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.

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    $\begingroup$ 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. $\endgroup$
    – Xiaowen Hu
    Mar 27, 2022 at 6:45
  • $\begingroup$ @XiaowenHu But $Li^*$ is already zero. I do not think that a zero object can become non zero after twisting. $\endgroup$
    – XT Chen
    Mar 27, 2022 at 6:46
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    $\begingroup$ @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! $\endgroup$
    – XT Chen
    Mar 27, 2022 at 7:06
  • $\begingroup$ Yes, exactly. Thank you for evoking this good example of Riemann-Hilbert correspondence. $\endgroup$
    – Xiaowen Hu
    Mar 27, 2022 at 13:33


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