Comparison Theorems

Remember the definition of the complement

$${U}^{\bot}:=\{f \in {W}^*={\rm Hom}_{R}{(W,R)}\vert \; f(u)=0 \; \forall u \in U\}$$

of a submodule $U$ in an $R$-module $W$ and the definition of the evaluation map

$${\rm Ev}_{W}:W \rightarrow {{W}^*}^*, \;\mbox{ given by } {\rm Ev}_{W}(x)(y):=y(x)\; x \in W, \; y \in {W}^*$$

In ${{{W}^*}^*}^*$ we have the following commutativity rule

$${\rm Ev}_{{W}^*}({U}^{\bot})={{\rm Ev}_{W}(U)}^{\bot}.$$ (3)

Turning to our special situation we first note that $tr(b[x,a])=tr(x[a,b])$ for all $x, a, b \in {\cal E}$. Since the bilinear form induced by $tr$ is nondegenerate, it follows:

$${\vartheta}_{tr}([x,a])(b)=tr(b[x,a])=0 \mbox{ for all } x \in{\cal E}\Longleftrightarrow [a,b]=0.$$ (4)

The following fundamental lemma of this section is easy to prove now.

Lemma 3.1   We have ${K(A)}^{\bot} ={\rm Ev}_{{\cal E}}(C(A))$ and ${{K(A)}^{\bot}}^{\bot} ={\rm Ev}_{{{\cal E}}^*}({C(A)}^{\bot})$

PROOF: According to the definition ${\rm Ev}_{{\cal E}}(b)\in {K(A)}^{\bot}$ if and only if ${\rm Ev}_{{\cal E}}(b)({\vartheta}_{tr}([x,a]))={\vartheta}_{tr}([x,a])(b)=0$ for all $x \in {\cal E}$ and $a\in A$. Applying (4) this is the case if and only if $[a,b]=0$ for all $a\in A$, thus if and only if $b \in C(A)$. The second equation follows from the first by use of equation (3).$\Box$

Remember that the dual module ${C}^*={\rm Hom}_{R}{(C,R)}$ of a coalgebra $C$ always posseses the structure of an algebra by use of the convolution product

$$\mu \nu :=(\mu \otimes \nu) \circ\Delta \;\;\mbox{ where } \mu ,\nu \in {C}^*.$$

Here, we have identified $\mu \otimes \nu$ with its image under the natural homomorphism ${C}^* \otimes {C}^* \rightarrow {(C \otimes C)}^*$. Note that this construction is functorial. In the special case $C={{\cal E}}^*$ we obtain an algebra structure on ${{{\cal E}}^*}^*$. Furthermore it is easy to show that the evaluation map ${\rm Ev}_{{\cal E}}:{\cal E}\rightarrow {{{\cal E}}^*}^*$ is an isomorphism of algebras.

Lemma 3.2   Let $C$ be a coalgebra together with an epimorphism $\pi: {{\cal E}}^*\rightarrow C$. Set $K:={\rm ker}(\pi)$. Then $\rho:={\rm Ev}_{{\cal E}}^{-1}\circ {\pi}^*:{C}^*\rightarrow {\cal E}$ is a monomorphism of $R$-algebras and ${\rm im}(\rho)={\rm Ev}_{{\cal E}}^{-1}({K}^{\bot})$.

PROOF: By functoriality the dual map ${\pi}^*:{C}^* \rightarrow {{{\cal E}}^*}^*$ is an algebra homomorphism. Since ${\rm Hom}_{R}{(-,R)}$ is exact on the right, ${\pi}^*$ is injective. One easily shows ${\rm im}({\pi}^*)={ K}^{\bot}$. This completes the proof, since ${\rm Ev}_{{\cal E}}$ is an algebra isomorphism as mentioned above.$\Box$

Theorem 3.3 (First Comparison Theorem)   The centralizer algebras $C(A)$ and the dual of the centralizer coalgebra ${M(A)}^*$ are isomorphic to each other. An isomorphism is given by $\rho$.

PROOF: This follows immediately from lemmas 3.1 and 3.2.$\Box$

Now, let us compare the dual of $C(A)$ with $M(A)$. To this aim we consider the dual map ${J}^*:{{\cal E}}^*\rightarrow {C(A)}^*$ of the inclusion $J:C(A)\hookrightarrow {\cal E}$. Because of ${\rm Ev}_{{{\cal E}}^*}(U)\subseteq {{U}^{\bot}}^{\bot}$ for arbitrary submodules $U$ and according to Lemma 3.1 we have

$$K(A)\subseteq{\rm Ev}_{{{\cal E}}^*}^{-1}({{K(A)}^{\bot}}^{\b... ...{\cal E}}^*}({C(A)}^{\bot}))= {C(A)}^{\bot} ={\rm ker}({J}^*).$$

Therefore ${J}^*$ factors to an $R$-module homomorphism

$$\theta :M(A)\rightarrow {C(A)}^*.$$

Lemma 3.4   The kernel of $\theta$ is precisely the torsion submodule of $M(A)$

PROOF: From general results of commutative algebra (cf. [Oe] Anhang A 1.2) it follows that the torsion submodule of $M(A)$ coincides with ${\rm Ev}_{{{\cal E}}^*}^{-1}({{K(A)}^{\bot}}^{\bot})/K(A)$. By the above calculations this is just the kernel ${C(A)}^{\bot}/K(A)$ of $\theta$.$\Box$

This immediately implies

Corollary 3.5 (Criterion of torsion freeness)   The following statements are equivalent:

(a)

$M(A)$ is torsion free

(b)

$K(A)={C(A)}^{\bot}$

(c)

$\theta$ is injective.

Remark 3.6   The map $\theta$ is surjective if and only if the extension group ${\rm Ext}_{R}^1({\cal E}/C(A), R)$ is trivial, in particular if $C(A)$ is a direct summand in ${\cal E}$.

At the beginning of this section we have constructed an algebra structure on the dual module of a coalgebra. Conversely we should obtain a coalgebra structure on the dual of an algebra $A$. Whereas this is not possible in general, it can be done under certain restrictions to the $R$-module structure of the algebra $A$. To be more precise, the natural $R$-homomorphism from ${A}^*\otimes {A}^*$ into ${(A\otimes A)}^*$ must be an isomorphism. This is the case if $A$ is projective and finitely generated. Therefore, under these circumstances the construction is always possible and functorial, that is: duals of algebra maps become coalgebra maps. Thus we obtain

Theorem 3.7 (Second Comparison Theorem)   Suppose the centralizer algebra $C(A)$ of $A$ is a direct summand in ${\cal E}$ as an $R$-module. Then ${C(A)}^*$ can be turned into a coalgebra. The map $\theta:M(A)\rightarrow {C(A)}^*$ is an epimorphism of coalgebras, whose kernel is just the torsion submodule of $M(A)$.