Anticommutation of variation $\delta$ and differential $d$

Question

In Quantum Fields and Strings: A Course for Mathematicians, it is said that variation $\delta$ and differential $d$ anticommute (this is only classical mechanics), which is very strange to me. This is in page 143-144:

If we deform $x$ we have $$\delta L = m \langle \dot x, \delta \dot x \rangle dt$$ $$= - m \langle \ddot x, \delta x \rangle dt - d\left(m \langle \dot x,\delta x\rangle \right)\,.$$ Here $\delta$ is the differential on the space $\mathcal F$ of trajectories $x$ of the particle, $d$ is the differential on $\mathbb R$, and the second minus sign arises since $\delta$ and $d$ anticommute on $\mathcal F \times \mathbb R$.

As far as I know, we should have $d(\delta x)=\delta(dx)$. It doesn't make sense to me why a "differential" with respect to the trajectory would anticommute with a differential with respect to time.

Answer 1

Perhaps the following comment is helpful: If $M=M^{\prime}\times M^{\prime\prime}$ is a product manifold then the exterior differential $\mathrm{d}=\mathrm{d}^{\prime}+\mathrm{d}^{\prime\prime}$ on $M$ is a sum of the exterior differentials on $M^{\prime}$ and $M^{\prime\prime}$. We can assign degree $(1,0)$ to $\mathrm{d}^{\prime}$ and degree $(0,1)$ to $\mathrm{d}^{\prime\prime}$ to form a bicomplex of differential forms. Note that in order for each of the 3 exterior differentials to square to zero, it is important that $\mathrm{d}^{\prime}$ and $\mathrm{d}^{\prime\prime}$ anticommute.

(A similar situation happens for the Dolbeault complex.)

The variational bicomplex $D=\delta+d$ that OP asks about is a very similar construction. We stress that $\delta$ is an exterior differential; not an infinitesimal variation in the heuristic physics sense.

Answer 2

To elaborate on Qmechanic's answer to show why anticommutation in a bigraded differential algebra is natural, consider a manifold $X$ and its exterior algebra $\Omega(X)$. Suppose that there is a bigrading on $\Omega(X)$ such that $$ \Omega(X)=\bigoplus_{(r,s)\in\mathbb Z^2}\Omega^{r,s}(X), $$ where the sum is a direct sum. Suppose furthermore that when restricted to any pure grade subspace $\Omega^{r,s}(X)$, the exterior derivative goes as $$ d:\Omega^{r,s}(X)\rightarrow\Omega^{r+1,s}(X)\oplus\Omega^{r,s+1}(X). $$ If the bigrading is compatible with the ordinary grading by degree in the sense that each differential form of pure bigrade also has pure "monograde" then this is natural.

We then define the splitting of the exterior derivative as $$ d=d_1 +d_2,\quad d_1=\pi^{r+1,s}\circ d,\quad d_2=\pi^{r,s+1}\circ d,$$ where $$ \pi^{r,s}:\Omega(X)\rightarrow\Omega^{r,s}(X) $$ is the projection - which is well-defined because the bigraded decomposition is a direct sum.

This decomposition is then well-defined on any element by extending linearly. The nilpotency condition $d^2=0$ of the original exterior derivative now gives $$ 0=d^2=\left(d_1+d_2\right)^2=d_1^2+d_2^2+d_1d_2+d_2d_1. $$ Suppose that we applied $d^2$ to an element of pure bigrade $(r,s)$. Then $d_1^2$ maps to $\Omega^{r+2,s}(X)$, $d_2^2$ maps to $\Omega^{r,s+2}(X)$, and both $d_1d_2$ and $d_2d_1$ maps to $\Omega^{r+1,s+1}(X)$. Because of the direct sum decomposition, these subspaces are disjoint and linearly independent, therefore $d_1^2$, $d_2^2$ and $d_1d_2+d_2d_1$ must separately vanish. The first two gives $$ d_1^2=0,\quad d_2^2=0, $$i.e. the derived operators $d_1$ and $d_2$ are also nilpotent differentials, and the third condition gives $$ d_1d_2=-d_2d_1, $$ which shows that $d_1$ and $d_2$ anticommute.