Notation : $r^2=x^2+y^2$.

Exercise : Define $$F_\sigma (x,y)= (f_\sigma (x,y),\frac{x^2-y^2}{2},xy)$$ where $$f_\sigma (x,y)=(1- \sigma^2 r^2)(x-\sigma x^3,y-\sigma y^3)$$

Define $ G_\sigma: \mathbb{R}^4\rightarrow \mathbb{R}^4$ by $$G_\sigma(x,y,a,b)=(dF_\sigma )_{(x,y)} (a,b) =( (df_\sigma)_{(x,y)}(a,b) ,R(x,y))$$ where $R$ is $\angle ((a,b),(1,0))$-rotation

Then there is a sequence of balls $U_\sigma$ and unit circle $S$ s.t. $\lim_{\sigma\rightarrow 0}\ {\rm diam}\ U_\sigma =0$ and $G_\sigma (U_\sigma \times S)$ has positive second fundamental form.

And $G_0 (U\times S)$, homeomorphic to solid torus, has no positive second fundamental form for any ball $U$ (These are obatined from direct computation)

Question :Intuitively, given saddle $F_0$, how can we find saddle $F_\sigma$ s.t. $G_\sigma (U_\sigma\times S)$ has a positive second fundamental form ?

**Reference :** On intrinsic geometry of surface in normed spaces - Burago and Ivanv

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Exercise (cf. Reference) : Hypersurface has a positive second fundamental form at $p$ if there is a linear function $L$ and $c>0$ s.t. $L(q)\leq L(p)- c|q-p|^2$ for all $q$ in a neighborhood of $p$

If $ F_0(x,y)=(x,y,\frac{x^2-y^2}{2},xy)$, then $dF_0\ U\times S$ has out normal $W$ at a point $(a,b,0,0),\ (a,b)\in S$.

$ \langle W,(a,b,0,0) -(a_0,b_0,0,0)\rangle \leq -c \{ |(a,b)-(a_0,b_0)|^2+|(x,y)|^2\}$ does not hold since there is a term $|(x,y)|^2$.

Hence we have a strategy to find $f_\sigma$ s.t. $| (df_\sigma)_{(x,y)}(a,b) -(a,b)| \leq O(x^2,xy,y^2) $