Existence and uniqueness of adjoints with respect to pairings












1












$begingroup$


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



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.






linear-algebra commutative-algebra category-theory adjoint-operators monoidal-categories







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


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



    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.






    linear-algebra commutative-algebra category-theory adjoint-operators monoidal-categories







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


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



      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.






      linear-algebra commutative-algebra category-theory adjoint-operators monoidal-categories







      share|cite|improve this question











      $endgroup$




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



      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.






      linear-algebra commutative-algebra category-theory adjoint-operators monoidal-categories




      linear-algebra commutative-algebra category-theory adjoint-operators monoidal-categories






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 24 '18 at 23:33







      Arrow

















      asked Dec 24 '18 at 23:04









      ArrowArrow

      5,18431446




      5,18431446






















          1 Answer
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          active

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          $begingroup$

          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 $g_V: Wto newcommandHom{operatorname{Hom}}Hom(V,L)$ be the obvious map. Now for any $f$, we can consider the map $g_f : Wto Hom(V,L)$ 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 $newcommandim{operatorname{im}}im g_V$, 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 $g_Vf^* = g_f$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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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          $begingroup$

          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 $g_V: Wto newcommandHom{operatorname{Hom}}Hom(V,L)$ be the obvious map. Now for any $f$, we can consider the map $g_f : Wto Hom(V,L)$ 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 $newcommandim{operatorname{im}}im g_V$, 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 $g_Vf^* = g_f$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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
















          1












          $begingroup$

          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 $g_V: Wto newcommandHom{operatorname{Hom}}Hom(V,L)$ be the obvious map. Now for any $f$, we can consider the map $g_f : Wto Hom(V,L)$ 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 $newcommandim{operatorname{im}}im g_V$, 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 $g_Vf^* = g_f$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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














          1












          1








          1





          $begingroup$

          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 $g_V: Wto newcommandHom{operatorname{Hom}}Hom(V,L)$ be the obvious map. Now for any $f$, we can consider the map $g_f : Wto Hom(V,L)$ 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 $newcommandim{operatorname{im}}im g_V$, 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 $g_Vf^* = g_f$.






          share|cite|improve this answer











          $endgroup$



          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 $g_V: Wto newcommandHom{operatorname{Hom}}Hom(V,L)$ be the obvious map. Now for any $f$, we can consider the map $g_f : Wto Hom(V,L)$ 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 $newcommandim{operatorname{im}}im g_V$, 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 $g_Vf^* = g_f$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 24 '18 at 23:49

























          answered Dec 24 '18 at 23:35









          jgonjgon

          15.1k32042




          15.1k32042












          • $begingroup$
            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


















          • $begingroup$
            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
















          $begingroup$
          So simple, so easy. Thanks!
          $endgroup$
          – Arrow
          Dec 24 '18 at 23:36




          $begingroup$
          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 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






          $begingroup$
          @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




          $begingroup$
          Very cool. Thank you!
          $endgroup$
          – Arrow
          Dec 24 '18 at 23:50


















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