the largest proper two-sided ideal. The third statement, according to her, follows from the second and the same Borel theorem. A detailed derivation is not present in the record.

Idempotents and Groups

The unit is the only idempotent in any group. If a monoid $M$ has an idempotent $e$ different from the unit, then $e$ lies in $M \setminus G$. Therefore, a monoid in which the unit is the only idempotent is a group.

The speaker gave an example: an algebraic group with an added boundary. Such situations occur in matrix monoids. She described the boundary as the set of non-invertible elements of the monoid.

Order on Idempotents

For idempotents $e$ and $f$, a relation is introduced: $e \le f$ if and only if $ef = fe = e$. This is a partial order. If $e \le f$, then $e = fef$, so $e$ lies in $fSf$, and the set $eS$ is contained in $fSf$.

Sets of the form $eS$ and $Se$ are closed, since they are defined by equations. For $e \le f$, the corresponding closed sets are nested in reverse order: $eS \subseteq fSf$. It follows that in every descending chain of closed sets there exists a minimal element among the idempotents.

For the existence of a maximal element, the speaker introduced other closed sets, which are nested in the order opposite to the order on idempotents. She noted that the proof was formulated without a complete check of all details.

The statement about the minimal element: an idempotent $e$ is minimal if and only if $eSe$ is a group. From the minimality of $e$ it follows that there are no other idempotents inside $eS$. If a monoid has no idempotents other than $e$, then by the previous point the monoid is a group.

Maximal submonoids containing the unit $e$ correspond to maximal subgroups. The speaker claims that there exists a bijection between the set $E(S)$ and the set of maximal subgroups. She also noted that every subgroup lies in some maximal submonoid with the same unit. For this, the set $eS \cap Se$ with unit $e$ is considered, which contains the group of invertible elements. The exact formulation of the correspondence is not recorded.

The Commutative Case

If $S$ is commutative, then the product of two idempotents is an idempotent. The minimal element is then unique. If there were two different minimal elements, then both would be smaller than every idempotent, which contradicts minimality.

For the finiteness of the number of idempotents in the commutative case, a homomorphism $S \to E(S)$ is considered. The preimage of each idempotent is a subsemigroup. If each preimage contains a finite number of idempotents, then the number of idempotents in each connected component is also finite. The speaker reduces the problem to a connected semigroup $S$, for which the group of invertible elements in $xS$ consists of a single element $x$. From this she draws the conclusion of finiteness, but the argument could not be completed in the record.

Transition of Idempotents under a Homomorphism

The image of an idempotent under a homomorphism of semigroups is an idempotent. The preimage of an idempotent is a subsemigroup. These properties are used in the proof of finiteness and in the description of connected components.

The Variety of Idempotents and Its Components

Let $M$ be a monoid and $G$ its group of invertible elements. The set of idempotents $E(M)$ is an affine variety. The speaker claims that its irreducible components are the conjugacy classes with respect to $G$, that is, the orbits of the action of $G$ on $E(M)$ by conjugation. The components do not intersect.

The dimensions of the components may differ. Points and lines occur as components, and the unit always forms an isolated point. The smoothness of the set of idempotents is proved via a morphism from a direct product of suitable varieties into a neighborhood of a given point, and by comparing dimensions of tangent spaces. The speaker indicated that this verification is cumbersome.

Smoothness of the set of idempotents is not a general property. If we take a non-smooth variety $X$ and define the multiplication $xy = y$, then every element becomes idempotent, and the set of idempotents coincides with $X$, that is, it is not smooth.

Examples in Matrices

In the matrix monoid $M_n$, the zero matrix is a minimal idempotent and the only minimal one, and the identity is the maximal one. The group of invertible elements in this monoid is $GL_n$. For matrices, sets of the form $eS$ and $Se$ for idempotents $e$ are considered, and their properties coincide with those described above.