Infinite-Dimensional Modular Lie Superalgebra Ω

Lemma 1. Let  k, k^′ ∈ Q  and  t ∈ ℕ₀. Then the following statements hold.

• (i)

  $$.

• (ii)

  If$$, then$$, where  i ∈ M₁.

• (iii)

  Let  x^ky^λξ^u ∈ Ω_[1]  and  t ≥ ht(x^ky^λξ^u). Then  𝒮_t(x^ky^λξ^u) ≥ 3.

Proof. (i) We see that

Note that$$. By the uniqueness of  p-adic expression, we have$$Thus

as desired.

(ii) If$$, then$$. We see that$$. So (ii) holds.

If$$and$$for  b ≥ 1, then we can assume that

Hence$$.

(a) If  t < b, then we get

(b) If  t = b  and  p > 3, then

(c) If  t > b, then

Thus (ii) holds.

(iii) By  |k| + 2k₁ + |u| = i + 2 ≥ 3, we have  ∑ (pad₀(k_i)p⁰ + pad₁(k_i)p¹ + ⋯) + 2pad₀(k₁) + |u| ≥ 3;   that is,  ∑ pad₀(k_i) + 2pad₀(k₁) + |u| ≥ 3.

Lemma 2. Let $$Then the following statements hold.

• (i)

  If$$, then$$.

• (ii)

  If$$, then$$.

• (iii)

  If$$, then$$.

Proof. (i) As $$, $$Then we have

By the equality above and Lemma 1, we get  pad(k^′ + (k − e₁)) = pad(k^′) + pad(k − e₁).  Thus Hence

(ii) The proof is completely analogous to (i).

(iii) For  i ∈ M₁, by assumption of this lemma, we have$$, which combined with Lemma 1 yield

For  i ∈ T, it is easily seen that$$Also by Lemma 1, we obtain

Hence Lemma 2 holds.

Lemma 3. Let$$and$$. Let$$be a nonzero summand of$$. Then$$.

Proof. By a direct computation, we obtain that

which satisfies the conditions of Lemma 2.

Lemma 4. Ω_[1]⊆nil(Ω).

Proof. Given  t ∈ ℕ, put

Clearly, we have  𝒮_t(x^ky^λξ^u) < l_t  for all standard basis element  x^ky^λξ^u  of  Ω.

Let  t ∈ ℕ  such that  t ≥ ht(z).  For any$$with$$, we have

Note that$$By using Lemma 3 repeatedly we see that$$. Hence  Ω_[1]⊆nil(Ω).

Lemma 5. (i) If$$where  f_t ∈ Ω_t, then  f_t ∈ nil(Ω).

(ii) If$$then  f₋₂ = 0.

(iii) If$$ then  f₋₁ = 0.

(iv)   $$.

(v)   $$.

Proof. (i) See Lemma  5 in [14].

(ii) By (i), we see that  f₋₂  is ad-nilpotent. If  f₋₂ ≠ 0, then$$for all  m > 0  and  λ ≠ 1. By a direct computation, we obtain$$, contradicting the nilpotency of  f₋₂. Hence  f₋₂ = 0, as desired.

(iii) Clearly,  f₋₁  is ad-nilpotent by virtue of (i). If  f₋₁ ≠ 0, then we can suppose that$$ where  γ_i ∈ 𝔽. Thus there exists some  γ_j ≠ 0.  By computation, we have

Similarly, we get  (ad$$, a contradiction. Consequently,  f₋₁ = 0.

(iv) Suppose that  f = f₀ + f_[1]  is an arbitrary element of nil$$, where$$. By (1) of this lemma, we see that$$. Hence nil$$and Nil$$.

Conversely, we have$$Nil$$by means of Lemma 4. Clearly, Nil$$. Thus Nil$$This shows that Nil$$, as desired.

(v) It is obvious that Nil($$($$Conversely, we assume that$$

Then by (ii) of this lemma,  f₋₂ = 0. Hence$$, where$$and$$. It follows from (iii) that  f₋₁ = 0. Now  f = f_[0]∈ nil$$. Noting that  Ω_[1]⊆ nil(Ω), we have  f = f_[0]⊆nil$$. Thus nil($$, and the assertion holds.

Lemma 6. Let  i, j ∈ M₁. Suppose that  x^ky^λξ^u  is an arbitrary standard element of  Ω. Then the following statements hold.

• (i)

  $$.

• (ii)

  If  [i] = [j]  and  i ≠ j, then$$.

• (iii)

  If  [i]≠[j]  and  j ≠ i^′, then$$.

• (iv)

  $$.

Proof. (i) By a direct computation, we get

and$$. It follows from the binomial theorem that

(ii) Also by a direct calculation, we have

Put $$. Obviously,

Hence$$.

(iii) The proof is completely analogous to (ii).

(iv) According to (i), we see that$$.

Lemma 7. Suppose that i, j, k ∈ T are different from each other, and  γ, χ ∈ 𝔽. Then

(i)  f = γξ_iξ_jy^λ + χξ_iξ_ky^λ ∈ nil(Ω)  for  γ² + χ² = 0.

(ii)$$.

Proof. (i) Set  x^ky^λξ^u  be an arbitrary standard element of  Ω. Then

Put Obviously,

Hence

Noting that  γ² + χ² = 0, we obtain

Similarly,  (adf) ⁴ = 0, and then  f = γξ_iξ_jy^λ + χξ_iξ_ky^λ ∈ nil(Ω).

(ii) Let  k ∈ T∖{i, j}.  It follows from (i) that

Hence$$.

Lemma 8. $$for  i ∈ M₁  and  j ∈ T.

Proof. By a direct computation, we obtain

Put$$and  C = −x_iy^λD_j. Observing  A² = B² = C² = 0  and  AB = BA = 0, we see that

Thus$$, as required.

Lemma 9. The following statements hold

• (i)

  Nil(Ω₀) = span _𝔽{x_ix_jy^λ, x_iξ_jy^λ, ξ_iξ_jy^λ∣i, j ∈ M₁ ∪ T}.

• (ii)

  For  q ≥ 3,$$.

• (iii)

  For  q = 2,$$.

Proof. (i) Suppose that$$is an arbitrary element of nil(Ω₀), where  γ₁ ∈ 𝔽. If  γ₁ ≠ 0, then  [γ₁x₁y^λ, y^λ] = γ₁(1 − λ)y^2λ. A direct calculation shows that$$. Thus  f  is not ad-nilpotent, contradicting the nilpotency of  f. Hence  γ₁ = 0  and  nil(Ω₀)⊆span _𝔽{x_ix_jy^λ, x_iξ_jy^λ, ξ_iξ_jy^λ∣i, j ∈ M₁ ∪ T}.  Obviously,  span _𝔽{x_ix_jy^λ, x_iξ_jy^λ, ξ_iξ_jy^λ∣i, j ∈ M₁ ∪ T}  is a subalgebra of  Ω, which yields Nil(Ω₀)⊆span _𝔽{x_ix_jy^λ, x_iξ_jy^λ, ξ_iξ_jy^λ∣i, j ∈ M₁ ∪ T}.

Conversely, we have  span _𝔽{x_ix_jy^λ, x_iξ_jy^λ, ξ_iξ_jy^λ∣i, j ∈ M₁ ∪ T}⊆Nil(Ω₀)  by virtue of Lemmas 6, 7, and 8. It follows that Nil(Ω₀) = span _𝔽{x_ix_jy^λ, x_iξ_jy^λ, ξ_iξ_jy^λ∣i, j ∈ M₁ ∪ T}.

(ii) By (i) of the lemma, we have

Conversely, the assertion$$follows from Lemmas 6 and 7.

(iii) Suppose$$ where  γ, γ_ij ∈ 𝔽. If  γ ≠ 0, then  adf(ξ_sy^λ) = [f, ξ_sy^λ] = γξ_r+1y^2λ  and  (adf) ²(ξ_sy^λ) = γ²ξ_sy^3λ. A direct calculation shows that  (adf) ^2m(ξ_sy^λ) = γ^2mξ_sy^2mλ ≠ 0. It follows that  f  is not ad-nilpotent. Thus  γ = 0. Then$$and nil$$Note that  span _𝔽{x_ix_jy^λ∣i, j ∈ M₁}  is a subalgebra of  Ω. Hence Nil$$and (iii) holds.

Lemma 10. (i)  ρ(Ω₀) = ℒ.

(ii) If  f ∈ nil(Ω₀), then  ρ(f)  is a nilpotent matrix.

Proof. (i) Let  i, j, k ∈ M₁ ∪ T. By computation, we have

Hence$$and  ρ(x₁) = (1 − μ_k − λ)E_s−2.  The other cases are treated similarly. Thus (i) holds.

(ii) As  f  is a nilpotent elements, ρ(f)  is a nilpotent liner transformation. Then by the definition of  ρ, we see that  ρ(f)  is a nilpotent matrix.

Lemma 11. If$$,  f ≠ 0, then there exists a$$such that$$.

Proof. By Lemma 9, we can assume that$$ where  γ_l, β_lt, χ_lt ∈ 𝔽.

Suppose  γ_i ≠ 0  for some  i ∈ M₁  and$$A direct calculation shows that

where every item of  h  does not contain  x_i. Then (ad$$. Thus  (ad$$,  for  all  n ∈ ℕ, which implies that  [f, z]  is not a nilpotent element.

If  γ_i = 0  for all  i ∈ M₁, then we let  β_ij ≠ 0  for some  i, j ∈ M₁  and$$. Also by computation, we have

where every item of h does not contain  x_i.  Similarly,  (ad$$,  for  all  n ∈ ℕ, and then  [f, z]  is not nilpotent.

Hence our assertion follows.

If  γ_i = 0  and  β_ij = 0  for all  i, j ∈ M₁, then  f = ∑_l,t∈T,l<tχ_ltξ_lξ_ty^λ ∈ span _𝔽{ξ_lξ_ty^λ∣l, t ∈ T, λ ∈ H}.  We see that  ρ(f)  is a antisymmetric nilpotent matrix. By Lemmas 9(ii) and 10, it is easy to see that there is  z ∈ span _𝔽{ξ_iξ_jy^λ∣i, j ∈ T}  such that  [ρ(f), ρ(z)]  is not a nilpotent matrix; that is,  [f, z]  is not an ad-nilpotent element. Hence our assertion holds.

Proposition 12. nilΩ = Ω_[1] ⊕ span_𝔽{x_ix_jy^λ, x_iξ_jy^λ, ξ_iξ_jy^λ∣λ ∈ H, i, j ∈ M₁ ∪ T}.

Proof. According to Lemma 4, we need only to determine all ad-nilpotent elements in  Ω₋₂ ⊕ Ω₋₁ ⊕ Ω₀. For any  n ∈ ℕ, a direct computation shows that

Lemmas 6, 7, and 8 imply that  x_ix_jy^λ,  x_iξ_jy^λ, and ξ_iξ_jy^λ  are ad-nilpotent for  i, j ∈ M₁ ∪ T. Hence our assertion holds.

Lemma 13. $$and  Ω_[0]  is invariant.

Proof. Firstly, we prove the inclusion$$. Lemmas 6–9 show that Nil$$, which combined with (iv) and (v) of Lemma 5, yield

Then

that is,$$.

Let us consider the converse inclusion. Suppose that$$, where  f₋₂ ∈ Ω₋₂  and  f_[−1] ∈ Ω_[−1]. If  f₋₂ ≠ 0, then$$, where  h ∈ Ω_[0]  and  γ ∈ 𝔽, a contradiction. Consequently,  f₋₂ = 0.

Now suppose  f = f₋₁ + f_[0], where$$If  y₋₁ ≠ 0  and  γ_j ≠ 0  for some  j ∈ M₁, then we have$$, where  h ∈ Ω_[0], a contradiction. Thus  f₋₁ = 0  and$$This proves the asserted inclusion.

By the proof above, we know that  Ω_[0]  is invariant.

Lemma 14. $$and  Ω_[1]  is invariant.

Proof. Let$$. Suppose$$. By Lemma 5,  f₋₂ = 0  and  f₋₁ = 0.  Then we can assume that  f = f₀ + f_[1] ∈ ℳ, where$$Let  f₀ ≠ 0.  Clearly,  f₀ ∈ nil(Ω₀).  Lemma 5(1) implies that$$. According to Lemma 11, there exists$$such that$$. Thus$$and  f_[1]  is not nilpotent, contradicting the result of Lemma 5. Hence  f₀ = 0  and$$; that is,$$.

Conversely,$$It is obvious that$$and the proof is complete.

Lemma 15. (i)  $$.

(ii)  $$.

(iii)  Ω_[−1] = {f ∈ Ω∣[f, Ω_[1]]⊆Ω_[0]}.

Proof. (i) Put$$Suppose  f = f₋₁ + f_[0], where$$and  γ_i ∈ 𝔽. Let  γ_j ≠ 0  for some one  j ∈ T. Then$$; that is,$$. This contradicts  f ∈ 𝒜. Thus  f₋₁ = 0.

Let  f = f₀ + f_[1]  be an arbitrary element of  𝒜, where$$and  γ_i,j ∈ 𝔽. If there is a  γ_it ≠ 0, then$$, contradicting  f ∈ 𝒜. Hence  f₀ = 0  and$$. Consequently,$$.

Conversely, let$$. Then$$, which shows that$$.

(ii) We first prove the inclusion$$. Suppose$$, where  ξ^u = ξ_iξ^v. Noting that$$and  x^ky^λξ^u = −2[ξ_i, x₁x^ky^λξ^v], we obtain$$The converse inclusion is obvious.

(iii) The proof is completely analogous to (ii).

Theorem 16. Ω  is transitive.

Proof. Assume the contrary. Suppose that there exists a nonzero  f ∈ Ω_l  such that  [f, Ω₋₁] = 0, where  l ∈ ℕ₀. Let  t  be the maximal exponent of  x₁  of all monomial expressions occurring in  f. Then we may assume that

where  χ, ω ∈ 𝔽. For any  j ∈ M₁, we have where  h  is the sum of summand that the exponent of  x₁  is less than  t. Since$$is linear independence,$$; that is, $$. For any  j ∈ T, we get

where  h  is the sum of summand that the exponent of  x₁  is less than  t. Thus$$; that is,  x^ky^λξ^u  does not contain  ξ_j  and$$.

If  t = 0, then  f ∈ Ω₋₂, contradicting  l ≥ 0. Let  t > 0. Then we can suppose

where$$For  j ∈ M₁, we have where  h  is the sum of summand that the exponent of  x₁  is less than  t − 1. Thus By $$ contains  x_j  and $$ dose not contain  x_j, we have  t = 0, contradicting  t > 0. Hence  Ω  is transitive, as required.

Theorem 17. Suppose that  Ω  and  Ω^′  are the Lie superalgebras of  Ω-type. If  φ  is an isomorphism of  Ω  onto  Ω^′, then$$ for all  i ≥ −2.

Proof. As the isomorphism  φ  is an even mapping, we have$$and$$. Hence$$. By Lemmas 13 and 14, we obtain

Now by virtue of Lemma 15, we get It follows that$$. Because$$, we have Since  Ω  is transitive by the lemma above, we have  Ω_[i+1] = {f ∈ Ω_[i]∣[f, Ω_[−1]]⊆Ω_[i]}  for all  i ≥ 0. It is easy to show that$$by induction on  i.

Theorem 18. Suppose that  ϕ  is an automorphism of  Ω. Then  ϕ(Ω_[i]) = Ω_[i]  for all i ≥ −2, that is; the filtration of  Ω  is invariant under the automorphism group of  Ω.

Proof. This is a direct consequence of Theorem 17.

Theorem 19. Let  Ω(r, m, q)  and  Ω(r^′, m^′, q^′)  be the Lie superalgebras of  Ω-type. Then  Ω(r, m, q)≅Ω(r^′, m^′, q^′)  if and only if  r = r^′, m = m^′, q = q^′.

Proof. The sufficient condition is obvious. We will prove the necessary condition. Assume that  φ : Ω(r, m, q) → Ω(r^′, m^′, q^′)  is an isomorphism of Lie superalgebra. According to Theorem 17, we have

Then  φ  induces an isomorphism of  ℤ₂-graded spaces It is easy to see that

We conclude that  m = m^′  by the dimension comparison.

Similarly, we also obtain an isomorphism of  ℤ₂-graded spaces:

and the quotient space  Ω_[−1]/Ω_[0]  is isomorphic to  Ω₋₁.  A comparison of dimensions shows that  r + q = r^′ + q^′. Note that  φ(Ω(r, m, q) _α) = Ω(r^′, m^′, q^′) _α, where  α ∈ ℤ₂.  It follows that  r = r^′  and  q = q^′.