How can the graph of a complex function be embedded in three dimensional space?
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In his book Visual Complex Analysis (an awesome book, by the way), Needham, on the topic of graphing complex functions, says that
Actually, the situation is not quite as hopeless as it seems. First,
note that although two-dimensional space is needed to draw the graph
of a real function $f$, the graph itself is only a one-dimensional
curve, meaning that only one real number (namely $x$) is needed to
identify each point within it. Likewise, altough four-dimensional
space is needed to draw the set of points with coordinates $(x,y,u,v)= (z, f(z))$, the graph itself is two-dimensional, meaning that only two real numbers (namely $x$ and $y$) are needed to identify each
point within it. Thus, intrinsically, the graph of a complex function
is merely a two-dimensional surface, and it is this susceptible to
visualization in ordinary three-dimensional space.
Is this actually possible? I've been thinking for a while about how one would graph a complex function using only three dimensions, but I can't find a way.
complex-analysis
asked Jan 3 '13 at 0:21
Javier
4,43252949
add a comment |
up vote
4
down vote
favorite
In his book Visual Complex Analysis (an awesome book, by the way), Needham, on the topic of graphing complex functions, says that
Actually, the situation is not quite as hopeless as it seems. First,
note that although two-dimensional space is needed to draw the graph
of a real function $f$, the graph itself is only a one-dimensional
curve, meaning that only one real number (namely $x$) is needed to
identify each point within it. Likewise, altough four-dimensional
space is needed to draw the set of points with coordinates $(x,y,u,v)= (z, f(z))$, the graph itself is two-dimensional, meaning that only two real numbers (namely $x$ and $y$) are needed to identify each
point within it. Thus, intrinsically, the graph of a complex function
is merely a two-dimensional surface, and it is this susceptible to
visualization in ordinary three-dimensional space.
Is this actually possible? I've been thinking for a while about how one would graph a complex function using only three dimensions, but I can't find a way.
complex-analysis
asked Jan 3 '13 at 0:21
Javier
4,43252949
It would be very strange for a book with that title to make such a claim and not back it up with actual examples of visualization...
– user53153
Jan 3 '13 at 0:33
@PavelM: He later says: "This approach will not be explored in this book, though the last three chapters in particular should prove helpful in understanding Riemann's original physical insights [...]". Maybe it's not as simple as it sounds.
– Javier
Jan 3 '13 at 0:59
@JoeHobbit: What do you mean? Could you give an example?
– Javier
Jul 21 '13 at 4:59
math.stackexchange.com/questions/301929/…
– Dale
Aug 10 '13 at 14:30
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
In his book Visual Complex Analysis (an awesome book, by the way), Needham, on the topic of graphing complex functions, says that
Actually, the situation is not quite as hopeless as it seems. First,
note that although two-dimensional space is needed to draw the graph
of a real function $f$, the graph itself is only a one-dimensional
curve, meaning that only one real number (namely $x$) is needed to
identify each point within it. Likewise, altough four-dimensional
space is needed to draw the set of points with coordinates $(x,y,u,v)= (z, f(z))$, the graph itself is two-dimensional, meaning that only two real numbers (namely $x$ and $y$) are needed to identify each
point within it. Thus, intrinsically, the graph of a complex function
is merely a two-dimensional surface, and it is this susceptible to
visualization in ordinary three-dimensional space.
Is this actually possible? I've been thinking for a while about how one would graph a complex function using only three dimensions, but I can't find a way.
complex-analysis
asked Jan 3 '13 at 0:21
Javier
4,43252949
In his book Visual Complex Analysis (an awesome book, by the way), Needham, on the topic of graphing complex functions, says that
Actually, the situation is not quite as hopeless as it seems. First,
note that although two-dimensional space is needed to draw the graph
of a real function $f$, the graph itself is only a one-dimensional
curve, meaning that only one real number (namely $x$) is needed to
identify each point within it. Likewise, altough four-dimensional
space is needed to draw the set of points with coordinates $(x,y,u,v)= (z, f(z))$, the graph itself is two-dimensional, meaning that only two real numbers (namely $x$ and $y$) are needed to identify each
point within it. Thus, intrinsically, the graph of a complex function
is merely a two-dimensional surface, and it is this susceptible to
visualization in ordinary three-dimensional space.
Is this actually possible? I've been thinking for a while about how one would graph a complex function using only three dimensions, but I can't find a way.
complex-analysis
complex-analysis
asked Jan 3 '13 at 0:21
Javier
4,43252949
asked Jan 3 '13 at 0:21
Javier
4,43252949
asked Jan 3 '13 at 0:21
Javier
4,43252949
asked Jan 3 '13 at 0:21
Javier
4,43252949
asked Jan 3 '13 at 0:21
Javier
4,43252949
4,43252949
It would be very strange for a book with that title to make such a claim and not back it up with actual examples of visualization...
– user53153
Jan 3 '13 at 0:33
@PavelM: He later says: "This approach will not be explored in this book, though the last three chapters in particular should prove helpful in understanding Riemann's original physical insights [...]". Maybe it's not as simple as it sounds.
– Javier
Jan 3 '13 at 0:59
@JoeHobbit: What do you mean? Could you give an example?
– Javier
Jul 21 '13 at 4:59
math.stackexchange.com/questions/301929/…
– Dale
Aug 10 '13 at 14:30
add a comment |
It would be very strange for a book with that title to make such a claim and not back it up with actual examples of visualization...
– user53153
Jan 3 '13 at 0:33
@PavelM: He later says: "This approach will not be explored in this book, though the last three chapters in particular should prove helpful in understanding Riemann's original physical insights [...]". Maybe it's not as simple as it sounds.
– Javier
Jan 3 '13 at 0:59
@JoeHobbit: What do you mean? Could you give an example?
– Javier
Jul 21 '13 at 4:59
math.stackexchange.com/questions/301929/…
– Dale
Aug 10 '13 at 14:30
It would be very strange for a book with that title to make such a claim and not back it up with actual examples of visualization...
– user53153
Jan 3 '13 at 0:33
It would be very strange for a book with that title to make such a claim and not back it up with actual examples of visualization...
– user53153
Jan 3 '13 at 0:33
@PavelM: He later says: "This approach will not be explored in this book, though the last three chapters in particular should prove helpful in understanding Riemann's original physical insights [...]". Maybe it's not as simple as it sounds.
– Javier
Jan 3 '13 at 0:59
@PavelM: He later says: "This approach will not be explored in this book, though the last three chapters in particular should prove helpful in understanding Riemann's original physical insights [...]". Maybe it's not as simple as it sounds.
– Javier
Jan 3 '13 at 0:59
@JoeHobbit: What do you mean? Could you give an example?
– Javier
Jul 21 '13 at 4:59
@JoeHobbit: What do you mean? Could you give an example?
– Javier
Jul 21 '13 at 4:59
math.stackexchange.com/questions/301929/…
– Dale
Aug 10 '13 at 14:30
math.stackexchange.com/questions/301929/…
– Dale
Aug 10 '13 at 14:30
add a comment |
1 Answer
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I think he means this: You have four coordinates $(x,y,u,v)$, but $u=u(x,y)$ and $v=v(x,y)$, so you only have two independent coordinates and you can (at least locally, but I think the surface needs to be orientable) embed the surface in three (even two) dimensions as an abstract manifold.
However, this embedding will not preserve the information of the function, just like the information of the function $y=f(x)$ is lost if you just look at a local embedding of $y$ into $mathbb{R}$.
answered Jan 3 '13 at 0:29
Espen Nielsen
3,000825
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protected by Asaf Karagila♦ Nov 24 at 7:00
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up vote
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I think he means this: You have four coordinates $(x,y,u,v)$, but $u=u(x,y)$ and $v=v(x,y)$, so you only have two independent coordinates and you can (at least locally, but I think the surface needs to be orientable) embed the surface in three (even two) dimensions as an abstract manifold.
However, this embedding will not preserve the information of the function, just like the information of the function $y=f(x)$ is lost if you just look at a local embedding of $y$ into $mathbb{R}$.
answered Jan 3 '13 at 0:29
Espen Nielsen
3,000825
add a comment |
up vote
3
down vote
I think he means this: You have four coordinates $(x,y,u,v)$, but $u=u(x,y)$ and $v=v(x,y)$, so you only have two independent coordinates and you can (at least locally, but I think the surface needs to be orientable) embed the surface in three (even two) dimensions as an abstract manifold.
However, this embedding will not preserve the information of the function, just like the information of the function $y=f(x)$ is lost if you just look at a local embedding of $y$ into $mathbb{R}$.
answered Jan 3 '13 at 0:29
Espen Nielsen
3,000825
add a comment |
up vote
3
down vote
up vote
3
down vote
I think he means this: You have four coordinates $(x,y,u,v)$, but $u=u(x,y)$ and $v=v(x,y)$, so you only have two independent coordinates and you can (at least locally, but I think the surface needs to be orientable) embed the surface in three (even two) dimensions as an abstract manifold.
However, this embedding will not preserve the information of the function, just like the information of the function $y=f(x)$ is lost if you just look at a local embedding of $y$ into $mathbb{R}$.
answered Jan 3 '13 at 0:29
Espen Nielsen
3,000825
I think he means this: You have four coordinates $(x,y,u,v)$, but $u=u(x,y)$ and $v=v(x,y)$, so you only have two independent coordinates and you can (at least locally, but I think the surface needs to be orientable) embed the surface in three (even two) dimensions as an abstract manifold.
However, this embedding will not preserve the information of the function, just like the information of the function $y=f(x)$ is lost if you just look at a local embedding of $y$ into $mathbb{R}$.
answered Jan 3 '13 at 0:29
Espen Nielsen
3,000825
answered Jan 3 '13 at 0:29
Espen Nielsen
3,000825
answered Jan 3 '13 at 0:29
Espen Nielsen
3,000825
answered Jan 3 '13 at 0:29
Espen Nielsen
3,000825
3,000825
add a comment |
add a comment |
protected by Asaf Karagila♦ Nov 24 at 7:00
Thank you for your interest in this question.
Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).
Would you like to answer one of these unanswered questions instead?
It would be very strange for a book with that title to make such a claim and not back it up with actual examples of visualization...
– user53153
Jan 3 '13 at 0:33
@PavelM: He later says: "This approach will not be explored in this book, though the last three chapters in particular should prove helpful in understanding Riemann's original physical insights [...]". Maybe it's not as simple as it sounds.
– Javier
Jan 3 '13 at 0:59
@JoeHobbit: What do you mean? Could you give an example?
– Javier
Jul 21 '13 at 4:59
math.stackexchange.com/questions/301929/…
– Dale
Aug 10 '13 at 14:30