# Partial Derivatives and Dependent Variables

Suppose I have a 4-index, real-valued tensor $\lambda_{ij}^{ab}$ where $i$ and $j$ are from some subspace $O$ of the whole numbers, and $a$ and $b$ are from some other subspace $V,$ which is disjunct from $O.$ Suppose also that we know that $\lambda$ is antisymmetric. If we impose the constraint that $\lambda$ is Hermitian (i.e. $\lambda_{ij}^{ab} = (\lambda_{ab}^{ij})^*=\lambda_{ab}^{ij}$ since $\lambda$ is real), and we have some function:

$E = \sum_{ij\in O, ab \in V} k_{ab}^{ij}\lambda_{ij}^{ab} + \sum_{ij\in O, ab \in V} k_{ij}^{ab}\lambda_{ab}^{ij}$

(where $k_{ij}^{ab}$ is some antisymmetric, Hermitian tensor independent of $\lambda$) and then we take the partial derivative of that function with respect to some specific $\lambda_{a^\prime b^\prime}^{i^\prime j^\prime}$, what do we get? What I don't understand is that if you make the substitution for the Hermitian constraint before differentiating, you can combine the two summations into one, but then differentiating gives you the same answer with a different constant, right?

The answer that the derivation I am reading gives is $k_{ij}^{ab}$, rather than $2k_{ij}^{ab}$. Could someone explain why?

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Cross-posted on MO: mathoverflow.net/questions/38058 – J. M. Sep 8 '10 at 14:15
Figured out the problem (see answer below). Feel free vote to delete if you want to avoid confusing future users – David Hollman Sep 8 '10 at 14:59

It turns out that $2k_{ij}^{ab}$ is, in fact, the correct answer; it's just that when the derivation was done, the next step was to set the derivative equal to zero, which allows you to divide through by 2, and so the original deriver still got the right answer. Sorry for the confusion!