Gamma Exceptional Set

Consider a homotopy

(1)
\begin{align} H(x,t) = (1-t)G(\x)+\gamma tF(x). \end{align}

Let $G = x^2-1$. Find all $\gamma\in\bC$ such that there exists $t\in\bR$ with a singular solution $x$ to $H(x,t)=0$

• for $F=x^2+2x-3$;
• for $F=x^2+ax+b$ ($a,b\in\bR$).
• What is the real dimension of the set of such $\gamma$?

Solution

(2)
\begin{eqnarray} \nonumber H(x,t) &=& (1-t)(x^2 - 1) + \gamma t (x^2 + 2x -3) \\ \nonumber &=& ((1-t) + \gamma t)x^2 +(2 \gamma t) x + 3\gamma t (1 - t) \in (\mathbb{C}[\gamma, t])[x] \ \end{eqnarray}

The solutions can be solved explicitly using the quadratic formula:

(3)
\begin{eqnarray} \nonumber x_{1,2}^{*} = \frac{-2 \gamma t \pm \sqrt{4 \gamma^{2} t^{2} - 4 (1 - t + \gamma t)(3 \gamma t - 3 \gamma t^{2})}}{2 (1 -t + \gamma t)} \ \end{eqnarray}

If a singular is to occur along the path, the the derivative $H^{'}(x,t)$ must also vanish.

(4)
\begin{eqnarray} \nonumber H^{'}(x,t) = 0 \rightarrow x^{*} = - \frac{\gamma t}{1 - t + \gamma t} \ \end{eqnarray}

Comparing the two necessary conditions, the discriminant must vanish:

(5)
\begin{eqnarray} \nonumber 4 \gamma^{2} t^{2} - 4 (1 - t + \gamma t)(3 \gamma t - 3 \gamma t^{2}) = 0 \ \end{eqnarray}

Solving explicitly for $\gamma$ we have:

(6)
\begin{eqnarray} \nonumber \gamma^{*}(t) = \frac{3(1-t)^{2}}{t(3t - 2)} \ \end{eqnarray}

That is, for $t \neq 0$ and $t \neq \frac{2}{3}$ we have a singularity for some $\gamma \in \mathbb{C}$. In fact, this illuminates the fact that $\gamma \in \mathbb{R}$ since $\gamma^{*}(t)$ is a real-valued function. The range of $\gamma^{*}(t)$ is $\{\gamma \in \mathbb{R} : \gamma \leq -3 \mbox{ or } \gamma \geq 0 \}$.