Today we will discuss the 3rd lecture (last Wednesday, Jan. 25, 2012) by J.-C. Yoccoz (corresponding to his 2011-2012 course on square-tiled surfaces), but before doing so, we will make a quick review of the material of this previous post.
Let be an origami associated to a finite group generated by two elements and , and a subgroup of containing no nontrivial normal subgroup. Denoting by the normalizer of in , we have that the automorphism group is . For the sake of the exposition, we will denote .
In this language, we saw that the absolute homology group has a natural -invariant decomposition as
where has dimension and has codimension (i.e., dimension ).
Finally, we computed an explicit formula for the multiplicity in of a -irreducible -representation of character , and we used this formula to prove that (for any origami and any ).
So, this is it as far as the quick revision is concerned. Below the fold the reader will find my notes for J.-C. Yoccoz 3rd lecture: his main goal in it was to show how to use the information (derived in the previous two lectures) about -representations to deduce useful facts about -representations. As it turns out, a significant part of this classical and, in particular, the representation theory facts we’re going to use (without proof) can be found in the books of J.-P. Serre and W. Fulton and J. Harris. Nevertheless, this lecture will contain some new facts (from the forthcoming paper by J.-C. Yoccoz, D. Zmiaikou and C.M.) concerning the specific case of representations related to origamis.
Let be a field. We have that
Denote by the set of isomorphism classes of representations defined over and -irreducible.
A theorem of Brauer (see Theorem 24 in Serre’s book) says that if contains all th roots of unity where for all , then .
In particular, one has that where is the algebraic closure of .
Recall that the Galois group acts on , and is identified to the orbit space of this action.
Let be such an orbit and denote by be its character (by thinking of as a -representation). We can write
where is an integer known as Schur index.
Let be a -space where acts with character . We denote by the commuting algebra of in (i.e., the elements of commuting with the action of on ). In general is a field (not necessarily commutative) such that
where is the center of .
Now we consider the decomposition
into isotypical components (i.e., as -module for some ). Observe that the endomorphisms of as -module have the form
In the sequel, we’ll treat exclusively the case :
Standing Assumption. From now on, .
In this situation, given , we have three possibilities:
- is real, that is, , and (and takes only real values);
- is complex, that is, , , (and takes some complex non-real value);
- is quaternionic, that is, , , (and takes only real values).
(Here, designs Hamilton’s quaternions)
In the specific case of origamis, recall that the absolute homology group comes equipped with a (symplectic) intersection form
Since acts on by automorphisms, we have that is -invariant. Furthermore, since and are orthogonal (with respect to ), the restriction of to is non-degenerate.
Proposition. The subspaces introduced above are mutually orthogonal with respect to . In particular, the restriction of to each is non-degenerate.
Proof. Since is -invariant, it defines an isomorphism (of -modules) between and its dual. On the other hand, for each , we have that
where is the dual of . Therefore, this isomorphism (induced by ) preserves each isotypical component.
This proposition says that it makes sense to define
In the sequel, we will discuss the structure of in the three possible cases ( real, complex or quaternionic) in terms of adequate invariant bilinear forms. Below, the treatment of the latter objects in the quaternionic case will follow these notes here of Y. Le Cornulier.
The -module irreducible module of type can be equipped with a -invariant scalar product , and, by irreducibility, is unique up to multiplication by a positive scalar.
Proposition. The multiplicity of in is even, i.e., . Also, there exists an isomorphism of -modules such that
Consequently, for , by writing
we have that the map is an isomorphism between and the usual symplectic group .
Proof. Fix an isomorphism of -modules and define
Note that defines an isomorphism of -modules
On the other hand, by using the -invariant scalar product , we can identify with . Hence, it follows that induces a linear map ,
such that the matrix is antisymmetric and invertible. Thus, we have that is even, say and, furthermore, we can choose an isomorphism of the form such that can be written in the standard (symplectic) form in the statement of the proposition. Finally, the last claim of the proposition is a direct verification left as an exercise to the reader.
– complex (, )-
We fix an isomorphism and we equip with the corresponding structure of -vector space. In this way, is the underlying space of a -irreducible representation of , and we can equip with a -invariant (positive-definite) Hermitian product (unique up to multiplication by a positive scalar).
Proposition. There exists an isomorphism of -modules and a pair of integers with such that
Consequently, for , by writing
(), we have that the map is an isomorphism between and the unitary group of the Hermitian form
Proof. Fix an isomorphism of -modules and write
It is not hard to check that the bilinear forms on are -invariant. It follows that are linear combinations of the real and imaginary parts of the (positive-definite) Hermitian form . Thus, the fact that implies that
and, a fortiori, . Hence,
defines a -invariant non-degenerate Hermitian form on . In particular, we can write
where and , and, since is non-degenerate, we can find an automorphism of -modules and a pair of integers with such that verifies
Finally, the verification of the last claim of the proposition is straightforward (and again we leave as an exercise).
– quaternionic (, )-
We fix an isomorphism . Recall that given an element , its complex conjugate is , and its norm is given by the formula . Also, the complex conjugation satisfies .
Using the isomorphism we can render into a right -vector space by imposing
for and .
Definition. A Hermitian form on a right -vector space is a map verifying:
- (linearity on the 2nd factor);
- (“usual” Hermitian condition).
Observe that the two conditions above imply (i.e., is antilinear in the 1st factor).
Example. The standard Hermitian form on is .
In general, given an Hermitian form , we can write:
Observe that is symmetric while are antisymmetric. Moreover, they verify the relations
allowing to express in terms of for any . For later use, we will focus on antisymmetric forms, e.g., for sake of concreteness, because in the setting of origamis we will be interested in producing Hermitian forms from the symplectic intersection form.That being said, observe that satisfies
because , and hence is derived from by conjugation by whenever (i.e., is a purely imaginary quaternion with unit norm.
Conversely, given a bilinear antisymmetric verifying (1) above, then
is a Hermitian form.
Next, we observe that, by irreducibility, has an unique (up to multiplication by a positive real number) -invariant positive-definite Hermitian form whose components , , , form a basis of the space of -invariant -bilinear forms (and, in particular, the space of symmetric -invariant -bilinear forms has dimension 1, while the space of antisymmetric -invariant -bilinear forms has dimension 3).
At this point, we are ready to state the proposition whose proof will occupy complete today’s discussion.
Proposition. There exists an isomorphism of -modules and a pair of integers with such that the Hermitian form corresponding to by the formula (2) above satisfies
Consequently, for , by writing
, we have that is an isomorphism between and the unitary group of the Hermitian form
We will show this proposition with the aid of the following three lemmas:
Lemma 1. Let be a right -vector space and assume that is an isotypical -module of quaternionic type. Let be a -invariant antisymmetric non-degenerate -bilinear form. Then, there exists and such that
Proof of Lemma 1. Otherwise, for all , so that, by bilinearity,
i.e., . Since is antisymmetric, we deduce that
Since is -invariant, it follows that
for all . Since is non-degenerate, we get that for all , that is, the action of factors through the action of a subgroup whose elements have order two, i.e., , and for all . Since for all , we have that is Abelian and, in particular, it doesn’t have representations of quaternionic type. Of course, since the action of on factors through , we conclude that can’t be a representation of quaternionic type, a contradiction with our hypothesis.
Lemma 2. Under the assumptions of Lemma 1, we can write
where the ‘s are -invariant, irreducible, and mutually orthogonal with respect to .
Proof of Lemma 2. We proceed by induction on . For there is nothing to prove. For , we choose satisfying the conclusion of Lemma 1, and we define . We have that is -invariant and isotypical, and is -invariant. In particular, since is irreducible, we also have that is non-degenerate (otherwise its annihilator would be a non-trivial -invariant subspace of ). In particular, we can write
where is -orthogonal to , and hence -invariant of multiplicity .
Lemma 3. Let be a -invariant antisymmetric -bilinear form on . Then, there exists , such that
Proof of Lemma 3. For a -invariant antisymmetric -bilinear form , we define its adjunction by the formula
The adjunction has the following properties:
- because it commutes with the action of ;
- because is antisymmetric;
- for all ;
- if , then because and , i.e., is purely imaginary with unit norm, implies and hence , so that is purely imaginary with unit norm as well.
In particular, we have that is an anti-involution (i.e., and ) preserving the space of purely imaginary quaternions. Since an anti-involution can’t act by the identity on the space of purely imaginary quaternions (as is not commutative), it follows that there exists a purely imaginary such that
Also, recall that denotes the -component of the (unique up to multiplication by a positive real number) -invariant positive-definite Hermitian product on . We saw that satisfies (1) above, that is,
so that its adjunction verifies
(Also, similar formulas hold for and in the place of )
Now we come back to the bilinear form of the statement of the lemma. As we saw, there exists a purely imaginary with unit norm such that
On the other hand, the reader can check that the conclusion (3) of the lemma is equivalent to .
The adjunction of (when ) can be computed as follows.
so that . In particular, by choosing , such that , we obtain that
It follows that preserves the subspace generated by and . This leaves us with three possibilities:
- the restriction of to is a reflection;
- the restriction of to is ;
- the restriction of to is .
The case of reflection is easily excluded as follows: up to conjugation, we may assume that has the form
and this last map is not an anti-involution (as it is the conjugation by and thus a “true” involution).
The case of can also be excluded because it forces to the the complex conjugation, and this can’t be the adjunction of a non-trivial antisymmetric form (but only of a symmetric one). For instance, given an antisymmetric form , we can write . If the adjunction of is the complex conjugation, we would deduce
Here, in the last equality, we used our explicit knowledge of the adjunctions (see (4) above). Since is a basis, one has that , a contradiction (as implies that but is not the complex conjugation as we can see from (4) above).
It follows that and . By combining this with (5) above, we conclude that , and, as we already observed, this is equivalent to (3), so that the lemma is proved.
At this stage, we can complete the proof of the proposition (and today’s post). By Lemma 2, we can write
where are irreducible and mutually orthogonal with respect to the symplectic intersection form . Using this we can build an isomorphism such that
where are -invariant antisymmetric non-zero forms on .
By Lemma 3, we can choose such that the isomorphism
where each verifies (3) above. In particular, we have that verifies (1) above and thus
is a -invariant Hermitian form. By uniqueness (up to real scalars), we have that
Finally, by permuting coordinates and performing appropriate scaling, we can replace by an isomorphism such that the numbers become and, therefore,
for a pair of integers with . Finally, the verification that the map where
is an isomorphism between and the unitary group of the standard Hermitian form is left as an exercise to the reader.