Let $Y_1=Y_2=\mathbb{P}^1$, $Y=Y_1\times Y_2$, $p_i:Y\rightarrow Y_i$, $i=1,2$ be a canonnical projections. How to compute explicitly the sheaf $p_{2*}p_1^*(\mathcal{O}_{Y_1}(1))$?
1 Answers
Using Proposition 9.3 of Hartshorne, $$(p_2)_*p_1^*\mathcal{O}_{\mathbb{P}^1}(1)\simeq f^*f_*\mathcal{O}_{\mathbb{P}^1}(1)$$ where $f:\mathbb{P}^1\to\mbox{Spec}(k)$ is the structural morphism. Now $$f_*\mathcal{O}_{\mathbb{P}^1}(1)\simeq H^0(\mathbb{P}^1,\mathcal{O}_{\mathbb{P}^1}(1))\simeq k^2=\mathcal{O}_{\mbox{Spec}(k)}\oplus\mathcal{O}_{\mbox{Spec}(k)},$$ and $$f^*f_*\mathcal{O}_{\mathbb{P}^1}(1)\simeq H^0(\mathbb{P}^1,\mathcal{O}_{\mathbb{P}^1}(1))\otimes_k\mathcal{O}_{\mathbb{P}^1}\simeq\mathcal{O}_{\mathbb{P}^1}\oplus\mathcal{O}_{\mathbb{P}^1}$$ Of course this can also be done by writing out the definition of the pullback and pushforward, finding the morphism on every open set, and proving it is an isomorphism on stalks.
Edit: This can be easily generalized: If $\mathcal{L}$ is a line bundle on $X\times X$ for any variety $X$, then $(p_2)_*p_1^*\mathcal{L}\simeq H^0(X,\mathcal{L})\otimes_k\mathcal{O}_X$.
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Thank you! And the result is not a linear bundle, right? – cmplx_variable Jan 12 '15 at 19:59
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I just edited it to make it more explicit. No, it is not a line bundle. – rfauffar Jan 12 '15 at 20:10
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Thanx. The last thing I didn't understand precisely is the meaning of $$f^(H^0(\mathbb{P}^1,\mathcal{O}{\mathbb{P}^1}(1)))\cong H^0(\mathbb{P}^1,\mathcal{O}{\mathbb{P}^1}(1))\otimes_k\mathcal{O}_{\mathbb{P}^1}.$$ Is it true in general that for $f:X\rightarrow Y$ and $Y$ affine and for any quasi-coherent sheaf $F$ on $X$ $$f^f_*F\cong H^0(X, F)^{\tilde{}}\otimes_{\mathcal{O}_Y}\mathcal{O}_X?$$ – cmplx_variable Jan 12 '15 at 22:04
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1This shouldn't be true in general. Here we use that the only non empty open set of $\mbox{Spec}(k)$ is $\mbox{Spec}(k)$, and so we get global sections. – rfauffar Jan 16 '15 at 11:31