Lemniscate of Gerono

The lemniscate of Gerono is given by the equation $x^4-x^2+y^2= 0$. Draw a
picture of it to see how this plane curve looks. Using YALMIP give an approximate
2-sos decomposition of $x + 1$ modulo the equation of the curve. Can you find an
exact one?

Use reduced moment matrices to write down the spectrahedral representation
of $TH_2 (I)$ where $I = <x^4-x^2+y^2>$. What can you say about $TH_1(I)$? Can you
decide whether $TH_2(I)$ coincides with the convex hull of the lemniscate?


Solution

Let $f = x^4-x^2+y^2$.
Draw the real variety $\mathcal{V}_{\mathbb{R}}(f)$:

Gerono.gif
(Thank you, Wolfram Alpha!)

Using Yalmip to approximate 2-sos decomposition of $x + 1$ modulo the equation of the curve:

We are looking for a decomposition of the following form:
$x+1 = \sum_i s_i(x)^2 + h(x) f(x)$
where $g(x)$ is a polynomial of degree $\leq2$.
This is equivalent to $x+1 - h(x) f(x)$ being a sum of squares.
Let us check if this is true using YALMIP. To make things easy, we choose $h(x)$ to be a constant and hope that it suffices:

sdpvar x y h
f=x^4-x^2+y^2
q=x+1
p=q+h*f
F=sos(p)
solvesos(F,h)
double(h)
sdisplay(sosd(F))

The solver reports numerical problems, but we still get an approximate decomposition that is indeed of degree 2:

ans =

    0.5000

ans = 

    '-0.9999999999-0.500009021*x+0.4999639732*x^2'
    '1.804270243e-05-0.499982016*x-0.4999909773*x^2'
    '0.7070749267*y'
    '2.637194987e-06-2.63724226e-06*x+2.637290158e-06*x^2'

Can you find an exact one?
Now we do some rounding and "guess" the following 2-sos decomposition:

(1)
\begin{align} s_1^2=(-1-1/2*x+1/2*x^2)^2 \\ s_2^2=(-1/2*x-1/2*x^2)^2 \\ s_3^2=1/2*y^2 \end{align}

Being lazy, we check with M2 if this is really a decomposition:

R=QQ[x,y]
f=x+1+1/2*(x^4-x^2+y^2)
s1=(-1-1/2*x+1/2*x^2)^2
s2=(-1/2*x-1/2*x^2)^2
s3=1/2*y^2

f==s1+s2+s3

TRUE!

Spectrahedral representation of $TH_1(I)$ and $TH_2(I)$
To write down the moment matrix, first choose a (theta-)basis $\mathcal{B}$ of $\mathbb{R}[x]/I$.
Since we are only interested in $TH_1(I)$ and $TH_2(I)$, we only have to consider basis elements of degree two and less. For any degree-respecting monomial ordering, the monomial $x^4$ will be the initial term, therefore all monomials of degree two and less are standard monomials. Let us choose $\mathcal{B}_2 = \{1, \ x, \ y, \ x^2, \ xy, \ y^2\}.$
The multiplication matrix indexed by $\mathcal{B}_2$ is the following matrix (remember, we have to reduce by $I$):

(2)
\begin{bmatrix} 1 & x & y & x^2 & xy & y^2 \\ x & x^2 & xy & x^3 & x^2y & xy^2 \\ y & xy & y^2 & x^2y & xy^2 & y^3 \\ x^2 & x^3 & x^2y & x^2y-y^3 & x^3y & x^2y^2 \\ xy & x^2y & xy^2 & x^3y & x^2y^2 & xy^3 \\ y^2 & xy^2 & y^3 & x^2y^2 & xy^3 & y^4 \end{bmatrix}

Linearizing with new variables $z_{ij}$ this yields the following moment matrix:

(3)
\begin{align} M_{\mathcal{B}_2} = \begin{bmatrix} 1 & z_{1,0} & z_{0,1} & z_{2,0} & z_{1,1} & z_{0,2}\\ z_{1,0} & z_{2,0} & z_{1,1} & z_{3,0}& z_{2,1}& z_{1,2} \\ z_{0,1} & z_{1,1} & z_{0,2} & z_{2,1} & z_{1,2} & z_{0,3} \\ z_{2,0} & z_{3,0} & z_{2,1} & z_{2,1}-z_{0,3} & z_{3,1} & z_{2,2} \\ z_{1,1} & z_{2,1} & z_{1,2} & z_{3,1} & z_{2,2} & z_{1,3} \\ z_{0,2} & z_{1,2} & z_{0,3} & z_{2,2} & z_{1,3} & z_{0,4} \end{bmatrix} \end{align}

Using the moment matrix, we can now describe the theta bodies of the lemniscate:

(4)
\begin{align} TH_1(I) = cl \left( \left\{ (z_{1,0},z_{0,1}) \ |\ \exists z \text{ with } \begin{bmatrix} 1 & z_{1,0} & z_{0,1} \\ z_{1,0} & z_{2,0} & z_{1,1} \\ z_{0,1} & z_{1,1} & z_{0,2} \\ \end{bmatrix} \succeq 0 \right\} \right) \end{align}

To see what $TH_1(I)$ looks like, let us translate the psd condition to conditions on the minors. we get the following inequalities:

(5)
\begin{align} z_{2,0} & \geq 0 \\ z_{0,2} & \geq 0 \\ z_{2,0}-z_{1,0}^2 & \geq 0 \\ z_{0,2}-z_{0,1}^2 & \geq 0 \\ z_{2,0}z_{0,2}-z_{1,1}^2 &\geq 0 \\ determinant = -z_{0,1}^2z_{2,0} + 2z_{1,0}z_{0,1}z_{1,1}-z_{1,0}^2z_{0,2}-z_{1,1}^2+z_{2,0}z_{0,2} & \geq 0 \end{align}

By making the natural choice

(6)
\begin{align} z_{2,0} = z_{1,0}^2 \quad z_{1,1} = z_{1,0}z_{0,1} \quad z_{0,2} = z_{0,1}^2 \end{align}

we see that there is a solution to these inequalities for every choice of $(z_{1,0}, z_{0,1})$.

For the second theta body, we have:

(7)
\begin{align} TH_2(I) = cl \left( \left\{ (z_{1,0},z_{0,1}) \ | \ \exists z \text{ with } M_{\mathcal{B}_2} \succeq 0 \right\} \right) \end{align}
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