A difficulty in understanding a part of a paragraph in P.41 in Guillemin & Pollack (2)
$begingroup$
The paragraph is given below:
But I do not understand:
1-In the forth line why we can not have the case $df_{x} =$ constant other than 0, could anyone explain this for me please?
2-In the sixth line how f is simply the first coordinate function, could anyone give me a concrete example for describing this please?
3-In the tenth line I could not understand why the authors said "But if $f(x)$ ia an extreme value, then obviously $f$ can not be a coordinate function near x" , could anyone explain this statement for me please may be by a concrete example?
thank!
general-topology differential-geometry algebraic-topology differential-topology
asked Dec 5 '18 at 3:33
hopefullyhopefully
317113
$endgroup$
|
show 5 more comments
$begingroup$
The paragraph is given below:
But I do not understand:
1-In the forth line why we can not have the case $df_{x} =$ constant other than 0, could anyone explain this for me please?
2-In the sixth line how f is simply the first coordinate function, could anyone give me a concrete example for describing this please?
3-In the tenth line I could not understand why the authors said "But if $f(x)$ ia an extreme value, then obviously $f$ can not be a coordinate function near x" , could anyone explain this statement for me please may be by a concrete example?
thank!
general-topology differential-geometry algebraic-topology differential-topology
asked Dec 5 '18 at 3:33
hopefullyhopefully
317113
$endgroup$
$begingroup$
1. $df_x$ is a linear map of vector spaces, so how many such constant maps could it be?
$endgroup$
– Randall
Dec 5 '18 at 3:42
$begingroup$
you mean $f$ is a linear map of vector spaces or $df$? @Randall
$endgroup$
– hopefully
Dec 5 '18 at 3:46
$begingroup$
Check the definitions: given $f: X to Y$ a smooth map, $df_x$ is a linear transformation $df_x: T_xM to T_{f(x)}N$.
$endgroup$
– Randall
Dec 5 '18 at 3:48
$begingroup$
So by this definition since the differential(derivative) of a linear map is the linear map itself .... I do not know the answer for your first question .... I am confused.@Randall
$endgroup$
– hopefully
Dec 5 '18 at 3:52
1
$begingroup$
You are confusing the concept of linear function from elementary mathematics with the concept of a linear transformation of vector spaces that occurs in linear algebra. There is a connection between the two, but they are not the same thing.
$endgroup$
– Justin Young
Dec 5 '18 at 15:01
|
show 5 more comments
$begingroup$
The paragraph is given below:
But I do not understand:
1-In the forth line why we can not have the case $df_{x} =$ constant other than 0, could anyone explain this for me please?
2-In the sixth line how f is simply the first coordinate function, could anyone give me a concrete example for describing this please?
3-In the tenth line I could not understand why the authors said "But if $f(x)$ ia an extreme value, then obviously $f$ can not be a coordinate function near x" , could anyone explain this statement for me please may be by a concrete example?
thank!
general-topology differential-geometry algebraic-topology differential-topology
asked Dec 5 '18 at 3:33
hopefullyhopefully
317113
$endgroup$
The paragraph is given below:
But I do not understand:
1-In the forth line why we can not have the case $df_{x} =$ constant other than 0, could anyone explain this for me please?
2-In the sixth line how f is simply the first coordinate function, could anyone give me a concrete example for describing this please?
3-In the tenth line I could not understand why the authors said "But if $f(x)$ ia an extreme value, then obviously $f$ can not be a coordinate function near x" , could anyone explain this statement for me please may be by a concrete example?
thank!
general-topology differential-geometry algebraic-topology differential-topology
general-topology differential-geometry algebraic-topology differential-topology
asked Dec 5 '18 at 3:33
hopefullyhopefully
317113
asked Dec 5 '18 at 3:33
hopefullyhopefully
317113
asked Dec 5 '18 at 3:33
hopefullyhopefully
317113
asked Dec 5 '18 at 3:33
hopefullyhopefully
317113
asked Dec 5 '18 at 3:33
hopefullyhopefully
317113
317113
$begingroup$
1. $df_x$ is a linear map of vector spaces, so how many such constant maps could it be?
$endgroup$
– Randall
Dec 5 '18 at 3:42
$begingroup$
you mean $f$ is a linear map of vector spaces or $df$? @Randall
$endgroup$
– hopefully
Dec 5 '18 at 3:46
$begingroup$
Check the definitions: given $f: X to Y$ a smooth map, $df_x$ is a linear transformation $df_x: T_xM to T_{f(x)}N$.
$endgroup$
– Randall
Dec 5 '18 at 3:48
$begingroup$
So by this definition since the differential(derivative) of a linear map is the linear map itself .... I do not know the answer for your first question .... I am confused.@Randall
$endgroup$
– hopefully
Dec 5 '18 at 3:52
1
$begingroup$
You are confusing the concept of linear function from elementary mathematics with the concept of a linear transformation of vector spaces that occurs in linear algebra. There is a connection between the two, but they are not the same thing.
$endgroup$
– Justin Young
Dec 5 '18 at 15:01
|
show 5 more comments
$begingroup$
1. $df_x$ is a linear map of vector spaces, so how many such constant maps could it be?
$endgroup$
– Randall
Dec 5 '18 at 3:42
$begingroup$
you mean $f$ is a linear map of vector spaces or $df$? @Randall
$endgroup$
– hopefully
Dec 5 '18 at 3:46
$begingroup$
Check the definitions: given $f: X to Y$ a smooth map, $df_x$ is a linear transformation $df_x: T_xM to T_{f(x)}N$.
$endgroup$
– Randall
Dec 5 '18 at 3:48
$begingroup$
So by this definition since the differential(derivative) of a linear map is the linear map itself .... I do not know the answer for your first question .... I am confused.@Randall
$endgroup$
– hopefully
Dec 5 '18 at 3:52
1
$begingroup$
You are confusing the concept of linear function from elementary mathematics with the concept of a linear transformation of vector spaces that occurs in linear algebra. There is a connection between the two, but they are not the same thing.
$endgroup$
– Justin Young
Dec 5 '18 at 15:01
$begingroup$
1. $df_x$ is a linear map of vector spaces, so how many such constant maps could it be?
$endgroup$
– Randall
Dec 5 '18 at 3:42
$begingroup$
1. $df_x$ is a linear map of vector spaces, so how many such constant maps could it be?
$endgroup$
– Randall
Dec 5 '18 at 3:42
$begingroup$
you mean $f$ is a linear map of vector spaces or $df$? @Randall
$endgroup$
– hopefully
Dec 5 '18 at 3:46
$begingroup$
you mean $f$ is a linear map of vector spaces or $df$? @Randall
$endgroup$
– hopefully
Dec 5 '18 at 3:46
$begingroup$
Check the definitions: given $f: X to Y$ a smooth map, $df_x$ is a linear transformation $df_x: T_xM to T_{f(x)}N$.
$endgroup$
– Randall
Dec 5 '18 at 3:48
$begingroup$
Check the definitions: given $f: X to Y$ a smooth map, $df_x$ is a linear transformation $df_x: T_xM to T_{f(x)}N$.
$endgroup$
– Randall
Dec 5 '18 at 3:48
$begingroup$
So by this definition since the differential(derivative) of a linear map is the linear map itself .... I do not know the answer for your first question .... I am confused.@Randall
$endgroup$
– hopefully
Dec 5 '18 at 3:52
$begingroup$
So by this definition since the differential(derivative) of a linear map is the linear map itself .... I do not know the answer for your first question .... I am confused.@Randall
$endgroup$
– hopefully
Dec 5 '18 at 3:52
1
1
$begingroup$
You are confusing the concept of linear function from elementary mathematics with the concept of a linear transformation of vector spaces that occurs in linear algebra. There is a connection between the two, but they are not the same thing.
$endgroup$
– Justin Young
Dec 5 '18 at 15:01
$begingroup$
You are confusing the concept of linear function from elementary mathematics with the concept of a linear transformation of vector spaces that occurs in linear algebra. There is a connection between the two, but they are not the same thing.
$endgroup$
– Justin Young
Dec 5 '18 at 15:01
|
show 5 more comments
1 Answer
1
active
oldest
votes
$begingroup$
(1) The only constant linear map is the zero map. (2) The claim in the sixth line is essentially the implicit function theorem. (3) Consider $f(x)=x^2$. No change of coordinates will turn this into $x$ near $0$.
answered Dec 5 '18 at 5:38
Kevin CarlsonKevin Carlson
32.8k23372
$endgroup$
$begingroup$
why the only constant linear map is the zero map?
$endgroup$
– hopefully
Dec 5 '18 at 6:30
$begingroup$
How it is the implicit function theorem?
$endgroup$
– hopefully
Dec 5 '18 at 6:43
$begingroup$
do not understand the answer of (3)
$endgroup$
– hopefully
Dec 5 '18 at 6:51
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
(1) The only constant linear map is the zero map. (2) The claim in the sixth line is essentially the implicit function theorem. (3) Consider $f(x)=x^2$. No change of coordinates will turn this into $x$ near $0$.
answered Dec 5 '18 at 5:38
Kevin CarlsonKevin Carlson
32.8k23372
$endgroup$
$begingroup$
why the only constant linear map is the zero map?
$endgroup$
– hopefully
Dec 5 '18 at 6:30
$begingroup$
How it is the implicit function theorem?
$endgroup$
– hopefully
Dec 5 '18 at 6:43
$begingroup$
do not understand the answer of (3)
$endgroup$
– hopefully
Dec 5 '18 at 6:51
add a comment |
$begingroup$
(1) The only constant linear map is the zero map. (2) The claim in the sixth line is essentially the implicit function theorem. (3) Consider $f(x)=x^2$. No change of coordinates will turn this into $x$ near $0$.
answered Dec 5 '18 at 5:38
Kevin CarlsonKevin Carlson
32.8k23372
$endgroup$
$begingroup$
why the only constant linear map is the zero map?
$endgroup$
– hopefully
Dec 5 '18 at 6:30
$begingroup$
How it is the implicit function theorem?
$endgroup$
– hopefully
Dec 5 '18 at 6:43
$begingroup$
do not understand the answer of (3)
$endgroup$
– hopefully
Dec 5 '18 at 6:51
add a comment |
$begingroup$
(1) The only constant linear map is the zero map. (2) The claim in the sixth line is essentially the implicit function theorem. (3) Consider $f(x)=x^2$. No change of coordinates will turn this into $x$ near $0$.
answered Dec 5 '18 at 5:38
Kevin CarlsonKevin Carlson
32.8k23372
$endgroup$
(1) The only constant linear map is the zero map. (2) The claim in the sixth line is essentially the implicit function theorem. (3) Consider $f(x)=x^2$. No change of coordinates will turn this into $x$ near $0$.
answered Dec 5 '18 at 5:38
Kevin CarlsonKevin Carlson
32.8k23372
answered Dec 5 '18 at 5:38
Kevin CarlsonKevin Carlson
32.8k23372
answered Dec 5 '18 at 5:38
Kevin CarlsonKevin Carlson
32.8k23372
answered Dec 5 '18 at 5:38
Kevin CarlsonKevin Carlson
32.8k23372
32.8k23372
$begingroup$
why the only constant linear map is the zero map?
$endgroup$
– hopefully
Dec 5 '18 at 6:30
$begingroup$
How it is the implicit function theorem?
$endgroup$
– hopefully
Dec 5 '18 at 6:43
$begingroup$
do not understand the answer of (3)
$endgroup$
– hopefully
Dec 5 '18 at 6:51
add a comment |
$begingroup$
why the only constant linear map is the zero map?
$endgroup$
– hopefully
Dec 5 '18 at 6:30
$begingroup$
How it is the implicit function theorem?
$endgroup$
– hopefully
Dec 5 '18 at 6:43
$begingroup$
do not understand the answer of (3)
$endgroup$
– hopefully
Dec 5 '18 at 6:51
$begingroup$
why the only constant linear map is the zero map?
$endgroup$
– hopefully
Dec 5 '18 at 6:30
$begingroup$
why the only constant linear map is the zero map?
$endgroup$
– hopefully
Dec 5 '18 at 6:30
$begingroup$
How it is the implicit function theorem?
$endgroup$
– hopefully
Dec 5 '18 at 6:43
$begingroup$
How it is the implicit function theorem?
$endgroup$
– hopefully
Dec 5 '18 at 6:43
$begingroup$
do not understand the answer of (3)
$endgroup$
– hopefully
Dec 5 '18 at 6:51
$begingroup$
do not understand the answer of (3)
$endgroup$
– hopefully
Dec 5 '18 at 6:51
add a comment |
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$begingroup$
1. $df_x$ is a linear map of vector spaces, so how many such constant maps could it be?
$endgroup$
– Randall
Dec 5 '18 at 3:42
$begingroup$
you mean $f$ is a linear map of vector spaces or $df$? @Randall
$endgroup$
– hopefully
Dec 5 '18 at 3:46
$begingroup$
Check the definitions: given $f: X to Y$ a smooth map, $df_x$ is a linear transformation $df_x: T_xM to T_{f(x)}N$.
$endgroup$
– Randall
Dec 5 '18 at 3:48
$begingroup$
So by this definition since the differential(derivative) of a linear map is the linear map itself .... I do not know the answer for your first question .... I am confused.@Randall
$endgroup$
– hopefully
Dec 5 '18 at 3:52
1
$begingroup$
You are confusing the concept of linear function from elementary mathematics with the concept of a linear transformation of vector spaces that occurs in linear algebra. There is a connection between the two, but they are not the same thing.
$endgroup$
– Justin Young
Dec 5 '18 at 15:01