Defining a partial derivative with respect to an antisymmetric tensor/matrix I'm looking at some nonlinear electrodynamics, and have been following a textbook which contains a primer on some of the stuff I'm interested in following up. However, I seem to have fallen at the first hurdle...
I won't provide a load of context, as I don't think it's necessary: my question is fundamentally about a function $M$ of the antisymmetric field strength tensor with components $F_{ab}$; antisymmetry tells us that $F_{ab} + F_{ba}=0$. The author tells us that the partial derivatives of any function $M$ with respect to the field strength are defined by
$\delta M = \frac{1}{2}\frac{\partial M}{\partial F_{ab}}\delta F_{ab}$,
where $\delta M$ and $\delta F_{\mu\nu}$ denote variations of the function $M$ and the field strength tensor, respectively. (The ultimate aim here is to vary an action involving such a function $M$ under an integral over an arbitrary number of spatial dimensions.)
My question is then: why are the partial derivatives defined in this way? Why the factor of $1/2$? I've been playing around with this, and while I suspect it's something to do with the antisymmetry property, I've not been able to convince myself why exactly this is the case. Any thoughts anyone has to offer on this would be much appreciated.
 A: $M$ looks like this: $M(u) = (F_{12}(u),\dots,F_{1n}(u),F_{23}(u),\dots F_{2n}(u)),\dots ,F_{nn}(u))$
$M(u) = (F_{12}(u),F_{21}(u),\dots) = (F_{12}(u),-F_{12}(u),\dots)$ doesn't make much sense.
then the first variation: $\delta M(u) = \frac{d}{d\epsilon} M(F_{12}(u+\epsilon v),\dots,F_{nn}(u+\epsilon v))\big|_{\epsilon = 0}$ 
using the chain rule:
$$\delta M(u) = \sum_{a=1}^n\sum_{b\leq a}\frac{\partial M}{\partial F_{ab}} \frac{d}{d\epsilon} F_{ab}(u+\epsilon v)\big|_{\epsilon = 0} = \sum_{a=1}^n\sum_{b\leq a}\frac{\partial M}{\partial F_{ab}} \delta F_{ab}(u)$$
therefore:
\begin{align}
\delta M &= \sum_{a=1}^n\sum_{b< a}\frac{\partial M}{\partial F_{ab}}\delta F_{ab}\\
&= \frac{1}{2} (\sum_{a=1}^n\sum_{b< a}\frac{\partial M}{\partial F_{ab}}\delta F_{ab} + \sum_{a=1}^n\sum_{b< a}\frac{\partial M}{\partial F_{ab}}\delta F_{ab})\\
&= \frac{1}{2} (\sum_{a=1}^n\sum_{b< a}\frac{\partial M}{\partial F_{ab}}\delta F_{ab} + \sum_{b=1}^n\sum_{a< b}\frac{\partial M}{\partial F_{ba}}\delta F_{ba})\\
&= \frac{1}{2} (\sum_{a=1}^n\sum_{b< a}\frac{\partial M}{\partial F_{ab}}\delta F_{ab} - \sum_{b=1}^n\sum_{a< b}\frac{\partial M}{\partial F_{ab}}\delta F_{ba})\\
&= \frac{1}{2} (\sum_{a=1}^n\sum_{b< a}\frac{\partial M}{\partial F_{ab}}\delta F_{ab} + \sum_{b=1}^n\sum_{a< b}\frac{\partial M}{\partial F_{ab}}\delta F_{ab})\\
&= \frac{1}{2} \sum_{a=1}^n\sum_{b =1}^n\frac{\partial M}{\partial F_{ab}}\delta F_{ab}\\
&= \frac{1}{2} \frac{\partial M}{\partial F_{ab}}\delta F_{ab}\\
\end{align}
Notice that because of the asymmetry $F_{aa} = -F_{aa} \Rightarrow F_{aa} = 0$
