= Second-order-cone optimality certificate
A curved <feasible set> can have the same certificate structure. Consider
$$
\min_{t,z}t\qquad\text{subject to}\qquad \sqrt{1+z^2}\leq t.
$$
Write $x=(t,z)^T$ and use the <second-order cone> $K=\{(s_0,s_1,s_2):s_0\geq\sqrt{s_1^2+s_2^2}\}$. In this coordinate ordering,
$$
A=\begin{pmatrix}1&0\\0&0\\0&1\end{pmatrix},\quad
b=\begin{pmatrix}0\\-1\\0\end{pmatrix},\quad
c=\begin{pmatrix}1\\0\end{pmatrix},\quad
Ax-b=(t,1,z)^T.
$$
<Self-duality of a second-order cone> makes $K^*=K$. The equality $A^Ty=c$ forces $y_0=1$ and $y_2=0$, leaving $-1\leq y_1\leq1$. Its dual <objective function> is $b^Ty=-y_1$, maximized at $y=(1,-1,0)^T$. Thus every <feasible point> satisfies $t\geq1$, and $(t,z)=(1,0)$ attains that bound: $\boxed{p_*=d_*=1}$.
\Image[/conic-second-order-primal-and-dual-slices.png]
{title=A curved primal constraint and an interval of dual certificates}
{description=The left panel is the slice $s_1=1$ of the <second-order cone>, so its boundary is $t=\sqrt{1+z^2}$. The right panel is the dual slice $y_0=1$, $y_2=0$: admissible $y_1$ form $[-1,1]$, and $-y_1$ reaches $1$ at $y_1=-1$. The optimal <slack variable> and <dual variable> lie on different cone boundary rays with zero <dot product>.}
The zero <dot product> is $(1,1,0)\cdot(1,-1,0)=0$. Individual coordinate products need not be nonnegative here: the cancellation $1-1=0$ occurs across coordinates. The valid <complementary slackness> condition is cone-level <orthogonality>, rather than the coordinatewise interpretation of the <nonnegative orthant>.
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