# Kulkarni-Nomizu product and identity on space of 2-forms

I am sure I am missing something stupid, but I have been on this for a while and I cannot get a factor of 2 right.

Let $E$ be a vector space with a non-degenerate quadratic form $q$. The Kulkarni-Nomizu product is the map $\bar\wedge: S^2E\times S^2E\rightarrow \Lambda^2 S^2 E$ given by $$h\bar\wedge k = (h_{ac}k_{bd}+k_{ac}h_{bd}-h_{ad}k_{bc}-k_{ad}h_{bc}) e^a\otimes e^b\otimes e^c\otimes e^d =\frac{1}{4}(h_{ac}k_{bd}+k_{ac}h_{bd}-h_{ad}k_{bc}-k_{ad}h_{bc}) (e^a\wedge e^b)\otimes (e^c\wedge e^d).$$

It is well known that, making use of the isomorphism $\otimes ^2 (\Lambda^2 E) \simeq \mathrm{End}(\Lambda^2E)$ induced by $q$, $$\frac{1}{2}q\bar\wedge q=\mathrm{Id}_{\Lambda^2E}.$$ However I always get a factor of 2 wrong no matter how I compute it. For example, for $\alpha$ a 2-form, $$\begin{split} \left( \frac{1}{2} q \bar\wedge q\right) (\alpha) & =\left( (\delta _a^c \delta_b^d - \delta _a^d \delta_b^c ) e ^a \otimes e ^b \otimes e _c \otimes e _d\right) \left( \alpha _{m n } e ^m \otimes e ^n \right) \\ & = (\delta _a^c \delta_b^d - \delta _a^d \delta_b^c ) \alpha _{ mn } \delta ^m _c \delta ^n _d \, e ^a \otimes e ^b = \alpha _{ cd } \, (e ^c \otimes e ^d- e ^d \otimes e ^c ) = \alpha _{ cd } \, e ^c \wedge e ^d= 2 \alpha \end{split}$$

Where is the mistake?

In the end, this seems to be just a matter of conventions in choosing the $q$-induced isomorphism $S^2\Lambda^2E \simeq \mathrm{End}(\Lambda^2 E)$. We get the right result by defining, for $T\in S^2\Lambda^2E$, $\alpha\in\Lambda^2E$ and denoting by $\tilde T$ the element associated to $T$ by the isomorphism, $$\tilde T (\alpha) =\frac{1}{2} T_{abcd} \alpha_{rs} q^{rc}q^{sd} =\frac{1}{2} T_{abcd} \alpha^{cd}.$$ Since $T$ is symmetric in the interchange of first and second pair of indices it does not matter on which pair we contract with $\alpha$. The contracted indices are skew, so the factor 1/2 is actually quite natural.