# Index theorem, UV and IR face of chiral anomaly

+ 7 like - 0 dislike
166 views

The index theorem in theory with fermions and gauge fields implies the relation between the index $n_{+}-n_{-}$ of Dirac operator and the integral $\nu$ over EM field chern characteristic class: $$\tag 1 n_{+} - n_{-} = \nu$$ Let's focus on 4D. The index theorem is obtained by computing anomalous jacobian $$J[\alpha] = \text{exp}\left[-2i\alpha \sum_{n = 1}^{N = \infty}\int d^{4}x_{E}\psi^{\dagger}_{n}\gamma_{5}\psi_{n}\right]$$ Here $n$ denotes the number of eigenfunction of the Dirac operator
$$D_{I}\gamma_{I}, \quad D_{I} \equiv i\partial_{I} - A_{I}$$

From the one side, this is bad defined quantity,
$$J[\alpha] \simeq \text{exp}\left[i\alpha \lim_{x \to y}\text{Tr}(\gamma_{5})\delta (x - y)\right],$$
so it requires the UV regularization. The explicit form of this regularization is fixed by the requirements of gauge and ''euclidean'' invariance, leading to introducing the function $f\left( \left(\frac{D_{I}\gamma_{I}}{M}\right)^{2}\right)$, with $M$ being the regularization parameter. From the other side, by using the regularization, it is not hard to show that the exponent is equal to the $-2i\alpha (n_{+}-n_{-})$. Since this number defines the difference of zero modes, it depends only on IR property of theory. Moreover, $\nu$ is also determined by the behavior of gauge fields on infinities, being IR defined number.

Because of this ''puzzle'', I want to ask: does the index theorem provide the relation between IR (zero modes, large scale topology) nature and UV (regularization required) nature of chiral anomaly? Precisely, I know the "spectral flow" interpretation of chiral anomaly, according to which an anomaly is the collective motion of chiral charge from UV world to IR one. Does the index theorem provide this interpretation?

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\varnothing$ysicsOverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.