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I got problems to understand a solution of the following problem.

Problem:

Let $d \in \mathbb N$, be $B_d := \{z \in \mathbb C^2|~z_1^d+z_2^2 = 0\}$ and $S^3 \subset \mathbb C^2\subset \mathbb R^4$ the unit sphere. Show that $B_d \cap S^3$ is a smooth manifold.

Solution:

$\phi:~(z_1,z_2)\mapsto z_1^d + z_2^2$ and therefore $B_d\subset \mathbb C^2$ has complex dimension 1 and real dimension 2 (1).

Since $S^3\subset \mathbb R^4$ has real dimension 3, the real dimension of the tangent-spaces are $dim_{\mathbb R}(TS^3) = 3$ and $dim_{\mathbb R}(TB_d) = 2$. So, it remains to check that $dim_{\mathbb R}(TS^3\cap TB_d) = 1$ (2). This means $TB_d\not\subset TS^3$. (3)

Since $TB_d \cong ker(D\phi)$, one finds that $T_zB_d =\lambda \begin{pmatrix} 2z_2\\-dz_1^{d-1} \end{pmatrix} $ and therefore $T_zB \not\subset T_zS^3 \Rightarrow \left<\lambda \begin{pmatrix} 2z_2\\-dz_1^{d-1} \end{pmatrix} ,\begin{pmatrix}z_1\\z_2\end{pmatrix} \right>\neq 0$. (4)

Questions

  1. Why is $B_d$ a smooth manifold since $0$ is not a regular value of $\phi$ ? The differential is vanishing at $(0,0)$.
  2. Why do I have to check that $dim_{\mathbb R}(TS^3\cap TB_d) = 1$? What is the stronger statement behind this?
  3. And why does $TB_d\not\subset TS^3$ imply that the dimension of the intersection has to be $1$?
  4. Why is this statement true? And what do i compare here? $T_zB_d$ and the normal-vector field on $S^3$?
  5. The last question: How do I proceed from here? In the exercise session they said it is trivial but I don't see it.

Thanks for your help!

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    If you consider the map $f:\Bbb C^2\to \Bbb C\times \Bbb R$, $(z_1,z_2)\mapsto (z_1^d+z_2^2,|z_1|^2+|z_2|^2)$, you can can check that $S^3\cap B_d=f^{-1}(0,1)$ and that the differential is surjective for every point in $S^3\cap B_d$---this avoids the potential problems at $(0,0)$ and you can compute the differential explicitly. – 2017-02-12
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    @Danu: Ah, thanks! This makes things way easier! But what about the provided solution? I still don't understand the different steps.. – 2017-02-12

1 Answers 1

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I think that a clear solution of the problem has been already given in the comments. I want only to say a few words about question number 2.

The general principle behind this thing is called transversality. Suppose that $A$ and $B$ are two smooth submanifolds of a given manifold $M$. We say that $A$ and $B$ are transversal in $M$ (and we write $A \pitchfork B $) if $$T_pM = T_pA+ T_pB$$ at each $p \in A \cap B$. It is an easy consequence of the implicit function theorem (exercise!) that if $A\pitchfork B$ then $A \cap B$ is a smooth submanifold of $M$. Furthermore, the Grassmann formula is satisfied $$\text{dim}(M)=\text{dim}(A)+\text{dim}(B)-\text{dim}(A \cap B) \ .$$

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    If you feel interested in the construction of the exercise there is an entire book about this: http://press.princeton.edu/titles/1570.html It is my favourite book!! – 2017-02-12
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    Ok. Now i get it! This is why the dimension of the intersection has to be one. Thank you! – 2017-02-13