Proof of theorem 6.1

Let us first reduce to showing that ${\bf B}_r$ is a set of generators for the $R$-module $A^{{\rm s}}_{R}(n,r)$. This can be done by the following proposition where notations from the proof of corollary 6.2 are used.

Proposition 7.1   $\vert{\bf B}_r\vert={\rm dim}_{{\mathbb{C}}}{(A_0(n,r))}$.

PROOF: We use [Do2] p. 74 ff. First the reader may check that our definition of $A_0(n)$ and $A_0(n,r)$ is identical to the one given there. According to [Do2] and 2.2c in [Do1] we have

$${\rm dim}_{{\mathbb{C}}}{(A_0(n,r))}= \sum_{\lambda\in \pi_0(n,r)} {\rm dim}_{{\mathbb{C}}}{(Y_0(\lambda))}^2$$ (16)

where $Y_0(\lambda):={\rm Ind}_{B_0}^{{\rm GSp}_{{\mathbb{C}}}(n)}({{\mathbb{C}}}_{\lambda})$ is the irreducible ${\rm GSp}_{{\mathbb{C}}}(n)$ module induced from the linear character ${{\mathbb{C}}}_{\lambda}$ of the Borel subgroup $B_0$ (notations taken from [Do2]). Here $\lambda$ runs through the set $\pi_0(n,r)$ of dominant weights corresponding to the irreducibles occuring in $V^{\otimes r}$. If $T_0$ denotes the maximal torus of ${\rm GSp}_{{\mathbb{C}}}(n)$ we may consider the weights $\lambda$ as the group-like elements in its coordinate ring. More precisely $\lambda\in \pi_0(n,r)$ is of the form

$$\lambda=x_{1 1}^{\mu_1}x_{2 2}^{\mu_2} \ldots x_{m m}^{\mu_m} d^l$$

as can be seen from the argumentation in [Do2]. Here, $\mu:=(\mu_1, \ldots , \mu_m)\in \Lambda^+(m, r-2l)$ is a partition of $r-2l$ in not more than $m$ parts and $0\leq l\leq \frac r2$ an integer. Restricting to the symplectic group ${\rm Sp}_{{\mathbb{C}}}(n)$ we have to set the coefficient of dilation $d$ equal to $1$. Thus the restriction of $\lambda$ to the maximal torus of ${\rm Sp}_{{\mathbb{C}}}(n)$ is just the dominant weight

$$\bar{\lambda}=x_{1 1}^{\mu_1}x_{2 2}^{\mu_2} \ldots x_{m m}^{\mu_m}$$

for the symplectic group itself. Furthermore, it is easy to show that restricting the ${\rm GSp}_{{\mathbb{C}}}(n)$-module structure of $Y_0(\lambda)$ to the symplectic group gives the module $\bar Y(\bar{\lambda}):= {\rm Ind}_{\bar B}^{{\rm Sp}_{{\mathbb{C}}}(n)}({{\mathbb{C}}}_{\bar{\lambda}})$ induced from the linear character ${{\mathbb{C}}}_{\bar{\lambda}}$ of the Borel subgroup $\bar B=B_0\cap {\rm Sp}_{{\mathbb{C}}}(n)$ of ${\rm Sp}_{{\mathbb{C}}}(n)$ (for details see [Oe], 3.3.3). But the dimension of the latter one is known to be the cardinality of $I_{\mu}^{\rm sym}$ (see [Do3] theorem 2.3 b for instance). Thus, we obtain

$${\rm dim}_{{\mathbb{C}}}{(A_0(n,r))}= \sum_{0\leq l\leq \fra... ...m, r-2l)} \vert I_{\mu}^{\rm sym} \vert^2=\vert{\bf B}_r\vert.$$

$\Box$

Observe that by theorem 4.1 the proof of 6.1 can be reduced to the case $R={\mathbb{Z}}$, since the definition of bideterminants over $R$ and ${\mathbb{Z}}$ respectively commutes with the isomorphism ${\mu}_{R}$ when $R$ is considered as a ${\mathbb{Z}}$-algebra. Now, suppose we have shown that ${\bf B}_r$ generates $A^{{\rm s}}_{{\mathbb{Z}}}(n,r)\subseteq A^{{\rm s}}_{{\mathbb{C}}}(n,r)$ as a ${\mathbb{Z}}$-module. Then the image of ${\bf B}_r$ in $A_0(n,r)$ under the epimorphism considered in the proof of 6.2 is a set of generators, too. By the above proposition it must be a basis of $A_0(n,r)$. Consequentely, there can't be any relations among the elements of ${\bf B}_r$, especially none with integer coefficients, giving the desired result.

The proof that ${\bf B}_r$ is indeed a set of generators will follow from a symplectic version of the famous straightening formula. For convenience of the reader we will first state the algorithm leading to the classical straightening formula. To do so, we put an order on the set $\Lambda^+(r)$ of partitions of $r$ writing $\lambda < \mu$ if the dual $\lambda'$ occurs before the dual $\mu'$ in the lexicographical order. In this order the fundamental weight $\omega_r:=(1,1, \ldots , 1)\in \Lambda^+(r, r)$ is the largest element, whereas $\alpha_r:=(r)\in \Lambda^+(1, r)$ is the smallest one. We abbreviate $A:=A_{R}(n,r)$ and define $A(>\lambda)$ resp. $A(\geq \lambda)$ to be the $R$-linear span in $A$ of all bideterminants $T^{\mu}({\bf i}:{\bf j})$ such that $\mu > \lambda$ resp. $\mu \geq \lambda$. For $\lambda=\omega_r$ we set $A(>\omega_r):=0$. Clearly $A=A(\geq \alpha_r)$.

Proposition 7.2 (Classical Straightening Algorithm)   Let $\lambda\in \Lambda^+(r)$ be a partition of $r$ and ${\bf j} \in I(n,r) \backslash I_{\lambda}^{}$. Then to each ${\bf k}\in I(n,r)$ satisfying ${\bf k}={\bf j}w$ ²for some $w \in {\cal S}_{r}$ and ${\bf k}\ll {\bf j}$ there is an element $a_{{\bf j}{\bf k}} \in R$ such that in $A$ the following congruence relation holds for all ${\bf i} \in I(n,r)$:

$$T^{\lambda}({\bf i}:{\bf j}) \equiv \sum_{{\bf k}\ll {\bf j... ...ambda}({\bf i}:{\bf k}) \; \; \mbox{ mod } \; \; A(>\lambda) .$$

Here, the order on $I(n,r)$ is the lexicographical one according to the given order on $\underline{n}$, in our case $\ll$. A proof of the proposition can be found for example in [Mr], 2.5.7.

Let us now state the symplectic analogue. First consider the algebra

$$A^{{\rm sh}}_{R}(n):=A^{{\rm s}}_{R}(n)/\left<d \right>$$

where $\left< d\right>$ is the ideal in $A^{{\rm s}}_{R}(n)$ generated by the coefficient of dilation. It is graded since $d$ is a homogenous element but not a bialgebra because an augmentation map is missing. In fact, it turns out that (in the case where $R=K$ is an algebraically closed field) it is the coordinate ring of the semigroup ${\rm SpH}_{K}(n):={\rm SpM}_{K}(n)\backslash {\rm GSp}_{K}(n)$ of noninvertible elements in the symplectic monoid (see remark ). Let us abbreviate its submodule of homogenous elements of degree $r$ by $A':=A^{{\rm sh}}_{R}(n,r)$ and define $A'(>\lambda)$ and $A'(\geq \lambda)$ in the same manner as above. Further, define a map $f:I(n,r)\rightarrow {\mathbb{N}}_0^m$ by $f({\bf i})=(a_1, \ldots , a_m)$, where

$$a_l:=\vert\{ j\in \underline{r}\vert\; i_j=l \;\mbox{ or } \; i_j=l' \}\vert,$$

and order ${\mathbb{N}}_0^m$ writing $(a_1,\ldots , a_m)<(b_1, \ldots , b_m)$ if and only if $(b_m, b_{m-1}, \ldots , b_1)$ appears before $(a_m, a_{m-1}, \ldots , a_1)$ in the lexicographical order. Next, we obtain an order $\lhd$ on ${\mathbb{N}}_0^m\times I(n,r)$ in a lexicographical way, as well:

$$(a, {\bf i})\lhd (b, {\bf j}) :\Longleftrightarrow a < b \mbox{ or } (a = b \mbox{ and } {\bf i} \ll {\bf j}).$$

Finally, this gives a new order $\lhd$ on $I(n,r)$ via the embedding $I(n,r)\hookrightarrow {\mathbb{N}}_0^m\times I(n,r)$ given by ${\bf i}\mapsto (f({\bf i}), {\bf i})$. Now we are able to state the symplectic straightening algorithm:

Proposition 7.3 (Symplectic Straightening Algorithm)   Let $\lambda\in \Lambda^+(r)$ be a partition of $r$ and ${\bf j} \in I(n,r) \backslash I_{\lambda}^{\rm sym}$. Then to each ${\bf k}\in I(n,r)$ satisfying ${\bf k}\lhd {\bf j}$ there is an element $a_{{\bf j}{\bf k}} \in R$ such that in $A'$ the following congruence relation holds for all ${\bf i} \in I(n,r)$:

$$T^{\lambda}({\bf i}:{\bf j}) \equiv \sum_{{\bf k}\lhd {\bf ... ...mbda}({\bf i}:{\bf k}) \; \; \mbox{ mod } \; \; A'(>\lambda) .$$

Before proving this, let us deduce that ${\bf B}_r$ is a set of generators for $A^{{\rm s}}_{R}(n,r)$. First note that multiplication by the coefficient of dilation $d$ leads to an exact sequence for $r>1$

$$A^{{\rm s}}_{R}(n,r-2)\stackrel{\cdot d}{\rightarrow } A^{{\rm s}}_{R}(n,r) \rightarrow A^{{\rm sh}}_{R}(n,r) \rightarrow 0.$$

Therefore, by induction on $r$ we can reduce to showing that

$$\{ T^{\lambda}({\bf i}:{\bf j})\vert\; \lambda \in \Lambda^+(m, r),\; {\bf i},{\bf j} \in I_{\lambda}^{\rm sym}\}$$

is a set of generators for $A'=A^{{\rm sh}}_{R}(n,r)$. For this claim it is enough to show that

$$\{ T^{\lambda}({\bf i}:{\bf j})\vert\;\; {\bf i},{\bf j} \in I_{\lambda}^{\rm sym}\}$$

is a set of generators of $A'(\geq \lambda)/A'(>\lambda)$ for each partition $\lambda$. This can be deduced from the straightening algorithm 7.3: Since $I(n,r)$ is a finite set, the elemination of multi-indices ${\bf j}$ not being $\lambda$-symplectic standard in an expression $T^{\lambda}({\bf i}:{\bf j})$ must terminate. This gives the straightening formula concerning the right-hand-side argument of $T^{\lambda}({\bf i}:{\bf j})$:

Corollary 7.4 (Symplectic Straightening Formula)   Let $\lambda\in \Lambda^+(r)$ be a partition of $r$ and ${\bf j} \in I(n,r)$. Then, to each ${\bf k}\in I_{\lambda}^{\rm sym}$ there is an element $a_{{\bf j}{\bf k}} \in R$, such that in $A'$ we have for all ${\bf i} \in I(n,r)$:

$$T^{\lambda}({\bf i}:{\bf j}) \equiv \sum_{{\bf k} \in I_{\la... ...T^{\lambda}({\bf i}:{\bf k}) \; \; \mbox{ mod } A'(>\lambda) .$$

Now, there is an algebra automorphism $\theta$ on $A_{R}(n)$ induced by matrix transposition and given by $\theta (x_{i j})=x_{j i}$ on generators. Of course it is an anti-automorphism of coalgebras. It can be readily seen that $\theta({\cal F})={\cal F}$ and $\theta (d)=d$. Therefore, it factors to an automorphism of $A^{{\rm sh}}_{R}(n)$ which will be denoted by the same symbol. From the definition of bideterminants we see $\theta(T^{\lambda}({\bf i}:{\bf j}))=T^{\lambda}({\bf j}:{\bf i})$.

Applying $\theta$ to the congruence relation in 7.4 we see that a non $\lambda$-symplectic standard entry ${\bf i}$ on the left-hand-side entry of a bideterminant can be eleminated, too, not affecting the right-hand-side entry. Thus, it must be possible to write $T^{\lambda}({\bf i}:{\bf j})$ as a sum of bideterminants $T^{\lambda}({\bf k}:{\bf l})$ modulo $A'(>\lambda)$ where ${\bf k}, {\bf l} \in I_{\lambda}^{\rm sym}$. Therefore, the proof of theorem 6.1 is finished as soon as proposition 7.3 is established.

Remark 7.5   A symplectic straightening formula similar to 7.4 can be obtained as a special case of a theorem by de Concini ([Co], theorem 2.4) where $m=2r$ (in the notation taken from there). The algebra denoted $A$ in that paper becomes the coordinate ring of the semigroup denoted ${\rm SpH}_{K}(n)$ above. But, note that this result is not strong enough for our purpose since we don't know wether $A=A^{{\rm sh}}_{K}(n)$ or not. On the other hand, the latter identity follows from theorem 6.1 in a similar way to corollary 6.2 using theorem 3.6 of [Co].

Also, the symplectic straightening formula is related to the treatment of symplectic Schur-modules in [Do3] and [Ia], section 6, as can be seen from the proof of lemma 8.1 below. Concerning the latter paper it should be noted that the algebra $A^{sp}_{K}(\overline n)$ defined there as the coordinate ring of the symplectic group itself is only filtered by $\sum_{t=0}^r A^{sp}_{K}(\overline n,t)$ but not graded by $A^{sp}_{K}(\overline n,r)$. In fact, it can be deduced from the above remark that $A^{{\rm sh}}_{K}(n)$ is the corresponding graded algebra, that is $A^{sp}_{K}(\overline n,r)=A^{{\rm sh}}_{K}(n,r)$ as $K$-vector spaces.