The case g_w ~= 0 is ill-specified.

[PierreQuinton]

In case the solution gives $|g_w|=0$, then we get division by $0$. In this case, we get that the lagrangian $\lambda$ is $0$, therefore the solution is the same as the unconstrained one. Going back to the original problem we are therefore solving $\max_{d\in\mathbb R^n} \min_{i\in [n]}\langle g_i, d\rangle$. Now it is not too hard to show that whenever there is $0< w\in \mathbb{R}^m$ such that $A^T w=0$ ( $A$ is the jacobian matrix), then for any $d\neq 0$, then there is $i$ such that $\langle g_i, d\rangle < 0$. This means that the $\min$ is $-\infty$ unless $d=0$, therefore the outer $\max$ is reached for $d=0$.

[Cranial-XIX]

Sorry that I do not see the problem here. Note that when $g_w = 0$ it has to be that $g_0 = 0$, as $g_w^\top g_0 \geq (1 - c) ||g_0||^2 \geq 0$. So it just means that CAGrad finds a stationary point of $L_0$, which is also a Pareto optimal point.

[PierreQuinton]

This holds only when $c\leq 1$, I guess you don't consider so much the case $c>1$, could you explain a bit why ?

If $c>1$, it is possible that $g_w=0$ and $g_0\neq 0$ in that case when dividing by $| g_w|$ the computation of CAGrad in your code will blow up, This is because $\lambda=0$ and setting $d=g_0+g_w/\lambda$ is a problem. Actually in the original paper, I think we are supposed to split the cases when this happens, luckily if we know that $\lambda=0$, then the problem has the same solution as the unconstrained one, therefore from what I said above, you can just pick $d=0$.

[Cranial-XIX]

Thanks for asking, I now understand what you mean. Yes, $\lambda = 0$ should be a separate case as the optimal $d$ form is different. Thanks for catching it.