Evaluación del Desempeño Humano

Recall that in linear algebra, a matrixM ∈GLn(k) is said to be unipotent if (M−1)n= 0 for somen∈_N. Thus, M has eigenvalues 1 and under some basisM can be identified with an upper triangular matrix with 1’s on the diagonal:

     1 ∗ ∗ . . . ∗ 0 1 ∗ . . . ∗ .. . ... ... . .. ... 0 0 0 . . . 1      .

LetUn denote the closed affine subgroup of GLn consisting of matrices with such a form; It is a closed affine subgroup since it is represented by the quotient ofk[xij,det−1] by extra relations making the diagonal 1 and the bottom triangle 0. As we did for diagonalisable groups, we can consider a subgroup M ⊆ GLn(k) of unipotent matrices. The following theorem gives us a neat characterisation of such groups:

Theorem 5.4.1 (Kolchin’s Theorem). Let M ⊆GLn(k) be a linear subgroup consisting of unipotent matrices. Then there exists a basis such that allM ∈ M is an element of Un(k).

5.4. UNIPOTENT GROUPS 61

Proof. [Wat12, pg. 62]

It is also clear that every matrix M ∈ Mcan be identified as an element in Un under the same basis if and only if there exists a non-zero vector v ∈V such that M(v) = v for all M ∈ M. Motivated by the theorem, we make the following definition:

Definition 5.4.2. An affine group G ∼= Spec(A) is unipotent if every representation of G contains a non-zero fixed vector. In terms of comodulesρ :V →V ⊗A, this is equivalent to saying that any comodule V of A has somev∈V non-zero such that ρ(v) =v⊗1.

The following proposition provides an equivalent formulation of unipotent affine groups through their representations:

Proposition 5.4.3. Let G ∼=Spec(A) be an affine group. Then G is unipotent if and only if for every finite-dimensional representation Φ : G → GLV ∼= GLn, Φ maps to Un under

some basis ofV.

Proof. We will prove the first statement by induction. It is obviously true when V is of dimension 1. Now assume that the statement is true for all representations of dimensionn

and letρ :V →V ⊗Abe a representation of dimensionn+ 1. Letvn+1 be the fixed vector

and let V0 := Span{vn+1}. Then ρ induces a comodule structure on V /V0. Since V /V0 is

of dimensionn, by the induction hypothesis there exists a basis{vi}ni=1 forV /V0 such that

ρ(vj) =P vi⊗aij with aij =      1, fori=j, 0, fori > j, ∗, otherwise.

Choosing vi ∈ V such that vi =vi+kivn+1, then {vi}ni=1+1 forms a basis of V. Under this

basis we have that

ρ(vj) = n X i=1 vi⊗aij+vn+1⊗ n X i=1 kiaij,

which shows that Φ maps toUn under this basis.

For the converse, let V be a representation ofG. Since every representation contains a finite-dimensional subrepresentation, we may assume thatV is finite-dimensional. Let{ei} be a basis for V with the representation Φ mapping into Un. Then ei is a non-zero fixed vector of V.

As every algebraic affine group G is isomorphic to some closed subgroup of GLn, one would then expect that G being unipotent is equivalent to G being is isomorphic to some closed subgroup of Un through some representation V, and indeed, this is true. To prove this we will need the notion of coconnectedness of Hopf algebras:

Definition 5.4.4. A Hopf algebra A is said to be coconnected if there exists a filtration

C0⊆C1 ⊆ · · · of A with linear subspacesCi such that

C0 =k, ∪i≥0Ci=A, and ∆(Cr)⊆ r

X

i=0

Proposition 5.4.5. Let G ∼= Spec(A) be an algebraic affine group. Then the following conditions are equivalent:

(i) G is unipotent;

(ii) G is isomorphic to some closed subgroup of Un for some n;

(iii) A is coconnected.

Proof. (i) ⇒ (ii) Since G is algebraic, we have a faithful finite-dimensional representation Φ :G →GLV ∼= GLn. IfG is unipotent, the previous proposition then tells us that Φ maps intoUnunder some basis of V and so is isomorphic to some closed subgroup of Un.

(ii) ⇒ (iii) Note that if B satisfy equation (11), then so does B/I by considering the quotient image of eachCi. So it is sufficient to prove this for the case whereG ∼=Un. This

gives us k[G] ∼= k[xij : i < j]/hxii−1i. We assign a weight of j−i to each xij, so that each monomialQ

xnij

ij with have weight

P

nij(j−i). DefineCr to be the subspace spanned by polynomials with weight less than or equal tor. Then we have C0 =k, A= ∪Cr and

CiCj ⊂Ci+j. Since ∆ is linear we can check that equation 11) holds for each monomial in

Cr. By the structure of Un, we get

∆(xij) =xij ⊗1 + 1⊗xij+ X i<k<j xik⊗xkj ⊆ j−i X r=0 Cr⊗Cj−i−r.

So equation (11) holds for any monomial of the form xij. With their weight as base cases, we now proceed by induction. Assume that it is true for monomials P and Qof weight n

andm. Then we get

∆(P Q) = ∆(P)∆(Q) ⊆ n X r=0 Cr⊗Cn−r ! _m X s=0 Cs⊗Cm−s ! ⊆X r,s (CrCs⊗Cn−rCm−s) ⊆X r,s (Cr+s⊗Cn+m−r−s).

Thus (iii) holds.

(iii)⇒(i) Assume thatAis coconnected and letρ:V →V ⊗Abe a comodule. Define

Vr:={v∈V :ρ(v)⊆V ⊗Cr}.

Then V =∪Vr. If there is a non-zero vector in V0, then ρ(v) = v0 ⊗1. By applying the

identity (id⊗)◦ρ = id we obtain v0 =v, so v is fixed. If we now show that Vr = 0 ⇒

Vr+1 = 0 for allr, then V0 can not be 0, which will complete the proof. We know that

(id⊗∆)(ρ(Vr+1)⊆(id⊗∆)(V ⊗Cr+1)⊆V ⊗ X i Ci⊗Cr+1−i ! .