TOPICS IN NON COMMUTATIVE ALGEBRAIC GEOMETRY AND CONGRUENT ZETA FUNCTIONS

(PART V). WEYL ALGEBRAS REVISITED

YOSHIFUMI TSUCHIMOTO

1. A review

Let us first recall the definition. (See Part I for details.)

Definition 1.1. Let k be a field. Let n be a positive integer. A Weyl algebra A_{n}(k) over a commutative ring k is an algebra over k generated by 2n elements {γ_{1}, γ_{2}, …, γ_{2n}} with the “canonical commutation relations”

(CCR)  [γ_{i}, γ_{j}](= γ_{i}γ_{j} − γ_{j}γ_{i}) = h_{ij}  (1 ≤ i, j ≤ 2n).

Where h is a non-degenerate anti-Hermitian 2n × 2n matrix of the following form.

(h_{ij}) = (\begin{matrix}0 & −1_{n} \\ 1_{n} & 0 \end{matrix}) .

Throughout this section, the letter h will always represent the matrix above and the letter h̄ will always represent the inverse matrix of h.

1.1. perfect field.

Definition 1.2. A field k of characteristic p ≠ 0 is called perfect if the Frobenius homomorphism

k ∋ x → x^{p} ∈ k

is surjective.

1.2. A matrix representation of Weyl algebra. From now on in this section we fix a prime number p and we assume k is a field of characteristic p. We will simply write A_{n} instead of A_{n}(k).

Let us define operators(matrices) {μ_{i}}_{i=1}^{2n} acting on p^{n}-dimensional vector space

V = k[x_{1}, x_{2}, …, x_{n}]/(x_{1}^{p}, x_{2}^{p}, … x_{n}^{p})

by

\begin{matrix}μ_{i} = multiplication by x_{i} \\ μ_{i+n} = ∂/∂x_{i}. \end{matrix}\} i = 1, 2, …, n.

Let

S_{n} = k[T_{1}, T_{2}, …, T_{2n−1}, T_{2n}]

be a polynomial ring of 2n-variables over k.

Then we have a faithful representation

Φ : A_{n} → M_{p^{n}}(S_{n})

of the Weyl algebra A_{n} by putting

Φ(γ_{i}) = T_{i} · 1_{p^{n}} + μ_{i}.

Furthermore, for any c = (c_{1}, c_{2}, …, c_{2n}) ∈ k^{2n}, we have by specialization the following representation of A_{n}.

Φ_{c} : A_{n} → M_{p^{n}}(k)

by putting

Φ_{c}(γ_{i}) = c_{i} · 1_{p^{n}} + μ_{i}.

We recall also that

Lemma 1.3. (Part I, Lemma 7.3) The center Z_{n} of A_{n} is isomorphic to a polynomial algebra with 2n indeterminates. Namely,

Z_{n} = k[γ_{1}^{p}, γ_{2}^{p}, …, γ_{2n}^{p}].

Lemma 1.4. (Part I, Corollary 7.8) Let k be an algebraically closed field of characteristic p ≠ 0. Then every finite dimensional irreducible representation α : A_{n}(k) → End_{k}(V) of A_{n}(k) is equivalent to a representation Φ_{c} for some c ∈ k^{2n}.

Lemma 1.5. Let k be a field of characteristic p. If k contains infinite number of elements, then:

⋂_{c∈k^{2n}} Ker(Φ_{c}) = 0.

Here is another thing we need to know.

Lemma 1.6. Φ gives a k-algebra isomorphism

Φ̂ : S_{n} ⊗_{Z_{n}} A_{n} ≅ M_{p^{n}}(S_{n}).

Proof. One may easily verify that we have a well-defined homomorphism given by

Φ̂ : S_{n} ⊗_{Z_{n}} A_{n} ∋ f ⊗ a ↦ fΦ(a) ∋ M_{p^{n}}(S_{n}).

For any j ∈ {1, 2, 3, …, 2n}, we have

Φ̂(1 ⊗ γ_{j} − T_{j} ⊗ 1) = μ_{j}.

So the image Image(Φ̂) contains μ_{j}. Since we know that {μ_{j}}_{j=1}^{2n} generates M_{p^{n}}(k) as k-algebra (Part I, Corollary 7.10) , we conclude that the map Φ̂ is surjective.

Now, both S_{n} ⊗_{Z_{n}} A_{n} and M_{p^{n}}(S_{n}) is free S_{n}-modules of rank p^{2n}. So the map Φ̂ is generically injective. (That means, if we take the quotient field Q(S_{n}) of S_{n} and consider

1_{Q(S_{n})} ⊗ Φ̂ : Q(S_{n}) ⊗_{S_{n}} (S_{n} ⊗_{Z_{n}} A_{n}) → M_{p^{n}}(Q(S_{n})).

then by an elementary theorem in linear algebra, we see that it is an isomorphism.)

Since S_{n} (a polynomial algebra over k) is an integral domain, S_{n} ⊗_{Z_{n}} A_{n} is Z_{n}-torsion free. (That means,

S_{n} ⊗_{Z_{n}} A_{n} → Q(S_{n}) ⊗_{S_{n}} (S_{n} ⊗_{Z_{n}} A_{n})

is injective.)

So we see that Φ̂ is injective.

□

1.3. Algebra endomorphisms and centers of Weyl algebras.

Lemma 1.7. Let k be a field of characteristic p ≠ 0. For any kalgebra endomorphism ϕ of A_{n}(k),

(1) Φ_{c} ∘ ϕ is a surjective homomorphism for any c ∈ k^{2n}.

(2) ϕ(Z_{n}(k)) ⊂ Z_{n}(k).

Proof. We may assume that k is an algebraically closed field.

(1) The composition ϕ_{c} = Φ_{c} ∘ ϕ is a representation of A_{n}. By Lemma 1.4 we see that any irreducible sub representation of ϕ_{c} is equivalent to Φc̄ for some c̄ ∈ k^{2n}. By a dimensional argument, we conclude that ϕ_{c} itself is equivalent to Φc̄. (In other words, there exists g_{c} ∈ GL_{n}(k) such that

ϕ_{c}(x) = g_{c}Φc̄(x)g_{c}^{−1} (∀x ∈ A_{n})

holds.) Thus ϕ_{c} is surjective as required.

(2) Let z ∈ Z_{n}, x ∈ A_{n}. For any c ∈ k^{2n}, we have (using the same notation as above)

[ϕ_{c}(z), ϕ_{c}(x)] = ϕ_{c}([z, x]) = 0

Since we know by (1) that ϕ_{c} is surjective, we see that ϕ_{c}(z) belongs to the center of M_{p^{n}}(k). In particular for any y ∈ A_{n}, we have

Φ_{c}([ϕ(z), y]) = [Φ_{c}(ϕ(z)), Φ_{c}(y)] = [ϕ_{c}(z), Φ_{c}(y)] = 0.

Thus

[ϕ(z), y] ∈ ⋂_{c} Ker(Φ_{c}) = 0.

So ϕ(z) ∈ Z(A_{n}) = Z_{n} as required.

Corollary 1.8. Let ϕ : A_{n} → A_{n} be a k-algebra endomorphism of A_{n}. Then by restriction we obtain a homomorphism

ϕ_{Z_{n}} : Z_{n} → Z_{n}.

Furthermore, if the base field k is perfect, then ϕ may be uniquely extended to its p-th root.

ϕ_{S_{n}} : S_{n} → S_{n}.

In precise, Let us write down ϕ_{Z_{n}} like

ϕ_{Z_{n}}(γ_{j}^{p}) = ∑_{I} f_{j,I}(γ^{p})^{I} (j = 1, 2, …, 2n).

Then ϕ_{S_{n}} is given by the following formula.

ϕ_{S_{n}}(T_{j}) = ∑_{I}(f_{j,I})^{1/p}T^{I} (j = 1, 2, …, 2n).

Here comes a geometric interpretation of endomorphisms of Weyl algebras.

Corollary 1.9. Let k be a perfect field of characteristic p ≠ 0. Let ϕ : A_{n} → A_{n} be a k-algebra endomorphism of A_{n}. Then we have a matrix valued function G ∈ GL_{p^{n}}(S_{n}) and a morphism f : Spec(S_{n}) → Spec(S_{n}) which enables the following diagram commute.

(*)  \begin{matrix}A_{n} & ⟶^{ϕ} & A_{n} \\ ^{Φ}↓ & & ^{Φ}↓ \\ M_{p^{n}}(S_{n}) & ⟶^{ϕ̄} & M_{p^{n}}(S_{n}) \end{matrix}

Where ϕ̄ is defined as

(**)  ϕ̄(x) = Gf^{∗}(x)G^{−1}.

We may write down the commutative diagram above as the following equation.

Φ(ϕ(a)) = Gf^{∗}(Φ(a))G^{−1}

Proof. Let us define

ϕ_{Z_{n}} : Z_{n} → Z_{n}

and

ϕ_{S_{n}} : S_{n} → S_{n}

as in the previous Corollary. We put f = Spec(ϕ_{S_{n}}).

We have an well-defined Z_{n}-algebra homomorphism

ϕ_{S_{n}} ⊗ ϕ : S_{n} ⊗_{Z_{n}} A_{n} → S_{n} ⊗_{Z_{n}} A_{n}.

By using the isomorphism in , we obtain a Z_{n}-homomorphism

ϕ̄ = Φ̂(ϕ_{S_{n}} ⊗ ϕ)Φ̂^{−1} : M_{p^{n}}(S_{n}) → M_{p^{n}}(S_{n})

which is compatible with ϕ in the sense that it satisfies the commutative diagram (*) of the statement.

It remains to prove that the map ϕ̄ is represented as (**). By pullback, we obtain an S_{n}-algebra homomorphism

ρ : M_{p^{n}}(S_{n}) ≅ S_{n} ⊗_{ϕ_{Sn},S_{n}} M_{p^{n}}(S_{n}) ∋ y ⊗ x ↦ yϕ̄(x) ∈ M_{p^{n}}(S_{n}).

Where the first isomorphism in the above line is the inverse of the following S_{n}-algebra homomorphism.

S_{n} ⊗_{ϕ_{Sn},S_{n}} M_{p^{n}}(S_{n}) ∋ y ⊗ x ↦ yf^{∗}(x) ∈ M_{p^{n}}(S_{n}).

by an argument similar to that in (I,Lemma 7.9), we see that there exists G ∈ GL_{p^{n}}(S_{n}) such that

ρ(x) = GxG^{−1} (∀y ∈ M_{p^{n}}(S_{n}))

holds. (See appendix for the detail.)

□

1.4. appendix.

Definition 1.10. (temporary) Let R be a commutative ring. A finitely generated R-module M is said to be projective if there exists an R-module M′ such that

M ⊕ M′ ≅ R^{⊕n}

for some n.

We shall argue homological algebra in much more detail later. So the definition here is meant to be minimum. See a book on homological algebra for “correct definition”.

Definition 1.11. Let R be a domain. The generic rank of M is the rank of Q(R) ⊗_{R} M over the quotient field Q(R).

Lemma 1.12. Let R be a commutative unique factorization domain(UFD). Suppose an R module M satisfies the following conditions.

(1) M is of generic rank 1.

(2) M is finitely generated.

(3) M is projective. Then M is R-free of rank 1.

Proof. (Essentially borrowed from [1]) Let K = Q(R) be the quotient field of R.

Since M is projective and R is an integral domain, M is torsion free. So

ι : M → M ⊗_{R} K

is injective. Since M is of generic rank 1, M ⊗_{R} K is isomorphic to K. as a K-module. We may thus assume that M ⊂ K. Since M is finitely generated, we may further assume that M ⊂ R.

Now, Let us paraphrase the condition that M being projective. First of all, the condition is equivalent to an existence of R-module homomorphisms

f : M → R^{n}, g : R^{n} → M

such that g ∘ f = id. Secondly, we may then represent f, g in matrix form.

f(m) = (f_{1}(m) ; f_{2}(m) ; ⋮ ; f_{n}(m)) , g((r_{1} ; r_{2} ; ⋮ ; r_{n})) = (g_{1}, g_{2}, g_{3}, …, g_{n}) (r_{1} ; r_{2} ; ⋮ ; r_{n})

f : M → R^{n}, g : R^{n} → M

such that g ∘ f = id.

Thirdly, each f_{i}, g_{i} is represented by a linear map from K to K. That means, by an element of K.

f = (a_{1} ; a_{2} ; ⋮ ; a_{n}) , g = (b_{1}, b_{2}, b_{3}, …, b_{n})

We may obtain several properties of {a_{i}}, {b_{j}}:

(i) g(R^{n}) ⊂ M implies b_{i} ∈ M ⊂ R(∀i).

(ii) g ∘ f : R^{n} → R^{n} implies a_{i}b_{j} ∈ R(∀i∀j).

(iii) g(R^{n}) ⊂ M implies b_{i} ∈ M(∀i).

(iv) g(R^{n}) = M implies ∑_{i} Rb_{i} = M.

(v) g ∘ f = id implies ∑_{i} b_{i}a_{i} = 1. Let us write

a_{i} = \frac{l_{i}}{m_{i}} (gcd(l_{i}, m_{i}) = 1).

R ∋ a_{i}b_{j} = \frac{l_{i}b_{j}}{m_{i}}

Since l_{i} and m_{i} are coprime, we conclude that b_{j} is divisible by m_{i}. Let us denote by l the largest common multiple of m_{1}, m_{2}, …, m_{n}.

l|b_{j}

By (iv) we see M ⊂ lR.

By (v) we see

l = l ∑_{i} b_{i}a_{i} = ∑_{i} b_{i}l_{i} \frac{l}{m_{i}} ∈ ∑_{i} b_{i}R ⊂ M.

Thus M ⊃ lR. So M = lR. □

Lemma 1.13. Let R be a commutative UFD. Then for any R-algebra homomorphism

ρ : M_{n}(R) → M_{n}(R),

there exits an element G ∈ GL_{n}(R) such that

ρ(x) = GxG^{−1}

holds for any x ∈ M_{n}(R).

Proof. Let us denote by e_{ij} ∈ M_{n}(R) the matrix element. Let us consider R-modules

M_{i} = ρ(e_{ii})R^{n} (i = 1, 2, 3, …, n).

Then by an argument similar to that in (I,Lemma 7.9), we see that

R^{n} = ⨁_{i=1}^{n} M_{i}.

and that the multiplication by ρ(e_{ji})

M_{i} →^{ρ(e_{ji}).} M_{j}

give isomorphisms between the modules. Hence we see easily that M_{0} satisfy the assumptions of the previous lemma. We conclude that M_{1} is freely generated by single element v_{1}. Then we put

v_{j} = ρ(e_{j1})v_{1}.

and

G = (v_{1}v_{2}v_{3} … v_{n}).

We may easily see that G plays the roles as expected. □

2. appendix 2

Proposition 2.1. Let d be a positive integer. Let (X, ω_{X}), (Y, ω_{Y}) be smooth symplectic algebraic varieties of dimension 2n over a field k. Let ϕ : X → Y be a symplectic morphism. That means, it is a morphism which preserves the symplectic structure :

ϕ^{∗}(ω_{Y}) = ω_{X}

Then the tangent map of ϕ is of full rank at every point P on X.

We make an effective use of the theory of the Pfaffian Pfaff(M) of a given matrix M. A good reference is in [2, XV,§9]. Especially important theorem we need to know is the following lemma.

Lemma 2.2. ([2, XV,Theorem 9.1]) Let R be a commutative ring. Let (m_{ij}) = M be an alternating matrix with g_{ij} ∈ R. Then

det(M) = (Pfaff(M))^{2}.

Furthermore, if C is an n × n matrix in R, then

Pfaff(CM^{t}C) = det(C) Pfaff(M).

Proof. (of Proposition 2.1) We may assume that k is algebraically closed and that P is a k-valued point. Let us represent the tangent map of f at P by T_{P} f. One may choose a local coordinate system x_{1}, …, x_{2n} on X around P such that the symplectic form ω_{X} at P is represented by the matrix h when expressed in terms of dx_{1}, …, dx_{2n}. Likewise one may choose a local coordinate system y_{1}, …, y_{2n} around f(P) such that the symplectic form ω_{Y} at f(P) is represented by the matrix h when expressed in terms of dy_{1}, …, dy_{2n}. Then using the base ∂/∂x_{1}, ∂/∂x_{2}, … ∂/∂x_{2n} ∂/∂y_{1}, ∂/∂y_{2}, … ∂/∂y_{2n}, T_{P} f may be identified with a 2n × 2n-matrix. Since by hypothesis T_{P} f reserves the symplectic form, we have

(^{t}T_{P} f)h(T_{P} f) = h.

Let us compare the Pfaffian of the both hand sides.

det(T_{P} f)^{2} · Pfaff(h) = Pfaff((^{t}T_{P} f)hT_{P} f) = Pfaff(h).

Since Pfaff(h) = 1 , we conclude that the determinant of T_{P} f should be equal to 1 or (-1). □

Note: It goes without saying that when X is connected, and if the coordinate systems x_{1}, …, x_{2n} and y_{1}, …, y_{2n} are able to chosen globally (for example if X, Y are affine space 𝔸^{2n} = Spec k[T_{1}, T_{2}, T_{3}, …, T_{2n}] with

ω = dT_{1} ∧ dT_{n+1} + dT_{2} ∧ dT_{n+2} + dT_{3} ∧ dT_{n+3} + ⋯ + dT_{n} ∧ dT_{2n} as the symplectic form), then the Jacobian of f should either be 1 on the whole of X or be −1 on the whole of X.

References

[1] T. Y. Lam, Serre’s conjecture (Lecture notes in mathematics 635), Springer Verlag, 1978.

[2] S. Lang, Algebra (graduate texts in mathematics), Springer Verlag, 2002.