# Concrete computation of pullback of the $D$-module $\mathbb{C}[t,t^{-1}]$

I have some problems in calculating some example explicitly. Consider $$\{0\} \overset{i}{\rightarrow} \mathbb{C} \overset{j}{\leftarrow} \mathbb{C}^*.$$ Then $$Rj_+\mathbb{C}[t,t^{-1}] = \mathbb{C}[t,t^{-1}] = D_\mathbb{C}/\partial_ttD_\mathbb{C}$$. Then you can calculate $$Li^*\mathbb{C}[t,t^{-1}]$$ by $$D_{\{0\}\rightarrow \mathbb{C}}\otimes^L_{i^{-1}D_{\mathbb{C}}}i^{-1}\mathbb{C}[t,t^{-1}]=\mathbb{C}[\partial_t]\overset{\text{right multiplication }\partial_tt}{\longrightarrow}\mathbb{C}[\partial_t].$$ Note that $$\partial_tt$$ maps $$\partial_t^n$$ to $$(n+1)\partial_t^n$$, which implies that $$D_{\{0\}\rightarrow \mathbb{C}}\otimes^L_{i^{-1}D_{\mathbb{C}}}i^{-1}\mathbb{C}[t,t^{-1}]=0$$.

However, by Riemann-Hilbert corrspondence, we can consider $$i^{-1}Rj_*\mathbb{C}_{\mathbb{C}^*}$$, which are $$\mathbb{C}$$ in cohomology degree $$0$$ and $$1$$, contradicts the calculation above. But I do not know where is the thing going wrong.

• In Riemann-Hilbert correspondence, the counterpart of $i^{-1}$ should be $Li^*$ twisted by the Verdier dual. After doing that, the calculations on both sides match. Mar 27 at 6:45
• @XiaowenHu But $Li^*$ is already zero. I do not think that a zero object can become non zero after twisting. Mar 27 at 6:46
• @XiaowenHu I see. In the Riemann-Hilbert, I should first twist $\mathbb{C}[t,t^{-1}]$ by Verdier dual, then apply $Li^*$. But then the differential map of the pullback complex should be right multiplication by $t\partial_t$. This gives the correct computation! Thanks! Mar 27 at 7:06
• Yes, exactly. Thank you for evoking this good example of Riemann-Hilbert correspondence. Mar 27 at 13:33