Can topological invariants be related to Noetherian charges?

Question

I recently attended a seminar on physical mathematics, and learned about some topological invariants, especially in 4D spaces.

These topological invariants are considered to be invariant under continuous deformation of spacetime.

I find these to be similar with Noetherian charge; they are conserved under some continuous transformations on the spacetime.

Are they related? If yes, then how are they related?

Answer 1

1. Usual Noether charges can be seen as coming from a conserved current one-form $j$, such that $\newcommand{\d}{\mathrm{d}}$ $\d\star j=0$. Then the Noether charge is the integral of $\star j$ on a codimension-one manifold: $$ Q(\Sigma_{d-1}) := \int_{\Sigma_{d-1}} \star j.$$
2. This has been extended to higher-form conserved currents, where the conserved current is now a $(p+1)$-form, so it is to be integrated on a codimension-$(p+1)$ manifold $$ Q(\Sigma_{d-p-1}) := \int_{\Sigma_{d-p-1}} \star j_{[p+1]}.$$ A symmetry giving rise to such a current is called a $p$-form symmetry.
3. Usually, topological invariants of a $d$-manifold, $X_d$ can be expressed as $$ \mathrm{inv}(X_d) = \int_{X_d} \rho_\text{inv}.$$ For example, in four dimensions the Euler invariant can be written as $$ \chi(X_4) = \frac{1}{32\pi^2}\int_{X_4} \epsilon^{abcd} R_{ab}\wedge R_{cd}$$ and the signature invariant can be written as $$ \sigma(X_4) = \frac{1}{24\pi^2}\int_{X_4} R^a_{\ b}\wedge R^b_{\ a}.$$ where $R^{a}_{\ b}$ is the curvature two-form $R^{a}_{\ b} = \d{\omega^a_{\ b}} + \omega^a_{\ c}\wedge\omega^c_{\ b}$.
4. This shows that in such cases topological invariants can be interpreted as the Noether charges of a $(-1)$-form symmetry, i.e. a symmetry where the conserved current is a zero-form, i.e. a function. $$\mathrm{inv}(X_d) \equiv Q(X_d) := \int_{X_d} \star j_{[0]}$$
5. Note, however, that while these topological invariants can be identified with $(-1)$-form symmetries, not all $(-1)$-form symmetries can be interpreted as topological invariants.