# For a LCI curve $C$ embedded inside a smooth surface $i : C\rightarrow X/k$, is $i^*T_X\otimes\omega_X = i^*\Omega_X$?

I'm reading Hartshorne's Deformation Theory book. In exercise 5.9(b), he claims:

Let $C$ be a reduced projective locally complete intersection curve embedded inside a smooth surface $X$ (over some alg. closed field $k$), as: $$i : C\hookrightarrow X$$ Tensoring the exact sequence $$0\rightarrow T_{C/k}\rightarrow i^*T_{X/k}\rightarrow N_{C/X}\rightarrow T^1_{C/k}\rightarrow 0$$ with $\omega_{X/k}$ (the dualizing sheaf of $X$, ie $\omega_{X/k} = \det\Omega_{X/k}$), we get a sequence $$0\rightarrow T_{C/k}\otimes\omega_{X/k}\rightarrow i^*\Omega_{X/k}\rightarrow \omega_{C/k}\rightarrow T^1_{C/k}\otimes\omega_{X/k}\rightarrow 0$$ here $T = \Omega^\vee$ denotes the relative tangent sheaf, and $N$ the relative normal sheaf. The extremal terms of the sequence obviously are obtained by tensoring with $\omega_{X/k}$. The third term is $\omega_{C/k}$ by essentially the definition of the dualizing sheaf of an LCI scheme, so my question concerns the second term -

Why is $i^*T_{X/k}\otimes\omega_{X/k}\cong i^*\Omega_{X/k}$? Here, I assume the left hand side should be viewed as $$i^*T_{X/k}\otimes\omega_{X/k} := T_{X/k}\otimes_{\mathcal{O}_X}\omega_{X/k}\otimes_{\mathcal{O}_X}\mathcal{O}_C = i^*\left(T_{X/k}\otimes\omega_{X/k}\right)$$

• On any smooth projective variety $X$, the perfect pairing $\Omega^1_{X/k} \otimes \Omega^{n-1}_{X/k} \to \omega_{X/k}$ realises $\Omega^{n-1}_{X/k}$ as $\mathscr Hom(\Omega^1_{X/k},\omega_{X/k}) = T_{X/k} \otimes \omega_{X/k}$. Mar 17, 2017 at 21:02
• @R.vanDobbendeBruyn Ah! Neat! What's a good reference for this pairing? Mar 17, 2017 at 21:09
• Hartshorne, Exercise II.5.16 (b). Let me write out some details and post it as an answer. Mar 17, 2017 at 21:10
• @R.vanDobbendeBruyn Fantastic! If you could post this as an answer I'd be happy to accept it and mark this question answered. Mar 17, 2017 at 21:13

Let $\mathscr F$ be a rank $n$ locally free sheaf on a ringed space $X$. Then the wedge product map $$\bigwedge^r \mathscr F \otimes \bigwedge^{n-r} \mathscr F \to \bigwedge^n \mathscr F$$ is a perfect pairing. Indeed, it suffices to check this locally, where one does this by choosing a basis $x_i$ for $\mathscr F$. Indeed, given $I = \{i_1 < \ldots < i_r\}$ with complement $I^\mathsf{c} = \{i_{r+1} < \ldots < i_n\}$, write $x_I = x_{i_1} \wedge \ldots \wedge x_{i_r}$, and write $\sigma_I \colon \{1,\ldots,n\} \to \{1,\ldots,n\}$ for the permutation $k \mapsto i_k$. Then the dual basis of $\{x_I\}_I$ is $\{\operatorname{sgn}(\sigma_I)\ x_{I^{\mathsf{c}}}\}_I$.
Applying this to $\Omega^1_{X/k}$ for $X$ smooth projective of dimension $n$, we get a perfect pairing $$\Omega^1_{X/k} \otimes \Omega^{n-1}_{X/k} \to \omega_{X/k},$$ realising $\Omega^{n-1}_{X/k}$ as $\mathscr Hom(\Omega^1_{X/k},\omega_{X/k}) = T_{X/k} \otimes \omega_{X/k}$.