Mixed Tensor Products, Capelli Berezinians, and Newton's Formula for $\mathfrak{gl}(m|n)$
Sidarth Erat Arun S. Kannan Shihan Kanungo

Mixed Tensor Products, Capelli Berezinians, and Newton's Formula for (m|n)]Mixed Tensor Products, Capelli Berezinians, and Newton's Formula for (m|n)

[2020]17B10, 17B35


S. Erat]Sidarth Erat
Sidarth Erat  La Jolla High School  La Jolla, CA 92037
sidarth.erat@gmail.com
A. S. Kannan]Arun S. Kannan
Arun S. Kannan  Chicago, IL 60606
arun_kannan@alum.mit.edu
S. Kanungo]Shihan Kanungo
Shihan Kanungo  Henry M. Gunn High School  Palo Alto, CA 94306
shihankanungo@gmail.com
	
§ ABSTRACT  
    In this paper, we extend the results of Grantcharov and Robitaille in 2021 on mixed tensor products and Capelli determinants to the superalgebra setting. Specifically, we construct a family of superalgebra homomorphisms φ_R : U(𝔤𝔩(m+1|n)) →𝒟'(m|n) ⊗ U(𝔤𝔩(m|n)) for a certain space of differential operators 𝒟'(m|n) indexed by a central element R of 𝒟'(m|n) ⊗ U(𝔤𝔩(m|n)). We then use this homomorphism to determine the image of Gelfand generators of the center of U(𝔤𝔩(m+1|n)). We achieve this by first relating φ_R to the corresponding Harish-Chandra homomorphisms and then proving a super-analog of Newton's formula for 𝔤𝔩(m) relating Capelli generators and Gelfand generators. We also use the homomorphism φ_R to obtain representations of U(𝔤𝔩(m+1|n)) from those of U(𝔤𝔩(m|n)), and find conditions under which these inflations are simple. Finally, we show that for a distinguished central element R_1 in 𝒟'(m|n)⊗ U(𝔤𝔩(m|n)), the kernel of φ_R_1 is the ideal of U(𝔤𝔩(m+1|n)) generated by the first Gelfand invariant G_1. 



[
    [
    September 9, 2024
=====================






§ INTRODUCTION






 §.§ The Lie algebra setting
 Mixed-tensor type modules, also known as tensor modules or modules of Shen and Larson (see <cit.>), are modules over the tensor product algebra 𝒟'(m) ⊗ U((m)), where 𝒟'(m) is a certain algebra of differential operators on the ring ℂ[t_0^± 1, t_1, …, t_m] and U((m)) is the universal enveloping algebra of the general linear Lie algebra (m). Questions related to mixed-tensor type modules have been well studied over the past 35 years (see <cit.> and references therein). Of interest to us is an algebra homomorphism 

    U((m+1)) →𝒟'(m) ⊗ U((m))

which enables one to inflate representations from U((m)) to U((m+1)) (see <cit.>). Although this map can be written down algebraically in a straightforward way, there exists a natural geometric interpretation. In particular, it can be thought of as a form of Beilinson-Bernstein localization for parabolic subgroups using a certain maximal parabolic subgroup P ⊂ SL(m+1) such that GL(m) ⊂ P and such that SL(m+1)/P ≅ P^m (see <cit.> and references therein for more details).

In <cit.>, the homomorphism U((m+1)) →𝒟'(m) ⊗ U((m)) is extended to U((m+1)). Moreover, the authors explicitly determine the kernel of this extension as well as the image of the Gelfand generators of the center Z(U((m))) of U((m)). Key ingredients in determining the image of the center are Capelli determinants, Capelli generators, and Newton's formula (for all of these, see <cit.> and <cit.>). Along the way, numerous interesting algebraic identities are also proven; it is expected that these will have implications for the representation theory of tensor modules.

Additionally and importantly to us, the notion of mixed-tensor type modules can be generalized to the setting of Lie superalgebras. Moreover the study of tensor modules in the super setting has received much interest recently (see <cit.> and references therein). This setting is the focus of this paper. In particular, we generalize the results of <cit.> to the super setting for the general Lie superalgebra (m+1|n). Most of the arguments therein generalize, although there are some subtleties that arise from the super setting.




 §.§ Results of this paper

First, we construct a superalgebra of differential operators '(m|n) that preserve degree on the superalgebra 𝒞[t_0^± 1, t_1, …, t_m] ⊗Λ_n, where Λ_n is the exterior algebra on n generators with odd parity. In Theorem <ref>, we define via explicit formulas a family of superalgebra homomorphisms 

    φ_R: U((m+1|n)) →𝒟'(m|n) ⊗ U((m|n))
 
indexed by central elements R of 𝒟'(m|n) ⊗ U((m|n)). Moreover, for m+1 ≠ n there is a distinguished homomorphism φ_R_1 for a particular central element R_1. The map φ_R_1 is an honest generalization of the map ρ in <cit.> and is “compatible" with a certain projection homomorphism π^g: U((m+1|n)) → U((m+1|n)) and the canonical inclusion in the opposite direction (see Proposition <ref>). 

Using the homomorphism φ_R, we can consider inflating irreducible representations of U(𝔤𝔩(m|n)) to U(𝔤𝔩(m+1|n)) by tensoring with a highest-weight module ℱ_a over 𝒟'(m|n) generated by t_0^a (such a module can be defined for any a ∈ℂ) and pulling back via φ_R. A partial criteria for when such representations are irreducible is the combined content of Theorems <ref> and <ref>. 

We then take a closer look at the restriction of φ_R to the center of U((m+1|n)). In particular, we show that φ_R interacts nicely with the Harish-Chandra homomorphism in Theorem <ref> and compute the image of the Gelfand generators G_k^(m+1|n) of the center in Theorem <ref>. To do the latter, we use the theory of Yangians. In particular, we define what is meant by a Capelli generator in the super setting (originally due to <cit.>) and then in Theorem <ref> we prove a super Newton's formula (analogous to <cit.>)[Based on Umeda's comments in <cit.>,  we expect that our formula is probably already known and seen as a consequence of the “quantum Liouville" formula for Yangians, but we could not find the result nor a proof in the literature and therefore include it in the paper here.], which relates the Capelli generators to the Gelfand generators using what we call a Capelli Berezinian. Then, by computing the images of the Capelli Berezinian under φ_R and applying the super Newton's formula, one can compute the images of Gelfand generators.  An alternative, more computational approach is given in Appendix <ref>.

Finally, we show that the kernel of φ_R_1 is generated by the first Gelfand invariant G_1 in Theorem <ref>. All our results, except Theorems <ref> and <ref>, specialize to those of <cit.> when n = 0.



 §.§ Organization of this paper
 In Section  <ref> we review some basics about general linear Lie superalgebras and establish some notation. In Section  <ref> we construct the homomorphism φ_R for R ∈ Z(𝒟'(m|n) ⊗ U((m|n))). In Section <ref>, we discuss inflations of representations via the homomorphism φ.  In Section <ref>, we relate the restriction of φ to the center of U((m|n)) to the corresponding Harish-Chandra homomorphisms. In Section <ref>, we establish the super Newton's formula and determine the images of the Gelfand generators under φ. Finally, in Section <ref>, we derive the kernel of the map φ, a partial generalization of the results in Section 6 of <cit.>.

In Appendix <ref>, we give the proof that φ_R is a homomorphism, which is a long but straightforward computation. In Appendix <ref>, we give some explicit formulas for the images under φ of a set of homogeneous elements of U((m+1|n)), using a direct computational approach. All the Gelfand generators can be written as sums of elements from this set, so we obtain an alternative proof of Theorem <ref> that does not rely on other sections.



 §.§ Future Directions

Let us discuss some further directions of inquiry. First of all, unlike the ordinary setting, we do not know of a geometric interpretation of the map φ but we believe the restriction of this map to U((m+1|n)) should arise from considering a certain “parabolic subgroup” 𝐏⊆ SL(m+1|n). In particular, 𝐏 should be described by the Harish-Chandra pair  (𝐏_0 , 𝔭), where the underlying even group 𝐏_0 ⊂𝐏 is given by P as above, and the Lie superalgebra 𝔭 of 𝐏 satisfies 𝔭≅ V ⊕𝔤𝔩(m|n) where V is the tautological representation of 𝔤𝔩(m|n) (see <cit.> for theory of Harish-Chandra pairs for supergroups).

Another question is whether there is some larger structure to the family of maps φ_R. In particular, is there an algebra structure on this set? If so, is there any representation-theoretic interpretation? 

Finally, in <cit.> a family of left pseudoinverses are constructed for φ. It would be interesting to see if something analogous can be done in our setting.


 §.§ Acknowledgments

This paper is the result of MIT PRIMES-USA, a program that provides high-school students an opportunity to engage in research-level mathematics and in which the second author mentored the first and third authors. The authors would like to thank the MIT PRIMES-USA program and its coordinators for providing the opportunity for this research experience. We would also like to thank Dimitar Grantcharov for suggesting the project idea and for useful discussions. Additionally, we thank Pavel Etingof, Siddhartha Sahi, Tanya Khovanova, and Thomas Rüd for their thoughts and comments.




§ PRELIMINARIES

We shall introduce some background for the Lie superalgebra (m|n) in this section. Throughout the paper we fix nonnegative integers m,n. 



 §.§ Basic definitions



For a Lie superalgebra 𝔞, by U(𝔞) we denote the universal enveloping superalgebra of 𝔞 and by Z(𝔞) the center of a superalgebra 𝔞. We will always use |·| to denote the parity of a purely homogeneous element, which is 0 if even and 1 if odd (viewed as elements in /2). We will work with general linear Lie superalgebra (m|n), which we define as follows. Let 

    I {1,…, m,1̅,…,n̅},


    Î{0}∪ I

where we impose the total order 

    0<1 < ⋯ < m < 1̅ < ⋯ < n̅.

We also define a parity function | · | on Î, where the parity of indices without an overline is 0 and that of indices with an overline is 1. Now, let ℂ^m+1|n denote the super vector space with basis {e_0, e_1,…, e_m, e_1̅, …, e_n̅}, where the parity of a basis vector is given by the parity of its index. The general linear Lie superalgebra (m+1|n) is by definition the space of all linear maps on ℂ^m+1|n and can be thought of as super matrices of the block form  


    [ [ A B; C D ] ],

where the matrix A is (m+1) × (m+1), the matrix B is (m+1) × n, the matrix C is n × (m+1), and the matrix D is n × n. The underlying even Lie algebra is (m+1) ⊕(n) and corresponds to matrices with B = 0 and C = 0, and the odd part corresponds to matrices such that A and D are zero. The Lie bracket on purely homogeneous elements x, y ∈(m+1|n) is given by 

    [x,y]  xy - (-1)^|x||y|yx

where the multiplication is usual matrix multiplication. The supertrace  of a matrix of the form in (<ref>)is defined to be  

    ([ [ A B; C D ] ]) = (A) - (D),

where  denotes the usual trace of a matrix. The supertrace is a Lie superalgebra homomorphism, so by (m+1|n) we will denote the special linear Lie superalgebra whose elements are the kernel of . Our standard basis for (m+1|n) will be given by the usual elementary matrices {e_ij}_i,j ∈Î. Finally, we will always view (m|n)  as the Lie sub-superalgebra of (m+1|n) spanned by  {e_ij}_i,j ∈ I. We emphasize this point as the reader may be confused by the notation in what is to come if otherwise unaware.
A basis for (m+1|n) is 

    {e_ij: i,j∈Î, i j}∪{e_00-(-1)^|i|e_ii: i∈ I}

and we shall view (m|n) as the elements (m+1|n) that also lie in (m|n) under the inclusion above. Some other important subalgebra for (m+1|n) will be _m+1|n, the Cartan subalgebra of diagonal matrices in (m+1|n), and 𝔫^-_m+1|n and 𝔫^+_m+1|n, the subalgebras of strictly lower and upper triangular matrices in (m+1|n), respectively. We define the analogs of these subalgebras for (m|n) in the obvious way.
For a square matrix A and a variable or constant v, the expression A+v should be understood as the sum of A and the scalar matrix of the same size as A having v on the diagonal. Finally, let δ_kl denote the Kronecker delta, which evaluates to 1 if k=l and 0 otherwise.

In Sections <ref> and in Appendices <ref> and <ref>, a and b denote elements of I. In all sections, i and j denote elements of I unless specified otherwise.


 §.§ Weights and Root Systems



Let {δ_0, δ_1,…,δ_m,δ_1̅,…, δ_n̅} denote the basis of 𝔥_m+1|n^* dual to the canonical basis {e_00, e_11,…, e_mm,e_1̅1̅,…, e_n̅n̅} of _m+1|n^*. Whenever convenient, for 1 ≤ i ≤ n, we will write 

    ϵ_i:=δ_i̅.

We will refer to elements of 𝔥_m+1|n^* as weights, and will often write any weight λ = ∑_i ∈Îλ_i δ_i as a tuple (λ_0,λ_1, …, λ_m, λ_1̅, …, λ_n̅). The bilinear form (·, ·): 𝔥_m+1|n×𝔥_m+1|n⟶ℂ given by the supertrace (x, y) str(xy) naturally induces a bilinear form on 𝔥_m+1|n^*, which will also be denoted by (·, ·). For i,j∈Î, we have

    (δ_i, δ_j) = (-1)^|i||j|δ_ij.

We call x ∈(m+1|n) a root vector if there exists a nonzero α∈_m+1|n^* if for all h ∈_m+1|n we have [h, x] = α(h)x. The weight α is called a root, and set of all roots is called the root system. It is well known that a root system for (m+1|n) is given by Φ{δ_i - δ_j}_i ≠ j ∈Î, with a root vector corresponding to δ_i - δ_j being e_ij. Let Φ_0̅ and Φ_1̅ be the even and odd roots in Φ, respectively, where a root is even if the corresponding root vector is even (resp. odd). It is easily seen that 

    Φ_0̅ = {δ_i - δ_j }_1 ≤ i ≠ j ≤ m∪{_i - _j }_1 ≤ i ≠ j ≤ n,
 

    Φ_ = {±(δ_i - _j)}_1 ≤ i ≤ m, 1 ≤ j ≤ n.
 
A root α∈Φ is said to be isotropic if (α, α) = 0. The set of odd roots coincides with the set of isotropic roots for the general linear Lie superalgebras. Let Φ^+ = {δ_i - δ_j |  i < j ∈Î} be a positive system, and restrict this notation to the even and odd roots in the obvious way. Lastly, let 𝒲_m+1|n≅ S_m+1× S_n denote the Weyl group of (m+1|n) with natural action on 𝔥_m+1|n^*, i.e, S_m+1 permutes {δ_0, …, δ_m} and S_n permutes {ϵ_1, …, ϵ_n}.

Furthermore, we can define for any α∈Φ_ the corresponding coroot α^∨∈𝔥_m+1|n such that 

    ⟨λ, α^∨⟩ = 2(λ, α)/(α, α)  ∀λ∈𝔥_m+1|n^*.

The simple reflection s_α acts on 𝔥_m+1|n^* as expected: s_α(λ) = λ - ⟨λ, α^∨⟩α. Define the Weyl vector ρ_m+1|n as follows:

    ρ_m+1|n∑_i=0^m (m-i+1)δ_i - ∑_j=1^n jϵ_j = (m+1,m,…, 1, -1,-2,…, -n).
 
Note that our Weyl vector is a shifted version of the standard definition, which is half the sum of the positive even roots minus half the sum of the positive odd roots, but all this changes is that the Harish-Chandra homomorphism gets shifted. A weight λ∈_m+1|n^* is said to be antidominant if ⟨λ + ρ_m+1|n, α^∨⟩∉ℤ_> 0 for all α∈Φ^+_0. An element λ∈_m+1|n^* is said to be typical (relative to Φ^+) if there is no positive isotropic α∈Φ_ such that (λ + ρ_m+1|n, α) = 0.


For a weight λ = (λ_0,…, λ_m, λ_1̅,…, λ_n̅) of (m+1|n), denote by M(λ) and L(λ) the Verma module of highest weight λ and its unique simple quotient, respectively. Here

    M(λ) = U((m+1|n))⊗_U(_m+1|n⊕^+_m+1|n)ℂ_λ,

where ℂ_λ is the one-dimensional weight representation of _m+1|n with weight λ extended trivially to ^+_m+1|n.
Finally (and very importantly to avoid confusion later on), to be consistent with our viewing (m|n) as the subalgebra of (m+1|n) spanned by (e_ij)_i,j∈ I, we restrict all of the above definitions to (m|n) in the obvious way so that everything is compatible with the inclusion (m|n) (m+1|n) (with the “extra" index being 0). For instance,

    
  * we will simply write a weight λ∈_m|n^* of (m|n) as an (m+n)-tuple, understanding that λ_0 = 0 under the inclusion _m|n^* 𝔥_m+1|n^*;
    
  * the Weyl group _m|n is S_m × S_n and the Weyl vector ρ_m|n is
    
    ρ_m|n (m,m-1,…, 1, -1,-2,…, -n);

    
  * a Verma module M(λ) and its simple quotient L(λ) can either mean such a (m+1|n)-module with highest weight λ or such a (m|n)-module with highest weight λ (not by restriction). Context will make it clear to which we refer (using a (m+n)-tuple will mean interpret it as a module over (m|n), for instance).

 


 §.§ Superalgebra of Differential Operators

In this subsection, we define the superalgebra '(m|n) of differential operators. 
First, we define the superalgebra of polynomials that '(m|n) operates on.

Define [t_0^± 1,t_1,…, t_m] ⊗Λ_n, where Λ_n is the n-generator exterior algebra with odd generators t_1̅,…, t_n̅. 
The defining relations of Λ_n are given by

    t_it_j = - t_jt_i

for i,j∈{1̅,…,n̅}. This anticommutativity, combined with the fact that [t_0^± 1,t_1,…, t_m] is commutative, means that  is supercommutative, i.e. that is x and y are homogeneous, then xy = (-1)^|x||y|yx. A basis for ℱ is given by vectors of the form t_0^k_0t_1^k_1⋯ t_m^k_mt_1̅^k_1̅⋯ t_n̅^k_n̅, where k_0 ∈ℤ, k_1, …, k_m ∈_≥ 0, and k_1̅, … k_n̅∈{0,1}. For convenience, we will drop the tensor symbol in the notation and also freely permute the t_i's in accordance with the supercommutativity rules.

Next, we define the operation ∂∂ t_i. We write ∂_i as shorthand for ∂∂ t_i. 
Consider an element t_0^k_0t_1^k_1⋯ t_m^k_mt_1̅^k_1̅⋯ t_n̅^k_n̅∈. To operate by ∂_i, we first move t_i^k_i to the front using supercommutativity. Then replace t_i^k_i by k_it_i^k_i-1. Extend to the whole of  by linearity and supercommutativity.

Then, we define the superalgebra of differential operators '(m|n) to be the superalgebra generated by: left multiplication by t_i/t_0 for i∈Î, where we will abuse notation and simply write t_i/t_0 for this operation; and t_0∂_i for i∈Î, which consists of left applying the i'th derivative and then left multiplication by t_0. The parity of these is operators is given by the parity of i ∈Î.  We define the element 

    ∑_i ∈Î(t_i/t_0)(t_0∂_i)  =∑_i∈Î t_i∂_i ∈'(m|n).

The following lemma collects some identities in '(m|n). The proofs are simple so we omit them.

    The superalgebra '(m|n) of differential operators on [t_0^± 1,t_1,…, t_m] ⊗Λ_n generated by t_i/t_0 for i∈ I and t_0∂_i for i∈Î satisfies the following properties:
    
        
  * We have 
        
    t_it_j   = (-1)^|i||j|t_jt_i,for i,j∈Î
    ∂_it_j    = (-1)^|i||j|t_j∂_i, for i, j∈Î, and i  j
    ∂_i∂_j   = (-1)^|i||j|∂_j∂_i. for i,j∈Î, and i j

        
  * We have
        
    [∂_a, t_a]=∂_at_a - (-1)^|a|t_a∂_a= 1

        for a∈Î.
        
  * The element  is central in '(m|n) and if p = t_0^e_0t_1^e_1⋯ t_m^e_mt_1̅^e_1̅⋯ t_n̅^e_n̅, then p=( p)p, where p = e_0 + e_1+⋯ + e_m + e_1̅+⋯ + e_n̅.
        
  * Z('(m|n)⊗ U((m|n))) = [] ⊗ Z(U((m|n))).
    




§ THE HOMOMORPHISM TEXT


In this section, we construct the homomorphism 
    φ_R: U((m+1|n))→'(m|n)⊗ U((m|n))
 that extends the homomorphism ρ in <cit.>. 


    For any element R in Z('(m|n)⊗ U((m|n)))= [] ⊗ Z(U((m|n))), the correspondence given by
    
    e_ab   ↦ t_a∂_b⊗ 1 + 1⊗ e_ab+δ_ab(-1)^|a||b|R,  a,b∈ I
    
            e_a0   ↦ t_a∂_0⊗ 1 -∑_j∈ I (-1)^|a||j|t_j/t_0⊗ e_aj, a∈ I
    
            e_0b   ↦ t_0∂_b⊗ 1, b∈ I
    
            e_00   ↦ t_0∂_0 ⊗ 1 +R

    extends by the universal property to a homomorphism φ_R: U((m+1|n))→'(m|n)⊗ U((m|n)).

We defer the proof to Appendix <ref> as it a long but straightforward verification of commutators. We will write φ as shorthand for φ_R if we are referring to an arbitrary R.
Now, let φ^s denote the restriction of φ_R to U((m+1|n)). The formulas for φ_R give us the following:

    The correspondence given by
    
    e_ab   ↦ t_a∂_b⊗ 1 + 1⊗ e_ab, a,b∈ I and a b
    
            e_a0   ↦ t_a∂_0⊗ 1 -∑_j∈ I (-1)^|a||j|t_j/t_0⊗ e_aj,  a∈ I
    
            e_0b   ↦ t_0∂_b⊗ 1,b∈ I
    
            e_00-(-1)^|a|e_aa   ↦ (t_0∂_0-(-1)^|a|t_a∂_a)⊗ 1 - 1⊗ (-1)^|a|e_aaa∈ I

    extends by the universal property to a homomorphism φ^s: U((m+1|n))→'(m|n)⊗ U((m|n)) (notice this doesn't depend on the choice of R). 

Let

    G_1^(m|n)∑_i∈ I e_ii,
       G_1^(m+1|n)∑_i∈Î e_ii

be the first Gelfand invariants of (m|n) and (m+1|n), respectively (for the full definition of the Gelfand invariants see Section <ref>). Note that they are both central in the corresponding universal enveloping algebras.
Now, let us define the following homomorphisms: 

    
  * the natural embedding 
    ι^s: U((m+1|n))→ U((m+1|n));
 
    
  * (for m+1 n) the projection 
    π^g: U((m+1|n))→ U((m+1|n))
 defined by 
    π^g(B)= B-1/m+1-n(B)G_1^(m+1|n)

    for B∈(m|n) and the universal property; 
    
  * and the map 
    ι^g: U((m|n))→ U((m+1|n))
 defined by ι^g(C) = C-(C)e_00 for C ∈(m|n) and the universal property.

For the following proposition, let us suppose m+1 n and set 

    R_1  -1/m-n+1(⊗ 1+1⊗ G_1^(m|n)).

From the formulas we deduce that φ_R_1 = φ^s ∘π^g (note that R_1 is a super-analog of the R_1 defined in <cit.>). Let us also define

    γ: U((m+1|n))→'(m|n)⊗ U((m+1|n))

by γ (1⊗ι^g) ∘φ^s. 

    For m+1 n, we have γ = (1 ⊗ι^g)φ^s, π^gι^s =, and 
    φ^sπ^g = φ_R_1, and all other relations that directly follow from these three; in that sense, the following diagram is commutative (in the category of superalgebras).

    

    [shorten >=1pt,node distance=4cm,on grid,auto]
      every state=[fill=rgb:black,1;white,10]
        (q_0)                  U((m+1|n));
        (q_1)  at (6,0)        '(m|n)⊗ U((m|n));
        (q_2)  at (6,-2.5)     '(m|n)⊗ U((m+1|n));
        (q_3)  at (0,-2.5)     U((m+1|n));
        (p_3)  at (-0.2,-2.25) ;
        (p_0)  at (-0.2,-0.15) ;
        
        [->,thick,-Stealth[width=5pt, length=10pt]]
        (q_0) edge node φ_R_1             (q_1)
              edge node π^g            (q_3)
        (q_1) edge node 1⊗ι^g (q_2)
        (q_3) edge node φ^s           (q_1)
              edge node γ           (q_2)
        (p_3) edge node ι^s          (p_0);
    
 
    


Most of the diagram still holds if instead of φ_R_1 we take any φ_R for any R ∈ Z(𝒟'(m|n) ⊗ U(𝔤𝔩(m|n))). The only change to the diagram is that the π^g arrow is deleted (and the assumption m+1 ≠ n can be relaxed).



§ INFLATING REPRESENTATIONS


In this section we use φ (recall that φ is shorthand for φ_R) to extend a simple representation of U((m|n)) to a representation of U((m+1|n)) and find conditions for when this new representation is simple.

For any a ∈ℂ, define the '(m|n)-module 

    _a {t_0^a-k_1-⋯ - k_n̅t_1^k_1⋯ t_n^k_m t_1̅^k_1̅⋯ t_n̅^k_n̅| k_1,…, k_m∈_≥ 0,  k_1̅,…, k_n̅∈{0,1}}.

This is a generalization of the _a defined in <cit.>, both in the sense that it is defined for superalgebras and in that we do not require a∈.
Note that _a = '(m|n)(t_0^a) (i.e. the set that results when we apply elements of '(m|n) to t_0^a) and that = a on _a. Given a weight λ of (m|n), we write _a⊗ L(λ) (and _a⊗ M(λ)) for both the representation of '(m|n)⊗ U((m|n)) and the representation  U((m+1|n)) via φ, and the meaning should be clear from context. We fix a highest weight vector v_λ of L(λ).

Let us now consider inflating the simple module L(λ) using φ. In particular, we investigate when the resulting module is simple.

    Let v_λ be the unique highest weight vector for L(λ), up to scaling. The vector t_0^a⊗ v_λ is the only highest weight vector (up to scaling) for _a⊗ L(λ), considered as a U((m+1|n))-module.


For 0<i<j, we have that e_ij acts as t_i∂_j⊗ 1 + 1⊗ e_ij. Now t_i∂_j(t_0^a) = 0 and e_ij acts on v_λ as 0 since i<j, so e_ij acts on t_0^a⊗ v_λ as 0. Also, for each j > 0, e_0j acts on t_0^a⊗ v_λ as t_0∂_j⊗ 1. Now t_0∂_j (t_0^a)=0, so e_0j also acts as 0. It follows that t_0^a⊗ v_λ is a highest weight vector.

Now suppose

    w = p_1⊗ w_1+⋯ p_k⊗ w_k

is a highest weight vector, where 
    p_c=t_0^a-k_1,c-k_2,c-⋯ - k_m,c-k_1̅,c-⋯ - k_n̅,ct_1^k_1,c⋯ t_n^k_m,c t_1̅^k_1̅,c⋯ t_n̅^k_n̅,c,
 where k_1,c,…, k_m,c∈_≥ 0 and k_1̅,c,…, k_n̅,c∈{0,1}. Furthermore assume the p_c are distinct and the w_c are nonzero. 
Now e_0j (for j∈ I), which acts as t_0∂_j⊗ 1, must act as zero on w. Pick the p_c that is t-lexicographically maximal (i.e. relative to the t_1-degree, t_2-degree, and so on). Then,  e_01^k_1,c⋯ e_0n̅^k_n̅,c acts as zero on all p_d⊗ w_d except for p_c⊗ w_c, on which it gives a nonzero result. Thus e_01^k_1,c⋯ e_0n̅^k_n̅,c does not act as zero on w, a contradiction unless p_c = t_0^a, in which case k=1 and we can write w = t_0^a⊗ w.

Then, for i<j, e_ij acts as t_i∂_j⊗ 1 + 1 ⊗ e_ij, and this must act as zero on w. First, note that the t_i∂_j⊗ 1 part acts as zero. Thus we must have e_ij acting on w as 0 for all i<j, i.e. w is a highest weight vector of L(λ), so w is a scalar multiple of v_λ. Thus, t_0^a ⊗ v_λ is the only the highest weight vector (up to scaling) of _a ⊗ L(λ), considered as a U((m+1|n))-module.


We can write R in the form R = ∑_iℰ^i ⊗ z_i for some z_i ∈ Z(U(𝔤𝔩(m|n))) by Lemma <ref>. A direct computation shows that R acts as a scalar on ℱ_a ⊗ L(λ), which we will call r:


    R·(∑_j f_j ⊗ v_j)    = (∑_iℰ^i ⊗ z_i)(∑_j f_j ⊗ v_j) = ∑_i,j ( f_j)^i f_j ⊗χ_λ(z_i)v_j 
       = (∑_i a^i χ_λ(z_i) )(∑_j f_j ⊗ v_j) = r(∑_j f_j ⊗ v_j),

where f_j ∈ℱ_a, v_j ∈ L(λ) are arbitrary, and χ_λ is the central character corresponding to λ (see Section <ref>). It follows that the weight for t_0^a⊗ v_λ is

    λ (a + r, λ_1 + r, …, λ_m + r, λ_1̅ - r, …, λ_n̅ -r).

Next, we find conditions on λ for t_0^a⊗ v_λ to generate the module _a⊗ L(λ). For i∈ I, define

    f(i) ∑_j<i (-1)^|j|,

where the sum runs over j∈ I with j<i (recall the ordering on I defined by 1<2<⋯ < m< 1̅<⋯ < n̅). Then f(1)=0, f(2)=1,…, f(m)=m-1, f(1̅) = m, f(2̅)= m-1,…, f(n̅) = m+1-n.


    Let λ be a (m|n)-weight. If a+f(i)-(-1)^|i|λ_i is not a nonnegative integer for all i ∈ I, then _a ⊗ L(λ) is a highest weight module.




Let N be the submodule generated by the highest weight vector t_0^a⊗ v_λ. We first claim that all monomials in ℱ_a, and thus all elements of ℱ_a, create an element of N when tensored with v_λ, the highest weight vector for L(λ). We proceed by induction down on the exponent of t_0. Specifically, assume that T ⊗ v_λ= t_0^a_0S ⊗ v_λ = t_0^a_0t_1^a_1⋯ t_n̅^a_n̅⊗ v_λ∈ N for a_0 + a_1+⋯  + a_n̅= a, and a_i∈ for |i|=0 and a_i∈{0,1} for |i|=1. We claim that (t_i/t_0 ⊗ 1)(T ⊗ v_λ) ∈ N for any i∈ I. To do this we induct on i.
First look at the case i = 1. We have

    e_10· (t_0^a_0S⊗ v_λ) = (t_1∂_0 ⊗ 1 - ∑_jt_j/t_0⊗ e_1j)(t_0^a_0S ⊗ v_λ) = a_0t_1t_0^a_0-1S ⊗ v_λ- t_1t_0^a_0-1S ⊗λ_1 v_λ.

which means (t_1/t_0⊗ 1)(T⊗ v_λ) is in N as long as a_0 - λ_1 0. But since a-λ_1 is not a nonnegative integer, and a-a_0 is a nonnegative integer, a_0-λ_1 0.
Thus the base case is proved.
For general i  0, note that

    e_i0· (t_0^bS ⊗ v_λ)    = (t_i∂_0 ⊗ 1 - ∑_j (-1)^|i||j|t_j/t_0⊗ e_ij)(t_0^bS ⊗ v_λ)
       = bt_it_0^b-1S ⊗ v_λ - (-1)^|i|λ_it_it_0^b-1S ⊗ v_λ
       =- ∑_j < i (-1)^|i||j| + (|i|+|j|)(|t_0^bS|)t_jt_0^b-1S ⊗ (e_ij· v_λ).

In addition, for 0<j<i, if |j| = 0, we have

    e_ij· (t_jt_0^b-1S ⊗ v_λ)    = (t_i∂_j ⊗ 1)(t_jt_0^b-1S ⊗ v_λ) + (1⊗ e_ij)(t_jt_0^b-1S ⊗ v_λ)
       =(1+a_j) t_it_0^b-1S ⊗ v_λ + (-1)^|i||j| + |j| + (|i|+|j|)(|t_0^bS|) (t_jt_0^b-1S ⊗ (e_ij· v_λ))
       = (1+a_j) t_it_0^b-1S ⊗ v_λ + (-1)^|i||j| + (|i|+|j|)(|t_0^bS|) (t_jt_0^b-1S ⊗ (e_ij· v_λ)).

If |j| = 1, then

    -e_ij· (t_jt_0^b-1S ⊗ v_λ)    = -(t_i∂_j ⊗ 1)(t_jt_0^b-1S ⊗ v_λ) - (1⊗ e_ij)(t_jt_0^b-1S ⊗ v_λ)
       =-(1-a_j) t_it_0^b-1S ⊗ v_λ - (-1)^|i||j| + |j| + (|i|+|j|)(|t_0^bS|) (t_jt_0^b-1S ⊗ (e_ij· v_λ))
       = -(1-a_j) t_it_0^b-1S ⊗ v_λ + (-1)^|i||j| + (|i|+|j|)(|t_0^bS|) (t_jt_0^b-1S ⊗ (e_ij· v_λ)).


since if a_j = 0, then (t_i∂_j)(t_jt_0^b-1S) = t_it_0^b-1 and if a_j=1, then t_jt_0^b-1S=0 since t_j^2 = 0.
Also, both of these are in N because t_jt_0^b-1S⊗ v_λ∈ N, from the inductive hypothesis.
Adding these equations (for all j<i) to the first gives us that

    (b+∑_j<i
     |j| = 0(1+a_j) + ∑_j<i
     |j| = 1 (a_j-1) - (-1)^|i|λ_i)t_it_0^b-1S⊗ v_λ∈ N.


Now we show that the expression in parentheses (call it A) is nonzero. Note

    A = b+∑_j<ia_j + f(i)-(-1)^|i|λ_i =  b+∑_j∈ Ia_j +f(i) - (-1)^|i|λ_i- K  = a+f(i) - (-1)^|i|λ_i- K,



where K is some nonnegative integer. Since a+f(i)-(-1)^|i|λ_i is not a nonnegative integer, it follows that A  0
so t_it_0^b-1S⊗ v_λ∈ N.

Now, we show that p⊗ (e_ij· v_λ)∈ N, for any p∈_a and i,j∈ I. We have

    e_ij· (p⊗ v_λ) = (t_i∂_j)(p)⊗ v_λ + p⊗ (e_ij· v_λ) + δ_ij(-1)^|i||j|R(p⊗ v_λ),

and this element is in N. Now R acts as r (see the discussion preceding (<ref>)), so R(p⊗ v_λ)∈ N and (t_i∂_j)(p)⊗ v_λ∈ N from the first part of the proof. Thus p⊗ (e_ij· v_λ) ∈ N for all p∈_a and i,j∈ I.
Repeating this process multiple times shows that p⊗ (e_i_1 j_1e_i_2j_2⋯ e_i_kj_k· v_λ)∈ N for any k. Since v_λ generates L(λ), this shows p⊗ w∈ N for any p∈_a and w∈ L(λ). Thus N = _a ⊗ L(λ).

Using the following theorem, we can use the previous theorem to find conditions when _a⊗ L(λ)≅ L(λ).

    For a weight λ of (m|n), we have _a⊗ L(λ)≅ L(λ) as (m+1|n)-modules if and only if _a⊗ L(λ) is generated by the highest weight vector t_0^a ⊗ v_λ.


    First note that by Proposition <ref>, showing _a⊗ L(λ)=L(λ) is equivalent to showing _a⊗ L(λ) is simple. Let N denote the submodule of _a ⊗ L(λ) generated by t_0^a ⊗ v_λ. If N_a⊗ L(λ), it is obvious that _a⊗ L(λ) cannot be simple. Now we prove the “if” direction.

    Since N=_a⊗ L(λ), we know t_0^a⊗ v_λ generates _a⊗ L(λ). Let x be a nonzero element of _a⊗ L(λ). Let N(x) be the module generated by x. We will show that t_0^a⊗ v_λ∈ N(x), so N(x)=_a⊗ L(λ), thereby showing _a⊗ L(λ) is simple, since x is an arbitrary element.

    Write
    
    x = c_1p_1⊗ w_1+⋯ c_kp_k⊗ w_k,

    where 
    p_c=t_0^a-k_1,c-k_2,c-⋯ - k_m,c-k_1̅,c-⋯ - k_n̅,ct_1^k_1,c⋯ t_n^k_m,c t_1̅^k_1̅,c⋯ t_n̅^k_n̅,c,
 where k_1,c,…, k_m,c∈_≥ 0 and k_1̅,c,…, k_n̅,c∈{0,1}, w_c = e_i_1,cj_1,ce_i_2,cj_2,c⋯ e_i_d,cj_d,c· v_λ, and finally c_1,… c_k∈_ 0. We claim that a nonzero element of the form 
    
    x' = c_1't_0^a⊗ w_1'+⋯ c_k''t_0^a⊗ w_k''

    exists in N(x). For x in the form (<ref>), we define
    
    S(x) =∑_i∈ I∑_ℓ=1^k k_i,ℓ.

    Note S(x) is defined uniquely for all x and S(x) is a nonnegative integer. Additionally, S(x')=0 if and only if x' is in the form (<ref>). If S(x) = 0, we are done.
    
    Otherwise, there exists k_j,ℓ>0 for some j,ℓ. Applying e_0j to x (where we have e_0j· x = (t_0∂_j)x), we see that S(e_0j· x)< S(x), so S(·) decreases under this action. Also, note that e_0j· x is nonzero since k_j,ℓ>0. If S(e_0j· x)=0, we are done. Otherwise we can do the same process to e_0j· x, finding e_0j' e_0j· x 0 with S(e_0j' e_0j· x)< S(e_0j· x). If S(e_0j' e_0j· x)=0 we are done. Otherwise we can just keep continuing this process, which must terminate since S is always an integer.    
    Then it follows that we can find x' ∈ N(x) which can be written as (<ref>). We can then write x' as
    
    x' = t_0^a⊗ w,

    where w∈_a⊗ L(λ) and w 0.

    Next, we show that if t_0^a⊗ w'∈ N(x), then t_0^a⊗ (e_ij· w')∈ N(x), for i,j∈ I. We have
    
    e_ij· (t_0^a⊗ w')    = (t_i∂_j)· (t_0^a⊗ w') + (1⊗ e_ij)· (t_0^a⊗ w') + δ_ij(-1)^|i||j|R· (t_0^a⊗ w')
       = ± t_0^a⊗ (e_ij· w') + s(t_0^a⊗ w')

    for a constant s∈. Subtracting s(t_0^a⊗ w'), we get that t_0^a⊗ (e_ij· w')∈ N(x).

    Let N'(w) be the submodule generated by w in L(λ).
    Since x' = t_0^a⊗ w∈ N(x), from the proceeding argument, it follows that t_0^a⊗ N'(w)⊂ N(x). But since L(λ) is simple, it follows that N'(w)=L(λ). In particular, v∈ N'(w), so t_0^a⊗ v_λ∈ N(x). Thus N(x)=_a⊗ L(λ), so _a⊗ L(λ) is simple.



    Let λ be a (m|n)-weight. If (λ+ρ_m+1|n, δ_0 - δ_i)∉_>0 for 1≤ i≤ m and (λ+ρ_m+1|n, δ_0 - δ_i)∉_≥ 0 for 1̅≤ i ≤n̅, then _a⊗ L(λ) is simple, which implies _a⊗ L(λ)≅ L(λ). 



    We show that the conditions given are equivalent to the conditions in Theorem <ref>. Letting ρ = ρ_m|n, note that f(i)+1 = ρ_0 - ρ_i for 1≤ i≤ m and f(i) = ρ_0 + ρ_i for 1̅≤ i≤n̅, where ρ = ρ_m+1|n. It follows that the condition a+f(i)-(-1)^|i|λ_i not being a nonnegative integer is equivalent to the conditions stated in the corollary. Then the result follows by Theorem <ref> and Theorem <ref>.


Let us recall the following fact: if λ is both antidominant and typical, then M(λ) is simple, see for reference <cit.> or <cit.>. Our result will be a consequence of Corollary <ref>. 



    Let _m|n(a) denote the set of all typical antidominant (m|n)-weights λ such that λ_i-(-1)^|i|a∉ for i∈ I. If λ∈_m|n(a), then _a⊗ M(λ), considered as a (m+1|n)-representation, is isomorphic to M(λ).
 

    Since λ is antidominant and typical, L(λ)=M(λ). 
    For 1≤ i<j≤ m or 1̅≤ i<j≤n̅, we have (λ+ρ_m+1|n)_i - (λ+ρ_m+1|n)_j = (λ+ρ_m|n)_i-(λ+ρ_m|n)_j∉_>0. Additionally, note that for 1≤ i≤ m, (λ + ρ_m+1|n)_0 - (λ + ρ_m+1|n)_i = a + i - λ_i ∉_>0 since a-λ_i ∉. Thus λ is antidominant. 

    For 1≤ i≤ m and 1̅≤ j≤n̅, we have 
    
    (λ + ρ_m|n,δ_i-δ_j) =(λ+ρ_m+1|n)_i + (λ+ρ_m+1|n)_j = (λ+ρ_m|n)_i + (λ+ρ_m|n)_j  0

    since λ is typical. Finally, for 1̅≤ j≤n̅, we have
    
    (λ + ρ_m|n,δ_0-δ_j) =(λ+ρ_m+1|n)_0 + (λ+ρ_m+1|n)_j = a+m+1 + λ_j -j  0

    because a+λ_j is not an integer. Therefore λ is typical as well as antidominant.
   
    Thus M(λ)=L(λ) is simple. Then note that λ satisfies the conditions in Corollary <ref>, so the result follows.


Now suppose ξ∈𝔥_m+1|n^* such that (ξ+ρ_m+1|n, α^∨) ∉_> 0 for α∈Φ_^+ and (ξ+ρ_m+1|n, α)∉_≥ 0 for α∈Φ_^+. Note that the first condition is simply that ξ is antidominant and the second implies ξ is typical. Then inductively apply Corollary <ref> to use the modules _a to “build” the representation L(ξ) for ξ∈_m+1|n^*. The conditions on ξ allow us to apply Corollary <ref> iteratively (with R=0) to get a map

    U((m+1|n)) →'(m|n)⊗⋯⊗'(0|n)⊗'(n-1|0) ⊗⋯⊗'(0|0)

that gives us a (m+1|n)-representation

    ℱ(ξ) _ξ_0⊗_ξ_1⊗⋯⊗_ξ_n̅

that is isomorphic to L(ξ).
(Note that when constructing the chain have to use the obvious isomorphism from U((0|n)) to U((n|0)) given by e_i̅j̅↦ e_ij). 
















But ξ is antidominant and typical, so ℱ(ξ) is isomorphic to M(ξ) as well. This applies to substantially more general setting than <ref>, and motivates the following conjecture:

There is an isomorphism of U((m+1|n))-modules ℱ(ξ)≅ M(ξ) for all weights ξ∈_m+1|n^*.

Further supporting evidence is that the modules have the same formal supercharacters, which is verified in a manner analogous to the proof of Theorem 4.2, page 6 in <cit.>.





§ IMAGES UNDER HARISH-CHANDRA HOMOMORPHISMS

In this section, we relate the restriction of φ to the center of U((m+1|n)) with the Harish-Chandra homomorphisms.

For M∈{m+1,m}, recall ρ_M|n = (M,…, 1, -1, …, -n). If b=(b_1,…, b_M, b_1̅,…, b_n̅)∈^M|n, then the evaluation homomorphism _b: [l_1,…, l_M,l_1̅,…, l_n̅]→ is defined by 
    _b(p)  p(b_1,…, b_M,b_1̅,…, b_n̅).


Every z∈ Z(U((m|n)) acts on L(λ) as χ_λ(z), where χ_λ(z) = _λ+ρ_m|n(χ_m|n(z)) and χ_m|n: Z(U((m|n)))→[ℓ_1,…,ℓ_m, ℓ_1̅,…, ℓ_n̅]^_m|n is the Harish-Chandra homomorphism (see Theorem 13.1.1 (a) in <cit.>). We similarly define χ_m+1|n: Z(U((m+1|n)))→[ℓ_0,…,ℓ_m, ℓ_1̅,…, ℓ_n̅]^_m+1|n.

Next, we define χ_0,m|n: [] ⊗ Z(U((m|n)))→[ℓ_0] ⊗[ℓ_1,…, ℓ_n̅]^_m|n by 
    χ_0,m|n(∑_i^i⊗ z_i) = ∑_i ℓ_0^i⊗χ_m|n(z_i).
 Set ℓ = χ_0,m|n(R). In the case when R is given by (<ref>), we have

    ℓ = -1/n+1( ∑_i ∈Îℓ_i - m(m+1)/2 + n(n+1)/2).

 
Note that the symmetric superalgebra S(𝔥_m|n) is isomorphic, in a canonical way, to the superalgebra of _m|n-invariant polynomials in [ℓ_1, …, ℓ_n̅]. Let I(𝔥_m|n) be the subalgebra of S(𝔥_m|n) consisting of all _m|n-invariant functions θ on 𝔥_m|n^* such that if α is an isotropic odd root and (λ, α)=0, then θ(λ)=θ(λ + tα) for all t∈. It is known that χ_m|n: Z(U((m|n)))→ I(𝔥_m|n) is an isomorphism (see Theorem 13.1.1 (b) in <cit.>). Note that this means χ_0, m|n, considered as a map from []⊗ Z(U((m|n))) to [ℓ_0]⊗ I(𝔥_m|n) , is an isomorphism as well.

For the remainder of this section, we consider the module _a only when a is an integer. Then note _m|n_m|n(a) = _m|n(0).

    The modules 
    = ⊕_a∈_a  and ⊕_λ∈_m|nM(λ)
 are faithful over '(m|n) and (m|n), respectively.


    We prove that both modules have trivial annihilators.

    If z∈ Z(U((m|n))) annihilates ⊕_λ∈_m|nM(λ), then χ_λ(z)=0 for all λ∈_m|n. Since χ_λ(z) = _λ + ρ_m|n(χ_m|n(z)), we have
    
    _λ+ρ_m|n(χ_m|n(z))=0

    for all λ∈_m|n. But this can only happen if χ_m|n(z)=0. Thus, we must have z = 0.

    Now suppose x∈'(m|n) annihilates  and x is nonzero. Write x as the sum of elements of the form
    
    p(t_0,…, t_n̅)·∂_0^b_0⋯∂_n̅^b_n̅

    where p(t_0,…, t_n̅) is some nonzero polynomial. Furthermore, assume that each term in the sum has a different (b_0,…, b_n̅). Pick the term that is ∂-lexicographically maximal (i.e. relative to the ∂_0-degree, the ∂_1-degree, and so on). Consider the polynomial t_0^b_0t_1^b_1⋯ t_n̅^b_n̅. Then the leximaximal property of the term we picked guarantees that all the other terms act as zero. Furthermore,
    
    (∂_0^b_0⋯∂_n̅^b_n̅)(t_0^b_0⋯ t_n̅^b_n̅) = c

    for a nonzero c∈. Therefore
    
    x· (t_0^b_0⋯ t_n̅^b_n̅)    = (p(t_0,…, t_n̅)·∂_0^b_0⋯∂_n̅^b_n̅ )(t_0^b_0⋯ t_n̅^b_n̅)
       = c· p(t_0,…, t_n̅) 0.

    This contradicts the fact that x annihilates . Therefore  has a trivial annihilator. 


Define

    τ(p(ℓ_0,…, ℓ_m,ℓ_1̅,…ℓ_n̅))  p(ℓ_0 + ℓ + m+1, ℓ_1 + ℓ, …, ℓ_m+ℓ, ℓ_1̅ - ℓ, …, ℓ_n̅-ℓ).

Then it is easy to see that τ is a homomorphism from I(𝔥_m+1|n) to [ℓ_0]⊗ I(𝔥_m|n).

    We have that
    
    φ|_Z(U((m+1|n))) = χ_0,m|n^-1τχ_m+1|n.

    In particular, φ(Z(U((m+1|n)))) is a subalgebra of []⊗ Z(U((m|n))), and the following diagram is commmutative:
    

    [shorten >=1pt,node distance=4cm,on grid,auto]
      every state=[fill=rgb:black,1;white,10]
        (q_0)                  Z(U((m+1|n)));
        (q_1)  at (6,0)        []⊗ Z(U((m|n)));
        (q_2)  at (6,-2.5)     [ℓ_0]⊗ I(𝔥_m|n);
        (q_3)  at (0,-2.5)     I(𝔥_m+1|n);
        (p_3)  at (-0.2,-2.25) ;
        (p_0)  at (-0.2,-0.15) ;
        
        [->,thick,-Stealth[width=5pt, length=10pt]]
        (q_0) edge node φ             (q_1)
              edge node χ_m+1|n       (q_3)
        (q_1) edge node χ_0,m|n   (q_2)
         (q_3)     edge node τ           (q_2);
    
    



    Suppose λ∈_m|n and a∈. Let M(a,λ) = _a⊗ M(λ), considered as a (m+1|n)-module. We show that φ(z) = χ_0,m|n^-1τχ_m+1|n(z) for z∈ Z(U((m+1|n))), as an identity of endomorphisms of M(a,λ). By Lemma <ref>, and by the fact that the tensor product of faithful modules is faithful, the module ⊕_a∈, λ∈_m|n M(a,λ) is faithful over '(m|n)⊗ U((m|n)). Then it will follow that φ = χ_0,m|n^-1τχ_m+1|n(z). 
    
    By Corollary <ref>, we have M(a,λ)≅ M(λ). Thus, every z∈ Z(U((m+1|n))) acts on M(a,λ) as χ_λ(z). Now set ξ = χ_0, m|n^-1τχ_m+1|n. Then, ξ(z) acts on M(a,λ) as χ_a, λ(ξ(z)), where χ_a, λ(∑^i ⊗ z'_i) ∑ a^i χ_λ(z'_i), because ^i ⊗ 1 acts on M(a,λ) as a^i. Thus we just need to show χ_a,λ(ξ(z)) =χ_λ(z)
    
    For a polynomial p(t_0,…, t_n̅), we have
    
    _a,λ+ρ_m|nτ(p)
     =    _a,λ+ρ_m|n p(ℓ_0 + ℓ + m+1, ℓ_1 + ℓ, ⋯, ℓ_m + ℓ, ℓ_1̅-ℓ, …, ℓ_n̅-ℓ)
    
            =   p(a+ r+m+1, ℓ_1 + r + m,…, ℓ_m + r +1, ℓ_1̅ - r -1 , …, ℓ_n̅ - r - n)
    
            =    _λ + ρ_m+1|n p

    (Note that here we set _a, λ+ρ_m|n(p)  p(a, λ_1 + m, ..., λ_m + 1, λ_1̅ - 1, ..., λ_1̅ - n) for all p).
    As a result, setting p = χ_m+1|n(z), and noting that χ_a, λ(^i ⊗ z') = _a, λ + ρ_m|n(χ_0, m|n(^i ⊗ z')), we have
    
    χ_a,λ(ξ(z))=χ_a,λχ_0, m|n^-1τχ_m+1|n(z) = _λ + ρ_m+1|n(χ_m+1|n(z))= χ_λ(z).

    This completes the proof.




§ SUPER VERSION OF NEWTON'S FORMULA AND CAPELLI-TYPE BEREZINIANS


We have now computed the image of the center, so we aim to describe the images of individual generators, as in <cit.>. In <cit.>, the images of the Gelfand generators are computed explicitly using a progression of identities related to Capelli generators, Capelli determinants, and an analogue of Newton's formula for (m) (see <cit.>). However, in the super setting, there is no known analog of these formulas. Additionally, another challenge in the super setting is that the center is not finitely generated, which makes it harder to apply the method in <cit.>. We appeal to the theory of Yangians to find identities for analogs of Capelli determinants (which we call Capelli Berezinians), including a new form of the super Newton's formula, which relates the Capelli Berezinians to Gelfand invariants. We then use this to compute the images of the latter under φ.



 §.§ Yangians for gl(m|n)

Let us recall the Yangian Y((m|n)) as defined in <cit.> (also see <cit.>). This is the _2-graded associative algebra over  with generators

    {T_ij^(r): i,j∈ I, r≥ 1}

and defining relations

    [T_ij^(r), T_kl^(s)] = (-1)^|i||j|+|i||k|+|j||k|∑_p=0^min(r,s)- 1 (T_kj^(p)T_il^(r+s-1-p)-T_kj^(r+s-1-p)T_il^(p)).

The _2-grading is given by |T_ij^(r)| = |i|+|j|.
We define the formal power series

    T_ij(u) δ_ij+ T_ij^(1)u^-1 + T_ij^(2)u^-2+⋯

and the matrix T(u)  [T_ij(u)]∈Mat_m|n(Y((m|n))). We can also identify T(u) with

    ∑_i,j∈ IT_ij(u)⊗ e_ij(-1)^|j|(|i|+1).

In general, we can identify an operator ∑ A_ij⊗ e_ij(-1)^|j|(|i|+1) in Y((m|n))[[ u^-1]] ⊗^m|n with the matrix [A_ij]. The extra signs are inserted to let the product of two matrices be calculated in the usual way. The Yangian is a Hopf algebra with comultiplication

    Δ: T_ij(u)↦∑_k∈ I T_ik(u)⊗ T_kj(u),

antipode S: T(u)↦ T^-1(u), and counit T(u)↦ 1. 

We define the power series with coefficients in the Yangian Y((m|n)) of (m|n) given by

    Z_h(u)  1+ (T(u+h) - T(u)/h· T^-1(u) )

for any h∈, where (A) denotes the supertrace of a matrix A. The h here has no relation to the h that appears in <cit.>. Define

    Z(u) lim_h → m-nZ_h(u).

For m  n, Z(u) = Z_m-n(u) while for m = n we have

    Z(u) = 1 + ((d/duT(u))T^-1(u)).

Note that the same series Z(u) was defined in <cit.>; however it was not written in terms of this limit construction. The coefficients of the series Z(u) are known to generate the center of Y((m|n)) (see <cit.> and <cit.>).

Let B(u) be the quantum Berezinian of T(u), defined by

    B(u) [T_ij(u+(m-n-j))]_i,j=1,… m×[T_ij^*(u-(m+n-j+1))]_i,j=m+1,…, m+n,

where [T_ij^*(u)]= T^*(u)= (T^-1(u))^st and A^st= [A_ji(-1)^|i|(|j|+1)] denotes matrix supertransposition. By the determinant of a matrix X = [X_ab]_a,b=1,…, N with noncommuting entries, we mean the sum

    X ∑_σ∈ S_N(σ) X_σ(1),1⋯ X_σ(N),N.

It is shown in <cit.> that

    Z(u)=B(u+1)/B(u),

which could be referred to as a “quantum Liouville formula”. Additionally, there is a projection homomorphism π_m|n: Y((m|n))→ U((m|n)) given by 

    T_ij(u)↦δ_ij + e_ij(-1)^|i|u^-1.




 §.§ Super Newton's formula

We can use these tools from Yangians to generalize the Newton's formula for (m) (see <cit.> and Theorem 7.1.3 in <cit.>) to (m|n). Let 

    ((-1)^|i|e_ij)_i,j∈ I

and define

    C_m|n(T)    [_ij-T-i+1]_i,j = 1,…,m×[(_i̅j̅-T-m+i)^*]_i,j=1,…, n
     =   ∑_σ∈ S_m(σ) ( - T)_σ(1),1⋯ ( -T-m+1)_σ(m),m
       ×∑_τ∈ S_n(τ)(( - T -m+1)^-1)^st_m+τ(1),m+1⋯ (( - T-(m-n))^-1)^st_m+τ(n),n

to be the Capelli Berezinian of (m|n).
It follows from the definitions that C_m|n(T) = Q(T) π_m|n(B(-T-(m-n)+1)) for the rational function Q(T) given by

    Q(T) = 
            T(T+1) ⋯ (T+(m-n-1))    for m> n
    
            1    for m=n
    
            (T-(n-m))^-1(T-(n-m-1))^-1⋯ (T-1)^-1   for n>m.

It is well known that the coefficients of π_m|n(B(T)) generate Z(U((m|n))). Hence the coefficients of C_m|n(T) also generate Z(U((m|n))). We call these coefficients the Capelli generators of Z(U((m|n))). Since the coefficients of C_m|n(T) are central, we can consider C_m|n as a function from the algebra of Laurent series in T with coefficients in U((m|n)) to itself.
The Newton's formulas proved in <cit.> and Theorem 7.1.3 of <cit.> relate the Capelli Berezinian (called the Capelli determinant when n=0) to the Gelfand invariants G_k^(m). We generalize these results to (m|n) using the same method as <cit.>.

The Gelfand invariants of (m|n) are given by

    G_k^(m|n)∑_i_1,…,i_k ∈Î (-1)^|i_2|+⋯ + |i_k|e_i_1i_2⊗ e_i_2i_3⊗⋯ e_i_ki_1

for k≥ 1, where the sum ranges over all k-tuples (i_1,…, i_k) with terms in I. It is known that the Gelfand invariants G_k for k=0,1, … generate Z(U((m|n))) (see for example <cit.>). We can also write G_k^(m|n) = ^k. Indeed, we have

    ^k    =∑_i∈ I (-1)^|i|(^k)_ii
       = ∑_i∈ I (-1)^|i|∑_i_2,…, i_k∈ I_ii_2_i_2i_3⋯_i_ki
       =∑_i_1,…, i_k∈ I (-1)^|i_1| + (|i_1|+⋯ + |i_k-1| + |i_k|)e_i_1i_2e_i_2i_3⋯ e_i_ki_1
       =∑_i_1,…, i_k∈ I (-1)^|i_2|+⋯ + |i_k| e_i_1i_2e_i_2i_3⋯ e_i_ki_1
       = G_k^(m|n).


    We have
    
    C_m|n(T-(m-n))/C_m|n(T-(m-n)+1) = 1- ∑_k=0^∞ G_k^(m|n)T^-1-k.



   
    Since C_m|n(T) = Q(T) π_m|n(B(-T-(m-n)+1)), (<ref>) and that G_0^(m|n)=m-n yields that the identity is equivalent to 
    
    π_m|n(B(-T+1))/π_m|n(B(-T)) = 1 + 1/-T+(m-n)∑_k=1^∞^k T^-k.

    Using (<ref>) and (<ref>), we get that

    π_m|n(Z(-u))    = lim_h → m-n1+( u^-1 - (u-h)^-1/h(1- u^-1)^-1)
       = lim_h →m-n1 - 1/u(u-h)∑_k=0^∞^k+1 u^-k
       =1 - 1/u-(m-n)∑_k=1^∞^k u^-k.

Then the identity follows from (<ref>).



 §.§ Images of Central Elements

Next, we find the image of the Capelli Berezinian under the Harish-Chandra homomorphism, which combined with Theorem <ref>, will let us find the image of the Capelli Berezinian under φ.

    We have
    
    χ_m|n(C_m|n(-T)) = (T+ℓ_1-m)⋯ (T+ℓ_m-m) (T-ℓ_1̅-m)^-1⋯ (T-ℓ_n̅-m)^-1.



    Note C_m|n(T) is the product of two parts, one part summing over σ∈ S_m, and one summing over τ∈ S_n. However, note that in both parts, only the summand corresponding to the identity permutation has a nonzero image under χ_m|n, because in both +T and
    
    ( + T)^-1 = T^-1- T^-2+ ^2 T^-3-⋯,

    the only elements that act as scalars on a Verma module M(λ) are along the diagonal [This is because of the fact that r_k^(m|n)(i,j) = ^k_ij acts as a scalar on a Verma module M(λ) if and only if i=j. For a definition of r_ij^(m|n), see Appendix <ref>. To prove this, one simply has to consider how r_k^(m|n)(i,j) raises/lowers the weights of the element it is acting on.]. The product of the two terms gives the result. 

Define C_φ(T) via the identity C_φ(φ(T)) = φ(C_m+1|n(T)). Since the coefficients of the Capelli determinant are central and φ maps Z(U((m+1|n))) to Z('(m|n)⊗ U((m|n))), the domain of C_φ(T) consists of Laurent series in φ(T) with coefficients in '(m|n)⊗ U((m|n)).
For convenience we will write T for φ(T).

    We have
    
    C_φ(T+R) = ( - T)C_m|n(T+1).

    Consequently, C_φ( + R)=0.


    The identity is equivalent to
    
    φ(C_m+1(-T)) = ( +T +R)C_m|n(-T-R+1).

    To prove this, we use use Proposition <ref>, the corresponding version for m+1|n, Theorem <ref>, and that χ_0,n(R)=ℓ.

Define

    R_2 = ( + m-n)⊗ 1,

and

    A_k(s)    = ∑_g=0^k-skg R^g R_2^k-s-g.

We can combine the previous theorem with the Newton's formula to find the images of the Gelfand invariants under φ.

    The following formula holds for all positive integers k:
    
    φ(G_k^(m+1|n))    = A_k(1)(⊗ 1) + (m+1-n)R^k + ∑_g=0^k-1kgR^g(1⊗ G_k-g^(m|n))
       =-∑_s=2^k (1⊗ G_s-1^(m|n))A_k(s).



    Theorem <ref> implies that
    
    C_φ(T-(m-n)-1)/C_φ(T-(m-n)) = (1-1/T-R-R_2)C_m|n(T-R-(m-n))/C_m|n(T-R-(m-n)+1).

    Applying φ to the Newton's formula for (m+1|n) gives
    
    C_φ(T-(m-n)-1)/C_φ(T-(m-n))   = 1-∑_k=0^∞φ(G_k^(m+1|n))T^-1-k
       = 1- (m-n+1)T^-1 - T^-1∑_k=1^∞φ(G_k^(m+1|n))T^-k.

Analogously we can express the right hand side of (<ref>) as a power series in T^-1. We complete the proof by comparing and computing the coefficients of T^-k on both sides.





§ KERNEL OF VARPHIR1

In this section, we find the kernel of the map φ_R_1. For this section, we write φ=φ_R_1 and G_1 = G_1^(m+1|n) for convenience.

    Let (G_1) be the two sided ideal in U((m+1|n)) generated by G_1. Then the kernel of φ is (G_1).


    Recall from Proposition <ref> that φ = π^g φ^s. The kernel of π^g is clearly (G_1), so we just need to show that φ^s is injective. We construct a left inverse for φ^s, considered as a Lie superalgebra homomorphism. In particular, this means we view U((m+1|n)) and U((m|n)) as Lie algebras with bracket given by [x,y] = xy-(-1)^|x||y|yx.
    
    
    Let ”(m|n) be the subalgebra of '(m|n) generated by {t_a∂_b : a,b∈Î, a b}∪{t_0∂_0 - (-1)^|a|t_a∂_a: a∈Î}. 
    Consider the homomorphism 1⊗ : ”(m|n) ⊗ U((m|n))→”(m|n), where  is the counit map of U((m|n)). In particular (1⊗)(p ⊗ x) = 0 for p∈”(m|n) and x∈(m|n) and (1⊗)(p⊗ 1) = p. 
    Consider the homomorphism (1⊗) ∘φ^s: U((m+1|n)) →”(m|n). Corollary <ref> gives us that this homomorphism is generated by
    
    e_ab   ↦ t_a∂_b for a b
    
            e_00-(-1)^|a|e_aa   ↦ t_0∂_0- (-1)^|a|t_a∂_a.

    We claim that this homomorphism has a left inverse ψ. Letting ”'(m|n) be the algebra generated by {t_a∂_b : a,b∈Î}, we first define ψ': ”'(m|n)→ U((m+1|n)) and then its restriction ψ: ”(m|n)→ U((m+1|n)) will be a left inverse to (1⊗)∘φ^s. The map ψ' is generated by
    
    t_a∂_b    ↦ e_ab

    for a,b∈Î. To check ψ is a homomorphism, we check ψ'([t_a∂_b, t_c∂_d]) = [e_ab,e_cd] for all a,b,c,d∈Î.

    If a d, b c, then we have
    
    [t_a∂_b, t_c∂_d]    = t_a∂_bt_c∂_d - (-1)^(|a|+|b|)(|c|+|d|) t_c∂_d t_a∂_b
       =(-1)^|b||c|t_a t_c ∂_b∂_d - (-1)^(|a||c| + |b||c| + |a||c|)t_ct_a ∂_d ∂_b
       = (-1)^|b||c|t_at_c ∂_b t_d- (-1)^|b||c|t_at_c ∂_b t_d = 0.

    Now suppose a d, b=c. Note [e_ab,e_bd] = e_ad. Thus, we show [t_a∂_b, t_b∂_d] = t_a∂_d. Indeed, we have
    
    [t_a∂_b, t_b∂_d]    = t_a∂_b t_b ∂_d - (-1)^(|a|+|b|)(|b|+|d|)t_b∂_d t_a∂_b
       = t_a∂_b t_b ∂_d -(-1)^|b| t_a t_b∂_b ∂_d
       = t_a∂_d.

    If a=d, b c, reversing the order gives the previous case.
    If a=d, b=c, then [e_ab, e_ba] = e_aa- (-1)^|a|+|b|e_bb. We also have
    
    [t_a∂_b, t_b∂_a]    = t_a∂_b t_b∂_a - (-1)^|a|+|b|t_b∂_a t_a∂_b 
       = t_a (1+(-1)^|b|t_b∂_b)∂_a - (-1)^|a|+|b| t_b∂_a t_a∂_b
       = t_a∂_a - (-1)^|a| + |b| t_b∂_b (∂_at_a - (-1)^|a|t_a∂_a)
       = t_a∂_a - (-1)^|a|+|b|t_b∂_b.

    Thus ψ', hence ψ is a homomorphism. It is easily checked that ψ∘ (1⊗) ∘φ^s=, so φ^s is injective. This completes the proof.


amsalpha
   




§ PROOF OF THEOREM 3

In this appendix we provide the proof of Theorem <ref>. 

    Clearly φ maps even elements to even elements and odd elements to odd elements. Thus, we only need to check that for basis elements x and y, we have
    
    [φ(x),φ(y)] = φ([x,y]).

    We have multiple cases to check. Here a,b,c,d will always denote elements of I. Before we start verifying the cases, note that since R is central, all brackets with it are zero.
    
        
  Case 1 [e_ab,e_ba]. 5pt 
        If a = b then both brackets in (<ref>) are obviously zero. 
        
        If a b then [e_ab,e_ba] = e_aa - (-1)^|a|+|b|e_bb. Thus, 
        
    φ([e_ab,e_ba])    = (t_a∂_a-(-1)^|a|+|b|t_b∂_b)⊗ 1 + 1⊗(e_aa-(-1)^|a|+|b|e_bb)

    Now we show
    
    [t_a∂_b, t_b∂_a] = t_a∂_a - (-1)^|a|+|b|t_b∂_b

    for a b.
    The left hand side is
    
    t_a∂_bt_b∂_a - (-1)^|a||b|t_b∂_at_a∂_b.

    Now ∂_bt_b = 1 + (-1)^|b|t_b∂_b, so t_a∂_bt_b∂_a = t_a∂_a + (-1)^|b|t_at_b∂_b∂_a. Using supercommutativity, this becomes t_a∂_a + (-1)^|b|t_a∂_a t_b∂_b. A similar process with the second term gives us t_b∂_at_a∂_b =t_b∂_b +(-1)^|a|t_a∂_at_b∂_b. Plugging these back into (<ref>), we get
    
    [t_a∂_b, t_b∂_a]    = t_a∂_a+ (-1)^|b|t_a∂_a t_b∂_b - (-1)^|a|+|b|t_b∂_b - (-1)^|b|t_a∂_a t_b∂_b
       =t_a∂_a - (-1)^|a|+|b|t_b∂_b.

    Thus,
    
    [φ(e_ab),φ(e_ba)]    = [t_a∂_b⊗ 1 + 1⊗ e_ab,t_b∂_a⊗ 1 + 1⊗ e_ba]
       = (t_a∂_a-(-1)^|a|+|b|t_b∂_b)⊗ 1 + 1⊗(e_aa-(-1)^|a|+|b|e_bb)
       = φ([e_ab,e_ba]),

    verifying (<ref>).

        
  Case 2 [e_ab,e_bc], a c5pt
        First, we show
        
    [t_a∂_b, t_b∂_c] = t_a∂_c.

        The left hand side is 
        
    t_a∂_bt_b∂_c - (-1)^(|a|+|b|)(|b| + |c|)t_b∂_ct_a∂_b.

        We have
        
    t_a∂_bt_b∂_c    = (-1)^|b||c|t_a∂_b ∂_c t_b
       = t_a∂_c ∂_bt_b.

        Also,
        
    t_b∂_ct_a∂_b    = (-1)^|b||c|∂_c t_bt_a∂_b
       = (-1)^|b||c| + |b||a|∂_ct_at_b ∂_b
       = (-1)^|b||c| + |b||a| + |a||c|t_a∂_c t_b∂_b.

        Thus (-1)^(|a|+|b|)(|b| + |c|)t_b∂_c t_a∂_b = (-1)^|b|^2t_a∂_c t_b∂_b = (-1)^|b|t_a∂_c t_b∂_c. Plugging in to (<ref>), we get
        
    [t_a∂_b, t_b∂_c] = t_a∂_c(∂_bt_b - (-1)^|b|t_b∂_b) = t_a∂_c.

        Then, we have
        
    [φ(e_ab), φ(e_bc)]    = [t_a∂_b ⊗ 1 + 1 ⊗ e_ab, t_b∂_c ⊗ 1 + 1 ⊗ e_bc] 
       = t_a∂_c + 1 ⊗ e_ac
       = φ(e_ac)  = φ([e_ab, e_bc]),

        as desired.
        

        
  Case 3 [e_ab,e_ca], c b5pt
        Swapping the order of the two terms gives us Case 2. Note that all signs picked up in the process get canceled.

        
  Case 4 [e_ab,e_cd], c b and a d5pt
        First, we show
        
    [t_a∂_b, t_c∂_d] = 0.

        The left hand side is
        
    t_a∂_bt_c∂_d - (-1)^(|a|+|b|)(|c|+|d|)t_c∂_dt_a∂_b.

        We have
        
    t_c∂_d t_a∂_b    = (-1)^|a|(|c|+|d|)t_at_c∂_d ∂_b
       = (-1)^|a|(|c|+|d|)+|b|(|c|+|d|)t_a∂_bt_c∂_d
       = (-1)^(|a|+|b|)(|c|+|d|)t_a∂_bt_c∂_d.

        Plugging back into (<ref>), we get the desired result. Thus, it follows that φ(e_ab) commutes with φ(e_cd). Hence, [φ(e_ab), φ(e_cd)] = 0 = φ([e_ab, e_cd]) as needed.

        
  Case 5 [e_ab, e_b0]5pt
        Since R bracketed with anything is zero, we have
        
    [φ(e_ab),φ(e_b0)]    = [t_a∂_b⊗ 1 + 1⊗ e_ab,t_b∂_0⊗ 1 -∑_j (-1)^|b||j|t_j/t_0⊗ e_bj]
       =[t_a∂_b⊗ 1,t_b∂_0⊗ 1] - (-1)^|b|[t_a∂_b⊗ 1 , t_b/t_0⊗ e_bb] 
       = - ∑_j  a (-1)^|b||j|[1⊗ e_ab, t_j/t_0⊗ e_bj] 
       =-(-1)^|a||b|[1⊗ e_ab,t_a/t_0⊗ e_ba]
       = t_a∂_0 ⊗ 1 - (-1)^|b|t_a/t_0⊗ e_bb +(-1)^|b|t_a/t_0⊗ e_bb
       = - ∑_j (-1)^|a||j|t_j/t_0⊗ e_aj
       = t_a∂_0 ⊗ 1 - ∑_j (-1)^|a||j|t_j/t_0⊗ e_aj
       = φ([e_ab,e_b0])
 as desired. We now prove each of the four brackets used here. That [t_a∂_b⊗ 1, t_b∂_0 ⊗ 1] = t_a∂_0⊗ 1 was proved in Case 2. Next, we have
        
    [t_a∂_b⊗ 1 , t_b/t_0⊗ e_bb]    = [t_a∂_b, t_b/t_0]⊗ e_bb

        since e_bb is even. We have
        
    [t_a∂_b, t_b/t_0]    = t_a∂_b t_b/t_0 - (-1)^|b|(|a|+|b|)t_b/t_0t_a∂_b
       = t_a/t_0∂_bt_b - (-1)^|b|(|a|+|b|)+ |a||b|t_a/t_0t_b∂_b
       = t_a/t_0(∂_bt_b - (-1)^|b|t_b∂_b) = t_a/t_0.

        This verifies the second bracket. The third bracket is
        
    [1⊗ e_ab, t_j/t_0⊗ e_bj]

        for j a. Note that e_bje_ab = 0. Thus the bracket is
        
    (-1)^(|a|+|b|)|j|t_j/t_0⊗ e_aj.

        Multiplying by (-1)^|b||j| gives
        
    (-1)^|a||j|t_j/t_0⊗ e_aj,

        as desired. The final bracket is
        
    [1⊗ e_ab,t_a/t_0⊗ e_ba]    = (-1)^|a|(|a|+|b|)t_a/t_0⊗ e_aa - (-1)^(|a|+|b|)(|b|)t_a/t_0⊗ e_bb.

        Multiplying by (-1)^|a||b|, we get
        
    (-1)^|a||a|t_a/t_0⊗ e_aa - (-1)^|b|t_a/t_0⊗ e_bb,

        as desired.

        
  Case 6 [e_ab, e_0a] 5pt
        We show
        
    [t_a∂_b, t_0∂_a] = -(-1)^(|a|+|b|)|a|t_0∂_b.

        The left hand side is
        
    t_a∂_bt_0∂_a - (-1)^(|a|+|b|)|a|t_0∂_at_a∂_b.

        We have t_a∂_bt_0∂_a = (-1)^|a||b|t_0∂_bt_a∂_a = (-1)^|a|(-1)^(|a|+|b|)|a|t_0∂_b t_a∂_a. Also t_0∂_at_a∂_b = t_0∂_b∂_at_a. Plugging these back into (<ref>), we get (<ref>). Then, we have
        
    [φ(e_ab), φ(e_0a)]    = [t_a∂_b ⊗ 1 + 1 ⊗ e_ab, t_0∂_a ⊗ 1] 
       = -t_0∂_b ⊗ 1 = φ(-e_0b) 
       = φ([e_ab, e_0a]),
as desired.
        
        
  Case 7 [e_ab, e_00]5pt
        We have [e_ab, e_00]=0, so φ([e_ab, e_00])=0. Note that t_0∂_0 commutes with t_a∂_b, so [t_a∂_b, t_0∂_0]=0. Since R is central, we have
        
    [φ(e_ab),φ(e_00)]    = [t_a∂_b⊗ 1 + 1⊗ e_ab+δ_ab(-1)^|a||b|R, t_0∂_0⊗ 1 +R]
       =0 = φ([e_ab,e_00]).

        
  Case 8 [e_a0,e_b0]5pt
        We have 
        
    [φ(e_a0), φ(e_b0)]    = [t_a∂_0 ⊗ 1 - ∑_j (-1)^|a||j|t_j/t_0⊗ e_aj, t_b∂_0 ⊗ 1 - ∑_j (-1)^|b||j|t_j/t_0⊗ e_bj].

        We evaluate each sub-bracket one by one. It is easily checked that [t_a∂_0⊗ 1, t_b∂_0⊗ 1]=0. We have
        
    [t_a∂_0 ⊗ 1, t_j/t_0⊗ e_bj]    = t_a∂_0 t_j/t_0⊗ e_bj - (-1)^|a||b|+ (|b| + |j|)|a|t_j/t_0t_a∂_0 ⊗ e_bj
       = t_at_j (∂_0 1/t_0)⊗ e_bj - t_at_j(1/t_0∂_0) ⊗ e_bj.

        Similar to the identity ∂_0t_0 - t_0∂_0 = 1, we have ∂_01/t_0-1/t_0∂_0 = -1/t_0^2, so the bracket evaluates to -t_at_j. Thus
        
    [t_a∂_0⊗ 1, - ∑_j(-1)^|b||j|t_j/t_0⊗ e_bj] = ∑_j(-1)^|b||j|t_at_j/t_0^2⊗ e_bj.

        Next we evaluate 
        
    [t_i/t_0⊗ e_ai, t_j/t_0⊗ e_bj]

        for i,j∈ I. If i b then e_aie_bj=0. If j a then e_bje_ai=0. Thus
        
    [t_i/t_0⊗ e_ai, t_j/t_0⊗ e_bj]    = δ_ib(-1)^|j|(|a|+|b|)t_bt_j/t_0^2⊗ e_aj - δ_aj(-1)^|a|||b| + (|b|+|a)|i|t_at_i/t_0^2⊗ e_bi.

        Thus
        
    [-∑_i(-1)^|a||i|t_j/t_0⊗ e_ai, -∑_j(-1)^|b||j|t_j/t_0⊗ e_bj]   = (-1)^|a||b|∑_j(-1)^|a||j|t_bt_j/t_0^2⊗ e_aj
       = -∑_j (-1)^|b||j|t_at_j/t_0^2⊗ e_bj.

        Finally, 
        
    [t_j/t_0⊗ e_aj, t_b∂_0⊗ 1]    = (-1)^(|a|+|j|)|b| + |j||b| t_bt_j(1/t_0∂_0) ⊗ e_aj - (-1)^|a||b| t_bt_j(∂_0 1/t_0)⊗ e_aj
       = (-1)^|a||b|t_bt_j/t_0^2⊗ e_aj.

        Thus
        
    [ ∑_j (-1)^|a||j|t_j/t_0⊗ e_aj, t_b∂_0 ⊗ 1] =  - (-1)^|a||b|∑_j (-1)^|a||j|t_bt_j/t_0^2⊗ e_aj.

        Adding everything back together, we see that everything cancels, so
        
    [φ(e_a0), φ(e_b0)] = 0 = φ([e_a0,e_b0]).


        
  Case 9 [e_a0,e_0b]5pt
        Now
        
    [φ(e_a0), φ(e_0b)]    = [t_a∂_0 ⊗ 1 - ∑_j (-1)^|a||j|t_j/t_0⊗ e_aj, t_0∂_b ⊗ 1]
       = [t_a∂_0, t_0∂_b] ⊗ 1 - ∑_j(-1)^|a||j|[t_j/t_0⊗ e_aj ,t_0∂_b ⊗ 1].

        But
        
    [t_j/t_0⊗ e_aj, t_0∂_b ⊗ 1]    = (t_j/t_0⊗ e_aj)(t_0∂_b ⊗ 1) - (-1)^|a||b|(t_0∂_b ⊗ 1)(t_j/t_0⊗ e_aj) 
       = (-1)^(|a|+|j|)|b|(t_j/t_0t_0∂_b) ⊗ e_aj - (-1)^|a||b|(t_0∂_b)(t_j/t_0) ⊗ e_aj
       = (-1)^|a||b|[(-1)^|j||b|(t_j∂_b) -(∂_bt_j)] ⊗ e_aj

        If j = b the expression in the brackets is
        
    -(∂_b t_b - (-1)^|b|t_b∂_b) = -1.

        If j b the expression in the brackets is
        
    (-1)^|j||b|(t_j∂_b)-(∂_bt_j) = (-1)^|j||b|(t_j∂_b)-(-1)^|j||b|(t_j∂_b) = 0.

        Plugging these back into the original expression, we get
        
    [φ(e_a0), φ(e_0b)]    = [t_a∂_0, t_0∂_b]⊗ 1 + 1⊗ e_ab.

        If a b, from Case 2 we have [t_a∂_0,t_0∂_b] = t_a∂_b, so
        
    [φ(e_a0), φ(e_0b)] = t_a∂_b⊗ 1 + 1⊗ e_ab = φ([e_a0,e_0b]).

        If a = b, we have
        
    [t_a∂_0, t_0∂_a]    = t_a∂_0t_0∂_a - (-1)^|a|t_0∂_a t_a∂_0
       =∂_0t_0 t_a∂_a - (-1)^|a| t_0∂_0 ∂_a t_a
       = ∂_0t_0 t_a∂_a - (-1)^|a|t_0∂_0 (1+(-1)^|a| t_a∂_a)
       = (-1)^|a|t_0∂_0 + (∂_0 t_0 - t_0∂_0)t_a∂_a
       = t_a∂_a - (-1)^|a|t_0∂_0.

        Thus
        
    [φ(e_a0), φ(e_0a)] = t_a∂_0 ⊗ 1 - (-1)^|a|t_0∂_0 ⊗ 1 + 1⊗ e_aa = φ(e_aa - (-1)^|a|e_00) = φ([e_a0,e_0a]).

        
        
  Case 10 [e_a0,e_00]5pt
        Since R is central,
        
    [φ(e_a0), φ(e_00)] = [t_a∂_0⊗ 1 -∑_j (-1)^|a||j|t_j/t_0⊗ e_aj, t_0∂_0 ⊗ 1]

        Now,
        
    [t_a∂_0, t_0∂_0] = (t_a∂_0)(t_0∂_0) - (t_0∂_0)(t_a∂_0) = t_a(∂_0t_0 - t_0∂_0) ∂_0 = t_a∂_0.

        Also,
        
    [t_j/t_0⊗ e_aj, t_0∂_0 ⊗ 1] = t_j(∂_0t_0- t_0 ∂_0)1/t_0⊗ e_aj = t_j/t_0⊗ e_aj.

        Thus
        
    [φ(e_a0), φ(e_00)] = t_a∂_0 ⊗ 1 - ∑_j (-1)^|a||j|t_j/t_0⊗ e_aj = φ([e_a0,e_00]).

        
  Case 11 [e_0a,e_0b]5pt
        We have 
        
    φ([e_0a,e_0b])=φ(0)=0.

        Also
        
    [φ(e_0a),φ(e_0b)]    = [t_0∂_a⊗ 1, t_0∂_b⊗ 1]
       = t_0^2 (∂_a∂_b - (-1)^|a||b|∂_b∂_a) ⊗ 1 
       = 0


        
  Case 12 [e_0b,e_00]5pt
        We have [e_0b,e_00]=-e_0b, so φ([e_0b,e_00])=-t_0∂_b ⊗ 1. 
        Since R is central, we also have
        
    [φ(e_0b),φ(e_00)]    = [t_0∂_b⊗ 1, t_0∂_0 ⊗ 1]
       = (-t_0∂_0t_0∂_b + t_0∂_bt_0∂_0)⊗ 1
       = -( t_0(∂_0t_0-t_0∂_0) ∂_b)⊗ 1
       = - t_0∂_b⊗ 1 = φ([e_0b, e_00]).

        
  Case 13 [e_0c, e_ab], c a5pt We have [e_0c,e_ab]=0 and
        
    [φ(e_0c),φ(e_ab)]    = [t_0∂_c⊗ 1, t_a∂_b ⊗ 1 + 1⊗ e_ab + δ_ab(-1)^|a||b|R]
       = [t_0∂_c, t_a∂_b] ⊗ 1.

        Now we have
        
    [t_0∂_c, t_a∂_b]   =t_0∂_c t_a∂_b - (-1)^|c|(|a|+|b|)t_a∂_b t_0∂_c
       = t_0∂_c t_a∂_b - (-1)^|c|(|a|+|b|)+ |c|(|a|+|b|)t_0∂_c t_a∂_b
       = 0.

        Thus [φ(e_0c), φ(e_ab)] = 0 = φ([e_0c, e_ab]).
        
  Case 14 [e_ab, e_c0], b c5pt We have [e_ab, e_c0]=0. Also,
        
    [φ(e_ab), φ(e_c0)]    = [t_a∂_b⊗ 1 + 1⊗ e_ab + δ_ab(-1)^|a||b|R, t_c∂_0 ⊗ 1- ∑_j(-1)^|c||j|t_j/t_0⊗ e_cj]
       =[t_a∂_b, t_c∂_0]⊗ 1-∑_j(-1)^|c||j|[t_a∂_b⊗ 1, t_j/t_0⊗ e_cj]
        - ∑_j(-1)^|c||j|[1⊗ e_ab, t_j/t_0⊗ e_cj].

        Now we evaluate the brackets. We have
        
    [t_a∂_b, t_c∂_0]    = t_a∂_bt_c∂_0 - (-1)^(|a|+|b|)|c| t_c∂_0 t_a∂_b
       = (-1)^(|a|+|b|)|c| t_c∂_0 t_a∂_b - (-1)^(|a|+|b|)|c| t_c∂_0 t_a∂_b
       = 0.

        Next, we have
        
    [t_a∂_b⊗ 1, t_j/t_0⊗ e_cj]    = t_a∂_b t_j/t_0⊗ e_cj - (-1)^(|a|+|b|)|c|(t_j/t_0⊗ e_cj)(t_a∂_b⊗ 1)
       = 1/t_0t_a∂_bt_j⊗ e_cj - (-1)^(|a|+|b|)|j|1/t_0 t_j t_a∂_b ⊗ 1.

        If j b, then this becomes
        
    1/t_0t_a∂_bt_j⊗ e_cj - 1/t_0t_a∂_bt_j⊗ e_cj = 0.

        If j = b this becomes
        
    t_a/t_0 (∂_bt_b - (-1)^|b|t_b∂_b)⊗ e_cb = t_a/t_0⊗ e_cb.

        Finally, we have
        
    [1⊗ e_ab, t_j/t_0⊗ e_cj]    = (-1)^(|a|+|b|)|j|t_j/t_0⊗ (e_abe_cj) - (-1)^(|a|+|b|)|c|t_j/t_0⊗ e_cje_ab
       =  - (-1)^(|a|+|b|)|c|t_j/t_0⊗ e_cje_ab

        Now e_cje_ab is nonzero if and only if j=a. When j=a, the above expression becomes
        
    (-1)^(|a|+|b|)|c|t_a/t_0⊗ e_cb.

        Plugging these back into the orginal expression, we get
        
    [φ(e_ab), φ(e_c0)]  = - (-1)^|c||b|t_a/t_0⊗ e_cj + (-1)^|c||a|+ (|a|+|b|)|c|t_a/t_0⊗ e_cb = 0 = φ([e_ab, e_c0]).


        
        
    
    This completes the verification and finishes the proof. 




§ FORMULAS FOR THE IMAGES OF CERTAIN ELEMENTS UNDER VARPHI

For the following section, we will write these Gelfand invariants in terms of the following special elements of U((m|n)). Set r_0^(m|n)(a,b)=δ_ab(-1)^|a||b|. For k≥ 0 and a,b∈ I, let

    r_k+1^(m|n)(a,b) = ∑_i_1,…, i_k∈ I (-1)^|i_1|+⋯ + |i_k|e_ai_1⊗ e_i_1i_2⊗⋯⊗ e_i_kb.

Then we have

    G_k^(m|n) = ∑_i∈ I  r_k^(m|n)(i,i)

for k≥ 1. We compute the images of the elements r_k^(m|n)(a, b) for all k, a, b. The proofs of these formulas are not dependent on the results in previous sections of this paper. In this way, we can obtain an alternative (computational) proof of Theorem <ref>.

First, notice the following identity:

    r_k+1^(m+1|n)(a,b) = ∑_i∈Î (-1)^|i|r_k^(m+1|n)(a,i)e_ib.

Indeed, we have

    ∑_i∈Î (-1)^|i|r_k^(m+1|n)(a,i)e_ib   = ∑_i∈Î∑_i_1,…, i_k-1∈Î (-1)^|i_1|+⋯+|i_k-1|+|i| e_ai_1⋯ e_i_k-1i e_ib
       = r_k+1^(m+1|n)(a,b).


Define 

    f_s(a,b)=∑_i∈ I(-1)^|i|t_a∂_i⊗ r_s-1^(m|n)(i,b).

Also note that we have f_1(a,b) = t_a∂_b.

To simplify the statement of the following theorem, we first introduce some more notation. Let 


    A_k(s)    = ∑_g=0^k-skg R^g R_2^k-s-g
    
        B(s)    = ∑_i,j∈ I (-1)^|i|(1+|j|)∂_i t_j ⊗ r_s-1^(m|n)(i,j)
    
        D_k(a,b)    = ∑_g=0^kkgR^g (1⊗ r_k-g^(m|n)(a,b))
    
        E_k(a)    = ∑_g=0^k-1(kgR^g∑_j>0 (-1)^|a||j|t_j/t_0⊗ r_k-g^(m|n)(a,j))
    
        F_k(a,b)    = ∑_s=1^k f_s(a,b)A_k(s).

Then, we can find the images of the r_k^(m+1|n).

    We have
    
    φ(r_k^(m+1|n)(a,b))    = F_k(a,b) + D_k(a,b)
    φ(r_k^(m+1|n)(a,0))    = A_k(1)(t_a∂_0⊗ 1) - E_k(a) - (t_a/t_0⊗ 1)∑_s=2^k A_k(s)B(s)
    φ(r_k^(m+1|n)(0,b))    = F_k(0,b)
    φ(r_k^(m+1|n)(0,0))    = A_k(1)(t_0∂_0⊗ 1) + R^k -∑_s=2^k A_k(s)B(s).




    We prove all four statements simultaneously by induction on k. The base case k = 1 follows from the definition of φ. Suppose the formulas in the statement of the Theorem are true for some positive integer k. Let us prove them for k + 1.

    First, consider the value of φ(r_k^(m+1|n)(a, b)):
    
    φ(r_k+1^(m|n)(a,b))    = φ(r_k^(m+1|n)(a,0))φ(e_0b) +∑_i∈ I(-1)^|i|φ(r_k^(m|n)(a,i))φ(e_ib)
       = (A_k(1)(t_a∂_0⊗ 1) - E_k(a) - (t_a/t_0⊗ 1)∑_s=2^k A_k(s)B(s))(t_0∂_b⊗ 1)
       =+∑_i∈ I(-1)^|i|(F_k(a,i)+D_k(a,i))(t_i∂_b⊗ 1 + 1⊗ e_ib + δ_ib(-1)^|i||b|R)
       = A_k(1)(t_a∂_0 t_0∂_b⊗ 1) 
       =- ∑_g=0^k-1(kgR^g∑_i∈ I(-1)^(|a|+|b|)|i| + |a||b|t_i∂_b⊗ r_k-g^(m|n)(a,i))
       =-(t_a/t_0⊗ 1)∑_s=2^kA_k(s)B(s)(t_0∂_b⊗ 1)+ ∑_i∈ I(-1)^|i|∑_s=1^k f_s(a,i)(t_i∂_b⊗ 1)A_k(s)
       =+ ∑_i∈ I∑_g=0^k kgR^g (-1)^(|a|+|b|)|i| + |a||b|(t_i∂_b⊗ r_k-g^(m|n)(a,i))
       =+ ∑_i∈ I∑_s=1^k((-1)^|i|f_s(a,i)(1⊗ e_ib)A_k(s)) 
       =+∑_i∈ I∑_g=0^kkgR^g (1⊗ (-1)^|i|r_k-g^(m|n)(a,i)e_ib)
       =+∑_s=1^k f_s(a,b)A_k(s)R + ∑_g=0^kkgR^g+1(1⊗ r_k-g^(m|n)(a,b)).

    Write this sum as X_1 - X_2 - X_3 + X_4 + X_5 + X_6 + X_7 + X_8 + X_9. Then 
    
    X_5 - X_2 = R^k(t_a∂_b⊗ 1).

    We have
    
    X_3    = (t_a/t_0⊗ 1)∑_s=2^kA_k(s)B(s)(t_0∂_b⊗ 1)
       = ∑_s=2^k ∑_i,j∈ I(-1)^|i|(1+|j|)t_a/t_0∂_i t_j ⊗ r_s-1^(m|n)(i,j)⊗ (t_0∂_b⊗ 1)A_k(s)
       =∑_s=2^k ∑_i,j∈ I (-1)^|i|+|i||j| + |i||b| + |j||b|t_a∂_i t_j ∂_b⊗ r_s-1^(m|n)(i,j)A_k(s)

    and
    
    X_4    = ∑_i∈ I(-1)^|i|∑_s=1^k f_s(a,i)(t_i∂_b⊗ 1)A_k(s)
       =∑_s=1^k ∑_i,j∈ I(-1)^|i|+|j|t_a∂_j ⊗ r_s-1^(m|n)(j,i)⊗ (t_i∂_b⊗ 1)A_k(s)
       =∑_s=1^k ∑_i,j∈ I(-1)^|j| + |j||i| + |j||b| + |i||b|t_a∂_j t_i ∂_b⊗ r_s-1^(m|n)(j,i)A_k(s)
       = ∑_s=1^k ∑_i,j∈ I (-1)^|i|+|i||j| + |i||b| + |j||b|t_a∂_i t_j ∂_b⊗ r_s-1^(m|n)(i,j)A_k(s).

    Thus
    
    X_4 - X_3 = (∑_i∈ I(-1)^|i|t_a∂_it_i ∂_b⊗ 1)A_k(1) = ((t_a∂_b ⊗ 1)(R_2)-t_a∂_0t_0∂_b ⊗ 1)A_k(1)

    as ∑_i∈Î(-1)^|i|t_a∂_it_i∂_b⊗ 1 = (t_a∂_b⊗ 1)R_2:
    
    ∑_i∈Î (-1)^|i|t_a∂_i t_i ∂_b    = ∑_i∈Î t_a ((-1)^|i|+t_i∂_i)∂_b
       = (t_a∂_b)(m+1-n) +∑_i∈Î
     i bt_at_i∂_i∂_b + t_at_b∂_b∂_b
       = (t_a∂_b)(+ m+1-n) - (t_a∂_b)(t_b∂_b)+ t_at_b∂_b∂_b.

    If b is even, then the last two terms become -t_a∂_b. If b is odd, the last term is zero, and it is easily verified that t_a∂_bt_b∂_b = t_a∂_b, Either way, we get -(t_a∂_b)(t_b∂_b)+t_at_b∂_b∂_b = -t_a∂_b. Plugging this back in proves the desired identity.
    Then
    
    X_1 + X_4 - X_3 = (t_a∂_b⊗ 1)A_k(1)R_2 = (t_a∂_b⊗ 1)(∑_g=0^k-1kgR^gR_2^k-g).

    Therefore we have
    
    X_1 + X_4 - X_3 + X_5 - X_2 = (t_a∂_b⊗ 1)(∑_g=0^kkgR^g R_2^k-g).

    Since ∑_i∈ I(-1)^|i|f_s(a,i)e_ib = f_s+1(a,b), we have
    
    X_6 = ∑_s=1^k f_s+1(a,b)A_k(s) = ∑_s= 2^k+1f_s(a,b)(∑_g=0^k+1-skgR^gR_2^k+1-s-g).

    Thus
    
    X_6 + X_1 + X_4 - X_3 + X_5 - X_2 = ∑_s=1^k+1f_s(a,b)(∑_g=0^k+1-skgR^g R_2^k+1-s-g).

    But
    
    X_8 = ∑_s=1^k(f_s(a,b)∑_g=1^k-s+1kg-1R^gR_2^k-s-g+1).

    Thus
    
    X_8 +X_6 + X_1 + X_4 - X_3 + X_5 - X_2 = ∑_s=1^k+1(f_s(a,b)∑_g=0^k+1-sk+1gR^g R_2^k+1-s-g) = F_k+1(a,b).

    Now, since ∑_i∈ I(-1)^|i|r_k-g^(m|n)(a,i)e_ib= r_k+1-g^(m|n)(a,b), we find that
    
    X_7 = ∑_g=0^kkgR^g(1⊗ r_k+1-g^(m|n)(a,b)).

    Also
    
    X_9 = ∑_g=1^kkg-1R^g(1⊗ r_k+1-g^(m|n)(a,b))

    so
    
    X_7+ X_9 = ∑_g=0^kk+1gR^g(1⊗ r_k+1-g^(m|n)(a,b)) = D_k+1(a, b).


    Therefore
    
    X_1 - X_2 - X_3 + X_4 + X_5 + X_6 + X_8 + (X_7 + X_9) = F_k+1(a, b) + D_k+1(a, b).

    This completes the proof of the inductive step for φ(r_k+1^(m+1|n)(a,b)).
    
    
    Next, we consider the value of φ(r_k^(m+1|n)(a,0)). We have
    
    φ(r_k+1^(m+1|n)(a,0))    = φ(r_k^(m+1|n)(a,0))φ(e_00) + ∑_i∈ I(-1)^|i|φ(r_k^(m+1|n)(a,i))φ(e_i0)
       = (A_k(1)(t_a∂_0⊗ 1)-E_k(a)- (t_a/t_0⊗ 1)∑_s=2^kA_k(s)B(s))(t_0∂_0⊗ 1 + R)
       =+∑_i∈ I(-1)^|i|(F_k(a,i)+D_k(a,i))(t_i∂_0⊗ 1 - ∑_j∈ I(-1)^|i||j|t_j/t_0⊗ e_ij)
       = A_k(1)(t_a∂_0t_0∂_0⊗ 1)-∑_g=0^k-1(kg∑_j∈ I(-1)^|a||j|t_j∂_0⊗ r_k-g^(m|n)(a,j)R^g)
       = - ∑_s=2^k A_k(s) (∑_i,j∈ I(-1)^|i|(1+|j|)t_a∂_i t_j ∂_0⊗ r_s-1^(m|n)(i,j))
       =+A_k(1)R(t_a∂_0⊗ 1) - ∑_g=0^k-1(kgR^g+1∑_j∈ I (-1)^|a||j|t_j/t_0⊗ r_k-g^(m|n)(a,j))
       =-(t_a/t_0⊗ 1)∑_s=2^kA_k(s)RB(s)
       =+∑_s=1^k(∑_i∈ I(-1)^|i|f_s(a,i)(t_i∂_0⊗ 1))A_k(s)
       =+ ∑_g=0^kkgR^g (∑_i∈ I(-1)^|a||i|t_i∂_0⊗ r_k-g^(m|n)(a,i))
       =-∑_s=1^k(∑_i,j∈ I(-1)^|i|(1+|j|)f_s(a,i) (t_j/t_0⊗ e_ij))A_k(s)
       =-∑_g=0^k kgR^g(∑_i,j∈ I(-1)^|a||j|t_j/t_0⊗ (-1)^|i|r^(m|n)_k-g(a,i)e_ij).

    Call this sum X_1 - X_2 - X_3 + X_4 - X_5 - X_6 + X_7 + X_8 - X_9 - X_10. We have 
    
    X_8- X_2 = R^k (-1)^|a|t_a∂_0 ⊗ r_0^(m|n)(a,a) = R^k(t_a∂_0 ⊗ 1).

    Also
    
    X_5 + X_10   = ∑_g=1^k kg-1R^g(∑_j∈ I (-1)^|a||j|t_j/t_0⊗ r_k+1-g^(m|n)(a,j))
       =+∑_g=0^k kgR^g(∑_j∈ I (-1)^|a||j|t_j/t_0⊗ r_k+1-g^(m|n)(a,j))
       = ∑_g=0^k k+1gR^g(∑_j∈ I (-1)^|a||j|t_j/t_0⊗ r_k+1-g^(m|n)(a,j))
       = E_k+1(a).

    Next, 
    
    X_9 = ∑_s=1^k(∑_i,j∈ I(-1)^|i|(1+|j|)f_s(a,i) (t_j/t_0⊗ e_ij))A_k(s).

    For any positive integer s, 
    
    ∑_i,j∈ I(-1)^|i|(1+|j|) f_s(a,i)(t_j/t_0⊗ e_ij)
     =   ∑_i,j∈ I(-1)^|i|(1+|j|)(∑_ℓ∈ I (-1)^|ℓ| t_a∂_ℓ⊗ r_s-1^(m|n)(ℓ, i))(t_j/t_0⊗ e_ij)
    
            =   ∑_j,ℓ∈ I (-1)^|ℓ|(1+|j|) t_a∂_ℓt_jt_0^-1⊗(∑_i∈ I(-1)^|i|r_s-1^(m|n)(ℓ, i) e_ij)
    
            =   (t_a/t_0⊗ 1)∑_j,ℓ∈ I(-1)^|ℓ|(1+|j|)∂_ℓ t_j⊗ r_s^(m, n)(ℓ, j)
    
            =   (t_a/t_0⊗ 1)B(s+1).

    Then
    
    X_9 =  (t_a/t_0⊗ 1) ∑_s=2^k+1B(s)(∑_g=0^k+1-skgR^g R_2^k+1-s-g).

    Also
    
    X_6 =  (t_a/t_0⊗ 1) ∑_s=2^kB(s)(∑_g=1^k+1-skg-1R^g R_2^k+1-s-g).

    Adding, we get
    
    X_6 + X_9 =   (t_a/t_0⊗ 1) ∑_s=2^kA_k+1(s)B(s).


    Next, for any positive integer s, 
    
    ∑_i∈ I (-1)^|i|f_s(a,i)(t_i∂_0⊗ 1)    = ∑_i,j∈ I(-1)^|i||j|t_a∂_j t_i∂_0⊗ r_s-1^(m|n)(j,i)
       =∑_i,j∈ I(-1)^|i||j|t_a∂_it_j∂_0⊗ r_s-1^(m|n)(i,j).

    Thus
    
    X_7 = ∑_s=1^k(∑_i,j∈ I(-1)^|i||j|t_a∂_it_j∂_0⊗ r_s-1^(m|n)(i,j))A_k(s).

    On the other hand,
    
    X_3 = ∑_s=2^k(∑_i,j∈ I(-1)^|i||j|t_a∂_it_j∂_0⊗ r_s-1^(m|n)(i,j))A_k(s)

    so
    
    X_7 - X_3 = A_k(1)(t_a∂_0⊗ 1)(R_2-t_0∂_0⊗ 1).

    Then
    
    X_7 - X_3 + X_1 = A_k(1)R_2(t_a∂_0⊗ 1) = (∑_g=0^k-1kgR^g R_2^k-g)(t_a∂_0⊗ 1).

    Thus
    
    X_7 - X_3 + X_1 + X_8 - X_2 = (∑_g=0^kkgR^g R_2^k-g)(t_a∂_0⊗ 1).

    Since
    
    X_4 = (∑_g=1^k kg-1R^g R_2^k-g)(t_a∂_0⊗ 1),

    we get
    
    X_7 - X_3 + X_1 + X_8 - X_2 + X_4 = (∑_g=0^kk+1gR^g R_2^k-g)(t_a∂_0⊗ 1) = A_k+1(1)(t_a∂_0⊗ 1).

    Thus
    
    φ(r_k+1^(m+1|n)(a,0))    = (X_7 - X_3 + X_1 + X_8 - X_2 + X_4) - (X_5 + X_10) - (X_6+X_9)
       = A_k+1(1)(t_a∂_0⊗ 1) - E_k+1(a) - (t_a/t_0⊗ 1)∑_s=2^k A_k+1(s)B(s).

    This completes the inductive step for r_k+1^(m|n)(a,0).


    Next, we consider the value of φ(r_k^(m+1|n)(0, b)). We have
    
    φ(r_k+1^(m+1|n)(0,b)) 
               = φ(r_k^(m+1|n)(0, 0))φ(e_0b) +∑_i∈ I(-1)^|i|φ(r_k^(m+1|n)(0,i))φ(e_ib)
       = (A_k(1)(t_0∂_0⊗ 1)+R^k - ∑_s=2^k A_k(s)B(s))(t_0∂_b⊗ 1) 
       = + ∑_i∈ I(-1)^|i| F_k(0,i)(t_i∂_b ⊗ 1 + 1⊗ e_ib+δ_ib(-1)^|i||b|R)
       =A_k(1)(t_0∂_0t_0∂_b⊗ 1) + R^k(t_0∂_b⊗ 1) - ∑_s=2^kA_k(s)B(s)(t_0∂_b⊗ 1) 
       =+∑_s=1^k∑_i∈ I (-1)^|i|f_s(0,i)(t_i∂_b⊗ 1)A_k(s)
       = + ∑_s=1^k∑_i∈ I(-1)^|i|f_s(0,i)(1⊗ e_ib)A_k(s)
            +∑_s=1^k f_s(0,b) A_k(s)R.

    Write this sum as X_1 +X_2-X_3+X_4+X_5+X_6. Now 
    
    X_3    = ∑_s=2^k ∑_i,j∈ I (-1)^|i|(1+|j|)(∂_i t_j ⊗ r_s-1^(m|n)(i,j))(t_0∂_b⊗ 1)A_k(s)
       = ∑_s=2^k ∑_i,j∈ I (-1)^|i|(1+|j|) + (|i|+|j|)|b|(t_0∂_i t_j ∂_b ⊗ r_s-1^(m|n)(i,j))A_k(s)
       = ∑_s=2^k ∑_i,j∈ I (-1)^(|i|+|j|)(|i|+|b|)(t_0∂_i t_j ∂_b ⊗ r_s-1^(m|n)(i,j))A_k(s).

    Also 
     
    X_4    = ∑_s=1^k∑_i,j∈ I  (-1)^|i|+|j|(t_0∂_j⊗ r^(m|n)(j,i) ⊗ t_i∂_b ⊗ 1)A_k(s)
       =∑_s=1^k∑_i,j∈ I (-1)^|i|+|j|+(|i|+|j|)(|i|+|b|) (t_0∂_jt_i∂_b⊗ r_s-1^(m|n)(j,i) )A_k(s).

    Now (-1)^|i|+|j|+(|i|+|j|)(|i|+|b|)= (-1)^|j| + |j||i|+|i||b|+|j||b|= (-1)^(|j|+|i|)(|j|+|b|). Then swapping the variable names i,j gives
    
    X_4 = ∑_s=1^k∑_i,j∈ I(-1)^(|i|+|j|)(|i|+|b|)(t_0∂_i t_j∂_b ⊗ r_s-1^(m|n)(i,j))A_k(s).

    Then X_4-X_3 simplifies to 
    
    (∑_i∈ I (-1)^|i|t_0∂_it_i∂_b⊗ 1)A_k(1).

    Then X_1+X_4-X_3 is
    
    (t_0∂_b⊗ 1)∑_g=0^k-1kgR^g R_2^k-g.

    Thus
    
    X_2 + X_1 + X_4 - X_3  = (t_0∂_b⊗ 1)(∑_g=0^k kgR^gR_2^k-g).

    Since ∑_i∈ I(-1)^|i|f_s(0,i)(1⊗ e_ib) = f_m+1(0,i), we have 
    
    X_5 = ∑_s=1^k f_s+1(0,b)A_k(s) = ∑_s=2^k+1 f_s(0,b)∑_g=0^k+1-skgR^gR_2^k+1-s-g.

    Thus
    
    X_5 + X_2 + X_1 + X_4 - X_3 = ∑_s=1^k+1 f_s(0,b) ∑_g=0^k+1-skgR^g R_2^k+1-s-g.

    Also
    
    X_6 = ∑_s=1^kf_s(0,b)∑_g=1^k+1-skg-1R^gR_2^k+1-s-g.

    Adding the previous two equations gives 
    
    X_1 + X_2 - X_3 + X_4 + X_5+ X_6 = ∑_s=1^k+1f_s(0,b)A_k+1(s)

    which completes the inductive step for the value of φ(r_k+1^(m+1|n)(0,b)).

    
    Finally, we consider the value of φ(r_k^(m+1|n)(0,0)). We have
    
    φ(r_k+1^(m+1|n)(0,0))    = φ(r_k^(m+1|n)(0,0))φ(e_00) + ∑_i∈ I(-1)^|i|φ(r_k^(m+1|n)(0,i)) φ(e_i,0)
       =(A_k(1)(t_0∂_0⊗ 1) + R^k - ∑_s=2^k A_k(s)B(s))(t_0∂_0 ⊗ 1+R)
       = + ∑_i∈ I(-1)^|i|F_k(0,i)(t_i∂_0 -∑_j∈ I (-1)^|i||j|t_j/t_0⊗ e_ij)
       =A_k(1)(t_0∂_0t_0∂_0⊗ 1) + R^k(t_0∂_0⊗ 1) -∑_s=2^k A_k(s)B(s)(t_0∂_0⊗ 1)
       = + A_k(1)R (t_0∂_0⊗ 1) + R^k+1-∑_s=2^k(A_k(s)RB(s))
       =+∑_i∈ I (-1)^|i|∑_s=1^kf_s(0,i)(t_i∂_0⊗ 1)A_k(s)
       =-∑_i∈ I(-1)^|i|∑_s=1^k f_s(0,i)A_k(s)∑_j∈ I(-1)^|i||j|t_j/t_0⊗ e_ij.

    Write this as X_1 + X_2 - X_3 + X_4 + X_5 - X_6 + X_7 - X_8. 

    Since 
    
    X_7    = ∑_s=1^k∑_i,j∈ I (-1)^|i|+|j|(t_0∂_j⊗ r_s-1^(m|n)(j,i)⊗ t_i∂_0⊗ 1)A_k(s)
       =∑_s=1^k∑_i,j∈ I(-1^|j|(1+|i|))∂_j t_i⊗ r_s-1^(m|n)(j,i) A_k(s)(t_0∂_0⊗ 1)
       =∑_s=1^k∑_i,j∈ I(-1^|i|(1+|j|))∂_i t_j⊗ r_s-1^(m|n)(i,j) A_k(s)(t_0∂_0⊗ 1)

    we have 
    
    X_7-X_3 = ∑_i∈ I(-1)^|i|(∂_i t_i⊗ 1)(t_0∂_0⊗ 1) A_k(1) = (R_2-(t_0∂_0⊗ 1))(t_0∂_0⊗ 1)A_k(1).

    Then, expanding out the A_k(s) terms gives
    
    X_7-X_3 + X_1 + X_2 = (∑_g=0^k kgR^gR_2^k-g)(t_0∂_0⊗ 1).

    But since X_4 = (∑_g=1^k kg-1R^g R_2^k-g)(t_0∂_0⊗ 1), 
    
    X_7 - X_3 + X_1 + X_2 + X_4 = ( ∑_g=0^k k+1gR^g R_2^k-g)(t_0∂_0⊗ 1) = A_k+1(1)(t_0∂_0⊗ 1).


    We have 
    
    X_8 = ∑_s=1^k(∑_i,j∈ I(-1)^|i|(1+|j|)f_s(0,i)t_j/t_0⊗ e_ij)A_k(s).

    

    For any positive integer s, 
    
    ∑_i,j∈ I(-1)^|i|(1+|j|)f_s(0,i)t_j/t_0⊗ e_ij   = ∑_i,j∈ I(-1)^|i|(1+|j|)∑_ℓ∈ I (-1)^|ℓ|t_0∂_ℓ⊗ r_s-1^(m|n)(ℓ, i) ⊗t_j/t_0⊗ e_ij
       =∑_i,j∑_ℓ∈ I(-1)^|ℓ|(1+ |j|)∂_ℓt_j⊗ (-1)^|i|( r_s-1^(m|n)(ℓ, i) e_ij)
       = ∑_j,ℓ∈ I(-1)^|ℓ|(1+|j|)∂_ℓt_j⊗ r_s^(m|n)(ℓ, j)
       =B(s+1).

    Then
    
    X_8 = ∑_s=2^k+1(∑_g=0^k+1-skgR^g R_2^k+1-s-g)B(s).

    Also,
    
    X_6 = ∑_s=2^k(∑_g=1^k+1-skg-1R^g R_2^k+1-s-g)B(s)

    so
    
    X_6 + X_8 = ∑_s=2^k+1A_k+1(s) B(s).

    Finally, X_5 = R^k+1, so combining the above gives
    
    X_1 + X_2 - X_3 + X_4 + X_5 - X_6 + X_7 - X_8 = A_k+1(1)(t_0∂_0⊗ 1) + R^k+1 - ∑_s=2^k+1A_k+1(s)B(s).

    This completes the inductive step for φ(r_k+1^(m+1|n)(0,0)), and hence the proof of the theorem.




