BV, BRST, and Homological Physics Interfaces: Examples and Counterexamples

The maps ${\mathrm{\partial }}_x$ and $x{\mathrm{\partial }}_y$ are derivations by the polynomial product rule, and $[{\mathrm{\partial }}_x,x{\mathrm{\partial }}_y]f={\mathrm{\partial }}_x(x{\mathrm{\partial }}_{y}f)-x{\mathrm{\partial }}_y({\mathrm{\partial }}_{x}f)={\mathrm{\partial }}_{y}f$. Since $[u,v]=0$ but ${\mathrm{\partial }}_y\mathrm{\ne }0$, $\rho$ is not a Lie algebra homomorphism and is not an admissible input in Classical Lie-action BRST cochain complex.

For the $0$-cochain $f=y$, [F1] gives $(df)(u)=0$ and $(df)(v)=x$. Applying [F1] again gives $d^{2}f(u,v)=\rho (u)(x)-\rho (v)(0)-0={\mathrm{\partial }}_{x}x=1$. Thus the formula fails to square to zero when the representation hypothesis is dropped, refuting the statement.

The operator $x\frac{d}{dx}$ is a derivation; for the one-dimensional abelian $\mathfrak{g}$, its assignment to $e$ is a representation because $[\rho (e),\rho (e)]=0=\rho ([e,e])$. Thus this is an admissible input for [F1].

Since ${\mathrm{\Lambda }}^0\mathfrak{g}=\mathbb{C}$, ${\mathrm{\Lambda }}^1\mathfrak{g}=\mathfrak{g}$, and ${\mathrm{\Lambda }}^n\mathfrak{g}=0$ for $n\ge 2$, the only nonzero cochains are $C^0\cong \mathbb{C}[x]$ and $C^1\cong \mathbb{C}[x]$; [F1] gives $df=x{f}^{\mathrm{\prime }}$ and $d:C^1\to C^2=0$.

For $f={\sum }_{r=0}^N{a}_r{x}^r$, $x{f}^{\mathrm{\prime }}={\sum }_{r=1}^{N}r{a}_r{x}^r$; characteristic zero makes each $r\ge 1$ invertible, so the kernel is $\mathbb{C}$ and $H_{\mathrm{B}\mathrm{R}\mathrm{S}\mathrm{T}}^0\cong \mathbb{C}$ by [F2].

The image of $x\frac{d}{dx}$ is exactly $x\mathbb{C}[x]$: every output has zero constant term and $x^r$ is the image of $x^r\mathrm{/}r$ for $r\ge 1$. Hence $H_{\mathrm{B}\mathrm{R}\mathrm{S}\mathrm{T}}^1\cong \mathbb{C}[x]\mathrm{/}x\mathbb{C}[x]\cong \mathbb{C}$, while $C^n=0$ gives $H^n=0$ for $n\ge 2$.

Every element is $f(x)+\theta g(x)$ with $f,g\in \mathbb{C}[x]$ and ${\theta }^2=0$; the polynomial quotient multiplication is associative and the assigned degrees make it graded commutative. For homogeneous factors, the ordinary product rule gives the Leibniz rule for ${\mathrm{\partial }}_x$; direct expansion gives ${\mathrm{\partial }}_{\theta }(f\theta j)=fj$, ${\mathrm{\partial }}_{\theta }(\theta gh)=gh$, and ${\mathrm{\partial }}_{\theta }((\theta g)(\theta h))=0=g(\theta h)-(\theta g)h$, so ${\mathrm{\partial }}_{\theta }$ is a degree-$+1$ graded derivation. Thus $\mathrm{\Delta }(f+\theta g)=g^{\mathrm{\prime }}(x)$ has degree $+1$, $\mathrm{\Delta }(1)=0$, and ${\mathrm{\Delta }}^2=0$.

For a homogeneous derivation $D$, $[D,m_a]=m_{D(a)}$; the composition identity gives $[\mathrm{\Delta },m_a]=m_{\mathrm{\Delta }a}+m_{{\mathrm{\partial }}_{\theta }a}{\mathrm{\partial }}_x+(-1{)}^{\mathrm{\mid }a\mathrm{\mid }}{m}_{{\mathrm{\partial }}_{x}a}{\mathrm{\partial }}_{\theta }$. Commuting once more yields $[[\mathrm{\Delta },m_a],m_b]=m_{{\mathrm{\partial }}_{\theta }a}{m}_{{\mathrm{\partial }}_{x}b}+(-1{)}^{\mathrm{\mid }a\mathrm{\mid }}{m}_{{\mathrm{\partial }}_{x}a}{m}_{{\mathrm{\partial }}_{\theta }b}$, a multiplication operator. Multiplication operators gradedly commute, so the third commutator with $m_c$ is zero and [F1]'s order condition holds.

Direct substitution in [F1] gives $\{f(x),\theta g(x)\}=f^{\mathrm{\prime }}(x)g(x)$ because $\mathrm{\Delta }(f\theta g)=(fg{)}^{\mathrm{\prime }}$, $\mathrm{\Delta }f=0$, and $\mathrm{\Delta }(\theta g)=g^{\mathrm{\prime }}$. Hence $\{S,\theta g\}=x^{2}g$ and $\{S,f\}=0$; also $S$ and $S^2$ depend only on $x$, so $\{S,S\}=0$ by the same formula. Thus $S$ satisfies the CME [F2].

Step 2.2 gives $d_S(f+\theta g)=x^{2}g$, which is independent of $\theta$ and is killed by $d_S$. Therefore $d_S^2=0$ directly, in agreement with [F3].