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Sep 30, 2013 at 9:19 comment added diverietti Your proof is more or less the proof I knew...
Sep 30, 2013 at 4:11 comment added ziggurism The fiber of $H\otimes \mathcal{O}(1)$ is $V/[x]\otimes [x]^*.$ So let $\sigma\in [x]^*$ and $v+[x]\in V/[x].$ Define the map to be $\sigma\otimes(v+[x])\mapsto \sigma(x)\cdot d\pi_x(v).$ Then show that this is well-defined as a function on $P(V)$, i.e. $\sigma(x)\cdot d\pi_x(v) = \sigma(\lambda x)\cdot d\pi_{\lambda x}(v).$ But is there a more elegant way to say it, i.e. not in terms of an arbitrarily chosen point $x$ in $V$?
Sep 30, 2013 at 3:37 comment added ziggurism Can you elaborate about how we derive $T_{P(V)}\simeq H\otimes \mathcal{O}(1)$ from the differential of the projection map? Are we to understand $d\pi_x$ as the linear map on a fiber of a bundle map $H\otimes \mathcal{O}(1)\to T_{P(V)}$?
Apr 30, 2013 at 8:56 history edited diverietti CC BY-SA 3.0
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Apr 29, 2013 at 17:59 vote accept Benjamin Schmidt
Apr 29, 2013 at 16:33 history edited diverietti CC BY-SA 3.0
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Apr 29, 2013 at 16:27 history answered diverietti CC BY-SA 3.0