What's the meaning of the underlined sentence in page 1 of GTM 102?
up vote
0
down vote
favorite
What's the meaning of the underlined sentence in page 1 of GTM 102?
lie-groups lie-algebras
asked Nov 22 at 16:44
Born to be proud
779510
add a comment |
up vote
0
down vote
favorite
What's the meaning of the underlined sentence in page 1 of GTM 102?
lie-groups lie-algebras
asked Nov 22 at 16:44
Born to be proud
779510
Does the terminology "algebra" here mean group,ring and module?
– Born to be proud
Nov 22 at 16:50
2
It's a vector algebra: it's a subspace of the complex vector space of all functions $UtoBbb C $ and it's closed under (pointwise) multiplication and contains the constant $1$.
– Berci
Nov 22 at 17:33
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
What's the meaning of the underlined sentence in page 1 of GTM 102?
lie-groups lie-algebras
asked Nov 22 at 16:44
Born to be proud
779510
What's the meaning of the underlined sentence in page 1 of GTM 102?
lie-groups lie-algebras
lie-groups lie-algebras
asked Nov 22 at 16:44
Born to be proud
779510
asked Nov 22 at 16:44
Born to be proud
779510
asked Nov 22 at 16:44
Born to be proud
779510
asked Nov 22 at 16:44
Born to be proud
779510
asked Nov 22 at 16:44
Born to be proud
779510
779510
Does the terminology "algebra" here mean group,ring and module?
– Born to be proud
Nov 22 at 16:50
2
It's a vector algebra: it's a subspace of the complex vector space of all functions $UtoBbb C $ and it's closed under (pointwise) multiplication and contains the constant $1$.
– Berci
Nov 22 at 17:33
add a comment |
Does the terminology "algebra" here mean group,ring and module?
– Born to be proud
Nov 22 at 16:50
2
It's a vector algebra: it's a subspace of the complex vector space of all functions $UtoBbb C $ and it's closed under (pointwise) multiplication and contains the constant $1$.
– Berci
Nov 22 at 17:33
Does the terminology "algebra" here mean group,ring and module?
– Born to be proud
Nov 22 at 16:50
Does the terminology "algebra" here mean group,ring and module?
– Born to be proud
Nov 22 at 16:50
2
2
It's a vector algebra: it's a subspace of the complex vector space of all functions $UtoBbb C $ and it's closed under (pointwise) multiplication and contains the constant $1$.
– Berci
Nov 22 at 17:33
It's a vector algebra: it's a subspace of the complex vector space of all functions $UtoBbb C $ and it's closed under (pointwise) multiplication and contains the constant $1$.
– Berci
Nov 22 at 17:33
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Let me start with the explicit definition of $mathcal D(U)$, i.e.
$$
mathcal D(U) = { f: U to mathbb C mid f text{~is smooth} }.
$$
Now, an algebra over the field $k=mathbb C$ is a vector space with an bilinear from
$$
cdot: mathcal D(U) times mathcal D(U) to mathcal D(U): (f,g) mapsto f cdot g.
$$
This bilinear form assigns a ring structure to $mathcal D(U)$ and is also a multiplication in this sense, but it is not the same as the scalar multiplication!
In our case, this map is defined pointwise. For $f,g in mathcal D(U)$, we define $f cdot g in mathcal D(U)$ by
$$
f cdot g : U to mathbb C: x mapsto (fcdot g)(x) := f(x) cdot g(x),
$$
where the last addition is the addition of complex numbers.
In addition, this map has to satisfy the axioms of a bilinear map, e.g. stuff like linearity with respect to scalars of the field $lambda (fcdot g) = (lambda f) cdot g$, for $lambda in k = mathbb C$ and the other axioms.
Since the functions of the differentiable structure are complex valued, these properties are satisfied.
The last statement $1 in mathcal D(U)$ asserts, that the constant function
$1: U to mathbb C: x mapsto 1$ is an element of the algebra $mathcal D(U)$.
It is also the unity element with respect to the ring multiplication $cdot$.
If you know a bit of algebraic geometry, you can see that the axioms in the book
point a little bit in the direction of sheafs.
Hence, topological objects like open set $U, V$ are assigned to rings $mathcal D(U)$ and $mathcal D(V)$, an we if $mathcal U subseteq mathcal V$, then we can find a (restriction) morphism $vert_U: mathcal D(V) to mathcal D(U)$, which is in this case just the restriction of the functions onto the smaller domain.
(See the wikipedia article for the complete definitions of sheafs.)
Hence, these a sheaf is a link between topology and algebraic strucutres.
And they allow to relate purely local with global data, therefore it is no surprise, that they are also important for manifolds.
(But I am not an expert about that! Don't trust me.)
Details about the vector space structure:
The addition
$$
+ : mathcal D(U) times mathcal D(U) to mathcal D(U): (f,g) mapsto f+g
$$
is also defined pointwise by $(f+g) (x) = f(x) + g(x) in mathbb C$.
And the scalar multiplication
$$mathbb C times mathcal D(U) to mathcal D(U): (lambda,f) mapsto lambda f,
$$
is defined via $(lambda f)(x) = lambda cdot f(x) in mathbb C$.
answered Nov 22 at 17:33
Steffen Plunder
508211
So $M$ is assumed to be smooth, isn't it?
– Born to be proud
Nov 22 at 18:03
Good point! I guess that you can only talk about $C^{infty}$ structures on smooth manifolds. (Since if $M$ is only a $C^k$ manifold, then there is no practical definition for $C^infty$ maps, since the charts may change the differentiability of the chart repesentation of a map. (i.e. for some charts it maybe is smooth, but for others it is only $C^k$, since the chart is only $C^k$)). But you could also talk about $C^k$ differentiable structures. (In fact, these structures can be used to define manifolds!)
– Steffen Plunder
Nov 22 at 18:09
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3009336%2fwhats-the-meaning-of-the-underlined-sentence-in-page-1-of-gtm-102%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Let me start with the explicit definition of $mathcal D(U)$, i.e.
$$
mathcal D(U) = { f: U to mathbb C mid f text{~is smooth} }.
$$
Now, an algebra over the field $k=mathbb C$ is a vector space with an bilinear from
$$
cdot: mathcal D(U) times mathcal D(U) to mathcal D(U): (f,g) mapsto f cdot g.
$$
This bilinear form assigns a ring structure to $mathcal D(U)$ and is also a multiplication in this sense, but it is not the same as the scalar multiplication!
In our case, this map is defined pointwise. For $f,g in mathcal D(U)$, we define $f cdot g in mathcal D(U)$ by
$$
f cdot g : U to mathbb C: x mapsto (fcdot g)(x) := f(x) cdot g(x),
$$
where the last addition is the addition of complex numbers.
In addition, this map has to satisfy the axioms of a bilinear map, e.g. stuff like linearity with respect to scalars of the field $lambda (fcdot g) = (lambda f) cdot g$, for $lambda in k = mathbb C$ and the other axioms.
Since the functions of the differentiable structure are complex valued, these properties are satisfied.
The last statement $1 in mathcal D(U)$ asserts, that the constant function
$1: U to mathbb C: x mapsto 1$ is an element of the algebra $mathcal D(U)$.
It is also the unity element with respect to the ring multiplication $cdot$.
If you know a bit of algebraic geometry, you can see that the axioms in the book
point a little bit in the direction of sheafs.
Hence, topological objects like open set $U, V$ are assigned to rings $mathcal D(U)$ and $mathcal D(V)$, an we if $mathcal U subseteq mathcal V$, then we can find a (restriction) morphism $vert_U: mathcal D(V) to mathcal D(U)$, which is in this case just the restriction of the functions onto the smaller domain.
(See the wikipedia article for the complete definitions of sheafs.)
Hence, these a sheaf is a link between topology and algebraic strucutres.
And they allow to relate purely local with global data, therefore it is no surprise, that they are also important for manifolds.
(But I am not an expert about that! Don't trust me.)
Details about the vector space structure:
The addition
$$
+ : mathcal D(U) times mathcal D(U) to mathcal D(U): (f,g) mapsto f+g
$$
is also defined pointwise by $(f+g) (x) = f(x) + g(x) in mathbb C$.
And the scalar multiplication
$$mathbb C times mathcal D(U) to mathcal D(U): (lambda,f) mapsto lambda f,
$$
is defined via $(lambda f)(x) = lambda cdot f(x) in mathbb C$.
answered Nov 22 at 17:33
Steffen Plunder
508211
So $M$ is assumed to be smooth, isn't it?
– Born to be proud
Nov 22 at 18:03
Good point! I guess that you can only talk about $C^{infty}$ structures on smooth manifolds. (Since if $M$ is only a $C^k$ manifold, then there is no practical definition for $C^infty$ maps, since the charts may change the differentiability of the chart repesentation of a map. (i.e. for some charts it maybe is smooth, but for others it is only $C^k$, since the chart is only $C^k$)). But you could also talk about $C^k$ differentiable structures. (In fact, these structures can be used to define manifolds!)
– Steffen Plunder
Nov 22 at 18:09
add a comment |
up vote
2
down vote
accepted
Let me start with the explicit definition of $mathcal D(U)$, i.e.
$$
mathcal D(U) = { f: U to mathbb C mid f text{~is smooth} }.
$$
Now, an algebra over the field $k=mathbb C$ is a vector space with an bilinear from
$$
cdot: mathcal D(U) times mathcal D(U) to mathcal D(U): (f,g) mapsto f cdot g.
$$
This bilinear form assigns a ring structure to $mathcal D(U)$ and is also a multiplication in this sense, but it is not the same as the scalar multiplication!
In our case, this map is defined pointwise. For $f,g in mathcal D(U)$, we define $f cdot g in mathcal D(U)$ by
$$
f cdot g : U to mathbb C: x mapsto (fcdot g)(x) := f(x) cdot g(x),
$$
where the last addition is the addition of complex numbers.
In addition, this map has to satisfy the axioms of a bilinear map, e.g. stuff like linearity with respect to scalars of the field $lambda (fcdot g) = (lambda f) cdot g$, for $lambda in k = mathbb C$ and the other axioms.
Since the functions of the differentiable structure are complex valued, these properties are satisfied.
The last statement $1 in mathcal D(U)$ asserts, that the constant function
$1: U to mathbb C: x mapsto 1$ is an element of the algebra $mathcal D(U)$.
It is also the unity element with respect to the ring multiplication $cdot$.
If you know a bit of algebraic geometry, you can see that the axioms in the book
point a little bit in the direction of sheafs.
Hence, topological objects like open set $U, V$ are assigned to rings $mathcal D(U)$ and $mathcal D(V)$, an we if $mathcal U subseteq mathcal V$, then we can find a (restriction) morphism $vert_U: mathcal D(V) to mathcal D(U)$, which is in this case just the restriction of the functions onto the smaller domain.
(See the wikipedia article for the complete definitions of sheafs.)
Hence, these a sheaf is a link between topology and algebraic strucutres.
And they allow to relate purely local with global data, therefore it is no surprise, that they are also important for manifolds.
(But I am not an expert about that! Don't trust me.)
Details about the vector space structure:
The addition
$$
+ : mathcal D(U) times mathcal D(U) to mathcal D(U): (f,g) mapsto f+g
$$
is also defined pointwise by $(f+g) (x) = f(x) + g(x) in mathbb C$.
And the scalar multiplication
$$mathbb C times mathcal D(U) to mathcal D(U): (lambda,f) mapsto lambda f,
$$
is defined via $(lambda f)(x) = lambda cdot f(x) in mathbb C$.
answered Nov 22 at 17:33
Steffen Plunder
508211
So $M$ is assumed to be smooth, isn't it?
– Born to be proud
Nov 22 at 18:03
Good point! I guess that you can only talk about $C^{infty}$ structures on smooth manifolds. (Since if $M$ is only a $C^k$ manifold, then there is no practical definition for $C^infty$ maps, since the charts may change the differentiability of the chart repesentation of a map. (i.e. for some charts it maybe is smooth, but for others it is only $C^k$, since the chart is only $C^k$)). But you could also talk about $C^k$ differentiable structures. (In fact, these structures can be used to define manifolds!)
– Steffen Plunder
Nov 22 at 18:09
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Let me start with the explicit definition of $mathcal D(U)$, i.e.
$$
mathcal D(U) = { f: U to mathbb C mid f text{~is smooth} }.
$$
Now, an algebra over the field $k=mathbb C$ is a vector space with an bilinear from
$$
cdot: mathcal D(U) times mathcal D(U) to mathcal D(U): (f,g) mapsto f cdot g.
$$
This bilinear form assigns a ring structure to $mathcal D(U)$ and is also a multiplication in this sense, but it is not the same as the scalar multiplication!
In our case, this map is defined pointwise. For $f,g in mathcal D(U)$, we define $f cdot g in mathcal D(U)$ by
$$
f cdot g : U to mathbb C: x mapsto (fcdot g)(x) := f(x) cdot g(x),
$$
where the last addition is the addition of complex numbers.
In addition, this map has to satisfy the axioms of a bilinear map, e.g. stuff like linearity with respect to scalars of the field $lambda (fcdot g) = (lambda f) cdot g$, for $lambda in k = mathbb C$ and the other axioms.
Since the functions of the differentiable structure are complex valued, these properties are satisfied.
The last statement $1 in mathcal D(U)$ asserts, that the constant function
$1: U to mathbb C: x mapsto 1$ is an element of the algebra $mathcal D(U)$.
It is also the unity element with respect to the ring multiplication $cdot$.
If you know a bit of algebraic geometry, you can see that the axioms in the book
point a little bit in the direction of sheafs.
Hence, topological objects like open set $U, V$ are assigned to rings $mathcal D(U)$ and $mathcal D(V)$, an we if $mathcal U subseteq mathcal V$, then we can find a (restriction) morphism $vert_U: mathcal D(V) to mathcal D(U)$, which is in this case just the restriction of the functions onto the smaller domain.
(See the wikipedia article for the complete definitions of sheafs.)
Hence, these a sheaf is a link between topology and algebraic strucutres.
And they allow to relate purely local with global data, therefore it is no surprise, that they are also important for manifolds.
(But I am not an expert about that! Don't trust me.)
Details about the vector space structure:
The addition
$$
+ : mathcal D(U) times mathcal D(U) to mathcal D(U): (f,g) mapsto f+g
$$
is also defined pointwise by $(f+g) (x) = f(x) + g(x) in mathbb C$.
And the scalar multiplication
$$mathbb C times mathcal D(U) to mathcal D(U): (lambda,f) mapsto lambda f,
$$
is defined via $(lambda f)(x) = lambda cdot f(x) in mathbb C$.
answered Nov 22 at 17:33
Steffen Plunder
508211
Let me start with the explicit definition of $mathcal D(U)$, i.e.
$$
mathcal D(U) = { f: U to mathbb C mid f text{~is smooth} }.
$$
Now, an algebra over the field $k=mathbb C$ is a vector space with an bilinear from
$$
cdot: mathcal D(U) times mathcal D(U) to mathcal D(U): (f,g) mapsto f cdot g.
$$
This bilinear form assigns a ring structure to $mathcal D(U)$ and is also a multiplication in this sense, but it is not the same as the scalar multiplication!
In our case, this map is defined pointwise. For $f,g in mathcal D(U)$, we define $f cdot g in mathcal D(U)$ by
$$
f cdot g : U to mathbb C: x mapsto (fcdot g)(x) := f(x) cdot g(x),
$$
where the last addition is the addition of complex numbers.
In addition, this map has to satisfy the axioms of a bilinear map, e.g. stuff like linearity with respect to scalars of the field $lambda (fcdot g) = (lambda f) cdot g$, for $lambda in k = mathbb C$ and the other axioms.
Since the functions of the differentiable structure are complex valued, these properties are satisfied.
The last statement $1 in mathcal D(U)$ asserts, that the constant function
$1: U to mathbb C: x mapsto 1$ is an element of the algebra $mathcal D(U)$.
It is also the unity element with respect to the ring multiplication $cdot$.
If you know a bit of algebraic geometry, you can see that the axioms in the book
point a little bit in the direction of sheafs.
Hence, topological objects like open set $U, V$ are assigned to rings $mathcal D(U)$ and $mathcal D(V)$, an we if $mathcal U subseteq mathcal V$, then we can find a (restriction) morphism $vert_U: mathcal D(V) to mathcal D(U)$, which is in this case just the restriction of the functions onto the smaller domain.
(See the wikipedia article for the complete definitions of sheafs.)
Hence, these a sheaf is a link between topology and algebraic strucutres.
And they allow to relate purely local with global data, therefore it is no surprise, that they are also important for manifolds.
(But I am not an expert about that! Don't trust me.)
Details about the vector space structure:
The addition
$$
+ : mathcal D(U) times mathcal D(U) to mathcal D(U): (f,g) mapsto f+g
$$
is also defined pointwise by $(f+g) (x) = f(x) + g(x) in mathbb C$.
And the scalar multiplication
$$mathbb C times mathcal D(U) to mathcal D(U): (lambda,f) mapsto lambda f,
$$
is defined via $(lambda f)(x) = lambda cdot f(x) in mathbb C$.
answered Nov 22 at 17:33
Steffen Plunder
508211
edited Nov 22 at 17:41
answered Nov 22 at 17:33
Steffen Plunder
508211
answered Nov 22 at 17:33
Steffen Plunder
508211
answered Nov 22 at 17:33
Steffen Plunder
508211
508211
So $M$ is assumed to be smooth, isn't it?
– Born to be proud
Nov 22 at 18:03
Good point! I guess that you can only talk about $C^{infty}$ structures on smooth manifolds. (Since if $M$ is only a $C^k$ manifold, then there is no practical definition for $C^infty$ maps, since the charts may change the differentiability of the chart repesentation of a map. (i.e. for some charts it maybe is smooth, but for others it is only $C^k$, since the chart is only $C^k$)). But you could also talk about $C^k$ differentiable structures. (In fact, these structures can be used to define manifolds!)
– Steffen Plunder
Nov 22 at 18:09
add a comment |
So $M$ is assumed to be smooth, isn't it?
– Born to be proud
Nov 22 at 18:03
Good point! I guess that you can only talk about $C^{infty}$ structures on smooth manifolds. (Since if $M$ is only a $C^k$ manifold, then there is no practical definition for $C^infty$ maps, since the charts may change the differentiability of the chart repesentation of a map. (i.e. for some charts it maybe is smooth, but for others it is only $C^k$, since the chart is only $C^k$)). But you could also talk about $C^k$ differentiable structures. (In fact, these structures can be used to define manifolds!)
– Steffen Plunder
Nov 22 at 18:09
So $M$ is assumed to be smooth, isn't it?
– Born to be proud
Nov 22 at 18:03
So $M$ is assumed to be smooth, isn't it?
– Born to be proud
Nov 22 at 18:03
Good point! I guess that you can only talk about $C^{infty}$ structures on smooth manifolds. (Since if $M$ is only a $C^k$ manifold, then there is no practical definition for $C^infty$ maps, since the charts may change the differentiability of the chart repesentation of a map. (i.e. for some charts it maybe is smooth, but for others it is only $C^k$, since the chart is only $C^k$)). But you could also talk about $C^k$ differentiable structures. (In fact, these structures can be used to define manifolds!)
– Steffen Plunder
Nov 22 at 18:09
Good point! I guess that you can only talk about $C^{infty}$ structures on smooth manifolds. (Since if $M$ is only a $C^k$ manifold, then there is no practical definition for $C^infty$ maps, since the charts may change the differentiability of the chart repesentation of a map. (i.e. for some charts it maybe is smooth, but for others it is only $C^k$, since the chart is only $C^k$)). But you could also talk about $C^k$ differentiable structures. (In fact, these structures can be used to define manifolds!)
– Steffen Plunder
Nov 22 at 18:09
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3009336%2fwhats-the-meaning-of-the-underlined-sentence-in-page-1-of-gtm-102%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Does the terminology "algebra" here mean group,ring and module?
– Born to be proud
Nov 22 at 16:50
2
It's a vector algebra: it's a subspace of the complex vector space of all functions $UtoBbb C $ and it's closed under (pointwise) multiplication and contains the constant $1$.
– Berci
Nov 22 at 17:33