Arveson’s version of the Gauss–Bonnet–Chern formula for Hilbert modules over the polynomial ring

Penghui Wang; Ruoyu Zhang; Zeyou Zhu

doi:10.1515/crelle-2025-0049

Article

Arveson’s version of the Gauss–Bonnet–Chern formula for Hilbert modules over the polynomial ring

Penghui Wang , Ruoyu Zhang and Zeyou Zhu

Published/Copyright: July 24, 2025

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From the journal Journal für die reine und angewandte Mathematik (Crelles Journal) Volume 2025 Issue 827

Abstract

In this paper, we complete the framework of Arveson’s version of the Gauss–Bonnet–Chern formula by proving that Arveson’s version of the Gauss–Bonnet–Chern formula holding true for the quotient module in the Drury–Arveson module is equivalent to the associated submodule being locally algebraic. Moreover, we establish the asymptotic Arveson curvature invariant and the asymptotic Euler characteristic for contractive Hilbert modules over the polynomial ring in infinitely many variables, and obtain the infinitely-many-variables analogue of Arveson’s version of the Gauss–Bonnet–Chern formula. Finally, we solve the finite defect problem for submodules of the Drury–Arveson module H 2 in infinitely many variables by proving that H 2 has no nontrivial submodules of finite rank.

Funding source: National Natural Science Foundation of China

Award Identifier / Grant number: 12271298

Award Identifier / Grant number: 11871308

Funding statement: This work is supported by National Natural Science Foundation of China: 12271298 and 11871308.

A Appendix

In this section, we will prove (2.14), Proposition 4.8 and Proposition 4.10.

Let ℋ be a 𝑑-contractive Hilbert module of finite rank. Suppose that dim ⁡ ran ⁡ Δ H = n . Let A = { e 1 , … , e n } be a linear basis of ran ⁡ Δ H . Without loss of generality, suppose that B = { e 1 , … , e k } is a maximal linearly independent set of { e 1 , … , e n } in the module M H . Then, for every k + 1 ≤ i ≤ n , { e 1 , … , e k , e i } is linearly dependent in the module M H ; hence there exist polynomials p 1 , i , … , p k , i and a nonzero polynomial p i , i such that

(A.1) ∑ j = 1 k p j , i ⁢ e j + p i , i ⁢ e i = 0 .

Write

C = { deg ⁡ p j , i : 1 ≤ j ≤ k , k + 1 ≤ i ≤ n } ∪ { deg ⁡ p i , i : k + 1 ≤ i ≤ n } ,

and

(A.2) n 1 = max ⁡ { r : r ∈ C } .

For m > n 1 , k + 1 ≤ i ≤ n , set

S m = { f ∈ P d : deg ⁡ f ≤ m } and S m , i = { f ⋅ p i , i : f ∈ P d , deg ⁡ f ≤ m − n 1 } .

It is evident that S m , i is a subspace of S m . We can thus have a subspace, denoted by S m , i ′ , such that

(A.3) S m = S m , i ′ ⊕ S m , i .

For m ≥ 1 , set

(A.4) q m ⁢ ( x ) = ( x + 1 ) ⁢ ( x + 2 ) ⁢ ⋯ ⁢ ( x + m ) m ! .

To prove (2.14), we need the following two lemmas.

Lemma A.1

For k + 1 ≤ i ≤ n ,

lim m → ∞ dim ⁡ { f ⋅ e i : f ∈ S m , i ′ } q d ⁢ ( m ) = 0 .

Proof

For m > n 1 , by (A.3),

dim ⁡ S m , i ′ = dim ⁡ S m − dim ⁡ S m , i = q d ⁢ ( m ) − q d ⁢ ( m − n 1 ) .

Therefore,

dim ⁡ { f ⋅ e i : f ∈ S m , i ′ } ≤ dim ⁡ S m , i ′ = q d ⁢ ( m ) − q d ⁢ ( m − n 1 ) .

By the definition of q d ⁢ ( x ) in (A.4), there is a positive constant C d such that

dim ⁡ { f ⋅ e i : f ∈ S m , i ′ } ≤ q d ⁢ ( m ) − q d ⁢ ( m − n 1 ) ≤ C d ⁢ m d − 1 .

This implies that

lim m → ∞ dim ⁡ span ⁡ { f ⋅ e i : f ∈ S m , i ′ } q d ⁢ ( m ) = 0 . ∎

For any α = ( α 1 , α 2 , … , α d ) ∈ N d , set

| α | = ∑ i = 1 d α i , α ! = α 1 ! ⁢ ⋯ ⁢ α d ! , and z α = z 1 α 1 ⁢ z 2 α 2 ⁢ ⋯ ⁢ z d α d , z ∈ B d .

Lemma A.2

For m ≥ 1 ,

dim ⁡ span ⁡ { f ⋅ ζ : f ∈ P d , deg ⁡ f ≤ m , ζ ∈ B } q d ⁢ ( m ) = k .

Proof

Set

F m = { z α : | α | ≤ m } and B m = { f ⋅ e : f ∈ F m , e ∈ B } .

It suffices to show that B m is a linear basis of span ⁡ { f ⋅ ζ : f ∈ P d , deg ⁡ f ≤ m , ζ ∈ B } . In fact, obviously,

span ⁡ B m = span ⁡ { f ⋅ ζ : f ∈ P d , deg ⁡ f ≤ m , ζ ∈ B } .

We only need to prove B m is linearly independent. Write

B m = { f j ⋅ e i : 1 ≤ j ≤ F m ♯ , 1 ≤ i ≤ k } ,

where F m ♯ is the number of the elements in F m . Assume that, for λ i , j ∈ C ,

∑ i = 1 k ∑ j = 1 F m ♯ λ i , j ⁢ f j ⁢ e i = 0 ;

then, from the fact that 𝐵 is a maximal linearly independent set of { e 1 , … , e n } in the module M H , ∑ j = 1 F m ♯ λ i , j ⁢ f j = 0 for 1 ≤ i ≤ k . Notice that F m is linearly independent; therefore, λ i , j = 0 .

It is easy to see that F m ♯ = q d ⁢ ( m ) and B m ♯ = k ⁢ q d ⁢ ( m ) . Then

dim ⁡ span ⁡ { f ⋅ ζ : f ∈ P d , deg ⁡ f ≤ m , ζ ∈ B } = B m ♯ = k ⁢ q d ⁢ ( m ) ,

which implies that

dim ⁡ span ⁡ { f ⋅ ζ : f ∈ P d , deg ⁡ f ≤ m , ζ ∈ B } q d ⁢ ( m ) = k . ∎

Under the above preparations, we can prove (2.14).

Proposition A.3

Let ℋ be a 𝑑-contractive Hilbert module of finite rank; then

χ ⁢ ( H ) = lim m → ∞ dim ⁡ M m q d ⁢ ( m ) = d ! ⁢ lim m → ∞ dim ⁡ M m m d ,

where M m = span ⁡ { f ⋅ ζ : f ∈ P d , deg ⁡ f ≤ m , ζ ∈ ran ⁡ Δ H } .

Proof

By (2.1), χ ⁢ ( H ) = rank ⁡ M H . Notice that { e i } i = 1 n generates M H ; then, by Lemma 2.3, χ ⁢ ( H ) = rank ⁡ ( M H ) = k . Hence it suffices to prove that

lim m → ∞ dim ⁡ M m q d ⁢ ( m ) = k .

Set

V i = { f ⋅ e i : f ∈ P d , deg ⁡ f ≤ m } , k + 1 ≤ i ≤ n .

For k + 1 ≤ i ≤ n , by (A.1) and (A.2), we have

{ f ⁢ p i , i ⋅ e i : f ∈ P d , deg ⁡ f ≤ m − n 1 } ⊆ span ⁡ { f ⋅ ζ : f ∈ P d , deg ⁡ f ≤ m , ζ ∈ B } ;

then

V i = { f ⋅ e i : f ∈ S m } = { f ⋅ e i : f ∈ S m , i ⊕ S m , i ′ } = { f ⋅ e i : f ∈ S m , i } + { f ⋅ e i : f ∈ S m , i ′ } = { f ⁢ p i , i ⋅ e i : f ∈ P d , deg ⁡ f ≤ m − n 1 } + { f ⋅ e i : f ∈ S m , i ′ } ⊆ span ⁡ { f ⋅ ζ : f ∈ P d , deg ⁡ f ≤ m , ζ ∈ B } + { f ⋅ e i : f ∈ S m , i ′ } .

Hence

span ⁡ { f ⋅ ζ : f ∈ P d , deg ⁡ f ≤ m , ζ ∈ ran ⁡ Δ H } = span ⁡ { f ⋅ ζ : f ∈ P d , deg ⁡ f ≤ m , ζ ∈ A } = span ⁡ { f ⋅ ζ : f ∈ P d , deg ⁡ f ≤ m , ζ ∈ B } + V k + 1 + ⋯ + V n ⊆ span ⁡ { f ⋅ ζ : f ∈ P d , deg ⁡ f ≤ m , ζ ∈ B } + { f ⋅ e k + 1 : f ∈ S m , k + 1 ′ } + ⋯ + { f ⋅ e n : f ∈ S m , n ′ } .

Therefore,

dim ⁡ span ⁡ { f ⋅ ζ : f ∈ P d , deg ⁡ f ≤ m , ζ ∈ B } ≤ dim ⁡ M m = dim ⁡ span ⁡ { f ⋅ ζ : f ∈ P d , deg ⁡ f ≤ m , ζ ∈ ran ⁡ Δ H } ≤ dim ⁡ span ⁡ { f ⋅ ζ : f ∈ P d , deg ⁡ f ≤ m , ζ ∈ B } + ∑ i = k + 1 n dim ⁡ { f ⋅ e i : f ∈ S m , i ′ } .

By Lemma A.1 and Lemma A.2, we have

lim m → ∞ dim ⁡ M m q d ⁢ ( m ) = k . ∎

Next, we will prove Proposition 4.8. The ideas and techniques come from [3]. Let ℋ be an 𝜔-contractive Hilbert module of finite rank. For simplicity, write R = ran ⁡ Δ H . Let H 2 ⁢ ( ∂ B m ) be the Hardy spaces on the unit ball, and let P H m 2 ⊗ R be the projection from H 2 ⊗ R onto H m 2 ⊗ R . Considering the linear operator A m : E → H 2 ⁢ ( ∂ B m ) ⊗ R defined by

A m ⁢ ζ = b m ⁢ P H m 2 ⊗ R ⁢ Φ ⁢ ( 1 ⊗ ζ ) , ζ ∈ E ,

where Φ is defined in (4.8) and

b m : H m 2 ⊗ R → H 2 ⁢ ( ∂ B m ) ⊗ R

is the natural inclusion mapping. To continue, we need some lemmas.

Lemma A.4

K m ⁢ ( H ) = rank ⁡ H − trace ⁡ ( A m ⁢ A m ∗ ) .

Proof

Let Φ ̂ = P H m 2 ⊗ R ⁢ Φ ⁢ P H m 2 ⊗ E . Then, for every ζ 1 ∈ E , ζ 2 ∈ R , z ( m ) ∈ B m ,

⟨ Φ ̂ ⁢ ( z ( m ) ) ⁢ ζ 1 , ζ 2 ⟩ = ⟨ Φ ̂ ⁢ ( 1 ⊗ ζ 1 ) , K z ( m ) ⊗ ζ 2 ⟩ = ⟨ P H m 2 ⊗ R ⁢ Φ ⁢ P H m 2 ⊗ E ⁢ ( 1 ⊗ ζ 1 ) , K z ( m ) ⊗ ζ 2 ⟩ = ⟨ Φ ⁢ ( 1 ⊗ ζ 1 ) , K z ( m ) ⊗ ζ 2 ⟩ = ⟨ Φ ⁢ ( z ( m ) ) ⁢ ζ 1 , ζ 2 ⟩ .

A m ⁢ ζ ⁢ ( z ( m ) ) = Φ ̂ ⁢ ( 1 ⊗ ζ ) ⁢ ( z ( m ) ) = Φ ̂ ⁢ ( z ( m ) ) ⁢ ζ = Φ ⁢ ( z ( m ) ) ⁢ ζ ,

and for almost every z ( m ) ∈ ∂ B m ,

trace ⁡ ( Φ ⁢ ( z ( m ) ) ⁢ Φ ⁢ ( z ( m ) ) ∗ ) = trace ⁡ ( Φ ⁢ ( z ( m ) ) ∗ ⁢ Φ ⁢ ( z ( m ) ) ) = ∑ n = 1 ∞ ∥ Φ ⁢ ( z ( m ) ) ⁢ e n ∥ 2 = ∑ n = 1 ∞ ∥ A m ⁢ e n ⁢ ( z 1 , … , z m ) ∥ 2 ,

where { e n } i = 1 ∞ is a basis of 𝐸. Integrating over ∂ B m , we have

∫ ∂ B m trace ⁡ ( Φ ⁢ ( z ( m ) ) ⁢ Φ ⁢ ( z ( m ) ) ∗ ) ⁢ d σ m = ∑ n = 1 ∞ ∫ ∂ B m ∥ A m ⁢ e n ⁢ ( z ( m ) ) ∥ 2 ⁢ d σ m = ∑ n = 1 ∞ ∥ A m ⁢ e n ∥ H 2 ⁢ ( ∂ B m ) ⊗ R 2 = trace ⁡ ( A m ∗ ⁢ A m ) = trace ⁡ ( A m ⁢ A m ∗ ) .

Then, by (4.5), (4.6) and (4.7),

K m ⁢ ( H ) = ∫ ∂ B m trace ⁡ ( I R − Φ ⁢ ( z ( m ) ) ⁢ Φ ⁢ ( z ( m ) ) ∗ ) ⁢ d σ m = rank ⁡ H − trace ⁡ ( A m ⁢ A m ∗ ) .

The proof is complete. ∎

Now, we define a linear map Γ m : B ⁢ ( H m 2 ⊗ R ) → B ⁢ ( H 2 ⁢ ( ∂ B m ) ⊗ R ) as follows:

Γ m ⁢ ( X ) = b m ⁢ X ⁢ b m ∗ .

Let Z 1 , … , Z m be the canonical operators of the Hardy module H 2 ⁢ ( ∂ B m ) ⊗ R . The linear map d ⁢ Γ m : B ⁢ ( H m 2 ⊗ R ) → B ⁢ ( H 2 ⁢ ( ∂ B m ) ⊗ R ) is defined as follows:

d ⁢ Γ m ⁢ ( X ) = Γ m ⁢ ( X ) − ∑ k = 1 m Z k ⁢ Γ m ⁢ ( X ) ⁢ Z k ∗ .

Lemma A.5

d ⁢ Γ m ⁢ ( P H m 2 ⊗ R ⁢ Φ ⁢ Φ ∗ ⁢ P H m 2 ⊗ R ) = A m ⁢ A m ∗ .

Proof

Notice that, for ξ ∈ H 2 ⁢ ( ∂ B m ) ⊗ R , η ∈ E ,

⟨ 1 ⊗ A m ∗ ⁢ ξ , 1 ⊗ η ⟩ = ⟨ A m ∗ ⁢ ξ , η ⟩ = ⟨ ξ , A m ⁢ η ⟩ = ⟨ ξ , b m ⁢ P H m 2 ⊗ R ⁢ Φ ⁢ P H m 2 ⊗ E ⁢ ( 1 ⊗ η ) ⟩ = ⟨ P H m 2 ⊗ E ⁢ Φ ∗ ⁢ P H m 2 ⊗ R ⁢ b m ⁢ ξ ∗ , 1 ⊗ η ⟩ .

It follows that

⟨ A m ⁢ A m ∗ ⁢ ξ , η ⟩ = ⟨ 1 ⊗ A m ∗ ⁢ ξ , 1 ⊗ A m ∗ ⁢ η ⟩ = ⟨ P H m 2 ⊗ E ⁢ Φ ∗ ⁢ P H m 2 ⊗ R ⁢ b m ∗ ⁢ ξ , 1 ⊗ A m ∗ ⁢ η ⟩ = ⟨ Φ ∗ ⁢ b m ∗ ⁢ ξ , 1 ⊗ A m ∗ ⁢ η ⟩ = ⟨ Φ ∗ ⁢ b m ∗ ⁢ ξ , ( E 0 ⊗ I E ) ⁢ ( 1 ⊗ A m ∗ ⁢ η ) ⟩ = ⟨ ( E 0 ⊗ I E ) ⁢ Φ ∗ ⁢ b m ∗ ⁢ ξ , 1 ⊗ A m ∗ ⁢ η ⟩ = ⟨ ( E 0 ⊗ I E ) ⁢ Φ ∗ ⁢ b m ∗ ⁢ ξ , P H m 2 ⊗ E ⁢ Φ ∗ ⁢ P H m 2 ⊗ R ⁢ b m ∗ ⁢ η ⟩ = ⟨ b m ⁢ P H m 2 ⊗ R ⁢ Φ ⁢ P H m 2 ⊗ E ⁢ ( E 0 ⊗ I E ) ⁢ Φ ∗ ⁢ b m ∗ ⁢ ξ , η ⟩ = ⟨ b m ⁢ P H m 2 ⊗ R ⁢ Φ ⁢ ( E 0 ⊗ I E ) ⁢ Φ ∗ ⁢ P H m 2 ⊗ R ⁢ b m ∗ ⁢ ξ , η ⟩ .

Thus, from the fact that Φ ∈ hom ⁡ ( H 2 ⊗ E , H 2 ⊗ R ) , we have

A m ⁢ A m ∗ = b m ⁢ P H m 2 ⊗ R ⁢ Φ ⁢ ( E 0 ⊗ I E ) ⁢ Φ ∗ ⁢ P H m 2 ⊗ R ⁢ b m ∗ = ( b m ⁢ P H m 2 ⊗ R ⁢ Φ ) ⁢ ( I H 2 ⊗ E − ∑ i = 1 ∞ T i ⁢ T i ∗ ) ⁢ ( b m ⁢ P H m 2 ⊗ R ⁢ Φ ) ∗ = ( b m ⁢ P H m 2 ⊗ R ⁢ Φ ) ⁢ ( b m ⁢ P H m 2 ⊗ R ⁢ Φ ) ∗ − ∑ i = 1 m ( b m ⁢ P H m 2 ⊗ R ⁢ Φ ) ⁢ T i ⁢ T i ∗ ⁢ ( b m ⁢ P H m 2 ⊗ R ⁢ Φ ) ∗ = Γ m ⁢ ( P H m 2 ⊗ R ⁢ Φ ⁢ Φ ∗ ⁢ P H m 2 ⊗ R ) − ∑ i = 1 m Z i ⁢ ( b m ⁢ P H m 2 ⊗ R ⁢ Φ ) ⁢ ( b m ⁢ P H m 2 ⊗ R ⁢ Φ ) ∗ ⁢ Z i ∗ = Γ m ⁢ ( P H m 2 ⊗ R ⁢ Φ ⁢ Φ ∗ ⁢ P H m 2 ⊗ R ) − ∑ i = 1 m Z i ⁢ Γ m ⁢ ( P H m 2 ⊗ R ⁢ Φ ⁢ Φ ∗ ⁢ P H m 2 ⊗ R ) ⁢ Z i ∗ = d ⁢ Γ m ⁢ ( P H m 2 ⊗ R ⁢ Φ ⁢ Φ ∗ ⁢ P H m 2 ⊗ R ) ,

where T i = S z i ⊗ I E .∎

Set ϕ m ⁢ ( X ) = ∑ k = 1 m T k ⁢ X ⁢ T k ∗ .

Proof of Proposition 4.8

By Lemma A.5 and [3, Theorem 3.10], we have

trace ⁡ ( A m ⁢ A m ∗ ) = trace ⁡ ( d ⁢ Γ m ⁢ ( P H m 2 ⊗ R ⁢ Φ ⁢ Φ ∗ ⁢ P H m 2 ⊗ R ) ) = rank ⁡ H ⁢ lim n → ∞ trace ⁡ ( P H m 2 ⊗ R ⁢ Φ ⁢ Φ ∗ ⁢ P H m 2 ⊗ R ⁢ E n , m ) trace ⁡ ( E n , m ) ,

where E n , m is the projection from H m 2 ⊗ R onto its subspace of homogeneous (vector-valued) functions of degree 𝑛. Since L ∗ ⁢ L + Φ ⁢ Φ ∗ = I , by Lemma A.4,

K m ⁢ ( H ) = rank ⁡ H − trace ⁡ ( A m ⁢ A m ∗ ) = rank ⁡ H ⁢ lim n → ∞ ( 1 − trace ⁡ ( P H m 2 ⊗ R ⁢ Φ ⁢ Φ ∗ ⁢ P H m 2 ⊗ R ⁢ E n , m ) trace ⁡ ( E n , m ) ) = rank ⁡ H ⁢ lim n → ∞ trace ⁡ ( P H m 2 ⊗ R ⁢ L ∗ ⁢ L ⁢ P H m 2 ⊗ R ⁢ E n , m ) trace ⁡ ( E n , m ) = lim n → ∞ trace ⁡ ( P H m 2 ⊗ R ⁢ L ∗ ⁢ L ⁢ P H m 2 ⊗ R ⁢ E n , m ) q m − 1 ⁢ ( n ) = lim n → ∞ trace ⁡ ( L ⁢ P H m 2 ⊗ R ⁢ E n , m ⁢ P H m 2 ⊗ R ⁢ L ∗ ) q m − 1 ⁢ ( n ) = lim n → ∞ trace ⁡ ( ϕ m n ⁢ ( Δ H 2 ) ) q m − 1 ⁢ ( n ) .

The last equation follows from the reasoning below. For η ∈ H , by Theorem 4.1,

L ∗ ⁢ η = ( ζ 0 , ζ 1 , … ) , ζ n = ∑ i 1 , i 2 , … , i n = 1 ∞ e i 1 ⊗ ⋯ ⊗ e i n ⊗ Δ H ⁢ T i n ∗ ⁢ ⋯ ⁢ T i 1 ∗ ⁢ η .

Let ζ m , n = ∑ i 1 , i 2 , … , i n = 1 m e i 1 ⊗ ⋯ ⊗ e i n ⊗ Δ H ⁢ T i n ∗ ⁢ ⋯ ⁢ T i 1 ∗ ⁢ η ; then

L ⁢ P H m 2 ⊗ R ⁢ E n , m ⁢ P H m 2 ⊗ R ⁢ L ∗ ⁢ η = L ⁢ P H m 2 ⊗ R ⁢ E n , m ⁢ P H m 2 ⊗ R ⁢ ( ζ 0 , ζ 1 , … ) = L ⁢ P H m 2 ⊗ R ⁢ E n , m ⁢ ( ζ m , 0 , ζ m , 1 , … ) = L ⁢ P H m 2 ⊗ R ⁢ ζ m , n = L ⁢ ζ m , n = ∑ i 1 , … , i n = 1 m T i 1 ⁢ ⋯ ⁢ T i n ⁢ Δ H 2 ⁢ T i n ∗ ⁢ ⋯ ⁢ T i 1 ∗ ⁢ η = ϕ m n ⁢ ( Δ H 2 ) ⁢ η .

Therefore,

K m ⁢ ( H ) = lim n → ∞ trace ⁡ ( ϕ m n ⁢ ( Δ H 2 ) ) q m − 1 ⁢ ( n ) = lim n → ∞ ∑ k = 0 n trace ⁡ ( ϕ m k ⁢ ( Δ H 2 ) ) ∑ k = 0 n q m − 1 ⁢ ( k ) = lim n → ∞ trace ⁡ ( ∑ k = 0 n ϕ m k ⁢ ( Δ H 2 ) ) q m ⁢ ( n ) . ∎

Finally, we will prove Proposition 4.10.

Proof of Proposition 4.10

Obviously,

M H m , n = span ⁡ { f ⋅ Δ H ⁢ ζ : f ∈ P m , deg ⁡ f ≤ n , ζ ∈ H } = span ⁡ { T i 1 ⁢ ⋯ ⁢ T i k ⁢ Δ H ⁢ ζ : ζ ∈ H , 1 ≤ i 1 , … , i k ≤ m , k = 0 , 1 , … , n }

is a linear submanifold of ℋ. Then

M H m , n ⟂ = { η ∈ H : ∑ k = 0 n ∑ i 1 = 1 m ⋯ ⁢ ∑ i k = 1 m | ⟨ T i 1 ⁢ ⋯ ⁢ T i k ⁢ Δ H ⁢ ζ , η ⟩ | = 0 ⁢ for all ⁢ ζ ∈ H } = { η ∈ H : ∑ k = 0 n ∑ i 1 = 1 m ⋯ ⁢ ∑ i k = 1 m ∥ Δ H ⁢ T i k ∗ ⁢ ⋯ ⁢ T i 1 ∗ ⁢ η ∥ 2 = 0 } = { η ∈ H : ∑ k = 0 n ⟨ ϕ m k ⁢ ( Δ H 2 ) ⁢ η , η ⟩ = 0 } = ker ⁡ ( ∑ k = 0 n ϕ m k ⁢ ( Δ H 2 ) ) .

Hence, by (4.13),

χ m ⁢ ( H ) = m ! ⁢ lim n → ∞ dim ⁡ M H m , n n m = m ! ⁢ lim n → ∞ dim ⁡ ran ⁡ ( ∑ k = 0 n ϕ m k ⁢ ( Δ H 2 ) ) n m . ∎

Acknowledgements

The authors thank the referees for helpful suggestions, which make this paper more readable.

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Received: 2025-02-17

Revised: 2025-06-19

Published Online: 2025-07-24

Published in Print: 2025-10-01

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