I am trying to get that $f(x_{0},...,x_{n+1})=x_{n+1}$ has $index_{(0,...,0,1)}=n$
Can you find my mistake or post a partial solution?
My attempt
I evaluate df using inverse of stereographic proj. (omitting (0,...,0,-1))
$d f(\phi^{-1})=(\dfrac{4x_{1}^{2}\sum_{i=1}^{n}x_{i}^{2}}{(1+\sum_{i=1}^{n}x_{i}^{2})^{2}},...,\dfrac{4x_{n}^{2}\sum_{i=1}^{n}x_{i}^{2}}{(1+\sum_{i=1}^{n}x_{i}^{2})^{2}})$
and so $(0,...,0,1)$ is a critical point.
Then the first two rows for Hessian:
$Hessian(a_{1i})=(\dfrac{(8x_{1}\sum_{i=1}^{n}x_{i}^{2}+8x_{1}^{3})(1+\sum_{i=1}^{n}x_{i}^{2})^{2}-16x_{1}^{3}\sum_{i=1}^{n}x_{i}^{2}(1+\sum_{i=1}^{n}x_{i}^{2})}{(1+\sum_{i=1}^{n}x_{i}^{2})^{4}},...,\dfrac{-16x_{n}^{2}x_{1}}{(1+\sum_{i=1}^{n}x_{i}^{2})^{3}})=(0,...,0)$
$Hessian(a_{2i})=(\dfrac{-16x_{1}^{2}x_{2}}{(1+\sum_{i=1}^{n}x_{i}^{2})^{3}},\dfrac{(8x_{2}\sum_{i=1}^{n}x_{i}^{2}+8x_{2}^{3})(1+\sum_{i=1}^{n}x_{i}^{2})^{2}-16x_{2}^{3}\sum_{i=1}^{n}x_{i}^{2}(1+\sum_{i=1}^{n}x_{i}^{2})}{(1+\sum_{i=1}^{n}x_{i}^{2})^{4}},...,\dfrac{-16x_{n}^{2}x_{2}}{(1+\sum_{i=1}^{n}x_{i}^{2})^{3}})=(0,...,0)$
So the Hessian is just zeroes.
But the $index_{(0,...,0,1)}=n$. I am supposed to get n-negative values in the diagonal.
thanks
In the first entry, you compute the derivative of a quotient $$ \dfrac{f'g - f g'}{g^2} $$ where $f = 4 x_1^2 \sum_i x_i^2$ and $g = (1 + \sum x_i^2)^2$.
The $f g'$ term should be
$$ \left(4 x_1^2 \sum_i x_i^2\right) \left( 2 (1 + \sum_i x_i^2) 2 x_1 \right) \\ 16 x_1^3 (\sum_i x_i^2 )(1 + \sum_i x_i^2). $$
Presumably by plugging in $n = 1$ into your formula above, you can see where you went off the rails.
You seem to have $8$ instead of $16$. (This is the only term I checked).
I don't see, however, why computing only the first row of the Hessian suffices to show you that the eigenvalues are not all negative -- can you explain your reasoning? Or am I supposed to see, from the first row, the pattern of the others? If so, I still don't see why the e-values can't all be negative. (Except that maybe your swapping 8 for 16 would ruin it...)
Later comment, now that you've fixed the $8$ vs $16$ thing:
Looking more closely at your work, let's consider the unit circle in 2-space. Stereographic projection from the south pole takes the unit disk to the upper hemisphere (in this case, the interval $[-1, 1]$ to the upper circle) by $$ (x, 0) \mapsto (\frac{2x}{1 + x^2} , \frac{1-x^2}{1 + x^2}) $$
The derivative you want to compute is \begin{align} d(f \circ \phi^{-1}) &= d(\frac{1-x^2}{1 + x^2})/dx \\ &= \frac{(1-x^2)' (1 + x^2) - (1-x^2)(1 + x^2)'}{(1 + x^2)^2}\\ &= \frac{(-2x) (1 + x^2) - (1-x^2)(2x)}{(1 + x^2)^2}\\ &= \frac{(-2x - 2x^3) - (2x - 2x^3)}{(1 + x^2)^2}\\ &= \frac{(-2x - 2x^3) + (-2x + 2x^3)}{(1 + x^2)^2}\\ &= \frac{- 4x}{(1 + x^2)^2} \end{align}
Since, near $x = 0$, this is just $-4x$ to first order, the second derivative at $0$ will be $-4$, and not $0$. Presumably by plugging in $n = 1$ to your formula above you can discover where you went off the rails.
Lemme go ahead and compute the second derivative:
\begin{align} H &= d(\frac{- 4x}{(1 + x^2)^2} )/dx \\ &= \frac{(- 4x)'(1 + x^2)^2 + 4x( (1+x^2)^2 )'}{ (1 + x^2)^4} \\ &= \frac{- 4(1 + x^2)^2 + 4x( 2(1+x^2)(2x) )}{ (1 + x^2)^4} \\ &= \frac{- 4(1 + x^2)^2 + 16x^2(1+x^2)}{ (1 + x^2)^4} \\ &= \frac{ (-4(1 + x^2) + 16x^2)(1+x^2)}{ (1 + x^2)^4} \\ &= \frac{ (-4 + 12x^2)(1+x^2)}{ (1 + x^2)^4} \end{align} Now plug in $x = 0$ to get \begin{align} H_0 &= \left. \frac{ (-4 + 12x^2)(1+x^2)}{ (1 + x^2)^4} \right|_{x=0} \\ &= \frac{ (-4 )(1)}{ (1)^4} \\ &= -4. \end{align}