Matrices with two distinct real eigenvalues retract onto a circle

Description. With $s=a+d$, $u=a-d$, $v=b+c$, $w=b-c$ (an invertible linear change of coordinates),
$$\mathcal{U}=\{(s,u,v,w):u^2+v^2>w^2\}=\mathbb{R}_s\times\mathcal{V},\qquad\mathcal{V}=\{(u,v,w):w^2<u^2+v^2\}.$$

Deformation retraction of $\mathcal{V}$. Define $H:\mathcal{V}\times[0,1]\to\mathcal{V}$,
$$H\big((u,v,w),t\big)=(u,v,(1-t)w).$$
It stays in $\mathcal{V}$: $\big((1-t)w\big)^2\le w^2<u^2+v^2$. Also $H_0=\mathrm{id}$, $H_1(u,v,w)=(u,v,0)$, and $H_t$ fixes the points $\mathcal{V}\cap\{w=0\}=\{(u,v,0):(u,v)\ne0\}$. So $\mathcal{V}$ deformation retracts onto
$$\{(u,v,0):(u,v)\ne(0,0)\}\cong\mathbb{R}^2\setminus\{0\}.$$

Further retraction. $\mathbb{R}^2\setminus\{0\}$ deformation retracts onto the unit circle by $(u,v)\mapsto\frac{(u,v)}{|(u,v)|^{t}}$-type radial homotopy. Hence $\mathcal{V}\simeq S^1$.

The factor $\mathbb{R}_s$. It is contractible, so $\mathcal{U}=\mathbb{R}\times\mathcal{V}\simeq\mathcal{V}\simeq S^1$.

Interpretation. A point on the circle is the pair $(u,v)/|(u,v)|$, the direction of the "eigenspace configuration": the matrix $\operatorname{diag}(\lambda_1,\lambda_2)$ conjugated by a rotation of angle $\theta$ gives $(u,v)=(\lambda_1-\lambda_2)(\cos2\theta,\sin2\theta)/2\cdot$-type, and a rotation by $\pi$ returns to the same matrix ($\theta\mapsto\theta+\pi$ gives $2\theta\mapsto2\theta+2\pi$) while $\theta\mapsto\theta+\frac\pi2$ swaps the eigenvalues, going halfway around the circle.

Remark. The fundamental group of $\mathcal{U}$ is $\mathbb{Z}$. By contrast the set $\{\operatorname{tr}^2<4\det\}$ has two contractible components (previous discriminant problem), so the "real" region is topologically richer than the "complex" one.