1
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Adelaide Dokras Aug 20 '15 at 0:12
  • 1
    $\begingroup$ 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". $\endgroup$ – Jeremy Rickard Aug 20 '15 at 0:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.