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Suppose that $f_t$ is a homotopic family of function on $R^k$. Show that if $f_0$ is Morse in some neighborhood of a compact set $K$, then so is every $f_t$ for $t$ is sufficiently small.

I know that $f_t$ is a homotopic family of function on $R^k$, so there is a function $F: R^k\times I \to R$ such that $F(x,0)=f_0(x)$ and $F(x,1)=f_t(x)$

$f_0$ is Morse in some neighborhood of a compact set $K$, so every critical point $x$ in this neighbor hood is non degeneration point. Further more, form the result of previous problem we have

$det(H)^2+\sum_{i=1}^k (\frac{\partial f_0 }{\partial x_i})^2>0$.

I need to show that $det(H)^2+\sum_{i=1}^k (\frac{\partial f_t }{\partial x_i})^2>0$.

But I don't know how.

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  • $\begingroup$ so you are sure that your function $f$ is $R^k \to R^k$? If it would be real valued, you should change $F: R^k \times I \to R$ and the rest would make sense and I could post an answer. $\endgroup$ – Daniel Valenzuela Oct 4 '14 at 16:19
  • $\begingroup$ yes, you are right, maybe I misunderstood the content of the question, again. $\endgroup$ – Diane Vanderwaif Oct 4 '14 at 17:21
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Look at the function $g = det(H_t)+ \sum_i (\partial_i f_t)^2 : R^k \times I \to R$. It is continous since $f$ is smooth as is the homotopy $F$. Hence $g^{-1}(\{R-0\})$ is open as the preimage of an open set. But it contains $K \times 0$ and therefore also a neighborhood. Use compactness to show that there exists $\epsilon$ such that $K \times [0,\epsilon)$ is contained in this preimage. Now we know that $f_t$ for $t< \epsilon$ has non vanishing $g(x,t)$ on $K$ and is therefore Morse on a neighborhood of $K$.

Should I explain more details? If yes where?

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  • $\begingroup$ can I use the same reasoning for $f_0$ is Morse in compact manifold $X$ instead of just some neighborhood of $K$? $\endgroup$ – Diane Vanderwaif Oct 6 '14 at 18:43
  • $\begingroup$ Sure this is no problem! $\endgroup$ – Daniel Valenzuela Oct 6 '14 at 18:57
  • $\begingroup$ thank you very much, you will receive the bounty in the next 23 hours. $\endgroup$ – Diane Vanderwaif Oct 6 '14 at 19:10

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