Mixing
When is the intersection of the hyperboloid with the sphere transverse?
For which values of $a$ does the hyperboloid $x^2+y^2-z^2=1$ intersect
the sphere $x^2+y^2+z^2=a$ transversally? What does the intersection
look like for different values of $a$?
What I can see is that if $sqrt a<1, $ then the intersection is empty and hence transverse. If $sqrt a=1iff a=1$, then the intersection points lie on the circle $x^2+y^2=1$ on the $z=0$ plane. I guess at those points the tangent spaces to the hyperboloid and the sphere coincide (they are 2-dim planes). Is that correct? How do I show it more rigorously? (I can use local parametrizations of those manifolds and then compute the image of the corresponding differentials, but this seems to be a huge hassle.) As for the case $sqrt a > 1$, I guess here the intersection is transverse, but again, how do I show it?
manifolds differential-topology smooth-manifolds transversality
asked Feb 20 '18 at 2:18
user500094
592212
|
show 9 more comments
For which values of $a$ does the hyperboloid $x^2+y^2-z^2=1$ intersect
the sphere $x^2+y^2+z^2=a$ transversally? What does the intersection
look like for different values of $a$?
What I can see is that if $sqrt a<1, $ then the intersection is empty and hence transverse. If $sqrt a=1iff a=1$, then the intersection points lie on the circle $x^2+y^2=1$ on the $z=0$ plane. I guess at those points the tangent spaces to the hyperboloid and the sphere coincide (they are 2-dim planes). Is that correct? How do I show it more rigorously? (I can use local parametrizations of those manifolds and then compute the image of the corresponding differentials, but this seems to be a huge hassle.) As for the case $sqrt a > 1$, I guess here the intersection is transverse, but again, how do I show it?
manifolds differential-topology smooth-manifolds transversality
asked Feb 20 '18 at 2:18
user500094
592212
1
Think about normal vectors! Tangent spaces coincide precisely when normal vectors are parallel.
– Ted Shifrin
Feb 20 '18 at 2:25
1
But why is this helpful? I imagine normal vectors as vectors orthogonal to tangent spaces, so it seems to me that in formal proofs everything boils down to tangent spaces in any case. Besides, Guillemin and Pollack don't even define normal vectors (at least I don't remember them talking about normal vectors).
– user500094
Feb 20 '18 at 2:30
1
No, they talk about the dual notion: Namely, $ker(df_x) = T_x M$ for $M=f^{-1}(c)$ a level set of $f$ (for a regular value $c$). So think about $df_x$ and $dg_x$ ...
– Ted Shifrin
Feb 20 '18 at 2:46
@TedShifrin So in the case $a=1$, the differentials of $f$ and $g$ at $x$ (where $x$ is any intersection point) have the same kernels (since a point in the intersection is of the form $(x,y,0)$), and thus the intersection is not transverse. In the case $a > 1$, the kernels will be different, and each of them will be 2-dimensional, so the intersection will be transverse?
– user500094
Feb 20 '18 at 18:14
This is correct. :)
– Ted Shifrin
Feb 20 '18 at 18:16
|
show 9 more comments
For which values of $a$ does the hyperboloid $x^2+y^2-z^2=1$ intersect
the sphere $x^2+y^2+z^2=a$ transversally? What does the intersection
look like for different values of $a$?
What I can see is that if $sqrt a<1, $ then the intersection is empty and hence transverse. If $sqrt a=1iff a=1$, then the intersection points lie on the circle $x^2+y^2=1$ on the $z=0$ plane. I guess at those points the tangent spaces to the hyperboloid and the sphere coincide (they are 2-dim planes). Is that correct? How do I show it more rigorously? (I can use local parametrizations of those manifolds and then compute the image of the corresponding differentials, but this seems to be a huge hassle.) As for the case $sqrt a > 1$, I guess here the intersection is transverse, but again, how do I show it?
manifolds differential-topology smooth-manifolds transversality
asked Feb 20 '18 at 2:18
user500094
592212
For which values of $a$ does the hyperboloid $x^2+y^2-z^2=1$ intersect
the sphere $x^2+y^2+z^2=a$ transversally? What does the intersection
look like for different values of $a$?
What I can see is that if $sqrt a<1, $ then the intersection is empty and hence transverse. If $sqrt a=1iff a=1$, then the intersection points lie on the circle $x^2+y^2=1$ on the $z=0$ plane. I guess at those points the tangent spaces to the hyperboloid and the sphere coincide (they are 2-dim planes). Is that correct? How do I show it more rigorously? (I can use local parametrizations of those manifolds and then compute the image of the corresponding differentials, but this seems to be a huge hassle.) As for the case $sqrt a > 1$, I guess here the intersection is transverse, but again, how do I show it?
manifolds differential-topology smooth-manifolds transversality
manifolds differential-topology smooth-manifolds transversality
asked Feb 20 '18 at 2:18
user500094
592212
asked Feb 20 '18 at 2:18
user500094
592212
asked Feb 20 '18 at 2:18
user500094
592212
asked Feb 20 '18 at 2:18
user500094
592212
asked Feb 20 '18 at 2:18
user500094
592212
592212
1
Think about normal vectors! Tangent spaces coincide precisely when normal vectors are parallel.
– Ted Shifrin
Feb 20 '18 at 2:25
1
But why is this helpful? I imagine normal vectors as vectors orthogonal to tangent spaces, so it seems to me that in formal proofs everything boils down to tangent spaces in any case. Besides, Guillemin and Pollack don't even define normal vectors (at least I don't remember them talking about normal vectors).
– user500094
Feb 20 '18 at 2:30
1
No, they talk about the dual notion: Namely, $ker(df_x) = T_x M$ for $M=f^{-1}(c)$ a level set of $f$ (for a regular value $c$). So think about $df_x$ and $dg_x$ ...
– Ted Shifrin
Feb 20 '18 at 2:46
@TedShifrin So in the case $a=1$, the differentials of $f$ and $g$ at $x$ (where $x$ is any intersection point) have the same kernels (since a point in the intersection is of the form $(x,y,0)$), and thus the intersection is not transverse. In the case $a > 1$, the kernels will be different, and each of them will be 2-dimensional, so the intersection will be transverse?
– user500094
Feb 20 '18 at 18:14
This is correct. :)
– Ted Shifrin
Feb 20 '18 at 18:16
|
show 9 more comments
1
Think about normal vectors! Tangent spaces coincide precisely when normal vectors are parallel.
– Ted Shifrin
Feb 20 '18 at 2:25
1
But why is this helpful? I imagine normal vectors as vectors orthogonal to tangent spaces, so it seems to me that in formal proofs everything boils down to tangent spaces in any case. Besides, Guillemin and Pollack don't even define normal vectors (at least I don't remember them talking about normal vectors).
– user500094
Feb 20 '18 at 2:30
1
No, they talk about the dual notion: Namely, $ker(df_x) = T_x M$ for $M=f^{-1}(c)$ a level set of $f$ (for a regular value $c$). So think about $df_x$ and $dg_x$ ...
– Ted Shifrin
Feb 20 '18 at 2:46
@TedShifrin So in the case $a=1$, the differentials of $f$ and $g$ at $x$ (where $x$ is any intersection point) have the same kernels (since a point in the intersection is of the form $(x,y,0)$), and thus the intersection is not transverse. In the case $a > 1$, the kernels will be different, and each of them will be 2-dimensional, so the intersection will be transverse?
– user500094
Feb 20 '18 at 18:14
This is correct. :)
– Ted Shifrin
Feb 20 '18 at 18:16
1
1
Think about normal vectors! Tangent spaces coincide precisely when normal vectors are parallel.
– Ted Shifrin
Feb 20 '18 at 2:25
Think about normal vectors! Tangent spaces coincide precisely when normal vectors are parallel.
– Ted Shifrin
Feb 20 '18 at 2:25
1
1
But why is this helpful? I imagine normal vectors as vectors orthogonal to tangent spaces, so it seems to me that in formal proofs everything boils down to tangent spaces in any case. Besides, Guillemin and Pollack don't even define normal vectors (at least I don't remember them talking about normal vectors).
– user500094
Feb 20 '18 at 2:30
But why is this helpful? I imagine normal vectors as vectors orthogonal to tangent spaces, so it seems to me that in formal proofs everything boils down to tangent spaces in any case. Besides, Guillemin and Pollack don't even define normal vectors (at least I don't remember them talking about normal vectors).
– user500094
Feb 20 '18 at 2:30
1
1
No, they talk about the dual notion: Namely, $ker(df_x) = T_x M$ for $M=f^{-1}(c)$ a level set of $f$ (for a regular value $c$). So think about $df_x$ and $dg_x$ ...
– Ted Shifrin
Feb 20 '18 at 2:46
No, they talk about the dual notion: Namely, $ker(df_x) = T_x M$ for $M=f^{-1}(c)$ a level set of $f$ (for a regular value $c$). So think about $df_x$ and $dg_x$ ...
– Ted Shifrin
Feb 20 '18 at 2:46
@TedShifrin So in the case $a=1$, the differentials of $f$ and $g$ at $x$ (where $x$ is any intersection point) have the same kernels (since a point in the intersection is of the form $(x,y,0)$), and thus the intersection is not transverse. In the case $a > 1$, the kernels will be different, and each of them will be 2-dimensional, so the intersection will be transverse?
– user500094
Feb 20 '18 at 18:14
@TedShifrin So in the case $a=1$, the differentials of $f$ and $g$ at $x$ (where $x$ is any intersection point) have the same kernels (since a point in the intersection is of the form $(x,y,0)$), and thus the intersection is not transverse. In the case $a > 1$, the kernels will be different, and each of them will be 2-dimensional, so the intersection will be transverse?
– user500094
Feb 20 '18 at 18:14
This is correct. :)
– Ted Shifrin
Feb 20 '18 at 18:16
This is correct. :)
– Ted Shifrin
Feb 20 '18 at 18:16
|
show 9 more comments
1 Answer
1
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For $a=1$ we obtain $$x^2+y^2=1\z=0$$therefore the intersection is the unit circle on XY plane centered at (0,0). Generally for $a>1$ we have $$z^2+1=a-z^2$$ therefore$$z=pm sqrt {a-1over 2}\ x^2+y^2={a+1over 2}$$
Also for a 2 sheet hyperboloid for example for $x^2+y^2-z^2=-1$ we must have $$z^2-1=a-z^2$$ or $z^2={a+1over 2}$ in which $age -1$. If so, we have $$x^2+y^2=z^2-1={a-1over 2}$$which is possible only if $age 1$.
Conclusion: the intersection of the two planes contains two unit circles parallel to XY plane centered at $(0,0,pmsqrt{a-1over 2})$ in the former case and $(0,0,pmsqrt{a+1over 2})$ in the latter case
Here is an image (simulated by MATLAB) indicating the two sheet hyperboloid case:
answered Nov 21 '18 at 18:01
Mostafa Ayaz
14.1k3937
what about the use of the Kernel mentioned in the above comments?
– hopefully
Nov 21 '18 at 18:18
Actually i don't know what Kernel is. Can you elaborate on it a little?
– Mostafa Ayaz
Nov 21 '18 at 18:20
why it is not transversal if $a=1$ as mentioned in the above comments?
– hopefully
Nov 21 '18 at 19:46
what if it is a 2 sheet hyperboloid?
– hopefully
Nov 22 '18 at 15:27
1
I added some more details. Soon I will add some images..
– Mostafa Ayaz
Nov 22 '18 at 16:32
|
show 2 more comments
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For $a=1$ we obtain $$x^2+y^2=1\z=0$$therefore the intersection is the unit circle on XY plane centered at (0,0). Generally for $a>1$ we have $$z^2+1=a-z^2$$ therefore$$z=pm sqrt {a-1over 2}\ x^2+y^2={a+1over 2}$$
Also for a 2 sheet hyperboloid for example for $x^2+y^2-z^2=-1$ we must have $$z^2-1=a-z^2$$ or $z^2={a+1over 2}$ in which $age -1$. If so, we have $$x^2+y^2=z^2-1={a-1over 2}$$which is possible only if $age 1$.
Conclusion: the intersection of the two planes contains two unit circles parallel to XY plane centered at $(0,0,pmsqrt{a-1over 2})$ in the former case and $(0,0,pmsqrt{a+1over 2})$ in the latter case
Here is an image (simulated by MATLAB) indicating the two sheet hyperboloid case:
answered Nov 21 '18 at 18:01
Mostafa Ayaz
14.1k3937
what about the use of the Kernel mentioned in the above comments?
– hopefully
Nov 21 '18 at 18:18
Actually i don't know what Kernel is. Can you elaborate on it a little?
– Mostafa Ayaz
Nov 21 '18 at 18:20
why it is not transversal if $a=1$ as mentioned in the above comments?
– hopefully
Nov 21 '18 at 19:46
what if it is a 2 sheet hyperboloid?
– hopefully
Nov 22 '18 at 15:27
1
I added some more details. Soon I will add some images..
– Mostafa Ayaz
Nov 22 '18 at 16:32
|
show 2 more comments
For $a=1$ we obtain $$x^2+y^2=1\z=0$$therefore the intersection is the unit circle on XY plane centered at (0,0). Generally for $a>1$ we have $$z^2+1=a-z^2$$ therefore$$z=pm sqrt {a-1over 2}\ x^2+y^2={a+1over 2}$$
Also for a 2 sheet hyperboloid for example for $x^2+y^2-z^2=-1$ we must have $$z^2-1=a-z^2$$ or $z^2={a+1over 2}$ in which $age -1$. If so, we have $$x^2+y^2=z^2-1={a-1over 2}$$which is possible only if $age 1$.
Conclusion: the intersection of the two planes contains two unit circles parallel to XY plane centered at $(0,0,pmsqrt{a-1over 2})$ in the former case and $(0,0,pmsqrt{a+1over 2})$ in the latter case
Here is an image (simulated by MATLAB) indicating the two sheet hyperboloid case:
answered Nov 21 '18 at 18:01
Mostafa Ayaz
14.1k3937
what about the use of the Kernel mentioned in the above comments?
– hopefully
Nov 21 '18 at 18:18
Actually i don't know what Kernel is. Can you elaborate on it a little?
– Mostafa Ayaz
Nov 21 '18 at 18:20
why it is not transversal if $a=1$ as mentioned in the above comments?
– hopefully
Nov 21 '18 at 19:46
what if it is a 2 sheet hyperboloid?
– hopefully
Nov 22 '18 at 15:27
1
I added some more details. Soon I will add some images..
– Mostafa Ayaz
Nov 22 '18 at 16:32
|
show 2 more comments
For $a=1$ we obtain $$x^2+y^2=1\z=0$$therefore the intersection is the unit circle on XY plane centered at (0,0). Generally for $a>1$ we have $$z^2+1=a-z^2$$ therefore$$z=pm sqrt {a-1over 2}\ x^2+y^2={a+1over 2}$$
Also for a 2 sheet hyperboloid for example for $x^2+y^2-z^2=-1$ we must have $$z^2-1=a-z^2$$ or $z^2={a+1over 2}$ in which $age -1$. If so, we have $$x^2+y^2=z^2-1={a-1over 2}$$which is possible only if $age 1$.
Conclusion: the intersection of the two planes contains two unit circles parallel to XY plane centered at $(0,0,pmsqrt{a-1over 2})$ in the former case and $(0,0,pmsqrt{a+1over 2})$ in the latter case
Here is an image (simulated by MATLAB) indicating the two sheet hyperboloid case:
answered Nov 21 '18 at 18:01
Mostafa Ayaz
14.1k3937
For $a=1$ we obtain $$x^2+y^2=1\z=0$$therefore the intersection is the unit circle on XY plane centered at (0,0). Generally for $a>1$ we have $$z^2+1=a-z^2$$ therefore$$z=pm sqrt {a-1over 2}\ x^2+y^2={a+1over 2}$$
Also for a 2 sheet hyperboloid for example for $x^2+y^2-z^2=-1$ we must have $$z^2-1=a-z^2$$ or $z^2={a+1over 2}$ in which $age -1$. If so, we have $$x^2+y^2=z^2-1={a-1over 2}$$which is possible only if $age 1$.
Conclusion: the intersection of the two planes contains two unit circles parallel to XY plane centered at $(0,0,pmsqrt{a-1over 2})$ in the former case and $(0,0,pmsqrt{a+1over 2})$ in the latter case
Here is an image (simulated by MATLAB) indicating the two sheet hyperboloid case:
answered Nov 21 '18 at 18:01
Mostafa Ayaz
14.1k3937
edited Nov 22 '18 at 16:41
answered Nov 21 '18 at 18:01
Mostafa Ayaz
14.1k3937
answered Nov 21 '18 at 18:01
Mostafa Ayaz
14.1k3937
answered Nov 21 '18 at 18:01
Mostafa Ayaz
14.1k3937
14.1k3937
what about the use of the Kernel mentioned in the above comments?
– hopefully
Nov 21 '18 at 18:18
Actually i don't know what Kernel is. Can you elaborate on it a little?
– Mostafa Ayaz
Nov 21 '18 at 18:20
why it is not transversal if $a=1$ as mentioned in the above comments?
– hopefully
Nov 21 '18 at 19:46
what if it is a 2 sheet hyperboloid?
– hopefully
Nov 22 '18 at 15:27
1
I added some more details. Soon I will add some images..
– Mostafa Ayaz
Nov 22 '18 at 16:32
|
show 2 more comments
what about the use of the Kernel mentioned in the above comments?
– hopefully
Nov 21 '18 at 18:18
Actually i don't know what Kernel is. Can you elaborate on it a little?
– Mostafa Ayaz
Nov 21 '18 at 18:20
why it is not transversal if $a=1$ as mentioned in the above comments?
– hopefully
Nov 21 '18 at 19:46
what if it is a 2 sheet hyperboloid?
– hopefully
Nov 22 '18 at 15:27
1
I added some more details. Soon I will add some images..
– Mostafa Ayaz
Nov 22 '18 at 16:32
what about the use of the Kernel mentioned in the above comments?
– hopefully
Nov 21 '18 at 18:18
what about the use of the Kernel mentioned in the above comments?
– hopefully
Nov 21 '18 at 18:18
Actually i don't know what Kernel is. Can you elaborate on it a little?
– Mostafa Ayaz
Nov 21 '18 at 18:20
Actually i don't know what Kernel is. Can you elaborate on it a little?
– Mostafa Ayaz
Nov 21 '18 at 18:20
why it is not transversal if $a=1$ as mentioned in the above comments?
– hopefully
Nov 21 '18 at 19:46
why it is not transversal if $a=1$ as mentioned in the above comments?
– hopefully
Nov 21 '18 at 19:46
what if it is a 2 sheet hyperboloid?
– hopefully
Nov 22 '18 at 15:27
what if it is a 2 sheet hyperboloid?
– hopefully
Nov 22 '18 at 15:27
1
1
I added some more details. Soon I will add some images..
– Mostafa Ayaz
Nov 22 '18 at 16:32
I added some more details. Soon I will add some images..
– Mostafa Ayaz
Nov 22 '18 at 16:32
|
show 2 more comments
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1
Think about normal vectors! Tangent spaces coincide precisely when normal vectors are parallel.
– Ted Shifrin
Feb 20 '18 at 2:25
1
But why is this helpful? I imagine normal vectors as vectors orthogonal to tangent spaces, so it seems to me that in formal proofs everything boils down to tangent spaces in any case. Besides, Guillemin and Pollack don't even define normal vectors (at least I don't remember them talking about normal vectors).
– user500094
Feb 20 '18 at 2:30
1
No, they talk about the dual notion: Namely, $ker(df_x) = T_x M$ for $M=f^{-1}(c)$ a level set of $f$ (for a regular value $c$). So think about $df_x$ and $dg_x$ ...
– Ted Shifrin
Feb 20 '18 at 2:46
@TedShifrin So in the case $a=1$, the differentials of $f$ and $g$ at $x$ (where $x$ is any intersection point) have the same kernels (since a point in the intersection is of the form $(x,y,0)$), and thus the intersection is not transverse. In the case $a > 1$, the kernels will be different, and each of them will be 2-dimensional, so the intersection will be transverse?
– user500094
Feb 20 '18 at 18:14
This is correct. :)
– Ted Shifrin
Feb 20 '18 at 18:16