TQFT- Adding a $Q$-exact term which is equal to the action itself

It is known that Witten-type topological quantum field theories (TQFT) are invariant when $Q$-exact terms are added to the classical action, where $Q$ is the BRST charge. But for these theories, the action itself is $Q$-exact, so what stops us from cancelling off the entire action?

To be precise, given the partition function $$Z=\int\mathcal{D}X\textrm{ }e^{-Q\psi}$$ of a TQFT, where $Q\psi$ is the $Q$-exact action, we may add a term $Q\chi$ to the action, which can be shown (upon expanding $e^{-Q\chi}$) to be inconsequential, using the fact that $$\langle Q O\rangle=0$$ for any operator $O$; i.e., we find $$\int\mathcal{D}X\textrm{ }e^{-(Q\psi+Q\chi)}=\int\mathcal{D}X\textrm{ }e^{-Q\psi}(1-Q\chi+\frac{1}{2}Q(\chi Q \chi)+\ldots)=\int\mathcal{D}X\textrm{ }e^{-Q\psi}.$$

But if we choose $\chi=-\psi$, this would mean that $$\int\mathcal{D}X=\int\mathcal{D}X\textrm{ }e^{-Q\psi},$$ which seems to be an absurd statement, since the LHS is a divergent quantity.

• You are only allowed to make deformations which preserve the convergence of the Euclidean path integral. If you start with a positive-definite action $S_E$, you may deform it by $t Q[V]$ for any $t>0$ provided that $Q[V]$ is also positive. In your example, you are deforming the action by a negative-definite term, which is why you're getting nonsense. – Elliot Schneider Aug 30 '17 at 15:36
• But aren't there $Q$-exact deformations which are negative-definite, yet preserve convergence? For example, with the action written as $Q\psi$, the deformation $-\frac{1}{2}Q\psi$ would result in the original action multiplied by $1/2$. This surely is convergent. – Mtheorist Aug 31 '17 at 10:41
• Sure, but you can't scale that deformation with an arbitrarily large coefficient, which is the usual strategy in localization. – Elliot Schneider Aug 31 '17 at 21:39

The statement that one may add a $Q$ exact term to the action $S\to S+Q\chi$ without altering the correlators is not always strictly true. One has to be careful to make a choice of $\chi$ that does not change the asymptotic behaviour of the action $S$ at the boundary of field space. The choice of $\chi=-\psi$ is clearly a choice that drastically alters the asymptotic behaviour of $S$. Furthermore, $\left<Q\mathcal{O}\right>$ only vanishes up to boundary terms in field space. Non-zero boundary terms would cause the standard arguments you made to fail.

Edit: This can be demonstrated by a finite dimensional example which can be found in section 3.22 of the lectures by G. Moore on Donaldson Invariants and 4-manifolds at: http://www.physics.rutgers.edu/~gmoore/SCGP-FourManifoldsNotes-2017.pdf.

Consider the supersymmetric integral $$\mathcal{Z}=\int d\mathcal{M}e^{-S},\quad S=\frac{1}{2}H^2+iHs(x)-i\bar{\psi}\frac{ds(x)}{dx}\psi,\quad d\mathcal{M}=\frac{dxdHd\psi d\bar{\psi}}{2\pi i},$$ where $s:\mathbb{R}\to\mathbb{R}$ is a real valued function satisfying $|s(x)|\to\infty$ as $|x|\to\infty$. The action is invariant under a supersymmetry with $$Qx=\psi,\quad Q\psi=0,\quad Q\bar{\psi}=H,\quad QH=0.$$ Furthermore, the action is $Q$-exact $$S=Q\Psi,\quad \Psi=\left(\frac{1}{2}\bar{\psi}H+i\bar{\psi}s(x)\right).$$

Upon integrating out $\psi,\bar{\psi}$ and the auxiliary $H$ we find that $\mathcal{Z}$ is given by $$\mathcal{Z}=\int^{\infty}_{-\infty}\frac{dx}{\sqrt{2\pi}}s'(x)e^{-\frac{1}{2}s(x)^2}=\deg(f)\int^{\infty}_{-\infty}\frac{dy}{\sqrt{2\pi}}e^{-\frac{1}{2}y^2}=\deg(f)=\sum_{z(s)}\frac{s'(x_0)}{|s'(x_0)|}$$ to evaluate the the integral we changed variables $f:x\mapsto y=s$ and $\deg(f)$ denotes the degree of that map, finally $z(s)=\{x_0|s(x_0)=0\}$ is the zero set of $s$. Note that upon integrating out $H$, i.e. setting $H=-is(x)$, $z(s)$ coincides precisely with the set of $Q$ fixed points. However, we could have considered deforming the action by a $Q$ exact term, say, $S\to S+Q(i\bar{\psi}t(x))$ and repeating the above steps yields that $$\mathcal{Z}_t=\int^{\infty}_{-\infty}\frac{dx}{\sqrt{2\pi}}(s'(x)+t'(x))e^{-\frac{1}{2}(s(x)+t(x))^2}\stackrel{?}{=}\mathcal{Z}.$$ It should be clear that there are choices of $t$ such that $\mathcal{Z}_t$ doesn't exist, we require a $t$ such that $|s(x)+t(x)|\to\infty$ as $|x|\to\infty$ still holds. Even if this holds true there are choices such that $\mathcal{Z}_t\neq\mathcal{Z}$, as a simple demonstration you could just choose $s(x)=x^2-2x+1$ and $t(x)=-x^2$ then $z(s)=\{1\}$ and $s'(1)=0$ hence $\mathcal{Z}=0$ on the other hand the zero locus $z(s+t)=\{1/2\}$ $s'(1/2)+t'(1/2)=-2$ hence $\mathcal{Z}_{t=-x^2}=-1$.

• Why is changing the asymptotic behavior of $S$ not allowed? Moore does not explain this. – Mtheorist Aug 25 '17 at 14:56
• Thank you for the clarification. By the way, how does one evaluate the integral? I have posted this as a question on MathStackexchange here - math.stackexchange.com/questions/2407571/…, but there is no answer yet. – Mtheorist Aug 31 '17 at 8:51
• I have added an extra step but the idea is to make a change of variables $f:x\mapsto y=s$. Since that map is generically not one-to-one we pick up an extra factor $\deg(f)$ which is just counting the number of preimages of a point $y$ with signs. – Thomas Bourton Sep 2 '17 at 15:13

The statement that the path integral is independent of gauge-fixing (fermion) comes with various caveats. Recall that one of the reasons that we gauge-fix is to avoid integrating over an ill-defined infinite gauge volume. In general, a gauge symmetry manifests itself as a zero eigenvalue in the Hessian (i.e. the second derivative) of the ungauge-fixed action. In other words, we should make sure that the Hessian of the gauge-fixed action is non-degenerate. In particular, choosing the full action (and thereby the Hessian) to be identically zero would violate that. See also my related Phys.SE answer here.