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Why is it that if $\omega$ and $\nu$ are covectors of some finite dimensional space $V$, we have $\omega\nu=\frac12(\omega\otimes \nu+\nu\otimes\omega)$? In general why is it true that the pointwise product of functionals is equal to the symmetrization of the tensor product of functionals?

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"Why is it that... we have $\frac12(\omega\otimes \nu+\nu\otimes\omega)$?" I think there is an "=something" missing? –  rschwieb Apr 27 '12 at 17:41
Thanks, fixed. :) –  Eric Gregor Apr 27 '12 at 17:50

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up vote 1 down vote accepted

In general you can write $\omega\otimes\nu=\frac12(\omega\otimes \nu+\nu\otimes\omega)+\frac12(\omega\otimes \nu-\nu\otimes\omega)$ to isolate the symmetric (left part of right hand side) from the antisymmetric (right part of right hand side) parts of the tensor. I guess you are working with functionals into a field, and pointwise multiplication is going to commute, so one should expect the result to be completely symmetric (making the right part of the right hand side zero.)

The same idea would hold for a longer pointwise product.

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