How can one best visualize two dimensional manifolds in $mathbbR^4$ (more specifically, $mathbbS^2 times mathbbR)$?

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I'm trying to "get a picture", so to speak, of hypersurfaces in $mathbbS^2 times mathbbR$. One example would be $left(dfraccos(u)sqrt1+u^2, dfracsin(u)sqrt1+u^2,dfracusqrt1+u^2, v right)$, where $-2pi leq u leq 2pi$ and $v in mathbbR$. Now, even though this is a 2-dimensional surface, I haven't found any way to really understand what it looks like. I'm aware this is kind of hard to answer, but any help is appreciated.




differential-geometry intuition surfaces visualization




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  • 2




    You can also think about $S^2timesmathbbR$ as punctured 3-space, so you could probably get a pretty good picture of your surface.
    – Steve D
    Jul 31 at 16:33










  • $S^2timesBbb R$ is the normal line bundle on $S^2$
    – Andrea Mori
    Jul 31 at 16:36






  • 1




    Maybe try to plot it in 3d, using the 4th coordinate as color?
    – zokomoko
    Jul 31 at 16:37










  • This is a graph of your first three coordinates; your surface is just (this) x (real line).
    – Steve D
    Jul 31 at 17:05






  • 1




    That just maps your real line to the positive reals, so you don't get multiple ways to specify the same point.
    – Steve D
    Jul 31 at 20:12














up vote
1
down vote

favorite












I'm trying to "get a picture", so to speak, of hypersurfaces in $mathbbS^2 times mathbbR$. One example would be $left(dfraccos(u)sqrt1+u^2, dfracsin(u)sqrt1+u^2,dfracusqrt1+u^2, v right)$, where $-2pi leq u leq 2pi$ and $v in mathbbR$. Now, even though this is a 2-dimensional surface, I haven't found any way to really understand what it looks like. I'm aware this is kind of hard to answer, but any help is appreciated.




differential-geometry intuition surfaces visualization




share|cite|improve this question

















  • 2




    You can also think about $S^2timesmathbbR$ as punctured 3-space, so you could probably get a pretty good picture of your surface.
    – Steve D
    Jul 31 at 16:33










  • $S^2timesBbb R$ is the normal line bundle on $S^2$
    – Andrea Mori
    Jul 31 at 16:36






  • 1




    Maybe try to plot it in 3d, using the 4th coordinate as color?
    – zokomoko
    Jul 31 at 16:37










  • This is a graph of your first three coordinates; your surface is just (this) x (real line).
    – Steve D
    Jul 31 at 17:05






  • 1




    That just maps your real line to the positive reals, so you don't get multiple ways to specify the same point.
    – Steve D
    Jul 31 at 20:12












up vote
1
down vote

favorite









up vote
1
down vote

favorite











I'm trying to "get a picture", so to speak, of hypersurfaces in $mathbbS^2 times mathbbR$. One example would be $left(dfraccos(u)sqrt1+u^2, dfracsin(u)sqrt1+u^2,dfracusqrt1+u^2, v right)$, where $-2pi leq u leq 2pi$ and $v in mathbbR$. Now, even though this is a 2-dimensional surface, I haven't found any way to really understand what it looks like. I'm aware this is kind of hard to answer, but any help is appreciated.




differential-geometry intuition surfaces visualization




share|cite|improve this question













I'm trying to "get a picture", so to speak, of hypersurfaces in $mathbbS^2 times mathbbR$. One example would be $left(dfraccos(u)sqrt1+u^2, dfracsin(u)sqrt1+u^2,dfracusqrt1+u^2, v right)$, where $-2pi leq u leq 2pi$ and $v in mathbbR$. Now, even though this is a 2-dimensional surface, I haven't found any way to really understand what it looks like. I'm aware this is kind of hard to answer, but any help is appreciated.





differential-geometry intuition surfaces visualization





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 31 at 16:42









John Ma

37.5k93669




37.5k93669









asked Jul 31 at 16:30









Matheus Andrade

587214




587214







  • 2




    You can also think about $S^2timesmathbbR$ as punctured 3-space, so you could probably get a pretty good picture of your surface.
    – Steve D
    Jul 31 at 16:33










  • $S^2timesBbb R$ is the normal line bundle on $S^2$
    – Andrea Mori
    Jul 31 at 16:36






  • 1




    Maybe try to plot it in 3d, using the 4th coordinate as color?
    – zokomoko
    Jul 31 at 16:37










  • This is a graph of your first three coordinates; your surface is just (this) x (real line).
    – Steve D
    Jul 31 at 17:05






  • 1




    That just maps your real line to the positive reals, so you don't get multiple ways to specify the same point.
    – Steve D
    Jul 31 at 20:12












  • 2




    You can also think about $S^2timesmathbbR$ as punctured 3-space, so you could probably get a pretty good picture of your surface.
    – Steve D
    Jul 31 at 16:33










  • $S^2timesBbb R$ is the normal line bundle on $S^2$
    – Andrea Mori
    Jul 31 at 16:36






  • 1




    Maybe try to plot it in 3d, using the 4th coordinate as color?
    – zokomoko
    Jul 31 at 16:37










  • This is a graph of your first three coordinates; your surface is just (this) x (real line).
    – Steve D
    Jul 31 at 17:05






  • 1




    That just maps your real line to the positive reals, so you don't get multiple ways to specify the same point.
    – Steve D
    Jul 31 at 20:12







2




2




You can also think about $S^2timesmathbbR$ as punctured 3-space, so you could probably get a pretty good picture of your surface.
– Steve D
Jul 31 at 16:33




You can also think about $S^2timesmathbbR$ as punctured 3-space, so you could probably get a pretty good picture of your surface.
– Steve D
Jul 31 at 16:33












$S^2timesBbb R$ is the normal line bundle on $S^2$
– Andrea Mori
Jul 31 at 16:36




$S^2timesBbb R$ is the normal line bundle on $S^2$
– Andrea Mori
Jul 31 at 16:36




1




1




Maybe try to plot it in 3d, using the 4th coordinate as color?
– zokomoko
Jul 31 at 16:37




Maybe try to plot it in 3d, using the 4th coordinate as color?
– zokomoko
Jul 31 at 16:37












This is a graph of your first three coordinates; your surface is just (this) x (real line).
– Steve D
Jul 31 at 17:05




This is a graph of your first three coordinates; your surface is just (this) x (real line).
– Steve D
Jul 31 at 17:05




1




1




That just maps your real line to the positive reals, so you don't get multiple ways to specify the same point.
– Steve D
Jul 31 at 20:12




That just maps your real line to the positive reals, so you don't get multiple ways to specify the same point.
– Steve D
Jul 31 at 20:12










1 Answer
1






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oldest

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up vote
1
down vote



accepted










Clearly $x^2+y^2+z^2=1$ over your surface, so this is a curve drawn on a $2$-sphere, times a real line. You could call this an infinite curtain.



The $z$-value of the curve keeps increasing, so this is a spring with varying radius$^star$, spiraling around the $z$-axis monotonically, times a real line.



$^star$The radius is $frac1sqrt1+4pi^2$ at the extremities, goes monotonically up to $1$ (when $z=0$), then symetrically back to $frac1sqrt1+4pi^2$






share|cite|improve this answer





















  • Can I do similar things to help me visualize these surfaces when my $z$ value is an arbitrary function $f(v)$? Let's say I wanted to picture the more complicated case: $$left(dfract^2sqrtt^4 + t^2 + e^2t,dfractsqrtt^4 + t^2 + e^2t, dfrace^tsqrtt^4 + t^2 + e^2t, v^2 + cos(v)log(v) right)$$. Is it doable?
    – Matheus Andrade
    Jul 31 at 16:56










  • @MatheusAndrade It is actually the exact same, all you need is to figure out the range of that function. Instead of a product with a real line, what you get is a product with that range. The reason is that the variable $v$ does not turn up in $x,y$ or $z$.
    – Arnaud Mortier
    Jul 31 at 17:12










  • I see. But how would I go about actually plotting it in something like Geogebra?
    – Matheus Andrade
    Jul 31 at 17:24










  • Oh... I think that's not possible really. Am I right?
    – Matheus Andrade
    Jul 31 at 17:27










  • You can't really plot $4$-dimensional things, in any case you would need a good deal of imagination.
    – Arnaud Mortier
    Jul 31 at 17:30










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1 Answer
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1 Answer
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active

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active

oldest

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up vote
1
down vote



accepted










Clearly $x^2+y^2+z^2=1$ over your surface, so this is a curve drawn on a $2$-sphere, times a real line. You could call this an infinite curtain.



The $z$-value of the curve keeps increasing, so this is a spring with varying radius$^star$, spiraling around the $z$-axis monotonically, times a real line.



$^star$The radius is $frac1sqrt1+4pi^2$ at the extremities, goes monotonically up to $1$ (when $z=0$), then symetrically back to $frac1sqrt1+4pi^2$






share|cite|improve this answer





















  • Can I do similar things to help me visualize these surfaces when my $z$ value is an arbitrary function $f(v)$? Let's say I wanted to picture the more complicated case: $$left(dfract^2sqrtt^4 + t^2 + e^2t,dfractsqrtt^4 + t^2 + e^2t, dfrace^tsqrtt^4 + t^2 + e^2t, v^2 + cos(v)log(v) right)$$. Is it doable?
    – Matheus Andrade
    Jul 31 at 16:56










  • @MatheusAndrade It is actually the exact same, all you need is to figure out the range of that function. Instead of a product with a real line, what you get is a product with that range. The reason is that the variable $v$ does not turn up in $x,y$ or $z$.
    – Arnaud Mortier
    Jul 31 at 17:12










  • I see. But how would I go about actually plotting it in something like Geogebra?
    – Matheus Andrade
    Jul 31 at 17:24










  • Oh... I think that's not possible really. Am I right?
    – Matheus Andrade
    Jul 31 at 17:27










  • You can't really plot $4$-dimensional things, in any case you would need a good deal of imagination.
    – Arnaud Mortier
    Jul 31 at 17:30














up vote
1
down vote



accepted










Clearly $x^2+y^2+z^2=1$ over your surface, so this is a curve drawn on a $2$-sphere, times a real line. You could call this an infinite curtain.



The $z$-value of the curve keeps increasing, so this is a spring with varying radius$^star$, spiraling around the $z$-axis monotonically, times a real line.



$^star$The radius is $frac1sqrt1+4pi^2$ at the extremities, goes monotonically up to $1$ (when $z=0$), then symetrically back to $frac1sqrt1+4pi^2$






share|cite|improve this answer





















  • Can I do similar things to help me visualize these surfaces when my $z$ value is an arbitrary function $f(v)$? Let's say I wanted to picture the more complicated case: $$left(dfract^2sqrtt^4 + t^2 + e^2t,dfractsqrtt^4 + t^2 + e^2t, dfrace^tsqrtt^4 + t^2 + e^2t, v^2 + cos(v)log(v) right)$$. Is it doable?
    – Matheus Andrade
    Jul 31 at 16:56










  • @MatheusAndrade It is actually the exact same, all you need is to figure out the range of that function. Instead of a product with a real line, what you get is a product with that range. The reason is that the variable $v$ does not turn up in $x,y$ or $z$.
    – Arnaud Mortier
    Jul 31 at 17:12










  • I see. But how would I go about actually plotting it in something like Geogebra?
    – Matheus Andrade
    Jul 31 at 17:24










  • Oh... I think that's not possible really. Am I right?
    – Matheus Andrade
    Jul 31 at 17:27










  • You can't really plot $4$-dimensional things, in any case you would need a good deal of imagination.
    – Arnaud Mortier
    Jul 31 at 17:30












up vote
1
down vote



accepted







up vote
1
down vote



accepted






Clearly $x^2+y^2+z^2=1$ over your surface, so this is a curve drawn on a $2$-sphere, times a real line. You could call this an infinite curtain.



The $z$-value of the curve keeps increasing, so this is a spring with varying radius$^star$, spiraling around the $z$-axis monotonically, times a real line.



$^star$The radius is $frac1sqrt1+4pi^2$ at the extremities, goes monotonically up to $1$ (when $z=0$), then symetrically back to $frac1sqrt1+4pi^2$






share|cite|improve this answer













Clearly $x^2+y^2+z^2=1$ over your surface, so this is a curve drawn on a $2$-sphere, times a real line. You could call this an infinite curtain.



The $z$-value of the curve keeps increasing, so this is a spring with varying radius$^star$, spiraling around the $z$-axis monotonically, times a real line.



$^star$The radius is $frac1sqrt1+4pi^2$ at the extremities, goes monotonically up to $1$ (when $z=0$), then symetrically back to $frac1sqrt1+4pi^2$







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 31 at 16:37









Arnaud Mortier

18k21958




18k21958











  • Can I do similar things to help me visualize these surfaces when my $z$ value is an arbitrary function $f(v)$? Let's say I wanted to picture the more complicated case: $$left(dfract^2sqrtt^4 + t^2 + e^2t,dfractsqrtt^4 + t^2 + e^2t, dfrace^tsqrtt^4 + t^2 + e^2t, v^2 + cos(v)log(v) right)$$. Is it doable?
    – Matheus Andrade
    Jul 31 at 16:56










  • @MatheusAndrade It is actually the exact same, all you need is to figure out the range of that function. Instead of a product with a real line, what you get is a product with that range. The reason is that the variable $v$ does not turn up in $x,y$ or $z$.
    – Arnaud Mortier
    Jul 31 at 17:12










  • I see. But how would I go about actually plotting it in something like Geogebra?
    – Matheus Andrade
    Jul 31 at 17:24










  • Oh... I think that's not possible really. Am I right?
    – Matheus Andrade
    Jul 31 at 17:27










  • You can't really plot $4$-dimensional things, in any case you would need a good deal of imagination.
    – Arnaud Mortier
    Jul 31 at 17:30
















  • Can I do similar things to help me visualize these surfaces when my $z$ value is an arbitrary function $f(v)$? Let's say I wanted to picture the more complicated case: $$left(dfract^2sqrtt^4 + t^2 + e^2t,dfractsqrtt^4 + t^2 + e^2t, dfrace^tsqrtt^4 + t^2 + e^2t, v^2 + cos(v)log(v) right)$$. Is it doable?
    – Matheus Andrade
    Jul 31 at 16:56










  • @MatheusAndrade It is actually the exact same, all you need is to figure out the range of that function. Instead of a product with a real line, what you get is a product with that range. The reason is that the variable $v$ does not turn up in $x,y$ or $z$.
    – Arnaud Mortier
    Jul 31 at 17:12










  • I see. But how would I go about actually plotting it in something like Geogebra?
    – Matheus Andrade
    Jul 31 at 17:24










  • Oh... I think that's not possible really. Am I right?
    – Matheus Andrade
    Jul 31 at 17:27










  • You can't really plot $4$-dimensional things, in any case you would need a good deal of imagination.
    – Arnaud Mortier
    Jul 31 at 17:30















Can I do similar things to help me visualize these surfaces when my $z$ value is an arbitrary function $f(v)$? Let's say I wanted to picture the more complicated case: $$left(dfract^2sqrtt^4 + t^2 + e^2t,dfractsqrtt^4 + t^2 + e^2t, dfrace^tsqrtt^4 + t^2 + e^2t, v^2 + cos(v)log(v) right)$$. Is it doable?
– Matheus Andrade
Jul 31 at 16:56




Can I do similar things to help me visualize these surfaces when my $z$ value is an arbitrary function $f(v)$? Let's say I wanted to picture the more complicated case: $$left(dfract^2sqrtt^4 + t^2 + e^2t,dfractsqrtt^4 + t^2 + e^2t, dfrace^tsqrtt^4 + t^2 + e^2t, v^2 + cos(v)log(v) right)$$. Is it doable?
– Matheus Andrade
Jul 31 at 16:56












@MatheusAndrade It is actually the exact same, all you need is to figure out the range of that function. Instead of a product with a real line, what you get is a product with that range. The reason is that the variable $v$ does not turn up in $x,y$ or $z$.
– Arnaud Mortier
Jul 31 at 17:12




@MatheusAndrade It is actually the exact same, all you need is to figure out the range of that function. Instead of a product with a real line, what you get is a product with that range. The reason is that the variable $v$ does not turn up in $x,y$ or $z$.
– Arnaud Mortier
Jul 31 at 17:12












I see. But how would I go about actually plotting it in something like Geogebra?
– Matheus Andrade
Jul 31 at 17:24




I see. But how would I go about actually plotting it in something like Geogebra?
– Matheus Andrade
Jul 31 at 17:24












Oh... I think that's not possible really. Am I right?
– Matheus Andrade
Jul 31 at 17:27




Oh... I think that's not possible really. Am I right?
– Matheus Andrade
Jul 31 at 17:27












You can't really plot $4$-dimensional things, in any case you would need a good deal of imagination.
– Arnaud Mortier
Jul 31 at 17:30




You can't really plot $4$-dimensional things, in any case you would need a good deal of imagination.
– Arnaud Mortier
Jul 31 at 17:30












 

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