What's the meaning of the underlined sentence in page 1 of GTM 102?
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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
782510
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
782510
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
782510
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
782510
asked Nov 22 at 16:44
Born to be proud
782510
asked Nov 22 at 16:44
Born to be proud
782510
asked Nov 22 at 16:44
Born to be proud
782510
asked Nov 22 at 16:44
Born to be proud
782510
782510
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 |
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1 Answer
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active
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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 |
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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