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Show that the manifold $N=\{(x,y,z)\in \mathbb{R}^3|x^2+y^2= 4\}$ is transverse to M. Identify the resulting manifold $N\cap M$.

My Attempt: Pardon me for something vacuous, as I am a beginner in differential topology.

All I know is that it's intersection is union of two disjoint circles. It's clear to me that $M$ is the equation of a torus. Now, substituting the value $x^2+y^2=4$ in the equation $M$, we get two circles each $x^2+y^2=4$ one at $z=1$ and the other at $z=-1$. By definition of transversality,we should have $T_p(M)+T_{p}(N)=T_{p}\mathbb{R}^3\simeq\mathbb{R}^3,\forall p\in M\cap N$. But then I don't understand how the tangent spaces spans $\mathbb{R}^3$ being parallel(I guess)??

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  • $\begingroup$ You have the tangent space of the cylinder and the tangent space of the torus. Each should be two dimensional. If they are transverse, then the two tangent spaces will cross eachother in an X-like way. That's what transverse means. In which case, two vectors from one of the tangent spaces and a third from the other should span 3 dimensions. $\endgroup$ – Prototank Mar 14 at 13:15
  • $\begingroup$ got you! Thanks :) $\endgroup$ – Infinity Mar 14 at 17:34
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To find whether or not intersection of these surfaces $T^2=\lbrace (x,y,z): (2-\sqrt{x^2+y^2})+z^2=1\rbrace$ and $C=\lbrace (x,y,z): x^2+y^2=4\rbrace$ is transverse, lets first look at the intersection. So

\begin{align}T^2\cap C&=\lbrace\ (x,y,z): x^2+y^2=4\text{ and }(2-\sqrt{x^2+y^2})+z^2=1\rbrace\\ &=\lbrace\ (x,y,z): x^2+y^2=4\text{ and }(2-\sqrt{4})+z^2=1\rbrace\\ &=\lbrace\ (x,y,z): x^2+y^2=4\text{ and } z^2=1\rbrace \end{align} so they intersect in circles of radius 2 at heights $z=\pm 1$. Now suppose that $(x,y,1)$ is on the top circle and $x$ is positive. Then we can locally parametrize $T^2$ near $x$ with $\phi$ by solving for $z$ and choosing the positive branch. This gives $\phi:U\to\mathbb{R}^3$ by $$\phi(x,y)=\left(x,y,\sqrt{1-(2-|r|)^2}\right)\quad (\text{where $r^2=x^2+y^2$})$$ Let $\psi:V\to\mathbb{R}^3$ be the local parametrization of $C$ around $(x,y,1)$ given by $$\psi(x,z)=(x,\sqrt{4-x^2},z)$$ Then the tangent space of $T^2$ at $(x,y,1)$ is the image of $d\phi_{(x,y)}$ and the tangent space of $C$ at $(x,y,1)$ is the image of $d\psi_{(x,1)}$. These are $$d\phi_{(x,y)}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \frac{x(2-|r|)}{|r|\sqrt{1-(2-|r|)^2}} & \frac{y(2-|r|)}{|r|\sqrt{1-(2-|r|)^2}} \\ \end{bmatrix}$$ and $$d\psi_{(x,1)}=\begin{bmatrix} 1 & 0 \\ \frac{-x}{\sqrt{1-x^2}} & 0 \\ 0 & 1 \\ \end{bmatrix}$$ But remember that $x^2+y^2=4$ for $(x,y,1)$. That is, $r=2$. So the above $d\phi_{(x,y)}$ becomes $$d\phi_{(x,y)}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0\\ \end{bmatrix}$$ So the spanning set is $\lbrace d\phi_{(x,y)}(\vec{i}), d\phi_{(x,y)}(\vec{j}), d\psi_{(x,1)}(\vec{j})\rbrace=\lbrace\vec{i},\vec{j},\vec{k}\rbrace$

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