$i=2$, consider the canonical exact sequence defining $K$ $$0 \rightarrow K \rightarrow {\mathcal S}_I \rightarrow {\mathcal R}_I \rightarrow 0.$$ As $K$ is the direct sum of the $S$-modules $\ker (\operatorname{Sym}^t_S(I) \to I^t)$ that are supported on $V(I)$, we have $H^i_{\mathfrak m}(K)=0$ for $i>\dim S/I$. It follows that $H^2_{\mathfrak m}({\mathcal S}_I) \simeq H^2_{\mathfrak m}({\mathcal R}_I)$. Now, $a^2({\mathcal R}_I)=\max_{t\geq 1}\{ a^2(I^t)-td\}=\max_{t\geq 1}\{ a^1(S/I^t)-td\}$. The surjective map $(S/I)(-td)^{\binom{t+3}{3}}\to I^t/I^{t+1}$ provides the inequality $a^1(I^t /I^{t+1}) \leq a^1(S/I)+td$, and it remains to estimate $a^1(S/I)$. By Lemma \[regdim0\], $$a^2({\mathcal R}_I)\leq a^1(S/I) - d\leq  d-4,$$ unless $\mu (I)=2$. As $\operatorname{indeg}(I^{sat})\leq d$ it follows that $a^2({\mathcal R}_I)\leq \nu_0-2$, unless $I=(f,g)$, in which case $a^2({\mathcal R}_I) =\nu_0-1$.

For $i=3$, $H^3_{\mathfrak m}(Z_0)_\mu =H^3_{\mathfrak m}(S)_\mu=0$ for all $\mu>-3$, hence $a^3({\mathcal S}_I)\leq -3$. Finally, $a^3({\mathcal S}_I)=\oplus_{t\geq 0} H^3_{\mathfrak m}(S)(td)$, hence $a^3({\mathcal S}_I)=-3$.

If $V(I)=\emptyset$ then $a^2 ({\mathcal S}_I)=-\infty$ and $a^1 ({\mathcal S}_I)=a^1(I)-d=\operatorname{reg}(S/I)-d$. Also, when $\dim(S/I)=1$ and $V(I)$ is locally a complete intersection, one has $a^2 ({\mathcal S}_I)=a^2(I)-d$ as above and $a^1 ({\mathcal S}_I)=\max\{ a^1(I)-d,a^1(I^2)-2d\}$. This shows in particular that $a^1 ({\mathcal S}_I)$ and $a^2 ({\mathcal S}_I)$ can be (very quickly) computed from $I$, using a dedicated software.

Fibers of the canonical projection of the graph of $\phi$ onto ${\mathcal S}$ {#sec:main}
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