Mixing
One-forms are dual to tangent vectors
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
In my class it was said that
"A tangent vector $X in T_p(mathbbR^n)$ acts on a one-form to give a real number"
and
"A one-form acts on a tangent vector to give a real number"
Now the 'tangent space' $T_p(mathbbR^n)$ is a $n$-dimensional vector space and the elements of $T_p(mathbbR^n)$ which we call tangent vectors are actually derivations, which are linear maps $w : C^infty(mathbbR^n) to mathbbR$ satisfying a product rule.
One-forms are elements of the dual vector space $T_p^*(mathbbR^n)$, which we call the cotangent space. They are by definition of a dual vector space, linear maps from $T_p(mathbbR^n)$ to $mathbbR$, e.g $f : T_p(mathbbR^n) to mathbbR$. From this it is easy to see that a one-form takes as input a tangent vector and outputs a real number.
However I'm having trouble seeing how a tangent vector (derivation) takes as input a one-form to output a real number since it's domain isn't even $T_p^*(mathbbR^n)$.
multivariable-calculus differential-geometry vector-spaces
asked Aug 2 at 16:08
Perturbative
3,45211039
add a comment |Â
up vote
2
down vote
favorite
In my class it was said that
"A tangent vector $X in T_p(mathbbR^n)$ acts on a one-form to give a real number"
and
"A one-form acts on a tangent vector to give a real number"
Now the 'tangent space' $T_p(mathbbR^n)$ is a $n$-dimensional vector space and the elements of $T_p(mathbbR^n)$ which we call tangent vectors are actually derivations, which are linear maps $w : C^infty(mathbbR^n) to mathbbR$ satisfying a product rule.
One-forms are elements of the dual vector space $T_p^*(mathbbR^n)$, which we call the cotangent space. They are by definition of a dual vector space, linear maps from $T_p(mathbbR^n)$ to $mathbbR$, e.g $f : T_p(mathbbR^n) to mathbbR$. From this it is easy to see that a one-form takes as input a tangent vector and outputs a real number.
However I'm having trouble seeing how a tangent vector (derivation) takes as input a one-form to output a real number since it's domain isn't even $T_p^*(mathbbR^n)$.
multivariable-calculus differential-geometry vector-spaces
asked Aug 2 at 16:08
Perturbative
3,45211039
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
In my class it was said that
"A tangent vector $X in T_p(mathbbR^n)$ acts on a one-form to give a real number"
and
"A one-form acts on a tangent vector to give a real number"
Now the 'tangent space' $T_p(mathbbR^n)$ is a $n$-dimensional vector space and the elements of $T_p(mathbbR^n)$ which we call tangent vectors are actually derivations, which are linear maps $w : C^infty(mathbbR^n) to mathbbR$ satisfying a product rule.
One-forms are elements of the dual vector space $T_p^*(mathbbR^n)$, which we call the cotangent space. They are by definition of a dual vector space, linear maps from $T_p(mathbbR^n)$ to $mathbbR$, e.g $f : T_p(mathbbR^n) to mathbbR$. From this it is easy to see that a one-form takes as input a tangent vector and outputs a real number.
However I'm having trouble seeing how a tangent vector (derivation) takes as input a one-form to output a real number since it's domain isn't even $T_p^*(mathbbR^n)$.
multivariable-calculus differential-geometry vector-spaces
asked Aug 2 at 16:08
Perturbative
3,45211039
In my class it was said that
"A tangent vector $X in T_p(mathbbR^n)$ acts on a one-form to give a real number"
and
"A one-form acts on a tangent vector to give a real number"
Now the 'tangent space' $T_p(mathbbR^n)$ is a $n$-dimensional vector space and the elements of $T_p(mathbbR^n)$ which we call tangent vectors are actually derivations, which are linear maps $w : C^infty(mathbbR^n) to mathbbR$ satisfying a product rule.
One-forms are elements of the dual vector space $T_p^*(mathbbR^n)$, which we call the cotangent space. They are by definition of a dual vector space, linear maps from $T_p(mathbbR^n)$ to $mathbbR$, e.g $f : T_p(mathbbR^n) to mathbbR$. From this it is easy to see that a one-form takes as input a tangent vector and outputs a real number.
However I'm having trouble seeing how a tangent vector (derivation) takes as input a one-form to output a real number since it's domain isn't even $T_p^*(mathbbR^n)$.
multivariable-calculus differential-geometry vector-spaces
asked Aug 2 at 16:08
Perturbative
3,45211039
asked Aug 2 at 16:08
Perturbative
3,45211039
asked Aug 2 at 16:08
Perturbative
3,45211039
asked Aug 2 at 16:08
Perturbative
3,45211039
3,45211039
add a comment |Â
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
1
down vote
"tangent vectors act on $1$-forms": there is a linear map form the linear space of tangent vectors to a linear space of $1$-forms (the bidual space of the tangent space) that takes $1$-forms to reals.
What is not stated explicitly is by what map this action happens.
In this case it is always meant the "evaluation of the $1$-form on the tangent vector" map:
$$T_p(mathbbR^n)to T^**_p(mathbbR^n); Xmapsto(fmapsto f(X))$$
answered Aug 2 at 16:33
trying
4,0311722
add a comment |Â
up vote
1
down vote
You are right, but given a linear space $X$ there is a natural embedding of $X$ into $X^**$ by reversing the application. Let $x in X$ and $f in X^*$ and define $omega_x in X^** $ by $omega_x(f) = f (x). $
answered Aug 2 at 16:40
md2perpe
5,6191921
add a comment |Â
up vote
1
down vote
You are putting too much significance on the word "act", taking it to mean that the object itself should necessarily (and strictly) be a function whose domain includes what they are supposed to act upon.
An analogy is the following: Imagine a ruler (among a set $A$ of rulers with different scaling) and a piece of string (among a set $B$ of strings of different lengths). When you fix a ruler, it "acts" upon the strings by measuring them, and returns a number (their length). When you fix a string, it "acts" upon the rulers by being measured by them, and returns a number (the length of the string according to the scalings). Rulers and objects are, clearly, not functions which take each other as arguments strictly speaking, but they can be interpreted as so in the above sense. This is what happens with tangent vectors.
In the above framework, we can see the mapping
beginalign*
A times B &to mathbbR \
(R,O) &to textLength of $O$ according to $R$.
endalign*
We now have two natural mappings $mathfrakf:A to mathcalF(B;mathbbR)$ and $mathfrakg:B to mathcalF(A;mathbbR)$, which are just fixing an input of the above mapping which takes two inputs.
Going back to the context of differential topology of your question (although this is essentially linear algebra), the above mappings translate to
$$mathfrakf:T_pM^* to mathrmHom(T_pM;mathbbR)=T_pM^*$$
and
$$mathfrakg:T_pM to mathrmHom(T_pM^*; mathbbR) =T_pM^**. $$
The first map is just the identity, whereas the second map is the standard embedding of a vector space in its bidual. If you want to take the word "act" very seriously and strictly, then the text is actually talking about $mathfrakg(X)$. But the usual advice is not to take words very seriously and strictly, particularly when a lot of identifications will happen, and when objects can have multiple roles and contexts.
answered Aug 2 at 22:12
Aloizio Macedo
22.5k23283
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
"tangent vectors act on $1$-forms": there is a linear map form the linear space of tangent vectors to a linear space of $1$-forms (the bidual space of the tangent space) that takes $1$-forms to reals.
What is not stated explicitly is by what map this action happens.
In this case it is always meant the "evaluation of the $1$-form on the tangent vector" map:
$$T_p(mathbbR^n)to T^**_p(mathbbR^n); Xmapsto(fmapsto f(X))$$
answered Aug 2 at 16:33
trying
4,0311722
add a comment |Â
up vote
1
down vote
"tangent vectors act on $1$-forms": there is a linear map form the linear space of tangent vectors to a linear space of $1$-forms (the bidual space of the tangent space) that takes $1$-forms to reals.
What is not stated explicitly is by what map this action happens.
In this case it is always meant the "evaluation of the $1$-form on the tangent vector" map:
$$T_p(mathbbR^n)to T^**_p(mathbbR^n); Xmapsto(fmapsto f(X))$$
answered Aug 2 at 16:33
trying
4,0311722
add a comment |Â
up vote
1
down vote
up vote
1
down vote
"tangent vectors act on $1$-forms": there is a linear map form the linear space of tangent vectors to a linear space of $1$-forms (the bidual space of the tangent space) that takes $1$-forms to reals.
What is not stated explicitly is by what map this action happens.
In this case it is always meant the "evaluation of the $1$-form on the tangent vector" map:
$$T_p(mathbbR^n)to T^**_p(mathbbR^n); Xmapsto(fmapsto f(X))$$
answered Aug 2 at 16:33
trying
4,0311722
"tangent vectors act on $1$-forms": there is a linear map form the linear space of tangent vectors to a linear space of $1$-forms (the bidual space of the tangent space) that takes $1$-forms to reals.
What is not stated explicitly is by what map this action happens.
In this case it is always meant the "evaluation of the $1$-form on the tangent vector" map:
$$T_p(mathbbR^n)to T^**_p(mathbbR^n); Xmapsto(fmapsto f(X))$$
answered Aug 2 at 16:33
trying
4,0311722
answered Aug 2 at 16:33
trying
4,0311722
answered Aug 2 at 16:33
trying
4,0311722
answered Aug 2 at 16:33
trying
4,0311722
4,0311722
add a comment |Â
add a comment |Â
up vote
1
down vote
You are right, but given a linear space $X$ there is a natural embedding of $X$ into $X^**$ by reversing the application. Let $x in X$ and $f in X^*$ and define $omega_x in X^** $ by $omega_x(f) = f (x). $
answered Aug 2 at 16:40
md2perpe
5,6191921
add a comment |Â
up vote
1
down vote
You are right, but given a linear space $X$ there is a natural embedding of $X$ into $X^**$ by reversing the application. Let $x in X$ and $f in X^*$ and define $omega_x in X^** $ by $omega_x(f) = f (x). $
answered Aug 2 at 16:40
md2perpe
5,6191921
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You are right, but given a linear space $X$ there is a natural embedding of $X$ into $X^**$ by reversing the application. Let $x in X$ and $f in X^*$ and define $omega_x in X^** $ by $omega_x(f) = f (x). $
answered Aug 2 at 16:40
md2perpe
5,6191921
You are right, but given a linear space $X$ there is a natural embedding of $X$ into $X^**$ by reversing the application. Let $x in X$ and $f in X^*$ and define $omega_x in X^** $ by $omega_x(f) = f (x). $
answered Aug 2 at 16:40
md2perpe
5,6191921
answered Aug 2 at 16:40
md2perpe
5,6191921
answered Aug 2 at 16:40
md2perpe
5,6191921
answered Aug 2 at 16:40
md2perpe
5,6191921
5,6191921
add a comment |Â
add a comment |Â
up vote
1
down vote
You are putting too much significance on the word "act", taking it to mean that the object itself should necessarily (and strictly) be a function whose domain includes what they are supposed to act upon.
An analogy is the following: Imagine a ruler (among a set $A$ of rulers with different scaling) and a piece of string (among a set $B$ of strings of different lengths). When you fix a ruler, it "acts" upon the strings by measuring them, and returns a number (their length). When you fix a string, it "acts" upon the rulers by being measured by them, and returns a number (the length of the string according to the scalings). Rulers and objects are, clearly, not functions which take each other as arguments strictly speaking, but they can be interpreted as so in the above sense. This is what happens with tangent vectors.
In the above framework, we can see the mapping
beginalign*
A times B &to mathbbR \
(R,O) &to textLength of $O$ according to $R$.
endalign*
We now have two natural mappings $mathfrakf:A to mathcalF(B;mathbbR)$ and $mathfrakg:B to mathcalF(A;mathbbR)$, which are just fixing an input of the above mapping which takes two inputs.
Going back to the context of differential topology of your question (although this is essentially linear algebra), the above mappings translate to
$$mathfrakf:T_pM^* to mathrmHom(T_pM;mathbbR)=T_pM^*$$
and
$$mathfrakg:T_pM to mathrmHom(T_pM^*; mathbbR) =T_pM^**. $$
The first map is just the identity, whereas the second map is the standard embedding of a vector space in its bidual. If you want to take the word "act" very seriously and strictly, then the text is actually talking about $mathfrakg(X)$. But the usual advice is not to take words very seriously and strictly, particularly when a lot of identifications will happen, and when objects can have multiple roles and contexts.
answered Aug 2 at 22:12
Aloizio Macedo
22.5k23283
add a comment |Â
up vote
1
down vote
You are putting too much significance on the word "act", taking it to mean that the object itself should necessarily (and strictly) be a function whose domain includes what they are supposed to act upon.
An analogy is the following: Imagine a ruler (among a set $A$ of rulers with different scaling) and a piece of string (among a set $B$ of strings of different lengths). When you fix a ruler, it "acts" upon the strings by measuring them, and returns a number (their length). When you fix a string, it "acts" upon the rulers by being measured by them, and returns a number (the length of the string according to the scalings). Rulers and objects are, clearly, not functions which take each other as arguments strictly speaking, but they can be interpreted as so in the above sense. This is what happens with tangent vectors.
In the above framework, we can see the mapping
beginalign*
A times B &to mathbbR \
(R,O) &to textLength of $O$ according to $R$.
endalign*
We now have two natural mappings $mathfrakf:A to mathcalF(B;mathbbR)$ and $mathfrakg:B to mathcalF(A;mathbbR)$, which are just fixing an input of the above mapping which takes two inputs.
Going back to the context of differential topology of your question (although this is essentially linear algebra), the above mappings translate to
$$mathfrakf:T_pM^* to mathrmHom(T_pM;mathbbR)=T_pM^*$$
and
$$mathfrakg:T_pM to mathrmHom(T_pM^*; mathbbR) =T_pM^**. $$
The first map is just the identity, whereas the second map is the standard embedding of a vector space in its bidual. If you want to take the word "act" very seriously and strictly, then the text is actually talking about $mathfrakg(X)$. But the usual advice is not to take words very seriously and strictly, particularly when a lot of identifications will happen, and when objects can have multiple roles and contexts.
answered Aug 2 at 22:12
Aloizio Macedo
22.5k23283
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You are putting too much significance on the word "act", taking it to mean that the object itself should necessarily (and strictly) be a function whose domain includes what they are supposed to act upon.
An analogy is the following: Imagine a ruler (among a set $A$ of rulers with different scaling) and a piece of string (among a set $B$ of strings of different lengths). When you fix a ruler, it "acts" upon the strings by measuring them, and returns a number (their length). When you fix a string, it "acts" upon the rulers by being measured by them, and returns a number (the length of the string according to the scalings). Rulers and objects are, clearly, not functions which take each other as arguments strictly speaking, but they can be interpreted as so in the above sense. This is what happens with tangent vectors.
In the above framework, we can see the mapping
beginalign*
A times B &to mathbbR \
(R,O) &to textLength of $O$ according to $R$.
endalign*
We now have two natural mappings $mathfrakf:A to mathcalF(B;mathbbR)$ and $mathfrakg:B to mathcalF(A;mathbbR)$, which are just fixing an input of the above mapping which takes two inputs.
Going back to the context of differential topology of your question (although this is essentially linear algebra), the above mappings translate to
$$mathfrakf:T_pM^* to mathrmHom(T_pM;mathbbR)=T_pM^*$$
and
$$mathfrakg:T_pM to mathrmHom(T_pM^*; mathbbR) =T_pM^**. $$
The first map is just the identity, whereas the second map is the standard embedding of a vector space in its bidual. If you want to take the word "act" very seriously and strictly, then the text is actually talking about $mathfrakg(X)$. But the usual advice is not to take words very seriously and strictly, particularly when a lot of identifications will happen, and when objects can have multiple roles and contexts.
answered Aug 2 at 22:12
Aloizio Macedo
22.5k23283
You are putting too much significance on the word "act", taking it to mean that the object itself should necessarily (and strictly) be a function whose domain includes what they are supposed to act upon.
An analogy is the following: Imagine a ruler (among a set $A$ of rulers with different scaling) and a piece of string (among a set $B$ of strings of different lengths). When you fix a ruler, it "acts" upon the strings by measuring them, and returns a number (their length). When you fix a string, it "acts" upon the rulers by being measured by them, and returns a number (the length of the string according to the scalings). Rulers and objects are, clearly, not functions which take each other as arguments strictly speaking, but they can be interpreted as so in the above sense. This is what happens with tangent vectors.
In the above framework, we can see the mapping
beginalign*
A times B &to mathbbR \
(R,O) &to textLength of $O$ according to $R$.
endalign*
We now have two natural mappings $mathfrakf:A to mathcalF(B;mathbbR)$ and $mathfrakg:B to mathcalF(A;mathbbR)$, which are just fixing an input of the above mapping which takes two inputs.
Going back to the context of differential topology of your question (although this is essentially linear algebra), the above mappings translate to
$$mathfrakf:T_pM^* to mathrmHom(T_pM;mathbbR)=T_pM^*$$
and
$$mathfrakg:T_pM to mathrmHom(T_pM^*; mathbbR) =T_pM^**. $$
The first map is just the identity, whereas the second map is the standard embedding of a vector space in its bidual. If you want to take the word "act" very seriously and strictly, then the text is actually talking about $mathfrakg(X)$. But the usual advice is not to take words very seriously and strictly, particularly when a lot of identifications will happen, and when objects can have multiple roles and contexts.
answered Aug 2 at 22:12
Aloizio Macedo
22.5k23283
answered Aug 2 at 22:12
Aloizio Macedo
22.5k23283
answered Aug 2 at 22:12
Aloizio Macedo
22.5k23283
answered Aug 2 at 22:12
Aloizio Macedo
22.5k23283
22.5k23283
add a comment |Â
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%2f2870237%2fone-forms-are-dual-to-tangent-vectors%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