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Take $L=S^2$. For any Lagrangian embedding of $S^2$ into a symplectic four-manifold, the symplectic form identifies the normal bundle with $T^{\ast }{S}^2$. The self-intersection formula and the Euler characteristic giveup to the harmless orientation sign. If a smooth isotopy displaced $L$, its final image would represent the same homology class but have intersection number zero with $L$, a contradiction. This is the smooth non-displaceability from self-intersection.

For a compact ambient example, equip $S^2\times S^2$ with $\omega \oplus (-\omega )$. Its diagonalis Lagrangian because the two summands cancel on $T\Delta$, and the preceding argument shows that it is not smoothly displaceable.

Now $m=2n$ and $N$ has rank $n$. Write $d=\mathrm{deg}X$, so the pushforward of the fundamental class isThe self-intersection formula and the calculation of $c(N)$ giveOn the other hand, pulling back $d{H}^n$ makes the left side $d{H}^n\cap [X]$. Integrating over $X$ givesSince $d>0$, cancellation yields

For a smooth projective algebraic curve $C$, the normal bundle of its diagonal in $C\times C$ is $TC$: on the diagonal the quotient of $TC\oplus TC$ by the diagonal subbundle is identified with $TC$ by taking the difference. The self-intersection formula therefore gives ${\Delta }_C^2=\mathrm{deg}TC=2-2g(C)$.