Left adjoint to the tensor functor
$begingroup$
Let $V$ be a vector space over a field $k$.
Define $V otimes - :$ Vect $to$ Vect the tensor endofunctor on the category of $k$ vector spaces.
Assume that $V otimes - $ preserves limits. It can be shown using an adjoint functor theorem that it has a left adjoint $F dashv Votimes-$.
It is known that every vector space $W$ can be written as a colimit of $k$, namely that $W cong oplus_{i in I}k$ where a basis of $W$ is indexed by $I$.
Show that $F$ is given by tensoring with some vector space.
Now, the way I understand this question is that we're trying to prove that for any vector space $W$, $F(W) = W otimes V_W$, for some vector space.
Maybe they mean something stronger, that $F(W) = W otimes Z$ for all $W$.
In order to prove something like that I'd need to find a canonical map $phi: W times V_W to F(W)$ with the universal property of tensor products. However I'm not seeing where such a map may come from. The universality can maybe follow after that by using the fact the $F$ must preserve colimits, and so $F(W)$ is a colimit as well.
I also have this information:
Hom$(F(W), Z) cong$ Hom$(W, Votimes Z)$ for any two vector spaces $Z,W$ - though this doesn't seem to supply such a map.
Any guidance?
category-theory limits-colimits adjoint-functors
asked Dec 29 '18 at 12:36
MariahMariah
1,8231718
$endgroup$
|
show 1 more comment
$begingroup$
Let $V$ be a vector space over a field $k$.
Define $V otimes - :$ Vect $to$ Vect the tensor endofunctor on the category of $k$ vector spaces.
Assume that $V otimes - $ preserves limits. It can be shown using an adjoint functor theorem that it has a left adjoint $F dashv Votimes-$.
It is known that every vector space $W$ can be written as a colimit of $k$, namely that $W cong oplus_{i in I}k$ where a basis of $W$ is indexed by $I$.
Show that $F$ is given by tensoring with some vector space.
Now, the way I understand this question is that we're trying to prove that for any vector space $W$, $F(W) = W otimes V_W$, for some vector space.
Maybe they mean something stronger, that $F(W) = W otimes Z$ for all $W$.
In order to prove something like that I'd need to find a canonical map $phi: W times V_W to F(W)$ with the universal property of tensor products. However I'm not seeing where such a map may come from. The universality can maybe follow after that by using the fact the $F$ must preserve colimits, and so $F(W)$ is a colimit as well.
I also have this information:
Hom$(F(W), Z) cong$ Hom$(W, Votimes Z)$ for any two vector spaces $Z,W$ - though this doesn't seem to supply such a map.
Any guidance?
category-theory limits-colimits adjoint-functors
asked Dec 29 '18 at 12:36
MariahMariah
1,8231718
$endgroup$
1
$begingroup$
Is this even true? If tensoring is a right adjoint then it preserves all limits/products, which usually isn’t true.
$endgroup$
– Randall
Dec 29 '18 at 14:30
$begingroup$
A field is flat, @Randall, so tensoring is left exact over a field, or am I missing something
$endgroup$
– Card_Trick
Dec 29 '18 at 15:10
1
$begingroup$
I’m sure things are fine for finite limits. Tensoring doesn’t usually respect, eg, infinite direct products. @Card_Trick
$endgroup$
– Randall
Dec 29 '18 at 15:29
$begingroup$
@Randall the right adjoint here preserves limits iff $V$ is finite dimensional. According to a later clause in this question.
$endgroup$
– Mariah
Dec 29 '18 at 15:48
$begingroup$
I didn’t see that $V$ is fd.
$endgroup$
– Randall
Dec 29 '18 at 15:58
|
show 1 more comment
$begingroup$
Let $V$ be a vector space over a field $k$.
Define $V otimes - :$ Vect $to$ Vect the tensor endofunctor on the category of $k$ vector spaces.
Assume that $V otimes - $ preserves limits. It can be shown using an adjoint functor theorem that it has a left adjoint $F dashv Votimes-$.
It is known that every vector space $W$ can be written as a colimit of $k$, namely that $W cong oplus_{i in I}k$ where a basis of $W$ is indexed by $I$.
Show that $F$ is given by tensoring with some vector space.
Now, the way I understand this question is that we're trying to prove that for any vector space $W$, $F(W) = W otimes V_W$, for some vector space.
Maybe they mean something stronger, that $F(W) = W otimes Z$ for all $W$.
In order to prove something like that I'd need to find a canonical map $phi: W times V_W to F(W)$ with the universal property of tensor products. However I'm not seeing where such a map may come from. The universality can maybe follow after that by using the fact the $F$ must preserve colimits, and so $F(W)$ is a colimit as well.
I also have this information:
Hom$(F(W), Z) cong$ Hom$(W, Votimes Z)$ for any two vector spaces $Z,W$ - though this doesn't seem to supply such a map.
Any guidance?
category-theory limits-colimits adjoint-functors
asked Dec 29 '18 at 12:36
MariahMariah
1,8231718
$endgroup$
Let $V$ be a vector space over a field $k$.
Define $V otimes - :$ Vect $to$ Vect the tensor endofunctor on the category of $k$ vector spaces.
Assume that $V otimes - $ preserves limits. It can be shown using an adjoint functor theorem that it has a left adjoint $F dashv Votimes-$.
It is known that every vector space $W$ can be written as a colimit of $k$, namely that $W cong oplus_{i in I}k$ where a basis of $W$ is indexed by $I$.
Show that $F$ is given by tensoring with some vector space.
Now, the way I understand this question is that we're trying to prove that for any vector space $W$, $F(W) = W otimes V_W$, for some vector space.
Maybe they mean something stronger, that $F(W) = W otimes Z$ for all $W$.
In order to prove something like that I'd need to find a canonical map $phi: W times V_W to F(W)$ with the universal property of tensor products. However I'm not seeing where such a map may come from. The universality can maybe follow after that by using the fact the $F$ must preserve colimits, and so $F(W)$ is a colimit as well.
I also have this information:
Hom$(F(W), Z) cong$ Hom$(W, Votimes Z)$ for any two vector spaces $Z,W$ - though this doesn't seem to supply such a map.
Any guidance?
category-theory limits-colimits adjoint-functors
category-theory limits-colimits adjoint-functors
asked Dec 29 '18 at 12:36
MariahMariah
1,8231718
asked Dec 29 '18 at 12:36
MariahMariah
1,8231718
edited Dec 29 '18 at 22:32
Mariah
asked Dec 29 '18 at 12:36
MariahMariah
1,8231718
asked Dec 29 '18 at 12:36
MariahMariah
1,8231718
asked Dec 29 '18 at 12:36
MariahMariah
1,8231718
1,8231718
1
$begingroup$
Is this even true? If tensoring is a right adjoint then it preserves all limits/products, which usually isn’t true.
$endgroup$
– Randall
Dec 29 '18 at 14:30
$begingroup$
A field is flat, @Randall, so tensoring is left exact over a field, or am I missing something
$endgroup$
– Card_Trick
Dec 29 '18 at 15:10
1
$begingroup$
I’m sure things are fine for finite limits. Tensoring doesn’t usually respect, eg, infinite direct products. @Card_Trick
$endgroup$
– Randall
Dec 29 '18 at 15:29
$begingroup$
@Randall the right adjoint here preserves limits iff $V$ is finite dimensional. According to a later clause in this question.
$endgroup$
– Mariah
Dec 29 '18 at 15:48
$begingroup$
I didn’t see that $V$ is fd.
$endgroup$
– Randall
Dec 29 '18 at 15:58
|
show 1 more comment
1
$begingroup$
Is this even true? If tensoring is a right adjoint then it preserves all limits/products, which usually isn’t true.
$endgroup$
– Randall
Dec 29 '18 at 14:30
$begingroup$
A field is flat, @Randall, so tensoring is left exact over a field, or am I missing something
$endgroup$
– Card_Trick
Dec 29 '18 at 15:10
1
$begingroup$
I’m sure things are fine for finite limits. Tensoring doesn’t usually respect, eg, infinite direct products. @Card_Trick
$endgroup$
– Randall
Dec 29 '18 at 15:29
$begingroup$
@Randall the right adjoint here preserves limits iff $V$ is finite dimensional. According to a later clause in this question.
$endgroup$
– Mariah
Dec 29 '18 at 15:48
$begingroup$
I didn’t see that $V$ is fd.
$endgroup$
– Randall
Dec 29 '18 at 15:58
1
1
$begingroup$
Is this even true? If tensoring is a right adjoint then it preserves all limits/products, which usually isn’t true.
$endgroup$
– Randall
Dec 29 '18 at 14:30
$begingroup$
Is this even true? If tensoring is a right adjoint then it preserves all limits/products, which usually isn’t true.
$endgroup$
– Randall
Dec 29 '18 at 14:30
$begingroup$
A field is flat, @Randall, so tensoring is left exact over a field, or am I missing something
$endgroup$
– Card_Trick
Dec 29 '18 at 15:10
$begingroup$
A field is flat, @Randall, so tensoring is left exact over a field, or am I missing something
$endgroup$
– Card_Trick
Dec 29 '18 at 15:10
1
1
$begingroup$
I’m sure things are fine for finite limits. Tensoring doesn’t usually respect, eg, infinite direct products. @Card_Trick
$endgroup$
– Randall
Dec 29 '18 at 15:29
$begingroup$
I’m sure things are fine for finite limits. Tensoring doesn’t usually respect, eg, infinite direct products. @Card_Trick
$endgroup$
– Randall
Dec 29 '18 at 15:29
$begingroup$
@Randall the right adjoint here preserves limits iff $V$ is finite dimensional. According to a later clause in this question.
$endgroup$
– Mariah
Dec 29 '18 at 15:48
$begingroup$
@Randall the right adjoint here preserves limits iff $V$ is finite dimensional. According to a later clause in this question.
$endgroup$
– Mariah
Dec 29 '18 at 15:48
$begingroup$
I didn’t see that $V$ is fd.
$endgroup$
– Randall
Dec 29 '18 at 15:58
$begingroup$
I didn’t see that $V$ is fd.
$endgroup$
– Randall
Dec 29 '18 at 15:58
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
To get started, the claim is indeed that $F$ must be given by tensoring with a fixed vector space $V^*$. Now if $W$ is a vector space of dimension $kappa$, then $F(W)$ is the sum of $kappa$ copies of $F(k)$. This is the same as $Wotimes F(k)$! Indeed, every cocontinuous endofunctor of vector spaces $F$ is naturally isomorphic, by the composition $F(W)cong oplus_kappa F(k)cong F(k)otimes W$, to tensoring with $F(k)$, and I hope this is enough to get you going on the full proof. So the involvement of $V$ is a red herring here.
answered Dec 30 '18 at 21:31
Kevin CarlsonKevin Carlson
33.5k23372
$endgroup$
$begingroup$
Hey Kevin, thank you! Can you provide a reference to why $oplus_{kappa}F(k) cong W otimes F(k)$? I guess I'm lacking some knowledge about tensor products..
$endgroup$
– Mariah
Dec 30 '18 at 22:48
$begingroup$
@Mariah Well, the simplest reason is that tensoring preserves colimits, as it has a right adjoint, namely hom. This is a very important fact, so I'd suggest proving it for yourself if it's unfamiliar, although it's also possible to show that tensoring preserves coproducts by an explicit manipulation with bases.
$endgroup$
– Kevin Carlson
Dec 30 '18 at 22:54
add a comment |
$begingroup$
$V otimes (-)$ has a left adjoint if and only if $V$ is finite-dimensional; for one direction see this answer. In the finite-dimensional case we have a natural isomorphism
$$V otimes (-) cong text{Hom}(V^{ast}, -)$$
and then by the usual tensor-hom adjunction, the left adjoint (naturally in $V$) is $V^{ast} otimes (-)$.
In the general case of modules the condition is that if $M$ is an $(R, S)$-bimodule then $M otimes_S (-)$ has a left adjoint if and only if $M$ is finitely presented projective as a right $S$-module, in which case its left adjoint is given by tensoring with $text{Hom}_S(M, S)$ over $R$; see this blog post.
answered Dec 29 '18 at 22:41
Qiaochu YuanQiaochu Yuan
281k32593938
$endgroup$
$begingroup$
More generally, tensoring with a module preserves products if and only if the module is finitely presented, so this is a necessary condition for the tensor functor to have a left adjoint.
$endgroup$
– egreg
Dec 29 '18 at 22:45
$begingroup$
Thanks for the answer :) We're not assuming $V$ is finite dimensional here though. They intend us to show that $F$ is given by tensoring using the assumption that $V otimes - $ preserves limits, hence has a left adjoint, and by using the fact that every vector space is a colimit of the field.Do you know how to approach this with these assumptions?
$endgroup$
– Mariah
Dec 29 '18 at 22:51
$begingroup$
@Mariah: I'm not assuming it. It's a consequence of the assumption that $V otimes (-)$ preserves limits. If you really don't want to use this fact, you can instead prove that every colimit-preserving endofunctor of $text{Vect}$ (in particular every left adjoint) is given by tensoring with some vector space.
$endgroup$
– Qiaochu Yuan
Dec 29 '18 at 22:51
1
$begingroup$
@QiaochuYuan This last part of your comment is what I hoped to receive an answer to: why is $F$ given by tensoring. I'm not given, and cannot yet assume, that $V$ is finite dimensional. I need to use the fact the $V otimes -$ preserves limits, hence has a left adjoint, and that we can write any vector space $W$ as a colimit of the field. Can you assist in helping me find out why such a functor is given by tensoring?
$endgroup$
– Mariah
Dec 29 '18 at 23:21
$begingroup$
Every vector space is a coproduct of copies of the underlying field $k$ (given by picking a basis) and a colimit-preserving functor preserves this coproduct.
$endgroup$
– Qiaochu Yuan
Dec 30 '18 at 1:23
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
To get started, the claim is indeed that $F$ must be given by tensoring with a fixed vector space $V^*$. Now if $W$ is a vector space of dimension $kappa$, then $F(W)$ is the sum of $kappa$ copies of $F(k)$. This is the same as $Wotimes F(k)$! Indeed, every cocontinuous endofunctor of vector spaces $F$ is naturally isomorphic, by the composition $F(W)cong oplus_kappa F(k)cong F(k)otimes W$, to tensoring with $F(k)$, and I hope this is enough to get you going on the full proof. So the involvement of $V$ is a red herring here.
answered Dec 30 '18 at 21:31
Kevin CarlsonKevin Carlson
33.5k23372
$endgroup$
$begingroup$
Hey Kevin, thank you! Can you provide a reference to why $oplus_{kappa}F(k) cong W otimes F(k)$? I guess I'm lacking some knowledge about tensor products..
$endgroup$
– Mariah
Dec 30 '18 at 22:48
$begingroup$
@Mariah Well, the simplest reason is that tensoring preserves colimits, as it has a right adjoint, namely hom. This is a very important fact, so I'd suggest proving it for yourself if it's unfamiliar, although it's also possible to show that tensoring preserves coproducts by an explicit manipulation with bases.
$endgroup$
– Kevin Carlson
Dec 30 '18 at 22:54
add a comment |
$begingroup$
To get started, the claim is indeed that $F$ must be given by tensoring with a fixed vector space $V^*$. Now if $W$ is a vector space of dimension $kappa$, then $F(W)$ is the sum of $kappa$ copies of $F(k)$. This is the same as $Wotimes F(k)$! Indeed, every cocontinuous endofunctor of vector spaces $F$ is naturally isomorphic, by the composition $F(W)cong oplus_kappa F(k)cong F(k)otimes W$, to tensoring with $F(k)$, and I hope this is enough to get you going on the full proof. So the involvement of $V$ is a red herring here.
answered Dec 30 '18 at 21:31
Kevin CarlsonKevin Carlson
33.5k23372
$endgroup$
$begingroup$
Hey Kevin, thank you! Can you provide a reference to why $oplus_{kappa}F(k) cong W otimes F(k)$? I guess I'm lacking some knowledge about tensor products..
$endgroup$
– Mariah
Dec 30 '18 at 22:48
$begingroup$
@Mariah Well, the simplest reason is that tensoring preserves colimits, as it has a right adjoint, namely hom. This is a very important fact, so I'd suggest proving it for yourself if it's unfamiliar, although it's also possible to show that tensoring preserves coproducts by an explicit manipulation with bases.
$endgroup$
– Kevin Carlson
Dec 30 '18 at 22:54
add a comment |
$begingroup$
To get started, the claim is indeed that $F$ must be given by tensoring with a fixed vector space $V^*$. Now if $W$ is a vector space of dimension $kappa$, then $F(W)$ is the sum of $kappa$ copies of $F(k)$. This is the same as $Wotimes F(k)$! Indeed, every cocontinuous endofunctor of vector spaces $F$ is naturally isomorphic, by the composition $F(W)cong oplus_kappa F(k)cong F(k)otimes W$, to tensoring with $F(k)$, and I hope this is enough to get you going on the full proof. So the involvement of $V$ is a red herring here.
answered Dec 30 '18 at 21:31
Kevin CarlsonKevin Carlson
33.5k23372
$endgroup$
To get started, the claim is indeed that $F$ must be given by tensoring with a fixed vector space $V^*$. Now if $W$ is a vector space of dimension $kappa$, then $F(W)$ is the sum of $kappa$ copies of $F(k)$. This is the same as $Wotimes F(k)$! Indeed, every cocontinuous endofunctor of vector spaces $F$ is naturally isomorphic, by the composition $F(W)cong oplus_kappa F(k)cong F(k)otimes W$, to tensoring with $F(k)$, and I hope this is enough to get you going on the full proof. So the involvement of $V$ is a red herring here.
answered Dec 30 '18 at 21:31
Kevin CarlsonKevin Carlson
33.5k23372
answered Dec 30 '18 at 21:31
Kevin CarlsonKevin Carlson
33.5k23372
answered Dec 30 '18 at 21:31
Kevin CarlsonKevin Carlson
33.5k23372
answered Dec 30 '18 at 21:31
Kevin CarlsonKevin Carlson
33.5k23372
33.5k23372
$begingroup$
Hey Kevin, thank you! Can you provide a reference to why $oplus_{kappa}F(k) cong W otimes F(k)$? I guess I'm lacking some knowledge about tensor products..
$endgroup$
– Mariah
Dec 30 '18 at 22:48
$begingroup$
@Mariah Well, the simplest reason is that tensoring preserves colimits, as it has a right adjoint, namely hom. This is a very important fact, so I'd suggest proving it for yourself if it's unfamiliar, although it's also possible to show that tensoring preserves coproducts by an explicit manipulation with bases.
$endgroup$
– Kevin Carlson
Dec 30 '18 at 22:54
add a comment |
$begingroup$
Hey Kevin, thank you! Can you provide a reference to why $oplus_{kappa}F(k) cong W otimes F(k)$? I guess I'm lacking some knowledge about tensor products..
$endgroup$
– Mariah
Dec 30 '18 at 22:48
$begingroup$
@Mariah Well, the simplest reason is that tensoring preserves colimits, as it has a right adjoint, namely hom. This is a very important fact, so I'd suggest proving it for yourself if it's unfamiliar, although it's also possible to show that tensoring preserves coproducts by an explicit manipulation with bases.
$endgroup$
– Kevin Carlson
Dec 30 '18 at 22:54
$begingroup$
Hey Kevin, thank you! Can you provide a reference to why $oplus_{kappa}F(k) cong W otimes F(k)$? I guess I'm lacking some knowledge about tensor products..
$endgroup$
– Mariah
Dec 30 '18 at 22:48
$begingroup$
Hey Kevin, thank you! Can you provide a reference to why $oplus_{kappa}F(k) cong W otimes F(k)$? I guess I'm lacking some knowledge about tensor products..
$endgroup$
– Mariah
Dec 30 '18 at 22:48
$begingroup$
@Mariah Well, the simplest reason is that tensoring preserves colimits, as it has a right adjoint, namely hom. This is a very important fact, so I'd suggest proving it for yourself if it's unfamiliar, although it's also possible to show that tensoring preserves coproducts by an explicit manipulation with bases.
$endgroup$
– Kevin Carlson
Dec 30 '18 at 22:54
$begingroup$
@Mariah Well, the simplest reason is that tensoring preserves colimits, as it has a right adjoint, namely hom. This is a very important fact, so I'd suggest proving it for yourself if it's unfamiliar, although it's also possible to show that tensoring preserves coproducts by an explicit manipulation with bases.
$endgroup$
– Kevin Carlson
Dec 30 '18 at 22:54
add a comment |
$begingroup$
$V otimes (-)$ has a left adjoint if and only if $V$ is finite-dimensional; for one direction see this answer. In the finite-dimensional case we have a natural isomorphism
$$V otimes (-) cong text{Hom}(V^{ast}, -)$$
and then by the usual tensor-hom adjunction, the left adjoint (naturally in $V$) is $V^{ast} otimes (-)$.
In the general case of modules the condition is that if $M$ is an $(R, S)$-bimodule then $M otimes_S (-)$ has a left adjoint if and only if $M$ is finitely presented projective as a right $S$-module, in which case its left adjoint is given by tensoring with $text{Hom}_S(M, S)$ over $R$; see this blog post.
answered Dec 29 '18 at 22:41
Qiaochu YuanQiaochu Yuan
281k32593938
$endgroup$
$begingroup$
More generally, tensoring with a module preserves products if and only if the module is finitely presented, so this is a necessary condition for the tensor functor to have a left adjoint.
$endgroup$
– egreg
Dec 29 '18 at 22:45
$begingroup$
Thanks for the answer :) We're not assuming $V$ is finite dimensional here though. They intend us to show that $F$ is given by tensoring using the assumption that $V otimes - $ preserves limits, hence has a left adjoint, and by using the fact that every vector space is a colimit of the field.Do you know how to approach this with these assumptions?
$endgroup$
– Mariah
Dec 29 '18 at 22:51
$begingroup$
@Mariah: I'm not assuming it. It's a consequence of the assumption that $V otimes (-)$ preserves limits. If you really don't want to use this fact, you can instead prove that every colimit-preserving endofunctor of $text{Vect}$ (in particular every left adjoint) is given by tensoring with some vector space.
$endgroup$
– Qiaochu Yuan
Dec 29 '18 at 22:51
1
$begingroup$
@QiaochuYuan This last part of your comment is what I hoped to receive an answer to: why is $F$ given by tensoring. I'm not given, and cannot yet assume, that $V$ is finite dimensional. I need to use the fact the $V otimes -$ preserves limits, hence has a left adjoint, and that we can write any vector space $W$ as a colimit of the field. Can you assist in helping me find out why such a functor is given by tensoring?
$endgroup$
– Mariah
Dec 29 '18 at 23:21
$begingroup$
Every vector space is a coproduct of copies of the underlying field $k$ (given by picking a basis) and a colimit-preserving functor preserves this coproduct.
$endgroup$
– Qiaochu Yuan
Dec 30 '18 at 1:23
add a comment |
$begingroup$
$V otimes (-)$ has a left adjoint if and only if $V$ is finite-dimensional; for one direction see this answer. In the finite-dimensional case we have a natural isomorphism
$$V otimes (-) cong text{Hom}(V^{ast}, -)$$
and then by the usual tensor-hom adjunction, the left adjoint (naturally in $V$) is $V^{ast} otimes (-)$.
In the general case of modules the condition is that if $M$ is an $(R, S)$-bimodule then $M otimes_S (-)$ has a left adjoint if and only if $M$ is finitely presented projective as a right $S$-module, in which case its left adjoint is given by tensoring with $text{Hom}_S(M, S)$ over $R$; see this blog post.
answered Dec 29 '18 at 22:41
Qiaochu YuanQiaochu Yuan
281k32593938
$endgroup$
$begingroup$
More generally, tensoring with a module preserves products if and only if the module is finitely presented, so this is a necessary condition for the tensor functor to have a left adjoint.
$endgroup$
– egreg
Dec 29 '18 at 22:45
$begingroup$
Thanks for the answer :) We're not assuming $V$ is finite dimensional here though. They intend us to show that $F$ is given by tensoring using the assumption that $V otimes - $ preserves limits, hence has a left adjoint, and by using the fact that every vector space is a colimit of the field.Do you know how to approach this with these assumptions?
$endgroup$
– Mariah
Dec 29 '18 at 22:51
$begingroup$
@Mariah: I'm not assuming it. It's a consequence of the assumption that $V otimes (-)$ preserves limits. If you really don't want to use this fact, you can instead prove that every colimit-preserving endofunctor of $text{Vect}$ (in particular every left adjoint) is given by tensoring with some vector space.
$endgroup$
– Qiaochu Yuan
Dec 29 '18 at 22:51
1
$begingroup$
@QiaochuYuan This last part of your comment is what I hoped to receive an answer to: why is $F$ given by tensoring. I'm not given, and cannot yet assume, that $V$ is finite dimensional. I need to use the fact the $V otimes -$ preserves limits, hence has a left adjoint, and that we can write any vector space $W$ as a colimit of the field. Can you assist in helping me find out why such a functor is given by tensoring?
$endgroup$
– Mariah
Dec 29 '18 at 23:21
$begingroup$
Every vector space is a coproduct of copies of the underlying field $k$ (given by picking a basis) and a colimit-preserving functor preserves this coproduct.
$endgroup$
– Qiaochu Yuan
Dec 30 '18 at 1:23
add a comment |
$begingroup$
$V otimes (-)$ has a left adjoint if and only if $V$ is finite-dimensional; for one direction see this answer. In the finite-dimensional case we have a natural isomorphism
$$V otimes (-) cong text{Hom}(V^{ast}, -)$$
and then by the usual tensor-hom adjunction, the left adjoint (naturally in $V$) is $V^{ast} otimes (-)$.
In the general case of modules the condition is that if $M$ is an $(R, S)$-bimodule then $M otimes_S (-)$ has a left adjoint if and only if $M$ is finitely presented projective as a right $S$-module, in which case its left adjoint is given by tensoring with $text{Hom}_S(M, S)$ over $R$; see this blog post.
answered Dec 29 '18 at 22:41
Qiaochu YuanQiaochu Yuan
281k32593938
$endgroup$
$V otimes (-)$ has a left adjoint if and only if $V$ is finite-dimensional; for one direction see this answer. In the finite-dimensional case we have a natural isomorphism
$$V otimes (-) cong text{Hom}(V^{ast}, -)$$
and then by the usual tensor-hom adjunction, the left adjoint (naturally in $V$) is $V^{ast} otimes (-)$.
In the general case of modules the condition is that if $M$ is an $(R, S)$-bimodule then $M otimes_S (-)$ has a left adjoint if and only if $M$ is finitely presented projective as a right $S$-module, in which case its left adjoint is given by tensoring with $text{Hom}_S(M, S)$ over $R$; see this blog post.
answered Dec 29 '18 at 22:41
Qiaochu YuanQiaochu Yuan
281k32593938
edited Dec 29 '18 at 22:51
answered Dec 29 '18 at 22:41
Qiaochu YuanQiaochu Yuan
281k32593938
answered Dec 29 '18 at 22:41
Qiaochu YuanQiaochu Yuan
281k32593938
answered Dec 29 '18 at 22:41
Qiaochu YuanQiaochu Yuan
281k32593938
281k32593938
$begingroup$
More generally, tensoring with a module preserves products if and only if the module is finitely presented, so this is a necessary condition for the tensor functor to have a left adjoint.
$endgroup$
– egreg
Dec 29 '18 at 22:45
$begingroup$
Thanks for the answer :) We're not assuming $V$ is finite dimensional here though. They intend us to show that $F$ is given by tensoring using the assumption that $V otimes - $ preserves limits, hence has a left adjoint, and by using the fact that every vector space is a colimit of the field.Do you know how to approach this with these assumptions?
$endgroup$
– Mariah
Dec 29 '18 at 22:51
$begingroup$
@Mariah: I'm not assuming it. It's a consequence of the assumption that $V otimes (-)$ preserves limits. If you really don't want to use this fact, you can instead prove that every colimit-preserving endofunctor of $text{Vect}$ (in particular every left adjoint) is given by tensoring with some vector space.
$endgroup$
– Qiaochu Yuan
Dec 29 '18 at 22:51
1
$begingroup$
@QiaochuYuan This last part of your comment is what I hoped to receive an answer to: why is $F$ given by tensoring. I'm not given, and cannot yet assume, that $V$ is finite dimensional. I need to use the fact the $V otimes -$ preserves limits, hence has a left adjoint, and that we can write any vector space $W$ as a colimit of the field. Can you assist in helping me find out why such a functor is given by tensoring?
$endgroup$
– Mariah
Dec 29 '18 at 23:21
$begingroup$
Every vector space is a coproduct of copies of the underlying field $k$ (given by picking a basis) and a colimit-preserving functor preserves this coproduct.
$endgroup$
– Qiaochu Yuan
Dec 30 '18 at 1:23
add a comment |
$begingroup$
More generally, tensoring with a module preserves products if and only if the module is finitely presented, so this is a necessary condition for the tensor functor to have a left adjoint.
$endgroup$
– egreg
Dec 29 '18 at 22:45
$begingroup$
Thanks for the answer :) We're not assuming $V$ is finite dimensional here though. They intend us to show that $F$ is given by tensoring using the assumption that $V otimes - $ preserves limits, hence has a left adjoint, and by using the fact that every vector space is a colimit of the field.Do you know how to approach this with these assumptions?
$endgroup$
– Mariah
Dec 29 '18 at 22:51
$begingroup$
@Mariah: I'm not assuming it. It's a consequence of the assumption that $V otimes (-)$ preserves limits. If you really don't want to use this fact, you can instead prove that every colimit-preserving endofunctor of $text{Vect}$ (in particular every left adjoint) is given by tensoring with some vector space.
$endgroup$
– Qiaochu Yuan
Dec 29 '18 at 22:51
1
$begingroup$
@QiaochuYuan This last part of your comment is what I hoped to receive an answer to: why is $F$ given by tensoring. I'm not given, and cannot yet assume, that $V$ is finite dimensional. I need to use the fact the $V otimes -$ preserves limits, hence has a left adjoint, and that we can write any vector space $W$ as a colimit of the field. Can you assist in helping me find out why such a functor is given by tensoring?
$endgroup$
– Mariah
Dec 29 '18 at 23:21
$begingroup$
Every vector space is a coproduct of copies of the underlying field $k$ (given by picking a basis) and a colimit-preserving functor preserves this coproduct.
$endgroup$
– Qiaochu Yuan
Dec 30 '18 at 1:23
$begingroup$
More generally, tensoring with a module preserves products if and only if the module is finitely presented, so this is a necessary condition for the tensor functor to have a left adjoint.
$endgroup$
– egreg
Dec 29 '18 at 22:45
$begingroup$
More generally, tensoring with a module preserves products if and only if the module is finitely presented, so this is a necessary condition for the tensor functor to have a left adjoint.
$endgroup$
– egreg
Dec 29 '18 at 22:45
$begingroup$
Thanks for the answer :) We're not assuming $V$ is finite dimensional here though. They intend us to show that $F$ is given by tensoring using the assumption that $V otimes - $ preserves limits, hence has a left adjoint, and by using the fact that every vector space is a colimit of the field.Do you know how to approach this with these assumptions?
$endgroup$
– Mariah
Dec 29 '18 at 22:51
$begingroup$
Thanks for the answer :) We're not assuming $V$ is finite dimensional here though. They intend us to show that $F$ is given by tensoring using the assumption that $V otimes - $ preserves limits, hence has a left adjoint, and by using the fact that every vector space is a colimit of the field.Do you know how to approach this with these assumptions?
$endgroup$
– Mariah
Dec 29 '18 at 22:51
$begingroup$
@Mariah: I'm not assuming it. It's a consequence of the assumption that $V otimes (-)$ preserves limits. If you really don't want to use this fact, you can instead prove that every colimit-preserving endofunctor of $text{Vect}$ (in particular every left adjoint) is given by tensoring with some vector space.
$endgroup$
– Qiaochu Yuan
Dec 29 '18 at 22:51
$begingroup$
@Mariah: I'm not assuming it. It's a consequence of the assumption that $V otimes (-)$ preserves limits. If you really don't want to use this fact, you can instead prove that every colimit-preserving endofunctor of $text{Vect}$ (in particular every left adjoint) is given by tensoring with some vector space.
$endgroup$
– Qiaochu Yuan
Dec 29 '18 at 22:51
1
1
$begingroup$
@QiaochuYuan This last part of your comment is what I hoped to receive an answer to: why is $F$ given by tensoring. I'm not given, and cannot yet assume, that $V$ is finite dimensional. I need to use the fact the $V otimes -$ preserves limits, hence has a left adjoint, and that we can write any vector space $W$ as a colimit of the field. Can you assist in helping me find out why such a functor is given by tensoring?
$endgroup$
– Mariah
Dec 29 '18 at 23:21
$begingroup$
@QiaochuYuan This last part of your comment is what I hoped to receive an answer to: why is $F$ given by tensoring. I'm not given, and cannot yet assume, that $V$ is finite dimensional. I need to use the fact the $V otimes -$ preserves limits, hence has a left adjoint, and that we can write any vector space $W$ as a colimit of the field. Can you assist in helping me find out why such a functor is given by tensoring?
$endgroup$
– Mariah
Dec 29 '18 at 23:21
$begingroup$
Every vector space is a coproduct of copies of the underlying field $k$ (given by picking a basis) and a colimit-preserving functor preserves this coproduct.
$endgroup$
– Qiaochu Yuan
Dec 30 '18 at 1:23
$begingroup$
Every vector space is a coproduct of copies of the underlying field $k$ (given by picking a basis) and a colimit-preserving functor preserves this coproduct.
$endgroup$
– Qiaochu Yuan
Dec 30 '18 at 1:23
add a comment |
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$begingroup$
Is this even true? If tensoring is a right adjoint then it preserves all limits/products, which usually isn’t true.
$endgroup$
– Randall
Dec 29 '18 at 14:30
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A field is flat, @Randall, so tensoring is left exact over a field, or am I missing something
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– Card_Trick
Dec 29 '18 at 15:10
1
$begingroup$
I’m sure things are fine for finite limits. Tensoring doesn’t usually respect, eg, infinite direct products. @Card_Trick
$endgroup$
– Randall
Dec 29 '18 at 15:29
$begingroup$
@Randall the right adjoint here preserves limits iff $V$ is finite dimensional. According to a later clause in this question.
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– Mariah
Dec 29 '18 at 15:48
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I didn’t see that $V$ is fd.
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– Randall
Dec 29 '18 at 15:58