Do we really mean “Cartesian Product of Vector Spaces” or is this just a naming convention?
$begingroup$
While approaching tensors, I faced what is considered to be the cartesian product of 2 vector spaces. Now as much as I understood what this practically mean, I was quite concerned by its definition.
My concern comes from how the cartesian product is defined, how a vector space is defined and how these definitions seem to collide in the statement" Cartesian Product of Vector Spaces".
Cartesian Product:
The Cartesian product is a mathematical operation that returns a set from multiple sets. That is, for sets A and B, the Cartesian product A × B is the set of all ordered pairs (a, b) where a ∈ A and b ∈ B....
Vector Space:
A vector space is defined as a quadruple (V,K,⊕,⊙) where V is a set of elements called vectors, K is a field** (K,+,⋅) (and we say that the vector space is a space over K) , ⊕ is a binary operation (called sum) on V such that (V,⊕) is a commutative group and ⊙:K×V→V is a scalar multiplication such that....
Now, a Vector Space is defined as a quadruple while Cartesian Product operates on sets. This is why, even if I understood that for Vector Space is often meant the set V, this created some confusion, at least for me.
I don't know if this naming-statement shortcut/convention is used with other mathematical structures being something common. However as a beginner trying to make sense of all definitions, and trying to understand how they all tie together and interrelate, I was just confused by it.
Are my intuitions right about what it practically mean to say " A Cartesian Product of Vector Spaces"?
Is this a common naming convention? If yes, why is it useful?
vector-spaces notation terminology convention
edited Dec 12 '18 at 10:48
Asaf Karagila♦
303k32429761
asked Dec 12 '18 at 10:35
Gabriele ScarlattiGabriele Scarlatti
311112
$endgroup$
add a comment |
$begingroup$
While approaching tensors, I faced what is considered to be the cartesian product of 2 vector spaces. Now as much as I understood what this practically mean, I was quite concerned by its definition.
My concern comes from how the cartesian product is defined, how a vector space is defined and how these definitions seem to collide in the statement" Cartesian Product of Vector Spaces".
Cartesian Product:
The Cartesian product is a mathematical operation that returns a set from multiple sets. That is, for sets A and B, the Cartesian product A × B is the set of all ordered pairs (a, b) where a ∈ A and b ∈ B....
Vector Space:
A vector space is defined as a quadruple (V,K,⊕,⊙) where V is a set of elements called vectors, K is a field** (K,+,⋅) (and we say that the vector space is a space over K) , ⊕ is a binary operation (called sum) on V such that (V,⊕) is a commutative group and ⊙:K×V→V is a scalar multiplication such that....
Now, a Vector Space is defined as a quadruple while Cartesian Product operates on sets. This is why, even if I understood that for Vector Space is often meant the set V, this created some confusion, at least for me.
I don't know if this naming-statement shortcut/convention is used with other mathematical structures being something common. However as a beginner trying to make sense of all definitions, and trying to understand how they all tie together and interrelate, I was just confused by it.
Are my intuitions right about what it practically mean to say " A Cartesian Product of Vector Spaces"?
Is this a common naming convention? If yes, why is it useful?
vector-spaces notation terminology convention
edited Dec 12 '18 at 10:48
Asaf Karagila♦
303k32429761
asked Dec 12 '18 at 10:35
Gabriele ScarlattiGabriele Scarlatti
311112
$endgroup$
$begingroup$
One may think of a vector space as a set $V$ "equipped with" certain additional structure, namely $mathbb K$, $oplus$, and $odot$. Then the Cartesian product is understood as acting on underlying sets themselves (and can then be equipped with a vector-space structure constructed from that of the two original vector spaces). You may enjoy reading the answers to the previous question "What do mathematicians mean by "equipped"?" for many examples of why this kind of terminological convention is very useful.
$endgroup$
– Rahul
Dec 12 '18 at 10:56
add a comment |
$begingroup$
While approaching tensors, I faced what is considered to be the cartesian product of 2 vector spaces. Now as much as I understood what this practically mean, I was quite concerned by its definition.
My concern comes from how the cartesian product is defined, how a vector space is defined and how these definitions seem to collide in the statement" Cartesian Product of Vector Spaces".
Cartesian Product:
The Cartesian product is a mathematical operation that returns a set from multiple sets. That is, for sets A and B, the Cartesian product A × B is the set of all ordered pairs (a, b) where a ∈ A and b ∈ B....
Vector Space:
A vector space is defined as a quadruple (V,K,⊕,⊙) where V is a set of elements called vectors, K is a field** (K,+,⋅) (and we say that the vector space is a space over K) , ⊕ is a binary operation (called sum) on V such that (V,⊕) is a commutative group and ⊙:K×V→V is a scalar multiplication such that....
Now, a Vector Space is defined as a quadruple while Cartesian Product operates on sets. This is why, even if I understood that for Vector Space is often meant the set V, this created some confusion, at least for me.
I don't know if this naming-statement shortcut/convention is used with other mathematical structures being something common. However as a beginner trying to make sense of all definitions, and trying to understand how they all tie together and interrelate, I was just confused by it.
Are my intuitions right about what it practically mean to say " A Cartesian Product of Vector Spaces"?
Is this a common naming convention? If yes, why is it useful?
vector-spaces notation terminology convention
edited Dec 12 '18 at 10:48
Asaf Karagila♦
303k32429761
asked Dec 12 '18 at 10:35
Gabriele ScarlattiGabriele Scarlatti
311112
$endgroup$
While approaching tensors, I faced what is considered to be the cartesian product of 2 vector spaces. Now as much as I understood what this practically mean, I was quite concerned by its definition.
My concern comes from how the cartesian product is defined, how a vector space is defined and how these definitions seem to collide in the statement" Cartesian Product of Vector Spaces".
Cartesian Product:
The Cartesian product is a mathematical operation that returns a set from multiple sets. That is, for sets A and B, the Cartesian product A × B is the set of all ordered pairs (a, b) where a ∈ A and b ∈ B....
Vector Space:
A vector space is defined as a quadruple (V,K,⊕,⊙) where V is a set of elements called vectors, K is a field** (K,+,⋅) (and we say that the vector space is a space over K) , ⊕ is a binary operation (called sum) on V such that (V,⊕) is a commutative group and ⊙:K×V→V is a scalar multiplication such that....
Now, a Vector Space is defined as a quadruple while Cartesian Product operates on sets. This is why, even if I understood that for Vector Space is often meant the set V, this created some confusion, at least for me.
I don't know if this naming-statement shortcut/convention is used with other mathematical structures being something common. However as a beginner trying to make sense of all definitions, and trying to understand how they all tie together and interrelate, I was just confused by it.
Are my intuitions right about what it practically mean to say " A Cartesian Product of Vector Spaces"?
Is this a common naming convention? If yes, why is it useful?
vector-spaces notation terminology convention
vector-spaces notation terminology convention
edited Dec 12 '18 at 10:48
Asaf Karagila♦
303k32429761
asked Dec 12 '18 at 10:35
Gabriele ScarlattiGabriele Scarlatti
311112
edited Dec 12 '18 at 10:48
Asaf Karagila♦
303k32429761
asked Dec 12 '18 at 10:35
Gabriele ScarlattiGabriele Scarlatti
311112
edited Dec 12 '18 at 10:48
Asaf Karagila♦
303k32429761
edited Dec 12 '18 at 10:48
Asaf Karagila♦
303k32429761
edited Dec 12 '18 at 10:48
Asaf Karagila♦
303k32429761
303k32429761
asked Dec 12 '18 at 10:35
Gabriele ScarlattiGabriele Scarlatti
311112
asked Dec 12 '18 at 10:35
Gabriele ScarlattiGabriele Scarlatti
311112
asked Dec 12 '18 at 10:35
Gabriele ScarlattiGabriele Scarlatti
311112
311112
$begingroup$
One may think of a vector space as a set $V$ "equipped with" certain additional structure, namely $mathbb K$, $oplus$, and $odot$. Then the Cartesian product is understood as acting on underlying sets themselves (and can then be equipped with a vector-space structure constructed from that of the two original vector spaces). You may enjoy reading the answers to the previous question "What do mathematicians mean by "equipped"?" for many examples of why this kind of terminological convention is very useful.
$endgroup$
– Rahul
Dec 12 '18 at 10:56
add a comment |
$begingroup$
One may think of a vector space as a set $V$ "equipped with" certain additional structure, namely $mathbb K$, $oplus$, and $odot$. Then the Cartesian product is understood as acting on underlying sets themselves (and can then be equipped with a vector-space structure constructed from that of the two original vector spaces). You may enjoy reading the answers to the previous question "What do mathematicians mean by "equipped"?" for many examples of why this kind of terminological convention is very useful.
$endgroup$
– Rahul
Dec 12 '18 at 10:56
$begingroup$
One may think of a vector space as a set $V$ "equipped with" certain additional structure, namely $mathbb K$, $oplus$, and $odot$. Then the Cartesian product is understood as acting on underlying sets themselves (and can then be equipped with a vector-space structure constructed from that of the two original vector spaces). You may enjoy reading the answers to the previous question "What do mathematicians mean by "equipped"?" for many examples of why this kind of terminological convention is very useful.
$endgroup$
– Rahul
Dec 12 '18 at 10:56
$begingroup$
One may think of a vector space as a set $V$ "equipped with" certain additional structure, namely $mathbb K$, $oplus$, and $odot$. Then the Cartesian product is understood as acting on underlying sets themselves (and can then be equipped with a vector-space structure constructed from that of the two original vector spaces). You may enjoy reading the answers to the previous question "What do mathematicians mean by "equipped"?" for many examples of why this kind of terminological convention is very useful.
$endgroup$
– Rahul
Dec 12 '18 at 10:56
add a comment |
1 Answer
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$begingroup$
Yes, this is a naming convention. It's quicker, and as clear as "The vector space induced by the Cartesian product of the underlying sets of these two spaces, with the associated pointwise structure defined on that product".
We are the humans who speak to each other about mathematics. So we define, literally as the meaning, what our terms are going to be. There is often a significant overlap and overloading of terms and symbols because it is convenient, and because it makes sense conceptually.
Exactly like when you say $Bbb QsubseteqBbb R$ you are not technically correct if $Bbb R$ is defined as Dedekind cuts over $Bbb Q$, but you are "sufficiently clear" (most of the time) as to what you mean by that (and when you're not clear enough, someone will ask to specify the meaning of that).
answered Dec 12 '18 at 10:47
Asaf Karagila♦Asaf Karagila
303k32429761
$endgroup$
add a comment |
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$begingroup$
Yes, this is a naming convention. It's quicker, and as clear as "The vector space induced by the Cartesian product of the underlying sets of these two spaces, with the associated pointwise structure defined on that product".
We are the humans who speak to each other about mathematics. So we define, literally as the meaning, what our terms are going to be. There is often a significant overlap and overloading of terms and symbols because it is convenient, and because it makes sense conceptually.
Exactly like when you say $Bbb QsubseteqBbb R$ you are not technically correct if $Bbb R$ is defined as Dedekind cuts over $Bbb Q$, but you are "sufficiently clear" (most of the time) as to what you mean by that (and when you're not clear enough, someone will ask to specify the meaning of that).
answered Dec 12 '18 at 10:47
Asaf Karagila♦Asaf Karagila
303k32429761
$endgroup$
add a comment |
$begingroup$
Yes, this is a naming convention. It's quicker, and as clear as "The vector space induced by the Cartesian product of the underlying sets of these two spaces, with the associated pointwise structure defined on that product".
We are the humans who speak to each other about mathematics. So we define, literally as the meaning, what our terms are going to be. There is often a significant overlap and overloading of terms and symbols because it is convenient, and because it makes sense conceptually.
Exactly like when you say $Bbb QsubseteqBbb R$ you are not technically correct if $Bbb R$ is defined as Dedekind cuts over $Bbb Q$, but you are "sufficiently clear" (most of the time) as to what you mean by that (and when you're not clear enough, someone will ask to specify the meaning of that).
answered Dec 12 '18 at 10:47
Asaf Karagila♦Asaf Karagila
303k32429761
$endgroup$
add a comment |
$begingroup$
Yes, this is a naming convention. It's quicker, and as clear as "The vector space induced by the Cartesian product of the underlying sets of these two spaces, with the associated pointwise structure defined on that product".
We are the humans who speak to each other about mathematics. So we define, literally as the meaning, what our terms are going to be. There is often a significant overlap and overloading of terms and symbols because it is convenient, and because it makes sense conceptually.
Exactly like when you say $Bbb QsubseteqBbb R$ you are not technically correct if $Bbb R$ is defined as Dedekind cuts over $Bbb Q$, but you are "sufficiently clear" (most of the time) as to what you mean by that (and when you're not clear enough, someone will ask to specify the meaning of that).
answered Dec 12 '18 at 10:47
Asaf Karagila♦Asaf Karagila
303k32429761
$endgroup$
Yes, this is a naming convention. It's quicker, and as clear as "The vector space induced by the Cartesian product of the underlying sets of these two spaces, with the associated pointwise structure defined on that product".
We are the humans who speak to each other about mathematics. So we define, literally as the meaning, what our terms are going to be. There is often a significant overlap and overloading of terms and symbols because it is convenient, and because it makes sense conceptually.
Exactly like when you say $Bbb QsubseteqBbb R$ you are not technically correct if $Bbb R$ is defined as Dedekind cuts over $Bbb Q$, but you are "sufficiently clear" (most of the time) as to what you mean by that (and when you're not clear enough, someone will ask to specify the meaning of that).
answered Dec 12 '18 at 10:47
Asaf Karagila♦Asaf Karagila
303k32429761
answered Dec 12 '18 at 10:47
Asaf Karagila♦Asaf Karagila
303k32429761
answered Dec 12 '18 at 10:47
Asaf Karagila♦Asaf Karagila
303k32429761
answered Dec 12 '18 at 10:47
Asaf Karagila♦Asaf Karagila
303k32429761
303k32429761
add a comment |
add a comment |
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$begingroup$
One may think of a vector space as a set $V$ "equipped with" certain additional structure, namely $mathbb K$, $oplus$, and $odot$. Then the Cartesian product is understood as acting on underlying sets themselves (and can then be equipped with a vector-space structure constructed from that of the two original vector spaces). You may enjoy reading the answers to the previous question "What do mathematicians mean by "equipped"?" for many examples of why this kind of terminological convention is very useful.
$endgroup$
– Rahul
Dec 12 '18 at 10:56