# 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 $\sqrt a=1\iff a=1$, 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?

• Think about normal vectors! Tangent spaces coincide precisely when normal vectors are parallel. Feb 20 '18 at 2:25
• 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). Feb 20 '18 at 2:30
• 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 $dg_x$ ... 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? Feb 20 '18 at 18:14
• This is correct. :) Feb 20 '18 at 18:16

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$$z=\pm \sqrt {a-1\over 2}\\ x^2+y^2={a+1\over 2}$$

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 $$z^2={a+1\over 2}$$ in which $$a\ge -1$$. If so, we have $$x^2+y^2=z^2-1={a-1\over 2}$$which is possible only if $$a\ge 1$$.

Conclusion: the intersection of the two planes contains two unit circles parallel to XY plane centered at $$(0,0,\pm\sqrt{a-1\over 2})$$ in the former case and $$(0,0,\pm\sqrt{a+1\over 2})$$ in the latter case

Here is an image (simulated by MATLAB) indicating the two sheet hyperboloid case:

• what about the use of the Kernel mentioned in the above comments? Nov 21 '18 at 18:18
• Actually i don't know what Kernel is. Can you elaborate on it a little? Nov 21 '18 at 18:20
• why it is not transversal if $a=1$ as mentioned in the above comments? Nov 21 '18 at 19:46
• what if it is a 2 sheet hyperboloid? Nov 22 '18 at 15:27
• I added some more details. Soon I will add some images.. Nov 22 '18 at 16:32