# The relationship of covectors to the symmetric tensor

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. :) – user21725 Apr 27 '12 at 17:50

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.)