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We are given smooth projective varieties $X$ and $Y$ over $k$. Suppoese $X\simeq Y$. Do we have $\omega_X\simeq \omega_Y$?

If this is the case, then for $\pi_1:\mathbb{P}(\mathcal{E})\rightarrow X$ and $\pi_2:\mathbb{P}(\mathcal{E}\otimes \mathcal{L})\rightarrow X$ one gets $\omega_{\mathcal{E}}=\pi^*_1\omega_X\otimes \pi^*_1(\wedge^r\mathcal{E})\otimes \mathcal{O}_{\mathcal{E}}(-r)\simeq\pi^*_2\omega_X\otimes \pi^*_2(\wedge^r(\mathcal{E}\otimes\mathcal{L}))\otimes \mathcal{O}_{\mathcal{E}\otimes\mathcal{L}}(-r)=\omega_{\mathcal{E}\otimes\mathcal{L}} $, since $\mathbb{P}(\mathcal{E})\simeq \mathbb{P}(\mathcal{E}\otimes \mathcal{L})$

Here $\mathcal{E}$ is locally free of rank $r$ and $\mathcal{L}$ invertible.

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    $\begingroup$ Yes, because the canonical bundle is intrinsically defined. $\endgroup$ – user64687 Jan 18 '14 at 10:28

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