Mixing
Making sense of defining tensor products $bigotimes V$
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
This question is making sense of a definition. Suppose $R$ is a commutative ring.
How does one make a meaningful definition of $bigotimes_1^k V_i$ where $V_i$ are $R$-modules? I know of the $k-$ary product definition. But I do not like it since it does not build up from the original operation $otimes$, which can be regarded as a bifunctor.
We could show there is a well defined isomoprhism, $(A otimes B) otimes C cong A otimes (B otimes C)$, $tau_A,B,C: (a otimes b) otimes c mapsto a otimes (b otimes c)$.
In spirit of this proof. Let us define the left-associated form of $k$ tensor products, $(cdots (V_j otimes V_j+1) cdots otimes V_k-1) otimes V_k := L_j^k(V)$ for $j <k$.
So I wish to define $bigotimes_1^k V_i$ as $L_1^k(V)$. How does one justify the "correctedness" of this definition?
My sketchy attempt:
Suppose we have a $n$-ary representation of $V_i$ with some parathensization, denoted $X$, by inductive hypothesis
$$X cong (L_1^m(V) ) otimes (L_m+1^n(V)) cong (L_1^m(V)) otimes ((L_m+1^n-1 (V) ) otimes V_n)$$
by inductive hypothesis and the $3$ case,
$$(L_1^m(V) otimes L_m+1^n-1(V)) otimes V_n cong (L_1^n-1(V)) otimes V_n cong L_1^n(V)$$
This reasoning seems unnatural, somehow the isomorphisms are obscured. That is, we could have chosen any isomoprhism $tau_A,B,C: (A otimes B) otimes C rightarrow A otimes (B otimes C)$.
Is there a categorical explanation of all of this?
category-theory modules definition tensor-products
edited Jul 28 at 16:11
Batominovski
23k22777
asked Jul 28 at 16:00
Cyryl L.
1,7362721
add a comment |Â
up vote
3
down vote
favorite
This question is making sense of a definition. Suppose $R$ is a commutative ring.
How does one make a meaningful definition of $bigotimes_1^k V_i$ where $V_i$ are $R$-modules? I know of the $k-$ary product definition. But I do not like it since it does not build up from the original operation $otimes$, which can be regarded as a bifunctor.
We could show there is a well defined isomoprhism, $(A otimes B) otimes C cong A otimes (B otimes C)$, $tau_A,B,C: (a otimes b) otimes c mapsto a otimes (b otimes c)$.
In spirit of this proof. Let us define the left-associated form of $k$ tensor products, $(cdots (V_j otimes V_j+1) cdots otimes V_k-1) otimes V_k := L_j^k(V)$ for $j <k$.
So I wish to define $bigotimes_1^k V_i$ as $L_1^k(V)$. How does one justify the "correctedness" of this definition?
My sketchy attempt:
Suppose we have a $n$-ary representation of $V_i$ with some parathensization, denoted $X$, by inductive hypothesis
$$X cong (L_1^m(V) ) otimes (L_m+1^n(V)) cong (L_1^m(V)) otimes ((L_m+1^n-1 (V) ) otimes V_n)$$
by inductive hypothesis and the $3$ case,
$$(L_1^m(V) otimes L_m+1^n-1(V)) otimes V_n cong (L_1^n-1(V)) otimes V_n cong L_1^n(V)$$
This reasoning seems unnatural, somehow the isomorphisms are obscured. That is, we could have chosen any isomoprhism $tau_A,B,C: (A otimes B) otimes C rightarrow A otimes (B otimes C)$.
Is there a categorical explanation of all of this?
category-theory modules definition tensor-products
edited Jul 28 at 16:11
Batominovski
23k22777
asked Jul 28 at 16:00
Cyryl L.
1,7362721
2
Are you fine when defining the $k$-fold cartesian product of sets as $(dots(X_1 times X_2) times X_3)dots X_k-1)times X_k$? Because the same problem appears: $(prod_1^k X_i) times (prod_k^n X_i)$ is not equal to $prod_1^n X_i$, it is only in bijection with it. And you might not even notice it most of the time, taking element $(x_1,dots,x_n)$ in it, without even thinking about the definition of such a tuple. If this is ok for you, then so should your problem too. (The formal justification is Hurkyl's answer.)
– Pece
Jul 28 at 17:44
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
This question is making sense of a definition. Suppose $R$ is a commutative ring.
How does one make a meaningful definition of $bigotimes_1^k V_i$ where $V_i$ are $R$-modules? I know of the $k-$ary product definition. But I do not like it since it does not build up from the original operation $otimes$, which can be regarded as a bifunctor.
We could show there is a well defined isomoprhism, $(A otimes B) otimes C cong A otimes (B otimes C)$, $tau_A,B,C: (a otimes b) otimes c mapsto a otimes (b otimes c)$.
In spirit of this proof. Let us define the left-associated form of $k$ tensor products, $(cdots (V_j otimes V_j+1) cdots otimes V_k-1) otimes V_k := L_j^k(V)$ for $j <k$.
So I wish to define $bigotimes_1^k V_i$ as $L_1^k(V)$. How does one justify the "correctedness" of this definition?
My sketchy attempt:
Suppose we have a $n$-ary representation of $V_i$ with some parathensization, denoted $X$, by inductive hypothesis
$$X cong (L_1^m(V) ) otimes (L_m+1^n(V)) cong (L_1^m(V)) otimes ((L_m+1^n-1 (V) ) otimes V_n)$$
by inductive hypothesis and the $3$ case,
$$(L_1^m(V) otimes L_m+1^n-1(V)) otimes V_n cong (L_1^n-1(V)) otimes V_n cong L_1^n(V)$$
This reasoning seems unnatural, somehow the isomorphisms are obscured. That is, we could have chosen any isomoprhism $tau_A,B,C: (A otimes B) otimes C rightarrow A otimes (B otimes C)$.
Is there a categorical explanation of all of this?
category-theory modules definition tensor-products
edited Jul 28 at 16:11
Batominovski
23k22777
asked Jul 28 at 16:00
Cyryl L.
1,7362721
This question is making sense of a definition. Suppose $R$ is a commutative ring.
How does one make a meaningful definition of $bigotimes_1^k V_i$ where $V_i$ are $R$-modules? I know of the $k-$ary product definition. But I do not like it since it does not build up from the original operation $otimes$, which can be regarded as a bifunctor.
We could show there is a well defined isomoprhism, $(A otimes B) otimes C cong A otimes (B otimes C)$, $tau_A,B,C: (a otimes b) otimes c mapsto a otimes (b otimes c)$.
In spirit of this proof. Let us define the left-associated form of $k$ tensor products, $(cdots (V_j otimes V_j+1) cdots otimes V_k-1) otimes V_k := L_j^k(V)$ for $j <k$.
So I wish to define $bigotimes_1^k V_i$ as $L_1^k(V)$. How does one justify the "correctedness" of this definition?
My sketchy attempt:
Suppose we have a $n$-ary representation of $V_i$ with some parathensization, denoted $X$, by inductive hypothesis
$$X cong (L_1^m(V) ) otimes (L_m+1^n(V)) cong (L_1^m(V)) otimes ((L_m+1^n-1 (V) ) otimes V_n)$$
by inductive hypothesis and the $3$ case,
$$(L_1^m(V) otimes L_m+1^n-1(V)) otimes V_n cong (L_1^n-1(V)) otimes V_n cong L_1^n(V)$$
This reasoning seems unnatural, somehow the isomorphisms are obscured. That is, we could have chosen any isomoprhism $tau_A,B,C: (A otimes B) otimes C rightarrow A otimes (B otimes C)$.
Is there a categorical explanation of all of this?
category-theory modules definition tensor-products
edited Jul 28 at 16:11
Batominovski
23k22777
asked Jul 28 at 16:00
Cyryl L.
1,7362721
edited Jul 28 at 16:11
Batominovski
23k22777
edited Jul 28 at 16:11
Batominovski
23k22777
edited Jul 28 at 16:11
Batominovski
23k22777
23k22777
asked Jul 28 at 16:00
Cyryl L.
1,7362721
asked Jul 28 at 16:00
Cyryl L.
1,7362721
asked Jul 28 at 16:00
Cyryl L.
1,7362721
1,7362721
2
Are you fine when defining the $k$-fold cartesian product of sets as $(dots(X_1 times X_2) times X_3)dots X_k-1)times X_k$? Because the same problem appears: $(prod_1^k X_i) times (prod_k^n X_i)$ is not equal to $prod_1^n X_i$, it is only in bijection with it. And you might not even notice it most of the time, taking element $(x_1,dots,x_n)$ in it, without even thinking about the definition of such a tuple. If this is ok for you, then so should your problem too. (The formal justification is Hurkyl's answer.)
– Pece
Jul 28 at 17:44
add a comment |Â
2
Are you fine when defining the $k$-fold cartesian product of sets as $(dots(X_1 times X_2) times X_3)dots X_k-1)times X_k$? Because the same problem appears: $(prod_1^k X_i) times (prod_k^n X_i)$ is not equal to $prod_1^n X_i$, it is only in bijection with it. And you might not even notice it most of the time, taking element $(x_1,dots,x_n)$ in it, without even thinking about the definition of such a tuple. If this is ok for you, then so should your problem too. (The formal justification is Hurkyl's answer.)
– Pece
Jul 28 at 17:44
2
2
Are you fine when defining the $k$-fold cartesian product of sets as $(dots(X_1 times X_2) times X_3)dots X_k-1)times X_k$? Because the same problem appears: $(prod_1^k X_i) times (prod_k^n X_i)$ is not equal to $prod_1^n X_i$, it is only in bijection with it. And you might not even notice it most of the time, taking element $(x_1,dots,x_n)$ in it, without even thinking about the definition of such a tuple. If this is ok for you, then so should your problem too. (The formal justification is Hurkyl's answer.)
– Pece
Jul 28 at 17:44
Are you fine when defining the $k$-fold cartesian product of sets as $(dots(X_1 times X_2) times X_3)dots X_k-1)times X_k$? Because the same problem appears: $(prod_1^k X_i) times (prod_k^n X_i)$ is not equal to $prod_1^n X_i$, it is only in bijection with it. And you might not even notice it most of the time, taking element $(x_1,dots,x_n)$ in it, without even thinking about the definition of such a tuple. If this is ok for you, then so should your problem too. (The formal justification is Hurkyl's answer.)
– Pece
Jul 28 at 17:44
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
There is the general notion of a monoidal category; it is a category equipped with:
- a unit object $I$
- a bifunctor $otimes$
- a canonical natural isomorphism between any two ways to paranthesize an expression and/or insert/remove copies of $I$
with the property that any way of combining the canonical natural isomorphisms via composition or tensor product gives you another canonical natural isomorphism.
As an example, there are canonical natural isomorphisms
$$mu : I otimes ((A otimes B) otimes C) to A otimes ((B otimes I) otimes C)$$
$$nu : ((A otimes B) otimes C) to A otimes ((B otimes I) otimes C)$$
$$omega : I otimes (A otimes ((B otimes I) otimes C)) to A otimes ((B otimes I) otimes C)$$
We can construct another natural isomorphism parallel to $mu$ via $omega cdot (I otimes nu)$; these are guaranteed to be the same natural isomorphism.
This coherence property means its safe to pick whatever parenthesization you like; if you ever need to use two different parenthesizations there's a canonical way to do the reinterpretation.
In fact, we can even say something stronger: every monoidal category turns out to be monoidally equivalent to a strict one — one where the tensor product is actually an associative operation and where tensoring with $I$ is the identity functor. (and for all of the coherence morphisms, we've chosen the identity transformations to be the canonical isomorphisms)
There's a big theorem that says all of the coherence properties of a monoidal category can be expressed just by using the associator, the two unitors, and a handful of identities relating them; definitions of monoidal category you're likely to encounter will be in that form.
answered Jul 28 at 16:45
Hurkyl
107k9112253
add a comment |Â
up vote
5
down vote
To me, the tensor product of modules $M_1,ldots, M_k$ over a commutative
ring $R$ is a module that represents the functor of $R$-multilinear maps
from $M_1timescdotstimes M_k$. This means that there is a natural
correspondence between $R$-multilinear maps $M_1timescdots times
M_kto N$ and $R$-module homomorphism: $bigotimes_1^k M_ito N$.
Write $textrmMult(M_1,ldots,M_k;N)$ for the $R$-module of multilinear
maps from $M_1timescdots times M_k$ to $N$. Then
$$textrmMult(M_1,ldots,M_k;N)cong
textrmHomleft(bigotimes_1^k-1M_i,textrmHom(M_k,N)right)
congtextrmBilleft(bigotimes_1^k-1M_i,M_k;Nright)$$naturally
in $N$.
This means that
$$bigotimes_1^k-1M_iotimes M_kcongbigotimes_1^k M_i$$
as both represent the same functor. Iterating this gives
your left-associated form. But you don't need to do this,
you could work from the right, or in some more arbitrary manner
instead. The point is, that all these representations of the
$k$-fold tensor product represent the same representable functor
and so they are all canonically isomorphic.
answered Jul 28 at 16:16
Lord Shark the Unknown
84.6k950111
Somewhat merging the two answers, look up Leinster's unbiased monoidal categories.
– Berci
Jul 29 at 22:50
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
There is the general notion of a monoidal category; it is a category equipped with:
- a unit object $I$
- a bifunctor $otimes$
- a canonical natural isomorphism between any two ways to paranthesize an expression and/or insert/remove copies of $I$
with the property that any way of combining the canonical natural isomorphisms via composition or tensor product gives you another canonical natural isomorphism.
As an example, there are canonical natural isomorphisms
$$mu : I otimes ((A otimes B) otimes C) to A otimes ((B otimes I) otimes C)$$
$$nu : ((A otimes B) otimes C) to A otimes ((B otimes I) otimes C)$$
$$omega : I otimes (A otimes ((B otimes I) otimes C)) to A otimes ((B otimes I) otimes C)$$
We can construct another natural isomorphism parallel to $mu$ via $omega cdot (I otimes nu)$; these are guaranteed to be the same natural isomorphism.
This coherence property means its safe to pick whatever parenthesization you like; if you ever need to use two different parenthesizations there's a canonical way to do the reinterpretation.
In fact, we can even say something stronger: every monoidal category turns out to be monoidally equivalent to a strict one — one where the tensor product is actually an associative operation and where tensoring with $I$ is the identity functor. (and for all of the coherence morphisms, we've chosen the identity transformations to be the canonical isomorphisms)
There's a big theorem that says all of the coherence properties of a monoidal category can be expressed just by using the associator, the two unitors, and a handful of identities relating them; definitions of monoidal category you're likely to encounter will be in that form.
answered Jul 28 at 16:45
Hurkyl
107k9112253
add a comment |Â
up vote
4
down vote
accepted
There is the general notion of a monoidal category; it is a category equipped with:
- a unit object $I$
- a bifunctor $otimes$
- a canonical natural isomorphism between any two ways to paranthesize an expression and/or insert/remove copies of $I$
with the property that any way of combining the canonical natural isomorphisms via composition or tensor product gives you another canonical natural isomorphism.
As an example, there are canonical natural isomorphisms
$$mu : I otimes ((A otimes B) otimes C) to A otimes ((B otimes I) otimes C)$$
$$nu : ((A otimes B) otimes C) to A otimes ((B otimes I) otimes C)$$
$$omega : I otimes (A otimes ((B otimes I) otimes C)) to A otimes ((B otimes I) otimes C)$$
We can construct another natural isomorphism parallel to $mu$ via $omega cdot (I otimes nu)$; these are guaranteed to be the same natural isomorphism.
This coherence property means its safe to pick whatever parenthesization you like; if you ever need to use two different parenthesizations there's a canonical way to do the reinterpretation.
In fact, we can even say something stronger: every monoidal category turns out to be monoidally equivalent to a strict one — one where the tensor product is actually an associative operation and where tensoring with $I$ is the identity functor. (and for all of the coherence morphisms, we've chosen the identity transformations to be the canonical isomorphisms)
There's a big theorem that says all of the coherence properties of a monoidal category can be expressed just by using the associator, the two unitors, and a handful of identities relating them; definitions of monoidal category you're likely to encounter will be in that form.
answered Jul 28 at 16:45
Hurkyl
107k9112253
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
There is the general notion of a monoidal category; it is a category equipped with:
- a unit object $I$
- a bifunctor $otimes$
- a canonical natural isomorphism between any two ways to paranthesize an expression and/or insert/remove copies of $I$
with the property that any way of combining the canonical natural isomorphisms via composition or tensor product gives you another canonical natural isomorphism.
As an example, there are canonical natural isomorphisms
$$mu : I otimes ((A otimes B) otimes C) to A otimes ((B otimes I) otimes C)$$
$$nu : ((A otimes B) otimes C) to A otimes ((B otimes I) otimes C)$$
$$omega : I otimes (A otimes ((B otimes I) otimes C)) to A otimes ((B otimes I) otimes C)$$
We can construct another natural isomorphism parallel to $mu$ via $omega cdot (I otimes nu)$; these are guaranteed to be the same natural isomorphism.
This coherence property means its safe to pick whatever parenthesization you like; if you ever need to use two different parenthesizations there's a canonical way to do the reinterpretation.
In fact, we can even say something stronger: every monoidal category turns out to be monoidally equivalent to a strict one — one where the tensor product is actually an associative operation and where tensoring with $I$ is the identity functor. (and for all of the coherence morphisms, we've chosen the identity transformations to be the canonical isomorphisms)
There's a big theorem that says all of the coherence properties of a monoidal category can be expressed just by using the associator, the two unitors, and a handful of identities relating them; definitions of monoidal category you're likely to encounter will be in that form.
answered Jul 28 at 16:45
Hurkyl
107k9112253
There is the general notion of a monoidal category; it is a category equipped with:
- a unit object $I$
- a bifunctor $otimes$
- a canonical natural isomorphism between any two ways to paranthesize an expression and/or insert/remove copies of $I$
with the property that any way of combining the canonical natural isomorphisms via composition or tensor product gives you another canonical natural isomorphism.
As an example, there are canonical natural isomorphisms
$$mu : I otimes ((A otimes B) otimes C) to A otimes ((B otimes I) otimes C)$$
$$nu : ((A otimes B) otimes C) to A otimes ((B otimes I) otimes C)$$
$$omega : I otimes (A otimes ((B otimes I) otimes C)) to A otimes ((B otimes I) otimes C)$$
We can construct another natural isomorphism parallel to $mu$ via $omega cdot (I otimes nu)$; these are guaranteed to be the same natural isomorphism.
This coherence property means its safe to pick whatever parenthesization you like; if you ever need to use two different parenthesizations there's a canonical way to do the reinterpretation.
In fact, we can even say something stronger: every monoidal category turns out to be monoidally equivalent to a strict one — one where the tensor product is actually an associative operation and where tensoring with $I$ is the identity functor. (and for all of the coherence morphisms, we've chosen the identity transformations to be the canonical isomorphisms)
There's a big theorem that says all of the coherence properties of a monoidal category can be expressed just by using the associator, the two unitors, and a handful of identities relating them; definitions of monoidal category you're likely to encounter will be in that form.
answered Jul 28 at 16:45
Hurkyl
107k9112253
edited Jul 28 at 19:33
answered Jul 28 at 16:45
Hurkyl
107k9112253
answered Jul 28 at 16:45
Hurkyl
107k9112253
answered Jul 28 at 16:45
Hurkyl
107k9112253
107k9112253
add a comment |Â
add a comment |Â
up vote
5
down vote
To me, the tensor product of modules $M_1,ldots, M_k$ over a commutative
ring $R$ is a module that represents the functor of $R$-multilinear maps
from $M_1timescdotstimes M_k$. This means that there is a natural
correspondence between $R$-multilinear maps $M_1timescdots times
M_kto N$ and $R$-module homomorphism: $bigotimes_1^k M_ito N$.
Write $textrmMult(M_1,ldots,M_k;N)$ for the $R$-module of multilinear
maps from $M_1timescdots times M_k$ to $N$. Then
$$textrmMult(M_1,ldots,M_k;N)cong
textrmHomleft(bigotimes_1^k-1M_i,textrmHom(M_k,N)right)
congtextrmBilleft(bigotimes_1^k-1M_i,M_k;Nright)$$naturally
in $N$.
This means that
$$bigotimes_1^k-1M_iotimes M_kcongbigotimes_1^k M_i$$
as both represent the same functor. Iterating this gives
your left-associated form. But you don't need to do this,
you could work from the right, or in some more arbitrary manner
instead. The point is, that all these representations of the
$k$-fold tensor product represent the same representable functor
and so they are all canonically isomorphic.
answered Jul 28 at 16:16
Lord Shark the Unknown
84.6k950111
Somewhat merging the two answers, look up Leinster's unbiased monoidal categories.
– Berci
Jul 29 at 22:50
add a comment |Â
up vote
5
down vote
To me, the tensor product of modules $M_1,ldots, M_k$ over a commutative
ring $R$ is a module that represents the functor of $R$-multilinear maps
from $M_1timescdotstimes M_k$. This means that there is a natural
correspondence between $R$-multilinear maps $M_1timescdots times
M_kto N$ and $R$-module homomorphism: $bigotimes_1^k M_ito N$.
Write $textrmMult(M_1,ldots,M_k;N)$ for the $R$-module of multilinear
maps from $M_1timescdots times M_k$ to $N$. Then
$$textrmMult(M_1,ldots,M_k;N)cong
textrmHomleft(bigotimes_1^k-1M_i,textrmHom(M_k,N)right)
congtextrmBilleft(bigotimes_1^k-1M_i,M_k;Nright)$$naturally
in $N$.
This means that
$$bigotimes_1^k-1M_iotimes M_kcongbigotimes_1^k M_i$$
as both represent the same functor. Iterating this gives
your left-associated form. But you don't need to do this,
you could work from the right, or in some more arbitrary manner
instead. The point is, that all these representations of the
$k$-fold tensor product represent the same representable functor
and so they are all canonically isomorphic.
answered Jul 28 at 16:16
Lord Shark the Unknown
84.6k950111
Somewhat merging the two answers, look up Leinster's unbiased monoidal categories.
– Berci
Jul 29 at 22:50
add a comment |Â
up vote
5
down vote
up vote
5
down vote
To me, the tensor product of modules $M_1,ldots, M_k$ over a commutative
ring $R$ is a module that represents the functor of $R$-multilinear maps
from $M_1timescdotstimes M_k$. This means that there is a natural
correspondence between $R$-multilinear maps $M_1timescdots times
M_kto N$ and $R$-module homomorphism: $bigotimes_1^k M_ito N$.
Write $textrmMult(M_1,ldots,M_k;N)$ for the $R$-module of multilinear
maps from $M_1timescdots times M_k$ to $N$. Then
$$textrmMult(M_1,ldots,M_k;N)cong
textrmHomleft(bigotimes_1^k-1M_i,textrmHom(M_k,N)right)
congtextrmBilleft(bigotimes_1^k-1M_i,M_k;Nright)$$naturally
in $N$.
This means that
$$bigotimes_1^k-1M_iotimes M_kcongbigotimes_1^k M_i$$
as both represent the same functor. Iterating this gives
your left-associated form. But you don't need to do this,
you could work from the right, or in some more arbitrary manner
instead. The point is, that all these representations of the
$k$-fold tensor product represent the same representable functor
and so they are all canonically isomorphic.
answered Jul 28 at 16:16
Lord Shark the Unknown
84.6k950111
To me, the tensor product of modules $M_1,ldots, M_k$ over a commutative
ring $R$ is a module that represents the functor of $R$-multilinear maps
from $M_1timescdotstimes M_k$. This means that there is a natural
correspondence between $R$-multilinear maps $M_1timescdots times
M_kto N$ and $R$-module homomorphism: $bigotimes_1^k M_ito N$.
Write $textrmMult(M_1,ldots,M_k;N)$ for the $R$-module of multilinear
maps from $M_1timescdots times M_k$ to $N$. Then
$$textrmMult(M_1,ldots,M_k;N)cong
textrmHomleft(bigotimes_1^k-1M_i,textrmHom(M_k,N)right)
congtextrmBilleft(bigotimes_1^k-1M_i,M_k;Nright)$$naturally
in $N$.
This means that
$$bigotimes_1^k-1M_iotimes M_kcongbigotimes_1^k M_i$$
as both represent the same functor. Iterating this gives
your left-associated form. But you don't need to do this,
you could work from the right, or in some more arbitrary manner
instead. The point is, that all these representations of the
$k$-fold tensor product represent the same representable functor
and so they are all canonically isomorphic.
answered Jul 28 at 16:16
Lord Shark the Unknown
84.6k950111
answered Jul 28 at 16:16
Lord Shark the Unknown
84.6k950111
answered Jul 28 at 16:16
Lord Shark the Unknown
84.6k950111
answered Jul 28 at 16:16
Lord Shark the Unknown
84.6k950111
84.6k950111
Somewhat merging the two answers, look up Leinster's unbiased monoidal categories.
– Berci
Jul 29 at 22:50
add a comment |Â
Somewhat merging the two answers, look up Leinster's unbiased monoidal categories.
– Berci
Jul 29 at 22:50
Somewhat merging the two answers, look up Leinster's unbiased monoidal categories.
– Berci
Jul 29 at 22:50
Somewhat merging the two answers, look up Leinster's unbiased monoidal categories.
– Berci
Jul 29 at 22:50
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2865363%2fmaking-sense-of-defining-tensor-products-bigotimes-v%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
2
Are you fine when defining the $k$-fold cartesian product of sets as $(dots(X_1 times X_2) times X_3)dots X_k-1)times X_k$? Because the same problem appears: $(prod_1^k X_i) times (prod_k^n X_i)$ is not equal to $prod_1^n X_i$, it is only in bijection with it. And you might not even notice it most of the time, taking element $(x_1,dots,x_n)$ in it, without even thinking about the definition of such a tuple. If this is ok for you, then so should your problem too. (The formal justification is Hurkyl's answer.)
– Pece
Jul 28 at 17:44