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Let $V,W,L$ be $R$-modules over a commutative ring $R$. A pairing is an $R$-linear map $V\otimes W\to L$. An adjoint of an endomorphism $f:V\to V$ w.r.t a pairing $V\otimes W\overset{g}{\to}L$ is an endomorphism $f^{\dagger ^g}:W\to W$ such that $g(f\otimes 1)=g(1\otimes f^{\dagger^g})$.

When do such things exist and when are they unique? That is, what assumptions are needed on the modules involved, the pairing $g$, and $f$ itself? What if we suppose $g$ is a perfect pairing? (Perhaps this questions is as easily answerable for nice monoidal categories.)

For instance the adjugate of a linear endomorphism can be defined as an adjoint with respect to a canonical pairing. I would simply like to understand which "dualizability" assumptions are needed and where.

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I'm going to denote adjoints by $*$ rather than $\dagger_g$, for notational simplicity.

First answer:

Yes if $g$ is a perfect pairing, then adjoints always exist and are unique.

Let's exploit the tensor-hom adjunction and let $g_V: W\to \newcommand\Hom{\operatorname{Hom}}\Hom(V,L)$ be the obvious map. Now for any $f$, we can consider the map $g_f : W\to \Hom(V,L)$ defined by $g_f(w) = g_V(w)\circ f$. Then in order for an adjoint to exist, we must be able to solve the equation $$ g_Vf^* = g_f.$$

Therefore if $g_V$ is an isomorphism (i.e. if $g$ is a perfect pairing), there is a unique $f^*$ satisfying the equation, $f^*=g_V^{-1}g_f$.

Second answer:

Let's generalize slightly. When can we solve $g_Vf^*=g_f$? Consider the following diagram $$\require{AMScd} \begin{CD} W @>g_f>> \Hom(V,L) \\ @Vf^*VV @| \\ W @>>g_V> \Hom(V,L) \end{CD} $$

Well, one answer to when we can find such a $f^*$ is if $g_V$ is surjective and $W$ is projective. In this case $f^*$ won't be unique. In fact, this is roughly the most general we can get, though.

To generalize this slightly, observe that the image of $g_f$ had better be a subset of the image of $g_V$, otherwise there is no way we can solve it. However if the image of $g_f$ is a subset of the image of $g_V$, then we can replace $\Hom(V,L)$ with $\newcommand\im{\operatorname{im}}\im g_V$, so that $g_V$ is now surjective to its image. Then as long as $W$ is projective, we can lift $g_f$ along $g_V$ to find $f^*$.

My final version of the second answer:

As long as $W$ is projective, and for every $w$, there exists $w'$ so that $g_V(w)\circ f = g_V(w')$, then there exists a (possibly not unique) "adjoint" $f^*$ solving $g_Vf^* = g_f$.

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  • $\begingroup$ So simple, so easy. Thanks! $\endgroup$ – Arrow Dec 24 '18 at 23:36
  • $\begingroup$ @Arrow in the case of perfect pairings, yes, I'm currently editing, since I think I should be able to make this slightly more general. $\endgroup$ – jgon Dec 24 '18 at 23:37
  • $\begingroup$ @Arrow, I generalized slightly. $\endgroup$ – jgon Dec 24 '18 at 23:49
  • $\begingroup$ Very cool. Thank you! $\endgroup$ – Arrow Dec 24 '18 at 23:50

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