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I am currently trying to understand the BV-formalism, which makes heavy use of the functional derivative.

Let us consider the functional derivative, as defined in for example its Wikipedia article. Let $F$ be a functional, i.e. a map from, say, $C^\infty(\mathbb{R})$ to $\mathbb{R}$, and suppose it may be written as $F[\phi] = \int f\big(x,\phi(x),\phi'(x),\dots,\phi^{(n)}(x)\big)\\,dx$ for some function $f$ which depends on the derivatives of $\phi$ up to order $n$. Then the functional derivative of $F$ is $\displaystyle \frac{\delta F}{\delta \phi} = \sum_{i=1}^n(-1)^i\frac{d^i}{dx^i}\frac{\partial f}{\partial \phi^{(i)}}$.

Now, my background is that of differential equations and differential geometry, i.e. jet spaces and variational calculus and the like. In that area, the latter operator, $\sum_{i}(-1)^i\frac{d^i}{dx^i}\frac{\partial}{\partial \phi^{(i)}}$, is well known; it is called the variational derivative. Summarizing, then, we seem to have that the functional derivative of a functional is the variational derivative of (one of its) densities.

Since the variational derivative involves lots of derivatives, it certainly does not satisfy the Leibniz rule, i.e. it is not a derivation. In various places, however, I've come across the statement that the functional derivative does satisfy the Leibniz rule. (That already seems unexpected to me: how can an operator which is so intimately connected to a decidedly non-derivation be a derivation?) There are various ways to prove it, but I would like to understand this fact in terms of the variational derivative, if possible. So: how can the Leibniz rule of the functional derivative related to variational derivative; can the former be expressed somehow in terms of the latter?

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1) Maybe you should provide a precise statement of the Leibniz rule you're referring to. 2) I don't see any difference between what you call the "functional derivative" and what is often called the "variational derivative". Either way, it tells you what the directional derivative $ \left. F[\phi + t\dot\phi]\right|_{t=0} $ is, where you've integrated by parts to shift all the derivatives off $\dot\phi$. –  Deane Yang Oct 15 '12 at 13:57
    
The difference is that the variational derivative (as I understand it anyway) acts on ordinary functions, such as $f$, by the operator described above; while the functional derivative acts on functionals, such as $F$. –  Sietse Ringers Oct 15 '12 at 14:05
    
And what is the variational derivative (acting on an ordinary function) used for? –  Deane Yang Oct 15 '12 at 14:15
    
The Leibniz rule probably holds on the level of the functionals, which leads to the question: is the product $F[\varphi]G[\varphi]$ of two such functionals still a functional which can be written as the integral over a function $F[\varphi]G[\varphi]=\int h(\ldots)$? –  Michael Bächtold Oct 15 '12 at 14:45
    
Deane: the variational derivative is defined on the space of functions on jet space more or less as a jet space-analog of the functional derivative, but without the actual functional aspect (as described above and below). In short, it is important in for example the geometry of jet spaces (in particular in the horizontal cohomology, where it comes from the de Rham differential along the fibers); and in mathematical physics, since one can express Euler-Lagrange equations in terms of it. –  Sietse Ringers Oct 16 '12 at 7:44

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Connection of functional derivative with variational derivative: $\frac{\delta}{\delta\phi(x)} F[\phi] = \frac{\delta F[\phi]}{\delta\phi}(x)$. Note that the variational derivative carries an extra coordinate variable dependence. It helps to make it explicit when there is similar confusion.

Functional derivative Leibniz rule: $\frac{\delta}{\delta\phi(x)} F[\phi] G[\phi] = \frac{\delta F[\phi]}{\delta\phi}(x) G[\phi] + F[\phi] \frac{\delta G[\phi]}{\delta\phi}(x)$.

Special case: $F_x[\phi] = \phi(x)$, $G_{i,y}[\phi] = (\partial_i\phi)(y)$, and $$\frac{\delta}{\delta\phi(z)} F_x[\phi] G_{i,y}[\phi] = \delta(x-z) (\partial_i\phi)(y) - \phi(x) \frac{d}{dz_i}\delta(y-z)$$.

Notice the distributional coefficients in the derivatives. There is no way to get away from them if you wish to consider $\phi(x)$ and such as functionals in their own right.

If you are interested in the BV formalism in the physics formalism, where the distinction between the functional and variational derivatives is barely remarked, I recommend the reviews by Henneaux and by Gomis, París and Samuel: doi:10.1016/0920-5632(90)90647-D, doi:10.1016/0370-1573(94)00112-G. If you are interested in the BV formalism purely from the point of view of jets, without bringing functionals into the picture, other than peripherally, I recommend the early paper of McCloud and this sequence of papers by Barnich, Brandt and Henneaux: arXiv:hep-th/9307022, arXiv:hep-th/9405109, arXiv:hep-th/9405194, arXiv:hep-th/0002245. If you are more interested in the BV formalism more from the functional point of view, with the appropriate level of functional analysis included, and with jets appearing only peripherally, I recommend the papers by Fredenhagen and Rejzner, as well as Rejzner's thesis: arXiv:1101.5112, arXiv:1110.5232, arXiv:1111.5130.

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I should have asked for an explanation for what "BV formalism" is, so thank you for answering this anyway. –  Deane Yang Oct 15 '12 at 15:36
    
Thank you, Igor, for this answer. It was not precisely what I was looking for, but that's to be expected because I don't think I managed to write down exactly what I was looking for. –  Sietse Ringers Oct 16 '12 at 7:36
    
I would welcome your favorite expository reference for one of the many various interpretations of `BV formalism'. –  Jim Stasheff Oct 14 '13 at 14:43
    
@JimStasheff, I would say that my personal favorite expository reference is the review article by Henneaux, which is already cited above. It conveys a lot of intuition and contains pretty much all the heuristics you need to make the constructions formal. –  Igor Khavkine Oct 15 '13 at 10:19

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