# Functor $M\otimes_B -$ is left and right adjoint to $\operatorname{Hom}_A(M,-)$ when there is a symmetrizing form for $A$?

Suppose $A$ is an $R$-algebra over a commutative ring $R$, which is finitely generated and projective as an $R$-module. A symmetrizing form is a map $t\in\operatorname{Hom}_R(A,R)$ such that $t(ab)=t(ba)$ for $a,b\in A$, and the map $$\hat{t}\colon A\to\operatorname{Hom}_R(A,R)$$ defined by $(\hat{t}(a))(b)=t(ab)$ for $a,b\in A$ is an isomorphism of $(A,A)$-bimodules.

I read that if $B$ is a symmetric $R$-algebra and $M$ and $(A,B)$-bimodule which is finitely generated and projective as an $A$-module and a right $B$-module, then $$\operatorname{Hom}_A(M,-)\simeq\operatorname{Hom}_A(M,A)\otimes_A-\simeq\operatorname{Hom}_R(M,R)\otimes_A -$$ this first being "canonical" and the second coming from $f\mapsto tf$, (I'm assuming $f\colon M\to A$ and $t$ the symmetrizing form), and $\operatorname{Hom}_B(\operatorname{Hom}_R(M,R),-)\simeq M\otimes_B -$ so that $M\otimes_B -$ is left and right adjoint to $\operatorname{Hom}_A(M,-)$.

Unfortunately, I'm not familiar with these isomorphisms. I know of the Tensor-Hom adjunction, but I think in this case that says $-\otimes_A M$ is left adjoint to $\operatorname{Hom}_B(M,-)$, so everything seems switched around from what I'm used to.

Can anyone explain in a little more detail why these are actually isomorphisms, and how they give the conclusion?

We're assuming that both $A$ and $B$ are symmetric $R$-algebras, which means that $A\cong\operatorname{Hom}_R(A,R)$ as $A$-bimodules, and similarly for $B$. And we're assuming that $_AM_B$ is an $(A,B)$-bimodule finitely generated and projective over $A$ and over $B$.

For left $A$-modules $_AX$ and $_AY$, there is a natural $R$-module homomorphism $$\operatorname{Hom}_A(X,A)\otimes_AY\to\operatorname{Hom}_A(X,Y)$$ given by $\varphi\otimes y\mapsto[x\mapsto\varphi(x)y]$. In the case that $_AX=_A\!\!A$ it is easy to check that this is an isomorphism (note that both sides are then naturally isomorphic to $Y$). Also, the class of modules $X$ for which it is an isomorphism is closed under taking direct sums and summands, so it is an isomorphism whenever $_AX$ is finitely generated and projective. In particular, it is an isomorphism when $_AX=_A\!\!M$, and in that case, by naturality, it is an isomorphism of left $B$-modules.

So $$\operatorname{Hom}_A(M,A)\otimes_A-\cong\operatorname{Hom}_A(M,-).$$

For the second isomorphism, just notice that, since $A$ is symmetric, there are isomorphisms of $(B,A)$-bimodules $$\operatorname{Hom}_A(M,A) \cong\operatorname{Hom}_A\left(M,\operatorname{Hom}_R(A,R)\right) \cong\operatorname{Hom}_R(A\otimes_AM,R) \cong\operatorname{Hom}_R(M,R).$$

Similarly $\operatorname{Hom}_B(M,B)\cong\operatorname{Hom}_R(M,R)$ as $(B,A)$-bimodules: i.e., all three natural definitions of the "dual" of $_AM_B$ are isomorphic.

Therefore $$\operatorname{Hom}_A(M,-)\cong\operatorname{Hom}_B(M,B)\otimes_A-,$$ and so has $\operatorname{Hom}_B\left(\operatorname{Hom}_B(M,B),-\right)$ as a right adjoint .

Finally, suppose that $S_B$ and $_BT$ are a right and a left $B$-module. Then there is a natural $R$-module homomorphism $$S\otimes_BT\to\operatorname{Hom}_B\left(\operatorname{Hom}_B(S,B),T\right)$$ given by $s\otimes t\mapsto[\theta\mapsto\theta(s)t]$. When $S_B=B_B$ it is easy to check that this is an isomorphism, and so it is an isomorphism whenever $S_B$ is finitely generated and projective, and in particular for $S=M$. So there is an isomorphism $$M\otimes_B-\cong\operatorname{Hom}_B\left(\operatorname{Hom}_B(M,B),-\right),$$ and so the right adjoint of $\operatorname{Hom}_A(M,-)$ found above is isomorphic to $M\otimes_B-$.

Of course, the fact that $M\otimes_B-$ is also a left adjoint of $\operatorname{Hom}_A(M,-)$ is just the standard $\otimes$-$\operatorname{Hom}$ adjunction.

• Thanks Jeremy. If $A$ is a symmetric $R$-algebra, is $A\cong\operatorname{Hom}_R(A,R)$ definition? The only notion of symmetric algebra I know about is the quotient of the tensor algebra for a vector space $V$, $\operatorname{Sym}(V)=T(V)/(v\otimes w-w\otimes v)$, but I'm not sure if that's what they're talking about. – Adelaide Dokras Aug 20 '15 at 0:12
• Yes, it means it's isomorphic to its $R$-dual as a bimodule, which is equivalent to having a symmetrizing form. The quotient of the tensor algebra is a different and unrelated use of "symmetric". – Jeremy Rickard Aug 20 '15 at 0:28