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Let $f\colon N^n\to M^{2n}$ be an immersion. Then, we can extend $f$ to $\bar{f}\colon E(\nu_g)\to M$ of the total space of the normal bundle.

Let $s_0\colon N\to E(\nu_g)$ be a zero section and $s_1$ be the section which is zero at only a finite number of points. Define an isotopy $s_t\colon N\to E(\nu_g)$ between $s_0$ and $s_1$.

Let $f_t=\bar{f}\circ s_t\colon N\to M$. Then, $f_0=f$ by our construction.

Suppose that $f_0(x)=f_0(y)$ for $x\neq y$ is it ture that there exists a unique $y_1\in N$ in the neighborhood of $y\in N$ such that $f_0(x)=f_0(y)=f_1(y_1)$?

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  • $\begingroup$ Either there is a misprint somewhere, or else (i) you never used your isotopy, and (ii) you're asking whether every double point of an immersion is also an intersection point of the immersion with its arbitrary generic pushoff, moreover a rather special intersection point. This would be a really strange question. My guess is that your last formula should read "$f_0(x)=f_1(y_1)$". That would be a reasonable question, if only not "too localized". $\endgroup$
    – Sergey Melikhov
    Apr 25, 2011 at 14:22
  • $\begingroup$ I agree with Sergey that there must be something amiss with the question as my example below shows. I think one expects a pair of intersections between $f_1$ and $f_0$. The double point of a generic immersion will be as $(x_1,\ldots, x_n, 0, \ldots, 0)$ $\cup$ $(0,\ldots, 0, x_{n+1}, \ldots, x_{2n})$. The pushoff in this nbhd can be parametrized as $(x_1,\ldots, x_n, -1, \ldots, -1)$ $\cup$ $(1,\ldots, 1, x_{n+1}, \ldots, x_{2n})$. $\endgroup$
    – Scott Carter
    Apr 25, 2011 at 14:54
  • $\begingroup$ The answer is clearly no as stated. But perhaps if you added some transversality assumptions (such as $f_0$ transverse to $f_1$) and restrict the size of your isotopy (so that $f_0(x)$ is sufficiently close to $f_1(x)$ in some metric) it may be true. $\endgroup$
    – Mark Grant
    Apr 25, 2011 at 17:39
  • $\begingroup$ Also, I think you need to say that your immersion of the normal bundle is injective on fibres? $\endgroup$
    – Mark Grant
    Apr 27, 2011 at 14:17

1 Answer 1

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No. Let $n=1$, and consider a figure 8, which for definiteness, has its top lobe in the 2nd quadrant and its bottom lobe in the 4th quadrant, with a double point at the origin. In a nbhd of the origin, the curve runs along the coordinate axes. Push the curve off the $y$-axis in the positive $x$ direction. Then the pushoff is outside the top lobe and inside the bottom lobe. For fun, you can introduce a pair of double points on the bottom lobe by a type II Reidemeister move. Those double points can be far away from the origin. The required double point of the pushoff can be placed at (1,-1), and the other manufactured double points are far away. I am sure this example works in all dimensions with two coordinate disks being joined by an annulus in, say $S^{2n-1}$.

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  • $\begingroup$ "The required double point of the pushoff". Did you mean something called $f_0(x)=f_0(y)$ which is supposed to be on the original immersion not on the pushoff? $\endgroup$
    – Sergey Melikhov
    Apr 25, 2011 at 14:20
  • $\begingroup$ @Sergey I edited my response. I didn't have a piece of paper handy to draw the picture. Since $f_0$ has a double point at $(0,0),$ the push off that I chose must also have a double point. This occurs at $(1,-1)$. $\endgroup$
    – Scott Carter
    Apr 25, 2011 at 14:48

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