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Jul 24, 2011 at 10:15 comment added Martin Brandenburg I didn't know that $\mathcal{F}^* \cong \Lambda^{d-1} \mathcal{F} \otimes (\Lambda^d \mathcal{F})^*$, this is very useful. Thanks!
Jul 24, 2011 at 9:31 vote accept Martin Brandenburg
Jul 24, 2011 at 9:12 history answered Torsten Ekedahl CC BY-SA 3.0