Pfaffian Identities, with Applications to Free Resolutions, Dg−Algebras, and Algebras with Straightening Law

PFAFFIAN IDENTITIES, WITH APPLICATIONS TO FREE RESOLUTIONS, DG−ALGEBRAS, AND ALGEBRAS WITH STRAIGHTENING LAW

Andrew R. Kustin

Abstract. A pfaffian identity plays a central role in the proof that the minimal resolution of a Huneke-Ulrich deviation two Gorenstein ring is a DGΓ−algebra. A second pfaffian identity is the patch which holds together two strands of the minimal resolution of a residual intersection of a grade three Gorenstein ideal. In this paper, we establish a family of pfaffian identities from which the two previously mentioned identities can be deduced as special cases. Our proof is by induction: the base case is an identity of binomial coefficients and the inductive step is a calculation from multilinear algebra.

Two unrelated problems in commutative algebra have recently been solved by finding the appropriate pfaffian identity. Fix a commutative noetherian ring R.For the first problem, let X2n×2n and Y1×2n be matrices with entries from R. Assume that X is an alternating matrix. Huneke and Ulrich [10] showed that if the ideal I = I1(YX) + Pf(X) has the maximum possible grade (namely, 2n − 1), then I is a perfect Gorenstein ideal. The minimal R−resolution M of A = R/I was found in [11]. Srinivasan [15] proved M is a DGΓ−algebra. The pfaffian identity [15, 4.3] is the key to showing that the proposed multiplication satisfies the differential property. The algebra structure on M was used in [12] to show that if (R, m,k)is a regular local ring of equicharacteristic zero, then the Poincaréseries

X∞ M A i PA (z)= dimk Tori (M,k)z i=0

is a rational function for all finitely generated A−modules M. For the second problem, let J be a residual intersection of a grade three Goren- stein ideal I. If the data is sufficiently generic, then the generators of J,the minimal resolution of R/J, the minimal resolution of “half” of the divisor class group of R/J, and the minimal resolution of each power Ik have all been found in [14]. Each resolution in [14] is obtained by patching together two complexes all of whose maps are linear and well understood. The patch involves maps of many different degrees and is much more difficult to understand. Pfaffian identity [14,

1991 Mathematics Subject Classiﬁcation. 15A69, 13C40, 13D25, 13F50. Key words and phrases. algebras with straightening law, deviation two Gorenstein ideal, DG- algebras, pfaﬃan, residual intersection. The author was supported in part by the National Science Foundation DMS–9106329.

1 2 ANDREW R. KUSTIN

3.2] shows that patched object is a complex. (Some of the results from [14] may be obtained using other techniques: the generators of J are also calculated in [13] and the ideals Ik are also resolved in [2].) The above mentioned pfaffian identities (see Propositions 5.1 and 4.1 in the present paper) appear to be somewhat similar, but the exact relationship between them is not immediately obvious. The existing proofs of these identities are ad hoc and unpleasant. In the present paper, we derive the two identities from a common result: identity [15, 4.3] is the left side of Theorem 3.1 when the data is arranged so that the right side consists of only one term; and [14, 3.2] is the right side of Theorem 3.1 when the left side has become trivial. The proof of Theorem 3.1 is an induction which depends on an identity of binomial coefficients (Lemma 1.3) and a fact from multilinear algebra (Lemma 2.4). In section 6 we apply our techniques to derive a straightening formula for pfaf- fians. We conclude with a brief section which shows that many familiar pfaffian identities are special cases of Theorem 3.1.

1. Binomial coefficients

We often consider binomial coefficients with negative parameters; consequently, we now recall the standard definition and properties of these objects. m Definition 1.1. For integers i and m, the binomial coefficient i is defined to be  m(m − 1) ···(m−i+1)  if 0

m Observation 1.2. (a) If 0 ≤ m

The identities in Lemma 1.3 form the base step in the proof of our main result. Corollary 1.4 is well known. It may be proved directly using an argument similar to the proof of Lemma 1.3 or it may be deduced from Lemma 1.3 after two applications of Observation 1.2 (e). PFAFFIAN IDENTITIES 3

Lemma 1.3. Let A, B,andCbe integers. If 0 ≤ A,then X B k A B − k + − A , (a) ( 1) C k k =( 1) A C and k∈Z + + X B − k A B − A − k . (b) ( 1) C − k k = C k∈Z

Proof. The proof of (a) proceeds by induction on A.IfA= 0, then the only nonzero B term on the left side occurs when k = 0, and this term is C ,whichisequaltothe right side. We now suppose that the result holds for a ﬁxed A for all values of B and C. Let X B k A X − k + +1 . = ( 1) C k k k∈Z + X X X Observation 1.2 (b) gives = 1 + 2,where X B k A X B k A X − k + X − k + . 1 = ( 1) C k k and 2 = ( 1) C k k − k∈Z + k∈Z + 1 The induction hypothesis gives B B +1 X =(−1)A and X =(−1)A+1 . 1 A + C 2 A+C+1 We conclude that B +1 B B X = X + X =(−1)A+1 − =(−1)A+1 , 1 2 A+1+C A+C A +1+C as desired. Replace k with A − k in order to deduce (b) from (a). Corollary 1.4. Let A, B,andCbe integers. If 0 ≤ A,then X B A A B + . C−k k = C k∈Z The following generalization of Observation 1.2 (c) is used in the proof of Corol- lary 3.3. Lemma 1.5. If A and B are integers, then A A −B − 1 = +(−1)A+B . A − B B −A − 1

Proof. Let A A −B − 1 T = ,T= and T =(−1)A+B . 1 A − B 2 B 3 −A − 1

If 0 ≤ A,thenT1 =T2 and T3 = 0. Henceforth, we assume that A ≤−1. If 0 ≤ B, then A − B ≤ A ≤−1andT1 = 0. Furthermore, in this case, 0 ≤ B − A − 1; consequently, A B − A − B − A − −B − B 1 B 1 A+B+1 1 T2 = =(−1) =(−1) =(−1) = −T3. B B −A − 1 −A − 1 Finally, we assume that A ≤−1andB≤−1. In this case, T2 = 0 and A −B − −B − A+B 1 A+B 1 T1 = =(−1) =(−1) = T3. A − B A − B −A − 1 4 ANDREW R. KUSTIN

2. Multilinear algebra Data 2.1. R F R− Let be a commutativeV noetherian ring, be a free module of ﬁnite rank, and ϕ be an element of 2 F. V V V • F ∗ • F •F We make much useV of the -module structureV on ,aswellastheV - • ∗ i j ∗ module structure on F . In particular, if ai ∈ F and bj ∈ F ,then

j^−i i^−j ∗ ai(bj)∈ F and bj (ai) ∈ F.

The following formulas are well known and not diﬃcult to establish; see, for example, [4, Proposition A.3]. V• V• Proposition 2.2. Adopt Data 2.1. Let a, b ∈ F and c ∈ F ∗ be homogeneous elements. V (a) If a ∈ 1 F ,then

(a(c)) (b)=a∧(c(b)) + (−1)1+deg cc(a ∧ b).

V F (b) If c ∈ rank F ∗,then

(a(c)) (b)=(−1)ν (b(c)) (a),

where ν = (rank F − deg a)(rank F − deg b).

Note. The value for ν which is given above is correct and is diﬀerent than the value given in [4]. V• We also make heavy use of the divided power structure on F .Ifais a V• V` homogeneous element of F of even degree, then, for each integer k, a(k) ∈ F , V` where ` = k · deg a. (Of course, if `<0 or rank F<`,then F= 0.) Some of the properties of divided powers are collected below; more information may be found in [4]. Proposition 2.3. Adopt Data 2.1. (a) If k and ` are integers, then k ` ϕ(k) ∧ ϕ(`) + ϕ(k+`). = k (b) If b ∈ F ∗ and k is an integer, then b ϕ(k) = b(ϕ) ∧ ϕ(k−1).

The following multilinear algebra calculation is the key to the induction step in the proof of our main result. Lemma 2.4. k ` b b V Adopt Data 2.1. Let and be integers. If 1 and are homogeneous • ∗ elements of F with deg b1 =1,then h i h i h i (k) (`) (k−1) (`+1) (1) (k−1) (`) b1 ϕ (b) (ϕ ) + b1 ϕ (b) (ϕ ) − b1(ϕ ) ∧ ϕ (b) (ϕ ) h i (k) (`) = ϕ (b1 ∧ b) (ϕ ). PFAFFIAN IDENTITIES 5 (1) (k−1) (`) Proof. We expand S1 = b1(ϕ ) ϕ (b) (ϕ ) two diﬀerent ways. On the one hand, Proposition 2.2 gives S1 = S2 + S3,where (1) (k−1) (`) 1+deg b (k−1) (1) (`) S2 =b1(ϕ ) ∧ ϕ (b) (ϕ )andS3=(−1) ϕ (b) b1(ϕ ) ∧ ϕ .

V• V• Proposition 2.3, together with the module action of F ∗ on F , yields 1+deg b (k−1) (`+1) 1+deg b (k−1) (`+1) S3 =(−1) ϕ (b) b1(ϕ =(−1) ϕ (b) ∧ b1 (ϕ ) h i h i (k−1) (`+1) (k−1) (`+1) = − b1 ∧ ϕ (b) (ϕ )=−b1 ϕ (b) (ϕ ) .

On the other hand, Proposition 2.2 gives (k) (k) (k) b1(ϕ ) (b)=b1∧ϕ (b)−ϕ (b1∧b); and therefore, h i h i (1) (k−1) (`) (k) (`) S1 = b1(ϕ ) ∧ ϕ (b) (ϕ )= b1(ϕ ) (b) (ϕ ) h i (k) (`) (k) (`) = b1 ϕ (b) (ϕ ) − ϕ (b1 ∧ b) (ϕ ).

The proof is completed by equating the two values for S1. The identities in section 3 are stated and proved in a coordinate free manner; however, some of the applications in sections 4 through 7 are derived from these identites by using bases. The hypothesis in these applications is that X is an alternating matrix with entries from the commutative noetherian ring R. We recover Data 2.1 as follows. Suppose that the matrix X =(xij)hasNrows and columns. Let e1,...,eN beabasisforafreeR−module F ;letε1,...,εN be the corresponding dual basis for F ∗; and let ϕ be the element X ϕ = xij ei ∧ ej 1≤i

e e ∧ e ∧ ...∧e ε ε ∧ ε ∧ ...∧ε . I = n1 n2 ni and I = n1 n2 ni

Calculation 2.5. If I is an i−tuple of integers, then

i(i−1) eI (εI )=εI(eI)=(−1) 2 .

Proof. Let fi =(e1∧e2∧... ∧ ei)(ε1 ∧ ε2 ∧ ... ∧ εi). It is clear that f1 =1 i i(i−1) and fi+1 =(−1) fi. It is also clear that fi =(−1) 2 solves this recurrence relation. 6 ANDREW R. KUSTIN

Let I =(n1,... ,ni)beaﬁxedi−tuple of integers. If the integers n1,...,ni’s are distinct, then σ(I) is the sign of the permutation which rearranges n1,...,ni into ascending order. If there is a repeat among the n’s, then σ(I) = 0. We say that I is an index set of size i if n1

J1∪J2=S |J1|=j1

to mean that the sum is taken over all j1−element subsets J1 of S, and the com- J S J J J plement of 1 in isV denoted 2. In the above sum, 1 and 2 are both taken to r ∗ be index sets. If b ∈ F ,thenb(eI) is the element X b e σ I I be e (2.6) ( I )= ( 1 2) ( I1) I2 I1∪I2=I |I1|=r

Vi−r of F .IfY=(yij)isamatrixandI1 and J are j−tuples of integers, then we Y I J Y take ( 1; ) to be the determinant of the submatrix of which consistsP of rows ∗ I1 and columns J. In particular, If f : F → F is a map with f(εj)= yijei and i J is a j−tuple of integers, then ^j X f ε Y I J e . ( )( J )= ( 1; ) I1 I1∪I2={1,...,N} |I1|=j

Thus, Calculation 2.5 shows that if |I| = |J| = j,then ! j ^ j(j−1) (2.7) εI ( f)(εJ ) = εI (eI )Y (I,J)=(−1) 2 Y (I; J).

We use the notation “XI ” to denote pfaﬃans; in particular,  X  the pfaﬃan of the principal submatrix of consisting

It follows that if i is an even integer, then X ϕ(i/2) X e = J1 J1 ; J1∪J2={1,...,N} |J1|=i and therefore, Calculation 2.5 shows that (i/2) i/2 (2.8) εI ϕ = εI (eI ) XI =(−1) XI . PFAFFIAN IDENTITIES 7

3. The main results The ﬁrst main result in this paper is Theorem 3.1. Everything in sections 4 through 7, with one exception, follows from Theorem 3.1 or one of its ﬁve Corol- laries. The one exception is Proposition 5.4, and this result is a consequence of Theorem 3.7, which is the other main result in the paper. Vd Theorem 3.1. Adopt Data 2.1. Let A, B, C,anddbe integers. If b ∈ F ∗, then X B + k X B + d − k (a) (−1)k b(ϕ(k)) ∧ ϕ(A−k) = (−1)A+k ϕ(k)(b) (ϕ(A−k)), C k A C − k k∈Z + k∈Z +

and X X k B − k k A−k d k B − A + k k A−k (b) (−1) b(ϕ( )) ∧ ϕ( ) = (−1) + ϕ( )(b) (ϕ( )). C − k C − d k k∈Z k∈Z +

Note. If k<0orA

The only non-zero term on the right side occurs when k = 0, and this term is A Yb(ϕ( )), where B Y =(−1)A . A + C If A ≤−1, then ϕ(A) =0.If0≤A, then Lemma 1.3 shows that X = Y . 0 Henceforth, we assume that 1 ≤ d. We take b = b1 ∧ b ,withdegb1 = 1 and deg b0 = d − 1. Let X B k X − k + b ϕ(k) ∧ ϕ(A−k). = ( 1) C k ( ) k∈Z + V ∗ • The element b1 of F acts like a graded derivation on F ; therefore,

(1) X = b1(S1) − b1(ϕ ) ∧ S2, where X B k S − k + b0 ϕ(k) ∧ ϕ(A−k) 1 = ( 1) C k ( ) and k∈Z + X B k S − k + b0 ϕ(k) ∧ ϕ(A−1−k). 2 = ( 1) C k ( ) k∈Z + 8 ANDREW R. KUSTIN

The induction hypothesis gives X B d − − k S − A − k + 1 ϕ(k) b0 ϕ(A−k) 1 =( 1) ( 1) A C − k ( ) ( )and k∈Z + X B d − − k S − A − k+1 + 1 ϕ(k) b0 ϕ(A−1−k) . 2=( 1) ( 1) A − C−k ( ) ( ) k∈Z 1+ S S0 S00 Use Observation 1.2 (b) in order to write 1 = 1 + 1 ,where X B d − k S0 − A − k + ϕ(k) b0 ϕ(A−k) 1 =( 1) ( 1) A C − k ( ) ( )and k∈Z + X B d − − k S00 − A − k+1 + 1 ϕ(k) b0 ϕ(A−k) . 1 =( 1) ( 1) A C − k − ( ) ( ) k∈Z + 1 An index shift yields X B d − k S00 − A − k + ϕ(k−1) b0 ϕ(A−k+1) 1 =( 1) ( 1) A C − k ( ) ( )and k∈Z + X B d − k S − A − k + ϕ(k−1) b0 ϕ(A−k) . 2=( 1) ( 1) A C − k ( ) ( ) k∈Z + We now have X b S 0 S 00 − b ϕ(1) ∧ S = 1 ( 1 + 1 ) 1( ) 2   b ϕ(k) b0 ϕ(A−k) 1 ( ) ( ) X B + d − k   − A − k  b ϕ(k−1) b0 ϕ(A−k+1)  . =( 1) ( 1) A C − k  + 1 ( ) ( )  k∈Z + (1) (k−1) 0 (A−k) − b1(ϕ ) ∧ ϕ (b ) (ϕ ) Apply Lemma 2.4 to complete the proof of (a). The proof of (b) is completely analogous to the proof of (a); we omit the details. In our ﬁrst variation of Theorem 3.1, we arrange the data so that only one term on the right side of the identity survives. Vd Corollary 3.2. Adopt Data 2.1. Let p, q,anddbe integers. If b ∈ F ∗,then X q − d − k − k+p 1+ b ϕ(k) ∧ϕ(p+q−k) ϕ(q) b ϕ(p) ,and (a) ( 1) k−p ( ) = ( ) ( ) k∈Z

 X p − − k − k+d+q 1 b ϕ(k) ∧ ϕ(p+q−k) ϕ(q) b ϕ(p) . (b) ( 1) d − q − k ( ) = ( ) ( ) k∈Z PFAFFIAN IDENTITIES 9

Proof. Apply Theorem 3.1 (a) with A replaced by p + q, B replaced by q − d − 1, and C replaced by −p, in order to see that X q − d − 1+k X q − 1 − k (−1)k b(ϕ(k))∧ϕ(p+q−k) = (−1)q+p+k ϕ(k)(b) (ϕ(p+q−k)). k−p q − k k∈Z k∈Z The proof of (a) is complete because right side of the above sum is zero unless q = k. Assertion (b) follows from Theorem 3.1 (b) with A replaced by p + q, B replaced by p − 1andCreplaced with d − q. An alternate proof of (b) can be obtained from (a) by way of Lemma 1.5 and Corollary 3.4. In our second variation of Theorem 3.1, we remove the “k” from the bottom part of the binomial coeﬃcients. Vd Corollary 3.3. Adopt Data 2.1. Let A, B, L,anddbe integers. If b ∈ F ∗, then X X k B − k k A−k d k B − A + k k A−k (a) (−1) b(ϕ( )) ∧ ϕ( ) = (−1) + ϕ( )(b) (ϕ( ))and L L d − A k∈Z k∈Z + X B + k X d + B − k (b) (−1)k b(ϕ(k)) ∧ ϕ(A−k) = (−1)A+k ϕ(k)(b) (ϕ(A−k)). L L d − A k∈Z k∈Z + Proof. We ﬁrst prove (a). Apply (b) of Theorem 3.1, with C replaced by B − L,in L R order to see that 1 = 1,where X B − k L − k b ϕ(k) ∧ ϕ(A−k) 1= ( 1) B − L − k ( ) and k∈Z X B − A k R − d − k + ϕ(k) b ϕ(A−k) . 1 =( 1) ( 1) B − L − d k ( ) ( ) k∈Z + Apply (a) of Theorem 3.1, with B replaced by L−B −1andCreplaced by −B −1, L R in order to see that 2 = 2,where X L − B − 1+k L = (−1)k b(ϕ(k))∧ϕ(A−k) and 2 −B−1+k k∈Z X L − B − d−k R − A+k 1+ ϕ(k) b ϕ(A−k) . 2 = ( 1) −B− A−k ( ) ( ) k∈Z 1+ Apply Lemma 1.5 (with A replaced by B −k and B replaced by B −L−k) in order to see that X B − k L − LL − k b ϕ(k) ∧ ϕ(A−k). 1 +( 1) 2 = ( 1) L ( ) k∈Z Lemma 1.5 (with A replaced by B − A + k and B replaced by B − L − d + k)shows that X B − A k R − LR − k+d + ϕ(k) b ϕ(A−k) . 1 +( 1) 2 = ( 1) d L − A ( ) ( ) k∈Z + The proof of (a) is complete. Identity (b) follows from (a), by way of Observation 1.2 (e); replace B with L − B − 1. (An alternate proof of Corollary 3.3 can be obtained by mimicking the proof of Theorem 3.1.) In our third variation of Theorem 3.1, we impose hypotheses which allow us to set the identities in Corollary 3.3 to zero, one side at a time. 10 ANDREW R. KUSTIN

Corollary 3.4. Adopt Data 2.1. Let A, B, d,andLbe integers, and let b be an Vd element of F ∗. (i) If A + L +1≤d,then X B k − k + ϕ(k) b ϕ(A−k) , ( 1) L ( ( ))( )=0 and k∈Z X B − k − k ϕ(k) b ϕ(A−k) . ( 1) L ( ( ))( )=0 k∈Z

(ii) If d ≤ A − L − 1,then X B − k − k b ϕ(k) ∧ ϕ(A−k) , ( 1) L ( ) =0 and k∈Z X B k − k + b ϕ(k) ∧ ϕ(A−k) . ( 1) L ( ) =0 k∈Z

Proof. Identity (a) from Corollary 3.3 gives X A B − k X B k − d+k + b ϕ(k) ∧ ϕ(A−k) − k + ϕ(k) b ϕ(A−k) ( 1) L − d A ( ) = ( 1) L ( ( ))( ); k∈Z + k∈Z A+B−k and the hypothesis ensures that L−d+A = 0. The other three assertions are established in the same way. In a similar manner, one can ﬁnd hypotheses which set each side of the identities in Theorem 3.1 to zero. Corollary 3.5, which plays a central role in section 6, is an example of this technique. Corollary 3.5. A B C d V Adopt Data 2.1. Let , , ,and be integers. Assume that b ∈ d F ∗ −1−d≤B≤A C−d− , .If 2 + 1 then X B k − k + b ϕ(k) ∧ ϕ(A−k) . ( 1) C k ( ) =0 k∈Z +

Proof. If S is the left side of the identity, then Theorem 3.1 shows that X B d − k S − A+k + ϕ(k) b ϕ(A−k) . = ( 1) A C − k ( ) ( ) k∈Z +

If 0 ≤ B +d−k, then the hypothesis guarantees that 0 ≤ B +d−k ≤ A+C −k −1, and the binomial coeﬃcient is zero. If B + d − k ≤−1, then the hypothesis also guarantees that deg ϕ(k)(b) ≤−1; thus, ϕ(k)(b) = 0. It follows that S =0. V• Our ﬁfth variation of Theorem 3.1 involves two elements of F ∗. PFAFFIAN IDENTITIES 11

Corollary 3.6. Adopt Data 2.1. Let A, B, d,andLbe integers, and let b and b0 V• Vd be homogeneous elements of F ∗ with b ∈ F ∗.IfA+L+1≤d,then X B k − k + b0 ϕ(k) b ϕ(A−k) . ( 1) L ( ) ( ) ( )=0 k∈Z

Proof. The proof is by induction on the degree of b0.Ifdegb0= 0, then the result 0 00 is contained in Corollary 3.4. Henceforth, we assume that b = b1 ∧ b ,where 00 0 deg b1 =1anddeg b =degb−1.Apply Proposition 2.2 in order to see that b0(ϕ(k)) (b)isequalto 00 (k) 00 (k) deg b00+1 00 (k) (b1 b (ϕ ) )(b)=b1∧ b (ϕ ) (b) +(−1) b (ϕ ) (b1 ∧ b). When this identity is applied to ϕ(A−k), we see that ( b0(ϕ(k)) (b))(ϕ(A−k))isequal to 00 (k) (A−k) deg b00+1 00 (k) (A−k) b1 b (ϕ ) (b) (ϕ ) +(−1) b (ϕ ) (b1 ∧ b) (ϕ ).

deg b00+1 It follows that the left side of the identity is equal to b1(S1)+(−1) S2,where X B k S − k + b00 ϕ(k) b ϕ(A−k) 1 = ( 1) L ( ) ( ) ( )and k∈Z X B k S −k + b00 ϕ(k) b ∧ b ϕ(A−k) . 2= (1) L ( ) ( 1 ) ( ) k∈Z

The induction hypothesis yields S1 = S2 = 0, because 00 0 1+A+L≤deg b

The second main result in this paper is concerned with the data of Corollary 3.6 when the hypothesis A + L +1≤dis not satisﬁed. Our proof of Theorem 3.7 is much like an argument from a class in Diﬀerential Equations: we show that two expressions are equal by verifying that they both are solutions of the same “Initial Value Problem”. Theorem 3.7. Adopt Data 2.1. Let ϕe: F ∗ → F be the homomorphism which is ∗ 0 given by ϕe(b1)=b1(ϕ)for all b1 ∈ F .Letdand A be integers and let b and b be V• homogeneous elements of F ∗.Ifdeg b = d,then ! X X ^d (−1)k b0(ϕ(k)) (b) (ϕ(A−k))= (−1)kϕ(A−d−k) ∧ ϕ(k)(b0) ( ϕe)(b) . k∈Z k∈Z

Proof. For each ﬁxed integer A, we deﬁne bilinear maps ^• ^• ^• ^• ^• ^• ∗ ∗ ∗ ∗ ΦA : F ⊕ F −→ F and ΨA : F ⊕ F −→ F. 12 ANDREW R. KUSTIN

V• If b and b0 are homogeneous elements of F ∗,then X 0 k 0 (k) (A−k) ΦA(b,b)= (−1) b (ϕ ) (b) (ϕ )and k∈Z ! X deg^b 0 k (A−k−deg b) (k) 0 ΨA(b,b)= (−1) ϕ ∧ ϕ (b ) ( ϕe)(b) . k∈Z 0 0 0 We prove that ΦA(b ,b)=ΨA(b,b) by induction on deg b . We must show ! deg^b (A−deg b) (3.8) ΦA(1,b)=ϕ ∧ ϕe (b),

! deg^b (A−deg b) (3.9) ΨA(1,b)=ϕ ∧ ϕe (b),

0 0 deg b0+1 0 (3.10) ΦA(b1 ∧ b ,b)=b1(ΦA(b ,b)) + (−1) ΦA(b ,b1 ∧b), and

0 0 deg b0+1 0 (3.11) ΨA(b1 ∧ b ,b)=b1(ΨA(b ,b)) + (−1) ΨA(b ,b1 ∧b).

∗ for all b1 ∈ F . (A) ∗ It is clear that ΦA(1, 1) = ϕ .Ifb1∈F , then Lemma 2.4 shows that

ΦA(1,b1∧b)=b1(ΦA(1,b))−b1(ΦA(1,b))+b1(ϕ)∧ΦA−1(1,b)=b1(ϕ)∧ΦA−1(1,b).

We conclude that (3.8) holds. Assertion (3.9) is obvious. The proof of Corollary 0 (k) (A−k) 3.6 shows that (b1 ∧ b )(ϕ ) (b) (ϕ )isequalto 0 0 (k) (A−k) deg b +1 0 (k) (A−k) b1 b(ϕ ) (b) (ϕ ) +(−1) b(ϕ ) (b1 ∧b) (ϕ ) and (3.10) holds. Vd Now we prove (3.11). Let d =degb, and let X = ( ϕe)(b) . We expand h i (k) 0 Y = b1(ϕ ) (b ) (X) two diﬀerent ways. On the one hand, Proposition 2.2 gives h i h i h i h i (k) 0 (k) 0 (k) 0 (k) 0 Y = b1 ∧ ϕ (b ) (X) − ϕ (b1 ∧ b ) (X)=b1 ϕ (b ) (X) − ϕ (b1 ∧ b ) (X).

On the other hand, h i h i (k−1) 0 (k−1) 0 Y = b1(ϕ) ∧ ϕ (b ) (X)= (b1(ϕ)) ϕ (b ) (X). PFAFFIAN IDENTITIES 13

Apply Proposition 2.2, once again, in order to see that (k−1) 0 deg b0+1 (k−1) 0 Y = b1(ϕ) ∧ ϕ (b ) (X)+(−1) ϕ (b ) b1(ϕ)∧X . (k) 0 Combine the two expansions of Y in order to obtain that ϕ (b1 ∧ b ) (X)isequal to h i h i 0 (k) 0 (k−1) 0 deg b +1 (k−1) 0 b1 ϕ (b ) (X) − b1(ϕ) ∧ ϕ (b ) (X) − (−1) ϕ (b ) b1(ϕ)∧X .

deg b0+1 Thus, S1 = S2 + S3 +(−1) S4,where X h i k (A−d−k) (k) 0 0 S1 = (−1) ϕ ∧ ϕ (b1 ∧ b ) (X)=ΨA(b1∧b,b), k∈Z X h i k (A−d−k) (k) 0 S2 = (−1) ϕ ∧ b1 ϕ (b ) (X) , k∈Z X X k (A−d−k) (k−1) 0 k (A−d−k) (k) 0 S3 = − (−1) ϕ ∧b1(ϕ)∧ ϕ (b ) (X)= (−1) b1(ϕ )∧ ϕ (b ) (X), k∈Z k∈Z and X k (A−d−k) (k−1) 0 S4 = − (−1) ϕ ∧ ϕ (b ) (b1(ϕ) ∧ X) k∈Z X k (A−d−1−k) (k) 0 = (−1) ϕ ∧ ϕ (b ) (b1(ϕ) ∧ X) . k∈Z

0 Observe that S2 + S3 = b1 (ΨA(b ,b)). Observe also that ! ^d d^+1 b1(ϕ) ∧ X = b1(ϕ) ∧ ( ϕe)(b) =( ϕe)(b1 ∧ b).

0 It follows that S4 =Ψ(b,b1 ∧b). Assertion (3.11) has been established and the proof is complete.

4. A new proof of the Kustin-Ulrich pfaffian identity We give a new proof of [14, 3.2]. The notation is explained in section 2. Proposition 4.1. Let X be an alternating matrix with entries from the commutative noetherian ring R.IfA, B, C, and D are index sets of size a, b, c,andd, respectively, with a + b + d even and a + b ≤ d − 2,then X X (a+c+t)/2 (−1) σ(EF)XACEXBCF =0. t∈Z E∪F =D a+c+teven |E|=t

V• Proof. If M is an integer and β, β0,andβ00 are homogeneous elements of F ∗, then Corollary 3.6 shows that X h i (−1)k β0(ϕ(k)) (β) (ϕ(M−k)) (β00) k∈Z 14 ANDREW R. KUSTIN

is the zero element of the ring R, provided

(4.2) deg β00 =2M−deg β − deg β0 and

(4.3) 1 + M ≤ deg β.

a+b+d 0 We may apply Corollary 3.6 with M = c + , β = εA, β = εC ∧ εD,and 00 2 β = εB ∧ εC . Indeed, (4.2) is satisﬁed because b + c =2M−(a)−(c+d); and (4.3) is equivalent to the hypothesis a + b +2≤d. It follows that X h i k (k) (M−k) (4.4) (−1) εA(ϕ ) (εC ∧ εD) (ϕ ) (εB ∧ εC )=0. k∈Z

We complete the proof by translating (4.4) back into the language of the original statement. For the time being, considerV k to be ﬁxed. Let r =2k−aand let Y (k) r represent the element εA(ϕ )of F. We evaluate h i (M−k) (4.5) (Y (εC ∧ εD))(ϕ ) (εB ∧ εC ) .

Recall, from (2.6), that Xr X X |C2|·|E| (4.6) Y (εC ∧ εD)= (−1) σ(C1C2)σ(EF)Y (εC1 ∧ εE) ∧ εC2 ∧ εF . j=0 C1∪C2=C E∪F =D |C1|=j |E|=r−j

Furthermore, in (4.6), 2(M − k)−|C2 |−|F | = |B|+|C|; consequently, when (4.6) is inserted into (4.4), we have h i ε ∧ ε ϕ(M−k) ε ∧ε ε ∧ε ∧ε ∧ε ϕ(M−k) . ( C2 F )( ) ( B C)=( B C C2 F)( )

C ε ∧ ε If 2 is nonempty, then C C2 = 0. It follows that (4.5) is equal to X (M−k) (4.7) σ(EF)Y (εC ∧ εE) · (εB ∧ εC ∧ εF )(ϕ ). E∪F =D |E|=r−c

Furthermore, in (4.7), we have deg Y =deg(εC ∧εE); therefore, (k) a (k) Y (εC ∧ εE )=(εC ∧εE)Y =(εC ∧εE) εA(ϕ ) =(−1) (εA ∧ εC ∧ εE )(ϕ ).

We now see that (4.4) is X X k (k) (M−k) (−1) σ(EF)(εA ∧εC ∧εE)(ϕ )·(εB ∧εC ∧εF)(ϕ )=0. k∈Z E∪F =D |E|=2k−a−c

Apply (2.8) to see that X X k (−1) σ(EF)XACEXBCF =0. k∈Z E∪F =D |E|=2k−a−c

k t+a+c Replace by 2 in order to complete the proof. PFAFFIAN IDENTITIES 15

5. Srinivasan’s identities

In this section we show that the identities of [15, 16] are special cases of our results. We begin by deducing [15, 4.3] from Theorem 3.1. Proposition 4.4 in [15] is an immediate consequence of Corollary 3.3; results 4.5, 4.6, 4.7, and 4.8 in [15] are immediate consequences of Lemma 1.3. Proposition 5.3 provides a new proof of [15, 4.2]. In Proposition 5.4, we give a new proof of Srinivasan’s decomposition [15, X X ·X X ·X I J page 447] of the pfaﬃan IJ into products of the form IJ1 J2 and I1 ( 2; ). Proposition 5.5 is a reformulation of identity (b) from [15, page 447]. Proposition 5.1. Adopt Data 2.1. Let F have rank 2n,andleti,j,andtbe V n Vt integers with t + j even. If ξ is an element of 2 F ∗ and a is an element of F , then n− t+j i+j ϕ( 2 ) ∧ a (ξ) (ϕ([[ 2 ]])) is equal to j i+t h i X n − t−w i−w 2 (n− 2 ) ([[ 2 ]]) j−w ϕ (ξ) (a) ∧ ϕ . w=0 2 j−w even

Proof. Write i = i0 + ε,wherei0+tis even and ε is either 0 or 1. Notice that i + j i0 + j i + t i0 + t i − w i0 − w (5.2) = , = , and = . 2 2 2 2 2 2

p i0+j q n− t+j d n−t In identity (b) from Corollary 3.2, replace with 2 , with 2 , with 2 , k n w−t and with + 2 , in order to see that

 0 n− t+j i +j ϕ( 2 )(b) (ϕ( 2 )) is equal to

 0 X j−w i +t w−j − n + − 1 w−t i0−w 2 2 2 (n+ 2 ) ( 2 ) (−1) j−w b(ϕ ) ∧ ϕ . w∈Z 2 j−w even

Observation 1.2 (e) yields

 0 0 j−w i +t i +t − n + − 1 j−w n − 2 2 2 2 j−w =(−1) j−w ; 2 2 and therefore,

 0 i +t t+j i0+j X n − w−t i0−w (n− 2 ) ( 2 ) 2 (n+ 2 ) ( 2 ) ϕ (b) (ϕ )= j−w b(ϕ ) ∧ ϕ . w∈Z 2 j−w even 16 ANDREW R. KUSTIN

If j

Proposition 5.3. ϕe F ∗ → F Adopt Data 2.1. Let : be the homomorphismV which ∗ d ∗ is given by ϕe(b1)=b1(ϕ)for all b1 ∈ F ,andletiand d be integers. If b ∈ F , then i− d ! [[X2 ]] 2î−d b(ϕ(i))= (−1)k+1b(ϕ(i−k)) ∧ ϕ(k) + ϕe ϕ(d−i)(b) . k=1 Proof. In the notation of (3.9), observe that ! 2î−d (d−i) (d−i) ϕe ϕ (b) =Ψ2i−d(1,ϕ (b)); therefore, Theorem 3.7 and Proposition 2.3 yield ! 2î−d X ϕe ϕ(d−i)(b) = (−1)k ϕ(k)(ϕ(d−i)(b)) (ϕ(2i−d−k)) k∈Z X k d − i − k + ϕ(k+d−i) b ϕ(2i−d−k) = ( 1) d − i ( ) ( ) k∈Z X ` − i − d+` ϕ(`) b ϕ(i−`) . =( 1) ( 1) d − i ( ) ( ) `∈Z PFAFFIAN IDENTITIES 17

Apply Corollary 3.3, with L = 0 and B = A = i, in order to see that ! 2^i−d X ϕe ϕ(d−i)(b) =(−1)i (−1)`b(ϕ(`)) ∧ ϕ(i−`) `∈Z X = (−1)kb(ϕ(i−k)) ∧ ϕ(k) k∈Z

i− d [[X2 ]] = (−1)kb(ϕ(i−k)) ∧ ϕ(k). k=0 k< ϕ(k) i − d

Recall, from section 2, that X(I2; J) is a minor of X. Proposition 5.4. Let X be an alternating matrix with entries from the commutative noetherian ring R.LetIand J be index sets of size i and j, respectively. If i + j is even, then

X t+j X j(j+1) X − 2 σ J J X ·X − 2 σ I I X · X I J . ( 1) ( 1 2) IJ1 J2 =( 1) ( 1 2) I1 ( 2; ) t∈Z J1∪J2=J I1∪I2=I t+i even |J1|=t |I1|=i−j

b0 ε b ε , A i+j Proof. Evaluate Theorem 3.7 at = I , = J and = 2 , in order to see that L = R,where X i+j k (k) ( −k) L= (−1) εI(ϕ ) (εJ) ϕ 2 and k∈Z ! j X i−j ^ k ( −k) (k) R = (−1) ϕ 2 ∧ ϕ (εI ) ( ϕe)(εJ ) . k∈Z Apply (2.6) to see that X X k k i+j −k L − σ J J ε ϕ( ) ε · ε ϕ( 2 ) . = ( 1) ( 1 2) I ( ) ( J1 ) J2 k∈Z J1∪J2=J |J1|=2k−i

We know, from (2.8), that ε ϕ(k) ε − kX − k+iX . I ( ) ( J1 )=( 1) J1I =( 1) IJ1

It follows that X X k i i+j L − + + 2 σ J J X · X . = ( 1) ( 1 2) IJ1 J2 k∈Z J1∪J2=J |J1|=2k−i 18 ANDREW R. KUSTIN

Let t =2k−iin order to see that L is equal to the left side of the announced identity. Notice that ! j i − j i−j i−j ^ ( −k) (k)

R k i−j R Thus, there is only one non-zero term in . Ifwewrite for 2 ,then is equal to ! ^j k (k) (−1) ϕ (εI ) ( ϕe)(εJ ) ! X ^j − k σ I I ϕ(k) ε ∧ ε ϕe ε . =( 1) ( 1 2) ( I1 ) I2 ( )( J ) I1∪I2=I |I1|=2k

Apply (2.8) to see that the ! X ^j R σ I I X · ε ϕe ε . = ( 1 2) I1 I2 ( )( J ) I1∪I2=I |I1|=2k

We know that ϕe : F ∗ → F is the map ! X X ϕe(εj)=εj(ϕ)=εj xpqep ∧ eq = −xijei. p

It follows, from (2.7), that, if A and B are index sets with |A| = |B| = r,then ! r ^ r(r+1) εA ( ϕe)(εB) =(−1) 2 X(A; B).

Thus, j(j+1) X R − 2 σ I I X · X I J . =( 1) ( 1 2) I1 ( 2; ) I1∪I2=I |I1|=i−j

If X has size N ×N and {1,...,N}is the disjoint union of A, B, C,andD,then one obtains Srinivasan’s version [15, page 447] of the following identity by letting J = B ∪ C and I = C ∪ D. Proposition 5.5. Let X be an alternating matrix with entries from the commutative noetherian ring R, t be an integer, and A, B,andCbe index sets of size a, b, and c, respectively. If t + b is even and a + c is even, then X X a−c X 2 σ A A XA B · XA BC σ C C XABC · XBC . ( 1 2) 1 2 = s+t−c ( 1 2) 1 2 A1∪A2=A s∈Z 2 C1∪C2=C |A1|=t s+a+b even |C1|=s PFAFFIAN IDENTITIES 19

ε ∧ ε b VProof. Apply Corollary 3.2 (a) to the element B C . Replace the element of d ∗ F with εA ∧ εB.Ifpand q are integers which satisfy 2q +2p=a+2b+c,then (q) (p) ϕ (εA ∧ εB) (ϕ ) (εB ∧ εC )and X q − a − b − k −k+p 1+ ε ∧ε ϕ(k) ∧ϕ(q+p−k) ε ∧ε (1) k−p ( A B)( ) ( B C) k∈Z are equal elements of R. The techniques which are employed in the proof of Propo- sition 4.1 yield that X

σ(A1A2)XA1B · XA2BC A1∪A2=A |A1|+b=2q is equal to X q − a − b − k X k+p 1+ (−1) σ(C C )XABC · XBC . k−p 1 2 1 2 k∈Z C1∪C2=C |C1|+a+b=2k

k s+a+b q t+b p a+b+c−t Replace with 2 , with 2 ,and with 2 . Use Observation 1.2 (e) to complete the proof.

6. An application to Algebras with Straightening Law Let X be an N ×N alternating matrix with indeterminate entries. Consider the poset (P, ≤)ofpfaﬃansofX. The order on P is given as follows. If r and s are even integers, a1 < ···

X ≤ X s ≤ r a ≤ b ≤ i ≤ s a1...ar b1...bs provided and i i for 1 .

It is well known (see, for example, [7]) that the polynomial ring Z[X]isanalgebra with straightening law on P over Z. (We use the language of [3].) In particular, for each pair of index sets A and B, there exist integers rCD such that X (6.1) XAXB = rCDXCXD, C,D where the sum is taken over all pairs of index sets C and D with

(6.2) XC ≤ XD and XC ≤ XA.

Most, but not all, of the above “straightening formula” is proved in [8, Lemma 6.2]. Indeed, [8, Lemma 6.2] establishes (6.1), where C and D vary over all pairs of index sets with XC ≤ XD. One must modify the argument of [8] in order to obtain (6.1), where C and D fulfill both of the requirements of (6.2). One version of this modification occurs in [6]. Suppose that |A| and |B| are even integers with |B|≤|A|. Let i(A, B) be the largest index i for which ak ≤ bk for all k with 1 ≤ k ≤ i.If|A|=Nor i(A, B)=|B|,thenXA ≤XB and XAXB = XAXB satisfies both (6.1) and (6.2). Day’s argument proceeds by induction on |A| and 20 ANDREW R. KUSTIN

i A, B i i A, B A A0 ∪ C0 B C0 ∪ B0 ( ). Let = ( ). It is convenient to write = 1 and = 2 , A0 a < ···

B : cλ+1 < ... < cλ+i

A: a1 < ... < ai

Let C represent the index set cλ+1 < ···

X 0 ≤ X X 0 ≤ X It is clear that A C1 A and A C1B1 A. Furthermore,

0 0 0 i(A, B)

Corollary 6.5. Adopt Data 6.4. If M and Q are integers with Q+M even, M +a even, −1 ≤ Q,andQ+M≤b+c−a−2,then X `+r+Q `c r+a−` − + 2 2 X `, r . ( 1) `+r−M ( )=0 r,`∈Z 2 `+r+a even

A a+b+c B Q−a C Proof. Apply Corollary 3.5 with replaced by 2 , replaced by 2 , re- − M+a d a b ε placed by 2 , replaced by ,and replaced by A. Since the hypothesis of Corollary 3.5 is equivalent to the present hypothesis, we conclude that

 Q−a h i X k c+a+b k 2 + (k) ( −k) (−1) εA(ϕ ) ∧ ϕ 2 (εB ∧ εC ) −M−a k k∈Z 2 + PFAFFIAN IDENTITIES 21

is the zero element of R. Calculations similar to those in section 4 yield h i c+a+b X c+a+b (k) ( −k) `c+`+a+ εA(ϕ ) ∧ ϕ 2 (εB ∧ εC )= (−1) 2 X(`, r). r, `∈Z r+`+a=2k

k r+`+a Replace by 2 to see that X r+`+Q a c+a+b `c r+a−` − + 2 − + 2 2 X `, r . ( 1) ( 1) r+`−M ( )=0 r, `∈Z 2 r+`+a even

Proposition 6.6. Adopt Data 6.4. If Q and M are integers with Q + M even, M + a even, b − 1 ≤ Q,andQ+M≤b+c−a−2,then X `−r+Q `c r+a−` − + 2 2 X `, r . ( 1) `−r−M ( )=0 r,`∈Z 2 `+r+a even